# Naked Science Forum

## Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: HankRearden on 19/07/2009 23:09:39

Title: If I give an object some potential energy, does its mass increase?
Post by: HankRearden on 19/07/2009 23:09:39
So matter can basically be thought of as just potential energy because matter can be turned into energy and theoretically vice-versa, does that mean when I lift up a plate and increase it's potential energy, i've technically increased its mass?
Title: Re: If I give an object some potential energy, does its mass increase?
Post by: Soul Surfer on 19/07/2009 23:34:26
no you've just increased it's potential energy
Title: Re: If I give an object some potential energy, does its mass increase?
Post by: HankRearden on 19/07/2009 23:39:18
but isn't matter just potential energy?

Title: Re: If I give an object some potential energy, does its mass increase?
Post by: Ethos on 20/07/2009 01:11:18
but isn't matter just potential energy?

Mass is not potential energy, it is condensed energy E=mc^2. And BTW, I have a problem with the term 'potential energy' anyway. I understand what physicists mean when they use this expression, I just think it is not an appropriate use of the word 'potential'.
Title: Re: If I give an object some potential energy, does its mass increase?
Post by: lightarrow on 20/07/2009 11:20:41
So matter can basically be thought of as just potential energy because matter can be turned into energy and theoretically vice-versa, does that mean when I lift up a plate and increase it's potential energy, i've technically increased its mass?
The mass is increased, but not the mass of the plate: the mass of the system Earth-plate.
Title: Re: If I give an object some potential energy, does its mass increase?
Post by: HankRearden on 21/07/2009 02:40:50
Good enough for me
Title: Re: If I give an object some potential energy, does its mass increase?
Post by: Stefanb on 22/07/2009 07:15:26
E(p)= m x g x h Right?
So yes, mass is involved with the equation. However, because other variables are involved, mass and potential energy are not the same and cannot be manipulated as such.
Title: Re: If I give an object some potential energy, does its mass increase?
Post by: wanhafizi on 23/07/2009 05:37:50
So matter can basically be thought of as just potential energy because matter can be turned into energy and theoretically vice-versa, does that mean when I lift up a plate and increase it's potential energy, i've technically increased its mass?

I saw a documentary of LHC experiment, they were saying, mass of the protons will increase as the energy increase. Yes, mass....
Title: Re: If I give an object some potential energy, does its mass increase?
Post by: lightarrow on 23/07/2009 09:45:15
I saw a documentary of LHC experiment, they were saying, mass of the protons will increase as the energy increase. Yes, mass....
That's false, even if you can find this concept even in some books.
http://www.thenakedscientists.com/forum/index.php?topic=16789.0
http://www.thenakedscientists.com/forum/index.php?topic=21363.0
Title: If I give an object some potential energy, does its mass increase?
Post by: Pmb on 26/07/2009 00:38:25
So matter can basically be thought of as just potential energy because matter can be turned into energy and theoretically vice-versa, does that mean when I lift up a plate and increase it's potential energy, i've technically increased its mass?
You're thinking of mass-energy. The term "potential energy" refers to a different kind of energy. There are two ways the potential energy of a body can change. One can change the internal potential energy of a body. That kind of change will cause the rest mass of the body to change. The same thing will happen if you were to change the internal kinetic energy of the body increases (e.g. if the particles which make up the body vibrate faster).

Another way to change the potential energy of a body is to change its position in a field. Depending on what one means by "mass" its also possible that the mass by changing body’s potential energy. If the field is a gravitational field and one changes the position of the body in the gravitational field then there will be change in the mass of the body. This mass is related to the body’s rest mass m_0 by (let T = proper time)

m = (dt/dT) m_0

For they typical kind of gravitational field (i.e. corresponding to a time-orthogonal spacetime)

dt/dT = 1/sqrt[ 1 + 2*Phi – v^2/c^2]

where Phi = gravitational potential

The mass is then

m = m_0/sqrt[ 1 + 2*Phi – v^2/c^2]

Don’t confuse this with proper mass (aka “rest mass”)

Wanhafizi wrote – “I saw a documentary of LHC experiment, they were saying, mass of the protons will increase as the energy increase.”

They were speaking of the inertial mass of the proton, not its proper mass. When particle physicists speak of mass they are often referring to proper mass. However when relativists use the term “mass” they can mean one of several things. For example, in the lecture notes from Alan Guth’s “The Early Universe” course at MIT he explains that light has mass according to its energy. In this context he’s referring to what other relativists refer to as “relativistic mass.” This kind of mass is often spoken of in general relativity. It also goes under various other names such as simply “mass”  but also inertial mass, relativistic mass, active gravitational mass and passive gravitational mass.
Title: If I give an object some potential energy, does its mass increase?
Post by: lightarrow on 26/07/2009 15:05:45
Welcome on this forum Pmb!
Title: If I give an object some potential energy, does its mass increase?
Post by: lyner on 26/07/2009 15:59:08
Yes, I'd endorse that.
Title: If I give an object some potential energy, does its mass increase?
Post by: Pmb on 27/07/2009 12:52:52
Thank you. It's nice to be here. :)
Title: If I give an object some potential energy, does its mass increase?
Post by: Farsight on 29/07/2009 23:04:43
So matter can basically be thought of as just potential energy because matter can be turned into energy and theoretically vice-versa, does that mean when I lift up a plate and increase it's potential energy, i've technically increased its mass?
The answer is yes, but there's a catch.

Mass is a measure of the amount of energy that is not moving in aggregate with respect to you. A photon has no mass because it is moving with respect to you, and you cannot normally vary the speed of that photon. But there is a way to make it not move in aggregate: you employ pair production to convert a +1022keV photon into an electron and a positron. Forget the positron and consider the electron, ignoring its relatively slow motion after pair production. The energy is no longer moving laterally at c, instead it's moving in a circular fashion which we label as electron spin. You can then accelerate the electron as a whole, and the resistance to motion is a measure of its mass. See what pmb was saying above about proper mass and relativistic mass. The proper mass of the electron is the amount of energy tied up as that electron. The kinetic energy is the extra energy you give it by making it move, and the relativistic mass is the total.

Back to the plate: imagine it's out in space, and is falling to earth. Ignore air resistance. Just before it hits the ground the plate is moving at a considerable velocity. It now has kinetic energy. So the total energy of the plate appears to be greater than that of the plate up in space. However, it isn't, because gravity is a pseudoforce. This is not generally understood, but the key to understanding it is to consider two masses that fall together and coalesce. Energy causes gravity, and the  gravity caused by these two masses does not increase as they coalesce. No net energy is being added to this system. In similar vein no net energy is added to the falling plate. The total energy in that plate travelling at 11 km/s near the surface of the earth is the same as the total energy of the plate when it was motionless up in space. So if you catch that plate and cool it down, the total energy of the plate is now less than that of the plate at altitude. The reason is quite obvious when you look at gravitational time dilation. Everything in that plate, be it atoms or electrons is now moving at a slightly reduced rate when compared with the plate up in space. You can't measure this locally because it affects your clocks too. But you can measure it non-locally by comparison, as demonstrated by the GPS clock adjustment. This reduced rate means the energy locked up in the cold motionless plate at ground level is less than that of a cold motionless plate up in space.

So if you lift that plate up, then if the temperature of the plate is the same, the total energy tied up as mass in that plate is increased. Thus the mass increases. But here's the catch: if I had some device that could lift a plate up into space without giving back that 11 km/s of kinetic energy, that plate is going to get awfully cold. And there's a wrinkle too: light is deflected by gravity twice as much as matter, and when you compare a blue-shifted 511keV photon with a falling electron, there's an imbalance. But I'll save that one for another day.

Hi Pete.
Title: If I give an object some potential energy, does its mass increase?
Post by: exton on 29/07/2009 23:45:26
The total energy in that plate travelling at 11 km/s near the surface of the earth is the same as the total energy of the plate when it was motionless up in space. So if you catch that plate and cool it down, the total energy of the plate is now less than that of the plate at altitude. The reason is quite obvious when you look at gravitational time dilation. Everything in that plate, be it atoms or electrons is now moving at a slightly reduced rate when compared with the plate up in space.

I don't follow. What's this talk of cooling things down - why did the plate heat up? Do you mean that we hypothetically catch the plate by turning its kinetic energy into thermal energy?
Title: If I give an object some potential energy, does its mass increase?
Post by: Pmb on 30/07/2009 02:24:04
Hi Farsight!! How are you?

Conversations about mass can get confusing because its almost guaranteed that different people will be using different definitions of mass. What I spoke about above is often referred to as relativistic mass. What Farsight is talking about is what is often referred to as rest mass. Farsight’s comments can only be applied to special circumstances. Namely for bodies not subjected to stress.
Title: If I give an object some potential energy, does its mass increase?
Post by: VernonNemitz on 31/07/2009 17:25:38

http://www.nemitz.net/vernon/STUBBED2.pdf

Title: If I give an object some potential energy, does its mass increase?
Post by: lightarrow on 31/07/2009 18:14:17
Back to the plate: imagine it's out in space, and is falling to earth. Ignore air resistance. Just before it hits the ground the plate is moving at a considerable velocity. It now has kinetic energy. So the total energy of the plate appears to be greater than that of the plate up in space. However, it isn't, because gravity is a pseudoforce.
Do you mean that, according to GR, gravitational field doesn't exist (since it's actually spacetime warping) and so that region of space cannot have energy (= cannot have mass)? And so when the plate falls, since its kinetic energy increases, its proper mass have to decrease, to keep its total energy constant?  [???]
Title: If I give an object some potential energy, does its mass increase?
Post by: Farsight on 01/08/2009 12:50:19
Quote from: exton
I don't follow. What's this talk of cooling things down - why did the plate heat up?
The plate doesn't have to heat up. If it was a steel plate and you're accelerating it and decelerating it with a railgun you're turning electrical energy into kinetic energy and vice versa. But kinetic energy is usually dispersed as heat. Plus damage and noise etc, but usually there's a lot of heat.

Quote from: exton
Do you mean that we hypothetically catch the plate by turning its kinetic energy into thermal energy?
Yes. Imagine you've got a metal container full of cold gas travelling at 11km/s through space. You catch the container and stop it with a rail gun. But those gas molecules are still doing 11km/s rattling around inside the container. So your gas is now hot. It's actually a little more than 11km/s because the average velocity of a gas molecule in air at room temperature and pressure is about 0.5km/s.

Quote from: Pmb
Hi Farsight!! How are you?
Pretty great thanks Pete. Of course, things could be moving faster and I could do with some more free time. For example the wife is off to Cheltenham all day with her sister leaving me holding the baby. But I'm as happy as Larry. And you?

Do you mean that, according to GR, gravitational field doesn't exist (since it's actually spacetime warping)...
Heck, not at all. If you're in a place where you fall down, you're in a gravitational field. It exists all right.

..and so that region of space cannot have energy (= cannot have mass)?
No. Space has an energy to it, but we don't think of this as mass. Mass is a property of something that resists being moved, and you can't exactly move a region of space from A to B.

And so when the plate falls, since its kinetic energy increases, its proper mass have to decrease, to keep its total energy constant?
Yes, that's what I'm saying. Think about two bodies m1 and m2. Imagine you're some way off in space, feeling the effect of their combined gravity. You're measuring the energy content of that two-body system. Now imagine they've fallen together and coalesced into one body M without losing any energy (think of them as being made of water or something). You don't feel any extra gravity, because no net energy has been added to that system by the two objects falling together. People say the kinetic energy has come from the potential energy of the gravitational field, but that's missing the trick. The gravity of the two-body system doesn't increase, and nor does it reduce. You can take this further by thinking of a spherical shell, where the gravity you experience is the same as what you'd get with a central point mass, see http://hyperphysics.phy-astr.gsu.edu/HBASE/Mechanics/sphshell.html. The important point is that gravity is a pseudoforce like Einstein said, because no energy is added to the system. A falling body doesn't feel any force, and no force is acting upon it. If you're in freefall you can't feel any force, because there isn't any. A falling plate is not accelerating. Instead it's "accelerating" in line with the principle of equivalence when it isn't falling any more. The kinetic energy of that falling plate didn't come from the earth via some magical mysterious action-at-a-distance force. It didn't come from the gravitational field either because that would also involves action-at-a-distance, and gravity is a local effect. It isn't a force it's a pseudoforce, so no energy is being delivered. Instead the kinetic energy comes from the plate itself. But it's very slight, compare 11 km/s to 299,792 km/s for an initial indication, or better still look at the GPS clock adjustment: http://en.wikipedia.org/wiki/Global_Positioning_System#Relativity. And do note that it's a scale-change which affects your measuring devices and everything else. Some might say if you can't measure any difference there is no difference, but if you take mass as a measure of the amount of energy tied up in an object, then conservation of energy is telling you something important here. All good stuff to think about.
Title: If I give an object some potential energy, does its mass increase?
Post by: lightarrow on 01/08/2009 13:10:55
Do you mean that, according to GR, gravitational field doesn't exist (since it's actually spacetime warping)...
Heck, not at all. If you're in a place where you fall down, you're in a gravitational field. It exists all right.

..and so that region of space cannot have energy (= cannot have mass)?
No. Space has an energy to it, but we don't think of this as mass. Mass is a property of something that resists being moved, and you can't exactly move a region of space from A to B.
Then I don't understand what mass you are talking about: in a region of space which is not moving (with respect to some frame S) and in which you have energy, you also have mass: m = E/c2. Example: an electrostatic field has mass (I mean, proper mass = invariant mass). When you give energy (let's say electromagnetic energy) to an hydrogen atom, for example, you also give mass to the system, which goes in the electromagnetic field; the proton's and electron masses don't vary at all.

If you tell me that a system of two still masses, let's say two still planets, do interact through a gravitational field, then you *have* to ascribe a mass to that field and any variation of the system's energy = system's mass, comes out to belong to the field.

In the second part of your post, however, you keep the focus on the fact that gravity is a 'pseudo' force, that is, that there's no force at all; then there is no field, so why do you say instead: "Heck, not at all. If you're in a place where you fall down, you're in a gravitational field. It exists all right."? Don't understand.

Quote
People say the kinetic energy has come from the potential energy of the gravitational field, but that's missing the trick. The gravity of the two-body system doesn't increase, and nor does it reduce.
That's not correct. A field's energy density goes as the square of the field, so you cannot simply sum the effects of the two masses. If the masses' configuration varies, the field's energy varies as well. When the two masses are closer, the field's energy increases *in absolute value* and since the gravitational field's energy is negative, it means that the field's total energy is decreased. This is the reason of the fact that total system's energy is conserved.
Title: If I give an object some potential energy, does its mass increase?
Post by: Pmb on 01/08/2009 22:04:44
The important point is that gravity is a pseudoforce like Einstein said, because no energy is added to the system.
Einstein never said that gravity was a pseudoforce. Einstein argued that the gravitational force is of the same nature as inertial forces. Some Newtonians argue that since there is no source of such a force that it’s not “real” and hence terms like pseudoforce, apparent force, and fictitious force were coined. But others disagreed with that notion and Einstein was one of them. Einstein argued that inertial forces should be thought of as being real. It’s also wrong to assume that there is no potential energy in the gravitational field or that there is no mass equivalence to that potential energy. Einstein’s field equations are non-linear because gravity itself is a source of gravity.

For more on inertial forces see my web page http://www.geocities.com/physics_world/gr/inertial_force.htm

Pete
Title: If I give an object some potential energy, does its mass increase?
Post by: lightarrow on 02/08/2009 12:20:43
For more on inertial forces see my web page http://www.geocities.com/physics_world/gr/inertial_force.htm
Nice site. Thank you Pmb.
Title: If I give an object some potential energy, does its mass increase?
Post by: Farsight on 02/08/2009 18:00:35
Then I don't understand what mass you are talking about: in a region of space which is not moving (with respect to some frame S) and in which you have energy, you also have mass: m = E/c2.
I know what you mean. For example if you have a mirrored box and introduce a photon, the mass of the box is increased in line with the photon's energy. This is because mass is a measure of the amount of energy that is not moving in aggregate with respect to you.

Example: an electrostatic field has mass (I mean, proper mass = invariant mass). When you give energy (let's say electromagnetic energy) to an hydrogen atom, for example, you also give mass to the system, which goes in the electromagnetic field; the proton's and electron masses don't vary at all.
Yes, an electrostatic field has mass. But the electron's electric field is part of what it is. It isn't some central point particle with some set mass surrounded by an electric field with some variable mass. If you examine an electron at rest, you will deem its mass/energy to be 511keV. If you put that electron into your mirrored box, the mass of the box will increase accordingly. If you replace the stationary electron with a fast-moving electron bouncing back and forth inside the box, the increase in mass is greater. If you then replace your box with a proton that keeps the electron local to itself, the faster-moving electron means your hydrogen atom has more mass. But the electric field hasn't increased. The electron and the proton still exhibit unit charge.

If you tell me that a system of two still masses, let's say two still planets, do interact through a gravitational field, then you *have* to ascribe a mass to that field and any variation of the system's energy = system's mass, comes out to belong to the field.
A region of empty space is not empty of energy, and OK, in this respect one can ascribe mass to it (he said grudgingly). A planet is surrounded by a region of space which exhibits a gravitational field, and there is energy "density" in that region of space. But the gravitational field itself is only there because there's a gradient in the energy density. No gradient means no gravitational field. You can see this very easily if you consider uniform space. There's no gravity at all.

In the second part of your post, however, you keep the focus on the fact that gravity is a 'pseudo' force, that is, that there's no force at all; then there is no field, so why do you say instead: "Heck, not at all. If you're in a place where you fall down, you're in a gravitational field. It exists all right."? Don't understand.
Try this: if you're in a windowless box in free-fall you can't feel any force and you can't feel any acceleration. As far as you're concerned there is no gravitational field, you're just floating in space in an inertial reference frame, and nothing is falling down at all. But if I snap my fingers and give you a window, you change your mind pronto. You switch your reference frame from the box to the ground, and you are utterly convinced of the existence of that gravitational field. But you still can't feel any force acting upon you.  I snap my fingers again to give you a soft landing, whereupon you do feel a force which intensifies then levels off, but persists. You feel the force when you're standing on the ground, not when you're falling down.

Quote from: Farsight
People say the kinetic energy has come from the potential energy of the gravitational field, but that's missing the trick. The gravity of the two-body system doesn't increase, and nor does it reduce.
That's not correct. A field's energy density goes as the square of the field, so you cannot simply sum the effects of the two masses. If the masses' configuration varies, the field's energy varies as well.
What you've said here is a restatement of your earlier position. Let me try to work things through to explain things better. You start with a universe full of nothing but space. This space has an inherent energy. This isn't always obvious, but it's there, and there's a fixed amount of this "vacuum energy". But because space is uniform, there is no gravity.

You'll be aware that if you combine two out-of-phase photons you're left with no photons, but conservation of energy tells you the energy has not vanished. Instead it's now in the local space. OK, now do this in reverse across the board, such that a portion of your vacuum energy is converted into photons. Your uniform space is now full of photons. It isn't quite as uniform as before, but it's still fairly uniform, and there is still no discernable gravity.

Now use pair production to convert a portion of your photons into electrons and protons, and antiprotons and positrons. Let's forget the antiprotons and positrons, and say the electrons and protons take the form of hydrogen, also forgetting about helium. OK, if this was absolutely uniform, there's still no discernable gravity. The universe would be a ball of energy including vacuum energy and photons and hydrogen (and neutrinos etc too but forget them as well). The point is that it's still fairly uniform, and there is no gravity. There's no space beyond this universe to hold it all in, and no gravity within to pull anything towards the centre, so it would keep on expanding just like it always has done.

But this hydrogen isn't uniform. There's irregularities and variations, and hence there's just a little bit of gravity to make it start clumping together. Then the hydrogen forms clouds and galaxies and suns and planets, and there's gravity galore. There's gravity every where you look. There is no place in this universe where there is no gravity. But the sum total energy in the universe hasn't changed a bit. All that's happened is that the variations in energy density have been intensified.

When the two masses are closer, the field's energy increases *in absolute value* and since the gravitational field's energy is negative, it means that the field's total energy is decreased. This is the reason of the fact that total system's energy is conserved
This sounds like a restatement too. The gravitational field is only there because the masses are there. And those masses are made of energy. They're made of positive energy, there's no such thing as negative mass, and no such thing as negative energy. Go back to your hydrogen, and then consider the electrons and protons. The electrostatic field of each is a region where we see an energy density diminishing in line with the inverse square rule. When you clump them together to make a planet, then whilst there's no net charge, the electrostatic energy densities don't cancel. There's still an electrostatic energy density diminishing in line with the inverse square rule. Hence the mass/energy of the planet "conditions the surrounding space" and because it isn't uniform, we've got a gravitational field.

There is a relative negative to gravity, in that time dilation means everything is going slower down here than it is up there. Imagine a spinning plate up in space. Let's say it's doing 1000rpm. Because it's spinning, its got more energy, so its got more mass. Now compare it with a similar spinning plate down on the planetary surface. It too is doing 1000rpm, but because of the time dilation, it's actually spinning slower than the plate up in space. Hence its got less mass. Now think of pair production and treat an electron as a "spinning photon", and compare two identical motionless plates one up in space and one down on the planetary surface. You can liken every electron in each as a kind of spinning plate. Those on the surface are spinning slower, so the plate on the surface has less energy and so less mass.

Einstein never said that gravity was a pseudoforce. Einstein argued that the gravitational force is of the same nature as inertial forces. Some Newtonians argue that since there is no source of such a force that it's not 'real' and hence terms like pseudoforce, apparent force, and fictitious force were coined. But others disagreed with that notion and Einstein was one of them. Einstein argued that inertial forces should be thought of as being real. It's also wrong to assume that there is no potential energy in the gravitational field or that there is no mass equivalence to that potential energy. Einstein's field equations are non-linear because gravity itself is a source of gravity. For more on inertial forces see my web page http://www.geocities.com/physics_world/gr/inertial_force.htm
Yes, I know about the non-linearity. Hmmn, maybe pseudoforce is the wrong word, shame I've used it elsewhere. You'll be aware that what I've been generally finding is that people say "Einstein said x" when actually he didn't. Hence I wince at the thought of being guilty of the same. I was just looking at what you said back in January:

"This assumes that what is referred to as a pseudoforce doesn't fall under the category of a force. The name pseudoforce is used by those scientist who choose to think/define of a force in way such that an inertial force is not a force. But that is a subjective way to do things. Einstein proposed that the gravitational force is identical in nature to inertial forces (like the Coriolis force or the centrifugal force). Since scientists had come to think of inertial forces as not "real" forces this led some scientists to conclude that gravity is not a "real" force. Hence leading scientists use the term pseudoforce to refer to the gravitational force. However this is not what Einstein's theory implies and is not how Einstein himself viewed it. Einstein considered gravity to be a "real" force and concluded that inertial forces like the Coriolis force is also a "real" force. In his book on relativity, Wolfgang Pauli points out that its perfectly fine to think of the gravitational force as either a real force or a "fake" force, so long as you think of the gravitational force as the same kind of force as an inertial force.

I hope we can all agree that gravity is not fake. If we think of a compressed spring, as in an airgun, we can consider the compression to represent potential energy. When we release the spring this potential energy is converted into kinetic energy and no longer persists. When we stand on the surface of the earth and look at the plate up in free space, we say it has gravitational potential energy. But (let's ignore the sun) it's just a plate in space. It isn't in a gravitational field. When the earth rolls up to the plate, the plate's potential energy is converted into kinetic energy, and no longer persists. But it isn't the gravitational field that had the potential energy, it was the plate. The gravitational field is not reduced when the plate falls, because energy causes gravity, and all we've done is converted some of the energy of the plate into kinetic energy, leaving the plate itself in a reduced-energy state with a slightly reduced mass. As a result of encountering that plate, the gravitational field of the earth is now slightly increased. At no point did the force of gravity add any energy to the earth-plate system, just as the centripetal force doesn't continually add energy to a ball whirled on a string.

Title: If I give an object some potential energy, does its mass increase?
Post by: lyner on 02/08/2009 19:21:17
Quote
You'll be aware that if you combine two out-of-phase photons you're left with no photons
I don't want to interrupt the flow but that statement begs the question.
"If you combine" implies that they are produced at the same place and at the same time - else there wouldn't be cancellation everywhere.  I have a feeling that violates Pauli, for a start*. If they are  heading for each other then yo will get a standing wave when they are actually crossing. No energy problems either way.
If you want to talk in terms of phase, then you are talking waves and if you produce cancellation of two waves in one direction, the energy pops up elsewhere, in another direction.

Edit * Because they would have to be produced by superposed fermions.
Title: If I give an object some potential energy, does its mass increase?
Post by: lightarrow on 02/08/2009 20:45:04
Example: an electrostatic field has mass (I mean, proper mass = invariant mass). When you give energy (let's say electromagnetic energy) to an hydrogen atom, for example, you also give mass to the system, which goes in the electromagnetic field; the proton's and electron masses don't vary at all.
Yes, an electrostatic field has mass. But the electron's electric field is part of what it is. It isn't some central point particle with some set mass surrounded by an electric field with some variable mass. If you examine an electron at rest, you will deem its mass/energy to be 511keV. If you put that electron into your mirrored box, the mass of the box will increase accordingly. If you replace the stationary electron with a fast-moving electron bouncing back and forth inside the box, the increase in mass is greater. If you then replace your box with a proton that keeps the electron local to itself, the faster-moving electron means your hydrogen atom has more mass.
Ok up to here.

Quote
But the electric field hasn't increased. The electron and the proton still exhibit unit charge.
1. The electromagnetic field  have to be evaluated after the atom has given away the surplus of energy due to the initially fast moving electron, for example by emission of photons.
2. After that, you will see that the electric field has decreased, with respect to when the electron and the proton were far apart; it means that the total energy of the field (in all 3D space) has decreased. If you make the computations, you see that it has decreased exactly of the electric potential energy variation. Where has this energy gone? With the photons carrying the surplus of energy.

Quote
A region of empty space is not empty of energy, and OK, in this respect one can ascribe mass to it (he said grudgingly). A planet is surrounded by a region of space which exhibits a gravitational field, and there is energy "density" in that region of space. But the gravitational field itself is only there because there's a gradient in the energy density. No gradient means no gravitational field. You can see this very easily if you consider uniform space. There's no gravity at all.
I sincerely don't understand this. Consider a region of space in the void with a uniform electric field E. The energy density is (1/2)ε0E2, so it's uniform as well, so its gradient is zero. Would you deduce that the field is zero?  [???]

Quote
That's not correct. A field's energy density goes as the square of the field, so you cannot simply sum the effects of the two masses. If the masses' configuration varies, the field's energy varies as well.
What you've said here is a restatement of your earlier position. Let me try to work things through to explain things better. You start with a universe full of nothing but space. This space has an inherent energy. This isn't always obvious, but it's there, and there's a fixed amount of this "vacuum energy". But because space is uniform, there is no gravity.
Why not? I see what you intend after in your post, but it doesn't seem correct to me. Consider a rubber band which is initially stretched in a large circumference and then released; in every point of the rubber band, in the band's frame of reference, elastic forces pull in two opposite directions, so the point doesn't move along the circumference, ok, but that's different from saying that elastic forces don't act on the band: it quickly contracts in the other dimension. Monodimensional 'people' living in the rubber band would see that distances from any two points in its universe are decreasing.

Quote
When the two masses are closer, the field's energy increases *in absolute value* and since the gravitational field's energy is negative, it means that the field's total energy is decreased. This is the reason of the fact that total system's energy is conserved
This sounds like a restatement too. The gravitational field is only there because the masses are there. And those masses are made of energy. They're made of positive energy, there's no such thing as negative mass, and no such thing as negative energy.
Can you prove it?
Title: If I give an object some potential energy, does its mass increase?
Post by: Pmb on 03/08/2009 03:04:29
For more on inertial forces see my web page http://www.geocities.com/physics_world/gr/inertial_force.htm
Nice site. Thank you Pmb.
That's my personal website so thank you kindly for the compliment.
This forum is a bit weird. When I type there is a very long time delay before the letters appear. It's too hard to type directly into the window so I use an editor and do a cut and paste. There's a lot going on here that I wish I could keep track of and respond to.
Quote from: Farsight
A region of empty space is not empty of energy, and OK, in this respect one can ascribe mass to it (he said grudgingly). A planet is surrounded by a region of space which exhibits a gravitational field, and there is energy "density" in that region of space. But the gravitational field itself is only there because there's a gradient in the energy density. No gradient means no gravitational field. You can see this very easily if you consider uniform space. There's no gravity at all.
MTW give the stress-energy-momentum tensor for the gravitational field as that derived by Landau and Lifshitz. On 465 they give an expression for it for a weak gravitational field. The energy density is given in terms of phi,j*phi,j. For a uniform gravitational field phi = gz/c^2 an so the energy density is uniform. The field is still there though, i.e. you'd still "fall down".
Quote from: Farsight
Try this: if you're in a windowless box in free-fall you can't feel any force and you can't feel any acceleration. As far as you're concerned there is no gravitational field, you're just floating in space in an inertial reference frame, and nothing is falling down at all. But if I snap my fingers and give you a window, you change your mind pronto. You switch your reference frame from the box to the ground, and you are utterly convinced of the existence of that gravitational field. But you still can't feel any force acting upon you.  I snap my fingers again to give you a soft landing, whereupon you do feel a force which intensifies then levels off, but persists. You feel the force when you're standing on the ground, not when you're falling down.
Any object only "feels" a force on it when there is stress imposed in the body by the force field. If you're falling in a uniform gravitational field you'd feel no force. But the same would happen for a charged body in a uniform electric field too. If you're in a non-uniform gravitational field then you'd feel tidal forces since they'd impose stresses in your body. It's odd that people don't make this obvious comparison with the electric field!
Quote from: Farsight
there's no such thing as negative mass
This is wrong. I think you said this because you use the term "mass" as another name for rest energy. That's a problem. In any case dark energy is due to the presence of negative active gravitational mass density. Any time you have repulsive gravity, such as in inflation and dark energy, then you're talking about negative active gravitational mass density. It's also conceivable that there can be negative inertial mass if a body could exist that could support large enough values of tension. You probably never thought of this because you associate of mass with rest energy. Negative inertial mass means that the momentum of a body is opposite to its velocity and that's quite possible in special relativity given large enough values of negative pressure (i.e. tension) compared to the bodies energy.

Pete
Title: If I give an object some potential energy, does its mass increase?
Post by: VernonNemitz on 03/08/2009 15:02:28
And so when the plate falls, since its kinetic energy increases, its proper mass have to decrease, to keep its total energy constant?
Yes, that's what I'm saying. Think about two bodies m1 and m2. Imagine you're some way off in space, feeling the effect of their combined gravity. You're measuring the energy content of that two-body system. Now imagine they've fallen together and coalesced into one body M without losing any energy (think of them as being made of water or something). You don't feel any extra gravity, because no net energy has been added to that system by the two objects falling together. People say the kinetic energy has come from the potential energy of the gravitational field, but that's missing the trick. The gravity of the two-body system doesn't increase, and nor does it reduce.
That is a "trick" scenario, since you have explicity stated the two bodies do not lose any energy as a result of their collision.  In actual fact, though, the collision will cause the temperature of the combined body to be warmer than the separate bodies, and as time passes, that extra temperature will radiate away, after which the total gravitation of the combined mass will be diminished a bit, compared to the original two-body system.  Therefore, the potential energy of the original two-body system can indeed be said to have existed as mass (somewhere in the system), which was converted to kinetic energy (heat) during the collision and radiated away after the collision.
Title: If I give an object some potential energy, does its mass increase?
Post by: Farsight on 03/08/2009 20:00:25
Your comment noted sophie. Perhaps we could talk about destructive interference and the (very interesting) evanescant wave on another thread.

1. The electromagnetic field have to be evaluated after the atom has given away the surplus of energy due to the initially fast moving electron, for example by emission of photons.
2. After that, you will see that the electric field has decreased, with respect to when the electron and the proton were far apart; it means that the total energy of the field (in all 3D space) has decreased. If you make the computations, you see that it has decreased exactly of the electric potential energy variation. Where has this energy gone? With the photons carrying the surplus of energy.
The fast-moving electron decelerates and emits a photon. The electron has lost the energy. But OK, the total energy of the field is now reduced because the electron's electric field is part of what it is.

Quote from: Farsight
You can see this very easily if you consider uniform space. There's no gravity at all.
I sincerely don't understand this. Consider a region of space in the void with a uniform electric field E. The energy density is (1/2)ε0E2, so it's uniform as well, so its gradient is zero. Would you deduce that the field is zero?
Not at all. But the gravitational field is not the same as an electric field. Consider an electron and a proton, and let them attract one another to form a hydrogen atom. Their fields mask one another, and the total energy will be reduced by photon emission to give the negative binding energy. It's a very different situation for two masses.

Quote from: Farsight
But because space is uniform, there is no gravity.
Why not? I see what you intend after in your post, but it doesn't seem correct to me. Consider a rubber band which is initially stretched in a large circumference and then released; in every point of the rubber band, in the band's frame of reference, elastic forces pull in two opposite directions, so the point doesn't move along the circumference, ok, but that's different from saying that elastic forces don't act on the band: it quickly contracts in the other dimension. Monodimensional 'people' living in the rubber band would see that distances from any two points in its universe are decreasing.
I don't have any issue with what you're saying here. Your analogy is fine by me, but let me flip things around to try and get across what I mean about uniformity and gravity. Replace the elastic band with a stressball. Squeeze it down in your fist, then let go and watch it expand. The people living in the stressball see distances increasing. Elastic forces are pushing outwards in all directions, and the universe is expanding. But if the elastic is uniform, the people in the stressball can detect no gravity. Now imagine that we repeat this, but this time after we've stitched up some small internal portions of the squeezeball. Now when we let go, the stitched-up regions can't expand. These are galaxies. Or suns and planets if you prefer. They're local concentrations of energy. The people in the squeezeball can now detect gravity. Note that these stitched-up regions did not make the squeezeball contract when we released it. In similar vein the universe did not contract when it was very dense. Instead it expanded, and continues to do so whilst initial small irregularities in the energy density become more intense.

Quote from: Farsight
there's no such thing as negative mass, and no such thing as negative energy.
Can you prove it?
No. But I can't prove the non-existence of tachyons or fairies either. Take a look at this article which suggests that negative mass is untenable: http://www.concentric.net/~pvb/negmass.html. It usefully goes on to talk about the negative energy of gravity, which is what I'm challenging.

That is a "trick" scenario, since you have explicity stated the two bodies do not lose any energy as a result of their collision.  In actual fact, though, the collision will cause the temperature of the combined body to be warmer than the separate bodies, and as time passes, that extra temperature will radiate away, after which the total gravitation of the combined mass will be diminished a bit, compared to the original two-body system. Therefore, the potential energy of the original two-body system can indeed be said to have existed as mass (somewhere in the system), which was converted to kinetic energy (heat) during the collision and radiated away after the collision.
I didn't mean it to be a trick at all.
Title: If I give an object some potential energy, does its mass increase?
Post by: Farsight on 04/08/2009 01:41:33
Quote from: Pmb
MTW give the stress-energy-momentum tensor for the gravitational field as that derived by Landau and Lifshitz. On 465 they give an expression for it for a weak gravitational field. The energy density is given in terms of phi,j*phi,j. For a uniform gravitational field phi = gz/c^2 an so the energy density is uniform. The field is still there though, i.e. you'd still "fall down".
I know Misner/Thorne/Wheeler is an authoritative text, but let's take a look at that energy density. We imagine ourselves to be in a room of modest proportions. We raise an object to the top of the room and drop it. We also drop an object from halfway up the room. We notice no difference in the acceleration of these dropped objects, and conclude that the force of gravity at all points within the room is the same as far as we can measure it. There's no discernible tidal force. Now we perform a Pound-Rebka experiment. We detect a blueshift and know that at the bottom of our room, clocks and all processes are running at a slightly reduced rate as compared to the middle of the room. We repeat higher up, and find the same reduced rate between the top and the middle. Again we conclude that our gravitational field is uniform. Now we crouch down to the floor and use the Casimir effect to get some measure of the vacuum energy at that location. It isn't ideal, see http://www.math.ucr.edu/home/baez/vacuum.html but it'll have to do for now. We get a positive reading. The space near the floor has some positive energy pressing the plates together. We raise our apparatus to the middle and take another positive reading, and another near the ceiling. All positive. There is no negative energy to be found. Some might contrive it by claiming that the plates are being pulled together by a tension between them, but we know it isn't so. All points in the room show positive energy. So where is our negative gravitational energy? Consider the Pound-Rebka photon. It was blueshifted down near the floor. It appeared to gain energy. Where from? From the earth? No, there is no magical mysterious action-at-a-distance. Whatever is affecting that photon is local, it's right here in this room. Did the energy come from the surrounding vacuum energy? If so, what happens when we point a laser at the floor, or drop a pile of plates? Do we detect energy being sucked out of our local space? No. So where does the energy come from? Think about that blueshifted Pound-Rebka photon. It's a wave in space, and it's moved down to a region where clocks run slower. That includes light-clocks. So it's moved into a region where light runs slower. Its frequency hasn't changed, what's changed is the environment it moved into, affecting clocks and electrons and iron atoms. And like the plate, that apparent extra energy came from the photon itself. It only looks like it changed frequency, because all measuring processes run slower down by the floor. Because there's a gradient in the energy density between the floor and the ceiling, and the gravitational energy density is a measure of the difference.

Quote from: Pmb
Any object only "feels" a force on it when there is stress imposed in the body by the force field. If you're falling in a uniform gravitational field you'd feel no force. But the same would happen for a charged body in a uniform electric field too. If you're in a non-uniform gravitational field then you'd feel tidal forces since they'd impose stresses in your body. It's odd that people don't make this obvious comparison with the electric field!
But a charged body consists of electrons and protons, and as soon as you apply the electrical field you've got stresses. Ah I get you, if you limit yourself to just electrons or just protons, fair enough, they don't "feel" the force. OK, point taken.

Quote from: Pmb
Quote from: Farsight
there's no such thing as negative mass
This is wrong. I think you said this because you use the term "mass" as another name for rest energy. That's a problem.
I do use it that way. IMHO rest mass is rest energy, only it's not actually at rest. It's just moving in some circular or back-and-forth type fashion so it isn't moving in aggregate with respect to me. Now, show me some negative mass!

Quote from: Pmb
In any case dark energy is due to the presence of negative active gravitational mass density. Any time you have repulsive gravity, such as in inflation and dark energy, then you're talking about negative active gravitational mass density.
I dispute this. IMHO in the early days of the universe the energy density was high, so processes within it would have run slowly just as they do in a region of high gravitational potential, hence the initial expansion would appear to be rapid. Nowadays the galaxies aren't pushing each other apart, the space between them is expanding because it's still under pressure. The dimensionality of energy is pressure x volume, and there's nothing negative that I can see. Nor is there anything negative in a region of uniform space. Introduce a sun, and you've introduced positive energy to one portion of that region. You've also introduced a gradient in the energy density into the region we call a gravitational field. The active gravitational mass is telling you how much energy is there. I can't find any negative energy or negative mass anywhere.

Quote from: Pmb
It's also conceivable that there can be negative inertial mass if a body could exist that could support large enough values of tension. You probably never thought of this because you associate mass with rest energy. Negative inertial mass means that the momentum of a body is opposite to its velocity and that's quite possible in special relativity given large enough values of negative pressure (i.e. tension) compared to the bodies energy.
I just can't see it Pete. Sorry.

Quote from: Pmb
I'll look out for them.

Sheesh, look at the time. I've got to go. Perhaps we should talk about why things fall down, starting with the curvilinear path of a photon. It's good to talk.
Title: If I give an object some potential energy, does its mass increase?
Post by: Pmb on 04/08/2009 06:43:55
Quote from: Farsight
I do use it that way. IMHO rest mass is rest energy, only it's not actually at rest. It's just moving in some circular or back-and-forth type fashion so it isn't moving in aggregate with respect to me. Now, show me some negative mass!
Let me get this straight. When you use the term “mass” you mean nothing more and nothing less than “rest mass” = “rest energy/c^2” and you use the symbol “m” to represent it, correct? If so then I know of no situation where there is negative rest energy so there is no example I know of where your “mass” is negative. I assume you use it in the expression for momentum, i.e. p = gamma*m*v, right?

There is a problem with this definition of mass. It can only be used in special cases, i.e. it can’t be used in general circumstances. Especially when it comes to the mass density of, say, a magnetic field or when you have a body under stress or if your speaking of the mass of a fluid which has pressure.
Quote from: Farsight
I dispute this.  IMHO in the early days of the universe the energy density was high, so processes within it would have run slowly just as they do in a region of high gravitational potential, hence the initial expansion would appear to be rapid. Nowadays the galaxies aren't pushing each other apart, the space between them is expanding because it's still under pressure.
I don’t understand what you mean by this? Please explain what pressure you are referring to. And what I said does not merely refer to the expansion of space but that it’s expanding at an accelerating rate.
Quote from: Farsight
The dimensionality of energy is pressure x volume, and there's nothing negative that I can see. Nor is there anything negative in a region of uniform space. Introduce a sun, and you've introduced positive energy to one portion of that region. You've also introduced a gradient in the energy density into the region we call a gravitational field. The active gravitational mass is telling you how much energy is there. I can't find any negative energy or negative mass anywhere.
Do you know what a relativist means when the speak of “negative active gravitational mass density”? I’ve provided a web page to describe this term

http://www.geocities.com/physics_world/gr/active_grav_mass.htm

First off nobody knows why this is happening so its no wonder you can’t find anything. The term “negative active gravitational mass density” means nothing more and nothing less than there is gravitational repulsion and that the source of gravity is less than zero. E.g. for a relativistic fluid with proper energy density u and pressure p, in the weak field limit Einstein’s field equation becomes  del^2 phi = 4pi*G/c^2(u + 3p). The term (u+3p) is called the “active gravitational mass density”. When it’s less than zero (i.e. “negative mass) there is gravitational repulsion. This happens when there is negative pressure (i.e. tension). This happens in a vacuum domain wall. The tension is so large as to overwhelm the energy density and this gives rise to a negative active gravitational mass density

Quote from: Farsight
I just can't see it Pete. Sorry.
All you have to do is look at the situation of a body under tension. I’ve worked out an example at
http://www.geocities.com/physics_world/sr/inertial_energy_vs_mass.htm

Express the stress-energy-momentum tensor in coordinates where a body is moving. Let the speed be small so that p = mv. This “m” is what you call mass, right? The situation I’ve described m is a function of the tension (i.e. negative pressure) in the body and if it’s large then M will be less than zero. But it is nor proportional to the bodies rest energy. That’s why I don’t use the concept of mass as rest energy/c^2.

Title: If I give an object some potential energy, does its mass increase?
Post by: lightarrow on 04/08/2009 10:50:02
I sincerely don't understand this. Consider a region of space in the void with a uniform electric field E. The energy density is (1/2)ε0E2, so it's uniform as well, so its gradient is zero. Would you deduce that the field is zero?
Not at all. But the gravitational field is not the same as an electric field. Consider an electron and a proton, and let them attract one another to form a hydrogen atom. Their fields mask one another, and the total energy will be reduced by photon emission to give the negative binding energy. It's a very different situation for two masses.
But it's not different, because the gravitational energy density is negative, not positive as in the electrostatic case. When two massive bodies come very close, the field is increased (differently from the electrostatic case of two opposite charges), but being the field's energy negative, that means that the field's energy decreased, exactly as in the electrostatic case, so the result is the same.

Quote
Quote from: Farsight
there's no such thing as negative mass, and no such thing as negative energy.
Can you prove it?
No. But I can't prove the non-existence of tachyons or fairies either. Take a look at this article which suggests that negative mass is untenable: http://www.concentric.net/~pvb/negmass.html. It usefully goes on to talk about the negative energy of gravity, which is what I'm challenging.
The article says that a body with negative mass should disintegrate (= reduce to zero any gradient of that mass), but it doesn't say that regions of space with negative energy and so negative mass cannot exist. If space is uniformly filled with negative energy, then you take 1 cubic metre of fixed and still region of space and it has negative mass, exactly in the same way as, if a fixed and still volume of space contains photons so it has a total energy E, that volume of space have mass m = E/c2.
Title: If I give an object some potential energy, does its mass increase?
Post by: VernonNemitz on 04/08/2009 20:18:42
I didn't mean it to be a trick at all.
Remember that the Title of this Thread is about a relationship between (gravitational) potential energy and mass.  If one wishes to argue that there is no such thing, then to ignore the real world and focus on an idealized scenario is "a trick" to support that argument.  Meanwhile, in the real world, gravitational potential energy exists, and can be converted into loose energy.  I described that in my last post.  Is it not logical that if some loose energy escapes a system, then the total mass of that system (since energy is equivalent to mass) must have diminished?  And is it not logical that if the loose energy appeared at the expense of potential energy, then that is, in effect, a conversion of mass (which will diminish when the energy escapes) into loose energy?  Therefore it is simply logical to conclude that gravitational potential energy, when it exists, exists in the form of a slight portion of the total mass of some gravitationally interacting system.

Perhaps I didn't say the preceding clearly enough.  We observe that mass and loose energy, such as photons, are interchangeable.  There are mass-spectrometer measurements that directly support this observation.  Some atoms can absorb a gamma ray and become "metastable" for a modest amount of time, after which the gammay ray is re-emitted.  But while existing in the metastable state, an atom can measurably have more mass than it does in the ground state.

Next, we observe potential energy transform into kinetic energy, when two masses gravitationally approach each other.  And we observe that if two masses collide after such an approach, their overall kinetic energy sort-of-transforms into the kinetic energy of vast numbers of individual atoms and molecules.

That's a pseudo-relativity thing; if a meteoroid can be imagined as consisting of lots of atoms inside an invisible and insubstantial container, and the meteoroid hits the Moon, then the just before impact, all of its molecules, relative to each other, are moving much like they would if the meteoroid was in stationary in Outer Space.  Here they just happen also to be simultaneously moving en masse toward the Moon, at a velocity none of them, relative to each other, can notice.  But when that invisible container impacts the Moon and stops, all the molecules keep on going, many separately!

Next, we are aware that for a hot massive body, ignoring any radiatings of loose energy, the kinetic energy of its molecules and atoms contribute to the total gravitational mass of the body.  In the case of our collided meteoroid, it has mostly become part of the Moon and is now contributing to the Moon's total gravitation.

Finally, we observe that indivdual energetic molecules or atoms can radiate photons during collisions with each other; kinetic energy transforms into radiant energy.  Per the first thing above in this chain of reasoning, mass must diminish as energy is loosed.  NET LOGICAL CONCLUSION: Gravitational potential energy, which is converted first to kinetic energy and then into radiant energy, can successfully be described as being part of the overall mass of the initial "system", prior to the conversion of potential energy into kinetic energy.
Title: If I give an object some potential energy, does its mass increase?
Post by: Farsight on 05/08/2009 00:15:28
Let me get this straight. When you use the term “mass” you mean nothing more and nothing less than “rest mass” = “rest energy/c^2” and you use the symbol “m” to represent it, correct?
Yes, I do.

If so then I know of no situation where there is negative rest energy so there is no example I know of where your “mass” is negative.
Me too.

I assume you use it in the expression for momentum, i.e. p = gamma*m*v, right?
No. There's a symmetry between momentum and inertia. If it's you moving and this thing is hard to shift, we're talking about inertia. But if it's the other thing moving and it's hard to stop, we're talking about momentum. And we're never too sure about who's actually doing the moving. Momentum is not mass x velocity. Mass is inertia. It's how you see momentum when you're moving instead of it.

There is a problem with this definition of mass. It can only be used in special cases, i.e. it can’t be used in general circumstances. Especially when it comes to the mass density of, say, a magnetic field or when you have a body under stress or if your speaking of the mass of a fluid which has pressure.
This problem is reflected in my reticence to ascribe a mass to a volume of space. This space has an inherent energy, so it has a mass-equivalent. But you can't move it. So there's no definable resistance to motion. By the way, there is no such thing as a magnetic field. There's a magnetic force, and an electromagnetic field, but not a magnetic field. This goes back to Minkowski's wrench in Raum und Zeit. Let's atart a new thread to discuss these fundamentals of electromagnetism

Quote from: Farsight
I dispute this. IMHO in the early days of the universe the energy density was high, so processes within it would have run slowly just as they do in a region of high gravitational potential, hence the initial expansion would appear to be rapid. Nowadays the galaxies aren't pushing each other apart, the space between them is expanding because it's still under pressure.
I don’t understand what you mean by this? Please explain what pressure you are referring to.
It's a fundamental pressure. Energy is pressure x volume, and space has a volume. When you look at bonds and electrons and photons, you end up thinking in terms of spatial pressure and Weyl gauge change. It's the sort of thing Tolman was referring to, and Einstein too, though it isn't a perfect fluid. It's better to think in terms of a solid. Let's come back to this one once we've got gravity nailed down.

