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but isn't matter just potential energy?

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

I saw a documentary of LHC experiment, they were saying, mass of the protons will increase as the energy increase. Yes, mass....

The total energy in that plate travelling at 11 km/s near the surface of the earth is the same as the total energy of the plate when it was motionless up in space. So if you catch that plate and cool it down, the total energy of the plate is now less than that of the plate at altitude. The reason is quite obvious when you look at gravitational time dilation. Everything in that plate, be it atoms or electrons is now moving at a slightly reduced rate when compared with the plate up in space.

Back to the plate: imagine it's out in space, and is falling to earth. Ignore air resistance. Just before it hits the ground the plate is moving at a considerable velocity. It now has kinetic energy. So the total energy of the plate appears to be greater than that of the plate up in space. However, it isn't, because gravity is a pseudoforce.

I don't follow. What's this talk of cooling things down - why did the plate heat up?

Do you mean that we hypothetically catch the plate by turning its kinetic energy into thermal energy?

Hi Farsight!! How are you?

Do you mean that, according to GR, gravitational field doesn't exist (since it's actually spacetime warping)...

..and so that region of space cannot have energy (= cannot have mass)?

And so when the plate falls, since its kinetic energy increases, its proper mass have to decrease, to keep its total energy constant?

Quote from: lightarrow on 31/07/2009 18:14:17Do you mean that, according to GR, gravitational field doesn't exist (since it's actually spacetime warping)...Heck, not at all. If you're in a place where you fall down, you're in a gravitational field. It exists all right. Quote from: lightarrow on 31/07/2009 18:14:17..and so that region of space cannot have energy (= cannot have mass)?No. Space has an energy to it, but we don't think of this as mass. Mass is a property of something that resists being moved, and you can't exactly move a region of space from A to B.

People say the kinetic energy has come from the potential energy of the gravitational field, but that's missing the trick. The gravity of the two-body system doesn't increase, and nor does it reduce.

The important point is that gravity is a pseudoforce like Einstein said, because no energy is added to the system.

For more on inertial forces see my web page ...sorry, you cannot view external links. To see them, please REGISTER or LOGIN

Then I don't understand what mass you are talking about: in a region of space which is not moving (with respect to some frame S) and in which you have energy, you also have mass: m = E/c^{2}.

Example: an electrostatic field has mass (I mean, proper mass = invariant mass). When you give energy (let's say electromagnetic energy) to an hydrogen atom, for example, you also give mass to the system, which goes in the electromagnetic field; the proton's and electron masses don't vary at all.

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

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

Quote from: FarsightPeople say the kinetic energy has come from the potential energy of the gravitational field, but that's missing the trick. The gravity of the two-body system doesn't increase, and nor does it reduce.That's not correct. A field's energy density goes as the square of the field, so you cannot simply sum the effects of the two masses. If the masses' configuration varies, the field's energy varies as well.

When the two masses are closer, the field's energy increases *in absolute value* and since the gravitational field's energy is negative, it means that the field's total energy is decreased. This is the reason of the fact that total system's energy is conserved

Einstein never said that gravity was a pseudoforce. Einstein argued that the gravitational force is of the same nature as inertial forces. Some Newtonians argue that since there is no source of such a force that it's not 'real' and hence terms like pseudoforce, apparent force, and fictitious force were coined. But others disagreed with that notion and Einstein was one of them. Einstein argued that inertial forces should be thought of as being real. It's also wrong to assume that there is no potential energy in the gravitational field or that there is no mass equivalence to that potential energy. Einstein's field equations are non-linear because gravity itself is a source of gravity. For more on inertial forces see my web page ...sorry, you cannot view external links. To see them, please REGISTER or LOGIN

You'll be aware that if you combine two out-of-phase photons you're left with no photons

Quote from: lightarrow on 01/08/2009 13:10:55Example: an electrostatic field has mass (I mean, proper mass = invariant mass). When you give energy (let's say electromagnetic energy) to an hydrogen atom, for example, you also give mass to the system, which goes in the electromagnetic field; the proton's and electron masses don't vary at all.Yes, an electrostatic field has mass. But the electron's electric field is part of what it is. It isn't some central point particle with some set mass surrounded by an electric field with some variable mass. If you examine an electron at rest, you will deem its mass/energy to be 511keV. If you put that electron into your mirrored box, the mass of the box will increase accordingly. If you replace the stationary electron with a fast-moving electron bouncing back and forth inside the box, the increase in mass is greater. If you then replace your box with a proton that keeps the electron local to itself, the faster-moving electron means your hydrogen atom has more mass.

