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Given that gravitational waves transmit energy, the curvature of spacetime must, itself, contain energy. How do you calculate this energy? Specifically, I'm most interested in how one would calculate this for a black hole.

Qmechanic
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Doradus
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I would suggest you look into the paper "Gravitational field energy density for spheres and black holes" by D. Lynden-Bell and J. Katz. Due to authors the whole gravitational energy is contained in the spacetime outside of a black hole, adding up to $Mc^{2}$. https://www.researchgate.net/publication/234222351

JanG
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In an astrophysical black hole, there can be infalling matter that has not yet reached the singularity. But in idealized models of black holes, one often considers a black hole in which there is no such matter still in the process of infalling. In this case, the entire mass-energy of the black hole is in the form of the energy content attributable to the curvature of the spacetime.

In his second, more recent answer, Jan Gogolin gives an r-dependent equation for the energy. This is not right. There is no way to attribute a local energy density to the curvature of a spacetime, so the only thing that is well-defined is the energy density of the entire black hole, as seen by a distant observer. This energy is what is conserved in general relativity for a spacetime that consists of an isolated body surrounded by infinite, empty space (called an asymptotically flat spacetime).

user321748
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    You are of course right. My simplified explanation could be misleading although the equation is correct. However, your sentence 'no way to attribute a local energy density to the curvature of a spacetime' leaves me at loss. Is that not Einstein field equation $G_{\mu \nu}=\kappa T_{\mu \nu}$ which connects these things? – JanG Dec 14 '21 at 12:10
  • Actually, that strikes at the heart of my question! I had hoped to learn how to calculate the energy of the curvature so I could check whether it equals mc² and therefore accounts for all the energy. – Doradus Dec 14 '21 at 20:34
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    @Doradus, helpful could be the question and answers from https://physics.stackexchange.com/questions/111573/is-space-time-a-special-form-of-energy . I – JanG Dec 15 '21 at 17:17
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In simply words, the energy of a black hole of mass $m$ reads $E=m c^{2} g_{00}= m c^2 (1-r_{S}/r)$. The more detailed derivation you can find here: https://physics.stackexchange.com/a/77280/281096. Thus, for a distant observer the black hole energy is $E=m c^{2}$, and for proximal, $r=r_{S}$, it amounts to $E=0$. Because behind the black hole event horizon there is no matter (no mass), one can interpret it as the statement that whole energy $m c^{2}$ is contained in the warped exterior spacetime. Be aware, that $r$ is not simply radial coordinate but a curvature radius of a 2-sphere, with surface area $A=4 \pi r^2$.

JanG
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