Quote from: Farsight
The dimensionality of energy is pressure x volume, and there's nothing negative that I can see. Nor is there anything negative in a region of uniform space. Introduce a sun, and you've introduced positive energy to one portion of that region. You've also introduced a gradient in the energy density into the region we call a gravitational field. The active gravitational mass is telling you how much energy is there. I can't find any negative energy or negative mass anywhere.
Do you know what a relativist means when the speak of “negative active gravitational mass density”? I’ve provided a web page to describe this term

http://www.geocities.com/physics_world/gr/active_grav_mass.htm
Yes, of course I do. I read everything on your excellent website more than two years ago. And I am very much a relativist. But the view shifts.

First off nobody knows why this is happening so its no wonder you can’t find anything. The term “negative active gravitational mass density” means nothing more and nothing less than there is gravitational repulsion and that the source of gravity is less than zero. E.g. for a relativistic fluid with proper energy density u and pressure p, in the weak field limit Einstein’s field equation becomes  del^2 phi = 4pi*G/c^2(u + 3p). The term (u+3p) is called the “active gravitational mass density”. When it’s less than zero (i.e. “negative mass) there is gravitational repulsion. This happens when there is negative pressure (i.e. tension). This happens in a vacuum domain wall. The tension is so large as to overwhelm the energy density and this gives rise to a negative active gravitational mass density
I'm sorry Pete, but a vacuum domain wall is an abstraction. We see no such thing. And nor do we see gravitational repulsion. The galaxies do not propel themselves apart by pushing against one another. Instead the space between them expands. Energy spreads. That's what energy does.

Quote from: Farsight
I just can't see it Pete. Sorry.
All you have to do is look at the situation of a body under tension. I’ve worked out an example at http://www.geocities.com/physics_world/sr/inertial_energy_vs_mass.htm
This example is akin to ascribing a mass to an air-filled balloon in water. When you move past it rapidly, and strike it, it resists motion. But this resistance is not due to the balloon itself, it's due to the surrounding water. As such it's akin to ascribing the presssure on the Casimir plates to a tension between them, and is reminiscent of the negative-energy gravity issue. Let's have a new thread on mass and meanwhile get back to that falling plate and why things fall down.

Express the stress-energy-momentum tensor in coordinates where a body is moving. Let the speed be small so that p = mv. This “m” is what you call mass, right?
No, see above. Mass is energy/momentum when it's not moving in aggregate with respect to you. A photon moves at c, and you can't make it go faster or slower, so mass does not apply. But employ pair production to split it, and make it move round and round at c, and a 511keV phoron is "going nowhere fast". Only now we call it an electron. When you then move it, p=mv momentum is measuring how much this energy that is "going nowhere fast", is going somewhere. I know this sounds unfamiliar, but that's how it is.

The situation I’ve described m is a function of the tension (i.e. negative pressure) in the body and if it’s large then M will be less than zero. But it is nor proportional to the bodies rest energy. That’s why I don’t use the concept of mass as rest energy/c^2.
You're describing something like the balloon in water. It's a hole in space, and there is no such thing. A black hole is the closest I can think of (see gravastar), but with a black hole we have a positive mass/energy that exerts a considerable gravitational field. There simply is nothing with any negative mass or negative energy. It makes as much sense as a negative length or a negative carpet.

LightArrow/Vernon: sorry, I've got to go. I'll catch up with you tomorrow.

Title: If I give an object some potential energy, does its mass increase?
Post by: Pmb on 05/08/2009 11:39:49

Farsight; when you defined rest mass = rest energy/c^2 all you did was give rest energy a new name with new units. Merely defining a term does not justify its name. You’d have to demonstrate how it has the properties associated with inertia. Even then you’d only be using your own pet notion of mass as it pertains to acceleration. In relativity mass is always defined as to how I relates to momentum. You also claimed that a domain wall is an abstraction but you didn’t justify that claim other than claiming, “we see no such thing.” Vacuum domain walls were left over from phase transitions in the early universe. Just because you don’t see then it doesn’t mean they don’t exist or never have existed. Heck, it wasn’t until recently that black holes were actually detected but we spoke of them very meaningfully before that. Same with cosmic strings. Heck, I can’t see the other side of the universe but I’m pretty sure its there. :) You might not approve of the use of the term “mass” in cosmology and relativity texts/journals, but it’s there.
Title: If I give an object some potential energy, does its mass increase?
Post by: Farsight on 05/08/2009 19:57:32
But it's not different, because the gravitational energy density is negative, not positive as in the electrostatic case. When two massive bodies come very close, the field is increased (differently from the electrostatic case of two opposite charges), but being the field's energy negative, that means that the field's energy decreased, exactly as in the electrostatic case, so the result is the same.
It's a very different situation for two masses as compared to an electron and a proton. Assigning a negative energy density to a gravitational field is trying to make the two things more similar than they are. This is why people think the kinetic energy of the falling plate comes from somewhere other than the plate. Stop and think it through right from the beginning. Start with a region of space. It has some given energy density, which is positive. Now take one plate. Examine it, see that it too consists of positive energy. Now put the plate in the space. It exhibits a small gravitational field. But point to any location in space around that plate, and nowhere is there a negative energy density. It's all positive. There's no escaping this. Reduce your plate to a single electron, and that electron has a very small gravitational field. And the energy density of the space surrounding that electron is decreasing. There is no negative energy density to be found.

The article says that a body with negative mass should disintegrate (= reduce to zero any gradient of that mass), but it doesn't say that regions of space with negative energy and so negative mass cannot exist.
Granted.

If space is uniformly filled with negative energy, then you take 1 cubic metre of fixed and still region of space and it has negative mass, exactly in the same way as, if a fixed and still volume of space contains photons so it has a total energy E, that volume of space have mass m = E/c2.
But space is not filled with negative energy. Even when empty of photons, it's "filled" with positive energy. And when you add a plate, or a planet, to that space, you're adding even more.

Remember that the Title of this Thread is about a relationship between (gravitational) potential energy and mass. If one wishes to argue that there is no such thing, then to ignore the real world and focus on an idealized scenario is "a trick" to support that argument.
I was merely trying to simplify the argument to demonstrate conservation of energy.

Meanwhile, in the real world, gravitational potential energy exists, and can be converted into loose energy. I described that in my last post. Is it not logical that if some loose energy escapes a system, then the total mass of that system (since energy is equivalent to mass) must have diminished?
Yes, absolutely.

And is it not logical that if the loose energy appeared at the expense of potential energy, then that is, in effect, a conversion of mass (which will diminish when the energy escapes) into loose energy?
Yes.

Therefore it is simply logical to conclude that gravitational potential energy, when it exists, exists in the form of a slight portion of the total mass of some gravitationally interacting system.
Yes. You've nearly got it. Think about a plate in space. There is no nearby planet. We can detect no measurable gravitational field. We know that the plate has a gravitational field, but it's so slight that we can't detect it. This plate does however have a measurable mass. After some period a planet comes on to the scene, and the plate falls towards the planet, acquiring a velocity of 11km/s and hence considerable kinetic energy. This is readily measurable. The plate hits the planet and this kinetic energy is radiated away into space. The mass of the planet+plate system is reduced by this amount. Now we examine the plate with a microscope. It appears to be at the same temperature as it was when it was in space, because we emitted all of its kinetic energy. But when we examine its vibrating atoms, we notice that they are doing so at a reduced rate due to gravitational time dilation. We see that the electron "orbital" motion and spin is similarly reduced in rate. The effect is ubiquitous, and we conclude that this plate has lost energy. We therefore deduce that the kinetic energy came from the plate

Farsight; when you defined rest mass = rest energy/c^2 all you did was give rest energy a new name with new units. Merely defining a term does not justify its name. You'd have to demonstrate how it has the properties associated with inertia. Even then you'd only be using your own pet notion of mass as it pertains to acceleration. In relativity mass is always defined as to how it relates to momentum.
That's not my definition, Pete. That's Einstein's. And I do relate mass to momentum. It's intimately related, there's a symmetry between them. Here's something I've written previously to give you a flavour of my stance on this:

The answer is all down to motion. Or the lack of it. You have to get relative, because motion is relative, and you have to think in terms of momentum and inertia. A 10kg cannonball is a tangible thing, you can hold it in the palm of your hand and feel the mass of it. And if it’s travelling at 1m/s it’s “got” kinetic energy as quantified by KE=½mv2. It’s also “got” momentum, as quantified by p=mv, and it’s hard to stop. Brace yourself, then apply some constant braking force by catching it in the midriff. Ooof, and you feel the energy/momentum. Kinetic energy is looking at this in terms of stopping distance, whilst momentum is looking at it in terms of stopping time. The measure we call momentum is conserved in the collision because your gut and the cannonball shared a mutual force for the same period of time. The measure we call kinetic energy isn’t conserved, because some of the mass in motion is redirected into your deformation and heat and bruises, all of which involve mass in motion, but scattered motion instead of the tidy motion of a mass moving relative to you. Or you moving relative to it, because all the while you were never too sure whether it was you moving or the cannonball.

When we consider that the cannonball is stationary and it’s you moving at 1m/s, the cannonball hasn’t got any kinetic energy or momentum, all it’s got is inertia. You’re attempting to move it and accelerate it to your velocity of 1m/s, and that makes it harder to start, not harder to stop. In this respect momentum and inertia can be considered as two different aspects of the same thing, and that thing is energy.

You also claimed that a domain wall is an abstraction but you didn't justify that claim other than claiming, "we see no such thing." Vacuum domain walls were left over from phase transitions in the early universe. Just because you don't see them it doesn't mean they don't exist or never have existed.
True. But you're trying to support something we don't actually see with something else that we don't actually see. I'm very particular about what's actually there and what we can actually observe.

Heck, it wasn't until recently that black holes were actually detected but we spoke of them very meaningfully before that.
Not a problem. Black holes go all the way back to Michell, J., 1784, "On the Means of discovering the Distance, Magnitude, etc. of the Fixed Stars, in consequence of the Diminution of the velocity of their Light, in case such a Diminution should be found to take place in any of them, and such Data should be procurred from Observations, as would be farther necessary for that Purpose", Philosophical Transactions, 74: 35-57.

Same with cosmic strings. Heck, I can't see the other side of the universe but I'm pretty sure its there. You might not approve of the use of the term "mass" in cosmology and relativity texts/journals, but it's there.
Come on Pete, cosmic strings just aren't in the same league as black holes. And why might I not approve of the term "mass" in relativity texts and journals? Again here's something else I've written previously, I didn't think I was out of line on this.

We use the word "mass" in many different ways, and it really doesn't help. The accepted definition of mass is rest mass, which is the same thing as invariant mass, intrinsic mass, and proper mass - it's defined as the total energy of a system divided by c2. There’s also active gravitational mass which tells you how much gravity the energy causes, and passive gravitational mass, which is a measure of how much an object is attracted by gravity. People also talk of inertial mass, which tells us how much force we need to apply to accelerate or decelerate an object. It doesn't apply for a photon because it travels at c, and you can't make it go faster or slower. Then there’s relativistic mass, which is just a measure of energy, which is why it applies to a photon. When you apply it to a cannonball travelling at 1000m/s, it’s a measure that combines the rest mass energy with the kinetic energy into total energy.
Title: If I give an object some potential energy, does its mass increase?
Post by: VernonNemitz on 06/08/2009 03:14:02
we conclude that this plate has lost energy. We therefore deduce that the kinetic energy came from the plate
Heh, you are now the one that has almost got it right.  Remember that Isaac Newton's insight started with the question, "Why doesn't the Earth fall up to the apple?"

As you probably know, Newton concluded that both the apple and the Earth fall toward each other, but the Earth's motion isn't so obvious because of its fantastically greater mass.  Nevertheless, BOTH the apple and the Earth (or in your own text, the plate and the planet) must lose mass, to account for the kinetic energy that can appear during the collision and be radiated away after the collision.

Furthermore, there is a small fly in the ointment, regarding the most simple interpretation of what I quoted from your text, and ratios of the masses of plate and planet, and the ratios of the loss of mass, as gravitational potential energy is converted into kinetic energy.

See, physicists like to switch "reference frames" a lot.  General Relativity makes it an easy way to simplify some of the math they do.  For example, consider an electron and a proton, separated by a great distance.  We know that if they get close together and form a hydrogen atom, some energy will be released in the process.  If we assume that the "system" of proton and electron lost mass as they got together, to explain the source of that released energy, then we need to be consistent with the "frame switching" that physicists like to do.

More specifically, the physicists can switch from a reference frame in which the proton and electron are very far apart, to another frame that is inside the volume of a hydrogen atom that is at the "ground state".  The masses of the two particles do not change, when they make that frame switch!!!

So, where did the released energy come from?  Physicists use a mathematical trick called "negative binding energy".  (OK, it is used more in nuclear physics than in atom-formation, but rest assured it CAN be used as I've described here.)  It is only a trick, though, used to allow calculations involving mass to come out the same in any reference frame.  An alternate trick is necessary if one wishes to stick with one reference frame and describe events in other frames, without actually switching to them.

That alternate trick involves paying attention to certain key ratios in different environments.  If the ratios stay the same, then the calculations can also yield consistent results across different frames.  For electron and proton, the proton has about 1836 times greater mass.  If they maintain that ratio (among others) both far apart and close together, then calculations describing their behavior will be consistent in both frames.

So, if we consider that the proton and electron lose mass as they approach each other, converting it to kinetic energy that will be radiated when they coalesce into a hydrogen atom, then it is necessary for the proton to lose 1836 times as much mass as the electron, during the event, for both to have the same mass ratio at the end of the event, as they had at the beginning.

The same logic applies to gravitational potential energy.  It means the planet must lose more mass than the plate, too!  The total quantities of mass we are talking about are immeasurably small in this case, of course.  But the logic demands it happen that way, for alternate-reference-frame descriptions to have mathematically consistent results without invoking negative binding energy.

For more details, including how it could be possible for Massive Object A to lose more mass than Light Object B, while B acquires lots more kinetic energy than A as they accelerate toward each other, see the essay I mentioned earlier in this Thread:  http://www.nemitz.net/vernon/STUBBED2.pdf
Title: If I give an object some potential energy, does its mass increase?
Post by: Pmb on 06/08/2009 12:37:17
Quote from: Farsight
That's not my definition, Pete. That's Einstein's.
Huh? Where did you get that impression from?

You didn’t address the concerns in my response, i.e. that merely defining a term does not justify its name. You'd have to demonstrate how it has the properties associated with inertia. I.e. if you choose to “define” mass as another name for rest energy then you have to justify it. I can call myself a pot, but that doesn’t justify you pushing me into an oven! I explained that in relativity mass is always defined in terms of momentum. All you did so far is to tell me how, in slow speeds, objets which have rest mass is “related” to momentum, i.e. presumably by p = mv. There is an important difference since if mass is defined in terms of momentum then anything which has momentum has mass. This does not hold with the definition you chose. And you didn’t specify under what conditions that pertains to. If you assert that it holds in all possible applications then you are mistaken since (1) only holds in special spacetime coordinates (i.e. those corresponding to in inertial frames only frames) (2) even then only when the object is not subject to  stress.

In any case this is all still off topic since none of it addresses the original question. If the OP (HankRearden) has no further questions then it seems to me that the conversation is over, at least for me
Title: If I give an object some potential energy, does its mass increase?
Post by: Farsight on 08/08/2009 08:45:44
we conclude that this plate has lost energy. We therefore deduce that the kinetic energy came from the plate
Heh, you are now the one that has almost got it right.  Remember that Isaac Newton's insight started with the question, "Why doesn't the Earth fall up to the apple?" As you probably know, Newton concluded that both the apple and the Earth fall toward each other, but the Earth's motion isn't so obvious because of its fantastically greater mass. Nevertheless, BOTH the apple and the Earth (or in your own text, the plate and the planet) must lose mass, to account for the kinetic energy that can appear during the collision and be radiated away after the collision.
I agree. If you had two plates falling towards one another, they would each lose some energy. But that's not to say the energy loss is divided 50:50 between the earth and the plate. As an aside re Newton, check out Opticks queries 30, 20, and 21, where you read this:

"Are not gross bodies and light convertible into one another? ...Doth not this aethereal medium in passing out of water, glass, crystal, and other compact and dense bodies in empty spaces, grow denser and denser by degrees, and by that means refract the rays of light not in a point, but by bending them gradually in curve lines? ...Is not this medium much rarer within the dense bodies of the Sun, stars, planets and comets, than in the empty celestial space between them? And in passing from them to great distances, doth it not grow denser and denser perpetually, and thereby cause the gravity of those great bodies towards one another, and of their parts towards the bodies; every body endeavouring to go from the denser parts of the medium towards the rarer?"

Furthermore, there is a small fly in the ointment, regarding the most simple interpretation of what I quoted from your text, and ratios of the masses of plate and planet, and the ratios of the loss of mass, as gravitational potential energy is converted into kinetic energy.

See, physicists like to switch "reference frames" a lot.  General Relativity makes it an easy way to simplify some of the math they do.  For example, consider an electron and a proton, separated by a great distance.  We know that if they get close together and form a hydrogen atom, some energy will be released in the process.  If we assume that the "system" of proton and electron lost mass as they got together, to explain the source of that released energy, then we need to be consistent with the "frame switching" that physicists like to do.

More specifically, the physicists can switch from a reference frame in which the proton and electron are very far apart, to another frame that is inside the volume of a hydrogen atom that is at the "ground state".  The masses of the two particles do not change, when they make that frame switch!!!

So, where did the released energy come from?  Physicists use a mathematical trick called "negative binding energy".  (OK, it is used more in nuclear physics than in atom-formation, but rest assured it CAN be used as I've described here.)  It is only a trick, though, used to allow calculations involving mass to come out the same in any reference frame.  An alternate trick is necessary if one wishes to stick with one reference frame and describe events in other frames, without actually switching to them.

That alternate trick involves paying attention to certain key ratios in different environments. If the ratios stay the same, then the calculations can also yield consistent results across different frames.  For electron and proton, the proton has about 1836 times greater mass.  If they maintain that ratio (among others) both far apart and close together, then calculations describing their behavior will be consistent in both frames.

So, if we consider that the proton and electron lose mass as they approach each other, converting it to kinetic energy that will be radiated when they coalesce into a hydrogen atom, then it is necessary for the proton to lose 1836 times as much mass as the electron, during the event, for both to have the same mass ratio at the end of the event, as they had at the beginning.
This isn't right. The kinetic energy is 1/2mv2, and if the larger mass is moving slower than the smaller mass the ratios are skewed.

The same logic applies to gravitational potential energy. It means the planet must lose more mass than the plate, too! The total quantities of mass we are talking about are immeasurably small in this case, of course. But the logic demands it happen that way, for alternate-reference-frame descriptions to have mathematically consistent results without invoking negative binding energy.
With the falling plate example we use the planet as our reference frame and we say that mass of the plate is so small that the planet's motion is not detectable. The plate's motion however is detectable. It's 11km/s. Once it's on the ground having lost its kinetic energy, the gravitational time dilation means everything moving in that plate, be it molecules or atoms or electrons or light, is moving slower than it was. That's where the energy came from.

For more details, including how it could be possible for Massive Object A to lose more mass than Light Object B, while B acquires lots more kinetic energy than A as they accelerate toward each other, see the essay I mentioned earlier in this Thread:  http://www.nemitz.net/vernon/STUBBED2.pdf
I'll check it out.
Title: If I give an object some potential energy, does its mass increase?
Post by: Farsight on 08/08/2009 09:04:48
Quote from: Farsight
That's not my definition, Pete. That's Einstein's.
Huh? Where did you get that impression from?
From Einstein's 1905 paper Does the Inertia of a Body Depend on its Energy Content?

You didn’t address the concerns in my response, i.e. that merely defining a term does not justify its name. You'd have to demonstrate how it has the properties associated with inertia. I.e. if you choose to “define” mass as another name for rest energy then you have to justify it.
It isn't my definition.

I can call myself a pot, but that doesn’t justify you pushing me into an oven! I explained that in relativity mass is always defined in terms of momentum. All you did so far is to tell me how, in slow speeds, objects which have rest mass is “related” to momentum, i.e. presumably by p = mv.
I wasn't referring to p=mv. A photon has momentum p=hf/c but it has no mass. If you employ pair production the photon is no longer moving at c, and the electron and the positron do have mass. Combine them via annihilation and now you have two photons, but they don't have mass. Inertia is momentum when it's you moving instead of the other thing.

There is an important difference since if mass is defined in terms of momentum then anything which has momentum has mass. This does not hold with the definition you chose. And you didn’t specify under what conditions that pertains to. If you assert that it holds in all possible applications then you are mistaken since (1) only holds in special spacetime coordinates (i.e. those corresponding to in inertial frames only frames) (2) even then only when the object is not subject to  stress.
I might have said more, but what's important is that a photon has momentum but it doesn't have mass. To give it mass, you have to be moving instead of the photon, and the photon always moves at c. Until you employ pair production, then its moving at c but going round and round as electron spin, so in aggregate it isn't moving.

In any case this is all still off topic since none of it addresses the original question. If the OP (HankRearden) has no further questions then it seems to me that the conversation is over, at least for me
The fallen plate has less energy once the kinetic energy has dispersed. So its mass is reduced. The lifted plate has been given more energy, so its mass is increased. Everything we've been talking about is relevant to the OP, and IMHO it's very interesting how a simple question can lead into a focussed conversation on both gravity and mass.
Title: If I give an object some potential energy, does its mass increase?
Post by: Pmb on 09/08/2009 03:04:20
Quote from: Farsight
From Einstein's 1905 paper Does the Inertia of a Body Depend on its Energy Content?
That was merely his first paper on the subject of mass, not his last. Over the years he refined the subject and developed relations for more and more general cases. It appears to me that you mistook his first word on the subject for his final word. For example see The Principle of Conservation of the Center of Gravity and the Inertia of Energy, Albert Einstein, Annalen der Physik, 20 (1906): 626-633. In this paper Einstein assigns a mass density to radiation. In still later work he developes an expression for the inertia of stress and finally states that mass is completely defined my the energy-momentum tensor. With that tensor one can prove all the properties that physicists attribute to mass, such as the fact that the inertial mass density of a gas is a function of pressure.
Title: If I give an object some potential energy, does its mass increase?
Post by: Raghavendra on 09/08/2009 09:21:07
No.. You can't increase the mass of a plate.
Title: If I give an object some potential energy, does its mass increase?
Post by: VernonNemitz on 10/08/2009 15:02:40
So, if we consider that the proton and electron lose mass as they approach each other, converting it to kinetic energy that will be radiated when they coalesce into a hydrogen atom, then it is necessary for the proton to lose 1836 times as much mass as the electron, during the event, for both to have the same mass ratio at the end of the event, as they had at the beginning.
This isn't right. The kinetic energy is 1/2mv2, and if the larger mass is moving slower than the smaller mass the ratios are skewed.
Do not confuse the kinetic energies of the interacting objects with their masses.  The key is to always remember that we are talking about INTERACTIONS.  The mass lost by Object A appears as the kinetic energy of Object B, and vice-versa.
With the falling plate example we use the planet as our reference frame and we say that mass of the plate is so small that the planet's motion is not detectable. The plate's motion however is detectable. It's 11km/s. Once it's on the ground having lost its kinetic energy, the gravitational time dilation means everything moving in that plate, be it molecules or atoms or electrons or light, is moving slower than it was. That's where the energy came from.
Certainly the energy that emanates from the collided bodies came from the kinetic energies of the bodies.  But we are talking about where the kinetic energy came from: mass.

To Raghavendra:
If we assume a constant vibration rate --a kind of velocity-- for the molecules of some resting mass, at various locations in a gravitational field gradient, then:
1. General Relativity is a well-verified theory; the longer lengths and the slower time associated with the depths of a gravity gradient are widely accepted by physicists.
2. Quantum Mechanics and the wave/particle duality are also well-verified; the association of momentum (and thus mass if velocity is constant) with length and frequency is also widely accepted.
3. Therefore a mass' magnitude must change with its resting place (still assuming constant molecular vibrations) in a gravity gradient, just as does its gravitational potential energy!
Title: If I give an object some potential energy, does its mass increase?
Post by: Farsight on 10/08/2009 19:02:09
Quote from: Pmb
That was merely his first paper on the subject of mass, not his last. Over the years he refined the subject and developed relations for more and more general cases. It appears to me that you mistook his first word on the subject for his final word. For example see The Principle of Conservation of the Center of Gravity and the Inertia of Energy, Albert Einstein, Annalen der Physik, 20 (1906): 626-633. In this paper Einstein assigns a mass density to radiation. In still later work he developes an expression for the inertia of stress and finally states that mass is completely defined by the energy-momentum tensor. With that tensor one can prove all the properties that physicists attribute to mass, such as the fact that the inertial mass density of a gas is a function of pressure.
No problem. The point is that I wasn't just making up that definition. The mass of the plate is a measure of its energy content. And the dimensionality of energy is stress x volume.

No. You can't increase the mass of a plate.
Take a planet sitting motionless in space. Take a plate, motionless on that planet. Fire that plate at 11km/s and give it enough kinetic energy to achieve escape velocity. Examine the plate later when it is motionless in space with respect to the planet. You gave it more energy. You gave it more mass.

Do not confuse the kinetic energies of the interacting objects with their masses. The key is to always remember that we are talking about INTERACTIONS. The mass lost by Object A appears as the kinetic energy of Object B, and vice-versa.
Sorry Vernon, we've got a plate in a gravitational field. There's is no magical mysterious action-at-a-distance between the plate and the earth. The earth doesn't lose mass because the plate is falling. There isn't time for the earth/plate interaction to occur. And the earth's gravitational field doesn't lose mass either. The energy in the surrounding region of space, where the gravitational field is, increases.

Certainly the energy that emanates from the collided bodies came from the kinetic energies of the bodies. But we are talking about where the kinetic energy came from: mass.
We agree that the kinetic energy comes from the mass. I'd hope your reply to Raghavendra might make you appreciate that you can take this a stage further to agree on which mass it comes from.

Let me outline why things fall down, because I think you'll find it useful:

Many people think of gravity as "curved spacetime", and think this is the cause of gravity. It isn't. There's a further issue in that people see the bowling-ball-rubber-sheet analogy, and think the curvature here is the curved spacetime. It isn't. This is depicting the way gravity reduces in line with the inverse square law:

(https://www.thenakedscientists.com/forum/proxy.php?request=http%3A%2F%2Fwww.science.psu.edu%2Fjournal%2FSummer2005%2Fimages%2FrubberSheet72.jpg&hash=5541f7f55134158bc50600b30d3df0b9)

Take a sample square from the picture above and zoom into it to give yourself a smaller square. Keep zooming in until you can't see any of that original curvature. The square you're looking at represents space, and whilst it exhibits no curvature, it isn't flat because it exhibits a gradient:

(https://www.thenakedscientists.com/forum/proxy.php?request=http%3A%2F%2Fwww.learnenglish.de%2FIMAGES%2FVocab%2FShapes%2Fparallelogram.gif&hash=4f691be935e4d00928ffd790302a81ef)

There is a gradient in the local properties of space that you can detect in a room when you drop a pencil or throw a ball. Roll a marble across the square and track its path over time. This path is curved. That's curved spacetime. It isn't the cause of gravity, the gradient is the cause. The curved spacetime is the effect.

Here's the analogy: You have a toy motor boat. You set it going up a boating lake, and it goes straight as a die. Its path doesn't curve. Now we repeat the experiment, only this time I've shovelled a truckful of powdered wallpaper paste all down the left hand side of the lake. We wait overnight, and in the morning we find we've got a viscosity gradient from left to right across the lake. This is an analogy for the energy density gradient in the space surrounding the planet. We set the toy motorboat going up the lake. This time it doesn't travel in a straight line, it curves to the left. This toy motorboat equates to a passing photon which is deflected by the gravitational field.

Now we retrieve our boat, and jam its rudder over to the left. We set it down on the lake with the motor running, and it travels round and round in a tight circle. But every time it's going up the lake, it veers over to the left ever so slightly. It also does this when it's coming round the other way. As a result, it gradually works its way over to the left. This toy motorboat equates to an electron falling straight down in a gravitational field. (I'll have to tell you more about pair production to explain why, but please take it as read for now). Note that the circling boat is travelling up or down the lake for only half of each cycle. For the other half it's travelling left and right. You can simplify the circle to a square to appreciate this. This is why matter is deflected half as much as light.

Finally, observe the boat when its has worked its way over to the left hand side of the lake. It's still going round in circles, but at a slower rate than it did, because of the increased viscosity. The reduced cycle rate is related to the rate of travel to the left. The cycling motion is being converted into lateral motion.

Think of your plate as a whole flotilla of boats. The reduced cycle rate represents reduced energy within the plate. The travel to the left represents the kinetic energy. If you discard the latter, you're left with a reduced mass in your plate. It's really simple once it clicks. There's nothing mysterious about it, and no negative mass or energy.
Title: If I give an object some potential energy, does its mass increase?
Post by: VernonNemitz on 11/08/2009 15:20:58
Do not confuse the kinetic energies of the interacting objects with their masses. The key is to always remember that we are talking about INTERACTIONS. The mass lost by Object A appears as the kinetic energy of Object B, and vice-versa.
Sorry Vernon, we've got a plate in a gravitational field. There's is no magical mysterious action-at-a-distance between the plate and the earth. The earth doesn't lose mass because the plate is falling. There isn't time for the earth/plate interaction to occur. And the earth's gravitational field doesn't lose mass either. The energy in the surrounding region of space, where the gravitational field is, increases.

Certainly the energy that emanates from the collided bodies came from the kinetic energies of the bodies. But we are talking about where the kinetic energy came from: mass.
We agree that the kinetic energy comes from the mass. I'd hope your reply to Raghavendra might make you appreciate that you can take this a stage further to agree on which mass it comes from.
I will continue to disagree with you, because at the end of YOUR scenario the two masses no longer have the same mass ratio they started with, which violates General Relativity's allowing of easy reference-frame-switching (in which masses don't change at all).  You have offered nothing at all to deal with that very significant problem!

Furthermore, GR isn't the Last Word on gravitation; Quantum Mechanics is going to eventually have a very significant "say" on the subject (has already had some; look up "Hawking Radiation"), and indeed there will be interactions, and time for interactions, when that "say" arrives in detail.  My argument that the mass lost by A appears as the kinetic energy of B, is based on the inevitability that QM will have its "say".
Title: If I give an object some potential energy, does its mass increase?
Post by: Pmb on 12/08/2009 16:30:07
No.. You can't increase the mass of a plate.
Responses like this are rarely helpful since they convey no understanding. It's already been agreed that (1) the proper mass does not change and (2) the relativistic mass does. Simply saying You can't increase the mass of a plate adds nothing to the conversation.
Title: If I give an object some potential energy, does its mass increase?
Post by: Pmb on 12/08/2009 17:08:41
Quote from: Farsight
The mass of the plate is a measure of its energy content. And the dimensionality of energy is stress x volume.
There is very serious/major flaw in that kind of logic, one which I addressed in the article I wrote which is located at http://arxiv.org/abs/0709.0687

In short, just because you can give something units of energy it doesn't mean that it really is the energy of something. For example; I can ,multiply any constant which has the units of energy by v^2/c^2 and add it to the real energy. The results have the units of energy but is a meaningless quantity.
Title: If I give an object some potential energy, does its mass increase?
Post by: VernonNemitz on 12/08/2009 18:52:53
...the article I wrote which is located at http://arxiv.org/abs/0709.0687

That looks like a reasonably nice article.  I saw a couple of typos, but no big deal.  And you left out my own personal sneaking-suspicion-definition of mass:  "All mass is potential-energy-stored-as-mass."  Heh!
Title: If I give an object some potential energy, does its mass increase?
Post by: Farsight on 18/08/2009 12:00:42
Sorry everybody, I've had a big weekend and have been out of action.

I will continue to disagree with you, because at the end of YOUR scenario the two masses no longer have the same mass ratio they started with, which violates General Relativity's allowing of easy reference-frame-switching (in which masses don't change at all).  You have offered nothing at all to deal with that very significant problem!
I've offered plenty, Vernon. The bottom line is that with the falling plate scenario, the earth has no detectable motion, whilst the plate does. 11km/s is darn pretty detectable.

Furthermore, GR isn't the Last Word on gravitation; Quantum Mechanics is going to eventually have a very significant "say" on the subject..
OK, maybe GR as we understand it now isn't quite the last word on gravitation, but I'm afraid quantum mechanics isn't going to have a significant say on the subject. You can't quantize gravity.

I'm afraid Hawking Radiation is hypothesis, not a fact.

..and indeed there will be interactions, and time for interactions, when that "say" arrives in detail.  My argument that the mass lost by A appears as the kinetic energy of B, is based on the inevitability that QM will have its "say".
With respect Vernon, you're clutching at straws with this. Stick to the observables and the logic.
Title: If I give an object some potential energy, does its mass increase?
Post by: Farsight on 18/08/2009 12:46:58
Quote from: Farsight
The mass of the plate is a measure of its energy content. And the dimensionality of energy is stress x volume.
There is very serious/major flaw in that kind of logic, one which I addressed in the article I wrote which is located at http://arxiv.org/abs/0709.0687

Conclusion
I have demonstrated above that there are omissions, specious arguments
and errors on both sides of the relativistic mass/proper mass debate that has left
some areas in this dispute empty. I have also demonstrated that the claim that
relativistic mass can be replaced by inertial energy is wrong in general. I gave
one example and showed how such thinking could lead to an apparent paradox.
I have also shown that the quantity which deserves the name mass more than
anything else is relativistic mass since it is that quantity which possesses all three
aspects of mass, i.e. inertial mass, passive gravitational mass and active
gravitational mass. Accusations that relativistic mass is confusing cannot serve as
a strike against relativistic mass since almost all aspects of relativity are
confusing until they are mastered and once mastered they’re not confusing at all...

I agree with your argument, but there's no flaw in what I said. Just a simplification, and a bending to convention. Look at pair production. A photon has energy E=hf and momentum p=hf/c, but it really has energy/momentum and there's a distance-based and a time-based measure. A photon is converted into an electron and a positron. The electron has spin and angular momentum. See http://en.wikipedia.org/wiki/Planck_constant for "The Planck constant has dimensions of energy multiplied by time, which are also the dimensions of action. In SI units, the Planck constant is expressed in joule seconds (J s). The dimensions may also be written as momentum multiplied by distance (N m s), which are also the dimensions of angular momentum". In simple terms the electron is a photon travelling in a circular path, so the photon is "going nowhere fast". Hence momentum now appears as inertia. If it’s side-on and moving relative to you, you'd see this circular path looking like a helical path. One full turn round the helix represents the relativistic mass, the total energy. The circular component of this represents the "rest" mass. The difference represents the kinetic energy. It tells you how fast the energy that's going nowhere fast, is going somewhere.

In short, just because you can give something units of energy it doesn't mean that it really is the energy of something. For example; I can multiply any constant which has the units of energy by v^2/c^2 and add it to the real energy. The results have the units of energy but is a meaningless quantity.
Quite so. What's the energy of a photon? What you measure depends on your relative motion. Move towards it fast, and it's blue-shifted, so it appears to have a higher energy. But when you move towards that photon it doesn't acquire any extra energy at all. It didn't change, you did. It's the same for the blue-shifted photon in the Pound-Rebka experiment. It has not gained any extra energy. And it's the same for the falling plate. The total energy of the falling plate at the surface is the same as that of the plate at rest at altitude.
Title: If I give an object some potential energy, does its mass increase?
Post by: VernonNemitz on 18/08/2009 16:45:57
To Farsight:

Just because you SAY you have offered an explanation about how, before a plate falls and afterward, the mass ratios of it-to-planet don't change, that doesn't mean you have actually done any such thing.  Therefore, until I observe such an explanation, I will continue to say that your description violates General Relativity's freedom to switch reference frames (your description of where the kinetic energy comes from).  Also, I forgot to say previously that there is one other problem with a falling object sacrificing its own mass to become its kinetic energy:  It would make black holes unlikely (they would be an asymptotic limit of a curve-of-accumulation of ever-smaller arriving masses!).

Next, just because you SAY that gravity can't be quantized, that doesn't mean you are right, not at all!  Certainly you can't prove such a silly claim, and lots of people have made stabs at concocting something sensible.  String theory, for example.  Sure, not useful for much, since it hasn't made any testable predictions.  But even I can devise a reasonable QM description of gravity: http://knol.google.com/k/vernon-nemitz/simple-quantum-gravitation/131braj0vi27a/2
Part of it can even be tested: http://www.halfbakery.com/idea/Gravity_20Waves2#1225479012

Regarding Hawking Radiation, there is some evidence that you are quite wrong there.  Remember all the fuss about potential black hole formation at large particle accelerator facilities?  The fuss has been around since well before the Large Hadron Collider was constructed, see: http://en.wikipedia.org/wiki/Relativistic_Heavy_Ion_Collider
The main argument for the safety of the accelerators involves natural cosmic rays, which can be vastly more energetic than we can currently dream about making (to say nothing of actually making).  In 4 billion years of getting zapped by them, either no black hole was ever created by any of those events, able to devour the Earth, or Hawking Radiation has been there to save the world.  The likelier explanation is Hawking Radiation.  ALSO, there have been "events" at the RHIC which may be interpreted as quantum-black-hole explosions (which again are only possible per Hawking Radiation): http://news.bbc.co.uk/2/hi/science/nature/4357613.st
And there is one other category of mystery event, something called a "Bosenova", which can occur in a Bose-Einstein Condensate:  http://www.npl.washington.edu/av/altvw108.html   I'm speculating, of course, but it shouldn't take much thought to see the possibility of many atoms in a BEC, able to exist at a single point, being equivalent to the singularity in a black hole.  So, why doesn't such a black hole form, and the atoms stay trapped inside?  How about Hawking Radiation (or an equivalent)?

Finally, I remind you of Aristotle, who was quite right about a number of things, and as a result led people to think that Authority, logic, and a MINIMUM number of observables was all that was needed to reach a valid conclusion.  But Aristotle was dead wrong about objects in motion, which was why Isaac Newton had to explictly specify his First Law of Motion, even though it is "built into" the Second Law.  Newton had to overthrow Aristotlean/Authoritarian nonsense.  And that means YOUR mere say-so isn't good enough, either.
Title: If I give an object some potential energy, does its mass increase?
Post by: Farsight on 18/08/2009 19:10:12
Just because you SAY you have offered an explanation about how, before a plate falls and afterward, the mass ratios of it-to-planet don't change, that doesn't mean you have actually done any such thing.  Therefore, until I observe such an explanation, I will continue to say that your description violates General Relativity's freedom to switch reference frames (your description of where the kinetic energy comes from). Also, I forgot to say previously that there is one other problem with a falling object sacrificing its own mass to become its kinetic energy: It would make black holes unlikely (they would be an asymptotic limit of a curve-of-accumulation of ever-smaller arriving masses!).
It doesn't make black holes unlikely. Look at it this way. Imagine you've got a large region of space with a black hole in it. Now introduce a billion-tonne asteriod and let it go. It falls and falls and falls towards the black hole, accelerating all the time. It's going really fast when it gets swallowed up by the black hole. Then the mass/energy of the black hole is increased. But it's only increased by the energy-equivalent of a billion tonnes, not by any more than that.

Next, just because you SAY that gravity can't be quantized, that doesn't mean you are right, not at all! Certainly you can't prove such a silly claim, and lots of people have made stabs at concocting something sensible. String theory, for example. Sure, not useful for much, since it hasn't made any testable predictions. But even I can devise a reasonable QM description of gravity: http://knol.google.com/k/vernon-nemitz/simple-quantum-gravitation/131braj0vi27a/2
Part of it can even be tested: http://www.halfbakery.com/idea/Gravity_20Waves2#1225479012
It really isn't a silly claim, Vernon. A photon exhibits energy, so it has an active gravitational mass. And when a photon approaches you it does so smoothly. That means the gravity increases smoothly. Thinking you can quantize gravity is the silly claim (though I didn't say that to Lee Smolin).

When I looked at your link I got Knol is currently unavailable and is undergoing maintenance. I have no issue with gravity waves. One can think of them as akin to a "supersize" photon.

Regarding Hawking Radiation, there is some evidence that you are quite wrong there. Remember all the fuss about potential black hole formation at large particle accelerator facilities?  The fuss has been around since well before the Large Hadron Collider was constructed, see: http://en.wikipedia.org/wiki/Relativistic_Heavy_Ion_Collider
The main argument for the safety of the accelerators involves natural cosmic rays, which can be vastly more energetic than we can currently dream about making (to say nothing of actually making). In 4 billion years of getting zapped by them, either no black hole was ever created by any of those events, able to devour the Earth, or Hawking Radiation has been there to save the world. The likelier explanation is Hawking Radiation. ALSO, there have been "events" at the RHIC which may be interpreted as quantum-black-hole explosions (which again are only possible per Hawking Radiation): http://news.bbc.co.uk/2/hi/science/nature/4357613.st
I get a 404 page not found when I follow that link. I know about RHIC and all this stuff. Hawking radiation isn't the "likelier" explanation, and the fact remains that nobody has seen any Hawking radiation. There's absolutely no evidence for it. It remains conjecture. A hypothesis.

And there is one other category of mystery event, something called a "Bosenova", which can occur in a Bose-Einstein Condensate:  http://www.npl.washington.edu/av/altvw108.html
I know this stuff like the back of my hand:

"In the 1920s, therefore, it came as a big surprise when experimental physicists discovered that the electron breaks this quantum rule by having a spin angular momentum that is ½ of an h/2p unit (i.e., spin ½).  This means that electrons (along with quarks, neutrinos, protons, neutrons, and many atoms) are NOT the same after a 360o rotation.  They must be rotated through two full turns or 720o before they return to their original states.  This peculiar spin behavior is not well understood, but it is nevertheless an accepted fact of quantum physics".

This is a depiction of the electron:

Follow the dark black line round and note the 720o degrees of rotation to get back to the original orientation and position. However the Bosenova is very interesting. You might have drawn my attention to something important there. Thanks.

I'm speculating, of course, but it shouldn't take much thought to see the possibility of many atoms in a BEC, able to exist at a single point, being equivalent to the singularity in a black hole. So, why doesn't such a black hole form, and the atoms stay trapped inside?  How about Hawking Radiation (or an equivalent)?
There is no singularity "in" a black hole. The singularity is at the event horizon. But we're getting off the point with this. There's no evidence for Hawking Radiation. It remains a speculation. The 11 kms of the falling plate is not a speculation, and nor is conservation of energy.

Finally, I remind you of Aristotle, who was quite right about a number of things, and as a result led people to think that Authority, logic, and a MINIMUM number of observables was all that was needed to reach a valid conclusion. But Aristotle was dead wrong about objects in motion, which was why Isaac Newton had to explictly specify his First Law of Motion, even though it is "built into" the Second Law. Newton had to overthrow Aristotlean/Authoritarian nonsense. And that means YOUR mere say-so isn't good enough, either.
I don't mean to give a mere say so, I mean to give evidence and careful logic and a rational argument. Sometimes I might not give enough. I admit I haven't given enough re the quantization of gravity, but we can always talk about it further.
Title: If I give an object some potential energy, does its mass increase?
Post by: VernonNemitz on 18/08/2009 20:26:01
I forgot to say previously that there is one other problem with a falling object sacrificing its own mass to become its kinetic energy: It would make black holes unlikely (they would be an asymptotic limit of a curve-of-accumulation of ever-smaller arriving masses!).
It doesn't make black holes unlikely. Look at it this way. Imagine you've got a large region of space with a black hole in it. Now introduce a billion-tonne asteriod and let it go. It falls and falls and falls towards the black hole, accelerating all the time. It's going really fast when it gets swallowed up by the black hole. Then the mass/energy of the black hole is increased. But it's only increased by the energy-equivalent of a billion tonnes, not by any more than that.
You are mistaking what I was talking about; you are assuming an already-existing black hole.  But try adding mass to a neutron star, and you will see what I mean; if there is time for the kinetic energy after impact to radiate away, then the neutron star will have difficulty becoming a black hole, no matter how much mass falls toward it.  (Note I did originally say "unlikely", not "impossible".)

... even I can devise a reasonable QM description of gravity: http://knol.google.com/k/vernon-nemitz/simple-quantum-gravitation/131braj0vi27a/2
Part of it can even be tested: http://www.halfbakery.com/idea/Gravity_20Waves2#1225479012
It really isn't a silly claim, Vernon. A photon exhibits energy, so it has an active gravitational mass. And when a photon approaches you it does so smoothly. That means the gravity increases smoothly. Thinking you can quantize gravity is the silly claim (though I didn't say that to Lee Smolin).
When I looked at your link I got Knol is currently unavailable and is undergoing maintenance.
Hmmm...the link worked just fine a few minutes ago when I tried it, just before writing this.
Yours is still a silly claim, since you are making it in apparent ignorance of how quantized gravitation MIGHT work.  For example a photon of ordinary light has multi-terahertz frequency; how do you know it is not interacting gravitationally at a similar rate?  The "fine-ness" of Planck's Constant is plenty to allow photon-motion to be smoothly curved in a gravitational field, just as it has been plenty to make other things look smooth at our macroscopic scale.

Regarding Hawking Radiation, there is some evidence that you are quite wrong there. Remember all the fuss about potential black hole formation at large particle accelerator facilities?  The fuss has been around since well before the Large Hadron Collider was constructed, see: http://en.wikipedia.org/wiki/Relativistic_Heavy_Ion_Collider
The main argument for the safety of the accelerators involves natural cosmic rays, which can be vastly more energetic than we can currently dream about making (to say nothing of actually making). In 4 billion years of getting zapped by them, either no black hole was ever created by any of those events, able to devour the Earth, or Hawking Radiation has been there to save the world. The likelier explanation is Hawking Radiation. ALSO, there have been "events" at the RHIC which may be interpreted as quantum-black-hole explosions (which again are only possible per Hawking Radiation): http://news.bbc.co.uk/2/hi/science/nature/4357613.st
I get a 404 page not found when I follow that link. I know about RHIC and all this stuff. Hawking radiation isn't the "likelier" explanation, and the fact remains that nobody has seen any Hawking radiation. There's absolutely no evidence for it. It remains conjecture. A hypothesis.
I said, in effect, The Evidence Is That Planet Earth Is Still Here, after 4 billion years of most-extreme-energy cosmic-ray collisions.  Are you claiming that quantum black holes are impossible, or that none can ever be produced by such a collision?  What IS your explanation for that Evidence?
Regarding the link, sorry, a typo crept into it during my previous editing.  Here:  http://news.bbc.co.uk/2/hi/science/nature/4357613.stm  We are talking about actual data here, that MAY match the theoretical description of Hawking Radiation.

There is no singularity "in" a black hole. The singularity is at the event horizon.
Now you are dead wrong.  Look up the so-called "Law of Cosmic Censorship".  The singularity is the mathematical point at the center of a black hole, toward which everything inside the event horizon is endlessly falling.  Note per YOUR claim that an object's mass diminishes as it falls/accelerates, its mass can fall to zero as it reaches the event horizon, and therefore it can reach light-speed and enter the body of the black hole.  There is no big pile-up of time-slowed stuff outside the event horizon, waiting to get in.  (And in any scenario that includes Hawking Radiation, there also will be quantum fluctuations of the event horizon such that anything having >0 mass just outside the event horizon can still be swallowed.  One moment it is outside; fluctuation; now it is inside, still falling and so unable to get out again.)

Title: If I give an object some potential energy, does its mass increase?
Post by: Farsight on 19/08/2009 17:15:59
You are mistaking what I was talking about; you are assuming an already-existing black hole.  But try adding mass to a neutron star, and you will see what I mean; if there is time for the kinetic energy after impact to radiate away, then the neutron star will have difficulty becoming a black hole, no matter how much mass falls toward it. (Note I did originally say "unlikely", not "impossible".)
OK, agreed. It ties in with what I was saying about the black hole.

