Question regarding GR wrt potential energy

[Buzz Bloom]

The motivation for this post comes from a discussion on another thread

https://www.physicsforums.com/threads/questions-regarding-mass-volume-x-density.851172/​

It seems worthwhile to me to try a simplification in this post to hopefully avoid the complications raised in the cited thread.

Assume an otherwise empty universe with a very thin spherical shell of radius R₀ at time t = 0, with a uniform density of dust particles. There is zero radiation, zero pressure, and zero temperature.

If it is reasonable to do so, make additional assumptions:

each dust particle is a very tiny sphere of one milligram mass;
there are N>>1 of these particle;
they are approximately arranged uniformly on the surface of the sphere; and
the fact that it is impossible to make the arrangement completely uniform (and therefore the density is not exactly uniform) is not relevant to the reasoning presented later in this post.​

If it is not reasonable to make these assumptions, I would appreciate a post explaining why so that I can try a restatement in a different and acceptably reasonable way.

There is a test particle of negligible mass in a circular orbit around the spherical shell at a radius of

R_t > R₀.​

Using Newtonian gravity, the orbital velocity is

v = √(N m G / R_t)​

GR gives an approximate modified velocity

v ~= √(N m G / (R_t - R_s))​

where R_s is the Swartzchild radius

R_s = 2 G N m / c².​

The source of the non-exactness for v is that the total gravitational mass of the shell is not exactly = N m. I understand that the dust particles in the shell interact with each other gravitationally, and this interaction creates an additional gravitational mass-energy M_g.

I would much appreciate some help in understanding how to calculate this additional mass-energy (hopefully without having to do any tensor calculus). This is the main intended focus for starting this thread.

The following is my naive attempt to make this calculation. I expect that it is probably wrong, but I am hopeful that I can be educated to understand why it is wrong.

If we ignore the needed added mass M_g, then what effect will result from the fact that all the particles in the shell for time t > 0 will be falling towards the center of the sphere? As the particles fall they acquire kinetic energy. Will the added kinetic energy has a gravitational effect on the orbit of the test particle?

I am guessing that the falling of the particles with increasing kinetic energy will not actually effect the orbit. If I am mistaken I hope someone will correct this guess. My reasoning is that the sum of kinetic and potential energy is a constant as the particles fall. The stationary shell at t=0 has potential energy and the addition of kinetic energy equals the reduction of potential energy.
With Newtonian gravity, the shell as a whole exerts an attraction on each particle as if the shell were replaced by a mass at the center equal to N m / 2. Therefore,

M_g = G N m / (2 c² R₀).
​

Some corrections may be needed.

First, the M_g mass based on the potential energy will also have a gravitation effect on the particles, so there will be second order potential energy effects, etc., adding to the above M_g.

Second, The calculation of M_g due to the potential energy may require a GR correction to the Newtonian calculation.

All comments are most welcome.

Regards,
Buzz

 

[pervect]

Because of a lack of pressure, the thin spherical shell is going to collapse - this is true in Newtonian mechanics as well as in GR.

I don't think I've seen the thin spherical shell collapse in MTW, but a solid spherical dust-cloud collapse is certainly in there, complete with a metric. As I recall is a time-reversed cosmological solution. I don't think there will be any real difference between the hollow spherical collapse and the solid spherical collapse in the exterior region.

Basically, what should happen is that there will be absolutely no effect on the orbit of a test particle outside the collapsing shell as the shell collapses. Assuming no cosmological constant, the metric outside the shell will be pure Schwarzschild, which seems to be the gist of your question as I understand it. Additionally, because the Schwarzschild metric is asymptotically flat and has no gravitational waves, you can use either/both of the ADM or the Bondi masses, which will be constant in the asymptotically flat space-time.

 

[Buzz Bloom]

Hi @pervect:

Thanks for your post.

I am wondering since you did not comment on my derivation of M_g that you intended this to mean that my derivation is OK. If so, do you also agree that the correction I described as the "first" correction is also correct? And is there also a need for the "second" correction? If so, what would this "second" correction look like?

I am trying to understand as best I can at an intuitive level some subtle aspects of mass-energy in GR before tackling the complexities of the ADM and Bondi energies as presented in Wikipedia, and perhaps later in MTW.

Regards,
Buzz

 

[PeterDonis]

The source of the non-exactness for v is that the total gravitational mass of the shell is not exactly = N m.

If you are defining "m" as the mass of a single dust particle as measured locally, then yes, this is correct; the total gravitational mass of the shell, measured as you describe, by putting a test particle in orbit about the shell, will be less than N m.

I understand that the dust particles in the shell interact with each other gravitationally, and this interaction creates an additional gravitational mass-energy Mg.

Only if you interpret "additional" to mean "negative". As above, the total gravitational mass of the shell--i.e., the total mass that appears in your equation for the orbital velocity of the test particle--is less than N m. The difference is called "gravitational binding energy"; it is the energy that would have to be extracted from the system in order to take N dust particles, each separated by a very large distance from all the others, and bring them together into a spherical shell with a finite radius that was at rest at some instant of time. (Note that if you let the dust particles fall inward starting from infinite radius, the gravitational binding energy is zero and the total mass is equal to N m. The reason the total mass of the shell in your scenario is less than N m is that you specified that the shell was at rest at a finite radius at some instant of time.)

 

[PeterDonis]

because the Schwarzschild metric is asymptotically flat and has no gravitational waves, you can use either/both of the ADM or the Bondi masses

There are complications here. Suppose we have N dust particles, each of rest mass m as measured locally, each separated by a very large distance from all the others (so their gravitational interaction is negligible), and all at rest relative to each other. We stipulate that the total configuration is spherically symmetric. This system will have ADM mass N m. It will also have Bondi mass N m, if we evaluate the Bondi mass at a point on future null infinity that is close to spacelike infinity (so no radiation has escaped).

Now suppose we bring all the dust particles together into a spherical shell with a finite radius and at rest at some instant of time. We have to extract energy from the dust particles in order to do that (for example, we could have them emit radiation that escapes to infinity). However, the ADM mass of the total system is still N m; the process of forming the shell can't change it. The Bondi mass, however, will change (provided we evaluate it at a point on future null infinity after the radiation has escaped); it will now be the value I described in my previous post (N m less the gravitational binding energy).

Note that, in doing all this, we are implicitly using a metric that is not the Schwarzschild metric everywhere; the spacetime has to have at least one region where the metric is something like the outgoing Vaidya metric (the region where the radiation is escaping). So if we put our test particle orbit far enough away (so that it is outside all of the dust particles both before and after they are brought together into the spherical shell), its orbit will change--it will see radiation passing by as the dust particles are brought together, and its orbital parameters will change to correspond to a smaller mass, a reduction from mass N m to mass N m minus the gravitational binding energy.

