Pressure and Newtonian potential energy

[Jonathan Scott]

Total pressure doesn't even make sense. It is an intensive property, not an extensive property. You don't sum it up to get a total pressure.

I simply mean the total integral of the pressure over the system in the way being previously discussed (as in the Komar mass expression and the volume integral of 3p in the Newtonian case for isotropic pressure). I used "total" just to make it clear the pressure was temporarily changing, not just moving, but I guess I could have left some more words into be more accurate.

 

[PeterDonis]

If I have a brick balanced on a brittle stick and the stick snaps, the brick will not have gained much velocity or moved very far by the time the pressure in the stick has reached zero.

But the contribution of the pressure in the stick to the source of gravity is also very small, so you don't actually know that the change in pressure (and the similar small change in energy density) is not balanced by the change in momentum density. You're just waving your hands and claiming it without actually doing the math. And, as I said before, your intuitive handwaving doesn't carry any weight if it tells you something that contradicts an exact mathematical theorem.

a change in pressure is matched by a change in the unbalanced force density

No, it isn't. A change in pressure is balanced by a change in momentum (and energy) density. See below.

This does not affect the divergence, provided that on average the spatial gradient of the pressure is unchanged, as the divergence only involves the gradient of the pressure, not the absolute value nor the rate of change with respect to time.

The divergence (more precisely the space components of the divergence) also include the spatial gradient of momentum density, which will be nonzero if the material is not at rest. So the pressure gradient cannot remain unchanged in this case.

The force density is the rate of change of momentum density with respect to time, so it enters into the divergence in the equation containing the time rate of change of energy density; it doesn't enter into the divergence equation where pressure appears at all, as you say. (And the rate of change of force density doesn't appear anywhere in this; the force density itself is a rate of change of an SET component.)

 

[Jonathan Scott]

Under the special case of spherical symmetry any internal changes in pressure are automatically compensated by internal changes in some other part of the SET (no divergence) such that the external field remains the same as per Birkhoff.

You're missing the point (and Tolman's point too). The divergence is not affected by pressure changes. As for my simple classical examples, a sudden pressure change can occur without any change to total energy or momentum. This will soon cause transfers of energy and momentum, but at the time it occurs the only thing which changes abruptly is that some material is being accelerated, and it takes time for any change in momentum and position to take effect. If the change is purely internal, then that acceleration will eventually be stopped and the system will get back into equilibrium with the same total integral of pressure as before.

 

[PeterDonis]

I simply mean the total integral of the pressure over the system in the way being previously discussed (as in the Komar mass expression and the volume integral of 3p in the Newtonian case for isotropic pressure).

And since we know the covariant divergence of the SET is zero, we can also show that the Komar mass of the system remains unchanged as long as we can find a "world-tube" containing the system such that no stress-energy crosses the tube in either direction.

 

[Jonathan Scott]

The divergence (more precisely the space components of the divergence) also include the spatial gradient of momentum density, which will be nonzero if the material is not at rest. So the pressure gradient cannot remain unchanged in this case.

If a support breaks, causing the pressure to drop, it IS still at rest initially.

At the microscopic level, I agree that what actually happens in a real material is that this takes a finite time to settle, but as I've said before, the amount of energy stored in a rigid stick or similar for a given compression force is related to the compression distance but the value of the integral of the pressure is equal to the same force times the length of the whole stick, so there is no way to transfer energy or momentum equivalent to the overall pressure integral.

 

[PeterDonis]

If a support breaks, causing the pressure to drop, it IS still at rest initially.

No, it isn't. Some parts of it are at rest, and other parts are not. You can't treat the support as a single object. You have to treat it as a continuous substance whose 4-velocity can vary from point to point.

I'll say it once more: your argument is based on intuitive hand-waving, but intuitive hand-waving doesn't count when it goes against an exact mathematical theorem. The exact mathematical theorem says that the change in internal pressure is balanced by something else to keep the external field the same (in the idealized case where the external field is the same). The fact that we have not yet figured out in this thread what that something else is does not mean the exact mathematical theorem is wrong. So if you are claiming that the exact mathematical theorem is wrong, but all you have to back that up is intuitive hand-waving, we will just have to close this thread as it will go nowhere.

 

[Jonathan Scott]

No, it isn't. Some parts of it are at rest, and other parts are not. You can't treat the support as a single object. You have to treat it as a continuous substance whose 4-velocity can vary from point to point.

You're going down to a microscopic level which is completely unnecessary. I was just using classical mechanics, as that's all that's necessary here.

It's true that suddenly releasing the support could for example cause a brief "ping" of oscillation. However, as I just said, the energy stored in the compression of the support, which would be the only source of any instant overall change of momentum or position, is related to the pressure integral in a similar proportion to the way that the length by which the support is compressed is related to the total length of the support, and it depends on the rigidity of the support, so it is independent of the pressure integral. So there is no way that anything the support can do could make up for the missing "source" term caused by the pressure dropping to around zero.

 

[PeterDonis]

You're going down to a microscopic level which is completely unnecessary.

Well, you are getting an answer that seems clearly wrong (that the source of gravity is not conserved) by not going down to a microscopic level.

I was just using classical mechanics, as that's all that's necessary here.

Not if you're going to claim that there is a paradox in GR. You can't do that based on an analysis that uses Newtonian mechanics.

there is no way that anything the support can do could make up for the missing "source" term caused by the pressure dropping to around zero

You don't know that, because you aren't doing the math; you're just waving your hands and claiming something that appears to contradict an exact mathematical theorem. I've already explained why that's not going to work.

At this point I am closing the thread since we are just repeating our positions and the discussion is going nowhere.