Event and cosmological horizons in black hole structures

Question

Sometimes, we face two horizons in the dS black hole. How do we determine the difference between the event horizon and the cosmological event horizon? How is it calculated, and what is the difference? How do we know which is the event horizon of our black hole?

Answer 1

Consider the Schwarzschild-de Sitter metric (SdS metric): $$ \mathrm{d}^2s =\left( 1 -\frac{2M}{r} -\frac{\Lambda}{3}r^2 \right)\mathrm{d}^2t +\left( 1 -\frac{2M}{r} -\frac{\Lambda}{3}r^2 \right)^{-1}\mathrm{d}^2r +r^2\mathrm{d}\Omega. $$ It becomes singular for solutions of the equation: $$ f(r) =\frac{\Lambda}{3}r^3-r+2M \;\stackrel{!}{=}\;0. $$ Since $f(0)=2M>0$ and $\lim_{r\rightarrow-\infty}f(r)=-\infty$, one solution will always be negative and hence not be of any interest (because we consider $r\geq 0$). Of the two remaining solutions, the smaller one will be the event horizon and the larger on will be the cosmological horizon.

Of course, the polynomial also allows for no more or only one more solution, which happens when the dip down is gradually lifted above the $x$-axis. But this doesn't happen due to the mass of the black hole having an upper limit (with which the Schwarzschild-de Sitter metric is also called Narai metric). If the event horizon and the cosmological horizon coincide, then the minium of the dip down is exactly on the $x$-axis. Hence there is a radius $r>0$ with $f(r)=0$ and: $$ f'(r) =\Lambda r^2-1 \;\stackrel{!}{=}\;0 $$ The latter condition yields $r=\sqrt{1/\Lambda}$ and put into the former condition further yields: $$ f\left(\sqrt{1/\Lambda}\right) =-\frac{2}{3}\sqrt{\frac{1}{\Lambda}} +2M \;\stackrel{!}{=}\;0 \Rightarrow M=\frac{1}{3}\sqrt{\frac{2}{\Lambda}}. $$ Hence the upper limit for the mass of a black hole in the Schwarzschild-de Sitter metric is: $$ M\leq\frac{1}{3}\sqrt{\frac{2}{\Lambda}} $$ and ensures exactly two horizons, the smaller event horizon and the larger cosmological horizon.