But the electric field hasn't increased. The electron and the proton still exhibit unit charge.

A region of empty space is not empty of energy, and OK, in this respect one can ascribe mass to it (he said grudgingly). A planet is surrounded by a region of space which exhibits a gravitational field, and there is energy "density" in that region of space. But the gravitational field itself is only there because there's a gradient in the energy density. No gradient means no gravitational field. You can see this very easily if you consider uniform space. There's no gravity at all.

Quote from: lightarrow on 01/08/2009 13:10:55That's not correct. A field's energy density goes as the square of the field, so you cannot simply sum the effects of the two masses. If the masses' configuration varies, the field's energy varies as well.What you've said here is a restatement of your earlier position. Let me try to work things through to explain things better. You start with a universe full of nothing but space. This space has an inherent energy. This isn't always obvious, but it's there, and there's a fixed amount of this "vacuum energy". But because space is uniform, there is no gravity.

That's not correct. A field's energy density goes as the square of the field, so you cannot simply sum the effects of the two masses. If the masses' configuration varies, the field's energy varies as well.

Quote from: lightarrow on 01/08/2009 13:10:55When the two masses are closer, the field's energy increases *in absolute value* and since the gravitational field's energy is negative, it means that the field's total energy is decreased. This is the reason of the fact that total system's energy is conservedThis sounds like a restatement too. The gravitational field is only there because the masses are there. And those masses are made of energy. They're made of positive energy, there's no such thing as negative mass, and no such thing as negative energy.

Quote from: Pmb on 01/08/2009 22:04:44For more on inertial forces see my web page ...sorry, you cannot view external links. To see them, please REGISTER or LOGINNice site. Thank you Pmb.

Try this: if you're in a windowless box in free-fall you can't feel any force and you can't feel any acceleration. As far as you're concerned there is no gravitational field, you're just floating in space in an inertial reference frame, and nothing is falling down at all. But if I snap my fingers and give you a window, you change your mind pronto. You switch your reference frame from the box to the ground, and you are utterly convinced of the existence of that gravitational field. But you still can't feel any force acting upon you. I snap my fingers again to give you a soft landing, whereupon you do feel a force which intensifies then levels off, but persists. You feel the force when you're standing on the ground, not when you're falling down.

there's no such thing as negative mass

Quote from: lightarrow on 31/07/2009 18:14:17And so when the plate falls, since its kinetic energy increases, its proper mass have to decrease, to keep its total energy constant?Yes, that's what I'm saying. Think about two bodies m1 and m2. Imagine you're some way off in space, feeling the effect of their combined gravity. You're measuring the energy content of that two-body system. Now imagine they've fallen together and coalesced into one body M without losing any energy (think of them as being made of water or something). You don't feel any extra gravity, because no net energy has been added to that system by the two objects falling together. People say the kinetic energy has come from the potential energy of the gravitational field, but that's missing the trick. The gravity of the two-body system doesn't increase, and nor does it reduce.

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

Quote from: FarsightYou can see this very easily if you consider uniform space. There's no gravity at all. I sincerely don't understand this. Consider a region of space in the void with a uniform electric field E. The energy density is (1/2)ε_{0}E^{2}, so it's uniform as well, so its gradient is zero. Would you deduce that the field is zero?

You can see this very easily if you consider uniform space. There's no gravity at all.

Quote from: FarsightBut because space is uniform, there is no gravity.Why not? I see what you intend after in your post, but it doesn't seem correct to me. Consider a rubber band which is initially stretched in a large circumference and then released; in every point of the rubber band, in the band's frame of reference, elastic forces pull in two opposite directions, so the point doesn't move along the circumference, ok, but that's different from saying that elastic forces don't act on the band: it quickly contracts in the other dimension. Monodimensional 'people' living in the rubber band would see that distances from any two points in its universe are decreasing.

