Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

145 submissions , 122 unreviewed
3,930 questions , 1,398 unanswered
4,862 answers , 20,637 comments
1,470 users with positive rep
502 active unimported users
More ...

Circular Wilson Loop in AdS/CFT

+ 2 like - 0 dislike
73 views

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 22, 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

+ 0 like - 0 dislike

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 ]

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar\varnothing$sicsOverflow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
To avoid this verification in future, please log in or register.




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...