In GR the curvature of spacetime "is gravity". This curvature is expressed via the Riemann tensor (or the Ricci tensor + Ricci scalar). The curvature is connected via the Einstein Field Equations with the Energy-Momentum tensor (which does not include gravitational energy).

First question: Why should the EMT even contain gravitational energy? Wouldn't this energy be proportional to curvature? This energy would than again cause additional curvature which would in turn increase the grav. energy which would again increase the curvature etc.

Second question: What is gravitational energy in GR? (a formula and/or picture would be nice)

(I already searched for similar question but could not find anything satisfying. I am not that familiar with Killing vectors, which seem to be sometimes related to this stuff.)

  • $\begingroup$ As noted by @AccidentalFourierTransform, it's notoriously hard to provide a definition of a "gravitational stress-energy tensor" in GR that is both locally defined and is actually a tensor. (It's easier to define a global notion of gravitational energy, though it's still not something you'd see in a first course on the subject.) There's a long review of the subject available here. $\endgroup$ Jan 3, 2016 at 15:34
  • $\begingroup$ A reminder to everyone that comments are not to be used for answering the question. $\endgroup$
    – David Z
    Jan 3, 2016 at 15:43
  • $\begingroup$ @ChrisCundy EFE are non-linear, which is often rephrased as "gravity is a source of gravity". But this has nothing to do with gravitational waves carrying energy. They do not. We can define quantities that formally look like some kind of potential energy, but strictly they are not energy as per the usual definition $\endgroup$ Jan 3, 2016 at 15:46
  • $\begingroup$ I thought gravitational waves do carry energy and via this they were observed. Some double neutron stars (or something like that) slow down their rotation because the emit g-waves which carry some energy with them (thus the two neutron stars rotate more slowly). The prediction and the experiment work out just fine as far as I know. $\endgroup$ Jan 3, 2016 at 16:05
  • $\begingroup$ Related: physics.stackexchange.com/q/41662/2451 , physics.stackexchange.com/q/160139/2451 , physics.stackexchange.com/q/190502/2451 and links therein. $\endgroup$
    – Qmechanic
    Jan 5, 2016 at 0:11

1 Answer 1



This answer raised some controversy, which I believe comes from the fact I didn't properly define what I mean by energy. In this post, whenever I say energy I refer to the canonical/Hilbert definition of the energy-momentum tensor, i.e., the functional derivative of the Lagrangian with respect to the metric. In GR, this is the energy that matters, because it directly enters in the EFE.

Note that this definition, in principle, has nothing to do with the classical interpretation of energy(=work), which is conserved along the trajectories of point particles. In GR we don't care about this latter energy, because it is not conserved in general (you need some Killing fields to exist, but OP asked us not to mention Killing fields).

Finally, in some semi-classical analysis of GR we can define some kind of potential energy (e.g., $2\phi(r)\sim 1+g_{00}(r)$), which does behaves like the classical interpretation of energy(=work). But this semi-cassical analysis, where you mix Newtonian and GR concepts, can only be used for weak or highly symmetric fields.

With this in mind, note that you can have bodies accelerating/slowing down because of the action of gravity. This doesn't contradict the fact that gravitational energy doesn't exist: when we say energy is conserved in GR, we mean $\nabla T=0$, but this energy is the canonical/Hilbert energy, not the kinetic+potential energy of the bodies! If you want to talk about kinetic+potential energy of a test body, you'll need Killing sooner or later!

First Question

Why should the EMT even contain gravitational energy? Wouldn't this energy be proportional to curvature? This energy would than again cause additional curvature which would in turn increase the grav. energy which would again increase the curvature etc.

The Einstein Field Equations tell you how curvature is related to energy:

$$ R_{\mu\nu}-\frac{1}{2}Rg_{\mu\nu}=T_{\mu\nu} $$ (where I use natural units $c=8\pi G=1$)

The l.h.s. is a measure of curvature, while the r.h.s. is a measure of the energy of the system. Once you know the r.h.s, you can solve these PDEs to find the metric (l.h.s).

Some examples:

  • You probably know from classical electromagnetism that the electromagnetic field carries energy (something like $\boldsymbol E^2+\boldsymbol B^2$ should look familiar). In tensor notation, the energy is written $$ T^{\mu\nu}=F^{\mu\alpha}F^\nu_\alpha-\frac{1}{4}g^{\mu\nu}F^2 $$ where $F$ is the electromagnetic strenth tensor (note that $T^{00}\propto \boldsymbol E^2+\boldsymbol B^2$).

  • If you have some matter around that you can model as a perfect fluid (e.g, negligible viscosity), then the formula for the energy is given by $$ T^{\mu\nu}=(\rho+p)u^\mu u^\nu+p g^{\mu\nu} $$ where $\rho,p$ are the mass density and the pressure of the fluid, and $u$ its velocity.