Of course, we could ignore all these complications and just stipulate that the system starts out with the dust particles at a finite radius at some instant of time, no radiation anywhere, and metric Schwarzschild outside that radius for all time. But then (even ignoring the issue of how the dust particles got to that point--they will have had to be in an expanding spherical shell before that instant of time, and we will end up having a spacetime with a white hole as well as a black hole, with all the attendant issues), we have no way of giving any physical meaning to the statement "the gravitational mass of the system is less than N m", because there was never any physical system with mass N m to compare it to.

 

[Buzz Bloom]

Hi Peter:
Thanks for your helpful posts.

I am wondering about the relationship between the concepts "binding energy" and "potential energy". It seems to me to be the same thing, at least in Newtonian physics. It may well be that physics vocabulary has changed since I studied undergraduate physics in the 1950s.

In Newtonian physics the energy E needed for a test particle at distance R from a point mass M to no longer be gravitationally bound to M is

E = (1/2) m v²​

where v is the escape velocity for the test particle. Therefore

E = (1/2) m (2 G M / R) = G M m / R,​

which is the potential energy (ignoring the sign).

Regards,
Buzz

 

[Jonathan Scott]

"Binding energy" is the potential energy within a single object (or system) due to its own components.

This particular example has an interesting complication in GR because there is no pressure involved, and because it is dynamic.

If one looks at the total energy of a static system in GR, then for each pair of particles which makes it up, their energy is effectively lowered by the time dilation due to the other, both ways, which means their total energy is decreased by twice their potential energy relative to each other. However, if the system is static, then the force between them times the distance between them is equal to their potential energy, but with a positive sign. This quantity is equal to the integral of the pressure due to the force between those two particles over the line between those particles. (Any non-gravitational forces must balance the gravitational forces over a complete plane for the system to remain static). If this pressure integral is treated as a contribution to the total energy (as in the "Komar Mass" expression) then the total energy matches the energy of the components minus the binding energy, as for Newtonian theory.

I personally find it very difficult to see how treating the pressure integral like energy could be valid, as the pressure can be temporarily changed, and can of course be zero temporarily in a situation like a non-interacting dust cloud. This certainly doesn't work in the Newtonian equivalent. This is the subject of a "Tolman Paradox" (not the only one, the "antitelephone" being more famous), where Tolman pointed out that it seems that GR predicted that changes in the internal structure of a star could abruptly affect its gravitational field.

An alternative view which does match up with the Newtonian view but seems to look at things very differently from the normal GR view is to assume that there is effectively some positive energy in the gravitational field, given in Newtonian terms by $(1/8\pi G) g^2$ for field $g$, in an analogous way to the energy density in an electrostatic field, as described for example in the article "Gravitational field energy density for spheres and black holes" by D Lynden-Bell and J Katz: http://adsabs.harvard.edu/full/1985MNRAS.213P..21L

I think that this "energy density of the field" is closely related to Einstein's original "gravitational energy" pseudotensor. It describes the effective energy as seen from a particular point of view, and is useful for comparison with the Newtonian viewpoint.

With this alternative view, the energy of each source mass is reduced by the time dilation of the potential, again making the total energy decrease by twice the Newtonian potential energy, but the energy of the field then adds back the same amount as the potential energy (which can be shown by integrating it over all space in a similar way to the standard electromagnetic energy calculation), making the overall energy decrease correctly match the Newtonian potential energy.

 

[PeterDonis]

I personally find it very difficult to see how treating the pressure integral like energy could be valid, as the pressure can be temporarily changed

But if it changes, then that will cause changes in other stress-energy tensor components, because of the local conservation law $\nabla_\mu T^{\mu \nu} = 0$. Also, you have to include in the analysis anything that is confining the matter under pressure (for example, if you have a gas inside a container, you have to include the stresses in the container walls). As you note, in a static system, including these items in the integral ensures that everything balances correctly.

The different feature of the OP's example is that the system is not static. That means the Komar mass can't even be defined, since it requires a timelike Killing vector field, which doesn't exist in the OP's example. It also means that concepts like "gravitational binding energy" are problematic in general; however, the OP's assumption of spherical symmetry avoids many of those problems, since we can describe his scenario using a combination of particular known solutions (Schwarzschild and outgoing Vaidya, as I posted previously) whose properties are simple enough to allow a thought experiment like the one I described--taking a bunch of very widely separated dust particles and bringing them together into a spherical shell at rest at an instant at a finite radius--to yield reasonable results.

 

[pervect]

I won't go into the details, due to a lack of time, and a feeling that this has been talked about before without much success, but I do want to point out that according to the textbooks (specifically MTW's Gravitation", localizing the energy of the gravitational field is, unfortunately not possible. See section $20.4, pg 466-467 of MTW's "Gravitation".

If one doesn't have the textbook, it's possible to use Google search to find the relative section. Searching for following rather snappy quote (use quotation marks to indicate to the search engine that it's an exact quote) should find the relevant section.

""No, the question is wrong. The motivation is wrong. The result is wrong. The idea is wrong"

It should come back as "Gravitation - Part 3 - Page 467 - Google Books Result".

 

[Buzz Bloom]

making the overall energy decrease correctly match the Newtonian potential energy.

Hi @Jonathan Scott:

If I interpret the above correctly, you are agreeing that the orbital velocity squared of the test particle would be
v² = (G M /(R-R_s))
where R is the radius of the test particle orbit, and R_s is the Shwartzchild radius corresponding to the total effective mass M_s of the shell
R_s = 2 G M_s / c².
The effective mass M is calculated by
M_s = M_N - M_g
where M_N is the Newtonian mass
M_N = N m
and
M_g is the mass equivalent of the gravitational binding energy of the shell
M_g = G N m / (R₀ c²)
where R₀ = the radius of the shell at time t = 0.

Therefore

Please let me know if I have misunderstood you.

Regards,
Buzz

 

[Jonathan Scott]

I have done no specific calculation, and I suspect that mixing the Newtonian approximation for kinetic energy with the Schwarzschild solution will produce inaccurate results.

What I was pointing out is that for a semi-Newtonian approximation for GR involving multiple masses, one can preserve Newtonian conservation of energy by assuming that the rest energy of each object is reduced by the local time dilation factor due to other masses (which reduces both sides by the potential energy of the relevant pair of masses), but that there is positive energy in the gravitational field (or more usefully in the difference between the squares of the fields of the individual masses and the square of their combined field) which balances it all out. With that scheme, any change in (relativistic) kinetic energy is equal and opposite to the change in potential energy to maintain a constant total.

