# Why is the temperature of a black hole given by $T=E/2S$?

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?

• Why would you think the first statement should be true? – mmeent Sep 27 at 14:59
• How does your comment help explaining the factor 2? – frauke Sep 27 at 17:43

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.