  • Etc

Anything that is present in the system generates gravity, and thus you should include its energy in the EFE: $T^{\mu\nu}=T_1^{\mu\nu}+T_2^{\mu\nu}+\cdots$, where each $T_i^{\mu\nu}$ is a different source of energy. But we dont include a term $\boldsymbol{T_i^{\mu\nu}}$ for curvature: gravity has no energy tensor:

$$ \text{r.h.s}=T_\text{EM}+T_\text{matter}+T_\text{quantum?}+\cdots+ \overbrace{ T_\text{gravity}}^{\text{NO!}} $$

There is no tensor for gravitational energy and there is no term for gravity in the r.h.s. of the the EFE. In the r.h.s. of the EFE you only include nongravitational forms of energy.

Note that when in vacuum (i.e. no matter/no radiation), the EFE are $$ R_{\mu\nu}-\frac{1}{2}Rg_{\mu\nu}=0 $$ You can clearly see that there is no term in the r.h.s., even though there is gravity.

Therefore, the answer to your first question is: the EMT should not contain gravitational energy.

Second Question

What is gravitational energy in GR? (a formula and/or picture would be nice)

In GR it doesn't make sense to speak of gravitational energy.

Short explanation: as you probably know, in GR gravity is not a true force$^1$. Therefore, there is no potential energy$^2$ associated to it.

Longer explanation: in GR energy is defined as $$ T^{\mu\nu}=\frac{-2}{\sqrt{|g|}}\frac{\delta \mathcal L_m}{\delta g^{\mu\nu}} $$ where $$ \mathcal L=\sqrt{|g|}\left(R+\mathcal L_m\right) $$ is the (full) Lagrangian of the theory, and $\mathcal L_m$ is the nongravitational part of the Lagrangian. As you can see, the energy is defined through the nongravitational part of the Lagrangian, which in turns means that it simply makes no sense to speak of the gravitational energy.

Note that it is possible to define a quantity that (supposedly) looks a lot like a gravitational energy$^3$: $$ t^{\mu\nu}=\frac{1}{2}Rg_{\mu\nu}-R_{\mu\nu}+\frac{1}{2g}[g(g^{\mu\nu}g^{\alpha\beta}-g^{\mu\alpha}g^{\nu\beta})]_{,\alpha\beta} $$ but it is not in general accepted as a true gravitational energy (it's more like a formal analogy, without much use AFAIK). Its not even a tensor. However, to see how it works, we can calculate $t^{00}$ in the Schwarzschild solution in the weak field (non-relativistic) limit: $$ g^{00}=-\left(1-\frac{2M}{r}\right) \qquad g^{ij}=\delta^{ij} $$ (where now I take $c=G=1$)

The "potential energy" of this metric is $$ t^{00}=\frac{1}{2g} \nabla^2 (g^2)=\nabla^2g+\frac{1}{2g}(\nabla g)^2 $$ where $g=g^{00}$. I wonder if this looks like a potential energy to you (IMHO, it does not).

EDIT: There is a certain point I'd like to discuss a bit further. It just happens that in EFE we don't include curvature-energy in the r.h.s. You point out that if this were the case, then the curvature would cause more curvature, and the new curvature would cause even more, etc. Is this possible? or is it just absurd?

Well, I think you will like a lot this lecture by Feynman (Vol II. Chap 23: Cavity Resonators). The r.h.s. of Maxwell equations include both $\boldsymbol E$ and $\boldsymbol B$, so in this case there exists the feedback you talk about: a certain electric field can be responsible for a magnetic field, which in turns generates more electric field, which is responsible for more magnetic field, ad infinitum. But the result is a convergent series, so everything works out just fine (I know this is kinda irrelevant for your question, but I think its neat, and I believe you might like it as well :) )

$^1$: "In general relativity, the effects of gravitation are ascribed to spacetime curvature instead of a force." - from Wikipedia.

$^2$: "In physics, potential energy is the energy that an object has due to its position in a force field [...]" - from Wikipedia.

$^3$: this quantity satisfies certain relations that energy usually satisfies, and it's constructed using just the metric tensor $g$.

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    $\begingroup$ So where does the kinetic energy of an accelerating body due to the "spacetime curvature" come from? $\endgroup$
    – Peter R
    Jan 4, 2016 at 20:40
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    $\begingroup$ @PeterR you are thinking classically. See, for example, this and this posts. They explain very clearly how "curvature makes things move". $\endgroup$ Jan 4, 2016 at 21:03
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    $\begingroup$ I know how curvature makes things move but there is still the issue of conservation of energy which has some tricky aspects when it comes to Generl Reltivity. GR is the last of the great Classical theories. $\endgroup$
    – Peter R
    Jan 4, 2016 at 21:18
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    $\begingroup$ Do gravitational waves carry energy? $\endgroup$
    – Peter R
    Jan 4, 2016 at 21:36
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    $\begingroup$ @PeterR thats a rather subtle issue, and in fact many authors disagree about it. To fit the question into my answer: gravitational waves don't carry the kind energy you would write in the r.h.s. of the Einstein Field Equations. Can we define other objects that reflect our intuition about what energy means? yes: but it is not a trivial task (and most of the time, it's a non-covariantly observer-dependent quantity). For more details, see Michael Seifert's comment above. $\endgroup$ Jan 4, 2016 at 22:00

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