Hmmm...the link worked just fine a few minutes ago when I tried it, just before writing this. Yours is still a silly claim, since you are making it in apparent ignorance of how quantized gravitation MIGHT work. For example a photon of ordinary light has multi-terahertz frequency; how do you know it is not interacting gravitationally at a similar rate? The "fine-ness" of Planck's Constant is plenty to allow photon-motion to be smoothly curved in a gravitational field, just as it has been plenty to make other things look smooth at our macroscopic scale.
Just take it from me that I know how this works. It might seem like a silly claim to you, but I promise you it isn't. I'll send you something to back this up.

I said, in effect, The Evidence Is That Planet Earth Is Still Here, after 4 billion years of most-extreme-energy cosmic-ray collisions.  Are you claiming that quantum black holes are impossible, or that none can ever be produced by such a collision?  What IS your explanation for that Evidence? Regarding the link, sorry, a typo crept into it during my previous editing.  Here:  http://news.bbc.co.uk/2/hi/science/nature/4357613.stm. We are talking about actual data here, that MAY match the theoretical description of Hawking Radiation.
Yes, I'm saying quantum black holes are impossible. None will be produced by such a collision, because the required extra dimensions do not exist. I followed your link then looked at the paper at arXiv. IMHO it's clutching at straws.

Now you are dead wrong. Look up the so-called "Law of Cosmic Censorship". The singularity is the mathematical point at the center of a black hole, toward which everything inside the event horizon is endlessly falling. Note per YOUR claim that an object's mass diminishes as it falls/accelerates, its mass can fall to zero as it reaches the event horizon, and therefore it can reach light-speed and enter the body of the black hole. There is no big pile-up of time-slowed stuff outside the event horizon, waiting to get in.

And in any scenario that includes Hawking Radiation, there also will be quantum fluctuations of the event horizon such that anything having >0 mass just outside the event horizon can still be swallowed. One moment it is outside; fluctuation; now it is inside, still falling and so unable to get out again.
Isn't it supposed to be the other way round, in that a fluctuation leaves a particle outside the event horizon whereupon it escapes as Hawking radiation, stealing mass/energy from the black hole. Besides, Hawking radiation isn't really relevant. We were talking about potential energy and a plate. When you raise the plate you give the plate that potential energy. When it falls, that potential energy is converted into kinetic energy. The kinetic energy comes from the plate. It's really simple. What you've been taught would leave you with an asteroid falling into a black hole and giving it more mass/energy than the asteroid had to begin with. That breaks the rules of conservation of energy.

Title: If I give an object some potential energy, does its mass increase?
Post by: VernonNemitz on 19/08/2009 18:41:01
Hmmm...the link worked just fine a few minutes ago when I tried it, just before writing this. Yours is still a silly claim, since you are making it in apparent ignorance of how quantized gravitation MIGHT work. For example a photon of ordinary light has multi-terahertz frequency; how do you know it is not interacting gravitationally at a similar rate? The "fine-ness" of Planck's Constant is plenty to allow photon-motion to be smoothly curved in a gravitational field, just as it has been plenty to make other things look smooth at our macroscopic scale.
Just take it from me that I know how this works. It might seem like a silly claim to you, but I promise you it isn't. I'll send you something to back this up.
You should post it for all to see, if possible.  You also still need to post something I mentioned a while back: an explanation for how the mass ratio of plate-to-planet does not change, as required to be consistent with General Relativity's allowing of reference-frame-swapping, when you want the kinetic energy of the plate to appear at the expense of the mass of the plate only, when it falls toward the planet.

Yes, I'm saying quantum black holes are impossible. None will be produced by such a collision, because the required extra dimensions do not exist.
I'm not aware that extra dimensions are required for quantum-sized black holes to exist.  Black holes of any size strictly depend on the amount of mass crammed into a given volume, and nothing more than that, to the best of my knowledge.  So, again you appear to be making an arbitrary Authoritarian statement, without backing it up.

... the law of cosmic censorship is just another hypothesis.
True, but the DEFINITION of that hypothesis specifies a distinction between a singularity (more specifically, a "naked singularity") and an event horizon, which is why you are dead wrong in saying they are the same thing; physicists had specified the distinction between the two things well before the "censorship law" was proposed.

And in any scenario that includes Hawking Radiation, there also will be quantum fluctuations of the event horizon such that anything having >0 mass just outside the event horizon can still be swallowed. One moment it is outside; fluctuation; now it is inside, still falling and so unable to get out again.
Isn't it supposed to be the other way round, in that a fluctuation leaves a particle outside the event horizon whereupon it escapes as Hawking radiation, stealing mass/energy from the black hole.
Hawking radiation is also about "virtual particles in the vacuum", not just event-horizon fluctuations.  The event horizon fluctuates to swallow one of a pair of virtual particles; the other becomes real and MIGHT escape the hole.  But I was not talking about virtual particles at all; I was talking about already-real particles that had closely approached the event horizon by falling toward the hole.  Anyone who thinks that gravitational time dilation and/or other effects will prevent the real particles from entering the hole is failing to take those same event-horizon-fluctuations into account.

Besides, Hawking radiation isn't really relevant. We were talking about potential energy and a plate. When you raise the plate you give the plate that potential energy. When it falls, that potential energy is converted into kinetic energy. The kinetic energy comes from the plate. It's really simple. What you've been taught would leave you with an asteroid falling into a black hole and giving it more mass/energy than the asteroid had to begin with. That breaks the rules of conservation of energy.
Start over, and be more specific:  When you raise the plate (to, say, a shelf) you put potential energy into the plate/planet SYSTEM.  We are agreed that the mass of the system must increase, to match the increased potential energy.  We are disagreeing about details of where the increased mass goes.  You say it goes into the plate; I say most of it goes into the planet.  When the plate falls off the shelf and acquires kinetic energy at the expense of potential-energy-stored-as-mass, NEITHER of us has a problem with the Law of Energy Conservation.  (And no, if the black hole loses mass that appears as the kinetic energy of the asteroid, it simply gets that mass back after swallowing the asteroid; no net increase happens, of the hole/asteroid system.  Also, in terms of Quantum Mechanics, there are two possible answers to the question "How do gravitons get out of a black hole?"  Do you know either of those answers?)

Regardless of the preceding, YOU still have a problem with General Relativity that I don't have; your scenario means the mass-ratio of plate-to-planet must change, in violation of the rules that allow reference-frame-swapping.  (By the way, I might mention that I held your opinion a number of years ago, and had to abandon it for the same reason I'm giving you now.  Some of that is described in the essay I mentioned much earlier in this Thread, http://www.nemitz.net/vernon/STUBBED2.pdf --it would be nice to get some feedback about that, thanks!)
Title: If I give an object some potential energy, does its mass increase?
Post by: Pmb on 20/08/2009 00:02:39
Quote from: Farsight
I know this stuff like the back of my hand:..
We've all heard that statement a lot over the years since it makes a nice analogy when we're talking about something that we know very well. use that saying alot. Being a physicist I'm quite curious by nature so I trued a little experiment one day. I sat on my hands and tried to remember the number of moles or marks that are on the back of my hands. It then occurred to me that I had a poor recollection of what the back of my hands look like. ROTFL!!!
Quote from: Farsight
There is no singularity "in" a black hole. The singularity is at the event horizon.
Can you explain this response to me in mode detail? It is not inconceivable that the mass distribution inside, say, a primordial black hole could have all its matter at the center of the black hole. It would be impossible for an outside observer to make this determination.
Title: If I give an object some potential energy, does its mass increase?
Post by: Farsight on 21/08/2009 13:48:03
You should post it for all to see, if possible.
It's too bulky, Vernon.

You also still need to post something I mentioned a while back: an explanation for how the mass ratio of plate-to-planet does not change, as required to be consistent with General Relativity's allowing of reference-frame-swapping, when you want the kinetic energy of the plate to appear at the expense of the mass of the plate only, when it falls toward the planet.
But it does change. However if you follow the plate down to the surface, everything that happens to the plate also happens to you, you rods and clocks, and all other measuring devices. For example if you use a spring-powered pushing devices to assess the mass of the plate, that loses energy too. So you won't measure a reduction in the mass of the plate. It's an "immersive scale change". Your reference frame has changed, along with everything in it, including you. You don't notice the changes locally, you only notice them when you do a comparison with afar and notice the gravitational time dilation. In your new reference frame everything is moving slower, but so are you and your clocks, so you don't notice it locally.

I'm not aware that extra dimensions are required for quantum-sized black holes to exist. Black holes of any size strictly depend on the amount of mass crammed into a given volume, and nothing more than that, to the best of my knowledge.  So, again you appear to be making an arbitrary Authoritarian statement, without backing it up.
I'm not. See http://en.wikipedia.org/wiki/Micro_black_hole where it says "In familiar three-dimensional gravity, the minimum energy of a microscopic black hole is 1019 GeV, which would have to be condensed into a region of approximate size 10-33 cm. This is far beyond the limits of any current technology. "

... the law of cosmic censorship is just another hypothesis.
True, but the DEFINITION of that hypothesis specifies a distinction between a singularity (more specifically, a "naked singularity") and an event horizon, which is why you are dead wrong in saying they are the same thing; physicists had specified the distinction between the two things well before the "censorship law" was proposed.
The event horison isn't the same thing as the "naked" singularity associated with cosmic censorship. It isn't a point, it's a surface.

Hawking radiation is also about "virtual particles in the vacuum", not just event-horizon fluctuations.  The event horizon fluctuates to swallow one of a pair of virtual particles; the other becomes real and MIGHT escape the hole.
If you talk about that, talk about the situation where it swallows both virtual particles. Then the black hole is eating the vacuum energy of space. And it grows.

But I was not talking about virtual particles at all; I was talking about already-real particles that had closely approached the event horizon by falling toward the hole.  Anyone who thinks that gravitational time dilation and/or other effects will prevent the real particles from entering the hole is failing to take those same event-horizon-fluctuations into account.
It's just more hypothesis, Vernon. The gravitational time dilation at the event horizon is total. The time dilation is infinite. Try fluctuating infinity.

Start over, and be more specific:  When you raise the plate (to, say, a shelf) you put potential energy into the plate/planet SYSTEM.
No, you don't. You're in that system. The total energy of the system hasn't changed. If it had, this system would exert more gravity than previously. It doesn't.

We are agreed that the mass of the system must increase, to match the increased potential energy.
OK, if I reached down into the system to lift that plate up on to the shelf, I've given the plate potential energy and I've increased the total energy of the plate/planet system.

We are disagreeing about details of where the increased mass goes.  You say it goes into the plate; I say most of it goes into the planet.  When the plate falls off the shelf and acquires kinetic energy at the expense of potential-energy-stored-as-mass, NEITHER of us has a problem with the Law of Energy Conservation.
Fair enough.

And no, if the black hole loses mass that appears as the kinetic energy of the asteroid...
That would require action-at-a-distance. You've got mass magically leaping across a million miles of space to appear as the kinetic energy of the asteroid? No, it can't be like that. And the space surrounding the black hole doesn't lose any mass either, for the same reason. There's only one place left, Vernon.

..it simply gets that mass back after swallowing the asteroid; no net increase happens, of the hole/asteroid system.
I agree that there's no net increase.

Also, in terms of Quantum Mechanics, there are two possible answers to the question "How do gravitons get out of a black hole?"  Do you know either of those answers?)
Yes, here's one re virtual gravitons that go faster than light and break the laws of physics. http://imagine.gsfc.nasa.gov/docs/ask_astro/answers/980601a.html. Note that gravitons aren't part of the standard model. Again they're hypothetical.

Regardless of the preceding, YOU still have a problem with General Relativity that I don't have; your scenario means the mass-ratio of plate-to-planet must change, in violation of the rules that allow reference-frame-swapping.
There are a number of issues with what's called the modern interpretation of general relativity, in that it isn't in line with Einstein. I like to think I am in line with Einstein, so please can you give details of where you get this assertion from?

By the way, I might mention that I held your opinion a number of years ago, and had to abandon it for the same reason I'm giving you now.  Some of that is described in the essay I mentioned much earlier in this Thread, http://www.nemitz.net/vernon/STUBBED2.pdf --it would be nice to get some feedback about that, thanks!)
I'll take a look at it.

Pete: I'll have a rummage via http://www.google.co.uk/search?sourceid=navclient&hl=en-GB&ie=UTF-8&rlz=1T4ADBF_en-GBGB240GB240&q=%22singularity+at+the+event+horizon%22 and get back to you later.
Title: If I give an object some potential energy, does its mass increase?
Post by: VernonNemitz on 21/08/2009 17:37:21
But it [mass ratio] does change. However if you follow the plate down to the surface, everything that happens to the plate also happens to you, you rods and clocks, and all other measuring devices. For example if you use a spring-powered pushing devices to assess the mass of the plate, that loses energy too. So you won't measure a reduction in the mass of the plate. It's an "immersive scale change". Your reference frame has changed, along with everything in it, including you. You don't notice the changes locally, you only notice them when you do a comparison with afar and notice the gravitational time dilation. In your new reference frame everything is moving slower, but so are you and your clocks, so you don't notice it locally.
Nice try, but no cigar.  Sure, all those things that fall together in a gravity field experience similar-ratio mass changes, but GR requires their ratio of mass, to that of the planet, ALSO to be unchanged.

Get away from gravity for a minute and consider the in-some-ways-similar ElectroMagnetic Force.  This force can cause an electron and a proton, initially distant from each other, to fall together and form a hydrogen atom, and release some energy in the process.  Potential-energy-stored-as-mass becomes kinetic energy (with the electon getting 1836 times as much KE as the proton) becomes radiant energy.  Physicists can study the two particle independently and measure their masses.  Those masses are what physicists use in Quantum Mechanics to compute the "orbitals" in a hydrogen atom.  Now, if mass has actually been lost, when the hydroge atom is compared to the separates constituents, how can those calculations be accurate? (And they are indeed very accurate!)  Answer: The RATIO of masses of electron to proton must be identical both in the separated situation and in the atomic situation.  That means that the proton needs to lose 1836 times as much mass as the electron, during their mutual fall, even though the electron acquires 1836 times the KE.  This is still not a huge amount of mass (13.6ev) altogether, and one might wonder if the calculations/measurements are really THAT accurate...but the same sort of thing happens to a much greater degree (very measurable!) when a proton and a neutron get together to form a deuterium nucleus under the influence of the Strong Nuclear Force.  Their mass ratios are not changed by the event!

I hold the view that concepts associated with "potential energy stored as mass" must be consistent across all physical forces, if ever it is to be possible to create a Grand Unified Field Theory.  And that means the constancy of mass ratios is a requirement for Gravitation, just as it is observed for the Strong Force, and possibly observed for the EM Force.  Your mere CLAIM that the ratios must change, during gravitational interactions, has no supporting evidence, simply because the amounts of mass that convert to energy are far too tiny to measure.  But my claim has consistency-with-other-forces on its side, not to mention its being required by General Relativity, to allow easy reference-frame-switching.

I see below you asked for a reference.  I'm sorry, but my reference was a UseNet discussion that is about a decade old now, when I asked for a review of my original 1995 "Stubbed T.O.E." essay, which described a falling object as converting its own mass into the kinetic energy it acquired --- and access to the discussion appears to no longer be available.  I thorougly rewrote the essay as a result of that discussion.

I'm not aware that extra dimensions are required for quantum-sized black holes to exist. Black holes of any size strictly depend on the amount of mass crammed into a given volume, and nothing more than that, to the best of my knowledge.  So, again you appear to be making an arbitrary Authoritarian statement, without backing it up.
I'm not. See http://en.wikipedia.org/wiki/Micro_black_hole where it says "In familiar three-dimensional gravity, the minimum energy of a microscopic black hole is 1019 GeV, which would have to be condensed into a region of approximate size 10-33 cm. This is far beyond the limits of any current technology. "
Nice, but that has nothing at all to do with "extra dimensions", which is what you originally said (instead of talking about magnitudes of ordinary dimensions).  Please try to be more precise in the future.  (Hmmmm.... I do wonder why so much fuss was raised, though, if the requirments are actually as extreme as indicated in the Wikipedia article.)

... the law of cosmic censorship is just another hypothesis.
True, but the DEFINITION of that hypothesis specifies a distinction between a singularity (more specifically, a "naked singularity") and an event horizon, which is why you are dead wrong in saying they are the same thing; physicists had specified the distinction between the two things well before the "censorship law" was proposed.
The event horison isn't the same thing as the "naked" singularity associated with cosmic censorship. It isn't a point, it's a surface.
In physics a singluarity is a place where the laws of physics no longer work, the VOLUME (not a surface) that INCLUDES a special mathematical point, where the thing exists that CAUSES the surrounding region to misbehave.  The proposed law of cosmic censorship keeps all such places INSIDE an event horizon (which is indeed a surface, despite being purely mathematical and immaterial).  That's why they are two different things, and why you are still dead wrong.

Hawking radiation is also about "virtual particles in the vacuum", not just event-horizon fluctuations.  The event horizon fluctuates to swallow one of a pair of virtual particles; the other becomes real and MIGHT escape the hole.
If you talk about that, talk about the situation where it swallows both virtual particles. Then the black hole is eating the vacuum energy of space. And it grows.
False.  The one that becomes real does so at the expense of the mass of the black hole (talk to Hawking about how).  That's why the hole evaporates if enough of them escape.  So, any that do not escape just give their real-ness back to the black hole, and energy-conservation is maintained.

But I was not talking about virtual particles at all; I was talking about already-real particles that had closely approached the event horizon by falling toward the hole.  Anyone who thinks that gravitational time dilation and/or other effects will prevent the real particles from entering the hole is failing to take those same event-horizon-fluctuations into account.
It's just more hypothesis, Vernon. The gravitational time dilation at the event horizon is total. The time dilation is infinite. Try fluctuating infinity.
Wrong again.  The fluctuations of the event horizon are due to the Uncertainty Principle, which is very very real.  That is, the mass of the black hole, however large, is still finite and still fluctuates due to Uncertainty, and therefore its event horizon, mathematically PRECISELY dependent upon the mass of the hole, fluctuates also.  Thus (QED) any infalling real massy particle that gets close enough WILL be swallowed by the hole.

Start over, and be more specific:  When you raise the plate (to, say, a shelf) you put potential energy into the plate/planet SYSTEM.
No, you don't. You're in that system. The total energy of the system hasn't changed. If it had, this system would exert more gravity than previously. It doesn't.
I repeat, we are talking about quantities of mass too tiny to measure such side-effects.  AND, we need not be part of the system to raise the plate, if you assume (an extremely unlikely but not totally irrational assumption) the plate can absorb a gravity wave that arrives from Outside and acquire kinetic energy thereby, and rise to a height in a gravitational field, similar to an electron absorbing a photon arriving from Outside and rising to a new level in an atomic electrostatic field.

And no, if the black hole loses mass that appears as the kinetic energy of the asteroid...
That would require action-at-a-distance. You've got mass magically leaping across a million miles of space to appear as the kinetic energy of the asteroid? No, it can't be like that. And the space surrounding the black hole doesn't lose any mass either, for the same reason. There's only one place left, Vernon.
Wrong again; your imagination is missing something.  As preparation, return to the mutually-distant electron-proton analogy.  QM describes exchanges of virtual photons between them as the basis of the EM Force between them.  While photons travel at light-speed and the distance can be considerable, there is nothing at all preventing lots of them to simultanously exist along the path between the two particles.  Focussing on the electron, it interacts with one virtual photon after another, that is already en-route between them.  It does not NEED to wait for a single virtual photon to go back-and-forth between them, before the next phase of the interaction happens.  QM for Gravitation can work the same way, no delay needed between a sequence of interactions.  Now getting to your mental block, each virtual-graviton travelling from the planet to the plate consists of POTENTIAL energy from the planet, so any that are absorbed by the plate means the plate can acquire that energy; it can become real only if planet loses mass (ditto for electron absorbing virtual photon and acquiring KE at expense of proton's mass).  Remember the virtual particle that becomes real at the expense of the mass of the black hole, if its companion particle is swallowed!  What was the connection there?  But even without referencing the Hawking Radiation hypothesis again, WE ALREADY HAVE PROVED "spooky action at a distance" is a Real Thing (see any recent development in Quantum Encryption); this is just more of the same.  What is your problem with that?  That the planet's mass should disappear due to the number of external absorptions?  Ah, but that is exactly balanced (most of the time) by the number of absorptions of virtual gravitons by the planet, from those external objects!  Any unbalance results in an acceleration, of course (so the planet and plate fall toward each other).

Also, in terms of Quantum Mechanics, there are two possible answers to the question "How do gravitons get out of a black hole?"  Do you know either of those answers?)
Yes, here's one re virtual gravitons that go faster than light and break the laws of physics. http://imagine.gsfc.nasa.gov/docs/ask_astro/answers/980601a.html. Note that gravitons aren't part of the standard model. Again they're hypothetical.
That's only one of the two possible explanations, and no, it is not necessary that the laws of physics be violated if a gravtion can travel faster than light.  I'll get back to that in a minute.

The standard model does not include gravitons simply because Physics does not have a Grand Unified Quantum Field Theory yet.  Duh....

The second possible explanation for how a graviton can get out of a black hole involves the "interaction cross section" of a graviton.  We know they must be able to interact with each other; that is a requirement for consistency with General Relativity, since the existence of a gravitational field counts as mass/energy that contributes to the gravitational field.  But "being able to interact" and "always interacting" are two different things.  A low-enough rate of mutual interaction can easily suffice to let vast numbers of virtual gravitons out of a black hole, no matter how fast or slow they travel.

As you have stated so many times, gravitons are hypothetical.  That means even if they exist, we don't know for sure what they are like.  So this variation of the hypothesis is as good as any (and may be better than most):  What if a graviton is not describable as 'energy in motion"?  See, "energy in motion", such as is a photon, and also is a common hypothesis about gravitons, is required to always move exactly at light-speed.  Meanwhile, "mass in motion" is allowed to have any speed less than light-speed, and "imaginary mass in motion", should it exist, is required to always move faster than light-speed.  You are aware, I think, that if tachyons exist, the laws of physics will not be violated?  Well, a graviton doesn't necessarily have to be any of those three things in motion, and like tachyons its speed does NOT have to be associated with a violation of Physics.  See my "Simple Quantum Gravitation" knol for the details (which actually talks about very-slow gravitons, not fast gravitons, though fast ones are not ruled out).
Title: If I give an object some potential energy, does its mass increase?
Post by: Pmb on 21/08/2009 19:01:30
Quote from: VernonNemitz
Yes, here's one re virtual gravitons that go faster than light and break the laws of physics.
Ouch! Virtual gravitons can't break the laws of nature. In fact nothing that can exist in nature can ever be said to break the laws of physics. When a virtual particle is moving faster than the speed of light it is not violating a law of nature. In fact it's quite consistent with it. When a virtual particle is moving FTL then it becomes a tachyon and the existamce of tachyons do not violate amu law of nature.
Title: If I give an object some potential energy, does its mass increase?
Post by: VernonNemitz on 21/08/2009 19:10:27
Quote from: VernonNemitz
Yes, here's one re virtual gravitons that go faster than light and break the laws of physics.
Ouch! Virtual gravitons can't break the laws of nature. In fact nothing that can exist in nature can ever be said to break the laws of physics. When a virtual particle is moving faster than the speed of light it is not violating a law of nature. In fact it's quite consistent with it. When a virtual particle is moving FTL then it becomes a tachyon and the existamce of tachyons do not violate amu law of nature.
Please be more careful with your quotes.  The above quoted text attributed to me was actually originated by Farsight in Msg 270745.
Title: If I give an object some potential energy, does its mass increase?
Post by: Pmb on 21/08/2009 19:38:26
There's no need to tell me to be careful since I'm always careful. I'm just not perfect and as such I make mistakes.
Title: If I give an object some potential energy, does its mass increase?
Post by: Farsight on 21/08/2009 19:50:43
Nice try, but no cigar. Sure, all those things that fall together in a gravity field experience similar-ratio mass changes, but GR requires their ratio of mass, to that of the planet, ALSO to be unchanged.
Give me a reference. As I said, there are significant issues with the way GR has been reinterpreted and is no longer in accord with Einstein's original. Pmb will tell you more about that.

Get away from gravity for a minute and consider the in-some-ways-similar ElectroMagnetic Force. This force can cause an electron and a proton, initially distant from each other, to fall together and form a hydrogen atom, and release some energy in the process.  Potential-energy-stored-as-mass becomes kinetic energy (with the electon getting 1836 times as much KE as the proton) becomes radiant energy.
Vernon, this isn't right. You must know that p=mv and KE=1/2mv2, and that force is rate of change of momentum.  If we share momentum equally between two bodies of different masses, the lighter mass ends up moving faster, but the v2 means the kinetic energy is not shared equally. Search on this to check it out: http://www.google.co.uk/search?sourceid=navclient&hl=en-GB&ie=UTF-8&rlz=1T4ADBF_en-GBGB240GB240&q=momentum+%22kinetic+energy%22+shared

Physicists can study the two particle independently and measure their masses. Those masses are what physicists use in Quantum Mechanics to compute the "orbitals" in a hydrogen atom. Now, if mass has actually been lost, when the hydrogen atom is compared to the separates constituents, how can those calculations be accurate? (And they are indeed very accurate!) Answer: The RATIO of masses of electron to proton must be identical both in the separated situation and in the atomic situation. That means that the proton needs to lose 1836 times as much mass as the electron, during their mutual fall, even though the electron acquires 1836 times the KE. This is still not a huge amount of mass (13.6ev) altogether, and one might wonder if the calculations/measurements are really THAT accurate...but the same sort of thing happens to a much greater degree (very measurable!) when a proton and a neutron get together to form a deuterium nucleus under the influence of the Strong Nuclear Force. Their mass ratios are not changed by the event!
It isn't as you describe. That binding energy and the emitted photon represents a loss of mass/energy, but it isn't true to say the proton loses 1836 times as much as the electron.

I hold the view that concepts associated with "potential energy stored as mass" must be consistent across all physical forces, if ever it is to be possible to create a Grand Unified Field Theory. And that means the constancy of mass ratios is a requirement for Gravitation, just as it is observed for the Strong Force, and possibly observed for the EM Force.
You need to look into this further, see above, and you have to appreciate that the forces aren't identical.

Your mere CLAIM that the ratios must change, during gravitational interactions, has no supporting evidence, simply because the amounts of mass that convert to energy are far too tiny to measure. But my claim has consistency-with-other-forces on its side, not to mention its being required by General Relativity, to allow easy reference-frame-switching.
It does have supporting evidence, but you're not understanding it because you don't appreciate what I mean by an "immersive scale change". What I'm saying is vindicated by the Pound-Rebka experiment and the GPS clock adjustment, both of which are extremely sensitive. Do you remember what I said about a spinning plate? If it's spinning extremely fast it has more energy. Move this into an extremely time-dilated environment, and it's spinning slower, and hence has lost energy. Then I said you should consider a plate to be made up of "tiny spinning plates" called electrons and protons. It's really simple Vernon, but it's not what you've been taught.

I see below you asked for a reference. I'm sorry, but my reference was a UseNet discussion that is about a decade old now, when I asked for a review of my original 1995 "Stubbed T.O.E." essay, which described a falling object as converting its own mass into the kinetic energy it acquired --- and access to the discussion appears to no longer be available. I thoroughly rewrote the essay as a result of that discussion.
Shame. You were on the right track then, and got talked out of it. The kinetic energy of the falling plate comes from somewhere. It maybe comes out of the planet. It maybe comes out of the space surrounding the planet. Or it maybe comes out of the plate. The first two options demand magical-mysterious action-at-a-distance and an undetectable transfer of energy through some invisible undetectable particles. There is no scientific evidence to support this.

Nice, but that has nothing at all to do with "extra dimensions", which is what you originally said (instead of talking about magnitudes of ordinary dimensions).  Please try to be more precise in the future.  (Hmmmm.... I do wonder why so much fuss was raised, though, if the requirments are actually as extreme as indicated in the Wikipedia article.)
I said I didn't think quantum black holes are going to get created in the LHC and I've given you ample reason.

In physics a singluarity is a place where the laws of physics no longer work, the VOLUME (not a surface) that INCLUDES a special mathematical point, where the thing exists that CAUSES the surrounding region to misbehave. The proposed law of cosmic censorship keeps all such places INSIDE an event horizon (which is indeed a surface, despite being purely mathematical and immaterial).  That's why they are two different things, and why you are still dead wrong.
You're being too argumentative, and you're defending your argument with too many hypotheticals.

False. The one that becomes real does so at the expense of the mass of the black hole (talk to Hawking about how). That's why the hole evaporates if enough of them escape. So, any that do not escape just give their real-ness back to the black hole, and energy-conservation is maintained.
I can only reiterate that a significant number of physicists consider Hawking radiation to be an unproven hypothesis. You can't use it to assert the truth of falsehood of something else.

Wrong again. The fluctuations of the event horizon are due to the Uncertainty Principle, which is very very real. That is, the mass of the black hole, however large, is still finite and still fluctuates due to Uncertainty, and therefore its event horizon, mathematically PRECISELY dependent upon the mass of the hole, fluctuates also.  Thus (QED) any infalling real massy particle that gets close enough WILL be swallowed by the hole.
This is not what the uncertainty principle says. Check it out: http://en.wikipedia.org/wiki/Uncertainty_principle

I repeat, we are talking about quantities of mass too tiny to measure such side-effects. AND, we need not be part of the system to raise the plate, if you assume (an extremely unlikely but not totally irrational assumption) the plate can absorb a gravity wave that arrives from Outside and acquire kinetic energy thereby, and rise to a height in a gravitational field, similar to an electron absorbing a photon arriving from Outside and rising to a new level in an atomic electrostatic field.
Anything arriving from outside the system adds energy to the system. When it lifts the plate, it gives the plate potential energy. The energy goes into the plate. Hence its mass is increased. Simple.

Wrong again; your imagination is missing something. As preparation, return to the mutually-distant electron-proton analogy. QM describes exchanges of virtual photons between them as the basis of the EM Force between them. While photons travel at light-speed and the distance can be considerable, there is nothing at all preventing lots of them to simultanously exist along the path between the two particles. Focussing on the electron, it interacts with one virtual photon after another, that is already en-route between them. It does not NEED to wait for a single virtual photon to go back-and-forth between them, before the next phase of the interaction happens.
And they're virtual. They aren't real. They're a calculation tool. Feynman didn't ascribe them the same sort of reality as you do. You know what they really are? Collectively they're something called the evanescant wave.

QM for Gravitation can work the same way, no delay needed between a sequence of interactions. Now getting to your mental block, each virtual-graviton travelling from the planet to the plate consists of POTENTIAL energy from the planet, so any that are absorbed by the plate means the plate can acquire that energy; it can become real only if planet loses mass (ditto for electron absorbing virtual photon and acquiring KE at expense of proton's mass). Remember the virtual particle that becomes real at the expense of the mass of the black hole, if its companion particle is swallowed! What was the connection there?  But even without referencing the Hawking Radiation hypothesis again, WE ALREADY HAVE PROVED "spooky action at a distance" is a Real Thing (see any recent development in Quantum Encryption); this is just more of the same. What is your problem with that?
The same as Newton's:

“That gravity should be innate, inherent, and essential to matter, so that one body may act upon another at a distance through a vacuum, without the mediation of anything else, by and through which their action and force may be conveyed from one to another, is to me so great an absurdity that I believe no man who has in philosophical matters a competent faculty of thinking can ever fall into it”

That the planet's mass should disappear due to the number of external absorptions? Ah, but that is exactly balanced (most of the time) by the number of absorptions of virtual gravitons by the planet, from those external objects! Any unbalance results in an acceleration, of course (so the planet and plate fall toward each other).
Gravitons remain hypothetical, and virtual gravitons even more so.

That's only one of the two possible explanations, and no, it is not necessary that the laws of physics be violated if a graviton can travel faster than light.  I'll get back to that in a minute.
I dispute that. And do note that I've got a couple of old friends backing me up on this. One is Isaac Newton, and the other is Albert Einstein.

The standard model does not include gravitons simply because Physics does not have a Grand Unified Quantum Field Theory yet.
And when it does, this grand unified theory will be one where "the concept of field is no longer appropriate". And it will not include gravitons.

The second possible explanation for how a graviton can get out of a black hole involves the "interaction cross section" of a graviton.  We know they must be able to interact with each other; that is a requirement for consistency with General Relativity, since the existence of a gravitational field counts as mass/energy that contributes to the gravitational field. But "being able to interact" and "always interacting" are two different things.  A low-enough rate of mutual interaction can easily suffice to let vast numbers of virtual gravitons out of a black hole, no matter how fast or slow they travel.
No, we don't "know" they interact with one another. We don't even know that they exist. It's all just hypothesis on top of hypothesis.

As you have stated so many times, gravitons are hypothetical. That means even if they exist, we don't know for sure what they are like. So this variation of the hypothesis is as good as any (and may be better than most): What if a graviton is not describable as 'energy in motion"?  See, "energy in motion", such as is a photon, and also is a common hypothesis about gravitons, is required to always move exactly at light-speed. Meanwhile, "mass in motion" is allowed to have any speed less than light-speed..
Good stuff Vernon. This is more like it.

..and "imaginary mass in motion", should it exist, is required to always move faster than light-speed. You are aware, I think, that if tachyons exist, the laws of physics will not be violated?  Well, a graviton doesn't necessarily have to be any of those three things in motion, and like tachyons its speed does NOT have to be associated with a violation of Physics.
Aaaargh!

See my "Simple Quantum Gravitation" knol for the details (which actually talks about very-slow gravitons, not fast gravitons, though fast ones are not ruled out).
Like I said, I'll check it out.

Vernon, these posts are getting too long. I'm giving you a great deal of my time here but you're going to extraordinary lengths to avoid the obvious: the kinetic energy of the falling plate comes from that plate. The plate had the potential energy. We gave it to the plate. Go with the flow of this. Once you do, everything gets a whole lot simpler. Apologies, Friday night is calling, and the wife. Gotta go.
Title: If I give an object some potential energy, does its mass increase?
Post by: Farsight on 24/08/2009 09:02:17
Pete: sorry I haven't back to you before now. Here's a little something on the Schwarzschild singularity.

http://everything2.com/title/Schwarzschild+metric

See where he says:

"Well, it is a little more serious than that in fact: a particle observed (from outside) to fall towards the Schwarzschild Singularity will appear to take an infinite time to reach it, whereas the particle itself will think that it did reach r=2M in a finite time. Something odd is definitely happening here, and the conclusion to be drawn is that time (t) is not a sensible coordinate to be dealing with in this region (there is a better coordinate system - see below)".

There isn't a better coordinate system. That better coordinate system is counting time on a stopped clock. Hence we get back to the original frozen star concept. See http://en.wikipedia.org/wiki/Black_hole and note this section:

"Because of this property, the collapsed stars were briefly known as "frozen stars,"[citation needed] because an outside observer would see the surface of the star frozen in time at the instant where its collapse takes it inside the Schwarzschild radius. This is a known property of modern black holes, but it must be emphasized that the light from the surface of the frozen star becomes redshifted very fast, turning the black hole black very quickly. Many physicists could not accept the idea of time standing still at the Schwarzschild radius, and there was little interest in the subject for over 20 years.

In 1958, David Finkelstein introduced the concept of the event horizon by presenting Eddington-Finkelstein coordinates, which enabled him to show that "The Schwarzschild surface r = 2 m is not a singularity, but that it acts as a perfect unidirectional membrane: causal influences can cross it in only one direction".[6] This did not strictly contradict Oppenheimer's results, but extended them to include the point of view of infalling observers. All theories up to this point, including Finkelstein's, covered only non-rotating black holes".

But time does stand still. Every observer in this universe will agree on it. The time dilation at the event horizon is infinite. The proper time of the infalling observer never ever happens. Never. Finkelstein was wrong. The black hole ends up more like a gravastar, see http://en.wikipedia.org/wiki/Gravastar. See where it says "This region is called a gravitational vacuum, because it is a void in the fabric of space and time". What you end up with is even more of a hole that the traditional black hole. The event horizon is the end of events, and the end of space. It's a spherical surface of infinite time dilation and infinite radial length contraction, and you can never get past it.
Title: If I give an object some potential energy, does its mass increase?
Post by: VernonNemitz on 24/08/2009 16:42:07
Several of your remarks in the last message stressed the hypothetical nature of certain things.  But one variety of thing is not hypothetical at all: virtual particles.  WHILE they exist, they are exactly as real as ordinary particles.  And their temporary existence can detectably though indirectly affect real things; see the "Casimir Effect" for details: http://en.wikipedia.org/wiki/Casimir_effect

If you want to argue that virtual particles are ignorable/merely-hypothetical, then you are fighting (a losing fight!!!) against the Uncertainty Principle.  The UP requires that even a volume of space that is absolutely empty of any real particles or energy must nevertheless have an Uncertain energy content.  Therefore virtual particles must exist; they are the form taken by those Uncertain energy-variations, and the Casimir Effect is the proof they exist everywhere and all the time.

Another aspect of "piling" hypotheses is presented by you in this statement:
Gravitons remain hypothetical, and virtual gravitons even more so.
Sorry, but if they exist at all, then both types, virtual and real, will exist.  Period.  Any particle that can exist at all can also exist virtually; any particle that exists virtually can become real if it absorbs the appropriate quantity/type of real energy.  There is no "hypothetical" AT ALL, regarding virtual existence; one of the best pieces of evidence involves gamma-ray photons that turn into particle-pairs; pair-production is maximized at certain energies called "resonant energies".  Resonant with what, eh?  Virtual particles!

Finally, from your message #270459 (Aug 19, 2009, 17:15:59), I quote (regarding quantum black holes):
"None will be produced by such a collision, because the required extra dimensions do not exist."
You have yet to show why any extra dimensions are required.  You have presented OTHER reasons why quantum black holes will not be created, which I find satisfactory.  But don't try to make me think you didn't say what you actually said!

After your faulty mind-set has been corrected, regarding the above matters, then we can move on to other aspects of the discussion.
Title: If I give an object some potential energy, does its mass increase?
Post by: VernonNemitz on 25/08/2009 18:55:20
Wrong again. The fluctuations of the event horizon are due to the Uncertainty Principle, which is very very real. That is, the mass of the black hole, however large, is still finite and still fluctuates due to Uncertainty, and therefore its event horizon, mathematically PRECISELY dependent upon the mass of the hole, fluctuates also.  Thus (QED) any infalling real massy particle that gets close enough WILL be swallowed by the hole.
This is not what the uncertainty principle says. Check it out: http://en.wikipedia.org/wiki/Uncertainty_principle
You need to see the proper part of that page:  http://en.wikipedia.org/wiki/Uncertainty_principle#Energy-time_uncertainty_principle
(want more evidence?  The units of Planck's Constant are in "erg-seconds", i.e., energy multiplied by time)
To the extent that mass is equivalent to energy, the mass of a black hole will indeed fluctuate on the very-small time scale.  Meanwhile, as I stated previously, the event horizon is purely mathematical in its description and is EXACTLY correlated with the mass of the hole.  Therefore if the mass fluctuates, so does the event horizon.  The hole will be perfectly able to gobble any close-enough thing, regardless of General Relativity's time-slowing effects.
Title: If I give an object some potential energy, does its mass increase?
Post by: Ron Hughes on 25/08/2009 21:24:46
Back to the original question. Please tell me if I'm wrong. Any particle, group of particles or molecules accelerated in any way will gain mass, not because of the new velocity but because of the acceleration.

I find it difficult to believe that a clock in a mathematical boundary of an event horizon is stopped. If I'm observing this clock from outside the horizon it may appear stopped to me but I also know that if I zoom down to the clock I will find it happily ticking along. Not only that, as we fall toward the center it will still be ticking. If I look back at the rest of Universe I will see everything moving very very fast. Gravity can escape from a BH because time dilation can escape.
Title: If I give an object some potential energy, does its mass increase?
Post by: VernonNemitz on 25/08/2009 23:55:23
Back to the original question. Please tell me if I'm wrong. Any particle, group of particles or molecules accelerated in any way will gain mass, not because of the new velocity but because of the acceleration.
Ummm, the original question was about potential energy, not about accelerated particles.

Regarding already-accelerated particles, they have extra mass as seen from the perspective of a relatively stationary frame of reference.  There is no way, from the perspective of the accelerated particles, to detect that increase in mass, though.  For example, consider that galaxies very far away have a spectrum that indicates they are moving at a large fraction of light-speed.  Pick one at random.  We might think its mass has increased as a result of its speed, relative to us.  On the other hand, US relative to IT could be considered as having increased in mass, instead!  Do you suddenly feel more massive as a result of that realization?  No?  Like I said, there is no way to tell, at least when constant velocities are involved.  You are right, though, in that it is the acceleration toward a high velocity, and especially the energy it takes to cause that, that is the ultimate source of relativistic mass.

Please note some of this discussion is about locations of objects inside a gravitational field gradient.  Objects that are allowed to fall in that gradient will of course accelerate and acquire kinetic energy and relativistic mass in consequence.  But that kinetic energy has to come from somewhere; the explanation being talked about here concerns mass as the "storage" form of the potential energy that becomes kinetic energy.

I find it difficult to believe that a clock in a mathematical boundary of an event horizon is stopped. If I'm observing this clock from outside the horizon it may appear stopped to me but I also know that if I zoom down to the clock I will find it happily ticking along. Not only that, as we fall toward the center it will still be ticking. If I look back at the rest of Universe I will see everything moving very very fast. Gravity can escape from a BH because time dilation can escape.

Not bad, except for that last bit.  In terms of General Relativity and curved space-time, a black hole is not a problem; ANY mass is just a mass causing curvature.  It is only Quantum Mechanics, should it be applied to explain gravity, that has a problem.  Its generic explanation for a force involves something called "exchange particles", and it is those particles that somehow have to get out of a black hole to interact with other things.
Title: If I give an object some potential energy, does its mass increase?
Post by: Ron Hughes on 26/08/2009 01:28:21
Time is relative to the observers frame of reference which we all know. A clock in another moving frame of reference slows with respect to our observer.If we had a one kg block setting on a frictionless surface and apply a force of 1kg.m/sec^2 it will accelerate at 1m/sec^2. The only way to increase this acceleration is to increase the applied force OR slow the observer's clock. Isn't this exactly what happens in a gravity well?”

I hope the reader will agree that any particle can be considered it's own clock ( an observer ) and that a clock even one nanometer higher in a gravity well will run slightly faster than a clock one nanometer lower. Suppose we had a frictionless tube, one meter long, whose inside diameter is just large enough to allow a diatomic hydrogen molecule to move freely up and down in the tube. We fill the tube with diatomic hydrogen at STP and stand the tube up perpendicular to the surface of the Earth. Each molecule is moving up and down in the tube colliding with the molecule above and below. Let’s observe the path of a particular molecule which we call B. The molecule above we call A and the one below we call C. We start watching B as it moves down the tube toward C with velocity d/t. The molecule B (clock/observer) will calculate it's momentum, at the instant before the collision with C, using it's clock which is running slightly slower than when it collided with A. B will find it's momentum at C to be greater than the momentum of the collision with A. Doesn't this suggest that gravity is strictly a function of time dilation? Inertial forces can be considered as the same process. Particles accelerated in free space exhibit the same time dilation which would explain why we can't tell the difference between inertial forces and gravitational forces.
Title: If I give an object some potential energy, does its mass increase?
Post by: VernonNemitz on 26/08/2009 13:51:40
Time is relative to the observers frame of reference which we all know. A clock in another moving frame of reference slows with respect to our observer.If we had a one kg block setting on a frictionless surface and apply a force of 1kg.m/sec^2 it will accelerate at 1m/sec^2. The only way to increase this acceleration is to increase the applied force OR slow the observer's clock. Isn't this exactly what happens in a gravity well?
Please remember that in General Relativity, gravity involves curvature not just of space only but of space-time, caused by the presence of a mass.  You are describing things associated with gravity's distortion of time.  And no, you cannot conclude from that, that gravity is a consequence of the distorted time.

Another problem that Quantum Mechanics has, with respect to trying to describe gravity, involves the need to provide "matching" explanations for such things as time dilation in a gravity well.  A simple explanation brings potential-energy-stored-as-mass into the picture.  An object that loses mass while falling into a gravity well, due to potential energy being converted to kinetic energy, is an object that will have less mass at the bottom of the well, after it hits bottom and its kinetic energy is lost/radiated-away.  And in QM a lesser mass can be associated with a lesser vibration rate (per the wave-particle duality), or, in terms of what you wrote that I didn't quote, a slower internal clock.  (Note Farsight and I have been arguing about some of the details of that idea, but not the main idea itself.)
Title: If I give an object some potential energy, does its mass increase?
Post by: Ron Hughes on 26/08/2009 16:07:31
So my post is wrong, not because of the logic but because it's not a QM explanation? Do you ever question the fact that QM must invent hypothetical particles to explain mass and gravity? When asked to explain the gravitational anomalies they invent dark matter and energy? The propagation of light requires virtual particles that have no proofs in reality? Pretty neat, they invent Qm and when it can't explain something just add a particle.Now if someone else pulls a stunt like that they are grasping at straws to maintain their theory. Suppose for a second that my post above is correct. It would mean that the value of F = ma is not the same everywhere. That would explain the Pioneer anomaly and the rotation curve of galaxies, but I guess one should never use Occam's razor in science. A quote from Newton, "We are to admit no more causes of natural things than such as are both true and sufficient to explain their appearances. Therefore, to the same natural effects we must, so far as possible, assign the same causes."
Title: If I give an object some potential energy, does its mass increase?
Post by: PhysBang on 26/08/2009 17:03:39
So my post is wrong, not because of the logic but because it's not a QM explanation? Do you ever question the fact that QM must invent hypothetical particles to explain mass and gravity? When asked to explain the gravitational anomalies they invent dark matter and energy? The propagation of light requires virtual particles that have no proofs in reality? Pretty neat, they invent Qm and when it can't explain something just add a particle.Now if someone else pulls a stunt like that they are grasping at straws to maintain their theory. Suppose for a second that my post above is correct. It would mean that the value of F = ma is not the same everywhere. That would explain the Pioneer anomaly and the rotation curve of galaxies, but I guess one should never use Occam's razor in science. A quote from Newton, "We are to admit no more causes of natural things than such as are both true and sufficient to explain their appearances. Therefore, to the same natural effects we must, so far as possible, assign the same causes."
And Newton applied this principle by identifying causes as those theoretical principles that we could get agreeing measurements of from multiple sources. Does you explanation have any measurement behind it? Do you explain the Pioneer anomaly and the differences in galaxy rotation curves with a deviation from Newtonian mechanics that has accurate and agreeing measurements from both phenomena?
Title: If I give an object some potential energy, does its mass increase?
Post by: VernonNemitz on 26/08/2009 17:28:21
So my post is wrong, not because of the logic but because it's not a QM explanation?
I'm keeping my explanations of GR stuff separate from QM stuff; please do not confuse them.  Your post was wrong partly because its logic fails to explain the source of time-dilation for different gravity wells.
Each molecule is moving up and down in the tube colliding with the molecule above and below. Let’s observe the path of a particular molecule which we call B. The molecule above we call A and the one below we call C. We start watching B as it moves down the tube toward C with velocity d/t. The molecule B (clock/observer) will calculate it's momentum, at the instant before the collision with C, using it's clock which is running slightly slower than when it collided with A. B will find it's momentum at C to be greater than the momentum of the collision with A.
Also, of course, there is the fact that for molecule B to reach A after colliding with C, it has to climb a gravitational field gradient.  You are aware that anything rising in such a gradient tends to lose velocity/momentum (converts to potential energy)?  So, no need to invoke time-dilation there as a cause for its lesser momentum (and that would be backward, anyway, since clocks tick FASTER higher-up in a gravity gradient).

Regarding dark matter and dark energy, these things are proposals resulting from analyzing lots of cosmological data.  Aspects of those proposals are subject to change as more data is gathered/analyzed.  To the extent that current data/analysis is real and correct, then that means we have observed some things that cannot be explained without modifying in some way the usage of existing theories such as General Relativity (for example, by putting a proposal by Einstein regarding a "cosmological constant" back into the theory).  And, adding new/oddball types of matter or energy are ways of extending the application of existing theory without actually modifying the theory itself.

