## A curved lagrangian in terms of a flat one

Let us consider the following gravitational background:

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

The Euler-Lagrange equations can be found to be:

and

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

This is a “T-duality” transformation. For the second equation we see that if then we get

with the flat-space Polyakov lagrangian in terms of the new coordinates :

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

Then, pluging this back into the classical action we get

Finally we can write

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 then we can write:

with

The lagrangian can be writen as

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.

Index Guysaid, on August 29, 2008 at 1:36 pmThis is true when the worldsheet is Lorentzian.

Index Guysaid, on August 31, 2008 at 7:46 pmFor an Euclidean worldsheet the factors in the lagrangian get a relative minus sign and the action gets an overall factor of .