 

[Buzz Bloom]

Hi @Jonathan Scott, @PeterDonis, and @pervect:

Jonathan has pointed out to me that my derivation of a formula for the orbital velocity of a test particle, given the assumptions for the thought experiment described in my post #1, is likely to be inaccurate. I am assuming that it should be possible to derive such an accurate formula, or if not, an equation from which a numerical calculation can be made with the accuracy limited only by the numerical methods used. I also gather from previous posts that none of you know any specific references that have explored this specific problem. I would much appreciate any specific help on how to go about developing such a formula or a numerically solvable equation.

Regards,
Buzz

 

[PeterDonis]

my derivation of a formula for the orbital velocity of a test particle, given the assumptions for the thought experiment described in my post #1, is likely to be inaccurate

For a test object in a circular orbit around a mass $M$, the correct relativistic formula for orbital velocity is $v^2 = GM / \left( R - R_\text{s} \right)$. If you check, you will see that this gives $v = c$ for $R = 3GM / c^2$, or $R = (3/2) R_\text{s}$. So there are no circular orbits possible inside that radius. (Also, it turns out that the circular orbits with $R < 6 GM / c^2$, or $R < 3 R_\text{s}$, are unstable against small perturbations). But I don't see that that creates any issue in your scenario.

So your formula for orbital velocity in terms of $M$ is correct; what you need to focus on is how $M$, the mass of the shell, is determined. It will be less than $N m$, as has already been discussed, and the difference can be interpreted as "gravitational binding energy". You might want to start by looking at the case of a spherical shell of matter being slowly lowered onto a static spherical mass like a planet. If the planet starts with mass $M$ and has radius $R$ ("radius" here means "circumference divided by $2 \pi$), and the shell starts at infinity with mass $m$ and is slowly lowered until it is at radius $R$, how much energy is extracted by the slow lowering process? If we call that energy $E$, then the final mass of the planet + shell should be $M + m - E$, so $E$ can be thought of as the "gravitational binding energy" of the shell. Then consider how the same process would work if the "planet" wasn't actually there, so $M = 0$.

 

[Buzz Bloom]

Hi Peter:

Thanks very much for your helpful post.

I intend to try to work out a formula for v₂, and If that fails, I will try to develop a spreadsheet to perform a numerical calculation. I will post the result when I finish and ask for someone to check it for correctness.

Regards,
Buzz

 

[Jonathan Scott]

Apologies for my not having read the calculation carefully enough when I suggested it might involve a Newtonian approximation, and my thanks to PeterDonis for picking that up.

I think you can probably start with a massless shell (which will not have any binding energy and will not cause any potential) then integrate adding mass to it. That will tell you what the total mass should be with the correct binding energy taken into account. However, that will not be equal to the normal GR way of adding up the total energy of a static object in that case because of the absence of pressure (and the dynamic situation), which is what I was trying to say before. I think that the GR way of integrating the total energy would be equivalent to multiplying the rest energy of the material which was used to create the shell by the time dilation factor corresponding to the potential at the surface of the shell, and that would be less than the original rest energy of the material by twice the potential energy. So it's not clear to me whether the M used in the Schwarzschild solution to describe what happens should be less than the original rest energy of the material by once or twice the potential energy, and if anyone can clarify that I'd be very interested to know.

 

[PeterDonis]

it's not clear to me whether the M used in the Schwarzschild solution to describe what happens should be less than the original rest energy of the material by once or twice the potential energy, and if anyone can clarify that I'd be very interested to know.

I know what I think the answer is, but I'd prefer to wait until Buzz has posted his calculations before giving it away. But the key idea, as I said before, is to imagine the shell being slowly lowered from infinity in a spherically symmetric manner, and ask how much energy is extracted during the slow lowering process. To fully match the scenario posed in the OP, we would imagine that, once the shell has been slowly lowered to radius $R_0$, it is then released into free fall, so at the instant of release, all of the dust particles in the shell are at rest relative to each other.

 

[Buzz Bloom]

I think that the GR way of integrating the total energy would be equivalent to multiplying the rest energy of the material which was used to create the shell by the time dilation factor corresponding to the potential at the surface of the shell, and that would be less than the original rest energy of the material by twice the potential energy.

Please explain "the time dilation factor corresponding to the potential at the surface of the shell". What is the math for this?

Regards,
Buzz

 

[Jonathan Scott]

To fully match the scenario posed in the OP, we would imagine that, once the shell has been slowly lowered to radius $R_0$, it is then released into free fall, so at the instant of release, all of the dust particles in the shell are at rest relative to each other.

But immediately before release, the integral of the pressure is equal to the binding energy and immediately after release (even before anything gets moving), the integral of the pressure is zero. That's what bothers me about this case.

 

[Jonathan Scott]

Please explain "the time dilation factor corresponding to the potential at the surface of the shell". What is the math for this?

Regards,
Buzz

That's simply the time factor in the Schwarzschild metric (proper time per coordinate time).

But I may be confused about that too; I remember an exercise in MTW about the binding energy which never seemed to make sense.

 

[PeterDonis]

immediately before release, the integral of the pressure is equal to the binding energy

Not if the shell has zero pressure. Imagine each of the dust particles in the shell being suspended by a string; the strings are used to slowly lower each dust particle (and extract energy from each particle while doing so); then, once all the particles are stationary at radius $r_0$, the particles are all released at the same instant. Then the shell has zero pressure throughout; the force keeping each particle from free-falling during the slow lowering process is the tension in each of the strings. When each particle is released, the tension in its string goes to zero at the release point; a wavelike disturbance then propagates up each string as the tension in each infinitesimal piece of the string goes to zero.

 

[Jonathan Scott]

Not if the shell has zero pressure. Imagine each of the dust particles in the shell being suspended by a string; the strings are used to slowly lower each dust particle (and extract energy from each particle while doing so); then, once all the particles are stationary at radius $r_0$, the particles are all released at the same instant. Then the shell has zero pressure throughout; the force keeping each particle from free-falling during the slow lowering process is the tension in each of the strings. When each particle is released, the tension in its string goes to zero at the release point; a wavelike disturbance then propagates up each string as the tension in each infinitesimal piece of the string goes to zero.

If you describe any self-contained piece of apparatus which performs that suspension, then you will find that the volume integral of the pressure over that apparatus is still equal to the binding energy, and it drops instantly to zero (well, at the speed of propagation of tension in the medium) when the shell is released.

