Question regarding GR wrt potential energy

[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.

 

[Buzz Bloom]

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.

Hi Jonathan:

Thanks very much for for your post.

I find myself still in the dark about how to proceed. I guess I will just have to accept that this concept is too difficult for me to understand.

I wonder why the issue of pressure is relevant to the scenario I proposed. I assumed that at t=0 pressure is zero, and I see no reason for it to become greater than zero for t>0. I understand that the system is dynamic and that the shell will move toward its center for t>0, and as that happens the potential energy at t=0 will gradually reduce as the kinetic energy increases from zero so that kinetic plus potential remains a constant. I do not see why that would prevent a calculation of potential energy at t=0.

Regarding the Tolman paradox. I found an article,

http://www.ejtp.com/articles/ejtpv6i21p1.pdf​

but it seems to have nothing to do with my scenario. Is there some other Tolman paradox you had in mind?

Regards,
Buzz

 

[PeterDonis]

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.

Let's look at this for the Komar mass integral I posted earlier. In order to obtain expressions that we can actually solve analytically, I'm going to assume (highly unrealistically) that the density $\rho$ is constant, independent of $r$. For this case, there are known closed-form solutions for $g_{tt}$, $g_{rr}$, and $p$ as functions of $r$, which are (note that $R$ is the radial coordinate of the surface of the object):

$$
g_{tt} = \frac{1}{4} \left( 3 \sqrt{1 - \frac{8}{3} \pi \rho R^2} - \sqrt{1 - \frac{8}{3} \pi \rho r^2} \right)^2
$$

$$
g_{rr} = \frac{1}{1 - \frac{8}{3} \pi \rho r^2}
$$

$$
p = \rho \frac{\sqrt{1 - \frac{8}{3} \pi \rho r^2} - \sqrt{1 - \frac{8}{3} \pi \rho R^2}}{3 \sqrt{1 - \frac{8}{3} \pi \rho R^2} - \sqrt{1 - \frac{8}{3} \pi \rho r^2}}
$$

Notice that the denominator of $p$ is equal to $2 \sqrt{g_{tt}}$ and the first term in the numerator (and the second term in the denominator) is equal to $1 / \sqrt{g_{rr}}$; so we can write the Komar mass integral as (note that we have expanded out the numerator and canceled terms):

$$
M = \int_0^R 4 \pi r^2 dr \sqrt{g_{rr} g_{tt}} \rho \frac{2}{2 \sqrt{g_{rr} g_{tt}}} = \int_0^R 4 \pi r^2 \rho dr = \frac{4}{3} \pi \rho R^3
$$

Now, in order to assess the gravitational binding energy, we need to know what to compare $M$ to. What we want is what the total mass of the matter forming the object would be if it were all moved out to rest at infinity and widely separated. If you think about it, it should be evident that this total mass is given by the following integral:

$$
M_0 = \int_0^R 4 \pi r^2 dr \sqrt{g_{rr}} \rho = \int_0^R 4 \pi \rho \frac{r^2}{\sqrt{1 - \frac{8}{3} \pi \rho r^2}}
$$

This doesn't have a handy closed-form solution as it stands, but we can see that it is larger than $M$, and if we go to the weak field limit where $\frac{8}{3} \pi \rho r^2 << 1$, so that we can expand the square root in the denominator, we obtain

$$
M_0 \approx \int_0^R 4 \pi r^2 \rho dr \left( 1 + \frac{4}{3} \pi \rho r^2 \right) = \frac{4}{3} \pi \rho R^3 + \frac{16}{15} \pi^2 \rho^2 R^5 = M \left( 1 + \frac{3}{5} \frac{M}{R} \right)
$$

So the binding energy in this limit is approximately $3/5$ of the "naive" Newtonian value of $M / R$ times the rest mass.

Now we need to assess the impact of the pressure term. The obvious way to do that is to evaluate the integral

$$
M_1 = \int_0^R 4 \pi r^2 dr \sqrt{g_{rr} g_{tt}} \rho = \int_0^R 4 \pi r^2 dr \rho \frac{1}{2} \frac{3 \sqrt{1 - \frac{8}{3} \pi \rho R^2} - \sqrt{1 - \frac{8}{3} \pi \rho r^2}}{\sqrt{1 - \frac{8}{3} \pi \rho r^2}} = \frac{1}{2} \left( 3 M_0 \sqrt{1 - \frac{2M}{R}} - M \right)
$$

In the weak field limit, we can expand the square root and discard terms beyond first order in $M / R$, and we obtain

$$
M_1 = \frac{1}{2} M \left[ 3 \left( 1 - \frac{2}{5} \frac{M}{R} \right) - 1 \right] = M \left( 1 - \frac{3}{5} \frac{M}{R} \right)
$$

So in this limit, yes, we can view the Komar mass integral as the outcome of the following process: take the mass $M_0$ at infinity, bring it all into a spherical shape with radius $R$, which in the absence of pressure would have mass $M_1$, and then support it against its own gravity with pressure, which adds back half of the difference $M_0 - M_1$ to obtain a final mass $M$. The only difference is the factor $3/5$.

