I'm trying to get the AdS solution to the circular wilson loop. The standard AdS metric is:

$ds^2 = \frac{L^2}{z^2}(\eta_{\mu \nu} dx^{\mu} dx^{\nu} + dz^2)$

If I take the circle of radius R at the x1,x2 plane I can choose polar coordinates:

$x_1 = R cos(\theta)$, $x_2 = R sin(\theta)$

$ds^2 = \frac{L^2}{z^2}(-dt^2 + dr^2 + r^2d\theta^2 + dx3^2 + dz^2)$

Now i want to find the area that minimizes the Nambu-Goto action:

$S_{NG} = \int d\sigma d\tau \sqrt{g}$

Where g is the usual pullback: $g_{ab} = G_{\mu \nu} \partial_a X^{\mu} \partial_b X^{\nu}$. Now my fields are $X^{\mu} = (t,r,\theta,x3,z(r))$ and I choose the gauge where: $\sigma = r$, $\tau = \theta$ from where i get:

$S_{NG} = \int dr d\theta \frac{L^2 r}{z^2} \sqrt{1 + z'^2}$

From where I see that the Hamiltonian is conserved and we get:

$H = \frac{-L^2 r}{z^2} \frac{1}{\sqrt{1+z'^2}}$

But the answer is $S_{NG}= \sqrt{\lambda} (\frac{R}{z_0}-1)$ and I don't know where the problem is.

This post imported from StackExchange Physics at 2016-07-07 18:18 (UTC), posted by SE-user Jasimud