The Bekenstein–Hawking entropy is defined by the relation


where $k_B$ is the Boltzmann constant and $A$ is the area of the black hole's event horizon in units of the Planck area $l^2_p$.

I am not familiar with black holes thermodynamics, so I have three relatively simple questions.

  1. Where does the proportionality constant $\frac{1}{4}$ come from and how do you interpret it?
    I tried to find a clear answer but always get confused.

  2. If you consider the black hole as a 2D disc or circle (not a 3D sphere). Would I have the same expression (in the same unit) and in particular, the same proportionality constant ?

  3. In the derivation of the formulae, is there any assumption on the shape of the black hole ? For instance that it is a sphere or whatever the shape.

  • $\begingroup$ With regard to your second question, the dependence on the radius would be the same if the black hole was circular or spherical, since we would only see a disc. The fundamental quantity is the area, because the maximum amount of information that can be stored in a region of space goes as the surface area of that space in Planck units. $\endgroup$
    – Physics
    Commented Jan 9, 2020 at 16:54
  • $\begingroup$ Thus you would say the formula for the entropy of a 2D black hole (assimilated as a circle or disc) would be the same $𝑆=k_b A/(4 l^2_p)$ but with $𝐴=\pi R^2$ ? $\endgroup$
    – Nath
    Commented Jan 9, 2020 at 20:44
  • $\begingroup$ Black holes don't really exist in 2+1-dimensional gravity. Vacuum solutions are flat in 2+1 dimensions. $\endgroup$
    – user4552
    Commented Jan 9, 2020 at 21:25
  • $\begingroup$ @Physics the dependence of the entropy$\sim$area on radius is manifestly different for a circle ($S^1$) and a 2-sphere ($S^2$).@Nath In the former case $S = A/4$ with $A = 2\pi R$. $\endgroup$
    – Dwagg
    Commented Jan 9, 2020 at 23:28

1 Answer 1

  1. While there are proofs of $S= A/ 4$ (in natural units $\ell_p=1$, $k_B=1$), I think a good interpretation of the $\frac14$ is an open nonproblem (a good interpretation being subjective of course). The proofs are unenlightening as to the $\frac14$.

  2. If the black hole horizon has the topology of a $d$-sphere $S^d$ then the area $\sim r^d$ where $r$ is the radius. For a 3D black hole (2 spatial + 1 time dimension) the horizon is an $S^1$, and the entropy is $\frac14 2\pi R = \frac12 \pi R$. There is an example of a 3D black hole known as the BTZ black hole which lives in asymptotically AdS space (not flat space, i.e. not a vacuum solution), and this relation holds true.

  3. No, it is quite general. For example, if there is a non-spherically symmetric background matter field, the horizon may deform slightly so that it is no longer a sphere; the entropy is still given by $\frac14$area of horizon. The formula also holds for spinning black holes (Kerr solution) where the horizon is oblate.

  • $\begingroup$ From your answers, to be sure that I understood correctly, can you confirm the following statement that, the Bekenstein–Hawking entropy as written in my post, that is $S = k_b A/(4l^2p)$, is a general relation that holds for any black hole horizon topology. The only thing that changes is implicitly hidden in $A$ the area, as you showed in your answers 2 and 3. $\endgroup$
    – Nath
    Commented Jan 10, 2020 at 0:15
  • $\begingroup$ That's right, with the minor correction that one must write $\ell_p^d$ for general dimension $d$. $\endgroup$
    – Dwagg
    Commented Jan 10, 2020 at 0:44

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