The simplest self-contained model is that the dust is initially in the form of a rigid shell when then crumbles isotropically. Again, you'll find that if you integrate the pressure over the volume of the shell the result is equal to the binding energy. (The pressure may locally include additional forces for purpose of rigidity, but all the forces which are not counteracted only by gravity must cancel out over any given plane otherwise the initial state will not be static).

I think your suggestion is equivalent to physically removing the apparatus containing the pressure term at the time of release. I presume that would mean that by Birkhoff's theorem, when the apparatus passes out of the the sphere containing the relevant orbit, the orbit would change accordingly.

However, for my suggestion of a crumbling shell, I don't know where the pressure term goes. If one considers the relationship of tension and elasticity in the material, it is true that some energy can be stored in the tension, but that energy is less the stiffer the material, and there is no reason to assume it can store anything which adds up to the binding energy. (You can in fact model the suspending material by a spring to compare the energy stored to the total potential energy; it turns out that because the maximum force is fixed as the force of attraction, you could only store enough energy in it if the initial length stretched to infinity).

 

[jartsa]

I just want to say something about things under stress in gravity field.

There exists an anisotropy of the coordinate-speed of light in a gravity field. So also coordinate speed of sound is anisotropic in any object in gravity field. So inside a static compressed rod in a gravity field there are more phonons moving upwards than downwards, which means the rod has momentum in the upwards direction in the frame where the rod is static.

So have you guys taken this into account?I'm talking about this anisotropy:
Time for light to travel down to black hole event horizon: finite
Time for light to travel up from black hole event horizon: infinite

 

[PeterDonis]

inside a static compressed rod in a gravity field there are more phonons moving upwards than downwards, which means the rod has momentum in the upwards direction in the frame where the rod is static.

No, this is not correct. Anisotropy in the speed of light (or sound, assuming you are correct about that--I don't think you can assume it, and you would need a detailed material model to derive any such result) does not mean more photons are moving up than down. Coordinate speeds are just coordinate speeds; they don't have any physical meaning in and of themselves.

 

[Buzz Bloom]

Hi Jonathan and Peter:

Thanks for your posts.

Unfortunately I am still in the dark about the mathematical form for the binding energy that goes with the scenario I am investigating. Can either of you help me with this?

I also don't understand why calculating the escape velocity of a single particle is the shell, and multiplying by Nm/c², fails to correctly calculate the binding energy of the shell at t=0. The equation I used to make this calculation may be incorrect because I failed to take into account the effect of the non-flat space on escape velocity. I was not able to find any GR based equation for escape velocity, although Wikipedia did include the equation I used for circular orbit velocity with a Schwartzschild metric correction.

Regards,
Buzz

 

[PeterDonis]

I think your suggestion is equivalent to physically removing the apparatus containing the pressure term at the time of release. I presume that would mean that by Birkhoff's theorem, when the apparatus passes out of the the sphere containing the relevant orbit, the orbit would change accordingly.

I think this is correct, yes. Basically, during the slow lowering process, the energy being removed from the dust is being stored in the apparatus. Then, when the dust is released, the stored energy is propagated outward, ultimately to infinity; and when it passes the finite radius of the orbit of the test object, that orbit will change to reflect the (reduced) mass of the dust shell alone, rather than dust + apparatus.

 

[Jonathan Scott]

I think this is correct, yes. Basically, during the slow lowering process, the energy being removed from the dust is being stored in the apparatus. Then, when the dust is released, the stored energy is propagated outward, ultimately to infinity; and when it passes the finite radius of the orbit of the test object, that orbit will change to reflect the (reduced) mass of the dust shell alone, rather than dust + apparatus.

But surely my model of a crumbling shell should give the same result but doesn't have a mechanism to remove energy?
I must admit I haven't looked into this in detail for some time, and I'm quite rusty on it. There must be some explanation in GR, but I don't know what it is.

 

[PeterDonis]

surely my model of a crumbling shell should give the same result but doesn't have a mechanism to remove energy?

First, how did the shell get to be static at that radius in the first place? Was it slowly lowered? Did it collapse from some larger radius, and then radiate away the kinetic energy from the collapse? If it radiated energy away before, it could do so again.

Second, any "crumbling shell" model has to explain where the pressure goes, locally, when it vanishes. It can't just disappear because that would violate $\nabla_\mu T^{\mu \nu} = 0$. It has to be transferred to some other stress-energy component locally, before it can be "removed" to infinity. In a typical collapse scenario--for example, a star that stops generating heat--it gets transferred to the inward momentum of the collapsing matter; then, as the collapsing matter particles collide with each other, it gets transferred to radiation that escapes to infinity.

 

[Jonathan Scott]

Second, any "crumbling shell" model has to explain where the pressure goes, locally, when it vanishes. It can't just disappear because that would violate $\nabla_\mu T^{\mu \nu} = 0$. It has to be transferred to some other stress-energy component locally, before it can be "removed" to infinity. In a typical collapse scenario--for example, a star that stops generating heat--it gets transferred to the inward momentum of the collapsing matter; then, as the collapsing matter particles collide with each other, it gets transferred to radiation that escapes to infinity.

Remember that the zero divergence means that energy and momentum are conserved (and continuous). It does not mean that pressure or stress is conserved, although there are obviously temporary transients in those if the pressure suddenly changes.
Perhaps the solution lies in the fact that the "pressure" in the tensor is actually simply a two-way flow of momentum, so if dust particles start moving past each other instead of pushing against each other that would still count as pressure on average, but that definitely doesn't change the total energy.

 

[PeterDonis]

Remember that the zero divergence means that energy and momentum are conserved (and continuous). It does not mean that pressure or stress is conserved, although there are obviously temporary transients in those if the pressure suddenly changes.

I didn't say pressure was "conserved"; I said its behavior is constrained by the zero divergence condition--that is what tells you "where the pressure goes" when it changes. The zero divergence condition applies to all stress-energy tensor components, not just energy and momentum. Pressure and stress are stress-energy tensor components. Try writing down an individual component of $\nabla_\mu T^{\mu \nu}$ where $\nu$ is a spatial index, not the time index (the latter is what gives what is normally viewed as energy-momentum conservation).

 

[Buzz Bloom]

I also don't understand why calculating the escape velocity of a single particle is the shell, and multiplying by Nm/c2, fails to correctly calculate the binding energy of the shell at t=0.

