# Stress and pressure as gravitational source

1. Oct 16, 2014

### Jonathan Scott

As I'm sure I've mentioned before, I've heard people say that stress and pressure terms in the energy tensor act as a gravitational source, but I don't understand how this could be, given that these terms can simply appear or disappear (in a non-stationary situation), unlike energy and momentum.

I understand that in GR we normally discuss either static cases or fluid equilibrium, where this problem does not occur, but surely the fact that pressure or stress can vanish in this way does not fit with what we know about gravitational sources.

For example, if two small masses are held apart in a stationary configuration by a couple of rigid rods end to end, then the integral of the pressure in the rods is mathematically equal to the potential energy of the configuration (both by Newtonian physics and via the GR Komar mass expression). If one rod then one slips past the other, then the pressure in the rods drops to zero instantly (well, technically, at the speed of sound in the rod), but the potential energy of the configuration is surely initially unchanged (at least in the Newtonian view), and one would also expect the gravitational field to be essentially unchanged at first.

In the Newtonian view the integral of the pressure in the static case is equal to the potential energy, but it cannot actually be the potential energy, as is illustrated by what happens when the rod slips. The corresponding opposite tension of the field through space is however unaffected, and remains equal to the potential energy even in a dynamic situation. However, although the total force through a given plane is known, the location of this potential energy is not predetermined; a simple mathematical solution would be to copy electrostatics and model the energy density of the field as $g^2/8\pi$ where $g$ is the Newtonian field. But the GR view does not include "energy density" in the field.

Have I missed something? As far as I can see, these terms could only be a gravitational source if they are replaced in dynamic situations by some other equal terms involving rates of change, in a similar way to mechanical simple harmonic motion.

2. Oct 17, 2014

### Jonathan Scott

Another way of looking at pressure is that it is force per area which is momentum transfer per time per area.

Imagine equal streams of particles (like small rubber balls) bouncing elastically off either side of a block and going back the way they came, creating a certain amount of pressure within the block.

Now take the block away, so the two streams simply pass each other in opposite directions. The overall momentum flow is unchanged, but there is no longer any "pressure" as such involved. Should this change the gravitational effect? And we can arbitrarily insert and remove the block from time to time, which changes the selection of which particles are coming away from that point, but does not affect the overall momentum flow.

So is it perhaps that the term in the stress-energy tensor should not be taken specifically to represent pressure but rather to represent transfer of momentum in general?

And does this help? In the case of the light rigid rod which slips, the momentum transfer within the material has definitely stopped at the same time that the pressure became zero, although in the Newtonian view the overall rate of momentum transfer to the falling object from the field has suddenly jumped up.

3. Oct 17, 2014

### Staff: Mentor

This "technically" is actually a big part of the key to the puzzle. The pressure does not change to zero instantly, which means the stress-energy tensor of the rod does not change instantly. It takes time to change. What else is happening during that time?

I think you should make up your mind whether you want to talk about the Newtonian view or the GR view. They're not the same, and the differences between them, ISTM, are important for this discussion. For one thing, in GR, you can't define potential energy in a non-stationary spacetime, and if the only source of gravity in the problem is the masses and the rods, then when the rods start moving, the spacetime is no longer stationary.

At first, yes--because the stress-energy tensor takes time to change. But as it changes, the gravitational field changes too.

Because there isn't any; the stress-energy tensor does not arise from the field (meaning the metric and its derivatives). It arises from whatever non-gravitational "stuff" is present. When the rods start moving, that stuff is changing, and the change in that stuff changes the field.

Yes, and that pressure contributes to the total stress-energy tensor of the system. The momentum flux also contributes, but it's a different term in the SET--it's one of the time-space terms, not one of the diagonal space-space terms.
But the stress-energy tensor is not, because you took away the block and the pressure within it. So the diagonal space-space terms of the SET are now zero, but the time-space terms that represent the momentum flux are unchanged.
No, it's that momentum flux is a different term than pressure. See above.

4. Oct 17, 2014

### Jonathan Scott

Thanks for having a go, but if it was that simple I wouldn't be asking.

