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  Circular Wilson Loop in AdS/CFT

+ 2 like - 0 dislike

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
asked May 21, 2016 in Theoretical Physics by Jasimud (35 points) [ no revision ]
You seem to be concerned about two different things. The Wilson loop $W = exp(\oint A_\mu dx^\mu)$ evaluates a gauge connection around a loop. You can do this in AdS. Your question seems to be concerned with the action of a particle or string in AdS spacetime.

This post imported from StackExchange Physics at 2016-07-07 18:18 (UTC), posted by SE-user Lawrence B. Crowell
In the AdS/CFT conjecture the proposal is that the Maldacena-Wilson Loop in SYM is dual to the string ending in the loop path. In the large N limit where supergravity holds one has to compute the minimal area of the string in order to get the vev

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

1 Answer

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The Nambu-Goto action (including normalization) is $$ S_{NG} = \frac{1}{2\pi \alpha'}\int dr d\theta \frac{L^2 r}{z^2} \sqrt{1 + z'^2}=\frac{L^2}{\alpha'}\int dr \frac{r}{z^2} \sqrt{1 + z'^2} $$ The corresponding Hamiltonian is not conserved as the Lagrangian is explicitly $r$-dependent. You can however see that a solution to the equations of motion $$ 0=\frac{d}{dr}\frac{\partial L}{\partial z'}-\frac{\partial L}{\partial z} $$ with the boundary condition $z(r=R)=0$ is given by $z(r)=\sqrt{R^2-r^2}$. Pluggin this back into the Nabu-Goto action (cutting off the integal at $z=z_0$, i.e. at $r=\sqrt{R^2-z_0^2}$ in order to regularize), leads to the following integral to be evaluated $$ S_{NG}=\frac{L^2}{\alpha'}\int_0^{\sqrt{R^2-z_0^2}}dr\,\frac{R r}{(R^2-r^2)^{3/2}}=\frac{L^2}{\alpha'}\left(\frac{R}{z_0}-1\right). $$ From the basic holographic relation $\frac{L^4}{\alpha'^2}=g_{YM}^2 N=\lambda$, we see that the prefactor is $\sqrt{\lambda}$

This post imported from StackExchange Physics at 2016-07-07 18:18 (UTC), posted by SE-user physicus
answered Jul 7, 2016 by physicus (105 points) [ no revision ]

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