I just noticed that I had carelessly made two errors in this sentence. Here is what I should have said:

I also don't understand why calculating the escape velocity of a single particle in the shell, and multiplying its square by Nm/c2, fails to correctly calculate the binding energy of the shell at t=0.
​

Regards,
Buzz

 

[Jonathan Scott]

I didn't say pressure was "conserved"; I said its behavior is constrained by the zero divergence condition--that is what tells you "where the pressure goes" when it changes. The zero divergence condition applies to all stress-energy tensor components, not just energy and momentum. Pressure and stress are stress-energy tensor components. Try writing down an individual component of $\nabla_\mu T^{\mu \nu}$ where $\nu$ is a spatial index, not the time index (the latter is what gives what is normally viewed as energy-momentum conservation).

That sounds like wishful thinking to me; I'd like to see a worked example, but I think I satisfied myself that nothing magic happens some time ago after we were discussing the Komar mass expression.
The divergence condition applies separately to each of the four components, energy and each component of momentum, and effectively says that the amount moving through the three spacelike walls of the infinitesimal box at each point is equal and opposite to the rate of change with respect to time within that box. It doesn't impose any condition that's not already there in Newtonian mechanics (except of course that the covariant aspect means that it is conserved relative to local space rather than the coordinate system).
Although there are transient effects of a wave of unbalanced pressure passing through the system, I'm not aware of anything which prevents the initial state having a certain amount of energy-momentum plus pressure and the final state having the same amount of energy-momentum but no pressure.
It may be possible to calculate the GR stuff in this case from first principles, but it seems to be beyond my time and skills at present.

 

[Jonathan Scott]

Although there are transient effects of a wave of unbalanced pressure passing through the system, I'm not aware of anything which prevents the initial state having a certain amount of energy-momentum plus pressure and the final state having the same amount of energy-momentum but no pressure.

I should add that of course there is a significant difference between those situations, in that the initial state is static, and in the final state the lack of pressure means that the situation is dynamic and acceleration must be present, although if the pressure change is sudden, the energy and momentum will not have had time to change significantly from the initial state.

 

[PeterDonis]

The divergence condition applies separately to each of the four components, energy and each component of momentum, and effectively says that the amount moving through the three spacelike walls of the infinitesimal box at each point is equal and opposite to the rate of change with respect to time within that box.

Exactly--it relates the quantity of one thing to the rate of change of something else. See below.

I'm not aware of anything which prevents the initial state having a certain amount of energy-momentum plus pressure and the final state having the same amount of energy-momentum but no pressure.

There's nothing which prevents that, as long as the change in pressure is properly related to the rate of change of something else. See below.

in the final state the lack of pressure means that the situation is dynamic and acceleration must be present, although if the pressure change is sudden, the energy and momentum will not have had time to change significantly from the initial state

The energy and momentum won't have had time to change significantly, but their rate of change will have to. That's what the covariant divergence equation says.

(Note that the description you gave implicitly uses a global, non-inertial coordinate chart, in which the energy/momentum is not changing in the initial state but is in the final state. In a local inertial frame, in the initial state, the pressure that is present is related to the energy flux--which is there because an infinitesimal piece of the matter, being under pressure, has nonzero proper acceleration and therefore is changing its energy/momentum as viewed in a local inertial frame. In the final state, the pressure is zero and the infinitesimal piece of matter is now in free fall, with zero change in energy/momentum as viewed in a local inertial frame. The covariant divergence equation requires the two changes--difference in pressure vs. difference in rate of change of energy/momentum--to be equal.)

 

[Jonathan Scott]

Now I'm confused. Thanks for playing, but I think you just moved the goalposts off the pitch, to use a soccer analogy.

The conventional assertion is that in GR, the diagonal pressure terms in the tensor effectively contribute to the curvature of space in the same way as energy, and that is what I find hard to believe, because the pressure can drop to zero (almost instantaneously) without any change in the energy-momentum. If Birkhoff's theorem covers this case, the external field should be unchanged by any internal spherically symmetrical changes, but the pressure can abruptly change while the other terms remain unchanged. I don't see how any rates of change relate to this effect at all.

A very similar point is the subject of one of Tolman's paradoxes, and in some other thread recently I saw a reference to a paper which attempts to resolve that and concludes, roughly, after expanding the GR tensors, that the pressure term doesn't actually affect the external field but something else does, having an equivalent effect. I tried to understand the paper at the time, but it seemed to be looking at things in a way which I didn't trust, so I couldn't trust the conclusion either.

I don't understand this well enough to find the answer at the moment, and I'm not even sure I'd recognize it if I saw it. However, I'll add it to my list of things to think about if I ever get the time before my brain totally wears out, and if I make any interesting progress I'll start a new thread on it.

 

[Buzz Bloom]

I would like to return the discussion to the questions I asked in post #1. The discussion has suggested that the equation I wrote (repeated below) for the self potential energy of a spherical dust shell with zero velocity, temperature and pressure might be wrong, but there have been no specific statements that it is definitely wrong.

M_g = G N m / (2 c² R₀).

Rather, much of the discussion has been about how the assumed t=0 configuration of the shell got to be that way during t<0.

Why is the history of the shell during t<0 relevant to a calculation of its potential energy at t=0? Why isn't the formula for mass equivalent M_g of the potential energy at t=0 correct?

I have searched the internet for any discussion of potential energy with respect to GR, or the Schwartzschild metric, and failed to find anything. The conclusion that seems plausible to me is that there is no GR change to the Newtonian calculation for potential energy with respect to a spherically symmetric mass distribution. If someone definitely know this to be wrong, I would much appreciate a correction.

Regards,
Buzz

 

[PeterDonis]

The conventional assertion is that in GR, the diagonal pressure terms in the tensor effectively contribute to the curvature of space in the same way as energy

More precisely, if you rearrange the terms in the EFE appropriately, they do. Briefly, if you move the trace term from the LHS to the RHS, and look at the 0-0 component of the EFE, you get an equation relating the 0-0 component of the Ricci tensor, which describes the initial inward acceleration of a small ball of test particles, to the sum of the energy density plus three times the pressure at the center of the ball (assuming isotropic pressure). John Baez describes this in more detail here:

http://math.ucr.edu/home/baez/einstein/node3.html

(The whole article is worth reading, btw.)

But this is all purely local; what you are talking about when you mention Birkhoff's theorem and Tolman's paradox is trying to integrate all this local stuff over a spacelike slice to get the total mass of a system. That's a different question from the question of what terms contribute to the EFE locally.

the pressure can drop to zero (almost instantaneously) without any change in the energy-momentum

What causes the pressure to drop to zero? Can it happen without anything else changing?