But because space is uniform, there is no gravity.

Quote from: Farsightthere's no such thing as negative mass, and no such thing as negative energy.Can you prove it?

there's no such thing as negative mass, and no such thing as negative energy.

That is a "trick" scenario, since you have explicity stated the two bodies do not lose any energy as a result of their collision. In actual fact, though, the collision will cause the temperature of the combined body to be warmer than the separate bodies, and as time passes, that extra temperature will radiate away, after which the total gravitation of the combined mass will be diminished a bit, compared to the original two-body system. Therefore, the potential energy of the original two-body system can indeed be said to have existed as mass (somewhere in the system), which was converted to kinetic energy (heat) during the collision and radiated away after the collision.

MTW give the stress-energy-momentum tensor for the gravitational field as that derived by Landau and Lifshitz. On 465 they give an expression for it for a weak gravitational field. The energy density is given in terms of phi,j*phi,j. For a uniform gravitational field phi = gz/c^2 an so the energy density is uniform. The field is still there though, i.e. you'd still "fall down".

Any object only "feels" a force on it when there is stress imposed in the body by the force field. If you're falling in a uniform gravitational field you'd feel no force. But the same would happen for a charged body in a uniform electric field too. If you're in a non-uniform gravitational field then you'd feel tidal forces since they'd impose stresses in your body. It's odd that people don't make this obvious comparison with the electric field!

Quote from: Farsightthere's no such thing as negative massThis is wrong. I think you said this because you use the term "mass" as another name for rest energy. That's a problem.

In any case dark energy is due to the presence of negative active gravitational mass density. Any time you have repulsive gravity, such as in inflation and dark energy, then you're talking about negative active gravitational mass density.

It's also conceivable that there can be negative inertial mass if a body could exist that could support large enough values of tension. You probably never thought of this because you associate mass with rest energy. Negative inertial mass means that the momentum of a body is opposite to its velocity and that's quite possible in special relativity given large enough values of negative pressure (i.e. tension) compared to the bodies energy.

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

I do use it that way. IMHO rest mass is rest energy, only it's not actually at rest. It's just moving in some circular or back-and-forth type fashion so it isn't moving in aggregate with respect to me. Now, show me some negative mass!

I dispute this. IMHO in the early days of the universe the energy density was high, so processes within it would have run slowly just as they do in a region of high gravitational potential, hence the initial expansion would appear to be rapid. Nowadays the galaxies aren't pushing each other apart, the space between them is expanding because it's still under pressure.

The dimensionality of energy is pressure x volume, and there's nothing negative that I can see. Nor is there anything negative in a region of uniform space. Introduce a sun, and you've introduced positive energy to one portion of that region. You've also introduced a gradient in the energy density into the region we call a gravitational field. The active gravitational mass is telling you how much energy is there. I can't find any negative energy or negative mass anywhere.

Quote from: lightarrow on 02/08/2009 20:45:04 I sincerely don't understand this. Consider a region of space in the void with a uniform electric field E. The energy density is (1/2)ε_{0}E^{2}, so it's uniform as well, so its gradient is zero. Would you deduce that the field is zero?Not at all. But the gravitational field is not the same as an electric field. Consider an electron and a proton, and let them attract one another to form a hydrogen atom. Their fields mask one another, and the total energy will be reduced by photon emission to give the negative binding energy. It's a very different situation for two masses.

I sincerely don't understand this. Consider a region of space in the void with a uniform electric field E. The energy density is (1/2)ε_{0}E^{2}, so it's uniform as well, so its gradient is zero. Would you deduce that the field is zero?

Quote from: lightarrow on 01/08/2009 13:10:55Quote from: Farsightthere's no such thing as negative mass, and no such thing as negative energy.Can you prove it?No. But I can't prove the non-existence of tachyons or fairies either. Take a look at this article which suggests that negative mass is untenable: ...sorry, you cannot view external links. To see them, please REGISTER or LOGIN. It usefully goes on to talk about the negative energy of gravity, which is what I'm challenging.

