If I give an object some potential energy, does its mass increase?

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Offline yor_on

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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 :)
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Offline Farsight

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...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.   

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Offline Geezer

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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.
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Offline Farsight

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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.     

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Offline Geezer

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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.
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Offline VernonNemitz

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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.

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Offline yor_on

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Farsight :)

It's about definitions.
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.
« Last Edit: 26/01/2010 15:39:44 by yor_on »
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Offline VernonNemitz

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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 [nofollow]) 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.

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Offline Farsight

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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 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?".
« Last Edit: 28/01/2010 12:29:39 by Farsight »

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Offline yor_on

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"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?


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Offline Farsight

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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.

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Offline yor_on

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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 :)
« Last Edit: 28/01/2010 15:58:25 by yor_on »
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Offline VernonNemitz

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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 [nofollow] 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 [nofollow] 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 [nofollow]. 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".)
« Last Edit: 28/01/2010 16:39:16 by VernonNemitz »

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Offline yor_on

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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 :)
« Last Edit: 31/01/2010 16:36:33 by yor_on »
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Offline yor_on

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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

--End of quote-

Take a look here too Fluids and pressure
 
"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 :)

« Last Edit: 01/02/2010 04:14:38 by yor_on »
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Offline Farsight

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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.

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Offline yor_on

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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 :)
 
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Offline VernonNemitz

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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).

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Offline Farsight

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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.

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Offline VernonNemitz

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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.
« Last Edit: 02/02/2010 10:28:14 by VernonNemitz »

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Offline yor_on

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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.
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Offline VernonNemitz

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"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.

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Offline yor_on

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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.
« Last Edit: 02/02/2010 23:43:43 by yor_on »
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Offline VernonNemitz

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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).

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Offline Farsight

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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.
« Last Edit: 03/02/2010 12:29:49 by Farsight »

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Offline VernonNemitz

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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.
« Last Edit: 03/02/2010 14:36:36 by VernonNemitz »

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Offline yor_on

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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' :)
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Offline Geezer

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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)
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Offline yor_on

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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?
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Offline Geezer

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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.
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Offline yor_on

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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 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:



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

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
Einstein_field_equations
Mass_in_general_relativity

« Last Edit: 04/02/2010 22:50:15 by yor_on »
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Offline Geezer

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Hey! That's a very helpful summary. Thank you sir.

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

 
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Offline VernonNemitz

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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).

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Offline yor_on

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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 :)
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Offline VernonNemitz

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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.

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Offline yor_on

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"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?
« Last Edit: 12/02/2010 21:02:02 by yor_on »
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Offline VernonNemitz

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...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!
« Last Edit: 16/02/2010 06:52:22 by VernonNemitz »

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Offline Farsight

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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.   
« Last Edit: 16/02/2010 11:47:56 by Farsight »

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Offline VernonNemitz

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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).

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Offline Farsight

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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.


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Offline VernonNemitz

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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 [nofollow] (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!
« Last Edit: 16/02/2010 21:15:38 by VernonNemitz »

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Offline yor_on

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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.
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Offline Farsight

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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. 

 

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Offline VernonNemitz

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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.)
« Last Edit: 17/02/2010 03:20:40 by VernonNemitz »

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Offline VernonNemitz

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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.

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Offline JP

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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).

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Offline VernonNemitz

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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.

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Offline Farsight

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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.

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Offline VernonNemitz

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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.

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Offline Farsight

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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.