Back to QM.  It has a long history of inventing particles and later discovering them (the first was the anti-electron or positron, but also see the stories behind neutrinos and pions and the Weak Force exchange particles and all the quarks).  Do not assume that just because a proposed particle is currently hypothetical, it must be ignorable.  When such a proposal is widely accepted --such as is currently true for the Higgs boson, which was accepted enough to get the Large Hadron Collider built specifically to find it-- there MUST be good reason for it (usually it's necessary for theoretical logical self-consistency).  If dark matter and/or dark energy exists (and the current interpretation of the data can be taken to mean: "it exists on a large scale and has certain large-scale properties"), then it should be relatively obvious that it would have small-scale properties, also.  QM, of course, is the essence of our study of small-scale properties.  Why shouldn't QM researchers propose new particles?
Title: If I give an object some potential energy, does its mass increase?
Post by: Ron Hughes on 26/08/2009 18:02:19
You are correct I don't, but if someone with the necessary skills doesn't try using the idea we will never know. QM uses the graviton, a hypothetical particle, my idea has none. Because the vast majority of scientists buy into the standard model that makes it right. In 1905 the vast majority thought time was a constant. The reason for time dilation, I suspect has something to do with the density of electromagnetic radiation around matter. An impossible test of that would be if we could take any matter down to absolute zero it's weight should be zero.
Title: If I give an object some potential energy, does its mass increase?
Post by: lightarrow on 26/08/2009 19:19:53
Back to the original question. Please tell me if I'm wrong. Any particle, group of particles or molecules accelerated in any way will gain mass
No, mass stays constant (it's also called "invariant" mass exactly for this reason).
Title: If I give an object some potential energy, does its mass increase?
Post by: lightarrow on 26/08/2009 19:23:54
Please note some of this discussion is about locations of objects inside a gravitational field gradient.  Objects that are allowed to fall in that gradient will of course accelerate and acquire kinetic energy and relativistic mass in consequence.
Of course they will accelerate even in the absence of any gradient, provided there is a gravitational field...
Title: If I give an object some potential energy, does its mass increase?
Post by: Farsight on 26/08/2009 19:28:25
Several of your remarks in the last message stressed the hypothetical nature of certain things.  But one variety of thing is not hypothetical at all: virtual particles.  WHILE they exist, they are exactly as real as ordinary particles.  And their temporary existence can detectably though indirectly affect real things; see the "Casimir Effect" for details: http://en.wikipedia.org/wiki/Casimir_effect
There's certainly something real there, Vernon. The Casimir effect is real. But those vitural particles are virtual. It's wrong to ascribe them the degree of reality that you do.

If you want to argue that virtual particles are ignorable/merely-hypothetical, then you are fighting (a losing fight!!!) against the Uncertainty Principle.  The UP requires that even a volume of space that is absolutely empty of any real particles or energy must nevertheless have an Uncertain energy content. Therefore virtual particles must exist; they are the form taken by those Uncertain energy-variations, and the Casimir Effect is the proof they exist everywhere and all the time.
Like I said, there's something there that's real, and that energy is real. But to claim that this is proof of virtual particles popping in and out of existence is on a par with saying it proves the existence of tiny dancing angels. Think of virtual particles as the evanescent wave.

Next, regarding piling hypothesis on hypothesis, that's not true.  It is perfectly straightforward logic, that if gravitons exist, they must be able to interact with each other. The key point is that a Quantum Mechanics theory for Gravitation must be able to yield results that are basically identical to the results of General Relativity; we know the GR results are quite realistic, and a QM theory cannot be less realistic and also be correct. So, if a gravitational field counts as a type of mass/energy that can add to a gravitational field, then in terms of virtual gravitons making up that field, the virtual gravitons must be sources of additional virtual gravitons, in order for that description to be consistent with the GR description.
Vernon, gravitons do not exist. A photon conveys energy. Energy causes gravity. There are no actual gravitons zipping back and forth between a photon and everything else. And nor are there any virtual gravitons. It's all hypothesis.

And Time Reversal Symmetry requires that anything that can emit something must also be able to absorb it (the essence of "interact").  Simple ironclad logic; only one hypothesis needed, the one that says it ought to be possible to devise a QM theory for Gravitation, involving gravitons, consistent with observations (not to mention GR).  Do note that gravitons are supposed to be the smallest possible type of gravity wave (the theoretical existence of that generic thing appears to be supported by the measured behavior of close-orbiting neutron stars: http://www.astro.cornell.edu/academics/courses/astro201/psr1913.htm ). Because of various Conservation Laws, gravity waves are not considered to be any more hypothetical than radio waves from your favorite AM station.
Radio waves are quite measurable. Photons are detectable. Gravity is detectable. Gravitons are not. And nor are virtual gravitons.

We know that those kinds of radio waves are made up of lots of photons; we can logically deduce, if things like Space and Time are themselves quantized, then there must be a minimum-size gravity wave; why NOT call it a "graviton"? (If spacetime is not quantized, then of course there would be not have to be a limit to smallness of a gravity wave; they would not need to exist as quanta.) (Getting off-track now, there is a Law Conservation of Information that physicists have been taking very seriously in recent decades, and one problem with that has involved black holes. If they can evaporate via Hawking radiation, how does the information that fell into them get out again?  http://www.space.com/news/hawking_bet_040716.html  --The answer relates to space-time being quantized. YOU probably won't be impressed if you think that Hawking Radiation can't exist; my only point here is that QM has survived another test of overall self-consistency --with the consequence that gravitons are more likely to actually exist than to be merely hypothetical.  For more on Information Conservation and quanta, try to get this book: http://www.librarything.com/work/378995 --and this link might count as a sample of the material: http://www.springerlink.com/content/b43670p553581tv8/  )
Again, Hawking radiation is hypothesis. We were talking about where the energy of a falling plate comes from. You're defending your stance that it doesn't come from the plate, with hypothesis after hypothesis, for which no evidence exists whatsoever. Your stance is not supported. Mine is, by the spinning plate, which spins at a reduced rate in a region of gravitational time dilation, the same applying to its component electrons and their spin. It's really very simple.

Another aspect of "piling" hypotheses is presented by you in this statement:
Gravitons remain hypothetical, and virtual gravitons even more so.
Sorry, but if they exist at all, then both types, virtual and real, will exist. Period. Any particle that can exist at all can also exist virtually; any particle that exists virtually can become real if it absorbs the appropriate quantity/type of real energy. There is no "hypothetical" AT ALL, regarding virtual existence.
It is hypothesis.

One of the best pieces of evidence involves gamma-ray photons that turn into particle-pairs; pair-production is maximized at certain energies called "resonant energies". Resonant with what, eh? Virtual particles!
It's no proof of virtual particles the way you think of them.

Finally, from your message #270459 (Aug 19, 2009, 17:15:59), I quote (regarding quantum black holes):
"None will be produced by such a collision, because the required extra dimensions do not exist."
You have yet to show why any extra dimensions are required. You have presented OTHER reasons why quantum black holes will not be created, which I find satisfactory. But don't try to make me think you didn't say what you actually said!
See http://en.wikipedia.org/wiki/Micro_black_hole and please do your own research. This extra dimensions thing isn't my claim.

After your faulty mind-set has been corrected, regarding the above matters, then we can move on to other aspects of the discussion.
My mindset isn't faulty, Vernon. The energy of the falling plate came from the plate.
Title: If I give an object some potential energy, does its mass increase?
Post by: Farsight on 26/08/2009 19:45:15
Of course they will accelerate even in the absence of any gradient, provided there is a gravitational field...
If a gravitational field is there, there's a gradient there, lightarrow. There has to be some form of gradient, otherwise things wouldn't fall down.

Ron: Quantum mechanics is good stuff. Read up on Quantum Electrodynamics and what Feynman actually said, and it's all perfectly fine. He stresses that the virtual particles are virtual, and nobody actually knows what's going on under the covers. But some people get this wrong and ascribe a physical reality to virtual particles that Feynman and others never ever intended. The problems are in the interpretation, and people then misapply their misunderstanding and try to take it too far. You can't quantize gravity, because a photon causes gravity and it doesn't approach you in steps. It approaches you smoothly.
Title: If I give an object some potential energy, does its mass increase?
Post by: Ron Hughes on 26/08/2009 20:57:10
farsight, I didn't quantize gravity anymore than I quantized time. I have suggested it is a function of time and time is a function of electromagnetic density. If anyone wanted to they could calculate the momentum of the diatomic hydrogen molecule B just before the collision with C and A. I don't know how to do that but I'm sure there are members who do.
Title: If I give an object some potential energy, does its mass increase?
Post by: VernonNemitz on 26/08/2009 23:17:56
Several of your remarks in the last message stressed the hypothetical nature of certain things.  But one variety of thing is not hypothetical at all: virtual particles.  WHILE they exist, they are exactly as real as ordinary particles.  And their temporary existence can detectably though indirectly affect real things; see the "Casimir Effect" for details: http://en.wikipedia.org/wiki/Casimir_effect
There's certainly something real there, Vernon. The Casimir effect is real. But those vitural particles are virtual. It's wrong to ascribe them the degree of reality that you do.
PROVE IT.  Because the definition of "virtual" that I use is "temporary".  Their existence, By Definition, is a violation of Energy Conservation --an allowed temporary violation.  Consider the difference between a modern very thin/lightweight folded-up inflatable-strut tent and its deployed state; the shelter the deployed tent provides is very real and can be very temporary.  Consequently we could call the deployed tent a "virtual shelter", right?  The very small folded-up tent merely has potential; it is normally considered to have zero shelter associated with it, in that state.  (I'm saying the shelter is virtual, not the tent.)

If you think "virtual" means something else, that's your problem, not mine.  I know exactly what I'm talking about in this context; temporary existence can be Very Real.  And that's why the Casimir Effect is a very real side-effect.

Like I said, there's something there that's real, and that energy is real. But to claim that this is proof of virtual particles popping in and out of existence is on a par with saying it proves the existence of tiny dancing angels. Think of virtual particles as the evanescent wave.
I don't need to think of that, when virtual particles are much better at explaining things.  Look up the history of the pi-meson (pion).  Predicted to have certain properties, to exist as temporary particles to explain why protons stay together in an atomic nucleus, the "real" form of that particle was later found to possess the specified properties.  What overcomes the electrostatic repulsion of protons if not virtual pions? (Certainly not virtual gluons; those are locked inside protons and charged pions, holding those particles' constituent quarks together.  --Oh, I forgot, since separated quarks have never been seen, you probably "dis" them as being merely hypothetical.  Nevertheless, they actually have been detected as individual particles (while not especially separated): http://en.wikipedia.org/wiki/Parton_(particle_physics) )  Your evanescent wave hasn't got a chance, to explain complex stable nuclei, and therefore is a wrong explanation.  Worthless.  So if you want to claim virtual particles aren't real enough, then you need a better alternative, also able to explain the existence of atoms more complex than hydrogen.

Vernon, gravitons do not exist.
PROVE IT.  You are essentially saying it will be forever impossible to devise a quantum theory of gravitation.

There are no actual gravitons zipping back and forth between a photon and everything else.
Your bald claims are totally worthless without supporting evidence.  You don't even have logical self-consistency on your side, as was pointed out to me years ago in that UseNet discussion, while I do now have logical self-consistency on my side (and possibly a tiny amount of evidence; have you read that "Simple Quantum Gravitation" knol yet?  Here's a related teaser: http://adsabs.harvard.edu/abs/2004AIPC..699.1138M ).

Gravity is detectable. Gravitons are not. And nor are virtual gravitons.
Gravitons are not yet detectable.  Unless you have proof they cannot exist, you would be foolish to say they will never ever be detectable.  Meanwhile, no physicist expects to directly detect any particle that exists virtually; doing so would be a violation of Energy Conservation, so why do you even bring that up?  Can you directly detect the virtual shelter of a folded-up tent (especially if you don't know that the thing is a tent)?

We were talking about where the energy of a falling plate comes from.
We were, indeed.  And while I mentioned this before, it bears repeating: General Relativity states that the behavior of a system does not change when the frame of reference is changed (the viewpoint used to describe the system).  Nuclear physicists use this to specify that the masses of proton and neutron in a deuterium nucleus are the same as their masses when they exist as separate particles, despite the fact the nucleus has less total mass than the sum of the individual particles.  Their math works; at the very least the mass ratios of the two particles are unchanged by fusing them together.  Ditto does similar math work for separated electron and proton, compared to a hydrogen atom in the ground state; their mass ratios are unchanged.  If you want the plate-and-planet system to behave differently, you need a valid reason.  And your inability to accept the Real Fact of "spooky action at a distance" is not a valid reason!!!

One of the best pieces of evidence involves gamma-ray photons that turn into particle-pairs; pair-production is maximized at certain energies called "resonant energies". Resonant with what, eh? Virtual particles!
It's no proof of virtual particles the way you think of them.
It certainly is.  A simple gamma of about 1.022Mev will not, all by itself, easily transform into a separated electron/anti-electron pair.  It will only do it in the presence of a strong electric field. http://geant4.cern.ch/UserDocumentation/UsersGuides/PhysicsReferenceManual/BackupVersions/V9.0/html/node27.html  Why?  If virtual particles are present everywhere and all the time, the gamma should be interacting with them all the time (heh, that's why the gamma --and any other photon-- only travels at light-speed and not faster!).  The electric field is needed to separate a pair of virtual particles that have absorbed the gamma and become temporarily detectable (temporary in the sense that if the electric field wasn't there, the two particles would mutually annihilate and the gamma ray would continue on its way, thanks to Momentum Conservation).

"None will be produced by such a collision, because the required extra dimensions do not exist."
You have yet to show why any extra dimensions are required. You have presented OTHER reasons why quantum black holes will not be created, which I find satisfactory. But don't try to make me think you didn't say what you actually said!
This extra dimensions thing isn't my claim.
Then why did you even bother to mention it, especially since (A) it is hypothetical and (B) you "dis" stuff that is hypothetical!?!?

After your faulty mind-set has been corrected, regarding the above matters, then we can move on to other aspects of the discussion.
Worth repeating.
Title: If I give an object some potential energy, does its mass increase?
Post by: VernonNemitz on 26/08/2009 23:23:58
You can't quantize gravity, because a photon causes gravity and it doesn't approach you in steps. It approaches you smoothly.
Bad logic.  I mentioned before the possibility that a photon might interact gravitationally at a high frequency; you would not be able to distinguish between a step-wise approach and a smooth approach, because the steps would be too small to notice.
Title: If I give an object some potential energy, does its mass increase?
Post by: Ron Hughes on 27/08/2009 02:53:56
Vernon when you say, "Prove it." that is somewhat like a bible thumper telling me I can't prove God doesn't exist therefore that is proof God does exist. The onus should be on the bible thumper to provide the proof of Gods existance.
Title: If I give an object some potential energy, does its mass increase?
Post by: VernonNemitz on 27/08/2009 13:46:28
Vernon when you say, "Prove it." that is somewhat like a bible thumper telling me I can't prove God doesn't exist therefore that is proof God does exist. The onus should be on the bible thumper to provide the proof of Gods existance.
And my last messages to Farsight indicate we have rather more evidence for the actuality of virtual particles than their absence.  It's not my problem if he can't accept the facts.  (With respect to my other challenge, why, there's an overview of one logical QM version of gravitation right here: http://knol.google.com/k/vernon-nemitz/simple-quantum-gravitation/131braj0vi27a/2 )

Perhaps you'd be interested in a purely philosophical Question, "Why is there Something rather than Nothing?" --which could apply to anything, God included, if one chooses to believe that God exists.

My answer to that Question is derived from a counter-Question, "If Nothingness was the only State, then what would there be to prevent Something from existing?"  Obviously the answer to this question is "Nothing", and therefore it would be paradoxical if Something did not exist.  Well, the virtual particles in the vacuum (also known as vacuum fluctuations or the vacuum self-energy) are an extremely close match to this philosophical view; they represent Something and Nothing near-simultaneously existing everywhere and all the time.  And some physicists consider the entire Universe to be just a giant temporary fluctuation in Nothing....
Title: If I give an object some potential energy, does its mass increase?
Post by: lightarrow on 27/08/2009 19:15:52
Of course they will accelerate even in the absence of any gradient, provided there is a gravitational field...
If a gravitational field is there, there's a gradient there, lightarrow. There has to be some form of gradient, otherwise things wouldn't fall down.
A gradient of potential, not of the field. Vernon was talking about a gravitational *field*. Maybe it could seem nitpicking for you, but it's always better to be, because people could confound itself (and this already happened on this forum sometimes ago exactly on this very subject).
Title: If I give an object some potential energy, does its mass increase?
Post by: VernonNemitz on 27/08/2009 20:25:46
Of course they will accelerate even in the absence of any gradient, provided there is a gravitational field...
If a gravitational field is there, there's a gradient there, lightarrow. There has to be some form of gradient, otherwise things wouldn't fall down.
A gradient of potential, not of the field. Vernon was talking about a gravitational *field*. Maybe it could seem nitpicking for you, but it's always better to be, because people could confound itself (and this already happened on this forum sometimes ago exactly on this very subject).
I'm pretty sure you will find that a field practically always includes a gradient --especially if it can be computed as if arising from a point-source (such as gravity from an object's center of mass).
Title: If I give an object some potential energy, does its mass increase?
Post by: lightarrow on 28/08/2009 08:35:08
I'm pretty sure you will find that a field practically always includes a gradient --especially if it can be computed as if arising from a point-source (such as gravity from an object's center of mass).
"Especially" in this case is a bit eufemistic  [;)].
Of course I was talking about a region of space small enough and distant enough from point-like sources so that the field could be considered as uniform (and don't tell me that, however, a mathematically totally uniform field is impossible, we are talking about physics; furthermore, an inertial field is equivalent to a gravitational one).
Title: If I give an object some potential energy, does its mass increase?
Post by: lyner on 28/08/2009 13:40:35
The reason that things 'fall down' is a gradient in potential , not a field gradient. It isn't hard to produce a region of uniform gravitational field even on Earth - things will still fall down in it.
Title: If I give an object some potential energy, does its mass increase?
Post by: VernonNemitz on 28/08/2009 14:12:19
Talk about nitpicking.  It seems to me that the practically-always-existing gradient of a field is equivalent to a gradient of potential.  If an external gravitational field wasn't present, affecting some particle, there would be no gravitational potential energy associated with the particle, in that field, and thus no potential to fall.  Two aspects of one thing.  Like gravitational mass and inertial mass are postulated to be two aspects of one thing.  Like salt is a necessary nutrient in small quantities, and deadly in large quantities; two aspects of the same thing.
Title: If I give an object some potential energy, does its mass increase?
Post by: lightarrow on 28/08/2009 16:58:06
Talk about nitpicking.  It seems to me that the practically-always-existing gradient of a field is equivalent to a gradient of potential.  If an external gravitational field wasn't present, affecting some particle, there would be no gravitational potential energy associated with the particle, in that field, and thus no potential to fall.  Two aspects of one thing.  Like gravitational mass and inertial mass are postulated to be two aspects of one thing.  Like salt is a necessary nutrient in small quantities, and deadly in large quantities; two aspects of the same thing.
Then, space and velocity (for example) are the same thing? They "only" differ because one is the derivative of the other...
Title: If I give an object some potential energy, does its mass increase?
Post by: VernonNemitz on 28/08/2009 18:26:46
Then, space and velocity (for example) are the same thing? They "only" differ because one is the derivative of the other...
That particular derivation requires including an additional thing, time.  Why aren't both a field gradient and a gradient-of-potential as static as Space?  A typical recreation-park slide (for children) has a gradient and by itself is typically considered to be quite static.  I suppose I'm missing something....
Title: If I give an object some potential energy, does its mass increase?
Post by: Farsight on 28/08/2009 19:25:51
PROVE IT. Because the definition of "virtual" that I use is "temporary". Their existence, By Definition, is a violation of Energy Conservation --an allowed temporary violation.
There's energy in an electron's electric field. It doesn't vary, it doesn't violate conservation of energy, and it isn't temporary. The mathematics of the attraction between the electron and the proton can be modelled using virtual particles, but saying that these are real but temporary particles popping in and out of existence is taking QED too far. It's not what Feynman intended. He intended them as accounting units. Read http://www.amazon.co.uk/QED-Strange-Theory-Penguin-Science/dp/0140125051 to check this out.

If you think "virtual" means something else, that's your problem, not mine.  I know exactly what I'm talking about in this context; temporary existence can be Very Real. And that's why the Casimir Effect is a very real side-effect.
Virtual means virtual. As in not real.

I don't need to think of that, when virtual particles are much better at explaining things.
But they aren't better at explaining things. When you fire a plate up into the sky at 12 km/s you give the plate some energy. You don't give it to the earth, or the earth's gravitational field, you give it to the plate, and that plate achieves escape velocity and leaves the system. When you reverse this scenario, you have to accede that that the energy of a falling plate comes from the plate. The idea that it comes from the earth via virtual gravitons travelling at superluminal velocities is not supported by any scientific evidence.

Look up the history of the pi-meson (pion).  Predicted to have certain properties, to exist as temporary particles to explain why protons stay together in an atomic nucleus, the "real" form of that particle was later found to possess the specified properties.  What overcomes the electrostatic repulsion of protons if not virtual pions? (Certainly not virtual gluons; those are locked inside protons and charged pions, holding those particles' constituent quarks together.
Pions have a lifetime of 8.4 x 10-17 seconds and consist of two quarks. Neutrons hold protons tegother, and I rather think we're getting off topic with this.

Oh, I forgot, since separated quarks have never been seen, you probably "dis" them as being merely hypothetical. Nevertheless, they actually have been detected as individual particles (while not especially separated): http://en.wikipedia.org/wiki/Parton_(particle_physics)
They're partons. Parts. Even Gell-Mann considered them to be subunits, see http://www.achievement.org/autodoc/page/gel0int-1. You want to get a handle on this? Look at the picture of the trefoil knot below:

(https://www.thenakedscientists.com/forum/proxy.php?request=http%3A%2F%2Fwww.home.no%2Ft-o-k%2FPOV-Ray%2FTrefoil_Knot.jpg&hash=f8f827a0fb7c888d7a859c69165ffb85)

Now start from the bottom left and follow round looking at crossing points. Ignore crossings-under. Only look at crossings-over. Now call out each crossing direction in terms of whether it's up or down.

Your evanescent wave hasn't got a chance, to explain complex stable nuclei, and therefore is a wrong explanation. Worthless. So if you want to claim virtual particles aren't real enough, then you need a better alternative, also able to explain the existence of atoms more complex than hydrogen.
It isn't "my" evanesent wave. The evanescent wave is also known as the near-field in radio transmitters. Check out http://en.wikipedia.org/wiki/Near_and_far_field and search on evanescent:

Rather, in the near field, it is sometimes useful to express the contributions as a sum of radiating fields combined with evanescent fields, where the latter are exponentially decaying with r. And in the source itself, or as soon as one enters a region of inhomogeneous materials, the multipole expansion is no longer valid and the full solution of Maxwell's equations is generally required.

In quantum mechanical terms, the far-field is due simply to radiation of classical photons. These remove energy from the transmitter whether they are immediately absorbed or not. By comparison, the near-field, if it must be seen in quantum terms, may be thought of being composed of virtual photons, which have a more evanescent existence.

You are essentially saying it will be forever impossible to devise a quantum theory of gravitation.
Yes, that's right. Gravity doesn't work like that.

Your bald claims are totally worthless without supporting evidence.  You don't even have logical self-consistency on your side, as was pointed out to me years ago in that UseNet discussion, while I do now have logical self-consistency on my side (and possibly a tiny amount of evidence;
Mine aren't the bald claims. There is no evidence for virtual gravitons. And I do have logical consistency on my side. The energy of the falling plate comes from the plate.

Have you read that "Simple Quantum Gravitation" knol yet?  Here's a related teaser: http://adsabs.harvard.edu/abs/2004AIPC..699.1138M ).
No, I haven't read it yet. Sorry. If you spent less time throwing up hypotheticals to avoid the simple logic of the falling plate, I'd have more time to do so. I had a quick look at that teaser and didn't like what I saw re the existence of mass fluctuations and their use in exotic propulsion schemes.

Gravitons are not yet detectable. Unless you have proof they cannot exist, you would be foolish to say they will never ever be detectable...
No Vernon, it's foolish to carry on year after year adhering to a hypothesis that has no supporting evidence.

..And your inability to accept the Real Fact of "spooky action at a distance" is not a valid reason!!!
It isn't a real fact. We have no evidence of virtual gravitons rattling around at superluminal velocities. A plate falls because there's a gradient in its local space. It's observable, via the Pound-Rebka experiment and GPS.

Now please, let's try to keep on topic and keep the posts briefer.

Ron: well said.
Title: If I give an object some potential energy, does its mass increase?
Post by: Farsight on 28/08/2009 20:08:55
All: take a fairly small region of space, such as a room, and there's no measurable gradient in the gravitational field strength in that room. Drop an object from the top of the room or the middle of the room, and you measure the same acceleration of 9.8 m/s. But the object falls down, so there is some kind of gradient in that room. Sophie called it a gradient in gravitational potential, but we can take a step further. Check out the Pound-Rebka experiment, which is extremely sensitive: http://en.wikipedia.org/wiki/Pound%E2%80%93Rebka_experiment. There's a detectable frequency shift. As a photon travels downwards, its observed frequency changes. Some ascribe this to virtual gravitons travelling instantly up from the ground, but it's far simpler to think of this as a change in the local properties of the space that the photon travels through. This is totally in line with Einstein, who said in 1920:

Mach’s idea finds its full development in the ether of the general theory of relativity. According to this theory the metrical qualities of the continuum of space-time differ in the environment of different points of space-time, and are partly conditioned by the matter existing outside of the territory under consideration.

This space-time variability of the reciprocal relations of the standards of space and time, or, perhaps, the recognition of the fact that “empty space” in its physical relation is neither homogeneous nor isotropic, compelling us to describe its state by ten functions (the gravitation potentials gμν), has, I think, finally disposed of the view that space is physically empty.

It's like you've got a mechanical clock and you plunge it into an oil bath. It now runs slower, because of the viscosity, because the environment has changed. But you can't see it, because you're like a clockwork man, and you've jumped in after it.
Title: If I give an object some potential energy, does its mass increase?
Post by: PhysBang on 28/08/2009 20:29:58
You want to get a handle on this? Look at the picture of the trefoil knot below:

Now start from the bottom left and follow round looking at crossing points. Ignore crossings-under. Only look at crossings-over. Now call out each crossing direction in terms of whether it's up or down.
This is embarrassing! Your justification for ignoring the mountains of evidence in favour of the standard model of particle physics is a picture of a knot? Can you explain what this knot has to do with any experiment ever performed? Can your knot produce the formulae of quantum mechanics?
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Mine aren't the bald claims. There is no evidence for virtual gravitons. And I do have logical consistency on my side. The energy of the falling plate comes from the plate.
Virtual gravitons aren't the question here. You have made claims that the quantization of gravity is impossible. Where is the evidence?
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If you spent less time throwing up hypotheticals to avoid the simple logic of the falling plate, I'd have more time to do so. I had a quick look at that teaser and didn't like what I saw re the existence of mass fluctuations and their use in exotic propulsion schemes.
The problem is, VN is at least attempting to address actual science that explains actual observation. Your theory of gravity cannot actually predict how fast a plate will fall. This is a serious, serious problem with your theory, as we can have no reason to believe in your theory and no basis to use your theory to reject anything.
Title: If I give an object some potential energy, does its mass increase?
Post by: Ron Hughes on 28/08/2009 20:35:31
As vernon says, " Gravity and inertia are two aspects of the same thing. " Time dilation.
Title: If I give an object some potential energy, does its mass increase?
Post by: PhysBang on 28/08/2009 20:35:50
Mach’s idea finds its full development in the ether of the general theory of relativity. According to this theory the metrical qualities of the continuum of space-time differ in the environment of different points of space-time, and are partly conditioned by the matter existing outside of the territory under consideration.

This space-time variability of the reciprocal relations of the standards of space and time, or, perhaps, the recognition of the fact that “empty space” in its physical relation is neither homogeneous nor isotropic, compelling us to describe its state by ten functions (the gravitation potentials gμν), has, I think, finally disposed of the view that space is physically empty.

This gradient in the gravitational potential can be considered to be a "pressure" gradient or a "density" gradient, but I don't like either word when talking about space. The best I can offer is a gradient in the energy density of space, or a gradient in vacuum energy. It's observable as a gradient in gravitational time dilation. Think about c. You will always measure c to be 299,792,458 m/s. But gravitational time dilation is only radial. So if you're measuring c using a horizontal apparatus, the metres don't change.
Let's stop there. Because what does Einstein say about those ten functions in the quotation above? He says that they determine the metric of spacetime. That means that the metres do change. You can't simply cherry-pick Einstein and prove whatever you want. If you want to invoke the 1920 work by Einstein, then you have to accept the mathematical theory that Einstein uses, the mathematical theory that you seemingly cannot understand and constantly contradict.
Title: If I give an object some potential energy, does its mass increase?
Post by: lyner on 28/08/2009 23:04:23
VernonNemitz
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Talk about nitpicking.  It seems to me that the practically-always-existing gradient of a field is equivalent to a gradient of potential.
I'm not nitpicking - I'm talking about definitions. If you can't stick to them, what chance have you got?
If you don't understand the terms then you shouldn't be using them to prove points.
Title: If I give an object some potential energy, does its mass increase?
Post by: Farsight on 29/08/2009 14:53:28
Stop carping, PhysBang, and get up to speed: http://arxiv.org/find/grp_physics/1/ti:+AND+quantum+knot/0/1/0/all/0/1. There is no evidence to support the quantization of gravity, and the evidence against it is simple: energy causes gravity, a photon conveys energy, and it doesn't approach you in steps. The resultant gravity rises smoothly. And this isn't my theory of gravity, it's Albert Einstein's. He talked about the variable speed of light and the equations of motion, not curved spacetime. Plus others have looked into it way before me, see http://arxiv.org/abs/0708.3518 and http://www.ag-physics.de/. And please, go and look up length contraction in the Schwarschild metric. It's radial. The metric changes, but not the transverse metre. By the way, as you said to everybody earlier, my name is John Duffield, are you Albert C Marshall formerly of Sandia?
Title: If I give an object some potential energy, does its mass increase?
Post by: PhysBang on 29/08/2009 16:04:45
Stop carping, PhysBang,
Pointing out that you consistently use cherry-picking of quotations in order to deceptively make your case is not carping. You malign the work of Einstein in order to inflate your own appearence, pure and simple.
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and get up to speed: http://arxiv.org/find/grp_physics/1/ti:+AND+quantum+knot/0/1/0/all/0/1.
There is absolutely nothing in those papers that relates to your use of knots other than the five letters in the word "knots". Show the math.
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There is no evidence to support the quantization of gravity, and the evidence against it is simple: energy causes gravity, a photon conveys energy, and it doesn't approach you in steps. The resultant gravity rises smoothly. And this isn't my theory of gravity, it's Albert Einstein's. He talked about the variable speed of light and the equations of motion, not curved spacetime.
"Curved spacetime" is the basis for Einstein's theory. It is the words that we use to refer to Einstein's use of Riemannian geometry. That you cannot understand this is not surprising, as you make no effort to actually learn Einstein's work.

And nobody is claiming that spacetime isn't continuous in General Relativity. Rather, they are claiming that we have to develop a new theory. Some types of successor theories are on the verge of being rejected by the evidence, others are not. So far, there is no evidence that strongly supports any successor theory.
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Plus others have looked into it way before me, see http://arxiv.org/abs/0708.3518 and http://www.ag-physics.de/.
That you are not alone in doing bad physics is not much of a defence.
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And please, go and look up length contraction in the Schwarschild metric. It's radial. The metric changes, but not the transverse metre.
OK, show us the math.
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By the way, as you said to everybody earlier, my name is John Duffield, are you Albert C Marshall formerly of Sandia?
Is your name a secret? You are trying to sell your book here, aren't you?
Title: If I give an object some potential energy, does its mass increase?
Post by: Farsight on 29/08/2009 17:26:31
You are carping, PhysBang, that's all you ever do. And I certainly don't malign Einstein whatsoever. Curved spacetime is not the basis for Einstein's theory, as we see when we actually read The Foundation of the General theory of Relativity, and as Pmb showed in the arXiv paper that you dismissed as crank, just as you dismiss anything unfamiliar, like you'll dismiss these trefoil HEP papers: http://arxiv.org/find/grp_physics/1/ti:+trefoil/0/1/0/all/0/1. And no, I'm not selling anything, I'm trying to discuss physics. On a discussion forum. If you don't want to, that's up to you, but stop trying to spoil the forum for everybody else. And I repeat: are you Albert C Marshall formerly of of Sandia?
Title: If I give an object some potential energy, does its mass increase?
Post by: AllenG on 29/08/2009 17:42:06
Gentlemen,  argue the physics and please leave the personal snipes out of the conversation.

Title: If I give an object some potential energy, does its mass increase?
Post by: Farsight on 29/08/2009 18:17:51
OK, let's get back to the original post and work it through:

So matter can basically be thought of as just potential energy because matter can be turned into energy and theoretically vice-versa, does that mean when I lift up a plate and increase it's potential energy, i've technically increased its mass?

Let's imagine we're on an earth-sized planet with no atmosphere, and that the plate is robust and isn't going to shatter. We use something like a long-barrelled cannon to fire it upwards with a velocity of 10km/s. Lightarrow thought this:

The mass is increased, but not the mass of the plate: the mass of the system Earth-plate.

However there's a problem here with conservation of energy. We used chemical energy in the propellant, and converted it into the plate's kinetic energy. We haven't added any energy to the Earth-plate system, and if we had, the gravity of that system would increase. The earth doesn't recoil in any measurable fashion, so the distribution of 1/2 mv2 kinetic energy is not symmetrical. That kinetic energy is definitely in the plate: 10km/s is very measurable.

As the plate rises it slows down, and we say the kinetic energy is being converted into potential energy. But because it started with 10km/s the plate doesn't have escape velocity. Hence there will come a point when all of the kinetic energy is converted into potential energy, and the plate stops momentarily.

At this point the Earth and the plate are at rest with respect to one another, just as they were before we fired the plate up into the sky. All of that 10km/s has been converted into potential energy, and the plate has more energy, and so more mass than previously. It isn't obvious where it's gone, but if you look at electron spin within the plate and then gravitational time dilation, all the electrons are spinning at a greater rate than they were. You can't measure anything locally because it's an immersive scale-change that also affects your measuring devices. You can only detect this via say the GPS clock adjustment.

People attempt to evade the fact that the potential energy is in the plate, and say it's in the gravitational field or the earth rather than the plate, but take a look at what happens when you fire the plate at 12km/s. Here the plate does have escape velocity, and it leaves the system, taking way all the original mass/energy of the plate along with all the kinetic energy we gave it. The mass/energy of the earth is diminished, and the gravity is so reduced. It's cut and dried: that potential energy went into the plate.

Title: If I give an object some potential energy, does its mass increase?
Post by: VernonNemitz on 30/08/2009 10:39:33
There's energy in an electron's electric field. It doesn't vary, it doesn't violate conservation of energy, and it isn't temporary.
That's an ordinary electron, possibly.  Certainly "conventionally".  I won't argue that point at this time, because it is NOT a virtual electron, and therefore irrelevant to the present discussion.

The mathematics of the attraction between the electron and the proton can be modelled using virtual particles, but saying that these are real but temporary particles popping in and out of existence is taking QED too far. It's not what Feynman intended. He intended them as accounting units.
So?  Who says Feynmann Is The Last Word on this subject?  Feynman proposed that anti-particles move backward through time, too, as a bookkeeping notion.  And a lot of physicists talk about "negative binding energy", another bookkeeping notion (I don't know who originated it), as if it was quite real.  If you are going to say that negative binding energy cannot be real simply because it is an accounting trick, what are you going to put in its place?  Your potential-energy-as-mass-of-plate notion does NOT fit the math that describes the behavior of nuclear particles in the Strong-Force field-gradient.  I see you have neglected to offer any other reason why the virtual-accounting trick can't be real, besides, duh, "Feynmann said so."  Whoop-te-do.  How does he know???  Not to mention, he died more than 20 years ago, and thus missed a lot of recent theoretical developments.  Maybe he would be now be saying some of the same things I've been saying to you.

Virtual means virtual. As in not real.
Wrong again.  Educate yourself: http://dictionary.reference.com/browse/virtual  --a thing can be called "virtual" if it has some of the virtues of something else.  But obviously it has to exist in some fashion to have any virtues at all!!!  The reality of that existence is all I need.  Also for your edification: http://en.wikipedia.org/wiki/Virtual_particle

But they [virtual particles] aren't better at explaining things. When you fire a plate up into the sky at 12 km/s you give the plate some energy. You don't give it to the earth, or the earth's gravitational field, you give it to the plate, and that plate achieves escape velocity and leaves the system.
Incomplete. It's initial velocity is reduced by about 11 km/s, before it escapes.  The kinetic energy associated with that velocity has become potential energy.  We are agreed that it takes the form of mass.  Well, if the Earth's gravity field sucked that KE out of the plate, during the escape, why shouldn't the Earth end up with most of that potential-energy-stored-as-mass?

When you reverse this scenario, you have to accede that that the energy of a falling plate comes from the plate.
I most certainly do not.  In QM terms we would be talking about an attractive force; the Earth pulls the plate toward it, more than the plate pulls the Earth.  How many times after you pulled on something, accelerating it, were you able to say that the pulled object was the source of the kinetic energy it acquired?

The idea that it comes from the earth via virtual gravitons travelling at superluminal velocities is not supported by any scientific evidence.
DO NOT TWIST WHAT I HAVE WRITTEN. Virtual gravitons don't have to be superluminal; they merely need to be "entangled" with their mass-of-origin.  If one is absorbed by some other mass, then it is the entanglement that allows the absorbing mass to acquire kinetic energy at the expense of the origin-mass.  The quantity transferred is equal to the energy of the virtual graviton at that point in its lifespan (its Uncertain energy diminishes at a rate ideally describable by the curve of the function 1/x).  It is Observed Fact that entanglement-events, when they are triggered, act instantaneously; it is the essence of "spooky action at a distance".  And yet the whole thing can make better sense if a virtual graviton is perceived as being "semi-real"; if it is absorbed, then the absorbing mass acquires its energy.  Simple.  And if it is never absorbed, the origin-mass loses nothing; that's the other side of the "semi-real" coin.

Neutrons hold protons together.
HOW???  --and no, we are not off-topic at all, since virtual pions can provide a detailed explanation for how protons can stay together in an atomic nucleus (thanks to present knowledge about quarks, heh!).  And, while neutrons are involved, the key point here is that virtual particles offer the best explanation.  If you want to say virtual particle don't exist, then you need to say much much more than merely, "Neutrons hold protons together."

They're partons. Parts.
They were observed parts, in electron-scattering experiments. That was my point. We can say that quarks are real parts of protons because we have observed real parts of protons.  Regarding that knot-picture, I distinctly see, per your "Only look at crossings-over" instruction, a left/right crossing-over.  Another worthless argument on your part, therefore, since it was supposely about up/down crossings only.

It isn't "my" evanesent wave. The evanescent wave is also known as the near-field in radio transmitters. Check out http://en.wikipedia.org/wiki/Near_and_far_field and search on evanescent:
I might need to partly take back some of what I wrote in my last message to you, since it sounds to me, from the description you quoted, that "evanescent" is synonymous with "virtual". (And since QM has the wave-particle duality, "evanescent wave" translates as "virtual particle" quite easily!)

You are essentially saying it will be forever impossible to devise a quantum theory of gravitation.
Yes, that's right. Gravity doesn't work like that.
Just because you say so, that doesn't mean it's true.  Evidence, please?

Mine aren't the bald claims.
Of course they are.  See previous quote.

There is no evidence for virtual gravitons.
Duh, it is premature to require such evidence, simply because there isn't an accepted Theory yet, that involves virtual gravitons.  YET, I said.  After there exists such a Theory, then will be the time to see what evidence might be gathered in support of it.  On the other hand, gravitation exists, does it not?  Why cannot that count as evidence?

And I do have logical consistency on my side.
HaHaHaHaHaHaHaHaHaHa!!!!!  See above, about pulling on something.

The energy of the falling plate comes from the plate.
Bad Logic As I've pointed out above and elsewhere.

If you spent less time throwing up hypotheticals to avoid the simple logic of the falling plate, I'd have more time to [read your stuff].
I'm a patient fellow. I don't require instant replies to what I write.  And the so-called "logic" of your falling plate is quite "ill".

I had a quick look at that teaser and didn't like what I saw re the existence of mass fluctuations and their use in exotic propulsion schemes.
Why? Because there might be more types of gravity waves out there, than are dreamt of in your philosophy?  Tough!  A propulsion scheme might be a simple as, "If I build a gravity-wave generator such that it emits waves moving in one direction, then Conservation of Momentum requires the machine to move in the other direction."

...it's foolish to carry on year after year adhering to a hypothesis that has no supporting evidence.
Sounds like your own hypothesis, much more so than mine.

A plate falls because there's a gradient in its local space. It's observable, via the Pound-Rebka experiment and GPS.
And a correct QM theory of gravitation will make predictions indistinguishable from GR's predictions. Whoop-te-do.  A gradient for virtual-graviton-intensity is as possible as the way the Inverse Square Law typically affects other large quantities of radiated things, like photons.  Since a mass doesn't truly radiate all its virtual gravitons from its "center of mass" (that's a bookkeeping trick!), a rotating mass will have most virtual gravitons leaving it at an angle, with effects equivalent to the twisted space that GR describes for a rotating mass.  And so on.  The passage of time can be slower in a gravity field if one first considers that "the passage of time" can be described as being associated with a particular rate of some particular interaction.  Well, in a gravity field that particular interaction isn't the only thing going on; the interacting parts are also interacting with lots of virtual gravitons.  If each particle can only do one interaction at a time, then it logically figures that the deeper in a gravity well, the more time a particle spends interacting gravitationally instead of with anything else; the rate of interaction with anything else has diminished, exactly as if time was passing more slowly, for those other interactions.  There is no way to tell the difference, between that and the time-slowing of GR.  And so on....
Title: If I give an object some potential energy, does its mass increase?
Post by: Farsight on 30/08/2009 14:17:12
So? Who says Feynmann Is The Last Word on this subject?
QED is a good theory, Feynman was a good guy. He was "the great explainer". Don't dismiss him so lightly.

Feynman proposed that anti-particles move backward through time, too, as a bookkeeping notion.
He wasn't far off. But the backward motion isn't through time, it's through space. An electron has a spin, and a chirality. The positron has the opposite chirality.

And a lot of physicists talk about "negative binding energy", another bookkeeping notion ...as if it was quite real. If you are going to say that negative binding energy cannot be real simply because it is an accounting trick, what are you going to put in its place?
Less of the positive energy. There is no such thing as negative distance or negative mass or negative energy.

Your potential-energy-as-mass-of-plate notion does NOT fit the math that describes the behavior of nuclear particles in the Strong-Force field-gradient.
Sure it doesn't. Gravity is different.

I see you have neglected to offer any other reason why the virtual-accounting trick can't be real, besides, duh, "Feynmann said so."  Whoop-te-do.  How does he know???  Not to mention, he died more than 20 years ago, and thus missed a lot of recent theoretical developments.  Maybe he would be now be saying some of the same things I've been saying to you.
No, he'd be saying the things I'm saying to you.

Wrong again.  Educate yourself: http://dictionary.reference.com/browse/virtual  --a thing can be called "virtual" if it has some of the virtues of something else.  But obviously it has to exist in some fashion to have any virtues at all!!!  The reality of that existence is all I need.  Also for your edification: http://en.wikipedia.org/wiki/Virtual_particle
I have educated myself. And whilst it doesn't square with some heavily-promoted thoeries that offer no predictions, I'm not alone, and I'm not wrong. Those virtual particles are as real as virtual pennies flying into your bank account when your employer credits your salary.

Incomplete. Its initial velocity is reduced by about 11 km/s, before it escapes. The kinetic energy associated with that velocity has become potential energy. We are agreed that it takes the form of mass. Well, if the Earth's gravity field sucked that KE out of the plate, during the escape, why shouldn't the Earth end up with most of that potential-energy-stored-as-mass?
Answer: the Earth's gravitational field is reduced when the plate departs. It has lost energy. When that plate flies through space and finds another Earth, it falls down and reaches 12km/s. And this new Earth gets all the energy that the old Earth lost.

I most certainly do not. In QM terms we would be talking about an attractive force; the Earth pulls the plate toward it, more than the plate pulls the Earth. How many times after you pulled on something, accelerating it, were you able to say that the pulled object was the source of the kinetic energy it acquired?
The earth doesn't pull on the plate. The plate falls down because the matter/energy of the Earth "conditions" the space around it, and this is subject to an inverse square law. That's relativity. This is what Einstein said. It means the local space around the plate is not uniform. There's a gradient in its properties. People call it the gravitational potential. The plate is made up of electrons that spin, plus other things, but electrons will suffice. Put this in inhomogeneous space, and the result is motion. Gravity. It means the whole concept of gravitons is wrong.

The idea that it comes from the earth via virtual gravitons travelling at superluminal velocities is not supported by any scientific evidence.
DO NOT TWIST WHAT I HAVE WRITTEN. Virtual gravitons don't have to be superluminal; they merely need to be "entangled" with their mass-of-origin. If one is absorbed by some other mass, then it is the entanglement that allows the absorbing mass to acquire kinetic energy at the expense of the origin-mass.  The quantity transferred is equal to the energy of the virtual graviton at that point in its lifespan (its Uncertain energy diminishes at a rate ideally describable by the curve of the function 1/x). It is Observed Fact that entanglement-events, when they are triggered, act instantaneously; it is the essence of "spooky action at a distance".
Einstein never liked this magical mysterious spooky action at a distance. It will be my pleasure to finally demonstrate how right he was. And Newton too. Newton didn't know about impedance so the "density" is backwards, but he was amazingly close:

"Doth not this aethereal medium in passing out of water, glass, crystal, and other compact and dense bodies in empty spaces, grow denser and denser by degrees, and by that means refract the rays of light not in a point, but by bending them gradually in curve lines? ...Is not this medium much rarer within the dense bodies of the Sun, stars, planets and comets, than in the empty celestial space between them? And in passing from them to great distances, doth it not grow denser and denser perpetually, and thereby cause the gravity of those great bodies towards one another, and of their parts towards the bodies; every body endeavouring to go from the denser parts of the medium towards the rarer?"

And yet the whole thing can make better sense if a virtual graviton is perceived as being "semi-real"; if it is absorbed, then the absorbing mass acquires its energy. Simple. And if it is never absorbed, the origin-mass loses nothing; that's the other side of the "semi-real" coin.
It makes no sense, Vernon. A photon travels at the speed of light, and it conveys energy. Energy causes gravity. The notion that a photon is always surrounded by a cloud of gravitons has been going for fifty years. There's still no evidence whatsoever.

If you want to say virtual particle don't exist, then you need to say much much more than merely, "Neutrons hold protons together."
There's the strong force which holds a proton together, and the residual strong force that holds a nucleus together. You need neutrons for this. Look at the elements and their isotopes. It's a whole new subject, let's talk about it on a different thread.

They were observed parts, in electron-scattering experiments. That was my point. We can say that quarks are real parts of protons because we have observed real parts of protons. Regarding that knot-picture, I distinctly see, per your "Only look at crossings-over" instruction, a left/right crossing-over.  Another worthless argument on your part, therefore, since it was supposely about up/down crossings only.
Look at it again, and search arXiv on trefoil: http://arxiv.org/find/grp_physics/1/ti:+trefoil/0/1/0/all/0/1. This isn't worthless, it's cutting edge.

I might need to partly take back some of what I wrote in my last message to you, since it sounds to me, from the description you quoted, that "evanescent" is synonymous with "virtual". (And since QM has the wave-particle duality, "evanescent wave" translates as "virtual particle" quite easily!)
Please do. You get a better conceptual understanding of QED this way. What's there isn't a whole crowd of transient photons flicking in and out of existence. What's there is a standing wave.

Just because you say so, that doesn't mean it's true. Evidence, please?
I'm giving it to you, and I'm offering you more, but you're dimissing it in favour of gravitons and arguing too much. Just look at the size of this post and all your boldings and italics.

Duh, it is premature to require such evidence, simply because there isn't an accepted Theory yet, that involves virtual gravitons. YET, I said.
After what, fifty years?

On the other hand, gravitation exists, does it not?  Why cannot that count as evidence?
Because it isn't evidence. Gravity is evidence for gravity, not gravitons.

Why? Because there might be more types of gravity waves out there, than are dreamt of in your philosophy? Tough! A propulsion scheme might be a simple as, "If I build a gravity-wave generator such that it emits waves moving in one direction, then Conservation of Momentum requires the machine to move in the other direction."
Because there's a better way to do it.

