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I did general relativity years ago at Uni and was just revising with the aid of Dirac''s brilliant book; the beauty of this book is that it is so thin and concise. On reading this book I find that I have a few questions regarding energy.

One thing I had not appreciated before was that the energy in the energy tensor only includes all energy excluding gravitational energy. Is this true? What is the evidence for this position? How could we know that this energy term actually excludes gravitational energy?

The only argument that I can see is that the energy in the energy tensor is not fully conserved, so you could infer that there is a missing energy term and that that energy is gravitational energy. But if you take the missing quantity and call it the gravitational energy, this quantity turns out not to be a tensor.

Hence, its form will in general always look different in at least some different coordinate systems regardless of whichever quantities you use to write it out in. This latter point might only have mathematical consequences, but does it have physical consequences?

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  • $\begingroup$ Related: Energy-momentum for gravity, which in turn is a duplicate of Why does no physical energy-momentum tensor exist for the gravitational field? $\endgroup$ – Chiral Anomaly Oct 20 at 1:57
  • $\begingroup$ "Gravitational nergy" is just another term for "spacetime curvature". It is already there in the equations on the left side. You can move it to the right side (since it's negative anyway) and see the equations as the total energy is zero. And indeed, the gravitational energy does curve spacetime just fine, for example, in the Schwarzschild and other vacuum solutions. $\endgroup$ – safesphere Oct 20 at 4:56
  • $\begingroup$ I see that @Void had beaten me to this point. $\endgroup$ – safesphere Oct 20 at 5:07
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I came across this passage in Misner, Thorne & Wheeler (20.4) where they first talk about the stress-energy pseudotensor of the gravitational field and how one might calculate a contribution to the local momentum vector... and then memorably state:

Right? No, the question is wrong. The motivation is wrong. The result is wrong. The idea is wrong.

To ask for the amount of electromagnetic energy and momentum in an element of 3-volume makes sense. First, there is one and only one formula for this quantity. Second, and more important, this energy-momentum in principle "has weight." It curves space. It serves as a source term on the righthand side of Einstein's field equations. It produces a relative geodesic deviation of two nearby world lines that pass through the region of space in question. It is observable. Not one of these properties does "local gravitational energy-momentum" possess. There is no unique formula for it, but a multitude of quite distinct formulas. ... Moreover, "local gravitational energy-momentum" has no weight. It does not curve space. It does not serve as a source term on the righthand side of Einstein's field equations. It does not produce any relative geodesic deviation of two nearby world lines that pass through the region of space in question. It is not observable.

As Ben said, one cannot speak of gravitational energy being specified at one point: at the very least it requires integrating over a 4-volume, but even that turns out to be fraught and one gets several competing versions.

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  • $\begingroup$ "It does not curve space" - Really? Then what does curve space in the Schwarzschild vacuum solution? "one cannot speak of gravitational energy being specified at one point" - There is no energy at one point either for any finite density regardless of whether or not "one can speak" of it. "one gets several competing versions" - So what? The inability of out model or interpretation to come up with a consistent description does not prevent Nature from endowing gravity with energy. $\endgroup$ – safesphere Oct 20 at 5:41
  • $\begingroup$ @safesphere - Note that "it" refers to the pseudotensor. And the numerous and incompatible definitions of globalish energy in GR like the ADM, Komar, and Bondi masses have been dealt with in several questions on this SE. $\endgroup$ – Anders Sandberg Oct 20 at 13:05
  • $\begingroup$ @safesphere wrote, "...Really? Then what does curve space in the Schwarzschild vacuum solution?" Good question! I suppose one answer is that since Schwarzschild is a vacuum solution, the Ricci tensor is zero. This spacetime is Ricci-flat, even though not Riemann-flat. $\endgroup$ – Colin MacLaurin Oct 23 at 2:35
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Well, first of all you should ask yourself what do you expect from the concept of energy and momentum. Or in other words, what is energy and momentum, really? You have a set of intuitions in mind, but the minimal requirement is that these are quantities that are 1) conserved, and 2) reduce to our "usual" definitions of energy and momentum in suitable limits.

Let us take a look at the Einstein equations $$G_{\mu\nu} = 8\pi T_{\mu\nu}$$ The Einstein tensor $G_{\mu\nu} = R_{\mu\nu} - R g_{\mu\nu}/2$ does have the property of being conserved $G^{\mu\nu}_{\;\;\;;\nu} = 0$, same as the stress-energy tensor. Furthermore, and this is a somewhat nuanced point, certain parts of it can be interpreted as gravitational energy in the weak-field limit.

For instance, in the post-Minkowski limit you have $g_{\mu\nu} = \eta_{\mu\nu} + \epsilon h^{(1)}_{\mu\nu} + \epsilon^2 h^{(2)}_{\mu\nu}+...$. At first order in $\epsilon$ the left-hand side of the Einstein equations just correspond to a linear operator acting on $h^{(1)}_{\mu\nu}$. At second order, however, you have the same operator acting on $h^{(2)}_{\mu\nu}$ and terms quadratic in $h^{(1)}_{\mu\nu}$ that can be interpreted as (minus) the stress-energy tensor of the field $h^{(1)}_{\mu\nu}$ (sourcing the gravitational field correction $h^{(2)}_{\mu\nu}$). So $G_{\mu\nu}$ (or $-G_{\mu\nu}/8\pi$) seems to naturally offer itself as a gravitational stress-energy tensor.

