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Circular Wilson Loop, minimal surface in AdS5

  1. Aug 1, 2013 #1
    1. The problem statement, all variables and given/known data
    Consider [itex]AdS_5[/itex] space in Poincaré coordinates with metric [itex]ds^2=\frac{dz^2+dx_\mu dx^\mu}{z^2}[/itex]. There is a circular Wilson Loop with Radius [itex]R[/itex] in the Minkowskian boundary of [itex]AdS_5[/itex]. We want to find the surface of minimal area in [itex]AdS_5[/itex] that has this loop as boundary.

    We choose to parametrize the surface with cylindircal coordinates: [itex]x_1=r\cos{\theta}, x_2=r\sin{\theta}, z=f(r)[/itex] and [itex]x_0=x_3=0[/itex].

    2. Relevant equations



    3. The attempt at a solution

    The area is
    [itex]\int_0^{2\pi}d\theta\int_0^R dr\; r\sqrt{\underset{\alpha\beta}{\det}{g_{mn}(\partial_\alpha X^m \partial_\beta X^n})}[/itex]
    [itex]=\int_0^{2\pi}d\theta\int_0^R dr\; r\frac{1}{z^2}\sqrt{{\det}\begin{pmatrix} (\partial_\theta x_1)^2+(\partial_\theta x_2)^2 & \partial_\theta x_1 \partial_r x_1+\partial_\theta x_2 \partial_r x_2 \\ \partial_r x_1 \partial_\theta x_1+\partial_r x_2 \partial_\theta x_2 & (\partial_r z)^2+(\partial_r x_1)^2+(\partial_r x_2)^2\end{pmatrix}}[/itex]
    [itex]=\int_0^{2\pi}d\theta\int_0^R dr\; r\frac{1}{f(r)^2}\sqrt{{\det}\begin{pmatrix} r^2 & 0 \\ 0 & 1+f'(r)^2 \end{pmatrix}}[/itex]
    [itex]=\int_0^{2\pi}d\theta\int_0^R dr\; \frac{r^2}{f(r)^2}\sqrt{1+f'(r)^2}[/itex]
    [itex]=2\pi\int_0^R dr\; \frac{r^2}{f(r)^2}\sqrt{1+f'(r)^2}[/itex]

    I have tried to solve this by treating [itex]\frac{r^2}{f(r)}\sqrt{1+f'(r)^2}[/itex] as Lagrangian and solving the equations of motion. But that just becomes incredibly messy. Is there a mistake up to this point? Does anyone know a nicer method to solve this? Do you know where to find the solution?

    Thanks for any support!
     
  2. jcsd
  3. Aug 11, 2013 #2
    I think in the last line you made a mistake, it should be r, not r^2.
     
  4. Aug 21, 2013 #3
    Could you tell me why it is [itex]r[/itex] and not [itex]r^2[/itex]? Don't I get one factor of [itex]r[/itex] from the Jacobian when I go to polar coordinates in the integral plus another factor of [itex]r[/itex] from the determinant in the square root?
     
  5. Aug 21, 2013 #4

    fzero

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    Your factor of the square root of the induced metric is precisely what leads to the Jacobian. The factor of ##r## coming from inside the square root is the factor ##r## that you usually find in the measure for polar coordinates.

    The solution is found by a conformal transformation in the original literature, which is reviewed in these lectures. If you want to solve the equation of motion directly, it will help to write it as the sum of 3 terms involving the expression ##f f' + r##. Then you can look for a solution where this expression is zero.
     
    Last edited: Aug 21, 2013
  6. Aug 22, 2013 #5
    Can u explain more about how to write it as the sum of 3terms involving the expression ##f f' + r##?
     
  7. Aug 22, 2013 #6

    fzero

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    Since this is the HW section, I really shouldn't post the equation of motion until the OP shows his attempt. In words, after expanding everything out, pulling out common factors and grouping by derivatives, you should find 5 terms. Two of these carry a factor of 2 in front and get split up into different groupings, so we can write the whole thing as 7 terms. We can then identify 3 terms corresponding to a factor of ##(ff'+r)'##. The remaining 4 terms can be written as two terms with a common factor of ##ff'+r##.
     
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