I've looked into this aspect before, and I don't see how it can be relevant. A small displacement wave passes down the rod, but the energy and/or momentum in that wave depends on the material and can be negligibly small compared with the original integral of the pressure, equal to the potential energy of the configuration. And this can finish happening long before any significant displacement of either the rods or the supported masses has occurred, so it has nothing to do with their displacements or velocities.

I think that even for GR purposes, the overall effective gravitational and inertial mass of the system should not be affected by something internal like this. So although I would agree the Komar mass is no longer valid in GR once the system is no longer static, this should not suddenly cause the external field to change.

5. Oct 17, 2014

### Jonathan Scott

As I understand it, the first row is momentum per volume and is equivalent to the first column, energy being moved per time, but the diagonal "pressure" fields are the rate at which momentum is being moved in its own direction (or the opposite direction, giving tension). In that sense, a two-way stream of particles is equivalent to pressure, and I suspect that "pressure" is really effectively a macroscopic term which could be modelled microscopically by a two-way flow of momentum. However, I can't find this in any text book, as they always seem to talk either about fluids or about dust!

6. Oct 17, 2014

### Staff: Mentor

But the change in the stress-energy tensor can't be negligible, because by hypothesis it is equal to the original pressure, which drops to zero after the wave has passed.

No, it can't, because the covariant divergence of the SET must be zero. If the pressure components of the SET drop to zero, some other components must become nonzero to compensate, or the covariant divergence will not be zero. The only other components that can change are the time-time component--the energy density--and the time-space components--momentum density/energy flux. These can only change if the rods and/or the masses move. So something must be moving by the time the pressure drops to zero.

It's true that all of these changes are going to be small on an absolute scale for ordinary rods and masses. But they have to be "significant" in terms of the problem you're posing, because without them conservation laws are violated.

I think it depends on what the internal change is. If, for example, the configuration of matter is spherically symmetric, and stays spherically symmetric as internal changes happen, then yes, the external field is unchanged by Birkhoff's theorem. But I don't know of any more general results that show that "internal" changes can't affect the external field.

Yes. (Actually the first column is energy flux, energy per unit area per unit time; if you work it out this turns out to be equivalent to momentum density.)

Yes. More precisely, it's force in a given direction per unit area perpendicular to that direction, and force is rate of change of momentum, so it's rate of change of momentum per unit area.

No, because there's no force being exerted. The two momentum densities might cancel out, but that's not the same as them being equivalent to pressure.

It is.

Microscopically, pressure is the aggregate result of individual interactions between, say, gas particles, or between gas particles and the walls of a container, where each interaction involves momentum exchange--a gas particle bounces off a wall, or two gas particles bounce off each other. Momentum by itself, with no momentum transfer between particles, does not result in pressure; it's just momentum.

GR textbooks that I've read don't go into the details about how pressure arises, true. I would expect to find a better discussion of it in a textbook on thermodynamics or statistical mechanics. Unfortunately I don't have any particular ones to recommend.

7. Oct 17, 2014

### Jonathan Scott

What is conserved is energy and momentum. Pressure determines how they are changing and flowing with respect to time and space; a sudden change in pressure doesn't violate those conservation laws.

8. Oct 17, 2014

### Jonathan Scott

I think that from considerations of microscopic momentum flow, the two-way stream should be indistinguishable from pressure if it transfers the same amount of momentum (in both directions) in the same amount of time. If you consider this as represented for example by streams of small elastic particles (say gas molecules) within a box, then the questions of whether they actually collide with one another within the box is irrelevant to the pressure at a given point within the box, which can be defined by the force on the sides of the box.

9. Oct 17, 2014

### Staff: Mentor

Not in the most general case. In the most general case, what is "conserved" is the entire stress-energy tensor, in the sense that its covariant divergence is zero. Conservation of energy-momentum is an approximation that works in scenarios where you can describe the system using a 4-momentum vector instead of a full stress-energy tensor (the conservation law is then just that the length of that 4-vector is invariant). If pressure is non-negligible, you can't do that.

If you compute the covariant divergence of the stress-energy tensor, you will see that it does violate the general conservation law, unless other components of the SET change to compensate.

Transfers momentum to what? In the two-way stream case, no momentum is being transferred at all; nothing is exerting any force on anything else.