Title: If I give an object some potential energy, does its mass increase?
Post by: PhysBang on 30/08/2009 15:21:54
So? Who says Feynmann Is The Last Word on this subject?
QED is a good theory, Feynman was a good guy. He was "the great explainer". Don't dismiss him so lightly.
QED is based on special relativity and the idea that light always moves at a constant speed. Why do you dismiss him so lightly?
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Newton didn't know about impedance so the "density" is backwards, but he was amazingly close:

"Doth not this aethereal medium in passing out of water, glass, crystal, and other compact and dense bodies in empty spaces, grow denser and denser by degrees, and by that means refract the rays of light not in a point, but by bending them gradually in curve lines? ...Is not this medium much rarer within the dense bodies of the Sun, stars, planets and comets, than in the empty celestial space between them? And in passing from them to great distances, doth it not grow denser and denser perpetually, and thereby cause the gravity of those great bodies towards one another, and of their parts towards the bodies; every body endeavouring to go from the denser parts of the medium towards the rarer?"
Why are you quoting from an alchemical text of Newton? Do you also believe that one can turn lead into gold? This is another example of dishonest cherry-picking, as you obviously know because you omitted the source of the text.

In order to prove that gravity has anything to do with such a density gradient as Newton describesm we need a mathematical account of it.
Quote
Look at it again, and search arXiv on trefoil: http://arxiv.org/find/grp_physics/1/ti:+trefoil/0/1/0/all/0/1. This isn't worthless, it's cutting edge.
Because random articles in the arXiv use the word "trefoil" does not make your use of the word any less useless. You have yet to demonstrate that your inane use of the knot bears any relationship to any of the papers in the arXiv. It is not evidence to simply steal concepts at random from existing papers.

You have done nothing to attempt to show your "better way" but cherry-pick semi-relevant statements and reference papers that you show no sign of understanding. Show us how your mathematical theory predicts things better than any possible quantization of gravity. Indeed, show us how your theory can actually describe the motion of a plate through the air, since so far there is no reason to believe that your theory can even do this.
Title: If I give an object some potential energy, does its mass increase?
Post by: lyner on 30/08/2009 17:36:29
PhysBAng
I think you can make the same points without being so unpleasant. People tend to stop reading once they get to an insult! That is if you actually want to make a point.
Title: If I give an object some potential energy, does its mass increase?
Post by: VernonNemitz on 31/08/2009 06:56:01
I don't have time to write a thorough message just now; I'll merely focus on the worst offenses.

QED is a good theory, Feynman was a good guy. He was "the great explainer". Don't dismiss him so lightly.
Were not you the one who said that he preferred to talk about evanescent waves?  Well, then, I'm not dismissing him at all, since evanescent waves and virtual particles are essentially the same thing, per the wave-particle duality and other definitions.

Feynman proposed that anti-particles move backward through time, too, as a bookkeeping notion.
He wasn't far off. But the backward motion isn't through time, it's through space. An electron has a spin, and a chirality. The positron has the opposite chirality.
http://en.wikipedia.org/wiki/Antiparticle --and do remember it is only a bookkeeping trick!

There is no such thing as negative distance or negative mass or negative energy.
Tsk, tsk, another worthless bald claim.  Remember Energy-Time Uncertainty and those allowed fluctuations in the vacuum?  Tell, me, please, why those fluctuations must only take place on the positive side of zero?  Do remember that the description of those fluctuations involves Planck's Constant, a positive number, but ONLY a number, not something that Controls Nature.  Thus a negative Planck's Constant could be perfectly suited for describing negative-energy fluctuations in the vacuum, should they exist.  You want to claim they can't possibly exist?  Tell us why!!!

Incomplete. Its initial velocity is reduced by about 11 km/s, before it escapes. The kinetic energy associated with that velocity has become potential energy. We are agreed that it takes the form of mass. Well, if the Earth's gravity field sucked that KE out of the plate, during the escape, why shouldn't the Earth end up with most of that potential-energy-stored-as-mass?
Answer: the Earth's gravitational field is reduced when the plate departs. It has lost energy. When that plate flies through space and finds another Earth, it falls down and reaches 12km/s. And this new Earth gets all the energy that the old Earth lost.
Really bad logic. The plate is considered to be separate from the Earth as soon as you start treating it separate from the Earth.  That means even when falling off a cliff on Earth, it is not part of the Earth; it is part of the Earth/plate SYSTEM.  Therefore the Earth does not lose the mass of the plate when the plate is given an escape velocity; only the Earth/plate system loses it.  And the Earth still sucked about 11kps of velocity and associated kinetic energy from that escaping plate; the plate most certainly does not have it while traversing interplanetary space.  All it has is the potential to fall down a different gravity well, but that well is defined by the planet at its bottom, not by the presence of the plate.  That is, the amount of potential energy that the plate can acquire by falling down that well is defined by the planet at the bottom of the well.  If the planet has the same mass as the Earth, then the potential energy the plate can acquire will be the same as if it could acquire in falling to Earth.

I most certainly do not. In QM terms  ...
The earth doesn't pull on the plate. The plate falls down because ...
I do understand the GR notion, but that is not what I was talking about.  All your GR blather does not change by one whit what I wrote about how QM could describe things.

A photon travels at the speed of light, and it conveys energy. Energy causes gravity. The notion that a photon is always surrounded by a cloud of gravitons has been going for fifty years. There's still no evidence whatsoever.
DUHHHH...that's because gravitation is the weakest force, 30 orders of magnitude weaker than the Weak Nuclear Force.  It's going to take time to develop the technology to detect such feebleness.  You seem to think no one should ever even bother.  And again your mere say-so that gravitons do not exist cannot make them not exist.

Just because you say so, that doesn't mean it's true. Evidence, please?
I'm giving it to you, and I'm offering you more, but you're dimissing it in favour of gravitons and arguing too much. Just look at the size of this post and all your boldings and italics.
And I'll continue to use boldings and italics and capitalizations to stress the point that so far your evidence has been worthless, so why should I want more of the same?
Title: If I give an object some potential energy, does its mass increase?
Post by: Farsight on 31/08/2009 14:13:55
QED is based on special relativity and the idea that light always moves at a constant speed. Why do you dismiss him so lightly?
Check your facts, PhysBang. Einstein wrote his quantum mechanics paper Concerning the generation and transformation of light from a heuristic point of view before On the Electrodynamics of Moving Bodies. The former was in Annalen der Physik issue 17, and it's what he won his Nobel prize for. The latter was in issue 19, and evolved into special relativity, which wasn't accepted by mainstream physicists until the late twenties. See Clifford M Will's "The Confrontation between GR and Experiment" section 2.1.2 for this: http://relativity.livingreviews.org/Articles/lrr-2006-3/

Why are you quoting from an alchemical text of Newton? Do you also believe that one can turn lead into gold? This is another example of dishonest cherry-picking, as you obviously know because you omitted the source of the text.
It isn't from an alchemical text. It's from Opticks, queries 20 and 21. Here it is again:

"Doth not this aethereal medium in passing out of water, glass, crystal, and other compact and dense bodies in empty spaces, grow denser and denser by degrees, and by that means refract the rays of light not in a point, but by bending them gradually in curve lines? ...Is not this medium much rarer within the dense bodies of the Sun, stars, planets and comets, than in the empty celestial space between them? And in passing from them to great distances, doth it not grow denser and denser perpetually, and thereby cause the gravity of those great bodies towards one another, and of their parts towards the bodies; every body endeavouring to go from the denser parts of the medium towards the rarer?"

Also see query 30, where Newton said Are not gross bodies and light convertible into one another? He anticipated pair production, so it's no small wonder he was interested in alchemy. A great man was Newton, even greater than most people realise. And besides, what's the problem transmuting lead into gold? See http://en.wikipedia.org/wiki/Nuclear_transmutation and note where it says:

"Ironically, it transpired that, under true nuclear transmutation, it is far easier to turn gold into lead than the reverse reaction, which was the one the alchemists had ardently pursued. Nuclear experiments have successfully transmuted lead into gold, but the expense far exceeds any gain".

In order to prove that gravity has anything to do with such a density gradient as Newton describes we need a mathematical account of it.
Look no further than The Foundation of the General Theory of Relativity. Einstein talks about pressure and density, not curved spacetime. Alternatively cast your net a little wider: http://www.google.co.uk/search?hl=en&rlz=1T4ADBF_en-GBGB240GB240&q=einstein+density+gravity&meta= and see http://en.wikipedia.org/wiki/Einstein%E2%80%93Cartan_theory for example. Search on "density". And when it comes to pressure, stress is the same thing as pressure, and we see "stress-energy" tensor everywhere we look.

Because random articles in the arXiv use the word "trefoil" does not make your use of the word any less useless. You have yet to demonstrate that your inane use of the knot bears any relationship to any of the papers in the arXiv. It is not evidence to simply steal concepts at random from existing papers.
I give you ample evidence that the trefoil is of serious interest, and now you accuse me of stealing ideas? See http://arxiv.org/PS_cache/hep-th/pdf/0602/0602098v1.pdf and note that the idea goes all the way back to Kelvin.

You have done nothing to attempt to show your "better way" but cherry-pick semi-relevant statements and reference papers that you show no sign of understanding. Show us how your mathematical theory predicts things better than any possible quantization of gravity. Indeed, show us how your theory can actually describe the motion of a plate through the air, since so far there is no reason to believe that your theory can even do this.
I reiterate, it isn't "my theory". It's a model, and I give copious acknowledgements to others who have been "studiously ignored". It's an analytical synthesis that joins the dots to point the way to the completion of the standard model. And as I've said to you before, the issue is in the interpretation of existing mathematics, not in the mathematics itself.
Title: If I give an object some potential energy, does its mass increase?
Post by: Farsight on 31/08/2009 14:36:13
Absolutely wrong.  See the last section of this article: http://en.wikipedia.org/wiki/Antiparticle --and do remember it is only a bookkeeping trick!
It isn't wrong Vernon. It's unfamiliar to you, but search arXiv on "chiral".

1638232]Tsk, tsk, another worthless bald claim.  Remember Energy-Time Uncertainty and those allowed fluctuations in the vacuum? Tell, me, please, why those fluctuations must only take place on the positive side of zero?  Do remember that the description of those fluctuations involves Planck's Constant, a positive number, but ONLY a number, not something that Controls Nature.  Thus a negative Planck's Constant could be perfectly suited for describing negative-energy fluctuations in the vacuum, should they exist. You want to claim they can't possibly exist?  Tell us why!!!
Show me a negative length. Or a negative mass. Or negative energy. You can't. The vacuum of space has a positive energy, all points within a gravitational field consists of space with positive energy.

1638232]Really bad logic. The plate is considered to be separate from the Earth as soon as you start treating it separate from the Earth.  That means even when falling off a cliff on Earth, it is not part of the Earth; it is part of the Earth/plate SYSTEM. Therefore the Earth does not lose the mass of the plate when the plate is given an escape velocity; only the Earth/plate system loses it.  And the Earth still sucked about 11kps of velocity and associated kinetic energy from that escaping plate; the plate most certainly does not have it while traversing interplanetary space.
Yes it does. Because when it finds another earth and falls down, it gives it all back.

1638232]All it has is the potential to fall down a different gravity well, but that well is defined by the planet at its bottom, not by the presence of the plate.  That is, the amount of potential energy that the plate can acquire by falling down that well is defined by the planet at the bottom of the well. If the planet has the same mass as the Earth, then the potential energy the plate can acquire will be the same as if it could acquire in falling to Earth.
And once it's fallen down and cooled off, it has less energy than it did in space, because the gravitational time dilation means that all the electrons etc in that plate are spinning at a reduced rate.

I do understand the GR notion, but that is not what I was talking about. All your GR blather does not change by one whit what I wrote about how QM could describe things.
I'm afraid it does Vernon. I'm sorry, but a modest reinterpretation of General Relativity that takes a step back to the original ends up demolishing all attempts to quantize gravity.

DUHHHH...that's because gravitation is the weakest force, 30 orders of magnitude weaker than the Weak Nuclear Force. It's going to take time to develop the technology to detect such feebleness. You seem to think no one should ever even bother. And again your mere say-so that gravitons do not exist cannot make them not exist.
After fifty years without evidence, it's time to kick the graviton into the long grass and pursue more fruitful avenues. But if you don't agree, let's just agree to differ.
Title: If I give an object some potential energy, does its mass increase?
Post by: PhysBang on 31/08/2009 16:05:41
Check your facts, PhysBang. Einstein wrote his quantum mechanics paper Concerning the generation and transformation of light from a heuristic point of view before On the Electrodynamics of Moving Bodies. The former was in Annalen der Physik issue 17, and it's what he won his Nobel prize for. The latter was in issue 19, and evolved into special relativity, which wasn't accepted by mainstream physicists until the late twenties. See Clifford M Will's "The Confrontation between GR and Experiment" section 2.1.2 for this: http://relativity.livingreviews.org/Articles/lrr-2006-3/
What does any of that have to do with QED?
Quote
Look no further than The Foundation of the General Theory of Relativity. Einstein talks about pressure and density, not curved spacetime.
If you would read far enough to get to the mathematics of the theory, you would discover that the actual pressure and density is assigned to the contents of the spacetime and define the stress-energy tensor. The stress-energy tensor defines the curvature of spacetime, which it the action of gravity. Why do you constantly avoid the actual science?
Quote
Alternatively cast your net a little wider: http://www.google.co.uk/search?hl=en&rlz=1T4ADBF_en-GBGB240GB240&q=einstein+density+gravity&meta= and see http://en.wikipedia.org/wiki/Einstein%E2%80%93Cartan_theory for example. Search on "density". And when it comes to pressure, stress is the same thing as pressure, and we see "stress-energy" tensor everywhere we look.
Again, you ignore how Einstein actually used the stress-energy tensor. Moving from a metric theory account to an affine theory account does not help your point here, it rather makes it worse. Affine theories rely on a more refined account of geometry in order to deliver the content of the theory, not less.
Quote
I give you ample evidence that the trefoil is of serious interest, and now you accuse me of stealing ideas? See http://arxiv.org/PS_cache/hep-th/pdf/0602/0602098v1.pdf and note that the idea goes all the way back to Kelvin.
The idea of using the word "trefoil" may go back that far, but there is no reason to believe that Kelvin's use of the word has anything at all to do with your use of the word. Indeed, as you seem to be using "trefoil" to refer to something to do with sub-atomic physics, it seems incredibly dubious that you can claim some support for your theorty from Kelvin's use of the word.
Quote
I reiterate, it isn't "my theory". It's a model, and I give copious acknowledgements to others who have been "studiously ignored". It's an analytical synthesis that joins the dots to point the way to the completion of the standard model. And as I've said to you before, the issue is in the interpretation of existing mathematics, not in the mathematics itself.
OK, show us how your "model" can be used to calculate the motion of a dinner plate.
Title: If I give an object some potential energy, does its mass increase?
Post by: VernonNemitz on 31/08/2009 16:52:56
Feynman proposed that anti-particles move backward through time, too, as a bookkeeping notion.
He wasn't far off. But the backward motion isn't through time, it's through space. An electron has a spin, and a chirality. The positron has the opposite chirality.
Absolutely wrong.  See the last section of this article: http://en.wikipedia.org/wiki/Antiparticle --and do remember it is only a bookkeeping trick!
It isn't wrong Vernon. It's unfamiliar to you, but search arXiv on "chiral".
You are still completely wrong.  Pay attention. I originally wrote: "Feynman proposed that anti-particles move backward through time" --and that is EXACTLY true.  All your blather about the MODERN interpretation of antiparticles does not change the Real Fact that Feynmann did indeed make that proposal, as a bookkeeping trick.  When you stop confusing what I am actually talking about with other stuff, then we will be in a position to communicate better.

1638232] Tsk, tsk, another worthless bald claim.  Remember Energy-Time Uncertainty and those allowed fluctuations in the vacuum? Tell, me, please, why those fluctuations must only take place on the positive side of zero?  Do remember that the description of those fluctuations involves Planck's Constant, a positive number, but ONLY a number, not something that Controls Nature.  Thus a negative Planck's Constant could be perfectly suited for describing negative-energy fluctuations in the vacuum, should they exist. You want to claim they can't possibly exist?  Tell us why!!!
Show me a negative length. Or a negative mass. Or negative energy. You can't. The vacuum of space has a positive energy, all points within a gravitational field consists of space with positive energy.
More worthless/bad logic.  By that logic:  Before the invention of the microscope: "Bacteria can't exist, because you can't show me any."  Before the construction of really large telescopes: "Planets can't exist around other stars, because you can't show me any."  Have you ever thought about what Space might be like OUTSIDE the influence of a gravitational field?  Have you even paid attention to Basic Geometry???  Here's your Lesson For The Day:  If you draw a flexible line on a plane, you can flex the line in two directions upon the plane.  If you construct a flexible plane in Space, you can flex that plane in either of two directions in Space.  And if you can Curve 3D Space in a 4th geometric dimension, then you can curve it in either of two directions in that 4th dimension.  Guess what?  Only one of those two directions is associated with ordinary mass and ordinary gravitation in GR!  The other direction is perfectly available for negative mass-energy.

Try Googling for this: [Feynmann "anything not forbidden is mandatory"] (the brackets delimit the query; don't include them).  You can take that as my primary rationale for saying such things as, "You would be foolish to claim something cannot exist if you cannot prove it is forbidden."  Can you prove negative mass/energy and gravitons are forbidden?  HAH!  I don't go so far as to say they must therefore exist; I merely say it is idiotic to deny the possibilities, so let's consider some details about those possibilities (which I've mostly done elsewhere).

1638232] Really bad logic. The plate is considered to be separate from the Earth as soon as you start treating it separate from the Earth.  That means even when falling off a cliff on Earth, it is not part of the Earth; it is part of the Earth/plate SYSTEM. Therefore the Earth does not lose the mass of the plate when the plate is given an escape velocity; only the Earth/plate system loses it.  And the Earth still sucked about 11kps of velocity and associated kinetic energy from that escaping plate; the plate most certainly does not have it while traversing interplanetary space.
Yes it does.
Your mere say-so is still utterly worthless.  The Earth-plate system loses the plate, not the Earth.  And the plate will be traversing interplanetary space at about 1kps (not the 12kps specified in an earlier message).  What math do you have to say that the plate will possess 11kps worth of kinetic energy?  Certainly not this utter balderdash:

Because when it finds another earth and falls down, it gives it all back.
The preceding is nonsense because when the plate finds another Earth, at that point we describe a new planet/plate system.  The plate doesn't "give" anything special to that system since by-description it is already part of that system!  And while certainly in this new system the plate will have considerable potential energy, the mass of the planet is absolutely required as part of the computation of how much potential energy has the plate; the mass of the planet affects the rate at which the plate can accelerate in its gravity field, after all.

And once it's fallen down and cooled off, it has less energy than it did in space ...
It is certainly true that the planet/plate system has less energy than before, and therefore less mass than before.  It most certainly does not mean all the kinetic energy that appeared did so at the expense of some of the mass of the plate only.  It doesn't even mean most of the KE that appeared was derived from the plate's mass.  It only means that mass from the system became KE.

I do understand the GR notion, but that is not what I was talking about. All your GR blather does not change by one whit what I wrote about how QM could describe things.
... a modest reinterpretation of General Relativity that takes a step back to the original ends up demolishing all attempts to quantize gravity.
Whoop-te-do.  What rationale is there, for taking such a backward reinterpretation?  I don't recall you ever offering one (not to mention that your conclusion could very well be utterly wrong, but I won't know until I see the rationale and have a chance to dissect the argument).  And, did you ever think about how much you will break GR in the process?  It has its current formulation for the very good reason that that is what most accurately matches the measurements we've made!

DUHHHH...that's because gravitation is the weakest force, 30 orders of magnitude weaker than the Weak Nuclear Force. It's going to take time to develop the technology to detect such feebleness. You seem to think no one should ever even bother. And again your mere say-so that gravitons do not exist cannot make them not exist.
After fifty years without evidence, it's time to kick the graviton into the long grass and pursue more fruitful avenues.
That's the attitude of a quitter.  Perhaps you don't understand what "30 orders of magnitude more difficult"  means.  This article http://en.wikipedia.org/wiki/Weak_interaction indicates that the Weak Force theory was developed about 1968, and the relevant particles were detected in 1983, a 15-year gap.  Well, you should start with the time that we have a QM theory for gravitation, and allow 15,000,000,000,000,000,000,000,000,000,000 years for detection of gravitons, before deciding to give up.

Title: If I give an object some potential energy, does its mass increase?
Post by: Farsight on 31/08/2009 18:07:50
If you would read far enough to get to the mathematics of the theory, you would discover that the actual pressure and density is assigned to the contents of the spacetime and define the stress-energy tensor. The stress-energy tensor defines the curvature of spacetime, which it the action of gravity. Why do you constantly avoid the actual science?
I'm not avoiding it, you are. Read the original. Einstein talks about curvilinear motion, not curved spacetime. And as per pmb's "Einstein's gravitational field", curved spacetime is not something Einstein agreed with. It's crystal clear that the modern interpretation of General relativity is different to Einstein's. You wouldn't read Pmb's paper remember, and dismissed it as the work of a crank. I'm sorry Physbang, but you're avoiding the actual science.

Again, you ignore how Einstein actually used the stress-energy tensor. Moving from a metric theory account to an affine theory account does not help your point here, it rather makes it worse. Affine theories rely on a more refined account of geometry in order to deliver the content of the theory, not less.
Come off it. I dealt with your issue perfectly adequately. Besides, you should recall from our previous conversations that the electric field is curved space, and that "refining the geometry" was always Einstein's aim.

The idea of using the word "trefoil" may go back that far, but there is no reason to believe that Kelvin's use of the word has anything at all to do with your use of the word. Indeed, as you seem to be using "trefoil" to refer to something to do with sub-atomic physics, it seems incredibly dubious that you can claim some support for your theory from Kelvin's use of the word.
I've responded to your issue, I've given you ample evidence, now accede the point.

OK, show us how your "model" can be used to calculate the motion of a dinner plate.
No, I won't because your only purpose is to waste my time. I'd start with simple Newtonian expressions and you'd carp and demand more, then I'd switch to General relativity, and you'll still carp, and all the while you're employing spoiler tactics whilst deliberately ducking the issue: the difference in the interpretation, not in the mathematics.

Now where's your apology for your "alchemy" insult to Isaac Newton? If you cannot accede even one single point I'm afraid it does you no favours.

VernonNimitz: what I talk about doesn't break GR, it unleashes it. And it's all in line with Feynman, Dirac, Schroedinger, Maxwell, Faraday, and even Newton. Plus others. But we aren't getting anywhere with our conversation, so let's just agree to differ.

Title: If I give an object some potential energy, does its mass increase?
Post by: lightarrow on 31/08/2009 18:10:21
Then, space and velocity (for example) are the same thing? They "only" differ because one is the derivative of the other...
That particular derivation requires including an additional thing, time.  Why aren't both a field gradient and a gradient-of-potential as static as Space?  A typical recreation-park slide (for children) has a gradient and by itself is typically considered to be quite static.  I suppose I'm missing something....
What you are missing is that you seems to pretend to be right for something that is blatantly wrong. I am not putting in discussion your knowledge of physics in general, everyone can make mistakes, but as soon as one discovers one, it's meaningless to try to "go round" the reasonings in order not to admit the mistake. A field is a field and a gradient is a gradient, that's all.
Title: If I give an object some potential energy, does its mass increase?
Post by: VernonNemitz on 31/08/2009 18:41:38
what I talk about doesn't break GR, it unleashes it. And it's all in line with Feynman, Dirac, Schroedinger, Maxwell, Faraday, and even Newton. Plus others.
Obviously you are wrong, since you disagree with Feynmann's assertion that anything not forbidden is mandatory.  Otherwise you would not be so adamant claiming that gravitons and negative mass-energy and virtual particles cannot exist; you have in no way shown why they should be forbidden to exist.
Title: If I give an object some potential energy, does its mass increase?
Post by: VernonNemitz on 31/08/2009 18:58:13
Then, space and velocity (for example) are the same thing? They "only" differ because one is the derivative of the other...
That particular derivation requires including an additional thing, time.  Why aren't both a field gradient and a gradient-of-potential as static as Space?  A typical recreation-park slide (for children) has a gradient and by itself is typically considered to be quite static.  I suppose I'm missing something....
What you are missing is that you seems to pretend to be right for something that is blatantly wrong. I am not putting in discussion your knowledge of physics in general, everyone can make mistakes, but as soon as one discovers one, it's meaningless to try to "go round" the reasonings in order not to admit the mistake. A field is a field and a gradient is a gradient, that's all.
And so far as I know, a full description of a field must always include a description of how its intensity changes with distance from the field-source, which is one type of gradient.  That's why I don't make a lot of distinction between the two.  Likewise, the full description of a slide should include mentioning its gradient, too.  In casual conversation (call it gossip if you like) between Persons A and B, about Person C's latest gaffe, it usually isn't necessary to specify that Person C is alive.  Well, while the conversations around here are usually somewhat more technical than casual, there are still assumptions that one can usually expect another to be familiar with (such as the slide having a gradient), lest we spend all our time on minutia (describing the slide's gradient) instead of the main topic.
Title: If I give an object some potential energy, does its mass increase?
Post by: PhysBang on 31/08/2009 21:05:44
]I'm not avoiding it, you are. Read the original. Einstein talks about curvilinear motion, not curved spacetime.
You just avoided the actual science. Whatever Einstein writes about, he also wrote a scientific theory called the General Theory of Relativity that describes physical laws through generally covariant equations in a pseudo-Riemannian manifold, i.e., in curved spacetime. Nothing you do will change that the actual content of Einstein's work is wedded to the way that the presence of matter and energy changes the geometry of space and time. To deny this is simply to point out one's own folly.
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And as per pmb's "Einstein's gravitational field", curved spacetime is not something Einstein agreed with.
If he didn't agree with it, then he shouldn't have made it the sum total of his theory. Can you show anything, any single thing in GR that doesn't make use of curved spacetime?
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Again, you ignore how Einstein actually used the stress-energy tensor. Moving from a metric theory account to an affine theory account does not help your point here, it rather makes it worse. Affine theories rely on a more refined account of geometry in order to deliver the content of the theory, not less.
Come off it. I dealt with your issue perfectly adequately. Besides, you should recall from our previous conversations that the electric field is curved space, and that "refining the geometry" was always Einstein's aim.
Please explain how the stress-energy tensor is used to create "curvilinear motion".
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The idea of using the word "trefoil" may go back that far, but there is no reason to believe that Kelvin's use of the word has anything at all to do with your use of the word. Indeed, as you seem to be using "trefoil" to refer to something to do with sub-atomic physics, it seems incredibly dubious that you can claim some support for your theory from Kelvin's use of the word.
I've responded to your issue, I've given you ample evidence, now accede the point.
Accede what point? How does Kelvin's work support your claims about subatomic physics?
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OK, show us how your "model" can be used to calculate the motion of a dinner plate.
No, I won't because your only purpose is to waste my time. I'd start with simple Newtonian expressions and you'd carp and demand more, then I'd switch to General relativity, and you'll still carp, and all the while you're employing spoiler tactics whilst deliberately ducking the issue: the difference in the interpretation, not in the mathematics.
I don't actually believe that you can work with either Newtonian mechanics or General Relativity, but what I want to see is how anything that you have in your "model" comes into play in the motion of the plate. If your "model" has no bearing on the motion of the plate, then your plate example is moot.
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Now where's your apology for your "alchemy" insult to Isaac Newton? If you cannot accede even one single point I'm afraid it does you no favours.
Newton did pursue alchemy. And he did pursue this is the queries appended to the Optics.
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VernonNimitz: what I talk about doesn't break GR, it unleashes it. And it's all in line with Feynman, Dirac, Schroedinger, Maxwell, Faraday, and even Newton. Plus others. But we aren't getting anywhere with our conversation, so let's just agree to differ.
The reason that nobody gets anywhere in discussion with you, anywhere in the internet, it that there is nowhere to go. As soon as someone starts asking questions, you merely start throwing around arguments from authority. Then someone points out that the argument from authority makes no sense and you stonewall in some other way. If you could just do a simple calculation, this might produce progress.
Title: If I give an object some potential energy, does its mass increase?
Post by: VernonNemitz on 01/09/2009 21:23:26
Farsight, I wish to expand upon this that I wrote earlier:
It is certainly true that the planet/plate system has less energy than before, and therefore less mass than before.  It most certainly does not mean all the kinetic energy that appeared did so at the expense of some of the mass of the plate only.  It doesn't even mean most of the KE that appeared was derived from the plate's mass.  It only means that mass from the system became KE.
(I'm mostly going to not bother talking about the conversion of potential energy into kinetic energy, and then the conversion of that into radiant energy that leaves the system --or the importation of energy into the system to raise the plate to a high altitude.  Just assume this happens in the background as we consider the plate as being stationary at different heights.)

In General Relativity it is a key fact that certain aspects of the behavior of the system must be unchanged when various transformations are done (such as a change to a different frame-of-reference).  So, if I want to compute the strength of the gravitational force between plate and planet when they are X kilometers apart, I need to use the same mass when they are 100 times as far apart, or when they are 1/100 as far apart.  But both of us are describing a change in potential-energy-stored-as-mass (and therefore a change in mass) when different distances are involved, how do we reconcile this?  The simplest way is to ignore that change; it is a extraordinarily small quantity of mass, after all, that we are talking about.

One could invoke the "negative binding energy" concept (bookkeeping trick), and this will certainly allow the two masses to be the same in regardless of the altitude of the plate, but then we would be getting away from the other basic idea, that potential energy is stored as mass.  Obviously, then, GR would take a small "hit" from its current standard formulations to accommodate this tiny tiny change (as described in prior paragraph).  My Question to you is, what "hit" would have the LEAST effect upon GR???

The "hit" I'm promoting is for the ratio of the two masses to not change.  The system stays comparable to the original/standard unchanging-mass scenario.  So, if the planet has a zillion times as much mass as the plate, and the plate loses a zillionth of a gram, as we reposition it from Distance 100X km to Distance (1/100)X km from the planet, that means the planet loses a zillion times as much, which is just one gram.  (It would have to be an impressively massive plate, for its potential energy at 100X km to be describable as equivalent to 1-and-1-zillionth grams of mass!)  At any height we reposition the plate, the planet is always a zillion times more massive.

The "hit" you are promoting is much more serious, literally unbalancing General Relativity, compared to what I'm promoting. I don't need to describe it since you have already done so in several messages here.  Instead I'm going to change the subject for a bit.

Here's an experiment you can try.  First, if you can find one, get a bicycle light-and-generator set.  The small generator mounts next to the rear wheel, and friction causes it to turn.  It powers the light bulb, of course.  After setting things up for proper operation, turn the bicycle upside-down so you can crank the pedal by hand.  The experiment begins by removing the light bulb.  Note how much effort it takes to crank the pedal.  Then put the light bulb back in, and note how much more effort it takes to crank the pedal at the same rate.  It is a peculiarity of electric-power-generation that the LOAD operates against the force turning the generator.  It is a sensible thing overall, since we know the load is using the result of the effort that goes into turning the generator, but have you ever wondered about just how the system "knows" the load is there?  "It's all connected by wires," you say?  Okay....  Now read this:

The scenario described there, in the initial message, is something that can actually be done, whether or not it actually has been done.  And yes, the power company will know if you attach a load to an appropriate inductive power-collector.  But my prior question still applies:  How does the system know the load is there?  No connection wires this time!

Part of the answer is "virtual photons" --identical to evanescent waves.  However, we both know they are only virtual and are not carrying real energy (else all power would radiate from power lines, and nobody would be getting anything out of the far end of the wires).  So, the rest of the answer requires involving the instantaneous operation of the entanglement phenomenon.  Only when a virtual photon encounters an appropriate absorbing wire, that's when power gets transferred and we can say that "induction" has happened.

I can imagine an experiment, possible only in the future, that could verify that instant transfer of power, where we can isolate this experiment from all Earthly influences.  We have to build a high-voltage AC power line in Outer Space, away from all gravity fields (say, the Oort Cloud).  The line itself doesn't need to be very long, say one kilometer, and it has a ordinary generator at one end and an ordinary load at the other (with appropriate transformers also in the system).  In the middle of this power line we create a box that the wires pass through.  Everthing is thoroughly insulated from electric discharge (necessary in vacuum of space, lest electrons exit the wires instead of travelling through them!!! --air acts as a fairly good transmission-line-insulator on Earth).  One side of the box is attached to a kind of conduit.  The box is made of metal and the conduit is a wave-guide.  We want this conduit to be quite long, say a thousand kilometers.  The purpose of the wave-guide is to fight the Inverse Square Law.  At the end of the wave-guide we put our power-sucking induction circuit, with an on/off switch.  We can now do some precision time-measurements.  We know a fraction of a second must pass for photons, whether real or virtual, to traverse the wave-guide.  We certainly have to wait at least that much, after power starts to flow in the power line, for any virtual photons to reach the inductor-circuit.  However, if afterward we now throw a switch to turn on the induction-circuit, do we instantly have power flowing or do we have to wait another fraction of a second?  Do we have to wait for some sort of 'signal' to get back to the power line, that we are drawing from it inductively?  I'm saying that we don't have to wait; we will instantly have power in the induction circuit when the switch is thrown, because we will be invoking the entanglement-effect; the virtual photons absorbed into the inductor are spookily entangled with the power line a thousand km away.  So, measurements made at the power line, of all power drawn from it, should immediately indicate that the distant switch was thrown.

Now back to gravitation, and some of the things I've described in other messages here.  If what I've described above can work for virtual photons (and to the best of my knowledge it does work that way), then it can also work for virtual gravitons being absorbed; the mass that radiated the virtual gravitons becomes diminished when the virtual gravitons are absorbed by another mass, and the absorbing mass acquires kinetic energy.  This will maintain the ratio of the two masses, no matter how much potential-energy-stored-as-mass gets converted into other forms.
Title: If I give an object some potential energy, does its mass increase?
Post by: Farsight on 06/09/2009 15:10:09
Farsight, I wish to expand upon this that I wrote earlier:
It is certainly true that the planet/plate system has less energy than before, and therefore less mass than before.  It most certainly does not mean all the kinetic energy that appeared did so at the expense of some of the mass of the plate only.  It doesn't even mean most of the KE that appeared was derived from the plate's mass.  It only means that mass from the system became KE.
(I'm mostly going to not bother talking about the conversion of potential energy into kinetic energy, and then the conversion of that into radiant energy that leaves the system --or the importation of energy into the system to raise the plate to a high altitude.  Just assume this happens in the background as we consider the plate as being stationary at different heights.)
OK, I'm listening.

In General Relativity it is a key fact that certain aspects of the behavior of the system must be unchanged when various transformations are done (such as a change to a different frame-of-reference). So, if I want to compute the strength of the gravitational force between plate and planet when they are X kilometers apart, I need to use the same mass when they are 100 times as far apart, or when they are 1/100 as far apart.
Hmmn. It's not a fact, Vernon, it's an assumption, and an approximation. Besides you shouldn't be computing the strength of gravitational force, you should be measuring it. But nevermind, let's see where this takes us.

But both of us are describing a change in potential-energy-stored-as-mass (and therefore a change in mass) when different distances are involved, how do we reconcile this? The simplest way is to ignore that change; it is a extraordinarily small quantity of mass, after all, that we are talking about.
It's relatively small, but only because there's a heck of a lot of energy tied up as matter. The energy we're talking about is responsible for that 11 km/s velocity of a plate falling from free space to the surface of our airless planet. It's too big to ignore.

One could invoke the "negative binding energy" concept (bookkeeping trick), and this will certainly allow the two masses to be the same in regardless of the altitude of the plate, but then we would be getting away from the other basic idea, that potential energy is stored as mass. Obviously, then, GR would take a small "hit" from its current standard formulations to accommodate this tiny tiny change (as described in prior paragraph). My Question to you is, what "hit" would have the LEAST effect upon GR???
I'm not sure Vernon, it's a complicated subject, see http://en.wikipedia.org/wiki/Mass_in_general_relativity:

"If two objects have the same mass, and we heat one of them up from an external source, does the heated object gain mass? If we put both objects on a sensitive enough balance, would the heated object weigh more than the unheated object? Would the heated object have a stronger gravitational field than the unheated object?

The answer to all of the above questions is yes. The hot object has more energy, so it weighs more and has a higher mass than the cold object. It will also have a higher gravitational field to go along with its higher mass, by the equivalence principle. (Carlip 1999)"

The "hit" I'm promoting is for the ratio of the two masses to not change.  The system stays comparable to the original/standard unchanging-mass scenario.  So, if the planet has a zillion times as much mass as the plate, and the plate loses a zillionth of a gram, as we reposition it from Distance 100X km to Distance (1/100)X km from the planet, that means the planet loses a zillion times as much, which is just one gram. (It would have to be an impressively massive plate, for its potential energy at 100X km to be describable as equivalent to 1-and-1-zillionth grams of mass!) At any height we reposition the plate, the planet is always a zillion times more massive.
I know, but this preference is flying in the face of the evidence. Strap your plate to a rocket and light the blue touch paper. You're giving 11km/s worth of energy to the plate, not the earth.

The "hit" you are promoting is much more serious, literally unbalancing General Relativity, compared to what I'm promoting. I don't need to describe it since you have already done so in several messages here. Instead I'm going to change the subject for a bit.
I'm not sure it does, and besides, even if it does, I don't take the view that some imperfection means it's all wrong. But anyhow, OK, changing the subject for a while.

Here's an experiment you can try.  First, if you can find one, get a bicycle light-and-generator set. The small generator mounts next to the rear wheel, and friction causes it to turn. It powers the light bulb, of course. After setting things up for proper operation, turn the bicycle upside-down so you can crank the pedal by hand.  The experiment begins by removing the light bulb. Note how much effort it takes to crank the pedal. Then put the light bulb back in, and note how much more effort it takes to crank the pedal at the same rate.
Sadly, I haven't got a bike. But I can imagine that the energy book has to balance, so if more energy is going out because the bulb is shining, more energy has to be going in.

It is a peculiarity of electric-power-generation that the LOAD operates against the force turning the generator. It is a sensible thing overall, since we know the load is using the result of the effort that goes into turning the generator, but have you ever wondered about just how the system "knows" the load is there? "It's all connected by wires," you say? Okay....  Now read this:
Good stuff. Induction, no problem.

The scenario described there, in the initial message, is something that can actually be done, whether or not it actually has been done.  And yes, the power company will know if you attach a load to an appropriate inductive power-collector.  But my prior question still applies:  How does the system know the load is there? No connection wires this time!
Ditto.

Part of the answer is "virtual photons" - identical to evanescent waves. However, we both know they are only virtual and are not carrying real energy (else all power would radiate from power lines, and nobody would be getting anything out of the far end of the wires).
If you don't tap the field you aren't draining it of any energy, but the energy in there is real, Vernon.

So, the rest of the answer requires involving the instantaneous operation of the entanglement phenomenon. Only when a virtual photon encounters an appropriate absorbing wire, that's when power gets transferred and we can say that "induction" has happened.
If we say a virtual photon is the same animal as an evanescent wave, we can agree about the encounter with the wire. But IMHO saying it's an instaneous operation of the entanglement phenomenon is taking it too far. It's just inductance. See http://en.wikipedia.org/wiki/Inductance.

I can imagine an experiment, possible only in the future, that could verify that instant transfer of power, where we can isolate this experiment from all Earthly influences.  We have to build a high-voltage AC power line in Outer Space, away from all gravity fields (say, the Oort Cloud).  The line itself doesn't need to be very long, say one kilometer, and it has a ordinary generator at one end and an ordinary load at the other (with appropriate transformers also in the system). In the middle of this power line we create a box that the wires pass through.  Everthing is thoroughly insulated from electric discharge (necessary in vacuum of space, lest electrons exit the wires instead of travelling through them!!! --air acts as a fairly good transmission-line-insulator on Earth). One side of the box is attached to a kind of conduit. The box is made of metal and the conduit is a wave-guide.  We want this conduit to be quite long, say a thousand kilometers. The purpose of the wave-guide is to fight the Inverse Square Law. At the end of the wave-guide we put our power-sucking induction circuit, with an on/off switch. We can now do some precision time-measurements. We know a fraction of a second must pass for photons, whether real or virtual, to traverse the wave-guide. We certainly have to wait at least that much, after power starts to flow in the power line, for any virtual photons to reach the inductor-circuit. However, if afterward we now throw a switch to turn on the induction-circuit, do we instantly have power flowing or do we have to wait another fraction of a second? Do we have to wait for some sort of 'signal' to get back to the power line, that we are drawing from it inductively? I'm saying that we don't have to wait; we will instantly have power in the induction circuit when the switch is thrown, because we will be invoking the entanglement-effect; the virtual photons absorbed into the inductor are spookily entangled with the power line a thousand km away. So, measurements made at the power line, of all power drawn from it, should immediately indicate that the distant switch was thrown.
I'm all for experiment. But I wouldn't be surprised if this hadn't already been done, see http://www.consultrsr.com/resources/eis/induct5.htm re a straight wire.

Now back to gravitation, and some of the things I've described in other messages here. If what I've described above can work for virtual photons (and to the best of my knowledge it does work that way), then it can also work for virtual gravitons being absorbed; the mass that radiated the virtual gravitons becomes diminished when the virtual gravitons are absorbed by another mass, and the absorbing mass acquires kinetic energy. This will maintain the ratio of the two masses, no matter how much potential-energy-stored-as-mass gets converted into other forms.
I'm sorry Vernon, it's stretching things too far. And it all comes back to that photon that causes gravity because energy causes gravity. It's a photon, it's the only thing that's there, and it's travelling at c. A photon is a wave, it conveys energy. The dimensionality of energy is pressure x volume, so think of an ocean wave. It's a pressure pulse. Think of the evanescent wave as a standing wave, like what you see around a bridge pier. It's like half a photon that isn't going anywhere. Photons are real, inductance is real, and the evanescent wave is real. You could say this means virtual photons are real, but it doesn't mean they're transient photons shooting around at hyperlight velocities - you can't shield inductance with a sheet of black paper. And there aren't any gravitons buzzing around these things. And no virtual gravitons either. There's no evidence to support this hypothesis after fifty years. See http://www.bbc.co.uk/radio4/history/inourtime/inourtime_20051124.shtml.

I'm sorry vernon, I think we'll just have to agree to differ.

Title: If I give an object some potential energy, does its mass increase?
Post by: Farsight on 06/09/2009 15:30:44
Quote from: Physbang
If he didn't agree with it, then he shouldn't have made it the sum total of his theory. Can you show anything, any single thing in GR that doesn't make use of curved spacetime?
Yes, The Foundation of the General Theory of Relativity. See the 3.5Mbyte pdf at http://www.alberteinstein.info/gallery/pdf/CP6Doc30_English_pp146-200.pdf. It's the original. And it doesn't even mention curved spacetime. That's because curved spacetime was popularized by Dicke in the sixties, see http://www.emis.de/journals/LRG/Articles/lrr-2006-3/articlesu1.html. And see this abstract: http://www.iop.org/EJ/abstract/0256-307X/25/5/014 for a hint of how important this is. Now do excuse me, you have no sincerity, and no argument whatsoever.
Title: If I give an object some potential energy, does its mass increase?
Post by: PhysBang on 06/09/2009 16:48:42
Quote from: Physbang
If he didn't agree with it, then he shouldn't have made it the sum total of his theory. Can you show anything, any single thing in GR that doesn't make use of curved spacetime?
Yes, The Foundation of the General Theory of Relativity. See the 3.5Mbyte pdf at http://www.alberteinstein.info/gallery/pdf/CP6Doc30_English_pp146-200.pdf. It's the original. And it doesn't even mention curved spacetime.
Again you fail. The entire theory offered there is in terms of curved spacetime. What is the use of Christoffel's equations in the work if not for curved spacetime? [See page 168, 177, 179 in particular.] What does Einstein mean on page 197 when he writes, "Thus Euclidean geometry does not hold even to a first approximation in the gravitational field..."?
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That's because curved spacetime was popularized by Dicke in the sixties, see http://www.emis.de/journals/LRG/Articles/lrr-2006-3/articlesu1.html.
That is an article on how the Weak Equivalence Principle was used to go beyond the curved spacetime of GR to produce a broader class of curved spacetime theories of which GR is one member. It does nothing to support any of your points.
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And see this abstract: http://www.iop.org/EJ/abstract/0256-307X/25/5/014 for a hint of how important this is.
Yes, it is always possible to cook up a theory. I doubt that author is the first to do what that paper has done and I doubt that the paper will go very far for the same reason that past papers haven't gotten far: they cannot explain the vast array of measurement evidence that is out there.
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Now do excuse me, you have no sincerity, and no argument whatsoever.
Of course I have no argument. All I am doing is pointing out how you consistently avoid discussing the mathematics and use references that have nothing to do with your point. Of course, if you can tell us how Kelvin's work supports your theory of subatomic physics, I guess I'll have to take that back.
Title: If I give an object some potential energy, does its mass increase?
Post by: VernonNemitz on 08/09/2009 16:59:10
In General Relativity it is a key fact that certain aspects of the behavior of the system must be unchanged when various transformations are done (such as a change to a different frame-of-reference). So, if I want to compute the strength of the gravitational force between plate and planet when they are X kilometers apart, I need to use the same mass when they are 100 times as far apart, or when they are 1/100 as far apart.
Hmmn. It's not a fact, Vernon, it's an assumption, and an approximation. Besides you shouldn't be computing the strength of gravitational force, you should be measuring it. But nevermind, let's see where this takes us.
Obviously measuring it could verify or refute the assumption, but at this time our instruments are not up to the task of doing it accurately enough.

The energy we're talking about is responsible for that 11 km/s velocity of a plate falling from free space to the surface of our airless planet. It's too big to ignore.
Wrong.  Even if the plate has a mass of a metric ton, its kinetic energy at 11kps is equivalent to only about 2/3 of one-thousandth of one gram of mass.  Do the math and see for yourself that under most ordinary conditions the change in mass is ignorably tiny --because, like I said, our measuring tools aren't yet able to measure gravitational force when such tiny differences are involved.

"The hot object has more energy, so it weighs more and has a higher mass than the cold object. It will also have a higher gravitational field to go along with its higher mass, by the equivalence principle. (Carlip 1999)"
That's real energy being talked about, not potential energy.  Irrelevant to this discussion, therefore, about potential-energy-stored-as-mass.

... this preference [for maintaining mass ratio] is flying in the face of the evidence. Strap your plate to a rocket and light the blue touch paper. You're giving 11km/s worth of energy to the plate, not the earth.
Wrong again, because that so-called "evidence" has nothing to do with gravitationAFTER you have specified giving this velocity and energy to the plate, the earth's gravitation sucks it away, and it is that interaction which, by description, logically means the earth can end up with energy converted to potential-energy-as-mass.  It is also consistent with the other description, of the earth pulling the plate back down; the energy associated with the pulling most logically comes from the earth's potential-energy-as-mass.  --And before you once again start talking about the role of geometry in gravitation, remember that you yourself wrote above that GR is an approximation of reality.  Therefore geometry does not have to have the Last Word on the subject, especially if QM can eventually generate equivalent descriptions of known/observed gravitational effects.

The "hit" you are promoting is much more serious, literally unbalancing General Relativity, compared to what I'm promoting.
I'm not sure it does, and besides, even if it does, I don't take the view that some imperfection means it's all wrong.
Then you haven't studied the Symmetry Principles of Quantum Mechanics well enough.  Physicists prefer theories that have balance, else they wouldn't have become upset when Parity Symmetry was violated, and wouldn't have put effort into talking about such things as "CPT Symmetry".

... it all comes back to that photon that causes gravity because energy causes gravity. It's a photon, it's the only thing that's there, and it's travelling at c. A photon is a wave, it conveys energy. The dimensionality of energy is pressure x volume, so think of an ocean wave. It's a pressure pulse.
Perhaps this is the fundamental source of your error.  Energy is no more only pressure x volume than it is only force x distance, just like the Uncertainty Principle is no more only about a momentum x position relationship; an energy x time relationship is just as valid.  Therefore the energy of a photon can be described in other terms than the one that limits the conclusion you have reached!

You could say ... virtual photons are real, but it doesn't mean they're transient photons shooting around at hyperlight velocities - you can't shield inductance with a sheet of black paper.
Nobody has ever said that virtual photons move faster than light, including me.  Do not confuse the QM entanglement phenomenon with anything else, please!  Also, FYI, the virtual photons associated with an electromagnetic field are long-wavelength; they can penetrate black paper as easily as radio waves.

