The cosmological event horizon found in a de Sitter universe has some interesting similarities to that of a black hole. For example, since we can find a temperature at the horizon, we are able to use the relation $S = A/4G$ to calculate the entropy of the horizon related to it's surface.

I am looking for a method of directly calculating the entropy, instead of using the famous relation given above. One example I found was mentioned in Les Houches Lectures on De Sitter Space (pages 17,18). Unfortunately, the method is given as an exercise, and the conclusion given is far from obvious (to me).

They start by looking at the static Schwarzschild-de Sitter metric in three dimensions given by $$ds^2 = -\left(1 - 8GE - r^2\right)\,dt^2 + \dfrac{dr^2}{1-8GE-r^2} + r^2 d\theta^2 $$

The paper states that finding a Green's function for $SdS_3$ by analytic continuation from the smooth Euclidian solution periodic in $\tau \rightarrow \tau + \dfrac{2 \pi i}{\sqrt{1 - 8GE}}$ will give all required answers to find the entropy (the authors calculate temperature invoking a thermodynamic identity and they finish off with $S$).

I have yet to find another paper that uses this derivation, but I am quite interested in the part missing. Is there someone who can shed a light on how finding such a Green's function leads to temperature and entropy?


1 Answer 1


I am looking for a method of directly calculating the entropy, instead of using the famous relation (...) how finding such a Green function leads to temperature/entropy?

What you need is Green's functions' period, i.e. period $\beta$ on imaginary time. You can calculate this merely by removing the 'singularity' at the horizon. (At least) for any metric of the type $$ds^2=-f(r)\,dt^2+f^{-1}(r)\,dr^2+r^2\,d\Omega^2\tag{1}$$ let $$f(r_+)=0\tag{2}$$ then if you look near $r_+$, say for $r=r_++\epsilon^2$, at the lowest order, $$f(r_++\epsilon^2)\simeq\epsilon^2f^\prime(r_+)\tag{3}$$ and thus the metric here looks like $$ds^2=-\epsilon^2f^\prime(r_+)\,dt^2+4\epsilon^2\left[\epsilon^2f^\prime(r_+)\right]^{-1}\,d\epsilon^2+(r_++\epsilon^2)^2\,d\Omega^2\tag{4}$$ Now, letting $t=i\tau$, \begin{align}ds^2&=\epsilon^2f^\prime(r_+)\,d\tau^2+4\epsilon^2\left[\epsilon^2f^\prime(r_+)\right]^{-1}\,dr^2+(r_++\epsilon^2)^2\,d\Omega^2\\&=\frac{4}{f^\prime(r_+)}\left[d\epsilon^2+\left(\frac{f^\prime(r_+)}{2}\right)^2\epsilon^2d\tau^2\right]+(r_++\epsilon^2)^2\,d\Omega^2\tag{5}\end{align} and so, in order to avoid a conical singularity, you should have $$\frac{\left|f^\prime(r_+)\right|}{2}\int_0^\beta\int_0^{r_+}\epsilon\,d\epsilon\,d\tau=\pi{r}_+^2\,\Longrightarrow\,\beta=\frac{4\pi}{\left|f^\prime(r_+)\right|}\tag{6}$$ In this case $f(r)=1-8GE-r^2$, and so $\left|f^\prime(r_+)\right|=2r_+$ where $r_+=\sqrt{1-8GE}$. Then you just use $T=\beta^{-1}$ and (71) on "Les Houches Lectures" to get temperature and entropy.

This is the easiest way I know; I openly avoided Green's functions as that seem to be another question, and one that someone else can answer better than me.


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