curvature tensor component capable of doing work on $T_{\mu \nu}$

I'm wondering what part of the curvature tensor is able to do work (and hence transfer energy) in matter. I'm wondering if this tensor: http://en.wikipedia.org/wiki/Stress-energy-momentum_pseudotensor satisfies that property

I want to understand the generic assertion that GR doesn't conserve energy, and which scenarios do conserve it

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GR conserves energy as well as any other field theory, it is just difficult to localize the energy at points in space. The stress-energy pseudotensor is the field energy in GR, and if you have an asymptotically flat background, it integrates to the gravitational contribution to the total energy. – Ron Maimon Nov 29 '11 at 18:02
@Ron Maimon: "GR conserves energy as well as any other field theory, it is just difficult to localize the energy at points in space." No, this is incorrect. MTW has a nice discussion of this on p. 457. GR does not have a conserved, scalar measure of mass-energy that applies to all spacetimes. "...if you have an asymptotically flat background, it integrates to the gravitational contribution to the total energy. " Asymptotically flat spacetimes are a special case. In such a spacetime, there are conserved measures of mass-energy such as the ADM and Bondi masses. – Ben Crowell Nov 29 '11 at 18:09
@Ben: My comment is correct, although MTW are also correct. The asymptotically flat backgrounds are an instructive special case, because they show that GR doesn't produce energy locally. The energy non-conservation in cosmology can be thought of as stuff coming in and out of the horizon. The electromagnetic stresses are also not conserved in the presence of a horizon. For example, if you have an electromagnetic wave coming out of a Rindler horizon. The only plus of EM vs. GR is that the pseudo stress energy is not a tensor, but this doesn't bother me one bit. – Ron Maimon Nov 29 '11 at 20:31
@Ron Maimon: "The energy non-conservation in cosmology can be thought of as stuff coming in and out of the horizon." No, this is completely wrong. For example, in a matter-dominated, closed universe, there is no particle horizon, but as discussed in the MTW reference you cannot define a conserved, scalar mass-energy in such a cosmology. – Ben Crowell Nov 30 '11 at 6:29
@Ben: This is incorrect--- in a matter dominated closed universe there is still a horizon, and the horizon goes out and comes back so that a light ray will only go once around the universe at the moment of collapse (assuming zero cosmological constant). I agree with MTW that you can't define a global conserved energy, I am saying something else, that within a causal patch, you can define a non-conserved "energy" using the pseudotensor. I know the difference between a mathematical "boundary" and the horizon, but GR doesn't have as big a problem with energy not being conserved as you think. – Ron Maimon Nov 30 '11 at 7:24

Here is a FAQ entry I wrote for physicsforums.

How does conservation of energy work in general relativity, and how does this apply to cosmology? What is the total mass-energy of the universe?

Conservation of energy doesn't apply to cosmology. General relativity doesn't have a conserved scalar mass-energy that can be defined in all spacetimes.[MTW] There is no standard way to define the total energy of the universe (regardless of whether the universe is spatially finite or infinite). There is not even any standard way to define the total mass-energy of the observable universe. There is no standard way to say whether or not mass-energy is conserved during cosmological expansion.

Note the repeated use of the word "standard" above. To amplify further on this point, there is a variety of possible ways to define mass-energy in general relativity. Some of these (Komar mass, ADM mass [Wald, p. 293], Bondi mass [Wald, p. 291]) are valid tensors, while others are things known as "pseudo-tensors" [Berman 1981]. Pseudo-tensors have various undesirable properties, such as coordinate-dependence.[Weiss] The tensorial definitions only apply to spacetimes that have certain special properties, such as asymptotic flatness or stationarity, and cosmological spacetimes don't have those properties. For certain pseudo-tensor definitions of mass-energy, the total energy of a closed universe can be calculated, and is zero.[Berman 2009] This does not mean that "the" energy of the universe is zero, especially since our universe is not closed.

One can also estimate certain quantities such as the sum of the rest masses of all the hydrogen atoms in the observable universe, which is something like 10^54 kg. Such an estimate is not the same thing as the total mass-energy of the observable universe (which can't even be defined). It is not the mass-energy measured by any observer in any particular state of motion, and it is not conserved.

MTW: Misner, Thorne, and Wheeler, Gravitation, 1973. See p. 457.

Berman 1981: M. Berman, unpublished M.Sc. thesis, 1981.

