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  Zero divergence of energy-momentum tensor and gravitational energy

+ 3 like - 0 dislike

Trying to teach myself general relativity and have just hit yet another confusion. I'm reading that in curved spacetime the energy-momentum tensor has zero divergence, ie


But that this doesn't imply the total conservation of energy and momentum as there is an additional source of energy (the gravitational field) that isn't included in the EMT. If that's the case, and if the EMT doesn't describe the total energy of system, how is it valid to use the tensor to describe various systems. For example, $$T^{\mu\nu}=0$$ for the Schwarzschild solution, or assuming spacetime is a perfect fluid in cosmology? How are these assumptions valid if they don't include the energy contribution of the gravitational field? Seems a bit of an elephant in the room-type situation. Or is that energy so small it can be ignored?

Thank you

This post imported from StackExchange Physics at 2014-03-22 17:20 (UCT), posted by SE-user Peter4075
asked May 5, 2012 in Theoretical Physics by Peter4075 (15 points) [ no revision ]

2 Answers

+ 5 like - 0 dislike

The actual reason why one can't interpret the equation $$ \nabla_\mu T^{\mu\nu}=0 $$ as a global conservation law is that it uses covariant derivatives. If a law like that were valid with partial derivatives, you could derive such a law. But there's a covariant derivative which is one of the technical ways to explain that general relativity in generic backgrounds doesn't preserve any energy:


The text above also explains other reasons why the conservation law disappears in cosmology.

However, despite the non-existence of a global (nonzero) conserved energy in general backgrounds, the tensor $T_{\mu\nu}$ is still well-defined. As twistor correctly writes, it quantifies the contribution to the energy and momentum from all matter fields (non-gravitational ones) and matter particles. And if you can approximate the background spacetime by a flat one, $g_{\mu\nu}=\eta_{\mu\nu}$, which is usually the case with a huge precision (in weak enough gravitational fields, locally, or if you replace local objects that heavily curve the spacetime, including black holes, by some effective $T$, using a very-long-distance effective description), then $\nabla$ may be replaced by $\partial$ in the flat Minkowski coordinates and the situation is reduced to that of special relativity and the "integral conservation law" may be restored.

This post imported from StackExchange Physics at 2014-03-22 17:20 (UCT), posted by SE-user Luboš Motl
answered May 5, 2012 by Luboš Motl (10,278 points) [ no revision ]
Most voted comments show all comments
thanks. I did try to read the link but it was a little over my head (my level - I've no idea what Hamiltonians or Lagrangians are!). I've been trying to understand the Schwarzshild metric and relativistic cosmology both of which are based on a particular energy-momentum tensor (zero and perfect fluid). I still don't see how that can be valid if the EMT doesn't include gravitational energy. I sense you answering my question in your final paragraph ("very-long-distance effective description") but, sorry, don't understand. Please don't worry about making your answer too simple!

This post imported from StackExchange Physics at 2014-03-22 17:20 (UCT), posted by SE-user Peter4075
Apologies if I am wrong but I think that your "I don't see how this can be valid" follows from your misunderstanding of the basic laws of general relativity. The basic equation of general relativity is Einstein's tensor equation, $R_{\alpha\beta}-Rg_{\alpha\beta}/2=-8\pi G T_{\alpha\beta}$. It's a beautiful equation that may be justified or derived by various methods and it's been experimentally tested. The $T$ on the right hand side only contains the non-gravitational energy; the gravitational energy is on the LHS and the equation essentially says that they have to exactly cancel!

This post imported from StackExchange Physics at 2014-03-22 17:20 (UCT), posted by SE-user Luboš Motl
You could also rewrite the equations in various equivalent ways, like putting everything on one side etc. And some of these ways could be also saying that the gravitational field affects itself and sources it. And it's partly true: Einstein's equations are nonlinear. But the most convenient formulation of the equations simply separates the terms to those with $R$, the curvature, that only depend on the metric tensor, and they go to the left side; and then all the terms that depend on particles, liquids, electromagnetic fields and other matter, ...

This post imported from StackExchange Physics at 2014-03-22 17:20 (UCT), posted by SE-user Luboš Motl
... and whose dependence on the metric tensor is simply - and these things go to the right hand side of Einstein's equations. Effectively, the right hand side contains the same stress-energy tensor that would be conserved in flat space of special relativity, without any gravity, and it's just translated to general coordinates in a curved spacetime. And it is this non-gravitational energy-momentum tensor that sources the Einstein's curvature tensor that we write on the left hand side.

This post imported from StackExchange Physics at 2014-03-22 17:20 (UCT), posted by SE-user Luboš Motl
Oh dear. I think I've got this seriously wrong. Please bear with me. I thought the rhs of the field equations (ie the EMT) described the total energy-momentum of a system (which I think is correct). I thought the lhs was a description of the curvature of spacetime caused by the rhs (which I think also is correct). But I never thought of the lhs as describing gravitational energy. (Thinking about it, both sides must of course have the same units, so if the rhs describes energy so must the lhs). I thought that the “additional source of energy”, ie gravitational energy, was missing

This post imported from StackExchange Physics at 2014-03-22 17:20 (UCT), posted by SE-user Peter4075
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The only sentence in which I mentioned Lagrangians was when I said that you may calculate the formula for the stress-energy tensor out of the Lagrangian. If you can't use the Lagrangians, you won't be able to derive the stress-energy tensor in this way and you may ignore this sentence. You will have to use an explicit God-given formula for $T$. But otherwise nothing whatsoever changes about the remaining things I said. For the equivalence of global and local continuity equation - conservation laws - see en.wikipedia.org/wiki/Continuity_equation . It uses $\partial$, not $\nabla$.

This post imported from StackExchange Physics at 2014-03-22 17:20 (UCT), posted by SE-user Luboš Motl
apologies for being a bit slow on the uptake. This stuff would be a lot easier if (a) I was smarter and (b) I had a degree in physics. I confess I've only just today found out what a continuity equation is! I've also just read a reasonably understandable on-line piece by Weiss and Baez titled, "Is energy conserved in general relativity". Their answer is an unambiguous, "In special cases, yes. In general — it depends on what you mean by 'energy', and what you mean by 'conserved'." Glad that's sorted out.

This post imported from StackExchange Physics at 2014-03-22 17:20 (UCT), posted by SE-user Peter4075
+ 3 like - 0 dislike

The stress-energy tensor describes the energy momentum content contributed by all the fields present with the exception of the gravitational field itself. However, just because it doesn't appear in the energy momentum tensor does not mean that gravity can't act, in some sense, as a source of gravity. The Einstein equations are nonlinear, and from this nonlinearity arises the possibility for gravity to "gravitate" in a consistent way.

Attempts to describe the energy-momentum of the gravitational field itself locally (as you would need to do if you wished to include a gravitational field contribution in the stress energy tensor) are known to run into problems. By locally I mean that we would try to write down a tensor field that encapsulates the gravitational field energy-momentum at each point.

This post imported from StackExchange Physics at 2014-03-22 17:20 (UCT), posted by SE-user twistor59
answered May 5, 2012 by twistor59 (2,500 points) [ no revision ]

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