The <Ryu–Takayanagi formula> gives the leading entropy of a static boundary region:
$$
S(D_1)=\frac{\operatorname{Area}(\gamma_{D_1})}{4G}.
$$
On the $t=0$ slice of Poincaré $\operatorname{AdS}_4$,
$$
ds^2=\frac{dz^2+d\rho^2+\rho^2d\theta^2}{z^2}.
$$
The <minimal surface> anchored on a boundary circle of radius $R_1$ is the hemisphere
$$
z^2+\rho^2=R_1^2.
$$
Cutting it off at $z=\epsilon$, its area is
$$
\begin{aligned}
A_\epsilon
&=2\pi R_1\int_0^{\sqrt{R_1^2-\epsilon^2}}
\frac{\rho\,d\rho}{(R_1^2-\rho^2)^{3/2}}\\
&=2\pi\left(\frac{R_1}{\epsilon}-1\right).
\end{aligned}
$$
Therefore
$$
\boxed{
S(D_1)=\frac{\pi R_1}{2G\epsilon}
-\frac{\pi}{2G}}.
$$
The universal requirement was the perimeter-law divergence $\pi R_1/(2G\epsilon)$; the finite constant depends on the stated pure $\operatorname{AdS}_4$ vacuum geometry and equals $-\pi/(2G)$ here. Since $G\sim N^{-3/2}$, both terms are of order $N^{3/2}$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 354 / 2 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-354/2/a/solution