10. Oct 17, 2014

### Jonathan Scott

No, I'm sticking on this one. The covariant divergence effectively gives you a 4-component local equation of continuity for the energy and for each spatial component of momentum. It's true that if one thing changes, another usually changes, but if for example the pressure drops to zero somewhere, that just means that any forces towards that point will no longer have anything opposing them so stuff which was previously being pushed back may start to accelerate.

If you have equal flows of material in opposite directions, then their contributions to momentum transfer across a plane (as in pressure) add together with the same sign. If one flow is transferring material with momentum in the +x direction in the +x direction, and the other is transferring material with momentum in the -x direction in the -x direction, then the rate of momentum transfer across the plane perpendicular to the x direction is the sum of the two flows. I'm now satisfied (from considering the microscopic model) that "pressure" in general is equivalent to this situation. It is the rate of transfer of momentum across a plane, regardless of whether this involves collisions or just stuff moving both ways.

11. Oct 17, 2014

### Staff: Mentor

Which will change the momentum, which will change the momentum density components of the SET. (It will also change the energy density, because stuff that was at rest is now moving; which means the energy flux components must also change, as of course they must anyway since the tensor is symmetric.)

For an appropriate definition of "transfer", yes. I was using it with a different definition (where "transfer" means "between different particles", not just "transported by the same particle"). See below.

Is it? Suppose we have two streams of non-interacting particles (the "dust" that GR textbooks love to use in models ;) ) moving in opposite directions, with no container or other confinement. The momentum flows add as above, but the pressure is zero, right?

12. Oct 17, 2014

### pervect

Staff Emeritus
Certain aspects of the problem have puzzled me as well, but my conclusion, based on my reading, is that the basic problem is defining what the "energy of the system" is. It appears to me that in general that there isn't any known solution to this problem at the current time, as has been mentioned more than once.

The Komar formula is simple and applies to static, equilibrium space-times. But as you note, as soon as you use a non-static space-time, the formula starts to give, as your examples illustrate, incorrect results. The not-totally satisfying solution is to apply the Komar formula only in equilibrium situations and realize that it fails if you apply it to non-equilibrium situations such as your examples.

A closely related issue, which I'm interested in but haven't found anything about in the literature, would be ask in a quasi-static system, one that is almost-but-not-quite static, can we safely apply the Komar formula in some sort of limit. Unfortunately, I haven't seen this discussed anywhere :(. Probably some sort of limit exists, for instance we know that the formula fails if we "pop" a pressurized balloon that has gravitationally significant pressure. The question is, would it work to some degree of accuracy if we slowly allow the balloon to expand, instead of popping it? But without anything more definite, we can't rigorously justify using the formula in any system that isn't perfectly static. This is an unpleasant situation, if we take it really seriously, we couldn't apply concepts of total mass to our solar system, for instance.

An interesting random thought just hit me, but it's probably too speculative and would derail the thread, I may start another thread if I decide it's not too speculative.

As a sort of an aside, rather than popping the pressurized balloon with a single prick, I envision popping it with a rather large amount of shaped charges, all timed to go off at the same time (in the static frame), to mostly get around the speed of sound issues. I don't have this idea fully fleshed out, as I write this I realize that perhaps I haven't totally gotten around the conceptual issues of the speed of sound being a factor. Certainly if we imagine the shell being perfectly rigid, there is no energy stored in it by putting it in tension or pressure - but of course that's not a realistic assumption.

Anyway, if we assume we have an asymptotically flat space-time, the bad news is that the Komar formula appears to give bad results for non-static systems, as above. But the good news is that in this particular case, at least, there ARE other formulas to the system that do apply to give us a concept of "system energy". These are the ADM and Bondi masses.

The bad news is that these masses are not functions of the stress energy tensor $T_{ab}$, i.e. they are not some sort of integral of the energy density and the momentum density. Rather they are defined from the properties of the metric. The Komar mass has an interesting formula in terms of $T_{ab}$ and the redshift factor, there isn't any such formulation (that I'm aware of) for the Bondi and ADM masses.

13. Oct 17, 2014

### Jonathan Scott

Although the pressure may be zero in the conventional sense, I think that to be consistent the relevant element of the stress-energy tensor should reflect the equivalent rate of momentum transfer. Of course, if there's no containment, there's nothing to turn the streams round and the dust will soon run out. This is similar to the situation just after a balloon has burst, where pressure is rapidly dissipating.

