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.

A curved lagrangian in terms of a flat one

Posted in Relativity, String Theory by Index Guy on August 22, 2008

Let us consider the following gravitational background:

$ds^2 = A(r)\eta_{ab}dx^a dx^b + B(r)dr^2,$

and the Polyakov lagrangian in conformal gauge with Euclidean Lorentzian signature:

$L_{p} = A \eta_{ab}\partial_{i}x^{a}\partial_{i}x^{b} + B\partial_{i}r\partial_{i}r.$

The Euler-Lagrange equations can be found to be:

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

We look at this and think how can we make these equations simpler. The first equation can be solved with:

$\partial_{i} y^{a} = A\epsilon_{ij}\partial_{j}x^{a}.$

This is a “T-duality” transformation. For the second equation we see that if $A = B^{-1}$ then we get

$\displaystyle\frac{d\left(\ln{B}\right)}{d r} = \displaystyle\frac{\partial_{i}^{2}r}{L_{f}(y^{a}, r)},$

with $L_{f}$ the flat-space Polyakov lagrangian in terms of the new coordinates $y^{a}$:

$L_{f}\left(y^{a},r\right) = \eta_{ab}\partial_{i}y^{a}\partial_{i}y^{b} + \partial_{i}r\partial_{i}r.$

We can integrate this equation to find an expression for $B$ in terms of the solutions of the equations of motion and the flat-space lagrangian:

$B(r) = B(r_{0}) \exp{\left(-2\displaystyle\int_{r_{0}}^{r}d\tilde{r}\displaystyle\frac{\partial^{2}\tilde{r}}{L_{f}(\tilde{y}^{a} ,\tilde{r})}\right)}.$

Then, pluging this back into the classical action we get

$S = \displaystyle\int d^2 u B(r_{0}) L_{f}(y^{a}, r) \exp{\left(-2\displaystyle\int_{r_{0}}^{r}d\tilde{r}\displaystyle\frac{\partial^{2}\tilde{r}}{L_{f}(\tilde{y}^{a}, \tilde{r})}\right)}.$

Finally we can write

$d\tilde{r} = \partial_{j}\tilde{r} du_{j},$

so we can write the expression in the exponential as a sum of integrals over the worldsheet coordinates.

On the other hand, if instead we have $A = B$ then we can write:

$A(r) = A(r_{0}) = \exp{\left(-2\displaystyle\int_{r_{0}}^{r}d\tilde{r}\displaystyle\frac{\partial^{2}\tilde{r}}{K_{f}(\tilde{x}^{a} ,\tilde{r})}\right)},$

with

$K_{f}(x^{a} ,r) = 2\partial_{i}r\partial_{i}r - L_{f}(x^{a} , r).$

The lagrangian can be writen as

$L_{p} = A(r_{0}) L_{f}(x^{a}, r) \exp{\left(-2\displaystyle\int_{r_{0}}^{r}d\tilde{r}\displaystyle\frac{\partial^{2}\tilde{r}}{K_{f}(\tilde{x}^{a} ,\tilde{r})}\right)}.$

We have found an expression for the classical lagragian in some (not-so) arbitrary gravitational background in terms of the coordinates and the flat-space lagrangian. The problem is we do not get any information about the solution for the equations of motion.

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.

Some progress towards something

Posted in Relativity, String Theory by Index Guy on August 4, 2008

On Thursday night I was reading Polyakov’s contribution to the book 50 years of Yang-Mills theory. This is my version of a bed time story. 😉

Polyakov mentions some of his earlier work and how he solved different problems. I actually got tired after a while, and decided to work and solve my own problem. I had been sort of running away from it with feelings of overwhelming difficulty. There is also a chance that reading this post on the (now dead) string coffee table provided some motivation. In any case, here we go…

Posted in AdS/CFT, Relativity, String Theory by Index Guy on July 13, 2008

I have become more comfortable with the “T-dual” transformation and now I am confident on what to consider.