Consider an example: we have a star, supported against its own gravity by the pressure created by its high temperature, which is a result of nuclear reactions in its core. Then, suddenly, the nuclear reactions in the core stop. What happens? Does pressure suddenly drop to zero throughout the star? Of course not. The fact that reactions have shut down doesn't instantaneously change the pressure. The pressure is kinetic; it's due to temperature, and the temperature doesn't just go away because nuclear reactions stopped. What stopping the nuclear reactions does is to change the rate of change of temperature, from zero to some negative value. (This shows up as a change in the rate of change of energy density, since temperature is a part of the energy density.) So the temperature in the core starts dropping. This causes the pressure in the core to drop, which causes the core to start imploding. So the pressure isn't even the first thing to change.

More generally, consider some static object with a certain energy density and pressure, at some instant of time in its rest frame. Now consider the same object, one instant of time later in its rest frame, and suppose its pressure has dropped to zero. What does that mean in spacetime terms? It means we have one spacelike slice through the object, with nonzero energy density and pressure; and the next spacelike slice, adjacent to the first one, with the same energy density but zero pressure.

Now if you just look at the EFE in the object's rest frame, you are right that there is no component that directly relates the time rate of change of pressure to anything else. The time rate of change of energy is related to the spatial rate of change of momentum, and the time rate of change of momentum is related to the spatial rates of change of pressure and stress.

But the EFE is a covariant equation; it doesn't just hold in the object's rest frame, it holds in every frame. That means it must hold in a frame in which the object is moving rapidly at a constant velocity; and in that frame, the sudden drop in pressure between the two spacelike slices I described above will show up as a spatial rate of change of pressure, not just a time rate of change. But there won't be any time rate of change of momentum to correspond, since the momentum of the system in this new frame is constant. So the postulated pair of spacelike slices does violate the EFE.

Now, can you show me a case of "sudden drop in pressure with no change in energy/momentum" that does not violate the EFE in some frame, as described above?

 

[Jonathan Scott]

John Baez describes this in more detail here:

Yes, I've been there before, and as far as I'm concerned it merely makes the problem more unavoidable.

What causes the pressure to drop to zero? Can it happen without anything else changing?

Consider a pair of small masses held apart statically by a light thin rigid rod. The integral of the pressure over the rod is then equal and opposite to the potential energy of the pair of masses. Now suppose the rod has a plane cut in it, with frictionless surfaces on either side, and displace one part slightly to the side so that the ends pass each other. The pressure then drops to zero in the rod, as a wave that propagates at something like the speed of sound in the rod, possibly followed by a little lengthwise oscillation in the rod. When the pressure drop reaches the mass, it starts to accelerate, but the location and amounts of energy and momentum are initially unchanged.

(Actually, I think that relative to typical coordinates the total energy remains unchanged as the masses accelerate, as in the Newtonian view, at least to first order, but the two masses start to gain equal and opposite momentum towards each other).

Similarly consider any more complex object as a larger number of small masses with some structure of rods between them. The forces and pressures involved between each pair of masses again add up to be equal and opposite to the potential energy.

It is easy to show that the actual energy of the tension in the rod cannot anywhere near add up to the missing "pressure" equivalent of potential energy, since the force is the same as the force between the masses but the displacement through which it does work is negligible, where it would have to be the same as the distance between the masses to do as much work as the potential energy.

 

[PeterDonis]

Now suppose the rod has a plane cut in it, with frictionless surfaces on either side, and displace one part slightly to the side so that the ends pass each other. The pressure then drops to zero in the rod

Drops to zero instantly? Try looking at this in a frame in which the rod is moving rapidly; it will be subject to the same objection I gave in my previous post.

What would actually happen is that the two pieces of the rod would start expanding, starting at the cut off ends, as soon as the ends were displaced sideways so that they weren't pushing against each other. The expansion would gradually reduce the pressure inside the rod, but only gradually; it wouldn't just instantly disappear. (Think of a spring that has been compressed and then is released; the pressure in the spring doesn't immediately disappear, it expands the spring and is reduced gradually in the process of expansion.) A wave of reduced pressure would also start propagating towards the masses, but by the time it reached the masses ("time" here means time in the center of mass frame of the system as a whole), the cut off ends of the rod would already be moving at significant velocity.

It is easy to show that the actual energy of the tension in the rod cannot anywhere near add up to the missing "pressure" equivalent of potential energy

Even when you integrate the tension over the entire length of the rod? (Also, the rod would be under compression, not tension.)

 

[Jonathan Scott]

I have searched the internet for any discussion of potential energy with respect to GR, or the Schwartzschild metric, and failed to find anything. The conclusion that seems plausible to me is that there is no GR change to the Newtonian calculation for potential energy with respect to a spherically symmetric mass distribution. If someone definitely know this to be wrong, I would much appreciate a correction.

If you have a known central mass and hence a specific static Schwarzschild metric, then the potential energy of a test particle can be consistently calculated by applying the time factor from the metric to convert its local rest energy to the rest energy as reduced by time dilation, which is consistent with the approach used to define energy and frequency at a distance in general. The difference is then equivalent to the Newtonian potential energy. If a particle is released into free fall, the total energy remains constant so the change in kinetic energy is equal and opposite to the change in potential energy.

However, if you have two or more source masses involved, that doesn't work at all. For weak fields we can assume that they each create their own potentials which approximately add, but the potential energy of the first mass due to the second and that of the second due to the first are equal, and together they add up to twice the correct Newtonian potential energy of the system as a whole. A Newtonian solution to this (similar to an approach used for electromagnetic energy) is to assume that there is energy in the field which is equal and opposite to the potential energy, which gives a nicely conserved flow of energy and momentum. However, this cannot be "real" energy for GR purposes, since its existence depends on the point of view. A free fall observer would not see any energy being transferred to them by a gravitational source.

In Einstein's Field Equations, the diagonal terms of the energy-momentum tensor, representing pressure, also appear to contribute to the gravitational effect, and in a single static system it can be shown (even in Newtonian theory) that the volume integral of the pressure over a complete system (integrating pressure over three perpendicular planes which sweep through the volume of the system) is equal to the potential energy, even though it is not technically energy. This means that the gravitational effect of a star in this case is equivalent to that of the mass of the matter that made it up minus the binding energy, matching the Newtonian picture. But this relies on counting the pressure as being "like" energy, which I don't think makes sense. In the Newtonian picture, the potential energy is initially unaffected if a system ceases to be static, for example because some support breaks, but the integral of the internal pressure changes abruptly.

What I'm not happy about is that pressure is not conserved, which suggests that the distant gravitational effect of the contents of a black box could fluctuate depending on what is happening inside it (and that's also the essence of the relevant Tolman paradox).