14. Oct 17, 2014

### Jonathan Scott

Thanks very much for a thoughtful post (and again for introducing me to the Komar mass many years ago).

I'm quite sure now that the amount of mechanic energy that can be stored in the material under tension or pressure is not particularly relevant to the problem, nor the finite speed of sound, mainly because these can vary with the material, but the overall result cannot. I think that one can assume any supporting structure to be arbitrarily light and rigid without any loss of generality.

In the GR cases which are actually used for calculation, involving fluids or solids, I get the impression that the pressure term in the SET effectively provides the correct Komar mass adjustment for the potential energy, making everything consistent. However, in a more complex dynamic system this pressure term can abruptly change, even becoming zero, as parts of the system start to accelerate towards one another. One would expect the effective "potential energy" to remain unchanged, at least initially, and for any change to the gravitational field to be much less significant than that which would be produced by the total source energy having decreased by the potential energy. Also, if the original support structure is reinstated quickly (or the falling parts bounce elastically back to their original locations) then the original pressure can be mostly reinstated just as fast as it went.

This apparent inconsistency really bothers me. It feels like something is on the wrong side of the equation, and that the extra term should be more like the Newtonian energy of the field, which always accounts for the potential energy of the configuration and can't flicker into and out of existence like the pressure.

I've been wondering about other answers, for example that there is some symmetry such that if the pressure drops to zero, then although this changes the local curvature, the curvature further out remains unaffected (like redistributing some mass locally). However, I can't see how this would work when something abruptly becomes zero over a large region. I had wondered whether applying Birkhoff's theorem to the spherical case could prove something about the effect of a sudden loss of pressure (e.g. spherical collapse of a previously rigid internal supporting shell) but I suspect that the situation is so symmetrical that the static pressure terms have no overall effect.

I've also been trying to see whether I can find some sort of answer in the Landau-Lifshitz pseudotensor, but although that gives some sort of answer about where the gravitational energy appears to be, I haven't yet been able to find a way to use it to answer the question about whether it makes sense for sudden changes in the pressure to cause sudden changes in the shape of space.

15. Oct 17, 2014

### Staff: Mentor

I still think you're being misled by trying to extend the concept of "potential energy" to non-stationary systems; but in your "rods and masses" scenario it may be difficult to disentangle that issue from others. Let me try to construct a scenario that is simpler to analyze (it is basically what you were suggesting about trying the spherical case).

Suppose we have a static object like a star, supported against its self-gravity by its internal pressure. Then, at some time $t = 0$ (in the coordinates in which the object is static prior to that time), the internal pressure rapidly drops to zero (for example, the star stops burning nuclear fuel). The star then collapses in a spherically symmetric fashion, with each internal piece of the star following an infalling geodesic for $t > 0$ (as in the Oppenheimer-Snyder model of collapse to a black hole--but we won't carry things that far).

Any observer exterior to the star (i.e., at a radius larger than the star's radius in the static period $t < 0$) will see no change at all in its external field, by Birkhoff's Theorem. So the question is, if the external mass of the star is correctly given by the Komar mass calculation at $t = - \epsilon$, with nonzero pressure, how can it possibly stay the same at $t = + \epsilon$, given that the pressure has dropped to zero, if pressure is supposed to contribute to the mass?

To restate the problem a bit more explicitly, at $t = - \epsilon$, the mass was $M = m + p - U$, where, heuristically, $m$ is the contribution from rest mass, $p$ is the contribution from pressure, and $U$ is the contribution from potential energy (and I've given it a minus sign to emphasize that it contributes in the opposite way). Then, at $t = + \epsilon$, we have $m$ basically unchanged, $U$ basically unchanged, but $p$ is now zero, so how can $M$ be the same?

I don't have an immediate intuitive answer to this question, and I don't have time to do any computations right now (the computations I would want to do would involve looking at the relevant components of the covariant divergence of the SET and the Einstein Field Equation before and after). But this scenario might at least focus in more easily on the root issue.

16. Oct 18, 2014

### Jonathan Scott

Yes, as I suggested before, looking at how Birkhoff's theorem hides the change in the spherical case may shed some light on this, but so far I'm not clear on how to do this.