There is probably a slick way to generalize this result by using the equation for hydrostatic equilibrium, but I'm not up to trying it right now.

 

[PeterDonis]

I wonder why the issue of pressure is relevant to the scenario I proposed.

It isn't directly relevant to your scenario, since you have specified that the pressure is zero. But it is relevant to figuring out how to calculate the binding energy in your scenario, because the only formulas we have to start with are formulas for static situations, in which pressure must be present.

I do not see why that would prevent a calculation of potential energy at t=0.

The problem is that the concept of "potential energy", in general, is not well-defined in a non-static situation in GR. I think it can be defined for your specific situation, but it requires some care.

 

[Jonathan Scott]

Regarding the Tolman paradox. I found an article,
http://www.ejtp.com/articles/ejtpv6i21p1.pdf but it seems to have nothing to do with my scenario. Is there some other Tolman paradox you had in mind?

That's the most well-known Tolman paradox, but the GR pressure one is a different paradox.

 

[Jonathan Scott]

...
So in this limit, yes, we can view the Komar mass integral as the outcome of the following process: take the mass $M_0$ at infinity, bring it all into a spherical shape with radius $R$, which in the absence of pressure would have mass $M_1$, and then support it against its own gravity with pressure, which adds back half of the difference $M_0 - M_1$ to obtain a final mass $M$. The only difference is the factor $3/5$.

Thank you for confirming that. The factor of 3/5 is correct for the Newtonian binding energy of a uniform sphere. See Wikipedia for the calculation: https://en.wikipedia.org/wiki/Gravitational_binding_energy

 

[Buzz Bloom]

That's the most well-known Tolman paradox, but the GR pressure one is a different paradox.

Hi Jonathan:

Can you cite a reference for this paradox. I found the following searching "Tolman paradox general relativity pressure".

https://en.wikipedia.org/wiki/Tolman–Oppenheimer–Volkoff_equation​

I found no discussion of a paradox there.

Regards,
Buzz

 

[Buzz Bloom]

Hi Peter:

Thank you for your answers to my questions.

It isn't directly relevant to your scenario, since you have specified that the pressure is zero. But it is relevant to figuring out how to calculate the binding energy in your scenario, because the only formulas we have to start with are formulas for static situations, in which pressure must be present.

Would the following initial condition work? Suppose each of the tiny spherical dust particles had a charge such that that their electrostatic force established a pressure which maintained the static position of the shell. Then at t=0, the charge vanishes.

The problem is that the concept of "potential energy", in general, is not well-defined in a non-static situation in GR. I think it can be defined for your specific situation, but it requires some care.

I accept this is true, but it seems very strange, since the concept is well defined in Newtonian physics in terms of the conservation of energy: potential plus kinetic equals zero. Are you saying that the care required for solving a dynamic GR relationship about energy includes a need to consider states preceding the initial conditions at some t=0?

Regards,
Buzz

 

[Jonathan Scott]

Hi Jonathan:

Can you cite a reference for this paradox. I found the following searching "Tolman paradox general relativity pressure".

https://en.wikipedia.org/wiki/Tolman–Oppenheimer–Volkoff_equation​

I found no discussion of a paradox there.

Regards,
Buzz

I think it is mentioned in both of the articles which I referenced earlier in this thread.

 

[Buzz Bloom]

I think it is mentioned in both of the articles which I referenced earlier in this thread.

Hi Jonathan:

Thank you for the citation. I have an interest in paradoxes, so I will try to read the discussion of the Tolman paradox in these articles. However, I expect to have difficulty in understanding the discussion.

My interest in paradoxes started many decades ago, and led me to what has become an aphorism for me:

The nature of reality is fundamentally paradoxical.​

Regards,
Buzz

 

[PeterDonis]

The factor of 3/5 is correct for the Newtonian binding energy of a uniform sphere.

Ah, thanks, I had forgotten that the Newtonian calculation for the sphere gives that factor as well. That's a good thing since the weak field limit should match the Newtonian answer.