Calculate interval entanglement from a geodesic

Write the quantity that the shortest curve minimizes

Set $t=0$ in Poincaré coordinates. The boundary direction is $x$ and depth $z>0$ increases into the bulk. Use the same length units for $x,z,L$. Put endpoints at $(-\ell/2,\epsilon)$ and $(\ell/2,\epsilon)$. Here $\ell$ means their horizontal separation on the cutoff surface. Hold $\epsilon>0$ fixed while finding the shortest bulk curve.

For a curve $z(x)$, a small proper length is $L\sqrt{dx^2+dz^2}/z$. The factor $1/z$ makes motion near the boundary expensive. A curve can travel inward to reduce the cost of horizontal motion. The competition between descent and horizontal distance determines its depth.

Source: João Penedones · Shinsei Ryu and Tadashi Takayanagi

Translation symmetry determines a circle

The integrand $F(z,z^{\prime})=\sqrt{1+(z^{\prime})^2}/z$ has no explicit dependence on $x$. Its Euler–Lagrange equation therefore has first integral $F-z^{\prime}\,\partial F/\partial z^{\prime}=\mathrm{constant}$. At maximum depth the slope vanishes. Call that depth $R$, so the constant is $1/R$.

Rearrange to obtain $z\sqrt{1+(z^{\prime})^2}=R$. Separating variables and using left-right symmetry gives $x^2+z^2=R^2$. Requiring the circle to pass through our cutoff anchors fixes $R=\sqrt{(\ell/2)^2+\epsilon^2}$. Starting instead with a circle of radius $\ell/2$ anchored at the coordinate boundary and then clipping it moves the cutoff endpoints inward. That is a different regulator convention.

Integrate along the arc

Substitute $x=R\cos\theta$ and $z=R\sin\theta$. The Euclidean line element becomes $R\,d\theta$. Dividing by depth gives $L\,d\theta/\sin\theta$. The endpoints have angles $\theta_0$ and $\pi-\theta_0$, with $\sin\theta_0=\epsilon/R$. An antiderivative is $\log\tan(\theta/2)$.

Evaluating both endpoints gives the exact expression below. The inverse hyperbolic sine is $\operatorname{arsinh}u=\log(u+\sqrt{1+u^2})$. Expanding for small $\epsilon/\ell$ produces the logarithmic leading term. The first correction is quadratic in the cutoff ratio.

Worked example: choose $L=1$, $\ell=2$ and $\epsilon=0.1$. The exact length is $2\operatorname{arsinh}10\simeq5.99645$; the leading approximation is $2\log20\simeq5.99146$. Their difference is about $0.00498$. Halving the cutoff reduces this error by roughly four, although the length itself grows by nearly $2\log2$.

Subtract before removing the cutoff

Give intervals of widths $\ell_1$ and $\ell_2$ the same cutoff. Their common term $-2L\log\epsilon$ cancels. The difference has a finite limit. At finite cutoff the exact difference still contains power corrections, so the pure logarithmic difference is only the limiting result.

Increasing the width from two to four gives a limiting length difference $2L\log2$. At $\epsilon=0.1$, the exact difference is about $1.38256L$, slightly below $1.38629L$. Numerical quadrature should reproduce the finite value; a cutoff scan tests convergence toward the limit.

When a length becomes an entropy

The static Ryu–Takayanagi prescription divides the length of a minimal curve homologous to the boundary interval by $4G_3$. Set $\hbar=1$. The three-dimensional Newton constant $G_3$ has length units, so the entropy is dimensionless. In an AdS3/CFT2 description governed by Einstein gravity, the central charge $c=3L/(2G_3)$ measures the scale of boundary degrees of freedom.

The leading vacuum interval entropy is then $(c/3)\log(\ell/\epsilon)$. Doubling the width changes it by $(c/3)\log2$. This interpretation assumes a semiclassical bulk, large central charge and a controlled gravitational approximation. Chapter 7 introduces bulk quantum corrections, which matter when interpreting a vanishing leading mutual information.

Source: Shinsei Ryu and Tadashi Takayanagi