However, as you may have already noticed, when $-G_{\mu\nu}/8\pi$ is moved to the right-hand side of the Einstein equations as a part of stress-energy, then the interpretation of the equations ends up being that the total stress-energy content of any point of space-time is exactly zero. Every order of the post-Minkowski expansion is really just enforcing that statement, at higher and higher order in $\epsilon$.

This may be elegant, but underwhelming, especially in vacuum space-times. In vacuum space-times, such as space-times of inspiraling and merging black holes, a lot can be obviously going on, while this interpretation is just telling us there is no stress-energy ever flowing, present, or exchanged. This is the reason why other notions of energy and momentum of the gravitational field have been introduced. But I am going to be honest, their real value is in keeping track of how the gravitating mass (and/or momentum) of a system evolve in time. After this violent process with outgoing gravitational radiation, how much will this system still attract me gravitationaly? These "nonlocal" formulas mentioned in the other posts here provide the answer. But beyond this pragmatic meaning, I would defer from interpreting them too deeply.

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  • $\begingroup$ Great answer +1, but why "underwhelming"? Quite the opposite, the vacuum spacetimes are a perfect demonstration how the gravitational energy (the curvature of spacetime) curves spacetime. In fact, matter (stress-energy) curves spacetime only where it is, but not outside in vacuum. For example, spacetime inside the Earth is curved by the mass of the Earth, but spacetime outside is curved by the curvature of spacetime, according to the Schwarzschild vacuum solution. $\endgroup$ – safesphere Oct 20 at 5:22
  • $\begingroup$ When people say the total energy of the universe including gravitational (pseudo-)energy is zero, they mean that the stress-energy at each point is canceled by gravitational (pseudo-)energy elsewhere, not identically at every point. It's the same as any other conservation law. If div φ vanishes then φ is conserved. If φ vanishes there's nothing to conserve. The negative gravitational energy is some combination of Weyl curvature and coordinate artifacts. The Ricci curvature is already accounted for in the stress-energy tensor – it is the stress-energy tensor. It doesn't cancel with itself. $\endgroup$ – benrg Oct 20 at 20:35
  • $\begingroup$ @benrg I agree with your statements, but what I am saying in my post does not have anything to do with cosmology, as I hope is obvious. $\endgroup$ – Void Oct 20 at 21:14
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The simplest way to see this is that by the equivalence principle, we can always let the gravitational field at any point have any value we like, including zero. Therefore there is no possibility of defining an energy density of the gravitational field at one point.

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  • $\begingroup$ I don't see how one implies the other. Gravity strength of a worldpoint is frame dependent but how does this prevent defining gravity field energy density? This density can be but need not be function of strength only, it can depend on its derivatives. Real gravity field cannot be eliminated to zero everywhere. $\endgroup$ – Ján Lalinský Oct 19 at 18:02
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    $\begingroup$ Using the same logic, we can always let the kinetic energy of any particle "have any value we like, including zero". So we must exclude the kinetic energy from the stress-energy tensor too. $\endgroup$ – safesphere Oct 20 at 5:51
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    $\begingroup$ @safesphere: No, this is incorrect for a couple of reasons. (1) We're talking about the energy density of a field here, not a particle. The electric and magnetic fields do have energy densities, and it is not possible to make the electromagnetic field density go to zero by a choice of the frame of reference. (2) The stress-energy tensor has scalar invariants, and these can be nonzero. A theory that assigns a stress-energy tensor to the gravitational field would have to provide some way of assigning a nonzero value to these invariants. $\endgroup$ – Ben Crowell Oct 20 at 18:55
  • $\begingroup$ Basically I agree with safesphere's objection. You can make the acceleration zero but you can't make the curvature zero. So you can rule out a covariant energy that's a function of acceleration but not one that's a function of curvature. And T exists. How do we know there isn't another similar tensor field that gives (the rest of) the gravitational energy? (I mean, I know there isn't, but how do you derive that from the equivalence principle?) $\endgroup$ – benrg Oct 21 at 0:55
  • $\begingroup$ @benrg How do you make the acceleration zero? Assuming by acceleration you mean proper acceleration. $\endgroup$ – MBN Oct 21 at 11:24
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There are various definitions of the total global mass-energy contained in a spacetime: the ADM mass, Bondi mass, etc. In present understanding, these require specific conditions such as symmetry in time, or asymptotic flatness.

Then there are an awful lot of quasi-local proposals. The idea is that, since you can't define the gravitational energy at a point, draw a small box around a region of spacetime, and define the gravitational energy inside it. For instance, Hawking's proposal examines how light rays exiting the box diverge; the motivation is mass-energy curves spacetime which affects the light. The proposal by Epp (& Mann & McGrath), studies the acceleration of the walls of the box. Another proposal by Bartnik is more mathematically motivated, I'm told. Most definitions agree for simple cases, apparently.

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