If the pressure or stress terms change suddenly but the external field remains initially unchanged, as might be expected from the parallels with Newtonian theory, this suggests that whatever happens as a result of the pressure or stress change has the same effect as the original terms. What actually happens is that there is an instant change in the acceleration, so if accelerating matter had the same field as the pressure which previously resisted that acceleration then that would answer the paradox. However, acceleration seems to be related to the rate of change of the SET, so one would not expect to see an instant change to reflect the loss of pressure or stress.

At the microscopic level, all pressure or shear stress can be modelled by a lattice of struts and strings which only transfer momentum along their lengths, and those in turn can be modelled by streams of particles on average (apart from the minor complication that particles which mediate attractive forces carry momentum in the opposite direction to their direction of travel).

17. Oct 18, 2014

### Staff: Mentor

I don't think so. Birkhoffs theorem is about spherical symmetry more than about the terms of the SET. It does not hold for other geometries.

I am not sure what you are confused about. Do you think that the SET is not the source of gravity in GR or do you think that pressure is not part of the SET?

18. Oct 18, 2014

### Staff: Mentor

just to clarify, I suggested the spherical scenario to try to "distill" the key question the OP is concerned about from other issues, because Birkhoff's theorem guarantees that, if the object collapses spherically when its pressure goes to zero (the example I gave was the star that exhausts its nuclear fuel), the external field remains the same, so whatever internal changes happen must sum to zero, so to speak, in their effect on the external field. I realize that in non-spherical scenarios, the changes do not necessarily sum to zero; but it seemed possibly helpful to start with the idealized case where they do.

19. Oct 18, 2014

### PAllen

This is now looking a lot like an 'inverse Tolman Paradox'. His was what happens if a reaction inside a shell suddenly increases the pressure (while everything is still confined by the shell). Clearly, an orbiting body shouldn't be influenced. How does this work out? Here we have sudden disappearance of pressure, and though the system in our case ceases to be equilibrium, we ask about a moment after disappearance, so change from equilibrium shouldn't be significant.

The following paper argues for a general conclusion that pressure terms, as integrated into total mass measured from a distance, cancel for equilibrium systems. They only establish it for a few specific cases, but argue that it is likely a general rule.

http://arxiv.org/abs/gr-qc/0510041

Then, the answer to our present mystery would seem to be that if pressure effects cancelled when there was equilibrium, there is then no change when they disappear and the rest of the system has little time to deviate from equilibrium state. They would then become significant as system diverged more from equilibrium, with elements acquiring relative motion, and changes in pressure and shear occurred (in a realistic case). In our idealized case, as Peter proposed the matter elements just collapse geodesically, one expects no change at all in mass at distance despite the non-stationary system.

20. Oct 19, 2014

### Jonathan Scott

I agree that this is an equivalent paradox; this is very interesting.

This doesn't make much sense, and the basic idea appears to be wrong, as it conflicts with the results of the Komar mass formula, although I haven't yet looked at the paper.

From either the Komar mass formula or the Newtonian expression, the integral of the pressure term for a system which is in equilibrium is equal and opposite to the Newtonian gravitational potential energy of the system (with a positive sign). Overall, the area integral of the pressure over any plane has to be zero if the system is in equilibrium. The only "force" which isn't counted in the SET is the gravitational "force", so the integral is that of the gravitational pressure over the volume, which is equal to the potential energy. If there is increased pressure inside, it is countered by a increased negative pressure (tension) in the containing shell, so the overall integral remains constant.

The overall effective mass (again from the Komar mass formula) is given by multiplying the local energy density by the time dilation factor (which effectively reduces the overall mass by exactly twice the potential energy) then adding back the effective mass from the pressure term (opposite and equal to the potential energy) so it is simply equal to the local measured energy minus the potential energy, as for the Newtonian view. Any internal changes, for example changing pressure or the radius of some internal shell, can result in changes both to the potential energy and to the local energy, but the total remains constant as usual.

So this says that whatever you do temporarily within a closed system, when things get back into equilibrium, the Komar mass formula still gives the same result as before, even if the equilibrium state is very different from the previous one. But we already knew that.

The interesting case is the non-equilibrium case, and we know that in such a case, the Komar mass formula isn't even approximately right, as pressure can for example suddenly drop to zero and then be restored without any significant change to the configuration.