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Naively, I would imagine that $$ T = \frac{E}{S} $$ However, for a black hole, $E=c^4 R / 2G$ and $S= A k c^3 / 4 G \hbar $, which yields $$ T = \frac{E}{2S} $$ Is there a simple explanation for the factor 2?

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  • $\begingroup$ Why would you think the first statement should be true? $\endgroup$ – mmeent Sep 27 at 14:59
  • $\begingroup$ How does your comment help explaining the factor 2? $\endgroup$ – frauke Sep 27 at 17:43
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Is there a simple explanation for the factor 2?

Yes. This is a consequence of a purely mathematical fact called Euler's homogeneous function theorem. For (nonrotating and uncharged) black holes the first law of black hole mechanics could be written as $$ dM = T dA, $$ But since the mass $M$ of a black hole is a homogeneous function of degree $\frac 12 $ in area $A$, the mass could be expressed as a bilinear form: $$ M = 2 \, T A . $$ This relationship could also be generalized to include angular momentum $J$ and electric charge $Q$: $$ M = 2\, T A+2\,\Omega J+ \Phi Q, $$ where $\Omega$ is the angular velocity and $\Phi$ is electromagnetic potential at the horizon (there is no $2$ in the last term). This formula is known as Smarr relation.

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