1
$\begingroup$

In order to compute the holographic entanglement entropy for a spherical subsystem in AdS using the Ryu-Takayanagi conjecture, one needs to solve the following second order nonlinear differential equation:

$$zr(z)r''(z) - (d-1)r(z)(r'(z))^3 - (d-2)z(r'(z))^2 - (d-1)r(z)r'(z) - (d-2)z = 0$$

and obtain the solution $r^2 + z^2 = l^2$, I need help in obtaining this solution.

N.B.: The equation (in a mildly different form), and its solution are given in hep-th/0605073

$\endgroup$
0
$\begingroup$

In the case of a disk $d=2$, the equation is simplified: $$zr(z)r''(z) - r(z)(r'(z))^3 - r(z)r'(z) = 0$$ We discard the obvious solution of the equation $r = 0$, then putting $r' = y$ we find $$zy'=y^3+y$$ This equation is easily integrated, and using the boundary condition $y(R) =\infty $, we find $$\frac {r'}{\sqrt {1+r'^2}}=\frac {z}{R}$$

Solving the equation for $r`$, we have $$r'=\pm \frac {z}{\sqrt {R^2-z^2}}$$ And find finally $$r=\pm \sqrt {R^2-z^2}$$ Note that in the article Aspects of Holographic Entanglement Entropy the authors cited another equation on p.34 $$ rzz′′ + (d−1)z(z′)^3 + (d−1)zz′ + dr(z′)^2 + dr = 0. $$ For this equation, the solution $z^2+r^2=R^2$ exists for any $d$.

$\endgroup$
  • $\begingroup$ Thank you, any suggestion on how to generalize to higher dimensions? $\endgroup$ – Sabyasachi Maulik May 12 at 17:12
  • $\begingroup$ @SabyasachiMaulik See update to my answer $\endgroup$ – Alex Trounev May 12 at 20:12
  • $\begingroup$ Can you elaborate on that please? $\endgroup$ – Sabyasachi Maulik May 13 at 2:47
  • $\begingroup$ @SabyasachiMaulik Use code r*z*z'' + (d \[Minus] 1) z (z')^3 + (d \[Minus] 1) z*z' + d*r (z')^2 + d*r /. {z -> Sqrt[R^2 - r^2], z' -> -(r/Sqrt[-r^2 + R^2]), z'' -> -(r^2/(-r^2 + R^2)^(3/2)) - 1/Sqrt[-r^2 + R^2]} // Simplify $\endgroup$ – Alex Trounev May 13 at 15:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.