And there aren't any gravitons buzzing around these things. And no virtual gravitons either. There's no evidence to support this hypothesis after fifty years.
It is premature to make such a rash statement.  Remember that description of 30-orders-of-magnitude-greater-difficulty in one of my previous posts?  Even taking into account the fact that was a straight-line comparison, instead of comparing the problem's difficulty to our exponential increase in technical abilities, it will still be a while before we can reliably tackle such a task as detecting gravitons.
Title: If I give an object some potential energy, does its mass increase?
Post by: yor_on on 16/01/2010 12:57:26
Vernon, I hold to that virtual particles 'exist' as you seem to do. Their interactions definitely prove it. That Feynman meant that we couldn't really say what they were, I take to mean that we can't really say what they are :). And we can't, but we can count on them and observe their interactions, whatever they might be.

Then I saw you support the Higgs field, sounds okay to me too. When it comes to the Higgs boson? Don't know, maybe the LHC will clear that up?

But now I see that you expect gravitons too? Isn't that contradictorily?
Or do you see string theory as encompassing all those descriptions?

==
Ah, pre/assuming that you lean towards it?
Title: If I give an object some potential energy, does its mass increase?
Post by: Geezer on 16/01/2010 20:06:32
Reading through this thread, it's not clear we ever answered the original question. Didn't seem like a very complicated question either.

Let's try this (thought!) experiment:

We construct a very large solar powered ion rocket, then send it to the Moon. We attach the rocket to the Moon and fire it up so that its thrust is always directed towards the Earth. Over time, this will increase the distance between the Moon and the Earth and, therefore, increase the potential energy of the Moon/Earth system.

What will happen?

Will this increase the mass of the Moon/Earth system or not? If it does increase the mass, I think we can expect to see a change in the Earth's orbit round the Sun.

We could also try a more extreme variation of the experiment where we slow down the Moon to the point where it crashes into Earth so that the potential energy is eliminated.

Will the mass of the new system be different from the old system, and will the orbit of the new system be different from the old system?

What do you think?
Title: If I give an object some potential energy, does its mass increase?
Post by: Farsight on 17/01/2010 14:33:49
I think yes. We're capturing energy from the sun and keeping it in the system. Mass is the measure of a system's energy content, so the mass of the system increases. Seems cut and dried.
Title: If I give an object some potential energy, does its mass increase?
Post by: Geezer on 17/01/2010 17:25:25
I think yes. We're capturing energy from the sun and keeping it in the system. Mass is the measure of a system's energy content, so the mass of the system increases. Seems cut and dried.

To you, perhaps.

So now what happens if we don't use a solar powered rocket, but use a conventional rocket that obtained its energy from the system? Does the mass of the Moon still increase (and, because they are part of the same system, presumably by the same logic the Earth's mass would also increase)?
Title: If I give an object some potential energy, does its mass increase?
Post by: yor_on on 17/01/2010 17:29:38
Difficult one, for me that is :)

I look at it this way. If you take Earth as an example and then consider the Earths surface to be the place of its 'strongest' gravitation (1G) With other directions downwards or upwards as becoming 'weaker' gravitationally seen.

And you then lift that 1 kg. plate 1 meter (around three Feet) you will, when you have finished, placed that same plate in a marginally weaker gravitational field.

As you lifted it you lost some energy, but did that energy go into the plate? You could argue that it did so as if you let it fall it would get a kinetic energy from its 1 meter fall to the ground.

But assume that you put it on a table, 1 meter of the ground. Would you then tell me that it now had an higher energy than before lifting it? Remember that the gravitation is 'weaker' up there too :)

It's here 'potential energy' comes in as i understand it. As gravity is like a well, or a hole, with Earths ground as the highest gravitational nominator. And as we know that all things want to move toward that highest gravity we can say that without that table the plate definitely would fall to the ground. So it contain a 'need' to move if you like.

But it does not contain more atoms due to the lift, neither does the atoms electrons, molecules, etc jiggle more due to that lift. In fact it should have the other effect as I see it, they should jiggle slightly less as the gravitation is less 1 meter above, and therefore also contain a slightly lesser vibrational energy than the plate had closer to the ground. (Think of a black hole and that plate near a such, to see what I mean)

That it want to move to the strongest gravitational impact is due to how space bends around mass. When you count on potential energy you take a different approach.

"Potential energy is energy stored in an object. This energy has the potential to do work. Gravity gives potential energy to an object. This potential energy is a result of gravity pulling downwards. The gravitational constant, g, is the acceleration of an object due to gravity. This acceleration is about 9.8 meters per second on earth. The formula for potential energy due to gravity is PE = mgh.

As the object gets closer to the ground, its potential energy decreases while its kinetic energy increases. The difference in potential energy is equal to the difference in kinetic energy. After one second, if the potential energy of an object fell ten units than its kinetic energy has risen ten units. Potential energy units are joules." potential energy (http://www.tjhsst.edu/~jleaf/tec/html/10/potent.htm)

There you treat the 'need' to find that gravitational balance down the gravity well as a real 'force' where the object moved 1 meter up have gained a 'real' possible energy, even if not measurable in the object itself. That energy is equivalent to the work done to lift it and will express itself as the plate meet the ground in kinetic energy.

So if I look at Geezers rocket working against the moon, forcing it away from its orbit, it has to transfer an energy as we move the moon further out from Earths and the Suns combined gravity well, or at least Earths.

But Geezers example have more variables in it as the rocket seem to be constantly working? so its complicated :)

On the other hand, I find most things complicated.

==
Even shoe lashes is complicated nowadays :)

Title: If I give an object some potential energy, does its mass increase?
Post by: Geezer on 17/01/2010 17:53:07
Difficult one, for me that is :)

I look at it this way. If you take Earth as an example and then consider the Earths surface to be the place of its 'strongest' gravitation (1G) With other directions downwards or upwards as becoming 'weaker' gravitationally seen.

And you then lift that 1 kg. plate 1 meter (around three Feet) you will, when you have finished, placed that same plate in a marginally weaker gravitational field.

As you lifted it you lost some energy, but did that energy go into the plate? You could argue that it did so as if you let it fall it would get a kinetic energy from it 1 meter fall to the ground.

But assume that you put it on a table, 1 meter of the ground. Would you then tell me that it now had an higher energy than before lifting it? Remember that the gravitation is 'weaker' up there too :)

It's here 'potential energy' comes in as i understand it. As gravity is like a well, or a hole, with Earths ground as the highest gravitational nominator. And as we know that all things want to move toward that highest gravity we can say that without that table the plate definitely would fall to the ground. So it contain a 'need' to move if you like.

But it does not contain more atoms due to the lift, neither does the atoms electrons, molecules, etc jiggle more due to that lift. In fact it should have the other effect as I see it, they should jiggle slightly less as the gravitation is less 1 meter above, and therefore also contain a slightly lesser vibrational energy than the plate had closer to the ground. (Think of a black hole and that plate near a such, to see what I mean)

That it want to move to the strongest gravitational impact is due to how space bends around mass. When you count on potential energy you take a different approach.

"Potential energy is energy stored in an object. This energy has the potential to do work. Gravity gives potential energy to an object. This potential energy is a result of gravity pulling downwards. The gravitational constant, g, is the acceleration of an object due to gravity. This acceleration is about 9.8 meters per second on earth. The formula for potential energy due to gravity is PE = mgh.

As the object gets closer to the ground, its potential energy decreases while its kinetic energy increases. The difference in potential energy is equal to the difference in kinetic energy. After one second, if the potential energy of an object fell ten units than its kinetic energy has risen ten units. Potential energy units are joules." potential energy (http://www.tjhsst.edu/~jleaf/tec/html/10/potent.htm)

There you treat the 'need' to find that gravitational balance down the gravity well as a real 'force' where the object moved 1 meter up have gained a 'real' possible energy, even if not measurable in the object itself. That energy is equivalent to the work done to lift it and will express itself as the plate meet the ground in kinetic energy.

So if I look at Geezers rocket working against the moon, forcing it away from its orbit, it has to transfer an energy as we move the moon further out from Earths and the Suns combined gravity well, or at least Earths.

But Geezers example have more variables in it as the rocket seem to be constantly working? so its complicated :)

On the other hand, I find most things complicated.

Yoron - The fundamental question is "Did the mass increase?". In this case, we are not changing the kinetic energy of the plate - it's stationary in both end states.

If the mass actually did increase it should not be too difficult to devise an experiment to prove that it did. For example, we could apply a certain amount of electrical energy to accelerated the mass. As the mass increases, the resultant kinetic energy will be reduced for the same amount of electrical energy.
Title: If I give an object some potential energy, does its mass increase?
Post by: yor_on on 17/01/2010 19:33:30
I agree Geezer, but it's perfectly correct to count on it as 'potential energy' as that fits in all kinds of mathematical models describing the universe.

But when testing 'matter' and see if that plate isolated, after being lifted one meter, have gotten an increased energy I would expect the oposite. And if the 'jiggling' is less one meter up, then the mass should decrease as that is kinetic energy.

So it depend on what you choose to define as a 'system' I guess?
Maybe I'm thinking wrong though?
Title: If I give an object some potential energy, does its mass increase?
Post by: Farsight on 17/01/2010 19:52:47
To you, perhaps. So now what happens if we don't use a solar powered rocket, but use a conventional rocket that obtained its energy from the system? Does the mass of the Moon still increase (and, because they are part of the same system, presumably by the same logic the Earth's mass would also increase)?
Again, it's cut and dried. If you use energy from the system to make one part of the system move faster, it somehow comes at the cost of making some other part of the system move slower. The moon is large compared to the earth, so it's more complicated than the cannonball. See http://en.wikipedia.org/wiki/Tidal_power and note where it says "This loss of energy has caused the rotation of the Earth to slow in the 4.5 billion years since formation. During the last 620 million years the period of rotation has increased from 21.9 hours to the 24 hours we see now; in this period the Earth has lost 17% of its rotational energy". The earth's rotation has slowed down, so you could say the earth has lost some mass. But you can't quite say "the moon" has gained some mass. Instead the subsystem that is the moon in its circular orbit has gained potential/kinetic energy, and hence mass.
Title: If I give an object some potential energy, does its mass increase?
Post by: Geezer on 17/01/2010 20:00:34
I agree Geezer, but it's perfectly correct to count on it as 'potential energy' as that fits in all kinds of mathematical models describing the universe.

But when testing 'matter' and see if that plate isolated, after being lifted one meter, have gotten an increased energy I would expect the oposite. And if the 'jiggling' is less one meter up, then the mass should decrease as that is kinetic energy.

So it depend on what you choose to define as a 'system' I guess?
Maybe I'm thinking wrong though?

I agree. There is an increase in potential energy, and the math seems to be sufficiently correct. The question was "does the mass increase?". If someone wants to conduct the experiment I suggested, or they already have done something similar, we could find out quite easily.

Regarding the system, once you define what is in your system, you can't arbitrarily remove components of it and say they are no longer part of the system, even although the effects they produce are minimal. If you take the plate from one meter above Earth to 10 billion meters above Earth then bring it back to one meter again, it will have exactly the same PE that it had before.

By "jiggling", are you referring to the kinetic energy of the molecules that make up the plate?
Title: If I give an object some potential energy, does its mass increase?
Post by: Geezer on 17/01/2010 20:11:36
To you, perhaps. So now what happens if we don't use a solar powered rocket, but use a conventional rocket that obtained its energy from the system? Does the mass of the Moon still increase (and, because they are part of the same system, presumably by the same logic the Earth's mass would also increase)?
Again, it's cut and dried. If you use energy from the system to make one part of the system move faster, it somehow comes at the cost of making some other part of the system move slower.

Who said anything about any part of the system getting faster? Are you saying there was, or was not, an increase in the mass of the Moon, the Earth, or the system in either or any of the cases?

If it's as "cut and dried" as you say, I'm sure you will be able to predict the outcome of the experiments.
Title: If I give an object some potential energy, does its mass increase?
Post by: yor_on on 17/01/2010 20:28:54
yeah, by 'jiggling' I meant the 'motion' atoms create in matter, same as a gas 'jiggles' more as it gets hotter. and my thought was that the closer you move matter to a gravitational well the more 'jiggling' you will find.

And in this case, by removing matter from the well. I think the 'jiggling' should decrease. And as the 'jiggling' should transform into kinetic energy I would expect that piece of matter to become lighter, as the well recedes even though I'm not entirely sure on it.

It's a little like the photon transferring what we call mass due to being constricted inside that perfectly mirrored box. But a little more possible to test, maybe :)

It's like a drum skin. If you beat on it very fast will that increase its mass. I think it will.

===

Or make a really big hole :)
Title: If I give an object some potential energy, does its mass increase?
Post by: Geezer on 17/01/2010 21:01:11
yeah, by 'jiggling' I meant the 'motion' atoms create in matter, same as a gas 'jiggles' more as it gets hotter. and my thought was that the closer you move matter to a gravitational well the more 'jiggling' you will find.

And in this case, by removing matter from the well. I think the 'jiggling' should decrease. And as the 'jiggling' should transform into kinetic energy I would expect that piece of matter to become lighter, as the well recedes even though I'm not entirely sure on it.

It's a little like the photon transferring what we call mass due to being constricted inside that perfectly mirrored box. But a little more possible to test, maybe :)

It's like a drum skin. If you beat on it very fast will that increase its mass. I think it will.

===

Or make a really big hole :)

I think if there was any increase in the kinetic energy of the molecules in the plate, it would simply get warmer, just as it does when you heat it in an oven. As I understand it, sticking plates in ovens does not increase their mass  [;D]

If it does, it would be quite easy to measure, or are we talking about a very minute increase in mass?
Title: If I give an object some potential energy, does its mass increase?
Post by: yor_on on 17/01/2010 22:33:43
Yeah, it's very small if it is so. I mean, it is the atoms higher 'bouncing' frequency that will create what mass there will be more.

Lightarrow could probably count on it and RD and Maddidus and JP and . . .
Thinking of it, probably most all of you here, except me of course :::)))
Awh

*Takes out his pocket calculator*

"One , two, three, four, and? Where was I?"

Hang on a minute Geezer, I think I'm getting close here :)
Title: If I give an object some potential energy, does its mass increase?
Post by: Geezer on 17/01/2010 23:04:51
Ah! Then you would be referring to non-rest mass associated with relativistic effects. But I don't think that applies in this situation. The plate is static relative to Earth before and after it was moved, so I think we'll find it has exactly the same mass in both states.

And even if there was a very small change in mass, I don't think we'd want people to get the idea that its potential energy was caused by a change in mass. I think that would be very misleading.
Title: If I give an object some potential energy, does its mass increase?
Post by: Farsight on 18/01/2010 09:28:02
Who said anything about any part of the system getting faster?
I thought you did. Check out the wiki “orbital mechanics” article for how you accelerate an object in the orbital direction to gain altitude.

Are you saying there was, or was not, an increase in the mass of the Moon, the Earth, or the system in either or any of the cases?
If you add energy to the system the system gains mass. If you don't, it doesn't. Mass is a measure of the energy content of a system, and overall the system isn’t moving with respect to you. However the earth and the moon are, so when you change their motion you wouldn’t say they've gained or lost mass. It’s akin to a gyroscope: if you spin it, you add energy, and when you consider it to be an overall system because you put it in a black box, you'd say it isn’t moving with respect to you, so you’d say it had more mass. But if you focussed on some moving portion of the gyroscope, you’d say it has more angular momentum rather than more mass.

If it's as "cut and dried" as you say, I'm sure you will be able to predict the outcome of the experiments.
Experiments are difficult to do. There's a lot of energy tied up in matter, for example only about 1 gram of matter was converted into energy by the Hiroshima bomb. So you can't realistically heat a container of gas and see that it weighs more. This is what Einstein said in his 1905 paper "Does the Inertia of a body depend upon its Energy content?". See http://www.fourmilab.ch/etexts/einstein/E_mc2/www/.
Title: If I give an object some potential energy, does its mass increase?
Post by: Farsight on 18/01/2010 09:33:32
But it does not contain more atoms due to the lift, neither does the atoms electrons, molecules, etc jiggle more due to that lift. In fact it should have the other effect as I see it, they should jiggle slightly less as the gravitation is less 1 meter above, and therefore also contain a slightly lesser vibrational energy than the plate had closer to the ground.
Imagine you used some aspect of this "jiggle" to operate a clock. Clock run slower nearer the ground, due to gravitational time dilation. That means the jiggle must be slower near the ground.
Title: If I give an object some potential energy, does its mass increase?
Post by: litespeed on 18/01/2010 16:11:42
GPS Satellite clocks run FASTER then their counterparts on the earth's surface, when the observer is on the earth next to the ground clock. This is because the orbiting satellites are under the influence of two contradictory relativistic effects that are not equal.

Specifically, these satellites are subject relativity in TWO ways. The larger of the two  increases the speed of the satellite clocks by about 45 microseconds per day, as observed from the ground. This is because the satellites are significantly further from the earth's gravitational center.

However, the satellite clocks are also effected by speed. They are traveling quite fast relative to a ground observer, and time is accordingly slowed down by about 7 microseconds per day. Subtract 7 microsecond from 45 microseconds and the clocks need to be synchronized such that the overall effect is to decrease 37 microseconds per day.

http://www.astronomy.ohio-state.edu/~pogge/Ast162/Unit5/gps.html

PS: I know of one case in which potential energy can be hugely decreased in a given mass. Gunpowder. Remove all the charcoal from the mix and replace it with exactly the same mass of sand. And of course vice versa.

I do not know whether the mass of GPS satellites is increased or decreased when sent into orbit. This is because I do not have the math training to know whether the mass is decreased in proportion to the increased distance from the earth. I do know it is increased by the velocity transferred to them by accelerative forces.

However, the potential KINETIC energy of the speedy satellites is vastly greater then the one on the ground.
Title: If I give an object some potential energy, does its mass increase?
Post by: VernonNemitz on 18/01/2010 18:28:20
Vernon, I hold to that virtual particles 'exist' as you seem to do. Their interactions definitely prove it. That Feynman meant that we couldn't really say what they were, I take to mean that we can't really say what they are :). And we can't, but we can count on them and observe their interactions, whatever they might be.

Then I saw you support the Higgs field, sounds okay to me too. When it comes to the Higgs boson? Don't know, maybe the LHC will clear that up?

But now I see that you expect gravitons too? Isn't that contradictorily?
Or do you see string theory as encompassing all those descriptions?

First, thank you for letting me know this Message Thread was active again.

I don't know a lot about the proposed Higgs particle, but I think I can say this relatively safely, "Any proposed particle, if it actually can exist, already exists among the random/temporary virtual particles in the vacuum.  The "Higgs field" would thus be, in crude terms, the percentage of all the vacuum-self-energy virtual particles that are Higgs particles.  Whatever percentage that is, it is apparently enough (for theory to match observation) that ordinary particles are always able to interact with Higgs particles, and as a result most of them appear to possess mass."
Title: If I give an object some potential energy, does its mass increase?
Post by: Geezer on 18/01/2010 19:12:13
Farsight:

We reposition the Moon relative to the Earth using a conventional rocket (using energy from within the system).

I don't think anyone would argue that we have not altered the potential energy of the system (although I would not be totally surprised if someone did  [:D])

There had to be a redistribution of mass within the system to accomplish this but;

Did the mass of the system change, and if it did, why did it change?

(BTW, no matter was converted into energy or vice versa during this process.)
Title: If I give an object some potential energy, does its mass increase?
Post by: yor_on on 18/01/2010 19:27:26
Gotta think on that one.

As always it seems to fall down to how defining that Higg's field.
You can define it as 'particles' but if you do you seem to need to have a 'soup' of them as inertia is expected to be existent everywhere (deep space too). Then again, if distance and time is 'elastics' then that seems to question the very grounds we stand on when discussing 'virtual particles'. Down there 'distance' as we see it might become something different which then would leave space :) for this 'soup' too.

I like to see it as a 'field', which doesn't define anything really? The reason why I do so is because I find 'forces' questionable, well not macroscopically then, but I differ between that and QM, and what might exist beyond that. And if we have 'isolated' Higgs bosons 'traveling' I suppose I will be wrong, or possibly that the 'transformations' between what's 'under QM', as I see 'virtual particles' to be, and our macroscopic world is more magical than I ever thought.

==

What I was thinking of Geezer when discussing that 'jiggling'?
Awh, did you have to ask me that one?

We have what we call 'invariant mass'. The definition for that is the mass that will be intrinsic to a object no matter  what frame of reference you place it in, space or a black hole.

So it can't be that one. As it is a result of the objects interaction in a gravitational field you might call it relativistic mass as you did, or maybe even momentum.

But if I would be right there, then you might have another problem :)
As gravity is everywhere, as far as I know, how do we get to the right 'invariant mass'?
Title: If I give an object some potential energy, does its mass increase?
Post by: yor_on on 18/01/2010 19:48:27
Farsight "Imagine you used some aspect of this "jiggle" to operate a clock. Clock run slower nearer the ground, due to gravitational time dilation. That means the jiggle must be slower near the ground."

You make a very good point there. Time slows down relative an 'outside' observer when closer to a gravitational object. The way I was thinking of it was in the terms of potential energy, and that energy becomes more the further away you move your object from that 'impact zone' where gravity is at its highest.

==
Within limits of course as at some point it will no longer feel that gravitation (like deep space, Lagrangian points) enough to have the 'urge to move'.
==

Then i took that definition and compared it to the object (matter) it 'operated' on, and suggested that as we moved matter away from gravity its 'jiggling' would become less. As that 'jiggling' is transformed into kinetic energy the objects mass also would become less.
==

So if I choose your interpretation. Am I right in understanding it as that you associate 'slowing time' relative the observer as having less energy?

Which I then understand to lead to that a black holes energy level is less the closer you come to it as 'time' slows down relative our observer?
===
Or am I misunderstanding your suggestion?

Never the less, any which way the plate will differ in mass which then would lead to the question. Where exactly do we define the 'proper mass' for an object?
:)

---
So, now at last it almost makes sense :)

Title: If I give an object some potential energy, does its mass increase?
Post by: VernonNemitz on 18/01/2010 23:33:50
So now what happens if we don't use a solar powered rocket, but use a conventional rocket that obtained its energy from the system? Does the mass of the Moon still increase (and, because they are part of the same system, presumably by the same logic the Earth's mass would also increase)?

Details matter.  What sort of rocket is that?  If for purposes of thought-experiment you have magically introduced it from outside the Earth-Moon System, then you have introduced/added a great deal of potential energy (in the form of the rocket's potential ability to accelerate something like the Moon) to that System.  It should then be obvious that after the rocket has done its work, some potential energy has become real energy that has been added to the System, and thus the mass (that is, total mass-energy) of the Earth-Moon System would be greater than before.

On another hand, if the entirety of the rocket is constructed from Moon-matter, then its process of operation will cause a large amount of Moon-matter in the form of rocket exhaust to leave the Earth-Moon System at high velocity, and very likely the total mass-energy of that System, after the rocket does its work, will have become less than it was originally.

So, what would you prefer?  A proper answer DOES depend on the details!
Title: If I give an object some potential energy, does its mass increase?
Post by: Geezer on 19/01/2010 01:24:04
A proper answer DOES depend on the details!

Indeed it does. Which part of "conventional rocket" did you fail to comprehend?

If you are concerned about matter leaving the system, let's ensure that the system is large enough to include the exhaust from the rocket.

Now, did the mass (rest mass, naturally) of the system increase, or did it not? (I think this was the point of the original question.)
Title: If I give an object some potential energy, does its mass increase?
Post by: Farsight on 22/01/2010 02:37:50
So if I choose your interpretation. Am I right in understanding it as that you associate 'slowing time' relative the observer as having less energy?
Yes. If you think of a single electron, it's got spin and angular momentum. This occurs at a higher rate up in space than it does down near the surface of a planet. There's an energy difference between the two, and conservation of energy tells you this energy difference has to go somewhere.

Which I then understand to lead to that a black holes energy level is less the closer you come to it as 'time' slows down relative our observer?
I'm thinking it's the other way around. It's hard to say much about the black hole itself, but if you consider a photon near to a black hole event horizon, it will look more energetic as you fall towards it. Its energy doesn't change at all, instead you're changing. All the electrons etc that make up your body and your clocks spin at reducing rate, so your measurement of time slows down. Hence the photon frequency appears to increase.

Never the less, any which way the plate will differ in mass which then would lead to the question.
Agreed.

Where exactly do we define the 'proper mass' for an object?
It's a tricky one. When you lift the plate you give it some energy, so its mass increases slightly. But then when you lift yourself, you give yourself more energy. Ditto for all your measuring devices. So when you measure the mass of the plate again, it looks like it hasn't changed. If we try to say we should define proper mass in a place that is totally free from the effects of gravity, we've got another problem because there is no such place.
Title: If I give an object some potential energy, does its mass increase?
Post by: Farsight on 22/01/2010 02:46:58
Farsight: We reposition the Moon relative to the Earth using a conventional rocket (using energy from within the system). I don't think anyone would argue that we have not altered the potential energy of the system (although I would not be totally surprised if someone did). There had to be a redistribution of mass within the system to accomplish this but;

Did the mass of the system change, and if it did, why did it change?

(BTW, no matter was converted into energy or vice versa during this process.)
No, the mass of the system didn't change. All we've done is redistributed the energy within the system. We haven't changed the energy of the system, and mass is a measure of the energy of the system.

I notice that later on you said "rest mass, naturally". It's important to note that:

"The rest mass of a composite system is not the sum of the rest masses of the parts, unless all the parts are at rest. The total mass of a composite system includes the kinetic energy and field energy in the system".

See http://en.wikipedia.org/wiki/Mass_in_special_relativity#The_mass_of_composite_systems for details.
Title: If I give an object some potential energy, does its mass increase?
Post by: litespeed on 22/01/2010 20:31:23
Farsight - You wrote: "Did the mass of the system change, and if it did, why did it change?"

Potential chemical energy was thrust vectored towards the moon. Both the moon and the exhaust gases were accelerated and gained mass by E=mc2, as calculated by a stationary observer.

Title: If I give an object some potential energy, does its mass increase?
Post by: Geezer on 22/01/2010 23:30:49
Farsight: We reposition the Moon relative to the Earth using a conventional rocket (using energy from within the system). I don't think anyone would argue that we have not altered the potential energy of the system (although I would not be totally surprised if someone did). There had to be a redistribution of mass within the system to accomplish this but;

Did the mass of the system change, and if it did, why did it change?

(BTW, no matter was converted into energy or vice versa during this process.)
No, the mass of the system didn't change. All we've done is redistributed the energy within the system. We haven't changed the energy of the system, and mass is a measure of the energy of the system.

I notice that later on you said "rest mass, naturally". It's important to note that:

"The rest mass of a composite system is not the sum of the rest masses of the parts, unless all the parts are at rest. The total mass of a composite system includes the kinetic energy and field energy in the system".

See http://en.wikipedia.org/wiki/Mass_in_special_relativity#The_mass_of_composite_systems for details.

I would agree with that.  Does "all the parts are at rest" mean "at rest" in relative terms? Is the Moon "at rest" relative to the Earth? I think it almost is, but I'm not sure.
Title: If I give an object some potential energy, does its mass increase?
Post by: Farsight on 23/01/2010 11:36:24
I don't think that was me, litespeed, but yep, no problem. Give some part of the system some kinetic energy, and it makes a greater contribution to the system mass.

Geezer, I'd say it almost is, but not quite. The trouble with all the scenarios involving kinetic energy or just heat is that the mass increase is so very slight. I was looking for some experiment where it's been detected, but couldn't find anything.
Title: If I give an object some potential energy, does its mass increase?
Post by: yor_on on 23/01/2010 17:44:23
Farsight.

"When you lift the plate you give it some energy, so its mass increases slightly. "

No, i don't agree to that one :)

What you create is a different balance between the plate and the 'highest gravitational point' in that same frame. The plate hasn't gained any measurable energy, as far as I know?

My idea is that if gravity can influence the 'jiggling' it should be less the further you move it from gravitational influences. And if that energy is converted to kinetic then it should read as if the plate is 'lighter' in that frame.

The potential energy we are discussing is the result of a known relation between the plate and gravity, not defined to the plate solely, but only as a relation.
Title: If I give an object some potential energy, does its mass increase?
Post by: Farsight on 23/01/2010 18:25:04
Tie your plate to a long string, yor-on, and give it almighty push. You did work on the plate, you gave it kinetic energy. So now it swings up to the top of its arc and pauses momentarily. Now freeze the frame and examine the situation. What happened to that kinetic energy? Where did it go? I'm sure we all agree it was converted into potential energy, but where is it? Did it escape up the string? No. Did it somehow leave the plate and move into the surrounding space, the region we call the gravitational field? We can't detect any experimental evidence for any energy leaving the plate. Besides, we know that if we push a plate away from the earth at 11.2 km/s, it has escape velocity, and takes the potential energy away with it. It has now escaped the earth's gravitational field, so there is no relation any more. There's only one conclusion you can draw from this: the potential energy is in the plate. Yes, "gravity influences the jiggling", but it makes it go slower, not faster. This is the only way the conservation of energy works, and gravitational time dilation is your proof.
Title: If I give an object some potential energy, does its mass increase?
Post by: yor_on on 23/01/2010 19:58:22
Nice idea Farsight :)

First Vernon.
Rereading you and the lightbulb (induction)

A very cool example is from India where a lot of power disappeared from the Swedish built power plants as it was transmitted.

After using helicopters with IR in the night (searching for sources of warmth) they found cables (coils) buried under the ground leading to several villages ::))

quite clever..

Now, if we go back to your example. First of all you're introducing a momentum here, and you are limiting the 'force' of the rotation by the string attached. But even so, if you're not accelerating the plate it will be in what's called a 'uniform motion'.

Thingies in a uniform motion have a momentum, but it's unmeasurable as long as you don't have an interaction with some other frame of reference. That differs it from acceleration in where you directly can notice if you have gotten a larger momentum.

So what you see as energy transfered to a specific part of your new system I still see as a relation, where the potential energy can vary from null to ? depending on what other part you compare it too.

You might see it this way. If you do this on Earth and then let the plate hit the ground that relation will express itself as a certain kinetic energy getting released in the interaction as 'energy' in general.

Let a neutron star be you rotating, let the 'string' be gravity (frame dragging) and your plate be a object drawn by that.

Now, let another object travel beside it, getting slowly closer. As they meet there will be very little kinetic energy released.

It's all an relation, depending on your choice of reference frame. And that's why I don't see it as the plate having gotten any specific energy from rotating/ being lifted.

And when resting on that table it will be relative Earth an 'unmoving part of it' lifted out of the gravity, therefore having slightly less 'jiggling' What one could argue is that the rotational frame it is in is slightly increased, therefore counter balancing that loss of 'jingling'. But as it is a uniform motion it won't matter as I see it.
Title: If I give an object some potential energy, does its mass increase?
Post by: Geezer on 24/01/2010 02:51:22
A very cool example is from India where a lot of power disappeared from the Swedish built power plants as it was transmitted.

After using helicopters with IR in the night (searching for sources of warmth) they found cables (coils) buried under the ground leading to several villages ::))

I suspect that story was invented by some Norwegians. They've been making up stories for many centuries, so they are pretty good at it.

I have heard it's quite common for unauthorized persons to tap into the power system in India, but I'm fairly sure the inductive coupling story is a nordic saga or an urban legend.
Title: If I give an object some potential energy, does its mass increase?
Post by: yor_on on 24/01/2010 10:08:47
I don't think so. If I'm right I read it in the papers there? But it's some years ago and when I look on the net I don't seem to find it? But it wouldn't surprise me at all. India and electricity have a very friendly and extremely 'casual relation'. It's only enough with one bright soul to set it up. I think it was 'Vattenfall' that was involved in solving it? I guess they would have had to pay big time if they hadn't solved it too :)
Title: If I give an object some potential energy, does its mass increase?
Post by: Farsight on 24/01/2010 14:13:39
...It's all an relation, depending on your choice of reference frame. And that's why I don't see it as the plate having gotten any specific energy from rotating/being lifted.
You gave the plate kinetic energy when you gave it a push this → way. You applied a force for a distance, and accelerated it. It definitely gained energy. However the pendulum string doesn't rob it of any energy. You can twirl a ball on a string and whilst there's considerable force on the string, there's no motion in the direction of the force, so no work is being done. But at the top of the pendulum swing, that kinetic energy has gone and the plate has potential energy instead. Where has it gone? You have to be evidential about this rather than relying on relation. It hasn't gone up the string, and there's no trace of it leaving the plate. So it has to be in the plate. It's quite easy to see where it is. Imagine it's a spinning plate, spinning at the speed of light. It's rigged up with lasers etc to configure a clock. At the top of the swing, the clock runs faster. At the bottom of the swing, the clock runs slower, because of gravitational time dilation. That means the plate must be spinning slower at the bottom. Now think of electron spin instead of an overall spinning plate, then treat this along with all other subatomic motion as your "jiggling", and everything works out neatly. All you have to do is put in a minus sign.
Title: If I give an object some potential energy, does its mass increase?
Post by: Geezer on 24/01/2010 21:04:14
You can twirl a ball on a string and whilst there's considerable force on the string, there's no motion in the direction of the force, so no work is being done. But at the top of the pendulum swing, that kinetic energy has gone and the plate has potential energy instead. Where has it gone? You have to be evidential about this rather than relying on relation. It hasn't gone up the string, and there's no trace of it leaving the plate. So it has to be in the plate. It's quite easy to see where it is.

The energy is not in the plate. It's in the system that comprises the plate, the earth, and the force that tends to accelerate them towards each other.
Title: If I give an object some potential energy, does its mass increase?
Post by: Farsight on 24/01/2010 22:40:31
I'm sorry geezer, but when that plate escapes the system, it takes the potential energy away with it. That's proof positive that the energy is in the plate. Try proving otherwise, and you'll find you simply can't. You'll have to resort to a "spring" that simply isn't there, and magical mysterious action-at-a-distance, which even Newton knew was false:

"That gravity should be innate, inherent, and essential to matter, so that one body may act upon another at a distance through a vacuum, without the mediation of anything else, by and through which their action and force may be conveyed from one to another, is to me so great an absurdity that I believe no man who has in philosophical matters a competent faculty of thinking can ever fall into it."

Gravity is a local phenomenum, not an action-at-distance effect. It operates through a local gradient in gμν, just as Einstein described it. Unfortunately very few people read the original General Relativity to understand what Einstein actually said. As an example, in 1916 Einstein wrote Relativity: The Special and General Theory (see http://www.gutenberg.org/etext/5001) where in section 22, the English translation reads:

"In the second place our result shows that, according to the general theory of relativity, the law of the constancy of the velocity of light in vacuo, which constitutes one of the two fundamental assumptions in the special theory of relativity and to which we have already frequently referred, cannot claim any unlimited validity. A curvature of rays of light can only take place when the velocity of propagation of light varies with position".

However when you look at the original German, what he actually said was die Ausbreitungsgeschwindigkeit des Lichtes mit dem Orte variiert. This translates to the speed of light varies with the locality. It's crystal clear he meant speed rather than a vector-quantity velocity, because he was referring to one of the postulates of special relativity - the one that said the speed of light is constant. Once you appreciate this, you get a totally different picture of gravity. Here's an analogy that hopefully conveys how it works:

Imagine a swimming pool. Every morning you swim from one end to the other in a straight line. In the dead of night I truck in a load of gelatine powder and tip it all down the left hand side. This starts diffusing across the breadth of the pool, imparting a viscosity gradient from left to right. The next morning when you go for your swim, something's not right, and you find that you're veering to the left. If you could see your wake, you'd notice it was curved. That's your curved spacetime, because the pool is the space round a planet, the viscosity gradient is Einstein's non-constant gμν, and you're a photon. As to how the gradient attracts matter, consider a single electron. We can make an electron along with a positron from light, via pair production. Since the electron also has spin, think of it as light trapped in a circular path. So if you're swimming round and round in circles, whenever you're swimming up or down the pool you're veering left. Hence you find yourself working over to the left. That's why things fall down.

Your leftward motion comes out of a reduced rate of sub-atomic circulatory motion or spin. The latter is yor-on's "jiggling". The rate is reduced near the surface of a planet where where gravitational potential is lower. Gravitational time dilation is the clear evidence for this reduced rate of motion, and we see it in for example the GPS clock adjustment. The gμν gradient "veers" internal sub-atomic motion which we call potential energy, into the macroscopic motion which we call kinetic energy. I'm not fooling you about this, and I can give you more Einstein references to support what I'm saying. See for example his 1911 paper "On the Influence of Gravitation on the Propagation of Light" where he says c=c0(1+Φ/c²). He got this somewhat back to front, but there are examples from 1912, 1913, 1914, and 1915 where you can see his ideas evolving into something wherein gravity is the result of a gradient in c caused in turn by energy "conditioning" the surrounding space. PM me and I'll send you pdf page images if you wish.

However, I cannot explain why all this, or his Leyden Address, isn't in the text books, or why it is not taught. Perhaps it's something to do with the way people who have been taught that "Einstein told us the speed of light is constant" have difficulty when confronted with the original material that says "Einstein told us the speed of light reduces in line with gravitational potential". Rather than examining the evidence as a rational scientist should, they tend to dismiss it, and thus the myth and mystery persist.
Title: If I give an object some potential energy, does its mass increase?
Post by: Geezer on 25/01/2010 00:46:07
I'm sorry geezer, but when that plate escapes the system, it takes the potential energy away with it.

I think I can see why you are having a problem with this. The plate can never escape the system; the force never goes to zero, it just gets smaller. If the plate can't escape the system, it can't "take the potential energy away with it".

If you remove something from a system, you have just defined a new system. All previous bets are cancelled.
Title: If I give an object some potential energy, does its mass increase?
Post by: VernonNemitz on 26/01/2010 15:13:33
Tie your plate to a long string, yor-on, and give it almighty push. You did work on the plate, you gave it kinetic energy. So now it swings up to the top of its arc and pauses momentarily. Now freeze the frame and examine the situation. What happened to that kinetic energy? Where did it go? I'm sure we all agree it was converted into potential energy, but where is it? Did it escape up the string? No. Did it somehow leave the plate and move into the surrounding space, the region we call the gravitational field? We can't detect any experimental evidence for any energy leaving the plate. Besides, we know that if we push a plate away from the earth at 11.2 km/s, it has escape velocity, and takes the potential energy away with it. It has now escaped the earth's gravitational field, so there is no relation any more. There's only one conclusion you can draw from this: the potential energy is in the plate. Yes, "gravity influences the jiggling", but it makes it go slower, not faster. This is the only way the conservation of energy works, and gravitational time dilation is your proof.
I see you haven't learned much from our previous conversation in this Thread.  I had some hope when you wrote this in another message:
No, the mass of the system didn't change. All we've done is redistributed the energy within the system. We haven't changed the energy of the system, and mass is a measure of the energy of the system.
You do realize, don't you, that when we are here on Earth using fossil-fuel energy to do stuff, we are mostly just redistributing energy within the Earthly system?  That includes lifting the plate --the plate is not initially so isolated from the Earth that that enormous part of the system can be ignored!  Which is one reason why the Earth actually ends up with a greater portion of chemical-energy-converted-to-kinetic-energy-converted-to-potential-energy-in-the-form-of-mass, than the plate.
Title: If I give an object some potential energy, does its mass increase?
Post by: yor_on on 26/01/2010 15:35:24
Farsight :)

As I see it that plate has been removed from a 'force' more or less acting uniformly (as far as I know?) on it, namely gravity. That's why I would expect the jiggling to be less. As a counterexample you might imagine what would happen to that plate if placed between two black holes slowly gravitating, equally trying to 'pull' the plate. Wouldn't you expect the 'jiggling' to become more there?

===
Ah, as seen from the reference frame of the plate naturally.
Title: If I give an object some potential energy, does its mass increase?
Post by: VernonNemitz on 26/01/2010 16:02:09
I'm sorry geezer, but when that plate escapes the system, it takes the potential energy away with it. That's proof positive that the energy is in the plate. Try proving otherwise, and you'll find you simply can't. You'll have to resort to a "spring" that simply isn't there, and magical mysterious action-at-a-distance, which even Newton knew was false:
Tsk, tsk, and Aristotle "knew" that effort always had to be expended to keep something in constant motion.  Whoop-te-do.
"That gravity should be innate, inherent, and essential to matter, so that one body may act upon another at a distance through a vacuum, without the mediation of anything else, by and through which their action and force may be conveyed from one to another, is to me so great an absurdity that I believe no man who has in philosophical matters a competent faculty of thinking can ever fall into it."
"There are more things in Heaven and Earth that are dreamt of in your philosophy."
Gravity is a local phenomenum, not an action-at-distance effect. It operates through a local gradient in gμν, just as Einstein described it. Unfortunately very few people read the original General Relativity to understand what Einstein actually said.
You are wrong, because gravity is an infinite-range phenomenon.  And, if Einstein knew so much, why didn't he end up creating the Grand Unified Field Theory that he wanted?  "It's not what you don't know that hurts [your efforts] so much as what you do know that ain't so."  Specifically, Einstein didn't like certain aspects of Quantum Mechanics, such as the Uncertainty Principle, and, in thinking it was flawed, handicapped himself.
As an example, in 1916 Einstein wrote Relativity: The Special and General Theory (see http://www.gutenberg.org/etext/5001) where in section 22, the English translation reads:
"In the second place our result shows that, according to the general theory of relativity, the law of the constancy of the velocity of light in vacuo, which constitutes one of the two fundamental assumptions in the special theory of relativity and to which we have already frequently referred, cannot claim any unlimited validity. A curvature of rays of light can only take place when the velocity of propagation of light varies with position".

However when you look at the original German, what he actually said was die Ausbreitungsgeschwindigkeit des Lichtes mit dem Orte variiert. This translates to the speed of light varies with the locality. It's crystal clear he meant speed rather than a vector-quantity velocity, because he was referring to one of the postulates of special relativity - the one that said the speed of light is constant. Once you appreciate this, you get a totally different picture of gravity.
Wrong.  No totally different picture needed.  Look up the Law of Refraction.  Whenever the medium changes, through which light passes, its speed and its direction is affected.  Very simple, very consistent, gravity included.  What changes in the "medium" of the vaccum, when gravitational field intensity increases, to cause light to go slower and to curve more?  Simple: the concentration of numbers of virtual gravitons (due to inverse square law), relative to the concentration of all other types of virtual particles in the vacuum.
Here's an analogy that hopefully conveys how it works:
Imagine a swimming pool. Every morning you swim from one end to the other in a straight line. In the dead of night I truck in a load of gelatine powder and tip it all down the left hand side. This starts diffusing across the breadth of the pool, imparting a viscosity gradient from left to right. The next morning when you go for your swim, something's not right, and you find that you're veering to the left. If you could see your wake, you'd notice it was curved. That's your curved spacetime, because the pool is the space round a planet, the viscosity gradient is Einstein's non-constant gμν, and you're a photon. As to how the gradient attracts matter, consider a single electron. We can make an electron along with a positron from light, via pair production. Since the electron also has spin, think of it as light trapped in a circular path. So if you're swimming round and round in circles, whenever you're swimming up or down the pool you're veering left. Hence you find yourself working over to the left. That's why things fall down.
Pair-production is possible because of the existence of virtual electrons and virtual positrons in the vacuum, with which an appropriate-energy photon can interact.  The great thing about Quantum Mechanics is that it does allow us to have a consistent picture, such that we don't need to invoke geometry to explain gravitation.
Your leftward motion comes out of a reduced rate of sub-atomic circulatory motion or spin. The latter is yor-on's "jiggling". The rate is reduced near the surface of a planet where where gravitational potential is lower. Gravitational time dilation is the clear evidence for this reduced rate of motion, and we see it in for example the GPS clock adjustment. The gμν gradient "veers" internal sub-atomic motion which we call potential energy, into the macroscopic motion which we call kinetic energy. I'm not fooling you about this, and I can give you more Einstein references to support what I'm saying. See for example his 1911 paper "On the Influence of Gravitation on the Propagation of Light" where he says c=c0(1+Φ/c²). He got this somewhat back to front, but there are examples from 1912, 1913, 1914, and 1915 where you can see his ideas evolving into something wherein gravity is the result of a gradient in c caused in turn by energy "conditioning" the surrounding space. PM me and I'll send you pdf page images if you wish.
And gravitational time dilation is just as easily explained by QM, in terms of interactions with gravitons.  The more something is spending time interacting with virtual gravitons, the less it is spending time interacting with anything else --and it is those other interactions that we use to measure the passage of Time.  Simple.
However, I cannot explain why all this, or his Leyden Address, isn't in the text books, or why it is not taught. Perhaps it's something to do with the way people who have been taught that "Einstein told us the speed of light is constant" have difficulty when confronted with the original material that says "Einstein told us the speed of light reduces in line with gravitational potential". Rather than examining the evidence as a rational scientist should, they tend to dismiss it, and thus the myth and mystery persist.
I'm pretty sure the lessened speed of light in a gravity field, relative to empty space, is taught in the advanced classes and other places.  Certainly I found out about it without reading original source material by Einstein.
Title: If I give an object some potential energy, does its mass increase?
Post by: Farsight on 28/01/2010 11:57:21
Yor_on: yep, it's about definitions. I wouldn't expect the jiggling to be increased between a couple of black holes. It's time dilated down there, everything's happening slower. But from the reference frame of the plate, like if you were in a black box along with the plate, you can't tell.

Tsk, tsk, and Aristotle "knew" that effort always had to be expended to keep something in constant motion. Whoop-te-do.
There isn't any spring Vernon. I raise a brick, and you can wave your hand underneath it. The spring is just not there. And there's no sign of gravitons either.

You are wrong, because gravity is an infinite-range phenomenon. And, if Einstein knew so much, why didn't he end up creating the Grand Unified Field Theory that he wanted?  "It's not what you don't know that hurts [your efforts] so much as what you do know that ain't so."  Specifically, Einstein didn't like certain aspects of Quantum Mechanics, such as the Uncertainty Principle, and, in thinking it was flawed, handicapped himself.
Yep, it's infinite in range, but things fall down because the local space they're in isn't uniform. Call it curvature or call it a non-constant guv, it doesn't matter. If the space was homogeneous there wouldn't be any detectable gravitational field. Einstein ran out of time, and see http://en.wikipedia.org/wiki/Bohr-Einstein_debates re his stance on quantum mechanics. I'd say he disliked the lack of underlying reality more than anything else.

Wrong. No totally different picture needed. Look up the Law of Refraction.  Whenever the medium changes, through which light passes, its speed and its direction is affected. Very simple, very consistent, gravity included. What changes in the "medium" of the vaccum, when gravitational field intensity increases, to cause light to go slower and to curve more? Simple: the concentration of numbers of virtual gravitons (due to inverse square law), relative to the concentration of all other types of virtual particles in the vacuum.
We have no evidence of virtual gravitons. Einstein talked of inhomogeneous space. According to relativity, what changes is the space itself.

Pair-production is possible because of the existence of virtual electrons and virtual positrons in the vacuum, with which an appropriate-energy photon can interact.  The great thing about Quantum Mechanics is that it does allow us to have a consistent picture, such that we don't need to invoke geometry to explain gravitation.
You should read . Feynman says they're virtual. Also read [url=http://www.iop.org/EJ/abstract/0295-5075/76/2/189]Evanescent modes are virtual photons (http://www.amazon.co.uk/QED-Strange-Theory-Penguin-Science/dp/0140125051[The Strange Theory of Light and Matter[/url) by A. A. Stahlhofen et al, Europhys. Lett. 76 189-195, 2006. The geometry is right back in there now. And let's not forget that nobody has succeeded in quantizing gravity.

And gravitational time dilation is just as easily explained by QM, in terms of interactions with gravitons.  The more something is spending time interacting with virtual gravitons, the less it is spending time interacting with anything else --and it is those other interactions that we use to measure the passage of Time. Simple.
We use light. Atomic clocks employ microwaves, look at the definition of the second.

I'm pretty sure the lessened speed of light in a gravity field, relative to empty space, is taught in the advanced classes and other places.  Certainly I found out about it without reading original source material by Einstein.
One sometimes finds mention of a reduced coordinate speed, but I'd say what tends to be taught is in line with http://www.desy.de/user/projects/Physics/Relativity/SpeedOfLight/speed_of_light.html. Maybe it would be better to continue this conversation on a thread I started entitled "Is Einstein's general relativity misunderstood?".
Title: If I give an object some potential energy, does its mass increase?
Post by: yor_on on 28/01/2010 13:51:04
"But from the reference frame of the plate, like if you were in a black box along with the plate, you can't tell. "

That's a really good idea Farsight. if I would be right you should be able to tell by the increased jiggling, inside our black box, if the gravity might be higher, as I then would expect it to vary with 'distance' to gravitation.

