Index Concordia

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.

Yet another candidate

Posted in Relativity by Index Guy on July 9, 2008

Today I found yet another space that looks promising. This one looks even more nicer than the previous cases. To obtain the metric one can start with the product of 3-d Minkowski and d-dimensional Minkowski space. The coordinates can be taken as

$C^{M} = (Z^A, z, Y^m),$

with $A$ a two-dimensional Minkowski index and $m$ a d-dimensional Minkowski index. Let us write the metric as:

$ds^2 = -2dZ_{+}dZ_{-} + dz + \eta_{mn}dY^{m}dY^{n}.$

Now one imposes the condition:

$C^2 = 0,$

and introduces “Poincaré coordinates” that satisfy this constrain. These can be defined as:

$Y^m = \displaystyle\frac{R}{r}y^m,$      $\sqrt{2}Z_{+} = \displaystyle\frac{R^2}{r},$      $\sqrt{2}Z_{-} = \displaystyle \frac{r z^2}{R^2} + \displaystyle\frac{\eta_{mn}y^m y^n}{r}.$

Then writing the metric in terms of these coordinates one obtains:

$ds^2 = \displaystyle\frac{R^2}{r^2}dy^2 + \frac{z^2}{r^2}dr^2 + \frac{2z}{r}drdz + dz^2.$

One can see that when we set $z = R$ we obtain exact d-dimensional anti-de Sitter space. This metric is very atractive. It contains more symmetry than the other cases that I have consider previously. Besides the first term being invariant under Poincaré transformations, one also has an invariance under the rescalling of the $(y^m , r)$ coordinates. I shall explore this space further.

Three roads to anti-de Sitter space

Posted in Relativity by Index Guy on July 8, 2008

I want to summarize the three different routes I am pursuing to reach anti-de Sitter space.

First is the direct route. This means to start with anti-de Sitter space in the first place. Alday and Maldacena followed this path. For future references the metric for adS space in Poincaré coordinates looks like

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

Second is to deform the exact adS metric by introducing some arbitrary function $A(r)$ of the fifth coordinate. Right now my main struggle is what form of the deformation should I use. Here are some favorites:

• $ds^2 = \displaystyle\frac{R^2 - A^2}{r^2}\left(\eta_{mn}dx^m dx^n + dr^2\right)$
• $ds^2 = \left(\displaystyle\frac{R^2 - A^2}{r^2}\right)dr^2 + \displaystyle\frac{R^2}{r^2}\eta_{mn}dx^m dx^n$
• $ds^2 = \displaystyle\frac{R^2}{r^2}\left(1 + A^2\right)dr^2 + \displaystyle\frac{R^2}{r^2}\eta_{mn}dx^m dx^n$

Lastly we can try something completely different. Along this road it was suggested to start with an extra spatial dimensions and impose some constraints. One gets a six-dimensional space. In Poincaré coordinates this looks like:

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

After the so-called “T-dual” transformation, the metric obtains the following form:

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

This latter approach is the most attractive in my humble opinion. It contains adS in the limit when the coordinate $z$ vanishes. We also have an invariance under a scale transformation on the coordinates $(x^m , r)$ or $(y^m, \rho).$ What am I suppose to do with this metric? I want to consider classical strings propagating in any of these particular backgrounds.