# Conservation of tensor pressure terms

1. Nov 15, 2008

### Jonathan Scott

In Einstein's field equations, the pressure terms on the diagonal of the stress-energy tensor contribute like energy density to the "Komar mass", and help determine the curvature and hence the gravitational effect. I believe that these terms represent "rate of exchange of x-component of momentum in the x-direction per unit area" and so on for y and z as well.

As far as I can see these terms do not appear to be the density of a conserved quantity. For example, one could have two masses held apart by a nearly massless rigid rod, then move the rod aside, and the pressure in the rod would just drop to zero, without "going anywhere". Is this correct or have I missed something? (With a more realistic description of the removal process, I suspect that for a real not quite rigid material there would be a small amount of ordinary energy which would turn into kinetic energy of flexing, oscillation or heat within the removed rod, but as far as I can see this energy could be negligible compared with the pressure times the volume).

If that is correct, I find it hard to understand how this term could really contribute either to the gravitational effect of the system or to the effective total energy.

2. Nov 15, 2008

### tiny-tim

Hi Jonathan!

no individual term, nor even the whole diagonal together, is capable of representing a physical characteristic, let alone a conserved one …

surely only the whole stress-energy tensor, or its trace, could do that?

3. Nov 15, 2008

### Jonathan Scott

In the case I quoted, as far as I can see the only significant terms in the tensor before we remove the rod are the energy density (within the masses) and the pressure component within the rod along in the direction of the rod, and afterwards the pressure in the rod is gone but everything else is initially essentially the same (well, there would be a negative change of pressure in the masses when the rod is removed, as the near edges of the masses would be attracted towards each other slightly more than the far edges, but that makes it even worse, and I'm assuming the masses are small).

To me, this seems to be a large quantity of "Komar energy" which has suddenly gone missing. I know the Komar scheme requires a static situation, but I'm wondering what it turned into.

4. Nov 15, 2008

### Naty1

I can't answer your question but have been working for about week or so to try to understand (a) the physical meaning of the ten independent components each of the Ricci and Weyl tensors and (b) how they got in the formulas in the first place.

I Googled KOMAR ENERGY and got what appear to be some interesting hits.

If anyone can post online references describing the various components of the Ricci and Weyl tensors I'd love to gain more of an understanding of their physical consequences/interpretations, as Jonathan is questioning here. Ricci Decomposition in Wikipedia is the kind of thing I'd like to find, but in more detail.

What I'm trying to figure out is how valid representations among the twenty independent tensor components got singled out from invalid ones: so much of mathematics does NOT represent the physcial world that we have to rely on experimentalists, right? Who's tested these pieces...when we don't even know if gravity holds over cosmic distances nor below a few tenths of a millimeter.

Post: "...no individual term, nor even the whole diagonal together, is capable of representing a physical characteristic, let alone a conserved one.."

If this is true, and I'm not doubting it just astonished, how the heck were all the pieces developed...how were they chosen if there was no physical basis for selection? fortuitious guesses?

Last edited: Nov 15, 2008
5. Nov 15, 2008

### Naty1

Jonathan....I stumbled across the following and thought of your question....
from wikipedia
http://en.wikipedia.org/wiki/Stress-energy_tensor

Is this relevant?

6. Nov 15, 2008

### Naty1

In my reading I also came across

http://people.hofstra.edu/Stefan_Waner/diff_geom/Sec12.html [Broken]

Definition 12.4 to the bottom of the section, Einstein's conservation law, may be relevant...or maybe too basic for your purposes..?? Hope it helps...

Last edited by a moderator: May 3, 2017
7. Nov 15, 2008

### Jonathan Scott

I'm aware that conservation doesn't take into account gravitational energy, which in some sense is "hidden" in the shape of space-time, but what I'm puzzled about is how pressure can pop out of existence without any apparent consequences. I'm assuming I've missed something, but I can't see what.

I'm aware there's a local conservation law for each component of momentum as well as energy, but I don't yet fully understand the situation in this case. I remember once evaluating the terms of the conservation law for the electromagnetic stress-energy tensor (and finding that it didn't seem to be covariant, as was later clarified by sources such as J W Butler). I haven't done the same for this tensor yet.

If the 11, 22, 33 terms really represent pressure perpendicular to the relevant planes, but things are at rest, then the momentum is flowing equally in both directions. That is, it appears to me that pressure in a static system defines the rate per unit area at which momentum in one direction is flowing one way AND the rate per unit area at which momentum in the opposite direction is flowing the other way. However, overall nothing is going anywhere!

8. Nov 15, 2008

### tiny-tim

If we're talking about pressure of an ordinary fluid or solid, then the "static" pressure in the x-direction is caused by change in momentum of molecule A and its friends moving in that direction, and bouncing off molecule B and its friends.

The momentum of A & Co flowing in the positive x-direction balances the momentum of B & Co flowing in the negative x-direction.

"overall nothing is going anywhere" depends on your point of view.

9. Nov 15, 2008

### Jonathan Scott

True. Perhaps I should I have said "overall no matter or energy is going anywhere".