As you imply here, time will. as seen from the frame of the plate, always have the 'same time as usual' meaning that if that plate ah, looked on his watch, a second would seem a second and all sequences would seem the same inside that black box no matter :) where it was, between black holes or on the moon. But if you think of dropping something inside that black box the situation change, depending on where you are.

Let's say that our brilliant plate :) now would measure a ball being dropped one meter inside that box, using his watch to time it under different gravities. will time and the fall be the same under different gravities. Like a 1G fall take 1.s (according to his watch inside that box)for 1.m and then according to this logic 1000 000G fall then also would take 1.s (according to his watch inside that box, sharing reference frame) falling 1.m?

That is, if I understood your reference correctly?

Title: If I give an object some potential energy, does its mass increase?
Post by: Farsight on 28/01/2010 14:21:02
That doesn't work, yor_on. Dropping an object and timing it tells you the steepness of the local gradient in guv, but it doesn't tell you how the guv down by a black hole compares to the guv up in space. You compare them by looking at the gravitational time dilation, and you can't do it locally. You start with two identical clocks. You leave one up in space. You take the other one down near a black hole, hang around for a while, then go back up to space and compare the clock readings. These clocks clock up motion. If they're mechanical clocks they clock up the motion of gears and sprockets. If they're light clocks they clock up the motion of light. If they're jiggle clocks they clock up internal subatomic motion, things like electron spin. It doesn't matter what type of clock you're using, you'll always see that gravitational time dilation, because things move slower in a gravitational field.
Title: If I give an object some potential energy, does its mass increase?
Post by: yor_on on 28/01/2010 14:31:16
Well Farsight. Inside that box, being in the same reference frame, you will get different values for gravity. So time and gravity differ to me, which seems to make my idea almost plausible :)

You see, as I saw your idea there you seemed to say that locally time always will be seen as proceeding as usual, which makes sense to me. But time as a concept is not gravity, with gravity not acting the same way as 'times arrow' does even though it is closely correlated, meaning that it will change with gravity.

===

That last one will only make sense from an observers frame of reference, of course. For you, being inside that box, f.ex moving towards the EV (event horizon) of a BH, time will always be the same, working 'as usual'. At no time will you ever notice time becoming 'slower' just because you're accelerating, or finding yourself placed in a BH :)
Title: If I give an object some potential energy, does its mass increase?
Post by: VernonNemitz on 28/01/2010 15:32:56
Tsk, tsk, and Aristotle "knew" that effort always had to be expended to keep something in constant motion. Whoop-te-do.
There isn't any spring Vernon. I raise a brick, and you can wave your hand underneath it. The spring is just not there. And there's no sign of gravitons either.
A worthless analogy.  You can wave your hand between two magnets, also, and there is no sign of a spring or the virtual photons passing between them, that explain the magnetic force between them.  Yet Q.E.D. is the most accurately-measured/verified theory in Physics, despite nobody every directly detecting any virtual photons.

You are wrong, because gravity is an infinite-range phenomenon. And, if Einstein knew so much, why didn't he end up creating the Grand Unified Field Theory that he wanted?  "It's not what you don't know that hurts [your efforts] so much as what you do know that ain't so."  Specifically, Einstein didn't like certain aspects of Quantum Mechanics, such as the Uncertainty Principle, and, in thinking it was flawed, handicapped himself.
Yep, it's infinite in range, but things fall down because the local space they're in isn't uniform. Call it curvature or call it a non-constant guv, it doesn't matter. If the space was homogeneous there wouldn't be any detectable gravitational field.
I certainly agree with that last statement, but the manner of how space can be inhomogenous is what we are arguing about here.  The Casimir effect is proof enough that empty space can be emptier in some regions than others (emptier of virtual particles, that is).  It should be obvious that if QM can describe gravitation, then the presence of a mass, in the vacuum, will mean that the vacuum now has in it virtual gravitons radiating from mass, in addition to the usual background noise of virtual particles.  There will naturally be an intensity gradient with distance from the mass, too, meaning that space will be inhomogenous in that region.  Therefore we can deduce gravitational effects, equivalent to invoking Geometry (but now consistent with the rest of QM).

Einstein ran out of time, and see http://en.wikipedia.org/wiki/Bohr-Einstein_debates re his stance on quantum mechanics. I'd say he disliked the lack of underlying reality more than anything else.
The effect, though, is that he made no significant effort to include Quantum Mechanics in his thinkings about gravity.  Instead, I get the impression he wanted to use Geometry to explain the other forces.

Wrong. No totally different picture needed. Look up the Law of Refraction.  Whenever the medium changes, through which light passes, its speed and its direction is affected. Very simple, very consistent, gravity included. What changes in the "medium" of the vaccum, when gravitational field intensity increases, to cause light to go slower and to curve more? Simple: the concentration of numbers of virtual gravitons (due to inverse square law), relative to the concentration of all other types of virtual particles in the vacuum.
We have no evidence of virtual gravitons. Einstein talked of inhomogeneous space. According to relativity, what changes is the space itself.
Again, a worthless argument, like saying in 1990 that planets outside the Solar system can't exist because there was no available evidence for them.

Pair-production is possible because of the existence of virtual electrons and virtual positrons in the vacuum, with which an appropriate-energy photon can interact.  The great thing about Quantum Mechanics is that it does allow us to have a consistent picture, such that we don't need to invoke geometry to explain gravitation.
You should read . Feynman says they're virtual. Also read [url=http://www.iop.org/EJ/abstract/0295-5075/76/2/189]Evanescent modes are virtual photons (http://www.amazon.co.uk/QED-Strange-Theory-Penguin-Science/dp/0140125051[The Strange Theory of Light and Matter[/url) by A. A. Stahlhofen et al, Europhys. Lett. 76 189-195, 2006. The geometry is right back in there now. And let's not forget that nobody has succeeded in quantizing gravity.
And in 1900 nobody had succeeded in building a heavier-than-air flying machine.  Whoop-te-do, another worthless argument.

And gravitational time dilation is just as easily explained by QM, in terms of interactions with gravitons.  The more something is spending time interacting with virtual gravitons, the less it is spending time interacting with anything else --and it is those other interactions that we use to measure the passage of Time. Simple.
We use light. Atomic clocks employ microwaves, look at the definition of the second.
You are failing to understand.  Every time-measuring device involves interactions between various components, and all those components are also interacting gravitationally with the Earth (or, say, pick some other source, like if the clock was sitting on a neutron star).  However, these interactions cannot be simultaneous; the whole point of a fundamental "quantum of time" is that exactly one interaction would be possible in that interval.  So, regardless of however-many time-quanta exist during "one second", the important thing to ask is, "What percentage of them are being devoted to gravitational interactions?"  Obviously more would be devoted to that if the clock was sitting on the neutron star, instead of Earth.  Which leaves less time-quanta available for other interactions, between components of the clock, and those are the interactions we rely on to tell us the time.  The net effect is that the clock ticks slower on the star than on Earth, even though the same number time-quanta per second occur in both places.  Even without gravitation, this effect can be seen when we think about interactions with the virtual particles in the vacuum; a fast-moving spaceship will encounter more of them per second than a slow-moving spaceship, and therefore the clock ticks slower on the fast ship.

I'm pretty sure the lessened speed of light in a gravity field, relative to empty space, is taught in the advanced classes and other places.  Certainly I found out about it without reading original source material by Einstein.
One sometimes finds mention of a reduced coordinate speed, but I'd say what tends to be taught is in line with http://www.desy.de/user/projects/Physics/Relativity/SpeedOfLight/speed_of_light.html. Maybe it would be better to continue this conversation on a thread I started entitled "Is Einstein's general relativity misunderstood?".
Nope.  Because our argument is not about GR so much as is about QM, and the ways that QM can be extended to also explain the things that GR explains, thereby making GR irrelevant.  (Not "wrong", because GR is quite good at what it does, but "irrelevant", as in "unnecessary".)
Title: If I give an object some potential energy, does its mass increase?
Post by: yor_on on 31/01/2010 16:27:13
Vernon. If you were the observer to a clock falling toward the event horizon. How would you explain the red shift and 'slowing of time' that clock will express, relative you observing? In form of your 'time quanta', that is.

Will they according to you, become 'more' extending 'times arrow' as the clock moves slower relative your frame, or do you see the time quanta as the same 'amount' as it would be if seen from the same perspective (frame) as that clock falling in?

I'm curious to how you see the concept of time quanta, as a 'predefined even if undefined' amount, or as something that comes into 'work' as needed? And I hope I made some sense asking too :)
Title: If I give an object some potential energy, does its mass increase?
Post by: yor_on on 01/02/2010 03:46:08
Farsight I'm still of two minds in my thought experiment.

"Einstein concluded in one of his papers that a particle's increase in inertial mass due to motion would result in an increase in its gravitational mass"

And we do know that acceleration is equivalent to gravity, and mass :)

And:

---Quote--

Does a particle's relativistic mass increase result in a corresponding increase in the gravity field generated by the particle?

Yes, the particle has an increase in energy (kinetic and mass increase) and under general relativity, it is energy that generates a gravity field, not just rest mass.

Dr. Eric Christian.
Nasa (http://helios.gsfc.nasa.gov/qa_gp_gr.html)

--End of quote-

Take a look here too Fluids and pressure (http://en.wikipedia.org/wiki/Mass_in_general_relativity#Questions.2C_answers.2C_and_simple_examples_of_mass_in_general_relativity.)

"One might also ask about the answers to this question if one assumed that one were asking about the mass as it is defined in special relativity rather than the Komar mass. If one assumes that the space-time is nearly Minkowskian, the special relativistic mass exists. In this case, the answer to the first question is still yes, but the second question cannot be answered without even more data. Because the system consisting only of the gas is not an isolated system, its mass is not invariant, and thus depends on the choice of observational frame.

A specific choice of observational frame (such as the rest frame of the system) must be specified in order to answer the second question. If the rest frame of the object is chosen, and special relativistic mass rather than Komar mass is assumed, the answer to the second question becomes yes. This problem illustrates some of the difficulties one faces when talking about the mass of non-isolated systems."

But then against it you might say that we added an potential energy by lifting it, which then is equivalent to an added mass :)

So it's still a question of from where we define it.

The question seems to me to become if you can translate gravity into energy without using a definition of first 'interacting' with something, as the ground f.ex for the release of 'potential energy'?

==

Then again, if you consider gravity to be the geometric twisting of 'SpaceTime' it shouldn't transfer any 'energy', possibly. But 'potential energy' is created just out of those 'twists'?

Awh :)

Title: If I give an object some potential energy, does its mass increase?
Post by: Farsight on 01/02/2010 15:48:38
That's good stuff, yor_on, but IMHO it's simpler than you think. Think about Einstein's 1905 paper. Now start with a container of hot gas up in space and call it your system. It's hot, and it emits infra-red radiation. Wait for the radient energy to dissipate, and once it's cooled down, the gas molecules are moving slower. There's less energy in the system so the system has lost some mass.

Now start again with another container of hot gas up in space, and instead of letting it cool down, let it fall down. Wait for the kinetic energy to dissipate, and even though the gas is still hot, the gravitational time dilation means the gas molecules are moving slower. There's less energy in the system so the system has lost some mass.
Title: If I give an object some potential energy, does its mass increase?
Post by: yor_on on 01/02/2010 17:23:46
Would be nice to be sure on that one :)

I still have some wondering left to do there :) It seems to come down to how you view gravity and its 'interaction' with matter. and there I am of the view that gravity is a geometry, not a force? But then we have VMO:s ending in singularities like super black holes?

I hasn't considered gravity as a 'force' before so I will gratefully accept all links from those seeing it as such :)

It's this elusive 'potential energy' irritating me. It's like an itch for the moment :)

" We seek him here, we seek him there, Those Frenchies seek him everywhere.
Is he in heaven? — Is he in hell? That damned, elusive Pimpernel "

Ah, potential energy that is... Not Pimpernel..
And not only the Frenchies :)

Title: If I give an object some potential energy, does its mass increase?
Post by: VernonNemitz on 01/02/2010 17:54:19
Vernon. If you were the observer to a clock falling toward the event horizon. How would you explain the red shift and 'slowing of time' that clock will express, relative you observing? In form of your 'time quanta', that is.

Will they according to you, become 'more' extending 'times arrow' as the clock moves slower relative your frame, or do you see the time quanta as the same 'amount' as it would be if seen from the same perspective (frame) as that clock falling in?

I'm curious to how you see the concept of time quanta, as a 'predefined even if undefined' amount, or as something that comes into 'work' as needed? And I hope I made some sense asking too :)
I won't claim that my ideas are fully fleshed out, on this topic.  However, if QM truly can explain everything, then some things need to be true everwhere, like the speed of light in a vacuum, far from any mass, is always the same speed.  That's a primary assumption for Relativity, from which much else logically follows.  Here I will assume that time is quantized and that it has a fixed size, which is described here:  http://www.physlink.com/Education/AskExperts/ae281.cfm?CFID=25253792&CFTOKEN=1591d858dcc3d8ce-89D3AB98-15C5-EE01-B9FC9A191E4C5E40

As I mentioned to Farsight, during one time-quantum only one interaction event should be possible.  It might be a gravitational interaction or an electromagnetic interaction or some other interaction.  The huge number of time-quanta per second allows the appearance of simultaneous interactions (not unlike a computer operating system on a single-core computer that does rapid task-switching, and appears to be doing multiple tasks simultaneously, even though it is actually doing just one thing at a time).  Anyway, it is interactions between components of a clock (any clock!) that we use to determine the passage of time.  Normally there are plenty of "spare" time-quanta for such things as gravitational interactions, such that we don't notice a clock running slower in an ordinary planetary gravitational field.  A very strong field, like that of a neutron star or a black hole (very similar those fields can be! --the escape velocity from a neutron star can easily be 70% of lightspeed) will simply overwhelm the available time-quanta with gravitational interactions, leaving few for interactions between clock-components (or other types of interactions, like those associated with age-ing).  So, the closer you get to the surface of a neutron star, or the event horizon of a black hole, the greater is this effect, and the farther away, the less is this effect.  We can CALL it "time slowing", but perhaps a more accurate way to say it is, "Fewer Universal Clock Cycles are being devoted to the task of running that merely mundane clock."

The red shift of light leaving the vicinity of a black hole doesn't need to have anything special to do with time-quanta.  This is instead straightforward energy conversion from one type (SOME of the photon's content) to another type (potential).  The difference between the blue photon that was created near the black hole, and the red photon we see escaping to outer space, is energy that became potential energy (if the photon was reflected by a mirror back to the black hole, it would gain that energy back, and be exactly as blue when it arrives at its origin, as when it left).
Title: If I give an object some potential energy, does its mass increase?
Post by: Farsight on 02/02/2010 00:09:43
I still have some wondering left to do there :) It seems to come down to how you view gravity and its 'interaction' with matter. and there I am of the view that gravity is a geometry, not a force? But then we have VMO:s ending in singularities like super black holes?
It's a bit like Chinatown, yor-on:

Evelyn Mulwray: She's my daughter.
[Gittes slaps Evelyn]
Jake Gittes: I said I want the truth!
Evelyn Mulwray: She's my sister...
[slap]
Evelyn Mulwray: She's my daughter...
[slap]
Evelyn Mulwray: My sister, my daughter.
[More slaps]
Jake Gittes: I said I want the truth!
Evelyn Mulwray: She's my sister AND my daughter!

It's "a geometry", and it's "a force". But it isn't like when you push something. It isn't a force that adds energy to an object. The thing is, electromagnetism is like this too.
Title: If I give an object some potential energy, does its mass increase?
Post by: VernonNemitz on 02/02/2010 09:45:26
Farsight, it occurs to me that another reason you are wrong about the plate acquiring significant potential-energy-as-mass relates to the photon I described at the end of my last message here.  The photon would come away from the black hole almost as blue as when it was created, if you were right about where the potential energy goes!  Therefore it is I that am correct; most of the energy that becomes potential, as a small body or photon escapes the gravitational field of a larger body, goes to increasing the mass-energy of the larger body, not the smaller body or photon.  Do remember that the most important thing about General Relativity is that it is about "mass-energy" regardless of form.  You can't have a double-standard, treating the plate one way and the photon another.
Title: If I give an object some potential energy, does its mass increase?
Post by: yor_on on 02/02/2010 16:34:55
Vernon, you seem to seek a time like geometry.

"The escape velocity from a neutron star can easily be 70% of lightspeed) will simply overwhelm the available time-quanta with gravitational interactions, leaving few for interactions between clock-components"

One might say that looking at a 'time quanta'  your way, translated into geometry, that time might have different 'depths'. As I find it more understandable that way than if thinking of that 'quanta' as a equal 'distance' of time, that then somehow would become 'split up' depending on gravitational influences.

If you instead allow it to have a 'depth' to it, you introduce a concept where 'time' if so would have one apparent property that we define as it's arrow moving us along, and then another, more or less 'hidden' property to us, describing the relation between gravity, acceleration, uniform motion, mass, energy and time.

The problem here would then be how to to define that hidden property? As a new 'dimension' perhaps? I'm sort of allergic to 'dimensions' for the moment as I find them poorly proofed experimentally.

Mathematically you will have a plethora of ideas, building on each other, creating one 'manifold' after another. But I still wish we could proof that dimensions are a singular property experimentally, before deciding that our mathematical treatment is the only one existing.

So if not another 'dimension' then what?
That's a really good question I think and to me it seems as an 'emergence'. As such you don't need to treat it as a object with several 'singular properties' which releases a lot of mine inability to understand things :)

As an 'emergence' it will contain a 'whole', just as SpaceTime seems to do when looking at its 'plasticity'. Which still leaves us the reason why we can't 'see' it, but to me that says more about how we view the world than any inability of times arrow (or time) to express itself. It gives us clear indications to how it works but in our need to 'split it down' into its smallest constituents we forget that what we define as 'existing' is limited to what we believe. And our 'beliefs' change constantly nowadays :)

As I see it of course, you might have another view that I missed totally.
Title: If I give an object some potential energy, does its mass increase?
Post by: VernonNemitz on 02/02/2010 19:46:29
"The escape velocity from a neutron star can easily be 70% of lightspeed) will simply overwhelm the available time-quanta with gravitational interactions, leaving few for interactions between clock-components"
You quoted out of context.
"A very strong field, like that of a neutron star or a black hole (very similar those fields can be! --the escape velocity from a neutron star can easily be 70% of lightspeed) will simply overwhelm the available time-quanta with gravitational interactions"
The proper extraction from that is:
"A very strong field, like that of a neutron star or a black hole will simply overwhelm the available time-quanta with gravitational interactions"
The parenthesized part was there merely to explain the similarity between the gravity fields of neutron stars and black holes.

It is sometimes said that "Time is what keeps everything from happening all at once."  Time is about CHANGE.  Ordinary geometry is purely static, not dynamic.  As long as this fundamental distinction exists, I won't be trying to equate Time with another geometric dimension.
Title: If I give an object some potential energy, does its mass increase?
Post by: yor_on on 02/02/2010 23:19:10
Sorry if you felt mis quoted there. But now you're losing me Vernon. How do you mean when you write "Time is about CHANGE.  Ordinary geometry is purely static, not dynamic.  As long as this fundamental distinction exists, I won't be trying to equate Time with another geometric dimension."

SpaceTime is no static geometry, not according to me at least?
As for that times arrow is about 'change' or as some say 'events' there is no doubt.
But the question I asked was how you thought about 'changes' taking shorter or longer 'time' to happen depending on your frame of observation.

If you think of time 'changes' and watch that clock (observing it) hang at the event horizon, there are going to be an awful lot of changes happening to you before that clock ever moves a inch :) you will have died long before that. And that was the crux of it to me. How you saw those 'time quanta' as differing between different observations from different 'frames of reference'. The clock tick as usual from its own perspective, but you, the earth and the universe is accelerated and blueshifted, changing at a ever increasing pace as it falls in.

So how do you equate those two observations with your 'time quanta'?
Consider A here as the the minute hand ticking one minute as seen from the frame of the in-falling clock.
Here is 'time quanta A' as seen from the frame of the clock, with its time 'as usual' so to speak
<-A->

Here it is from the perspective of you observing the clock.

<-------------------A------------------->

That 'time quanta', lifted from the perspective of the clock, will now contain a infinite amount of 'changes/events' from your perspective observing. You can travel home to Earth and come back without that clock hand ever moved according to you.

Hope this clarified my question.
Title: If I give an object some potential energy, does its mass increase?
Post by: VernonNemitz on 03/02/2010 06:03:25
SpaceTime is no static geometry, not according to me at least?  As for that times arrow is about 'change' or as some say 'events' there is no doubt.  But the question I asked was how you thought about 'changes' taking shorter or longer 'time' to happen depending on your frame of observation.
I said that my ideas on this topic were not fully fleshed out.  But certainly time is the thing that allows one geometrical arrangement to become a different geometrical arrangement.  One particle can at most have one interaction during one time quantum.  This is not the same thing as saying that one time quantum only allows one interaction in the Universe to take place.  The time quantum affects the entire universe at once; different interactions happen in different locations during that quantum of time.  The type of interaction that each particle engages in, during that quantum of time, depends on the particle's neighbors, not on the time quantum.

If you think of time 'changes' and watch that clock (observing it) hang at the event horizon, there are going to be an awful lot of changes happening to you before that clock ever moves a inch :) you will have died long before that. And that was the crux of it to me. How you saw those 'time quanta' as differing between different observations from different 'frames of reference'. The clock tick as usual from its own perspective, but you, the earth and the universe is accelerated and blueshifted, changing at a ever increasing pace as it falls in.
I don't see any time quantum as being different from any other time quantum.  Each one is the same, Universe-wide.  The interactions that occur, at different places in the universe, depend only on what neighbors each particle is interacting with.
A clock that has no other neighbors will have all its parts interacting only with each other during each time quantum.  A clock with lots of neighbors will have at least some of its parts interacting with those neigbhors, during each time quantum.  The net effect is that the isolated clock runs faster, because it takes more time-quanta for the crowded clock to do the same purely internal interactions as the isolated clock.

So how do you equate those two observations with your 'time quanta'?  Consider A here as the the minute hand ticking one minute as seen from the frame of the in-falling clock.  Here is 'time quanta A' as seen from the frame of the clock, with its time 'as usual' so to speak
<-A->

Here it is from the perspective of you observing the clock.

<-------------------A------------------->

That 'time quanta', lifted from the perspective of the clock, will now contain a infinite amount of 'changes/events' from your perspective observing. You can travel home to Earth and come back without that clock hand ever moved according to you.
No infinity; the available evidence strongly suggests the Universe is finite.  (Example, an infinite universe would be associated with infinite gravitation, and we'd all be in a black hole as a result.)  The reference frames you describe are ignoring the available neighbors, for interacting.  Remember that even a vacuum is full of virtual particles with which interactions can occur, and a speeding reference frame can encounter more of those virtual particles than a stationary reference frame, in any specified number of time quanta (the speeding frame has more neighbors).
Title: If I give an object some potential energy, does its mass increase?
Post by: Farsight on 03/02/2010 12:25:25
Farsight, it occurs to me that another reason you are wrong about the plate acquiring significant potential-energy-as-mass relates to the photon I described at the end of my last message here.  The photon would come away from the black hole almost as blue as when it was created, if you were right about where the potential energy goes!  Therefore it is I that am correct; most of the energy that becomes potential, as a small body or photon escapes the gravitational field of a larger body, goes to increasing the mass-energy of the larger body, not the smaller body or photon.  Do remember that the most important thing about General Relativity is that it is about "mass-energy" regardless of form.  You can't have a double-standard, treating the plate one way and the photon another.
I don't treat them differently Vernon. The important  thing is that a photon doesn't actually change energy when it enters a gravitational field. There's no evidence of any energy transfer into the photon via unseen particles, the photon is the only particle there, and we must abide by conservation of energy. Yes, the photon appears to gain energy, for example it's measurably blue-shifted. But the frequency hasn't actually changed. Your measuring devices have, because they're subject to gravitational time dilation. Your clocks run slower, so you see the frequency as increased. It's a little like the way something feels warmer if you're cold.
Title: If I give an object some potential energy, does its mass increase?
Post by: VernonNemitz on 03/02/2010 14:31:16
Farsight, it occurs to me that another reason you are wrong about the plate acquiring significant potential-energy-as-mass relates to the photon I described at the end of my last message here.  The photon would come away from the black hole almost as blue as when it was created, if you were right about where the potential energy goes!  Therefore it is I that am correct; most of the energy that becomes potential, as a small body or photon escapes the gravitational field of a larger body, goes to increasing the mass-energy of the larger body, not the smaller body or photon.  Do remember that the most important thing about General Relativity is that it is about "mass-energy" regardless of form.  You can't have a double-standard, treating the plate one way and the photon another.
I don't treat them differently Vernon. The important  thing is that a photon doesn't actually change energy when it enters a gravitational field. There's no evidence of any energy transfer into the photon via unseen particles, the photon is the only particle there, and we must abide by conservation of energy. Yes, the photon appears to gain energy, for example it's measurably blue-shifted. But the frequency hasn't actually changed. Your measuring devices have, because they're subject to gravitational time dilation. Your clocks run slower, so you see the frequency as increased. It's a little like the way something feels warmer if you're cold.
You do indeed treat them differently, because for the plate you continue to say it gains mass when it is forced out of a gravitational field.  Dude, mass-energy is mass-energy, and G.R. makes NO distinction about the FORM in which it appears; all forms interact gravitationally, equally-in-manner (differently depending only on magnitude).  This means you are employing a double-standard, between your description of the plate and your description of the photon.  Just imagine an extremely energetic photon having the same magnitude of mass-energy as the just-accelerated plate, at the surface of a neutron star.  When both exit the star they must still have mass-energy equal to each other.  So, if you want to claim the mass of the plate has increased, in acquiring potential energy, then you must also claim the photon has become equivalently more energetic, also.  Since you don't, it means you are violating the basic principles of General Relativity; it means you don't know what you are talking about.
Title: If I give an object some potential energy, does its mass increase?
Post by: yor_on on 03/02/2010 15:54:49
Okay Vernon, I'm still not sure how you see it, but you're at the start of it as I understood. Virtual particles are defined as being outside Planck time and GR says that there is no measurable 'energy' to space as I understands it. Let's put it like this, if space had an measurable energy then it would have to be a 'medium' too as I see it. Doesn't mean you can't have 'virtual particles' and 'vacuum energy', as long as they don't make any measurable 'dent' of their own at SpaceTime. That we see indirect evidence seems to be allowable by GR though?

I'm looking forward to see you 'flesh your ideas out' Vernon. I'm also wondering about 'times arrow' :)
Title: If I give an object some potential energy, does its mass increase?
Post by: Geezer on 04/02/2010 06:21:24
Hi!

We seem to be drifting off topic just a tad here.

Anyone is welcome to propose a new theory under that heading, but it might be best if we try to answer the topic question in terms of well accepted science.

Or should we simply lock the thread?

Geezer (Mod)
Title: If I give an object some potential energy, does its mass increase?
Post by: yor_on on 04/02/2010 15:56:21
Well Geezer, you're right :)

Thinking of my idea of if the plate would 'jiggle' less if lifted up on that table instead of being left at the ground made me reevaluate how I understand gravity. As I said I believe it to be a 'geometry'. And as such it's not a 'force', but it still doesn't explain how it can be in all points (matter and space) and also work from each one of those points as if that would be the definite 'center'. If you look at Earth we say that in the middle of it, gravity will be 'nulled' as the 'mass' around that specific point will take out all gravitational forces (as I understand it). So gravity act both as each point would be its center as well as being able to add those points together to create an additive 'force'. And yes, suddenly I'm back to call it a force. ain't I :)

The thing here is, how can a bare geometry be additive?
What allows it that?
Title: If I give an object some potential energy, does its mass increase?
Post by: Geezer on 04/02/2010 20:07:27
Thanks Yoron  [;D]

There are several posts on this topic and I'm reasonably confident I'm not going to trawl through the lot, so could someone perhaps try to summarize for a poor old geezer where we are?

Something along the lines of:

According to classical mechanics, mass is not added because of  - insert formula here
According to GR, mass is added because of - insert formula here
According to so-and-so's theory, mass turns into a white rabbit because of - insert formula here

Something along those lines might help someone less "skilled-in-the-art" to understand what we're on about here.
Title: If I give an object some potential energy, does its mass increase?
Post by: yor_on on 04/02/2010 22:26:30
The white rabbit I might be able to help with? As for defining the math for mass in general relativity?
I'm afraid that that would be a thread all of its own :)

---Quote---

Generalizing this definition to general relativity, however, is problematic; in fact, it turns out to be impossible to find a general definition for a system's total mass (or energy). The main reason for this is that "gravitational field energy" is not a part of the energy-momentum tensor; instead, what might be identified as the contribution of the gravitational field to a total energy is part of the Einstein tensor on the other side of Einstein's equation (and, as such, a consequence of these equations' non-linearity). While in certain situation it is possible to rewrite the equations so that part of the "gravitational energy" now stands alongside the other source terms in the form of the Stress-energy-momentum pseudotensor, this separation is not true for all observers, and there is no general definition for obtaining it.

------end of quote---

Just as a by side if you look at this idea of using 'Stress-energy-momentum pseudotensors' Michael Weiss and John Baez (http://math.ucr.edu/home/baez/physics/Relativity/GR/energy_gr.html) have this to say about such.

"Mathematicians invented tensors precisely to meet this sort of demand -- if a tensor equation holds in one coordinate system, it holds in all.  Pseudo-tensors are not tensors (surprise!), and this alone raises eyebrows in some circles.  In GR, one must always guard against mistaking artifacts of a particular coordinate system for real physical effects.  (See the FAQ entry on black holes for some examples.) These pseudo-tensors have some rather strange properties.  If you choose the "wrong" coordinates, they are non-zero even in flat empty spacetime.  By another choice of coordinates, they can be made zero at any chosen point, even in a spacetime full of gravitational radiation.  For these reasons, most physicists who work in general relativity do not believe the pseudo-tensors give a good local definition of energy density, although their integrals are sometimes useful as a measure of total energy."

Now, what the heck did this mean "In fact, it turns out to be impossible to find a general definition for a system's total mass (or energy). The main reason for this is that "gravitational field energy" is not a part of the energy-momentum tensor;" ?

Well, 'gravitational field energy' I take to be gravity when seen/treated as a field.
And the 'energy-momentum tensor'? Hold on to your hats now.

--Quote(s)----

The stress-energy tensor (sometimes stress-energy-momentum tensor) is a tensor quantity (A generalization of the concept of a vector) in physics that describes the density and flux of energy and momentum in spacetime, generalizing the stress tensor of Newtonian physics. It is an attribute of matter, radiation, and non-gravitational force fields. The stress-energy tensor is the source of the gravitational field in the Einstein field equations of general relativity, just as mass is the source of such a field in Newtonian gravity.

=
And
=

The Einstein field equations (EFE) or Einstein's equations are a set of ten equations in Einstein's theory of general relativity which describe the fundamental interaction of gravitation as a result of spacetime being curved by matter and energy.[1] First published by Albert Einstein in 1915[2] as a tensor equation, the EFE equate spacetime curvature (expressed by the Einstein tensor) with the energy and momentum within that spacetime (expressed by the stress-energy tensor).

---End of quote(s)---

Now;

"The math behind general relativity is called Einstein Field Equations. They are equations of the coupled hyperbolic-elliptic nonlinear partial differential type, which, in plain English, means that they are really, really hard. Einstein himself recognized the mathematical difficulties of general relativity as "very serious." He predicted it as being the primary hindrance of general relativity's development. The equation can be stated in a "symbolic form" that isn't very useful. Here it is:

(https://www.thenakedscientists.com/forum/proxy.php?request=http%3A%2F%2Flibrary.thinkquest.org%2FC008537%2Frelativity%2Fmath%2Fm12.gif&hash=172e2dce7d28b7e8e8e1a635d81f6fff)

It doesn't mean much to us, but you can see on the left of the equal sign the stuff that describes the curvature of space-time. On the right is the matter within space-time, and how it behaves. " Taken from Relativity-Math (http://library.thinkquest.org/C008537/relativity/math/math.html)

Okay, the real problem with defining a mass in general relativity, as I understands it, have to do with that it's okay to labor with different definitions for 'mass' as long as we are doing it from a spatial infinity, that is, not trying to define it locally.

---Quote---

Since the 1970s, physicists and mathematicians have worked on the more ambitious endeavor of defining suitable quasi-local quantities, such as the mass of an isolated system defined using only quantities defined within a finite region of space containing that system. However, while there is a variety of proposed definitions such as the Hawking energy, the Geroch energy or Penrose's quasi-local energy-momentum based on twistor methods, the field is still in flux.

---End of quote-----

So we don't seem to have any stringent mathematical definition for mass locally in General Relativity as I understands it? And if I'm wrong :) Correct it please..

Mathematics_of_general_relativity (http://en.wikipedia.org/wiki/Mathematics_of_general_relativity)
Einstein_field_equations (http://en.wikipedia.org/wiki/Einstein_field_equations)
Mass_in_general_relativity (http://en.wikipedia.org/wiki/Mass_in_general_relativity)

Title: If I give an object some potential energy, does its mass increase?
Post by: Geezer on 05/02/2010 05:41:28
Hey! That's a very helpful summary. Thank you sir.

As you suggest, it's certainly a great target to stick on the wall.

Title: If I give an object some potential energy, does its mass increase?
Post by: VernonNemitz on 05/02/2010 18:16:59
Just in case there was some misunderstanding about all the G.R. stuff a couple of posts ago, I'd merely like to point out that in G.R. the gravitational field is considered to be a form of energy, which contributes to the intensity of the gravitational field.  This contribution is used to explain some observed extra motion in Mercury's orbit, that classical gravitational physics could not explain.  But by inspection, it leads to an infinte series: if gravitation itself can enhance gravitation, then where do the enhancements end? Well, we know that some infinite series can be added up to a finite sum, and obviously G.R. has to do that (and Q.M. will need to do the same if it is ever to explain gravitation).  I think this is the cause of the mis-match in the equations that were being described in that earlier post.  You can't have mass without also having the gravitational field of that mass, and the associated self-enhancement.  So what magnitude of "mass" do you want to describe?  The initial mass without the gravity field, or the mass that includes the equivalence of the immediatedly-associated gravitational field, or the mass that includes both that immediate gravitational field and the first level of enhanced field, or...?

Meanwhile, every mass is located within the gravity well of some other mass!  In nuclear physics they talk about "negative binding energy" a lot; it is a trick that allows potential energy to be kept distinct from the mass of a particle.  Two particles closely interacting via the Strong Nuclear Force have a lot more negative binding energy than two widely-separated particles (but the separated particles have more potential energy).  Inside some other body's gravitational field, a mass can be associated with different amounts of potential energy (depending on location), and G.R. needs to take this into account, also, when trying to describe the mass of a particular object.

Personally, I think that some of the headache can be solved by simply saying that potential energy always takes the form of mass.  Farsight says the same thing, but we disagree on details (how some potential energy is divided between two interacting masses).
Title: If I give an object some potential energy, does its mass increase?
Post by: yor_on on 06/02/2010 18:49:09
Yes Vernon, gravitational waves seems to be some sort of energy?
But It seems to be so that both statements exist side by side?
Yep, my headache is getting worse again :)
Title: If I give an object some potential energy, does its mass increase?
Post by: VernonNemitz on 08/02/2010 15:29:18
Yes Vernon, gravitational waves seems to be some sort of energy?
But It seems to be so that both statements exist side by side?
Yep, my headache is getting worse again :)
Do not confuse a gravitational field with a gravitational wave; they are not identical things, like an electromagnetic field and an electromagnetic wave are not identical things (you don't confuse the field surrounding a magnet with a radio wave, do you?).  Both have energy, though; both are different forms of energy.  In Quantum Mechanics the two things are related (radio waves consist of real-energy photons and a magnetic field consists of virtual-energy photons), and should QM one day describe gravity adequately, a similar description could be expected regarding gravitational waves and fields.  General Relativity does not offer such a close relationship, though; the gravitational field is simply a fixed region of curved space, and a gravitational wave is a moving ripple of space.
Title: If I give an object some potential energy, does its mass increase?
Post by: yor_on on 12/02/2010 20:43:38
"General Relativity does not offer such a close relationship, though; the gravitational field is simply a fixed region of curved space, and a gravitational wave is a moving ripple of space."

Lovely expression Vernon, I quite agree. And as all ideas of 'energy' seems to build on a transformation can we say that gravity 'transforms' anything? Well, that gravitational wave might change (transform) your geometry but it still seems that from any chosen frame you will be the same, 'internally' in that frame, so to speak. So how about 'potential energy' then? What the heck is that 'energy' we're talking about. Isn't it just an abstraction describing a relation?
Title: If I give an object some potential energy, does its mass increase?
Post by: VernonNemitz on 16/02/2010 06:44:05
...as all ideas of 'energy' seems to build on a transformation can we say that gravity 'transforms' anything?
Try walking around on the surface of a neutron star, and your shape will be transformed significantly.  Something like a billion gees, there.  To properly answer your question we probably have to be sure we are talking about exactly the same thing.  What is "gravity"?  For this Message I will start by choosing to define that as, "It is a thing that can tend to keep all sorts of slow masses in close proximity to each other."  Thus an Answer to your question appears: a role that involves the word "keep" does not automatically include "change" or "transform".  We thus realize that the definition is incomplete; the more mass is involved, the more powerfully gravity can keep objects in the vicinity of each other.  Further, we know of an additional incompleteness in the definition, because we know that gravity can eventually cause initially distant things to collide.  THAT means that "change" does indeed need to be involved in the definition!  Newton's equation for gravity is a most excellent first-approximation definition.  And, despite more modern definitions, gravity can be called a "force" simply because it can cause masses to accelerate --the essence of "force" is the acceleration of a mass, after all.  Not even Einstein could change the definition of "force" in Physics.
So how about 'potential energy' then? What the heck is that 'energy' we're talking about. Isn't it just an abstraction describing a relation?
You can't get something for nothing, that we know of.  The acceleration of a mass is always associated with a change in the kinetic energy of the mass.  If it increases, we need to know where it comes from, and if it decreases, we need to know where it goes.  "Potential" energy is just a way to talk about "storage".  It gives us a place to say, "There is where the kinetic energy went!" when we hurl a rock upwards toward some lofty height, and see it slow down to a vertical velocity of zero.  And of course it gives us a place to say, "There is where the kinetic energy came from" when we see the rock acquiring an increasing downward velocity.

The whole point of this entire Message Thread is the attempt to answer a question regarding the FORM of "potential" energy.  What are the properties of that "storage system"?  Newton offered a nice explanation, which I shall phrase as: "Keeping the gravity gradient in mind, with distance from the center of mass of the main gravitating body, potential energy can be represented as a simple magnitude of its lofty height above the surface of that mass."  Since different masses can have different gravity gradients, two identical objects at equal heights above those different masses can have different amounts of potential energy.

My personal objection to that explantion has three parts.  First, we know by direct measurement that when potential energy is associated with the Strong Nuclear Force, the Weak Nuclear Force, and the ElectroMagnetic Force, the form of that potential energy is always "mass".  Second, Newton did not know that E=mc2.  And third, physicists desire to Unify/simplify the descriptions of the Natural Forces, including "Gravity".  This makes it very very easy for me to be willing to assume that with respect to Gravity, potential energy also takes the form of mass.

However, if the overall logic is to be completely consistent (and as I've pointed out to Farsight in various messages), The Answer To This Message Thread's Main Topic Question ("If I give an object some potential energy [presumably by lofting it], does its mass increase?") would then be: "Yes and no.  First, due to E=mc2, the initially-supplied kinetic energy, that becomes lofty potential energy, only becomes an exceedingly tiny quantity of mass in most environments influenced by Gravity (so if you can't measure an increase in mass, can you really say that its mass increased?).  Second, most of the mass that does appear needs to become proportionately more-added-to the main source of the gravitational field, than added-to the lofted object (making the amount of increase for the lofted object even less measurable)."  If the overall logic is to be completely consistent, I repeat!
Title: If I give an object some potential energy, does its mass increase?
Post by: Farsight on 16/02/2010 11:46:21
I'm glad we seem to be agreed on that the potential energy goes into mass, as slight as it might be. But Vernon, you need to look at that distribution again. Think about ½mv² in the context of an explosion. Consider a 1kg cannonball fired by a 1000kg cannon. Conservation of momentum p=mv tells you the cannonball velocity is 1000 times the cannon recoil velocity. Now apply ½mv² and you see that more of the energy goes into the cannonball. Now embed the cannon firmly into the solid rock of the earth (mass 5.9736 × 1024 kg), and fire the cannonball upwards at 11.2 km/s. Most of the energy goes into the cannonball, and since 11.2 km/s gives it escape velocity, this energy is then lost from the earth system.
Title: If I give an object some potential energy, does its mass increase?
Post by: VernonNemitz on 16/02/2010 14:19:33
Farsight, I see you have ignored my message of 03/02/2010 14:31:16 (slightly edited in copying below).  As I wrote earlier today, "If the overall logic is to be completely consistent", then I'm talking about consistency with other known gravitational effects, where General Relativity is better at describing things than Newton.
Dude, mass-energy is mass-energy, and G.R. makes NO distinction about the FORM in which it appears; all forms interact gravitationally, equally-in-manner (differently depending only on magnitude).  This means you are employing a double-standard, between your description of the plate and your description of [a] photon.  Just imagine an extremely energetic photon having the same magnitude of mass-energy as the just-accelerated [cannonball], at the surface of a neutron star.  When both exit the star they must still have mass-energy equal to each other.  So, if you want to claim the mass of the [cannonball] has increased, in acquiring potential energy, then you must also claim the photon has become equivalently more energetic, also.  Since you don't, it means you are violating the basic principles of General Relativity; it means you don't know what you are talking about.
In other words, Farsight, you are not being consistent!  At the moment of acceleration, the mass-energy of the cannonball is increased significantly, due to kinetic energy being added to it.  Let us now imagine two photons, one (Photon A) equal in energy to the mass of the cannonball just prior to it being accelerated, and one (Photon B) equal to the total mass-energy of the cannonball just after its acceleration.  Upon being declared to have basically escaped the gravity field of the neutron star, we know that the two photons are seriously red-shifted (lost a lot of energy), and that the cannonball is moving lots slower (also lost a lot of energy).  But Photon B must still have the same total mass-energy as the cannonball.  You cannot say that all the original kinetic energy of the ball, that has now mostly become potential energy, has become extra mass for the ball, since you know you cannot say an equivalent thing for Photon B.  You can say that the energy of Photon B is a bit more than the original energy of Photon A, when it was at the surface of the neutron star (and was equal in energy to the mass of the not-yet-accelerated ball).  But only a bit more; the majority of energy-that-became-potential-energy-in-the-form-of-mass had to appear somewhere else:  It was added to the mass of neutron star, extracted from both the photons' original energy and from the accelerated cannonball's energy.  In simple-minded terms, as I explained in another message quite some time ago, we can say that gravitational interactions literally suck energy from escaping objects, and give energy to falling objects (by "pulling" on them, a thing typically associated with, yes, "force", and the-supplying-of-energy).  This is known as "balance", and is again consistent; it is even consistent with the idea of Gravitation being an aspect of a system (two interacting objects).  Your description, Farsight, basically breaks the system as soon as you apply sufficient kinetic energy to a mass, not after the mass has physically escaped.  (Why?  Because you want all the applied kinetic energy to be/stay part of the accelerated mass only!)  Sorry, but the system doesn't break that easily; Photon B and the cannonball will have the same mass-energy after escaping the star (--or the Earth, but the effects are much less obvious here).
Title: If I give an object some potential energy, does its mass increase?
Post by: Farsight on 16/02/2010 15:48:31
Farsight, I see you have ignored my message of 03/02/2010 14:31:16 (slightly edited in copying below).
Sorry, I simply missed it.

As I wrote earlier today, "If the overall logic is to be completely consistent", then I'm talking about consistency with other known gravitational effects, where General Relativity is better at describing things than Newton.
You know I agree with that. Let's have a careful look at what you're saying then.

Dude, mass-energy is mass-energy, and G.R. makes NO distinction about the FORM in which it appears; all forms interact gravitationally, equally-in-manner (differently depending only on magnitude).
I'd say energy is energy, and it can appear in a form that confers mass.

This means you are employing a double-standard, between your description of the plate and your description of [a] photon.
I'm not. But let's press on.

Just imagine an extremely energetic photon having the same magnitude of mass-energy as the just-accelerated [cannonball], at the surface of a neutron star.  When both exit the star they must still have mass-energy equal to each other.
I think I see your problem. The total energy of the moving cannonball comprises the energy locked up as its mass, plus the kinetic energy. When it has escaped the neutron star, the total energy is unchanged. The same is true of the photon.

So, if you want to claim the mass of the [cannonball] has increased, in acquiring potential energy, then you must also claim the photon has become equivalently more energetic, also. Since you don't, it means you are violating the basic principles of General Relativity...
I think the phrase mass-energy has caused confusion. We say "mass is invariant", and that the mass of a cannonball moving at 11.2 km/s is the same as that of a motionless cannonball at the same location. The total energy for the moving cannonball is however greater.

In other words, Farsight, you are not being consistent! At the moment of acceleration, the mass-energy of the cannonball is increased significantly, due to kinetic energy being added to it.
The total energy or relativistic mass is increased significantly, but the mass is not.

Let us now imagine two photons, one (Photon A) equal in energy to the mass of the cannonball just prior to it being accelerated, and one (Photon B) equal to the total mass-energy of the cannonball just after its acceleration.
OK.

Upon being declared to have basically escaped the gravity field of the neutron star, we know that the two photons are seriously red-shifted (lost a lot of energy), and that the cannonball is moving lots slower (also lost a lot of energy).
Wrong. Conservation of energy applies. If they lost energy, where did it go? The cannonball kinetic energy has been transformed into cannonball potential energy. We know it's cannonball potential energy, because if that cannonball escapes the system, it takes it away.

But Photon B must still have the same total mass-energy as the cannonball.
Absolutely correct.

You cannot say that all the original kinetic energy of the ball, that has now mostly become potential energy, has become extra mass for the ball, since you know you cannot say an equivalent thing for Photon B.
I know I can. A photon does not lose energy when it climbs out of a gravitational field. It's appears red-shifted only because the environment is no longer subject to gravitational time dilation. Its frequency hasn't changed.

You can say that the energy of Photon B is a bit more than the original energy of Photon A, when it was at the surface of the neutron star (and was equal in energy to the mass of the not-yet-accelerated ball).  But only a bit more; the majority of energy-that-became-potential-energy-in-the-form-of-mass had to appear somewhere else:  It was added to the mass of neutron star, extracted from both the photons' original energy and from the accelerated cannonball's energy.  In simple-minded terms, as I explained in another message quite some time ago, we can say that gravitational interactions literally suck energy from escaping objects, and give energy to falling objects (by "pulling" on them, a thing typically associated with, yes, "force", and the-supplying-of-energy).  This is known as "balance", and is again consistent; it is even consistent with the idea of Gravitation being an aspect of a system (two interacting objects).
That's wrong Vernon. That's the diference between Newtonian mechanics and relativity.

Now, take a look at that example I gave you which explains why most of the energy goes into the cannonball rather than the Earth.

Title: If I give an object some potential energy, does its mass increase?
Post by: VernonNemitz on 16/02/2010 17:26:21
Just imagine an extremely energetic photon having the same magnitude of mass-energy as the just-accelerated [cannonball], at the surface of a neutron star.  When both exit the star they must still have mass-energy equal to each other.
I think I see your problem. The total energy of the moving cannonball comprises the energy locked up as its mass, plus the kinetic energy. When it has escaped the neutron star, the total energy is unchanged. The same is true of the photon.
Wildly false, and the core of our disagreement.  In more detail, consider a just-created photon at the surface of a neutron star; it contains some X amount of energy.  If the photon travels sideways and is absorbed by a measuring instrument, we detect X amount of energy.  But if the photon climbs out of the gravity well, it is red-shifted; an instrument will not detect all of that X amount of energy.  See this classic physics joke:
http://books.google.com/books?id=UGGhM2XKE_0C&pg=PA6&lpg=PA6&dq=%22soaked+in+a+gravitational+field%22&source=bl&ots=o6DWAZarZM&sig=nitPtT1Paa_loBYnLR3i2beP2PY&hl=en&ei=dc16S73oOYaN8AaOhoX0CQ&sa=X&oi=book_result&ct=result&resnum=1&ved=0CAcQ6AEwAA#v=onepage&q=%22soaked%20in%20a%20gravitational%20field%22&f=false (scroll to see a bit more at the top of the page, after clicking the link)

Such "soaking" is exactly equivalent to a cannonball shot out of the gravity well; its observed total mass-energy after escaping is less than its total when just-accelerated.  You might want to claim that the total has remained the same, by having its initial kinetic energy become extra mass (potential-energy-stored-as-mass) as it arose/slowed, but there is no equivalent argument for the massless photon, and that is why your claim is invalid.

