[Pmb]

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

[lightarrow]

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

Nice site. Thank you Pmb.

[Farsight]

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.   

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.     
   

[lyner]

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.

[lightarrow]

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.

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.

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?  []

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.

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?

[Pmb]

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.

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".

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!

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.


If you want to learn more about this analyze the stress-energy-momentum tensor for continuous media. E.g. see either Tolman or Rindler.

Pete

[VernonNemitz]

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.

[Farsight]

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.

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. 

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.  

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.   

[Farsight]

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.   

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.

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!

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.   

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.

If you want to learn more about this analyze the stress-energy-momentum tensor for continuous media. E.g. see either Tolman or Rindler.

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.

[Pmb]

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.

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.

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

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.

[lightarrow]

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.

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.

[VernonNemitz]

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.

[Farsight]

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

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.

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.

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.
 

[Pmb]

A few last comments:

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.

[Farsight]

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.

[VernonNemitz]

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

[Pmb]

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

[Farsight]

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.

[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. 

[Pmb]

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.