Exact adS can be described by the metric

$ds^2 = \displaystyle\frac{R^2}{r^2} \left(\eta_{mn}dy^m dy^n + dr^2\right).$

After the T-transformation the metric returns to the same form but now $r = R^2 / \rho$,

$ds^2 = \displaystyle\frac{R^2}{\rho^2} \left(\eta_{mn}dx^m dx^n + d\rho^2\right).$

Both of these spaces can be described by embedding a hyper-surface in $R_{2,d}$. They both have scale invariance.

There are two extra-dimensional spaces that one can obtain by embedding a hyper-surface in $R_{3,d}$ and reduce to adS in some limit.

Case I: $C^2 = 0$

For this case anti-de Sitter space is obtained by setting the extra coordinate equal to $R$. The metric in the original Poincaré coordinates looks like:

$ds^2 = \displaystyle\frac{R^2}{r^2} \eta_{mn}dy^m dy^n + \displaystyle\frac{z^2}{r^2}dr^2 + dz^2 + 2\frac{z}{r}dzdr.$

After the T-transformation (leaving the $z$ coordinate intact) one obtains

$ds^2 = \displaystyle\frac{R^2}{\rho^2} \eta_{mn}dx^m dx^n + \displaystyle\frac{z^2}{\rho^2}d\rho^2 + dz^2 - 2\frac{z}{\rho}dz d\rho.$

Case II: $C^2 = -R^2$

Now adS is reached when the extra coordinate vanishes. In Poincaré coordinates,

$\displaystyle\frac{R^2}{r^2} \left(\eta_{mn}dy^m dy^n + dr^2\right) + \displaystyle\frac{z^2}{r^2}dr^2 + dz^2 + 2\frac{z}{r}dzdr.$

While in the T-coordinates we obtain

$\displaystyle\frac{R^2}{r^2} \left(\eta_{mn}dx^m dx^n + dr^2\right) + \displaystyle\frac{z^2}{r^2}dr^2 + dz^2 - 2\frac{z}{r}dzdr.$

Note that both of these cases have scale-invariance in the coordinates $(r, y)$ and $(\rho, x)$. These two cases are atractive because they can be constructed straight from a higher-dimensional reduction. On the other hand one can just modify the adS metric and see what happens. Here are some of the modifications.

• Soft wall: $ds^2 = \displaystyle\frac{R^2}{r^2} \eta_{mn}dy^m dy^n + \displaystyle\frac{R^2}{r^2}\left( 1 + A^2(r)\right)dr^2$
• Hard wall: $ds^2 = \displaystyle\frac{R^2}{r^2} \left(\eta_{mn}dy^m dy^n + dr^2\right)\theta(r - r_{0})$
• Cutoff: $ds^2 = \displaystyle\frac{R^2}{r^2} \eta_{mn}dy^m dy^n + \displaystyle\frac{R^2}{r^2 - a^2}dr^2$

I might not want to keep the scale invariance so obvious for the $y$ coordinates.

A certain background for strings (II)

Posted in AdS/CFT, Relativity, String Theory by Index Guy on June 25, 2008

Today I will approach the problem I started here from a different angle. In the work of Alday and Maldacena, the authors used the following parameterization when calculating the action near a cusp:

$X^{0} = e^{\tau} \sinh{\sigma}$   ,   $X^{1} = e^{\tau}\cosh{\sigma}$   and   $r = e^{\tau} w(\tau).$

In this form the scale and boost transformations are more explicit. But these coordinate come for the case of exact anti-de Sitter space and I want to consider the metric $G_{mn} = Y^{2}\eta_{mn}$ with

$Y^2 = \displaystyle\frac{R^2- A^2}{r^2} ,$

with $A$ being a function of the extra (fifth) coordinate only. This space will not have the same isometries as AdS, so we should not use the same parameterizations mentioned above. By in the UV limit we should obtain AdS space, so the parameterization should not be that different.

Maybe I should start with the 3d case. The metric goes as

$ds^2 = Y^2 \left( - dX_{0}^2 + dX_{1}^2 + dr^2\right).$

We expect the parameterization of the worldsheet to change. But what if we keep it the same and look for a constant $w$ solution? The action will still diverge. [To be started…]