I didn't mean it to be a trick at all.

Let me get this straight. When you use the term “mass” you mean nothing more and nothing less than “rest mass” = “rest energy/c^2” and you use the symbol “m” to represent it, correct?

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

I assume you use it in the expression for momentum, i.e. p = gamma*m*v, right?

There is a problem with this definition of mass. It can only be used in special cases, i.e. it can’t be used in general circumstances. Especially when it comes to the mass density of, say, a magnetic field or when you have a body under stress or if your speaking of the mass of a fluid which has pressure.

Quote from: FarsightI dispute this. IMHO in the early days of the universe the energy density was high, so processes within it would have run slowly just as they do in a region of high gravitational potential, hence the initial expansion would appear to be rapid. Nowadays the galaxies aren't pushing each other apart, the space between them is expanding because it's still under pressure. I don’t understand what you mean by this? Please explain what pressure you are referring to.

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

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

Quote from: Farsight I just can't see it Pete. Sorry. All you have to do is look at the situation of a body under tension. I’ve worked out an example at ...sorry, you cannot view external links. To see them, please REGISTER or LOGIN

I just can't see it Pete. Sorry.

Express the stress-energy-momentum tensor in coordinates where a body is moving. Let the speed be small so that p = mv. This “m” is what you call mass, right?

The situation I’ve described m is a function of the tension (i.e. negative pressure) in the body and if it’s large then M will be less than zero. But it is nor proportional to the bodies rest energy. That’s why I don’t use the concept of mass as rest energy/c^2.

But it's not different, because the gravitational energy density is negative, not positive as in the electrostatic case. When two massive bodies come very close, the field is increased (differently from the electrostatic case of two opposite charges), but being the field's energy negative, that means that the field's energy decreased, exactly as in the electrostatic case, so the result is the same.

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

If space is uniformly filled with negative energy, then you take 1 cubic metre of fixed and still region of space and it has negative mass, exactly in the same way as, if a fixed and still volume of space contains photons so it has a total energy E, that volume of space have mass m = E/c^{2}.

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

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

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

Therefore it is simply logical to conclude that gravitational potential energy, when it exists, exists in the form of a slight portion of the total mass of some gravitationally interacting system.

Farsight; when you defined rest mass = rest energy/c^2 all you did was give rest energy a new name with new units. Merely defining a term does not justify its name. You'd have to demonstrate how it has the properties associated with inertia. Even then you'd only be using your own pet notion of mass as it pertains to acceleration. In relativity mass is always defined as to how it relates to momentum.

You also claimed that a domain wall is an abstraction but you didn't justify that claim other than claiming, "we see no such thing." Vacuum domain walls were left over from phase transitions in the early universe. Just because you don't see them it doesn't mean they don't exist or never have existed.

Heck, it wasn't until recently that black holes were actually detected but we spoke of them very meaningfully before that.

Same with cosmic strings. Heck, I can't see the other side of the universe but I'm pretty sure its there. You might not approve of the use of the term "mass" in cosmology and relativity texts/journals, but it's there.

we conclude that this plate has lost energy. We therefore deduce that the kinetic energy came from the plate.

That's not my definition, Pete. That's Einstein's.

Quote from: Farsight on 05/08/2009 19:57:32 we conclude that this plate has lost energy. We therefore deduce that the kinetic energy came from the plate. Heh, you are now the one that has almost got it right. Remember that Isaac Newton's insight started with the question, "Why doesn't the Earth fall up to the apple?" As you probably know, Newton concluded that both the apple and the Earth fall toward each other, but the Earth's motion isn't so obvious because of its fantastically greater mass. Nevertheless, BOTH the apple and the Earth (or in your own text, the plate and the planet) must lose mass, to account for the kinetic energy that can appear during the collision and be radiated away after the collision.

