Index Concordia

Solving equations of motions in some gravity background

Posted in AdS/CFT, String Theory by Index Guy on August 31, 2008

I would like to consider the gravity background:

ds^2 = A dx^2 + B dr^2 ,

with the case that A = B^{-1}. We saw previously that the equations of motion were given by:

\partial_{i}\left(A\partial_{i}x^a\right) = 0     and     -2\partial_{i}^{2}r = \displaystyle\frac{d \left(\ln{B}\right)}{d r}\left(-\eta_{ab}\partial_{i}y^{a}\partial_{i}y^{b} +\partial_{i}r\partial_{i}r\right).

Now we take the warp factor to have the form B = r^{\beta}. Then we have

-2 r \partial_{i}^{2}r = \beta\displaystyle\left(-\eta_{ab}\partial_{i}y^{a}\partial_{i}y^{b} +\partial_{i}r\partial_{i}r\right).

We will introduce a new symbol,

\eta_{ab}\partial_{i}y^{a}\partial_{i}y^{b} = -W^{2}     with \displaystyle W(u_{1}, u_{2}) a function of the worldsheet coordinates.

The differential equation now looks like:

2 r \partial_{i}^{2}r + \beta \partial_{i}r\partial_{i}r = \beta W^{2}.

We now assume that r can be factored into functions of each coordinate, r = T(u_{1}) S(u_{2}). Then we can solve this PDE when

k^2 = \displaystyle\frac{\beta W^2}{2 T^{2} S^{2}}     with k a constant.

Separation of variables gives the following differential equation:

\displaystyle\frac{d^2 Y}{du_{i}^2} + \alpha \left(\displaystyle\frac{d Y}{d u_{i}}\right)^{2} = k_{i}^{2},

with

\alpha = 1 + \displaystyle\frac{\beta}{2}     and     T = \exp{Y}.

Inserting this into Mathematica gives:

Y(u) = C_{1} + \displaystyle\frac{1}{\alpha}\ln{\left(\cosh{\left(k_{i}\sqrt{\alpha}\left[u_{i} + C_{2}\right]\right)}\right)}.     which means     T(u) = C_{1} \left(\cosh{\left(k_{i}\sqrt{\alpha}\left[u_{i} + C_{2}\right]\right)} \right)^{1/\alpha}.

Notice that the case \beta = -2 is interesting: for the constraints we have used the solution is an exponential function of a quadratic polynomial.

2 comments

2 Responses

Subscribe to comments with RSS.

  1. Index Guy said, on September 2, 2008 at 6:35 pm

    This should be more correct if one takes W to be a constant and keeps all the r dependence explicit. In that case the PDE looks interesting, but hard to solve.

    Reply
  2. Index Guy said, on September 10, 2008 at 7:54 pm

    This is all non-sense.

    Reply

Leave a Reply