Choose a constant-time boundary interval whose endpoints differ by $\pi$ in $\phi$. Its <Ryu–Takayanagi formula> surface is the diameter through $r=0$. Cutting it off at $r_c=\pi/2-\epsilon$, its length is
$$
\ell=2R_{\rm AdS}\int_0^{r_c}\frac{dr}{\cos r}
=2R_{\rm AdS}\log(\sec r_c+\tan r_c)
=2R_{\rm AdS}\log\frac2\epsilon+O(\epsilon^2).
$$
The <holographic entanglement entropy> is therefore
$$
S=\frac\ell{4G}
=\frac{R_{\rm AdS}}{2G}\log\epsilon^{-1}+\text{finite}
=\boxed{\frac c3\log\epsilon^{-1}+\text{finite},}
$$
where the <Brown--Henneaux central charge> $c=3R_{\rm AdS}/(2G)$ was used.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 354 / 2 / a / Solution

{c}
{title2=$c=3R_{\rm AdS}/(2G)$}

The Brown--Henneaux central charge is the central charge of the two-dimensional conformal field theory dual to Einstein gravity on $\operatorname{AdS}_3$:
$$
c=\frac{3R_{\rm AdS}}{2G_3}.
$$


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