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)},


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.


2 Responses

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  1. Index Guy said, on August 29, 2008 at 1:36 pm

    This is true when the worldsheet is Lorentzian.

  2. Index Guy said, on August 31, 2008 at 7:46 pm

    For an Euclidean worldsheet the factors in the lagrangian get a relative minus sign and the action gets an overall factor of i.

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