Furthermore, there is a small fly in the ointment, regarding the most simple interpretation of what I quoted from your text, and ratios of the masses of plate and planet, and the ratios of the loss of mass, as gravitational potential energy is converted into kinetic energy.See, physicists like to switch "reference frames" a lot. General Relativity makes it an easy way to simplify some of the math they do. For example, consider an electron and a proton, separated by a great distance. We know that if they get close together and form a hydrogen atom, some energy will be released in the process. If we assume that the "system" of proton and electron lost mass as they got together, to explain the source of that released energy, then we need to be consistent with the "frame switching" that physicists like to do.More specifically, the physicists can switch from a reference frame in which the proton and electron are very far apart, to another frame that is inside the volume of a hydrogen atom that is at the "ground state". The masses of the two particles do not change, when they make that frame switch!!!So, where did the released energy come from? Physicists use a mathematical trick called "negative binding energy". (OK, it is used more in nuclear physics than in atom-formation, but rest assured it CAN be used as I've described here.) It is only a trick, though, used to allow calculations involving mass to come out the same in any reference frame. An alternate trick is necessary if one wishes to stick with one reference frame and describe events in other frames, without actually switching to them.That alternate trick involves paying attention to certain key ratios in different environments. If the ratios stay the same, then the calculations can also yield consistent results across different frames. For electron and proton, the proton has about 1836 times greater mass. If they maintain that ratio (among others) both far apart and close together, then calculations describing their behavior will be consistent in both frames.So, if we consider that the proton and electron lose mass as they approach each other, converting it to kinetic energy that will be radiated when they coalesce into a hydrogen atom, then it is necessary for the proton to lose 1836 times as much mass as the electron, during the event, for both to have the same mass ratio at the end of the event, as they had at the beginning.

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

For more details, including how it could be possible for Massive Object A to lose more mass than Light Object B, while B acquires lots more kinetic energy than A as they accelerate toward each other, see the essay I mentioned earlier in this Thread: ...sorry, you cannot view external links. To see them, please REGISTER or LOGIN

Quote from: FarsightThat's not my definition, Pete. That's Einstein's.Huh? Where did you get that impression from?

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

I can call myself a pot, but that doesn’t justify you pushing me into an oven! I explained that in relativity mass is always defined in terms of momentum. All you did so far is to tell me how, in slow speeds, objects which have rest mass is “related” to momentum, i.e. presumably by p = mv.

There is an important difference since if mass is defined in terms of momentum then anything which has momentum has mass. This does not hold with the definition you chose. And you didn’t specify under what conditions that pertains to. If you assert that it holds in all possible applications then you are mistaken since (1) only holds in special spacetime coordinates (i.e. those corresponding to in inertial frames only frames) (2) even then only when the object is not subject to stress.

In any case this is all still off topic since none of it addresses the original question. If the OP (HankRearden) has no further questions then it seems to me that the conversation is over, at least for me

From Einstein's 1905 paper Does the Inertia of a Body Depend on its Energy Content?

Quote from: VernonNemitz on 06/08/2009 03:14:02So, if we consider that the proton and electron lose mass as they approach each other, converting it to kinetic energy that will be radiated when they coalesce into a hydrogen atom, then it is necessary for the proton to lose 1836 times as much mass as the electron, during the event, for both to have the same mass ratio at the end of the event, as they had at the beginning.This isn't right. The kinetic energy is 1/2mv^{2}, and if the larger mass is moving slower than the smaller mass the ratios are skewed.

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

With the falling plate example we use the planet as our reference frame and we say that mass of the plate is so small that the planet's motion is not detectable. The plate's motion however is detectable. It's 11km/s. Once it's on the ground having lost its kinetic energy, the gravitational time dilation means everything moving in that plate, be it molecules or atoms or electrons or light, is moving slower than it was. That's where the energy came from.

That was merely his first paper on the subject of mass, not his last. Over the years he refined the subject and developed relations for more and more general cases. It appears to me that you mistook his first word on the subject for his final word. For example see The Principle of Conservation of the Center of Gravity and the Inertia of Energy, Albert Einstein, Annalen der Physik, 20 (1906): 626-633. In this paper Einstein assigns a mass density to radiation. In still later work he developes an expression for the inertia of stress and finally states that mass is completely defined by the energy-momentum tensor. With that tensor one can prove all the properties that physicists attribute to mass, such as the fact that the inertial mass density of a gas is a function of pressure.

