1

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: https://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

Qmechanic
  • 201,751
lurscher
  • 14,423
  • 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
  • 4
    @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. –  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
  • 1
    @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. –  Nov 30 '11 at 06:29
  • @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 07:24
  • I feel that i've not made the quality question i would have hope for this conundrum. I'm not sure at this point if its worth to ask another question or just edit this one; but the main concern i have with "energy is not conserved in GR" statements is if the statement allows for perpetual motion machines or there is just some "gotcha 22" that does avoid them. – lurscher Nov 30 '11 at 17:40

2 Answers2

2

Here is a FAQ entry I wrote for physicsforums.com.

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, On the Zero-Energy Universe. Int J Theor Phys, 48, 3278–3286

Weiss and Baez, "Is Energy Conserved in General Relativity?"

Wald, General Relativity, 1984

  • 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
  • very well, but does this means that spacetime curvature can do infinite work on matter fields? does it means that it can't? – lurscher Nov 29 '11 at 18:21
  • 1
    @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. –  Nov 30 '11 at 06:20
  • @lurscher: Sorry, I don't follow you. Why work? Why infinite work? I don't understand how this connects to your original question. –  Nov 30 '11 at 06:21
  • @Ben: Of course you can't add the stress-energy tensor, that's why you define the stress-energy pseudo-tensor. The stress energy pseudotensor is something that you can add up, because it is defined relative to a full coordinate system. It is Einstein's definition of stress-energy in gravitation, and it is absolutely additive. Perhaps it is not covered in Misner-Thorne-Wheeler, I don't know, but it is well known anyway. – Ron Maimon Nov 30 '11 at 07:34
  • @Ben, perpetual motion machines are impossible as long as two conditions hold 1) energy is conserved 2) entropy increases or stays the same. So my question is: if energy is not conserved in GR, then where are the gravitational perpetual motion machines? – lurscher Nov 30 '11 at 12:22
  • 1
    @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] –  Nov 30 '11 at 16:36
  • 1
    [...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. –  Nov 30 '11 at 16:37
  • @lurscher: "where are the gravitational perpetual motion machines?" There are conserved scalar measures of mass-energy in GR for the special case of an asymptotically flat spacetime. The spacetime in and around our solar system is extremely well approximated as being asymptotically flat. Only when you go to cosmological distance scales does the lack of asymptotic flatness become an issue. Therefore this property of GR cannot be exploited to create a perpetual motion machine unless the machine is much bigger than the solar system. –  Nov 30 '11 at 16:40
  • @Ben, ok, but your point is that is in principle possible to extract infinite energy from asymptotic non-flatness. I wonder if this is something widely acknowledged or is a point of contemption? – lurscher Nov 30 '11 at 16:50
  • @Ben: I agree that there is no unique pseudotensor for gravity, but this is the same issue as the ordinary stress tensor for a field theory. There is a Belinfante construction to make the angular momentum work, and the gravitational analog is the Landau-Lifschitz pseudotensor. There are other pseudotensors, but they all share the property that they give a conserved coordinate-dependent stress energy in GR, and this is enough to rule out perpetual motion and the like. Einstein came up with the pseudotensor to describe stress energy in gravity, including the gravitational field, not the local T. – Ron Maimon Nov 30 '11 at 18:50
  • @lurscher: The asymptotic non-flatness can be used for a sort-of perpetual motion, in that the whole universe is here when less of it was here earlier. This can be thought of as stress energy coming from the cosmological horizon. The violations of energy conservation in GR are best thought of as coming from edges of the sort provided by cosmological horizons. – Ron Maimon Nov 30 '11 at 18:51
  • 1
    @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. –  Nov 30 '11 at 21:20
  • @lurscher: "Ben, ok, but your point is that is in principle possible to extract infinite energy from asymptotic non-flatness. I wonder if this is something widely acknowledged or is a point of contemption?" OK, looking back at my comment I can see how you might construe it that way, but I didn't intend to make any claim as broad as that. I meant that it isn't possible to violate conservation of energy in a certain way. I didn't mean to make any affirmative statements about what was possible, nor did I say anything about infinite energy.[...continued...] –  Nov 30 '11 at 21:32
  • [...continued...] It is not possible to make an affirmative statement based on GR about the possibility of constructing an extremely large perpetual motion machine whose size would be cosmological. The reason is that we can't even define the notion of a perpetual motion machine in GR in a cosmological spacetime. The important fundamental fact here is that GR simply doesn't provide any definition of energy that applies to cosmological spacetimes. (Specifically, there is no uniquely defined, conserved, scalar measure of mass-energy.) –  Nov 30 '11 at 21:36
  • That is really weird and unexpected. Need to wrap my head around what does this means – lurscher Nov 30 '11 at 21:37
  • @Ben:they are not incorrect, you just never worked out the pseudotensor. It's counterintuitive, and many people rejected it completely. But it is a sensible point of view regarding energy in GR. – Ron Maimon Nov 30 '11 at 22:00
2

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.

Ron Maimon
  • 1
  • 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
  • @lurshcer: sure, it's not flat cosmologically, but were surrounded by a horizon, and who knows what comes out of that. Inside a given region, the pseud-stress-energy-tensor of gravity gives a good description of energy and momentum flows in a coordinate dependent way. – Ron Maimon Nov 29 '11 at 20:33
  • 1
    "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. –  Nov 30 '11 at 06:37
  • 1
    "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. –  Nov 30 '11 at 06:39
  • @Ben: I meant non-flat asymptotically flat, as should have been obvious from both leading and trailing context, but I'll make it explicit. – Ron Maimon Nov 30 '11 at 07:20
  • @Ben: It is absurd to think that I am confusing the mathematician's boundary with a cosmological horizon. The horizon is a physicist boundary, but not a mathematical boundary. Mathematician's boundaries have no place in GR, and should not be mentioned. A causal patch description of cosmology allows you to talk about the energy in the patch. It is not global to a global time slice, and it is locally conserved, but coordinate dependent. The pseudotensor is the way to talk about it, but it is not globally well defined. – Ron Maimon Nov 30 '11 at 07:28
  • 1
    @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. –  Nov 30 '11 at 21:25
  • 1
    @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
  • I don't expect this question be without any controversy, after all energy conservation is not one of the areas in GR that is widely accepted to be settled. But let just try to discuss the points technically and avoid throwing punts to each others. Can we move this to chat? – lurscher Dec 01 '11 at 19:59
  • I don't want to reignite this discussion but I agree with Ron that energy conservation holds in cosmology and is correctly understood with the use of pseudotensors or more modern covariant methods. I have written an article http://vixra.org/abs/1305.0034 which retutes all the objections I have come across. – Philip Gibbs - inactive May 06 '13 at 16:29