Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 354 / 2

On the $t=0$ slice, rotational symmetry lets the surface be written $z=z(r)$. The inversion symmetry fixing the boundary circle extends to an isometry of the Poincare half-space whose fixed set isThe unique extremal surface anchored at $r=R,z=0$ must be invariant under that isometry. Hence the Ryu–Takayanagi surface for a disk is the hemisphereOne may verify directly that it extremizes the area functional $2\pi \int r{z}^{-2}\sqrt{1+z^{\prime 2}}dr$.

Regulate the surface at $z=ϵ$, so $r_{\max }=\sqrt{R^2-{ϵ}^2}$. On the hemisphere, $\sqrt{1+z^{\prime 2}}=R/z$, and its area isThe Ryu–Takayanagi formula therefore givesThe first term is the ultraviolet area-law divergence proportional to the boundary circle's length; the finite constant is universal for the vacuum of a three-dimensional conformal field theory.

Translation invariance in $y$ lets the surface be written $z=z(x)$. Per unit $y$ length, its area and entropy functionals areThe boundary conditions are $z(\pm R)=0$, understood with an ultraviolet cutoff, and reflection symmetry gives $z^{\prime }(0)=0$ at $z(0)=z_{\max }$.

The Lagrangian $\mathcal{L}=z^{-2}\sqrt{1+z^{\prime 2}}$ has no explicit $x$ dependence, so its Beltrami identity is conserved:At the turning point $z=z_{\max }$ and $z^{\prime }=0$, the constant is $z_{\max }^{-2}$. Rearrangement gives

For the disk, the hemisphere reachesFor half of the strip surface, invert the first-order equation and integrate from the midpoint to the boundary:The beta function integral isThereforeThe strip's holographic entropy surface lies deeper in the bulk than that of a disk with the same half-width or radius.