No. You can't increase the mass of a plate.

Do not confuse the kinetic energies of the interacting objects with their masses. The key is to always remember that we are talking about INTERACTIONS. The mass lost by Object A appears as the kinetic energy of Object B, and vice-versa.

Certainly the energy that emanates from the collided bodies came from the kinetic energies of the bodies. But we are talking about where the kinetic energy came from: mass.

Quote from: VernonNemitz on 10/08/2009 15:02:40Do not confuse the kinetic energies of the interacting objects with their masses. The key is to always remember that we are talking about INTERACTIONS. The mass lost by Object A appears as the kinetic energy of Object B, and vice-versa.Sorry Vernon, we've got a plate in a gravitational field. There's is no magical mysterious action-at-a-distance between the plate and the earth. The earth doesn't lose mass because the plate is falling. There isn't time for the earth/plate interaction to occur. And the earth's gravitational field doesn't lose mass either. The energy in the surrounding region of space, where the gravitational field is, increases. Quote from: VernonNemitz on 10/08/2009 15:02:40Certainly the energy that emanates from the collided bodies came from the kinetic energies of the bodies. But we are talking about where the kinetic energy came from: mass.We agree that the kinetic energy comes from the mass. I'd hope your reply to Raghavendra might make you appreciate that you can take this a stage further to agree on which mass it comes from.

No.. You can't increase the mass of a plate.

The mass of the plate is a measure of its energy content. And the dimensionality of energy is stress x volume.

...the article I wrote which is located at http://arxiv.org/abs/0709.0687 [nofollow]

I will continue to disagree with you, because at the end of YOUR scenario the two masses no longer have the same mass ratio they started with, which violates General Relativity's allowing of easy reference-frame-switching (in which masses don't change at all). You have offered nothing at all to deal with that very significant problem!

Furthermore, GR isn't the Last Word on gravitation; Quantum Mechanics is going to eventually have a very significant "say" on the subject..

(has already had some; look up "Hawking Radiation")

..and indeed there will be interactions, and time for interactions, when that "say" arrives in detail. My argument that the mass lost by A appears as the kinetic energy of B, is based on the inevitability that QM will have its "say".

Quote from: FarsightThe mass of the plate is a measure of its energy content. And the dimensionality of energy is stress x volume.There is very serious/major flaw in that kind of logic, one which I addressed in the article I wrote which is located at ...sorry, you cannot view external links. To see them, please REGISTER or LOGIN for "The Planck constant has dimensions of energy multiplied by time, which are also the dimensions of action. In SI units, the Planck constant is expressed in joule seconds (J s). The dimensions may also be written as momentum multiplied by distance (N m s), which are also the dimensions of angular momentum". In simple terms the electron is a photon travelling in a circular path, so the photon is "going nowhere fast". Hence momentum now appears as inertia. If it’s side-on and moving relative to you, you'd see this circular path looking like a helical path. One full turn round the helix represents the relativistic mass, the total energy. The circular component of this represents the "rest" mass. The difference represents the kinetic energy. It tells you how fast the energy that's going nowhere fast, is going somewhere. Quote from: Pmb on 12/08/2009 17:08:41In short, just because you can give something units of energy it doesn't mean that it really is the energy of something. For example; I can multiply any constant which has the units of energy by v^2/c^2 and add it to the real energy. The results have the units of energy but is a meaningless quantity.Quite so. What's the energy of a photon? What you measure depends on your relative motion. Move towards it fast, and it's blue-shifted, so it appears to have a higher energy. But when you move towards that photon it doesn't acquire any extra energy at all. It didn't change, you did. It's the same for the blue-shifted photon in the Pound-Rebka experiment. It has not gained any extra energy. And it's the same for the falling plate. The total energy of the falling plate at the surface is the same as that of the plate at rest at altitude.

In short, just because you can give something units of energy it doesn't mean that it really is the energy of something. For example; I can multiply any constant which has the units of energy by v^2/c^2 and add it to the real energy. The results have the units of energy but is a meaningless quantity.