Some people (for example Lynden-Bell and Katz) consider that one can select a physically meaningful coordinate system and describe gravitational energy in that frame using a quantity very similar to the Newtonian field energy, which also matches Einstein's own gravitational energy pseudotensor. That seems to make a lot of sense to me, but I'm not sure it can be fitted in consistently with the concept of pressure also providing a source term.

So as far as I'm concerned, you are describing a system which runs directly into that Tolman paradox, because it is spherically symmetric but not static and involves more than one mass (because of the binding energy), and I don't know a consistent way to describe the energy in that case.

 

[Jonathan Scott]

Drops to zero instantly? Try looking at this in a frame in which the rod is moving rapidly; it will be subject to the same objection I gave in my previous post.

I don't see why that's relevant. This is all simple Newtonian physics.

A wave propagates at around the speed of sound in the rod, which is essentially instantaneous compared with the rate at which the object will fall. Speed in the rod is not relevant; it is assumed to be very light, but rigid, so it will not get very far, and can remain attached to its mass without changing the physics significantly.

Even when you integrate the tension over the entire length of the rod? (Also, the rod would be under compression, not tension.)

The energy stored in any compression of the rod is equal to the integral of the force applied times the distance through which the rod was compressed as a result, which can be made as small as you like by increasing the rigidity of the rod. In contrast, the potential energy is equal to the same force times the distance between the masses.

 

[PeterDonis]

This is all simple Newtonian physics.

Do you think the pressure drops to zero instantly in Newtonian physics? If so, please show your work. I think it drops gradually as the cut off ends of the rod expand, as I described in my previous post.

Anyway, I thought we were discussing the EFE and the covariant divergence condition, which are relativistic, not Newtonian. That's what my comment about having to be valid in every frame applies to.

A wave propagates at around the speed of sound in the rod, which is essentially instantaneous compared with the rate at which the object will fall

But not compared to the rate at which the cut off ends of the rod will expand once they are moved sideways.

 

[PeterDonis]

The energy stored in any compression of the rod is equal to the integral of the force applied times the distance through which the rod was compressed as a result, which can be made as small as you like by increasing the rigidity of the rod. In contrast, the potential energy is equal to the same force times the distance between the masses.

The comparison you are making here is not valid.

In the first sentence above, you are implicitly considering a process where we start with the two masses and an uncompressed rod, and let the masses compress the rod until equilibrium is reached. The compression of the rod is achieved by applying a certain force through a certain distance, doing a certain amount of work on the rod; the work done is equal to the change in potential energy between the two masses due to decreasing their separation by the distance through which the rod is compressed. In the limit in which the distance is much smaller than the total separation of the two masses (i.e., the total length of the rod), the force required to do this work is equal to the gradient of the potential energy at that separation.

In the second sentence, however, you are implicitly considering a process in which the two masses start out in contact, and enough work is done on them to separate them by a distance equal to the total length of the rod in the equilibrium described above. If we assume that the rod itself is doing the work, then the rod has to expand from essentially zero length to the length required to separate the masses in that equilibrium. That means the rod has to start out with a much greater amount of energy stored in it than it will have in equilibrium; all that additional energy gets transferred into the potential energy between the masses at the final separation. And, of course, the force required to accomplish the separation will be much larger when the separation is zero than when the separation is close to the final value; so this much larger value of energy arises because a much larger average force is being applied, as well as because it is applied through a much larger distance. But in any case, this much larger value is irrelevant because it is associated with a different process than the one above, so there's no reason to expect it to be the same as the energy associated with that process. The only thing that should be the same between the two is the final energy stored in the rod, because the final equilibrium state, by hypothesis, is the same in both cases.

In other words, trying to compare the actual compression energy stored in the rod with the potential energy between the masses is a red herring to begin with, because potential energy isn't "stored" in anything material. In fact, potential energy is only well-defined in the first place in a static situation; talking about the potential energy between the two masses as they are separated, i.e., in a non-static situation, is hand-wavy to start with, because it isn't really well-defined. If you actually look at how mass integrals like the Komar mass are done, they don't include any "potential energy" term. (I'll go further into that in a separate post.)

 

[Jonathan Scott]

Do you think the pressure drops to zero instantly in Newtonian physics? If so, please show your work. I think it drops gradually as the cut off ends of the rod expand, as I described in my previous post.

Depending on the physical properties of the rod, I might expect the end of the rod to overshoot and oscillate, with various waves propagating back and forth along it briefly. However, I consider this detail to be irrelevant as the total energy being dissipated in the rod is tiny compared with the potential energy, for the reasons I already gave, so I really don't think the details of how the energy of compression of the rod disperses are relevant. I'd expect there to be various longitudinal oscillations carrying away the energy of the compression.

 

[Jonathan Scott]

In other words, trying to compare the actual compression energy stored in the rod with the potential energy between the masses is a red herring to begin with, because potential energy isn't "stored" in anything material.

I don't think this is something I would ever have disagreed with. I'm just pointing out that there isn't anything available of the appropriate magnitude to replace the pressure integral in the static situation, which is exactly equal in magnitude to the potential energy.

 

[PeterDonis]

Depending on the physical properties of the rod, I might expect the end of the rod to overshoot and oscillate

As I suggested before, consider the rod as a spring under compression. Do you think that, if you compress a spring and then release it, it might not spring back, if the "physical properties" of the spring aren't set up right?

 

[Jonathan Scott]

In the second sentence, however, you are implicitly considering a process in which the two masses start out in contact, and enough work is done on them to separate them by a distance equal to the total length of the rod in the equilibrium described above.

I was not suggesting anything of the kind. Mathematically the potential energy $GmM/r$ is equal to the force $GmM/r^2$ times the distance $r$ between the masses. That means it is obviously a lot more than the same force $GmM/r^2$ times a tiny distance.

 

[Jonathan Scott]

As I suggested before, consider the rod as a spring under compression. Do you think that, if you compress a spring and then release it, it might not spring back, if the "physical properties" of the spring aren't set up right?

I don't see what this has to do with the situation, but if you like you can imagine the rod replaced by a perfect spring initially in compression, which will then oscillate when released, but still the internal pressure will on average be zero as there is no external force on the end.

 

[PeterDonis]

I'm just pointing out that there isn't anything available of the appropriate magnitude to replace the pressure integral in the static situation, which is exactly equal in magnitude to the potential energy.

Let's look at the actual Komar mass integral, which is the one that shows the issue we're discussing. For the case of a spherically symmetric, static mass distribution, that integral looks like this [note: edited to correct the factor under the square root]:

$$
M = \int 4 \pi r^2 dr \sqrt{g_{rr} g_{tt}} \left( \rho + 3 p \right)
$$

The exact form of $g_{tt}$ will depend on the exact form of $\rho$ and $p$ as functions of $r$.