Berman 2009: M. Berman, Int J Theor Phys, http://www.springerlink.com/content/357757q4g88144p0/

Weiss and Baez, "Is Energy Conserved in General Relativity?," http://math.ucr.edu/home/baez/physics/Relativity/GR/energy_gr.html

Wald, General Relativity, 1984

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The stress-energy pseudotensor is conserved, and its coordinate dependence only shifts the stress-energy around to different places, without altering the total mass-energy. – Ron Maimon Nov 29 '11 at 18:16
@Ron Maimon: "The stress-energy pseudotensor is conserved[...]" This is incorrect. First off, stress-energy is a tensor, not a pseudotensor. And the point of the MTW reference is that you can't even add up the stress-energy tensor, because parallel transport is path-dependent. – Ben Crowell Nov 30 '11 at 6:20
@Ron Maimon: "The stress energy pseudotensor is [...] Einstein's definition of stress-energy in gravitation[...]" No, this is incorrect. You're claiming that it's uniquely defined, when in fact there are many different energy pseudotensors that have been defined. There is an Einstein pseudotensor, but it is not a good or uniquely defined choice of such a pseudotensor, in particular because it's incompatible with defining a conserved ang. mom.. There are lots of other energy pseudotensors, e.g., the Landau-Lifshitz one. None of these is "the" stress-energy pseudotensor. [continued] – Ben Crowell Nov 30 '11 at 16:36
[...continued...] You're also incorrect to describe the Einstein pseudotensor as "Einstein's definition of stress-energy in gravitation." Einstein's definition of stress-energy in gravitation is the stress-energy tensor, not the Einstein tensor. – Ben Crowell Nov 30 '11 at 16:37
@Ron: Your statements are simply completely incorrect from A to Z. I don't know where you came up with this idiosyncratic notion about energy nonconservation in GR being related to horizons or edges. That's simply nonsense. – Ben Crowell Nov 30 '11 at 21:20

All asymptotically flat solutions conserve energy in GR. Asymptotically flat means that the spacetime is flat at infinity. It is also true for many other asymptotically stationary solutions, like cases where the spacetime at infinity is a quotient of euclidean space, like a cone. The cases where there is no conserved energy are best interpreted as cases where energy is coming in or leaving at the edge of spacetime.

Any non-flat asymptotically flat solution of GR has a positive total energy, by the positive mass theorem. So any localized curvature pattern has a total energy which is completely determined by the asymptotic falloff of the metric tensor at asymptotic distances. The difference from flatness goes like $M/r$ in the time and space components, where M is the total energy (c=1).

This energy can always be used to do work, if you have an infinite reservoir at zero entropy to dump entropy into. You can, for example, dump the curvature into a black hole, increasing its mass by M, and then run a heat-engine with the reservoir using the Hawking radiation.

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in any case, our space-time is not asymptotically flat right? only the space part of it is actually flat (at cosmological scales)? – lurscher Nov 29 '11 at 18:45
"The cases where there is no conserved energy are best interpreted as cases where energy is coming in or leaving at the edge of spacetime." This is incorrect. Comparing this with your other incorrect remark about horizons, it sounds like you don't understand the difference between a horizon within a spacetime manifold and a boundary (edge) of a manifold with boundary. Cosmological spacetimes are manifolds, not manifolds with boundaries, and energy is not conserved in them. Since they don't have edges, they are counterexamples to your claim. – Ben Crowell Nov 30 '11 at 6:37
"Any non-flat solution of GR has a positive total energy, by the positive mass theorem." This is incorrect. For a counterexample, consider the type of cosmological spacetime discussed in the MTW reference I gave. Such a spacetime is not flat, and, as discussed by MTW, it does not have a well defined total energy. – Ben Crowell Nov 30 '11 at 6:39
@Ron: You seem to have read through a lot of abstruse papers and learned to throw around various fancy terms without understanding what the terms actually mean. I'm sure that makes you sound impressive to people who don't know enough to realize that essentially every statement you're making is incorrect. What you're missing is any understanding of the fundamentals of GR. – Ben Crowell Nov 30 '11 at 21:25
@Ben: a little knowledge is often worse than no knowledge at all. Instead of making foolish statements, please just work out the Noether theorem for GR in a fixed coordinate system. You will get a coordinate object which is the pseudotensor. I read about it (twenty years ago) worked it out myself, I understand it well, it is worked out several places, and it does not conflict with GR. – Ron Maimon Nov 30 '11 at 22:05