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Lightlike geodesic in AdS5xS5, plane wave background

  1. Aug 31, 2013 #1
    1. The problem statement, all variables and given/known data

    My question is about a step in the lecture notes [http://arxiv.org/abs/hep-th/0307101] on page 6, and it is probably quite trivial:

    I want to see why a lightlike particle in [itex]AdS_5\times S^5[/itex] sees the metric as plane wave background. The metric is
    [itex]ds^2=R^2(-dt^2 \cosh^2\rho+d\rho^2+\sinh^2\rho \,d\Omega_3^2[/itex][itex]+d\psi^2\cos^2\theta+d\theta^2+\sin^2\theta\,\Omega_3'^2)[/itex]

    In order to study the metric close to a lightlike geodesic we make the follwoing change of coordinates:
    [itex]{x}^+=\frac{1}{2\mu}(t+\psi), {x}^-=\frac{\mu R^2}{2}(t-\psi), \rho=\frac{r}{R}, \theta=\frac{y}{R}[/itex]

    I am supposed to get in the [itex]R\to\infty[/itex] limit
    [itex]ds^2=R^2(-\mu^2(dx^+)^2+\mu^2(dx^+)^2)+(-2dx^+dx^--\mu^2r^2(dx^+)^2+dr^2+r^2d\Omega_3^2[/itex] [itex]-2dx^+dx^--\mu^2y^2(dx^+)^2+dy^2+y^2d\Omega'{}_3^2)+\mathcal{O}(R^{-2})[/itex]

    This is not the final result, but from there on I know how to continue.


    2. Relevant equations

    [itex]\cosh x=1+\frac{1}{2}x^2+\mathcal{O}(x^4), \cos x=1-\frac{1}{2}x^2+\mathcal{O}(x^4)[/itex]
    [itex]\Rightarrow \cosh^2 x = 1+x^2+\mathcal{O}(x^4), \cosh^2 x = 1-x^2+\mathcal{O}(x^4)[/itex]


    3. The attempt at a solution

    I can expand in [itex]\rho, \theta[/itex], since they will be small in the [itex]R \to \infty[/itex] limit:
    [itex]ds^2=R^2(-dt^2 \cosh^2\rho+d\rho^2+\sinh^2\rho \,d\Omega_3^2+d\psi^2\cos^2\theta+d\theta^2+\sin^2\theta\,\Omega'{}_3^2)[/itex]
    [itex]=R^2(-dt^2(1+\rho^2)+d\rho^2+\rho^2d\Omega_3^2+d\psi^2(1-\theta^2)+d\theta^2+\theta^2d\Omega'{}_3^2)+\mathcal{O}(R^{-2})[/itex]
    [itex]=R^2(-dt^2+d\psi^2)+(-dt^2r^2+dr^2+r^2d\Omega_3^2-d\psi^2y^2+dy^2+y^2d\Omega'{}_3^2=+\mathcal{O}(R^{-2})[/itex]

    Now I use:
    [itex] dx^+dx^-=\frac{1}{2\mu}(dt+d\psi)\frac{\mu R^2}{2}(dt-d\psi)=\frac{R^2}{4}(dt^2-d\psi^2)[/itex]
    So the first term above is [itex]R^2(-dt^2+d\psi^2)=-4dx^+dx^-[/itex].

    However, I do not know where all the [itex](dx^+)^2[/itex] in the solution are coming from, since
    [itex](dx^+)^2=\frac{1}{4\mu^2}(dt^2+2dt\,d\psi+d\psi^2)[/itex]
    Where do these mixed [itex]dt\,d\psi[/itex] terms come from?
     
  2. jcsd
  3. Sep 1, 2013 #2

    fzero

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    They come from quadratic terms in the expansion of cos and cosh that are finite in the scaling limit.
     
  4. Sep 1, 2013 #3
    Thank you, but could you be a bit more precise, please. From comparing the given result and what I got so far I should be able to show:
    [itex]-dt^2r^2-d\psi^2y^2=R^2(-\mu^2(dx^+)^2+\mu^2(dx^+)^2)-\mu^2r^2(dx^+)^2-\mu^2y^2(dx^+)^2[/itex]
    The left hand sinde are those quadratic terms from the expansion of [itex]\cos[/itex] and [itex]\cosh[/itex]. Since the first part of the right hand side is 0 anyways, I don't mind to much that I don't get that one, but I still don't see the equality:
    [itex]-\mu^2r^2(dx^+)^2-\mu^2y^2(dx^+)^2=-\frac{r^2+y^2}{4}(dt^2+2dt\,d\psi+d\psi^2)[/itex]
    Why is this equal to [itex]-r^2dt^2-y^2d\psi^2[/itex]?
     
  5. Sep 2, 2013 #4

    fzero

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    You are making the mistake of working backwards from the answer, when there are terms that get dropped in the limit. You can't obtain these original terms from just the finite part, hence your confusion. Express ##t,\psi## in terms of ##x^\pm## and just write down what

    $$-dt^2r^2-d\psi^2y^2$$

    works out to be.
     
  6. Sep 2, 2013 #5
    Perfect, thanks!
     
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