Notice that there is no "potential energy" term. There is, however, the extra factor under the square root sign, which is there because of the variation in "potential" with radius. It turns out that the effect of this factor, which reduces the value of the integral, exactly cancels the effect of adding the pressure, so the final integral is the same as if we just did

$$
M = \int 4 \pi r^2 dr \rho
$$

In other words, we get the same answer as we would if we just "naively" did the integration the way we would in Newtonian physics.

Now, what happens if, by some process, the pressure of this system vanishes? We'll defer (for a bit) the question of how fast it can vanish (it can't do so instantaneously, for the reasons I've given in previous posts). The point is, as long as the process is self-contained, i.e., no energy escapes to infinity, it can't change the value of $M$. So how does $M$ stay the same if $p$ becomes zero?

The answer is that the above integral is not valid for a non-static system. The factor under the square root is derived from the norm of the timelike Killing vector field of the spacetime, and in a non-static system, such as a star that is collapsing because its internal pressure has been removed by the stoppage of nuclear reactions in its core, there is no timelike Killing vector field. (At the instant that the change begins, there is "almost" a timelike KVF--but at that instant, the pressure can't have instantaneously dropped to zero; what can change in an instant, as I've pointed out before, is that the constraint that is keeping the system static can be removed. But removing that constraint doesn't instantly remove the pressure; it just allows the system to start moving to remove the pressure. The pressure won't actually be zero until there has been enough motion to remove it, by which point the non-staticity will be significant.)

So how can we even evaluate $M$? If the system is asymptotically flat, we can use the ADM mass (or the Bondi mass, but for the case of no radiation escaping to infinity, they're the same). The ADM mass does not look at the stress-energy tensor; it just looks at the metric coefficients, and it looks at them in a way that is not sensitive to the details of what is changing in the interior of the system; all it is really testing is what the metric looks like far away, to a test object in orbit about the system. By hypothesis, that won't change in the above scenario, and that is why we can say $M$ stays the same even though changes are happening in the interior of the system.

On this view, then, there really isn't a way to compute $M$ by adding up local contributions from all the individual pieces of a system if the system is not static. In other words, on this view, there is no way to say, once the system ceases being static, "where the energy is stored" locally; all you can do is say that $M$, the total energy as viewed from far away, doesn't change. But many people find that unsatisfactory, so they try to find pseudotensors of one sort or another that capture "energy stored in the gravitational field". The key limitation of all of these, for someone who likes the fact that GR expresses everything in covariant form, is that defining any of these pseudotensors requires picking a particular frame, and only works in that frame. But for the case under discussion, that shouldn't be a serious limitation, since there is already a natural frame to pick: the one in which the system starts out static. If everything is self-contained, that frame should still work as the "center of mass" frame of the system, even when it is no longer static.

I have not seen an analysis of this specific case using the pseudotensor method, but I expect that such an analysis would be interesting.

 

[PeterDonis]

if you like you can imagine the rod replaced by a perfect spring initially in compression, which will then oscillate when released, but still the internal pressure will on average be zero as there is no external force on the end

The internal pressure will on average be zero once a new equilibrium has been reached. The oscillations aren't important--assume that the spring is critically damped so once it is released it just goes to its new equilibrium position and stays there (as seen in the center of mass frame of the rod--more precisely, of each piece of the rod individually). The point is, the new equilibrium position is different from the original one, and it takes time for the spring to go from one to the other. The time it takes is of the same order as the change in length of the spring (or rod) divided by the speed of sound in the material. That is, it is of the same order as the travel time of the wave disturbance that travels through the rod when the constraint is released. And in that time, the two pieces of the rod, as viewed in the center of mass frame of the system as a whole, go from static to moving--the energy that was stored in the compression of the rod gets converted to kinetic energy of the pieces of the rod.

Both of the above times are much shorter than the time it will take the two masses to come back together, but that's fine. The key point is that the energy that was stored in the compression of the rod goes somewhere: into the motion of the pieces. At the end of the time described above, each piece of the rod will be moving towards the opposite mass, and the two masses will just be starting to fall together.

You might object that there is still a lot of potential energy in the system at this point--the energy that is now manifest as kinetic energy in the pieces of the rod is much smaller than the potential energy that will become kinetic energy of the two masses once they have fallen together. That is true (at least, given the assumptions we have been implicitly making about the relative magnitudes of the two--it might well be that those assumptions would require the rod to be made of exotic matter, violating the energy conditions, but that's a separate issue that I don't think affects this discussion). But all that potential energy was never "stored" in the rod to begin with. It wasn't "stored" anywhere, except in the curvature of spacetime if you want to look at it that way.

Consider a slightly different case: two masses which continually bounce off each other, rise, stop, and then fall together again to bounce once more. (This is basically the two masses and rod example with the rod removed--assume the maximum separation of the two masses is the same as the length of the rod in the version with the rod present.) What will the mass M of the whole system be? Will it just be the sum of the two rest masses? No. Will it be the sum of the two rest masses, minus some static "potential energy"--say the gravitational binding energy of a system that would be formed if the two masses were combined into a single static mass? No. It will be some value in between the two; $M$ could be viewed, heuristically, as consisting of three terms: the sum of the two rest masses, minus the static binding energy, plus the extra potential/kinetic energy due to the bouncing (entirely potential energy at maximum separation, entirely kinetic at minimum separation, a combination of the two in between). Where is that extra energy "stored"? Nowhere tangible. Possibly an analysis using pseudotensors, of the kind I mentioned in my previous post, might offer an intuitively plausible interpretation; but that's the best that can be done. There is no one "right" answer; the strict GR answer is that the question isn't well-defined in the first place.

 

[Jonathan Scott]

I'll have to look into my own notes on the Komar mass integral some time; I believe there are some complications relating to the choice of coordinates. The result which I'd previously understood was that the time dilation factor effectively resulted in exactly double the binding energy decrease (for reasons previously mentioned) but adding in the pressure compensated for one of those, giving the local mass minus the binding energy, matching the Newtonian view.

Anyway, when I previously tried to find something to resolve this, instead I found it to be a controversial and unresolved subject. Here are two of the papers which I previously found:

"Einstein's Gravity Under Pressure", Astrophys.Space Sci.321:151-156,2009 and http://arxiv.org/abs/0705.0825
"Pressure as a source of gravity", Phys.Rev. D72 (2005) 124003 and http://arxiv.org/abs/gr-qc/0510041

As far as I'm concerned, the original poster's question falls straight into this unresolved and apparently self-contradictory area which has long puzzled me, and I'm giving up on it for now.