Now I'm aware you might want to argue that I'm talking about measurements in two different reference frames, at the surface of the neutron star and out in Space far away from it.  Such an argument is partly a "red herring", because it fails to take into account one key detail.  To see it, let's use the same measuring device in both frames.  Sure, at first glance you might think the device that measured X photon-energy the surface of the star will also measure X photon-energy, for the same photon, far away from the star (per relativity; both the device and the photon are affected simultaneously/together/synchronously) --but the device was stationary at the surface of the star!  You have to add a lot of kinetic energy to the measuring device, to get it away from the star!  When you do that, you are not adding energy to the photon (no need; since we know the photon can escape).  The synchronicity between the photon and the measuring device is now broken.  Thus, far from the star, the total mass-energy of the measuring device can (in my view "A") be considered to have diminished to approximately its original value, and therefore can detect some red-shift (loss of energy) of the photon, or (in your view "B") be considered to have not diminished, and therefore can detect an even larger red-shift for the photon!  (That's relativity in actuality!)  Note that either way, there is a red-shift measurable!  In short, your assumptions are faulty (especially the one about the photon's energy not changing), and therefore your argument is worthless.

Conservation of energy applies. If they lost energy, where did it go? The cannonball kinetic energy has been transformed into cannonball potential energy. We know it's cannonball potential energy, because if that cannonball escapes the system, it takes it away.
Conservation of energy most certainly applies, and you are once again ignoring something I wrote to you a while back.
Really bad logic. The [cannonball] is considered to be separate from the Earth as soon as you start treating it separate from the Earth.  That means even when falling off a cliff on Earth, it is not part of the Earth; it is part of the Earth/[cannonball] SYSTEM.  Therefore the Earth does not lose the mass of the [cannonball] when the [cannonball] is given an escape velocity; only the Earth/[cannonball] system loses it.  And the Earth still sucked about 11kps of velocity and associated kinetic energy from that escaping [cannonball]; the [cannonball] most certainly does not have it while traversing interplanetary space.  All it has is the potential to fall down a different gravity well, but that well is defined by the planet at its bottom, not by the presence of the [cannonball].  That is, the amount of potential energy that the [cannonball] can acquire by falling down that well is defined by the planet at the bottom of the well.  If the planet has the same mass as the Earth, then the potential energy the [cannonball] can acquire will be the same as if it could acquire in falling to Earth.
This is why the cannonball and the photon can both lose energy, which becomes potential-energy-stored-as-extra-mass-of-the-main-gravitating-body.  And the next such body that either encounters will have the potential energy to give away, to anything falling toward that body!
Title: If I give an object some potential energy, does its mass increase?
Post by: yor_on on 16/02/2010 18:37:47
Vernon I wasn't discussing a Black Hole there was I? I thought I was commenting on a gravitational wave :) Which we have a reasonable expectation to exist. As for "I'm glad we seem to be agreed on that the potential energy goes into mass, as slight as it might be." Farsight? When did I do that? I thought I called it a relation, just as I see gravity. Relations all of them :)

As for what Gravity is Vernon. To me it's the geodesics of SpaceTime.
Don't get why you read me different here?

Hope I cleared that up now.
Title: If I give an object some potential energy, does its mass increase?
Post by: Farsight on 16/02/2010 23:02:13
Wildly false, and the core of our disagreement.  In more detail, consider a just-created photon at the surface of a neutron star; it contains some X amount of energy.  If the photon travels sideways and is absorbed by a measuring instrument, we detect X amount of energy.  But if the photon climbs out of the gravity well, it is red-shifted; an instrument will not detect all of that X amount of energy...
Because the instrument is in a different environment. It isn't subject to gravitational time dilation, so it of necessity measures the photon frequency as reduced.

...Such "soaking" is exactly equivalent to a cannonball shot out of the gravity well; its observed total mass-energy after escaping is less than its total when just-accelerated. You might want to claim that the total has remained the same, by having its initial kinetic energy become extra mass (potential-energy-stored-as-mass) as it arose/slowed, but there is no equivalent argument for the massless photon, and that is why your claim is invalid.
It isn't invalid. Conservation of energy says it isn't.

Now I'm aware you might want to argue that I'm talking about measurements in two different reference frames, at the surface of the neutron star and out in Space far away from it.
Yes, that's the crux of it.

Such an argument is partly a "red herring", because it fails to take into account one key detail. To see it, let's use the same measuring device in both frames. Sure, at first glance you might think the device that measured X photon-energy the surface of the star will also measure X photon-energy, for the same photon, far away from the star (per relativity; both the device and the photon are affected simultaneously/together/synchronously) - but the device was stationary at the surface of the star! You have to add a lot of kinetic energy to the measuring device, to get it away from the star!
If that device is at the surface, then to see the photon, the photon has to come back down. Sorry.

When you do that, you are not adding energy to the photon (no need; since we know the photon can escape). The synchronicity between the photon and the measuring device is now broken. Thus, far from the star, the total mass-energy of the measuring device can (in my view "A") be considered to have diminished to approximately its original value, and therefore can detect some red-shift (loss of energy) of the photon, or (in your view "B") be considered to have not diminished, and therefore can detect an even larger red-shift for the photon! (That's relativity in actuality!) Note that either way, there is a red-shift measurable! In short, your assumptions are faulty (especially the one about the photon's energy not changing), and therefore your argument is worthless.
It isn't. Yes, we measure a redshift, but we know our instruments are affected by gravitational time dilation. You're not accounting for it.

Conservation of energy most certainly applies, and you are once again ignoring something I wrote to you a while back.
Really bad logic. The [cannonball] is considered to be separate from the Earth as soon as you start treating it separate from the Earth.  That means even when falling off a cliff on Earth, it is not part of the Earth; it is part of the Earth/[cannonball] SYSTEM.  Therefore the Earth does not lose the mass of the [cannonball] when the [cannonball] is given an escape velocity; only the Earth/[cannonball] system loses it. And the Earth still sucked about 11kps of velocity and associated kinetic energy from that escaping [cannonball]; the [cannonball] most certainly does not have it while traversing interplanetary space.  All it has is the potential to fall down a different gravity well, but that well is defined by the planet at its bottom, not by the presence of the [cannonball]. That is, the amount of potential energy that the [cannonball] can acquire by falling down that well is defined by the planet at the bottom of the well. If the planet has the same mass as the Earth, then the potential energy the [cannonball] can acquire will be the same as if it could acquire in falling to Earth.
This is why the cannonball and the photon can both lose energy, which becomes potential-energy-stored-as-extra-mass-of-the-main-gravitating-body. And the next such body that either encounters will have the potential energy to give away, to anything falling toward that body!
Vernon, no. When you give that cannonball its 11.2 km/s of kinetic energy, it has more total energy than it had when it was motionless. All of this total energy is lost from the earth/cannonball system when the cannonball departs that system. The earth does not retain this energy, otherwise where does the 11.2km/s of kinetic energy come from when the cannonball falls down to another Earth? This other earth gains the mass of the cannonball along with that kinetic energy. As a result the gravitational field of this new earth will be increased. Conservation of energy, Vernon. Don't forget it.

Title: If I give an object some potential energy, does its mass increase?
Post by: VernonNemitz on 17/02/2010 00:18:01
Such an argument is partly a "red herring", because it fails to take into account one key detail. To see it, let's use the same measuring device in both frames. Sure, at first glance you might think the device that measured X photon-energy the surface of the star will also measure X photon-energy, for the same photon, far away from the star (per relativity; both the device and the photon are affected simultaneously/together/synchronously) - but the device was stationary at the surface of the star! You have to add a lot of kinetic energy to the measuring device, to get it away from the star!
If that device is at the surface, then to see the photon, the photon has to come back down. Sorry.
You STILL are ignoring what I wrote!  Pay attention!
In more detail, consider a just-created photon at the surface of a neutron star; it contains some X amount of energy.  If the photon travels sideways and is absorbed by a measuring instrument, we detect X amount of energy.  But if the photon climbs out of the gravity well, it is red-shifted; an instrument will not detect all of that X amount of energy.
Care to try again, Farsight, with a less-obviously ridiculous remark?  (There's no reason for me to write more until after I know you fully understand the starting point of this thought-experiment --which, currently, and obviously, you don't, not in the slightest.)
Title: If I give an object some potential energy, does its mass increase?
Post by: VernonNemitz on 17/02/2010 19:23:21
A photon does not lose energy when it climbs out of a gravitational field. It's appears red-shifted only because the environment is no longer subject to gravitational time dilation. Its frequency hasn't changed.
Here's another thing about which you are wrong, if less obviously so.  Interpretations matter!  Consider that Shrodinger's Wave Equation relates the momentum of a particle of matter with a wavelength.  And I assume you know that the momentum of a particle can depend on its mass, not just its velocity.

SO:  If we are agreed that a cannonball or plate has less mass at the bottom of a gravity well, than at the top, then the Wave Equation automatically associates every particle of that mass (at the bottom of the well) with longer wavelengths, as if "time dilation" was in effect.  In other words, you don't need to assume that time dilation is an intrinsic part of a gravity well; it is a simple logical consequence --a mere interpretation!-- associated with the idea of potential-energy-as-mass being converted to other forms (and finally radiated away) when every particle of matter at the bottom of a gravity well got there and cooled down after colliding.

Well, if gravitational time dilation is just an interpretation, what of an incoming photon?  Obviously all it need do is also be affected, acquiring energy as it falls into a gravity well (just like happens to everything else that falls in; G.R. makes no distinction regarding the form of mass-energy, remember!), and losing it as it climbs out again.  A matter of interpretation, some might say --but yours is absolutely flawed because you are trying to create a distinction between photons and other forms of mass-energy, with respect to General Relativity).  Therefore the photon looks blue-shifted when arriving, and is seen by an instrument at the bottom of the well, because it is blue-shifted, and it looks red-shifted when leaving, and is seen by an instrument at the top of the well, because it is red-shifted.
Title: If I give an object some potential energy, does its mass increase?
Post by: JP on 18/02/2010 02:47:31
So to come back to the original topic, I see that only a couple of posts have mentioned the stress-energy tensor, which is the tool needed to describe how an object bends space-time.  The point is that you need the whole tensor to make the equations of GR invariant under a transformation of coordinates (i.e. reference frame), so it seems to me that you can't just pull out the "mass" (which should be contained in the energy part of the tensor) and say that's a physically meaningful quantity in all reference frames.  You need the whole tensor or else you're dealing with a quantity that is only valid in your particular frame.

And if you want to also include gravitational energy in this discussion, you have to go further and define a psuedotensor, as yor_on pointed out (which isn't reference frame invariant).
Title: If I give an object some potential energy, does its mass increase?
Post by: VernonNemitz on 19/02/2010 14:32:12
So to come back to the original topic, I see that only a couple of posts have mentioned the stress-energy tensor, which is the tool needed to describe how an object bends space-time.  The point is that you need the whole tensor to make the equations of GR invariant under a transformation of coordinates (i.e. reference frame), so it seems to me that you can't just pull out the "mass" (which should be contained in the energy part of the tensor) and say that's a physically meaningful quantity in all reference frames.  You need the whole tensor or else you're dealing with a quantity that is only valid in your particular frame.
We aren't dealing with just one mass here.  There is the mass of the main gravitating body, of course, and for it the appropriate tensors would be necessary.  But for a small body such as a plate or a cannonball, we've been discussing how it behaves in the field of the other mass, and haven't been paying attention to its own (obviously relatively trivial) gravitational field.  So, the phrase, "you can't just pull out the 'mass'" is a bit misleading, since we are not at all talking about pulling out any of the mass of the main gravitating body; we are discussing a system of two interacting bodies, one of which is so small as to have a basically ignorable gravity field, and it's mass (or more accurately, mass-energy) is the only one being "pulled out".  And a very-much-tinier piece of mass energy is actually the main thing being argued about by Farsight and myself --the mass-equivalent of the potential-energy-associated-with-height of a plate or cannonball.  Its gravitational contribution is even more ignorable.
Title: If I give an object some potential energy, does its mass increase?
Post by: Farsight on 21/02/2010 13:09:45
You STILL are ignoring what I wrote!  Pay attention! ...
Let's just agree to differ, Vernon.

Yes, mass in general relativity is rather slippery, JP.
Title: If I give an object some potential energy, does its mass increase?
Post by: VernonNemitz on 22/02/2010 15:54:38
Let's just agree to differ, Vernon.
That's just a feeble excuse to let yourself keep thinking you have a valid argument.  Today I'm going to address one of your more fundamental errors.  This is the notion that energy cannot be undetectably transferred between objects; that is, after all, your primary excuse for erroneously thinking that Conservation of Energy requires all the kinetic energy that might be given to a plate or a cannonball to be retained by those objects, albeit in some form other than actual motion, when climing out of a gravity well.

The example-of-the-moment is the "simple" collision of one billiard ball against another, on a pool table.  Turn on your "mind's eye ultramicroscope" and look at the event really closely.  We see the surface of each billiard ball as consisting of atoms, clouds of electrons surrounding atomic nuclei.  More, we "see" the electric fields of those electrons, repelling other electrons.  The whole surface of each billiard ball is a thin layer of electrostatic repulsion; at a distance each ball electrically neutral, of course, but up-close-and-personal the story is very different!  (You are aware, are you not, that the electric force is something like 1040 times stronger than the gravitational force?)

Obviously, when the first billiard ball gets close enough to the other, those repulsive surface fields start to interact.  The balls never actually touch!  They simply get close enough for those mutually repelling electric fields to cause each ball as-a-whole to do the things that we typically observe at the ordinary macroscopic level.  And as you know, energy and momentum can be transferred during that collision event --across a distance!  Sure, the distance here is very tiny, but that is simply due to the generic cancellation of opposite electric fields, as seen from a distance, of protons and electrons.  It is necessary that ordinary physical objects approach very closely for the fine details (non-cancelled fields!) to be able to play a role in events.

The point I'm attempting to make is simple: A tiny distance for energy transferrance still qualifies as a "distance".  You erroneously think that energy can be transferred only when the distance is zero, between interacting bodies.  You are wrong because all ordinary interactions take place across distances!  The fact that usually the relevant distances are very tiny is totally irrelevant --the only reason those distances are tiny is because of charge-cancellation, as described above.  We don't normally encounter any field-cancellation of gravitation; therefore masses can interact at significant distances, transferring energy and momentum between them.

Perhaps instead of the above I should have just mentioned the phrase "gravitational slingshot"; according to your erroneous thinking, regarding Energy Conservation, the phenomenon that NASA has used on a number of space probe missions, to increase probe velocity, should never ever be possible.
Title: If I give an object some potential energy, does its mass increase?
Post by: Farsight on 22/02/2010 18:42:39
I mentioned the gravitational slingshot when we were talking about gravity and work, Vernon. See http://www.thenakedscientists.com/forum/index.php?topic=27444.msg291021#msg291021. A better example for "never actually touch" would have been a magnet. The reason a nail is attracted to a magnet is because its in a magnetic field, which is local to the nail. It's similar for a gravitional field. A field has energy, and is a non-uniform disposition of the space/energy caused by one object that has some effect upon another. All I can do is reiterate that when you fire a cannonball up into the sky at 11.2 km/s it escapes the Earth's gravitational field, which is then diminished. You really should look at the example I gave which explains why the energy goes into the cannonball rather than the far more massive Earth which has no detectable recoil.
Title: If I give an object some potential energy, does its mass increase?
Post by: VernonNemitz on 22/02/2010 20:37:26
I mentioned the gravitational slingshot when we were talking about gravity and work, Vernon. See http://www.thenakedscientists.com/forum/index.php?topic=27444.msg291021#msg291021.
So, you are indulging in hypocrisy? to claim that energy can transfer during a gravitational slingshot, but not transfer for other events?  Tsk, tsk.  Such inconsistency makes your argument even more worthless.

A better example for "never actually touch" would have been a magnet.
Wrong again.  It is not a better example because the point of my last message was about how "touching" is actually only an illusion.  Thus, to insist that energy can only be transferred when objects are in contact is to be wrong, since most interacting and energy-transferring objects are not actually engaging in the kind of contact you are imagining (a prime exception would be chemical reactions, where two molecules combine, but not relevant here).

The reason a nail is attracted to a magnet is because its in a magnetic field, which is local to the nail. It's similar for a gravitional field. A field has energy, and is a non-uniform disposition of the space/energy caused by one object that has some effect upon another.
Agreed, yet when a proton and an electron start out separated, and approach and interact and merge to form a hydrogen atom (radiating some energy), not only do physicists recognize that the atom has less mass-energy than the separated particles, they also are confident that the two particles maintain their mass ratio throughout the event.  Ditto for initially-separated proton and neutron combining to form a deuteron; their mass ratio stays constant.  I apply this to gravitation, keeping things consistent; you don't, because of your unwillingness to accept that energy can be undetectably tranferred across distance in more cases than those involving magnets or gravitational slingshots.

All I can do is reiterate that when you fire a cannonball up into the sky at 11.2 km/s it escapes the Earth's gravitational field, which is then diminished.
All you can do is repeat the same nonsense as before (but, I see, with less detail than before).  Yes, the velocity and kinetic energy of the cannonball is reduced during an escape from a gravitational field, and potential energy increases somewhere in the system while that happens.  Your feeble attempt to confuse the description of that system is noted and rejected.  We most certainly do not combine the mass of the cannonball with the mass of the main gravitating body, when computing the strength of the gravitational field that you are intending the cannonball to escape.  The system consists of separate masses (else, by definition, the cannonball would not be escaping!).

You really should look at the example I gave which explains why the energy goes into the cannonball rather than the far more massive Earth which has no detectable recoil.
If I hadn't seen that or an equivalent example months ago, I wouldn't have started this discussion with you, pointing out the flaws in that explanation.  For example, the "no detectable recoil" is a red herring, since it is irrelevant; we are talking about gravitational potential energy being converted into an immeasurably-small (with current instruments) amount of mass, after all!  Who cares if planetary or stellar recoil is also immeasurable with current instruments?!!  I don't, and neither should you.  What matters in a hypothesis is consistency, throughout its own details, and also with other known observations.  Which you haven't got, not at all, in your hypothesis.
Title: If I give an object some potential energy, does its mass increase?
Post by: Geezer on 23/02/2010 06:53:13
Can we consider the question in terms of the Earth and an object of significant mass, for example, 10% of the mass of the Earth?

Although the forces involved might be a little greater, I would not think the underlying physics would differ in the slightest. A model that only works for cannonballs of negligible mass (relative to the mass of the Earth) does not strike me as terribly viable.
Title: If I give an object some potential energy, does its mass increase?
Post by: VernonNemitz on 23/02/2010 07:36:47
Can we consider the question in terms of the Earth and an object of significant mass, for example, 10% of the mass of the Earth?  Although the forces involved might be a little greater, I would not think the underlying physics would differ in the slightest. A model that only works for cannonballs of negligible mass (relative to the mass of the Earth) does not strike me as terribly viable.
In general, though, it is "accepted" to initially construct models that are not complicated --complications should be added after one is convinced that a basic notion works.  In this case the basic notion is that gravitational potential energy can be described as part of the overall mass of a System (not just "mass-energy" but actual simple mass).  Given 400+ years of associating potential energy with height-in-a-gravity-gradient, you perhaps can see some wisdom in introduce Change in a small way rather than a big way.  So, the change I've been proposing is specifically designed to be compatible-with/similar-to descriptions of potential energy in terms of the other Forces known to Physics.  And yes, if a proton and neutron, having similar masses, can maintain their mass ratio from the moment of an initial Strong Force interaction when separate, to final merging to form a deuteron, and releasing a significant amount of "binding energy", then I have no qualms about talking similarly about planetary-magnitude masses interacting similarly (though merely gravitationally).  The main difference will be a smearing of said planetary bodies during the final moment of the merge, something the proton and neutron don't do!  Which is why such a version of the event might be called "hairy"  [O8)] and offer a rationale to stick with the simple cannonball or plate, for now.
Title: If I give an object some potential energy, does its mass increase?
Post by: Geezer on 23/02/2010 07:48:26
Thanks Vernon. But surely it's the same system whether it's a cannonball or a moon (and I think it was you that pointed out it is a system). If the system is somehow "different" because the masses involved happen to be different, I have a really hard time believing we understand any of this.

EDIT: Come to think of it, if we need a simplifying assumption, let's assume the object has the same mass as the Earth. The original question imposed no limitations on the relative masses.
Title: If I give an object some potential energy, does its mass increase?
Post by: Farsight on 23/02/2010 08:58:12
A 10% mass is better Geezer. Consider some impulse that separates a 100kg object m1 and a 10kg object m2. Conservation of momentum means m1v1 = m2v2 so we know that the velocity v2 of m2 from the point of origin is ten times v1. Let's say they are 100 m/s and 10 m/s. We can then work out the kinetic energy of each object using KE = ½mv2:

For m1 we get: ½ * 100 * 10  * 10  = 5000  Joules
For m2 we get: ½ * 10  * 100 * 100 = 50000 Joules

The energy is not evenly distributed. It's very simple stuff, and we do understand it. When you alter the mass ratios such that one is very much bigger than the other, very little of the kinetic energy goes into the larger mass. See http://www.physicsclassroom.com/Class/energy/U5L1b.cfm re gravitational potential energy - there is no mechanism by which the kinetic energy of a vertically-fired cannonball is transferred to the earth or the earth's gravitational field as the cannonball slows down and reaches its final height.
Title: If I give an object some potential energy, does its mass increase?
Post by: VernonNemitz on 23/02/2010 15:33:31
When you alter the mass ratios such that one is very much bigger than the other, very little of the kinetic energy goes into the larger mass. See http://www.physicsclassroom.com/Class/energy/U5L1b.cfm re gravitational potential energy - there is no mechanism by which the kinetic energy of a vertically-fired cannonball is transferred to the earth or the earth's gravitational field as the cannonball slows down and reaches its final height.
Tsk, tsk, the mechanism is called "gravity".  AND you are being deceptive; it is not kinetic energy that transfers so much as energy that transfers; please remember that we are talking about a conversion process, here!  Thus, kinetic energy of the smaller body gets converted to potential energy which appears as mass of the gravitating body.  If the two interacting bodies were more equal in mass, then the larger body would also have notice-able momentum and kinetic energy, and it would notice-ably slow, also, as the distance between the two bodies reaches a maximum.  Its kinetic energy becomes potential-energy-in-the-form-of-mass that is added to the mass of the smaller body (because, obviously, both bodies are pulling each other toward a final stop; it is only fair and balanced that if the large body sucks energy out of the smaller one, then the smaller one is doing the same to the larger one).  The net result is that the mass ratio of the two bodies remains constant throughout the event, just like other interacting masses do with respect to the other Forces in Physics.  So, Farsight, you are being silly to claim there is "no mechanism" for energy transfer, since there most certainly is for the Strong, Weak, and ElectroMagnetic forces.  Why shouldn't it work for Gravitation, also?  Or are you going to start claiming that the proton-to-neutron mass ratio is different inside a deuterium nucleus than when the two particles are widely separated?  Good luck convincing the Quantum ChromoDynamics particle physicists that have been measuring masses to high accuracy (and matching to theory) for decades!

Not to mention there is one other side-effect of claiming that the potential energy of an object is only part of its own mass.  This side-effect relates to neutron stars, and the accumulation of mass upon them.  It can be shown that in order for a falling body to reach 70+% of lightspeed, the terminal velocity of a typical neutron star, most of the original mass of the falling body will have to be converted into kinetic energy.  For a bigger neutron star, say 90%c terminal velocity, even more of the original mass of the falling body will have to be converted into kinetic energy.  After impact and energy-dissipation/escape, only a small amount of original mass actually gets added to the totality of the star.  Note that the bigger the star gets, the less mass gets added!  The thing called a "black hole" would be impossible to reach, an "asymptotic" endpoint of the curve of accumulation of ever-smaller arriving masses.  But black holes do exist.  Therefore you are wrong, Farsight, one more way (your hypothesis leads to a prediction that does not match observation).

Geezer, perhaps I wasn't clear enough in my previous message.  Two bodies with significant gravitation tend to merge to become a single body.  This makes it difficult to continue to think of them as being two bodies, when comparing before-and-after properties.  That's my only objection to using two large masses.
Title: If I give an object some potential energy, does its mass increase?
Post by: Geezer on 23/02/2010 19:15:52
there is no mechanism by which the kinetic energy of a vertically-fired cannonball is transferred to the earth or the earth's gravitational field as the cannonball slows down and reaches its final height.

Farsight, I'm afraid that is incorrect. Just because the changes in KE and PE of the Earth are negligible when firing a cannonball, that does not mean they do not exist.

Rerun your test with two equal masses and you will discover that they both behave identically. Only the magnitudes of the effects vary with mass. The physics is the same, regardless of the relative masses.

Furthermore, enough with pejorative comments like "it's very simple stuff".
Title: If I give an object some potential energy, does its mass increase?
Post by: Farsight on 24/02/2010 17:47:36
Vernon, the kinetic energy becomes potential energy, but it's cannonball potential energy. It isn't transferred out of the cannonball. I've never claimed that black holes do not exist, just that they're better described by the Weinberg Field Interpretation rather than the Misner/Thorne/Wheeler Geometric Interpretation. This doesn't invalidate what I've been saying, not at all. Note that what I'm saying is no hypothesis, it's in line with educational establishments such as http://www.physicsclassroom.com/Class/energy/U5L1b.cfm.

Geezer, the mass of the earth is 5.9742 × 1024 kilograms. Run the numbers and you'll see that the cannonball gets the lion's share of the kinetic energy. There really is no mechanism by which this is  transferred to the earth or the earth's gravitational field. If you beg to differ, please explain it or provide some educational reference that does. The "simple stuff" comment wasn't perjorative, the arithmentic of momentum and kinetic energy really is straightforward.
Title: If I give an object some potential energy, does its mass increase?
Post by: VernonNemitz on 24/02/2010 18:40:51
Vernon, the kinetic energy becomes potential energy, but it's cannonball potential energy.
Nope; the cannonball is part of a system, and so is the potential energy part of that system.  Geezer has one important thing right about requesting us to talk about a system in which the masses are more nearly equal to each other; you will be unable to assign the potential energy to just one part of it.

It isn't transferred out of the cannonball.
Just because you say so, that does not mean you are right.  Neither of us is that good.  But at least my position is consistent with respect to the other three Forces in Physics, and all the experimental observations about how they act.  (The advantage is that the other forces are much stronger than gravitation; therefore much more potential energy and potential-energy-as-mass can be involved in their interactions --enough to be measured, such as the proton-neutron mass ratio I've previously mentioned.)  You have yet to explain why picking an inconsistent position makes sense.  The only thing you appear to have claimed is, basically, "I must be right because I don't understand the mechanism that makes any alternative explanation possible."  Which is a worthless argument.

I've never claimed that black holes do not exist, ...
Nor did I claim that you made any such claim.  What I did claim (in different words) was that your hypothesis has logical consequences that do not match reality, and therefore your hypothesis must be erroneous.

Note that what I'm saying is no hypothesis, it's in line with educational establishments such as http://www.physicsclassroom.com/Class/energy/U5L1b.cfm.
Hardly!  I took a look at that page, and it clearly presents the conventional view that gravitational potential energy is defined in terms of a two-body system, not in terms of just one of those bodies.  Equally clearly, it is a hypothesis on both our parts to take the view that gravitational potential energy actually takes the form of mass, and to take the view that the field-gradient-of-one-body-and-height-of-second-body description is a workable and convenient bookkeeping trick, but not actually true (kind of like Feynmann's proposal that antiparticles move backward in time is a workable and convenient bookkeeping trick, but nobody who uses it seriously thinks it is actually true).  We even need that conventional explanation to compute how much potential-energy-as-mass we are talking about!  We only disagree on how that mass gets distributed in the system....
Title: If I give an object some potential energy, does its mass increase?
Post by: Geezer on 25/02/2010 01:03:20
Geezer has one important thing right about requesting us to talk about a system in which the masses are more nearly equal to each other; you will be unable to assign the potential energy to just one part of it.

Thanks Vernon. Yes, that's precisely my point.

Interactions between massive bodies make it very apparent that they are part of a common gravitational system. When the the ratio of the masses becomes very great we can disregard the KE and PE effects on the greater mass, which is the way we teach terrestrial mechanics/dynamics (as in the link that Farsight provided).

The measureable effects on the larger mass may asymptotically approach zero, but that does not mean they do not exist, it just means they are difficult to measure.
Title: If I give an object some potential energy, does its mass increase?
Post by: Farsight on 25/02/2010 11:03:13
Working through the numbers with equal masses does not assist with understanding here. Let's work the numbers with a 1kg cannonball fired upwards at 11,200 m/s. We’re ignoring air resistance, so this is escape velocity. The mass of the earth is 5.9742 × 1024kg, so conservation of momentum means:

m1v1                   = m2v2
m1v1                   = 1 * 11,200
5.9742 × 1024 x v2= 11,200
v2                       = 11,200 / 5.9742 * 1024
v2                       = 11.2 / 5.9742 * 1021
v2                       = 1.8747 * 10-21

Working out the kinetic energy via KE = ½mv2:

For Earth we get: ½ * 5.9742 × 1024 * 1.8742 * 10-20 * 1.8742 * 10-20 =  1.0492 * 10-17  Joules
For the cannonball we get: ½ * 11,200 * 11,200 = 6.72 x 107 Joules

Please check my arithmetic, but this says that the Earth acquires roughly one 10-25th of the energy acquired by the cannonball. Since the cannonball has escape velocity it leaves the earth’s gravitational field and escapes the system. Conservation of energy tells us the cannonball takes the energy away with it, which can also deduce by imagining a new cannonball poised ready to fall to earth. Alternatively we can envisage the earth as a large collection of cannonballs being fired upwards in succession. If these fired cannonballs did not remove their kinetic energy, and if instead this kinetic energy was transferred to the residual cannonballs, the collective gravitational field of the latter would have to rise in proportion to their mass. This is because energy causes gravity, and mass only causes gravity because of the energy content - a rapidly spinning mass would exert more gravity than one that was motionless. Ergo the kinetic energy of the fired cannonball is converted into cannonball potential energy, and is then removed into the vastness of space.
Title: If I give an object some potential energy, does its mass increase?
Post by: VernonNemitz on 25/02/2010 14:59:56
Since the cannonball has escape velocity it leaves the earth’s gravitational field and escapes the system. Conservation of energy tells us the cannonball takes the energy away with it ...
Your error is obvious. The kinetic energy of the cannonball approaches zero, due to gravity-caused slowing of the cannonball's velocity as its height increases, before the cannonball escapes the system.  Conservation of energy therefore involves the system, and not the cannonball only.  Thus, as kinetic energy converts to potential energy, the system acquires that potential energy, not the cannonball only.  If we choose to think that that potential energy appears as mass, then we are now free to divide up that mass between both bodies of the interacting system, in the manner that makes the most sense, in terms of other and similar observations --such as a kinetically-energized electron climbing out of the electrostatic field of a proton, and escaping.  Quantum ElectroDynamics, the most accurately-measured/supported theory in Physics, indicates that the overall mass ratio of the two bodies does not change throughout the event, so we have no reason to say things must work differently with respect to gravitation.
Title: If I give an object some potential energy, does its mass increase?
Post by: Geezer on 25/02/2010 18:46:05
Farsight, I think I see the source of the problem.

You said:

Since the cannonball has escape velocity it leaves the earth's gravitational field and escapes the system.

The problem with that statement is that, even though the cannonball has escape velocity, it does not mean it has left the Earth's gravitational field.

Gravitational fields don't stop at some arbitrary boundary. They just get weaker.

At escape velocity the cannonball and the Earth are still mutually attracting each other, but the KE of the cannonball is great enough that the Earth is not able to "pull it back".

Consequently, it's not valid to say that the cannonball has escaped the system. In fact, I don't think it's possible to define any point at which the cannonball has left the system. No matter how hard we try, there is always going to be some interaction. It may be an incredibly weak interaction of course, but it will not go to zero.

You also said, assertively, "Working through the numbers with equal masses does not assist with understanding here."

Why not? Because you said so? If you are going to dismiss another posters suggestion, unless you subscribe to "proof by loud assertion", it is customary to provide some reasoning.
Title: If I give an object some potential energy, does its mass increase?
Post by: VernonNemitz on 25/02/2010 19:05:11
Gravitational fields don't stop at some arbitrary boundary. They just get weaker.  At escape velocity the cannonball and the Earth are still mutually attracting each other, but the KE of the cannonball is great enough that the Earth is not able to "pull it back".  Consequently, it's not valid to say that the cannonball has escaped the system. In fact, I don't think it's possible to define any point at which the cannonball has left the system. No matter how hard we try, there is always going to be some interaction. It may be an incredibly weak interaction of course, but it will not go to zero.
Your theory is exactly correct, but what matters in the human realm is the practicality of things.  That means we indeed can specify a place where for all practical purposes we can say the Earth is not significantly affecting an object that has left its immediate vicinity.  I'm going to arbitrarily pick 2 Astronomical Units for this distance (the Asteroid Belt is about 1.8 AU from Earth).  If we want to talk about an object that has "escaped", then the velocity of this object can be computed to be (initial velocity) minus (escape velocity), pretty simple, don't you think?  Certainly an escaped object will always have at least that velocity...and at a distance from Earth such as 2 AU, the actual velocity will be close enough to that simply-computed velocity so as not to make a practical difference, over the short term at the very least, and likely also over the middle-term.
Title: If I give an object some potential energy, does its mass increase?
Post by: Geezer on 25/02/2010 20:18:51
Vernon, of course, for practical purposes what you say is quite true. But for something to "leave a system" we need to be able to define a boundary where the influences exerted by the system no longer exist. I don't believe that's possible in this case.

Just because the cannonball gets farther from Earth, that does not mean its potential energy with respect to Earth is eliminated.

If we temporarily relocate Earth to a region of space where all other gravitational influences are negligible and transport the cannonball to any distance from Earth, then "park" it in space, I maintain it will return to Earth. It might take a long time to get there, but if there are no other gravitational influences to redirect it, I'm pretty sure that's where it's going to go.

Other than the minor inconvenience associated with temporarily relocating Earth, this seems like a perfectly practical demonstration of the effect.
Title: If I give an object some potential energy, does its mass increase?
Post by: VernonNemitz on 25/02/2010 21:07:50
Geezer, what I was trying to say was that if we can specify a place where for practical purposes a cannonball or other object is not being significantly gravitationally affected by the Earth, then we can also say that the object, in that place, has for practical purposes escaped the system that had previously included both the Earth and the cannonball or other object.  I agree that technically there is never any escape outside of infinite distance, but since that is not achievable, a practical distance will have to do in its place --and can do in its place, over the short-term or middle-term.
Title: If I give an object some potential energy, does its mass increase?
Post by: Geezer on 25/02/2010 22:41:37
Vernon, I think that depends on what we are trying to do.

If we only want to calculate practical effects, then I agree with you. However, if we are trying to describe a general case where we are allowed to use terms like "leaves the system", we have to be able to unambiguously define when something crosses some boundary and actually "leaves the system". If we cannot create an unambiguous definition, the term has no meaning.

As we know of no limit to the extent of gravitational fields, I do not understand how it we can possibly ever create such a definition.

To say that something has effectively left the system, or has virtually left the system, reminds me of the girl who said to her father,

"But Daddy, I'm only a wee bit pregnant."
Title: If I give an object some potential energy, does its mass increase?
Post by: Farsight on 25/02/2010 23:28:07
Vernon, there is no mechanism by which the cannonball kinetic/potential energy leaves the cannonball. You might claim it does, but you have no mathematics and no logic to support that hypothesis. Quantum ElectroDynamics and "the mass ratio of the two bodies" is not appropriate when the cannonball gets 1025 times the energy that the earth gets.

Geezer: when an object has escape velocity, as it gets further away its velocity reduces because the Earth's gravity pulls it back. But the velocity reduces less than the gravity reduces. So it always gets further away. It never comes back.

Now come on guys, the kinetic energy of the cannonball is vastly greater than the kinetic energy of the earth. This is no proof by assertion, it's a mathematical proof. And there is absolutely no logic and no evidence to support the assertion that the energy you gave to the cannonball is somehow transferred to the earth or the earth's gravitational field. The cannonball gets away, and whilst that kinetic energy is converted into potential energy, it gets away too. It's obvious that it does, because a billion years later the cannonball finds a new earth, and falls down at 11.2km/s. It gives this new earth all the energy lost to the old one.

I don't know what morte I can say to explain this. I've given you the logic, the references, the arithmetic. There's nothing else I can give.
Title: If I give an object some potential energy, does its mass increase?
Post by: yor_on on 25/02/2010 23:32:52

'Potential energy' will differ on the choice of system, or frames you compare against. There is always frames of rest relative 'potential energy' and there will always be a 'potential energy' to objects at rest with each other when arbitrarily compared to something outside that frame.

That means that every object you can think of, at all times, will have a variety of 'potential energy' simultaneously depending on your choice of 'system', don't it?

Now if we changed that to 'gravitational impact' instead :)
Title: If I give an object some potential energy, does its mass increase?
Post by: Geezer on 26/02/2010 00:36:42
Geezer: when an object has escape velocity, as it gets further away its velocity reduces because the Earth's gravity pulls it back. But the velocity reduces less than the gravity reduces. So it always gets further away. It never comes back.

In the case where the object achieves escape velocity, I agree with you, it never comes back, and I did not say otherwise. But that does not mean it gave up all of its potential energy. That's the critical distinction.

To prove that it has not given up its potential energy, I proposed my slightly impractical experiment. In that experiment, we eliminated the kinetic energy of the object and put it in some stationary position (with respect to the Earth). We then wait for a long time and observe what happens. I believe it will accelerate towards Earth. Why would it not?

If it does, it still has potential energy with respect to the Earth. If it does not, it has no potential energy with respect to the Earth.

Anyway, back to my point. When an object achieves escape velocity, does that mean, or does that not mean, that it has given up its potential energy? I believe that is where you are confused. Let's try to answer one question at a time.
Title: If I give an object some potential energy, does its mass increase?
Post by: Farsight on 26/02/2010 14:19:15
I re-read what you said Geezer, and now understand what you meant. Apologies for misunderstanding. I'd say when an object achieves escape velocity, it has not given up its potential energy. Can I add though that this potential energy is only relative to us rather than something absolute. If we were on the surface of a different planet with a different gravitational field, we'd say the object had some different amount of potential energy.
Title: If I give an object some potential energy, does its mass increase?
Post by: VernonNemitz on 26/02/2010 17:25:13
Geezer, physicists typically work with both "ideal" cases and "real" cases.  They know the difference and aren't afraid to say which case they are talking about in any given situation.  The first two of Newton's Laws of Motion, for example, represent ideal cases in which there is no air resistance or other forms of friction.  So, in the case of an object that we know has an escape velocity from some large gravitating body, there is no reason why we cannot talk about the object as having actually escaped from that body; we can mentally remove the body from our discussions of the escaped object, even if we have to pretend that the object has reached infinite distance from the large gravitating body (because we know it can, given infinite time).

Farsight, stop saying the same nonsense over and over again.  I do have logic and data and math on my side:  The way the other Forces in Physics work proves there is a mechanism by which mass ratios can be maintained, during the conversion of kinetic energy into potential energy, and vice-versa.  And physicists want all the Forces to have similar descriptions, per the goal of Grand Unification.  Not to mention you have no proof --or even evidence!-- that the mechanism that works for the other forces (exchanges of virtual particles) cannot possibly also exist for Gravitation.  All you have is an unsupported claim, with Bad Logic behind it (since one logical consequence of your claim is that black holes can't exist, failing to match Reality).  You can't even be consistent with General Relativity!

More specifically, consider two rather large masses that are stationary and very far apart, and we measure their masses.  Now teleport one so it is very near the other, and still stationary (magically keeping them from getting any closer or losing their spherical shapes).  We both agree that a lot of potential energy has to be removed from the system, to allow such a description (so let's pretend a nice big gravity wave carries it away, like an electon emitting a photon when it gets near a proton).  GR says that when we switch reference frames, certain aspects of the system must be unchanged.  In this case we are switching from a frame in which the masses are far apart, to a frame in which they are nearby.  Some physics purists will go so far as to say that the masses themselves must be unchanged by both the teleport and the frame-switch; that's why the concept of Negative Binding Energy was introduced into nuclear physics, and could be included in the operation of any of the other Forces in Physics.  Well, obviously if the masses themselves are unchanged, their mass ratio is unchanged!  So, if we want to claim that potential energy exists in the form of mass, and if we want to be at least a little consistent with General Relativity, then we have to accept that when potential-energy-as-mass is converted into something else, during the interaction of two massive objects, their mass ratio must stay constant.  If the larger mass has twice the magnitude of the smaller, then it must always lose or gain twice as much potential energy as the smaller, during any Natural-Force interaction (not just Gravitational).  And if it has 1030 times as much mass, it must gain or lose 1030 times as much potential energy.  Period.

How dare you claim your version of the hypothesis is consistent with GR, when it so obviously is not!  Also, I did indeed describe in fair detail what the mechanism for keeping mass ratios constant could be.  That means you are wrong twice, to claim there is no such mechanism.  Quantum Mechanics, which will one day be used to describe gravitation, makes the task easy!

Next, as I have stated in other messages here, there is the simple English-language logic involving efforts and forces and "pulling":  If I pull something toward me (with either a rope or a gravity field) I am expending energy to do it, and this energy appears as kinetic energy of the thing I pull.  It most certainly was my potential energy that became that kinetic energy!  Your feeble attempt to invalidate that description by claiming that Gravity is not a Force, due to Einstein's description, matters not at all, since I have stated that I am relying on Quantum Mechanics, in which Gravity will be called a Force!

Finally, there is the nonsense of insisting that potential energy escapes a system, simply because you assume that's the only way it can become part of a new system.  In actual fact, though, the defining of a system is all that you need, to automatically define certain things about potential energy in that system.  Thus, a cannonball that escapes the Earth can leave most of its lost kinetic energy behind as a tiny amount of Earth-mass, because if the cannonball encounters a new Earth, defined by you as having the same mass as the one left behind-- the definition of that new Earth-cannonball system will include potential-energy-as-planet-mass that the cannonball can acquire as kinetic energy, by falling toward it (and lose should it escape again).  In one sense the two interpretations are equal; Energy is Conserved either way.  But your assumption violates General Relativity, and mine doesn't, since in my scenario the mass ratio stays constant.
Title: If I give an object some potential energy, does its mass increase?
Post by: Geezer on 26/02/2010 20:02:03
I re-read what you said Geezer, and now understand what you meant. Apologies for misunderstanding. I'd say when an object achieves escape velocity, it has not given up its potential energy. Can I add though that this potential energy is only relative to us rather than something absolute. If we were on the surface of a different planet with a different gravitational field, we'd say the object had some different amount of potential energy.

Farsigh, Yes. I completely agree with you.

I suppose it keeps on increasing it's PE while its distance increases from its origin, but only to a very small extent because the gravitational field is so weak.

I'm interested to know how much energy was spent in accelerating the Earth when the projectile was launched. I have to believe there was some. I suppose I'll have to do some math to find out!

Title: If I give an object some potential energy, does its mass increase?
Post by: Geezer on 26/02/2010 20:26:40
Geezer, physicists typically work with both "ideal" cases and "real" cases.  They know the difference and aren't afraid to say which case they are talking about in any given situation.  The first two of Newton's Laws of Motion, for example, represent ideal cases in which there is no air resistance or other forms of friction.  So, in the case of an object that we know has an escape velocity from some large gravitating body, there is no reason why we cannot talk about the object as having actually escaped from that body; we can mentally remove the body from our discussions of the escaped object, even if we have to pretend that the object has reached infinite distance from the large gravitating body (because we know it can, given infinite time).

Vernon, apparently you missed my point. When physicists describe theory they try to do it in a way that is entirely consistent in all situations. That means they can describe the general case. Then, and only then, can they make practical approximations so that insignificant factors can be ignored in a particular case.

There is nothing inconsistent about Newton's Laws of Motion. It's up to those who use them to factor in the influence of things like air resistance.

BTW, kindly stop shouting at Farsight. TNS does not approve of "Proof by loud assertion".
Title: If I give an object some potential energy, does its mass increase?
Post by: Farsight on 02/03/2010 00:11:51
Thanks Geezer. Yes, the PE keeps increasing with distance. And yes, there is some energy that accelerates the Earth. See the post above (http://www.thenakedscientists.com/forum/index.php?topic=24337.msg301163#msg301163) and check it with your own numbers. Like I said, I might have slipped up on the arithmetic, but the rule of thumb is that the more mismatched the masses are, the more energy goes into the smaller mass. See http://en.wikipedia.org/wiki/Recoil for a bit of backup on this. There would be some relativity adjustments necessary if you were dealing with objects moving at significant fractions of c, but nothing that changes the overall sense.
Title: If I give an object some potential energy, does its mass increase?
Post by: Tintin_Triton on 02/03/2010 08:20:20
So matter can basically be thought of as just potential energy because matter can be turned into energy and theoretically vice-versa, does that mean when I lift up a plate and increase it's potential energy, i've technically increased its mass?
The mass is increased, but not the mass of the plate: the mass of the system Earth-plate.

I think most of us may agree with this statement. I think he forgot to mention "Gravitational Potential Energy"

But I could never find a matching description or definition for potential energy. Is it always relative or inclusive for a system?
Title: If I give an object some potential energy, does its mass increase?
Post by: VernonNemitz on 02/03/2010 14:37:01
Geezer, I was under the impression that "shouting" in a message usually involves capital letters.  Meanwhile, italics is normally used to stress some part of a sentence, but with so many different fonts out there, it can be difficult sometimes to notice that some part of some text has been italicised.  So I like to add bolding to make sure it stands out.  My main beef with Farsight is not so much that there are problems with some of his ideas, so much as he appears to completely ignore any data that exposes those problems with his ideas.

...the rule of thumb is that the more mismatched the masses are, the more energy goes into the smaller mass.
Agreed, so long as we are talking only about the initial injection of kinetic energy into a system.  This has nothing to do with the process that converts kinetic energy, in a system, into potential energy, due to the influence of Gravitation or some other Natural Force.  And it is that conversion process, consistently for at least three of the four known Natural Forces (the Weak Force does not appear to either attract or repel), which can maintain mass ratios of the interacting bodies in the system.
Title: If I give an object some potential energy, does its mass increase?
Post by: Geezer on 03/03/2010 05:36:44
Vernon, you could try underlining instead. To me, bold has a connotation of loud, although it is, admittedly, not as loud as all caps.

Farsight, what do you think? You're on the receiving end.
Title: If I give an object some potential energy, does its mass increase?
Post by: yor_on on 04/03/2010 08:01:05
Potential energy is as simple as you want to make it. If you remember that it is a description of a relation between two objects, where your choice of objects always will be arbitrarily made, and that you simultaneously can have an arbitrary number of such relations relevant and valid for the same object  relative all other, you should be able to understand what it is.

It's not an 'energy' 'massed' at any specific point, it is an relation expressed when comparing two frames against each other. What you can talk about is expending energy, like acceleration. When you lift/move that plate you accelerate it, ever so little. When you after moving it f.ex place it on a table the movement/acceleration ends and you have an object at rest with the rest of its frame f.ex Earth. There is no new mass assembled in the object due to your moving it before. If the table dissolves and the object accelerate again, as observed/defined by an observer at rest with Earth, it will express an new relation with Earth as it meet it. That 'acceleration' is also a free fall where you could see it as that it's Earth coming to meet it. You can expend energy, like when lifting it against gravity, and you can define different relations (potential energy) for any object, be it at rest relative something or not, and that simultaneously, depending on your choice of comparison.

So either you will have to define potential 'whatever' energy as something not definable as a real sum as it will change depend on what you compare it too, or you will have to look at that plate reverently, understanding that in it it have all possible 'potential energies' there is, depending on your choice of comparison.

The third way is to pretend that 'potential energy' somehow becomes 'localized' by me when 'defining a system' which then seems to give the observer a Copernican outlook, not that surprising considering how short time we've been discussing 'relativity'.