# Index Concordia

## Minimal area in arbitrary background (I)

Posted in Gravity, String Theory by Index Guy on August 7, 2008

$ds^2 = A\eta_{mn}dx^m dx^n + Bdr^2 ,$

where the functions $A$ and $B$ are functions of the extra coordinate $r$. The case of anti-de Sitter space corresponds to

$A = B = \displaystyle\frac{R^2}{r^2}.$

Under the “T-duality” transformation

$\partial_{a}y^m = i A \epsilon_{ab}\partial_{b}x^m$   and   $\rho = \displaystyle\frac{R^2}{r}$

the metric in the dual space takes the form

$ds^2 = \tilde{A}\eta_{mn}dy^m dy^n + \tilde{B}d\rho^2 ,$

but now

$\tilde{A} = \left[A(\rho)\right]^{-1}$   and   $\tilde{B} = \displaystyle\frac{R^4}{\rho^4}B(\rho).$

Since we can always bring the metric in this form, we will just consider the initial case and see what can we do with it. Note that it could be the case that the change of variables between $(r,\rho)$ could be of the general form

$\rho = Ar$   which means   $dr^2 = \left(r\displaystyle\frac{\partial A}{\partial r} + A\right)^{-2}d\rho^2 .$

This case could be more complicated… For now we will just stick with the first change of variables introduced.

I will consider a classical string, and I will set one of the (spatial) coordinates equal to zero and work with a static gauge: the other two spatial coordinates describe the worldsheet of the strings. I am interested in a configuration very similar to that on the Alday-Maldacena paper. The one thing I could change is to allow the polygon where the worldsheet end to be time-like instead of light-like. This is achieved by setting

$x_0 = \pm \Upsilon_{1,2} x_{1,2},$

with $\Upsilon_{1,2}$ being a positive real number that is greater than unity. Our metric has signature $(-++...+)$.

In anticipation we will define the following symbols:

$x_0 \equiv X ,$     $X_{j} = \partial_{j}X ,$     $r_{j} = \partial_{j}r ,$     $F_{ij} = r_{i}X_{j}-r_{j}X_{i},$

with the index $j$ taking values of 1 and 2. Our coordinate vector in the static gauge looks like:

$C^M = \left(X, u_1, u_2, 0, r\right).$

The Nambu-Goto action involves the determinant of the induced metric,

$-G = \left(\partial_{1} C \cdot \partial_{2} C\right)^2 - \left(\partial_{1}C\right)^2 \left(\partial_{2}C\right)^2 .$

Evaluating we get:

$\left(\partial_{1} C \cdot \partial_{2} C\right)^2 = -AX_{1}X_{2} + Br_{1}r_{2} ,$     $\left(\partial_{1}C\right)^2 = A\left(1 - X_{1}^2 \right)+Br_{1}^{2}$     and     $\left(\partial_{2}C\right)^2 = A\left(1 - X_{2}^2 \right)+Br_{2}^{2}.$

Then, after plugging all these into the expression for the determinant we get:

$-G = -A\left[ A\left(1 - X_{1}^2 - X_{2}^2 \right) + B\left(r_{1}^2 + r_{2}^2 - F_{12}^2\right) \right].$

We see a factor of $i$ popping out of the action. This tells us that the worldsheet has Euclidean signature.

For convenience we will define

$J = A\left(1 - X_{1}^2 - X_{2}^2 \right) + B\left(r_{1}^2 + r_{2}^2 - F_{12}^2\right) .$

The lagrangian then looks like:

$L_{NG} = i \sqrt{AJ}.$

Let us now look at the equations of motion for the fields $X$ and $r$. The first field is a bit more simpler:

$\displaystyle\frac{\partial L_{NG}}{\partial X} = 0$     which implies     $\partial_{j}\displaystyle\frac{\partial L_{NG}}{\partial X_{j}} = 0.$

This is the expression for a two-dimensional vector that has vanishing divergence. Then it follows that the components of this vector must be of the form:

$V_{1} = \partial_{2}Y \equiv Y_2$     and     $V_{2} = -\partial_{1}Y \equiv -Y_1 ,$

for some other field $Y$ that is a function of both worldsheet coordinates. Note that adding a constant to $Y$ does not affect the equations of motion, so it may correspond to some coordinate with a shift symmetry. Let us look at the conjugate momenta:

$\displaystyle\frac{\partial L_{NG}}{\partial X_1} \equiv Y_2 = K\left(-AX_1 + BF_{12}r_2 \right) ,$     and     $\displaystyle\frac{\partial L_{NG}}{\partial X_2} \equiv -Y_1 = -K\left(AX_2 + BF_{12}r_1 \right) .$

We have denoted

$K = \sqrt{\displaystyle\frac{A}{J}}.$

One can see that the ratio of the conjugate momenta has a nice form:

$\displaystyle\frac{Y_2}{Y_1} = \displaystyle\frac{BF_{12}r_2 - AX_1}{BF_{12}r_1 + AX_2}.$

One can also use this expression to write the ratio of the warp functions in terms of derivatives of the fields:

$\displaystyle\frac{A}{B} = \displaystyle\frac{F_{12}\left(r_2 Y_1 - r_1 Y_2 \right)}{Y_1 X_1 + Y_2 X_2}.$

We see that the metric would blow up when the fields $(X,Y)$ have orthogonal gradients.