Index Concordia

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…

Let us consider the following metric in 5-dimensional spacetime:

ds^2 = A(r)\eta_{mn}dx^m dx^n + B(r)dr^2 .

We have coordinates C^M = \left(x_0 , x_1 ,x_2 , x_3 ,r\right) and we will work in a static gauge where a worldsheet will be parametrized by the two spacetime coordinates x_1, x_2. We will set x_3 = 0.

Then taking derivatives of the coordinate vector we have:

\partial_{1} C^M = \left(\partial_{1}x_0, 1, 0, 0, \partial_{1}r\right)   and   \partial_{2} C^M = \left(\partial_{2}x_0, 0, 1, 0, \partial_{2}r\right).

Then, the ingredients that go into the Nambu-Goto lagrangian are:

\left(\partial_{1}C\right)^2 = -A\left[\left(\partial_{1}x_{0}\right)^2 - 1\right]+B\left(\partial_{1}r\right)^2    and   \left(\partial_{2}C\right)^2 = -A\left[\left(\partial_{2}x_{0}\right)^2 - 1\right]+B\left(\partial_{2}r\right)^2 .

The cross term goes as:

\partial_{1}C \cdot \partial_{2}C = -A\partial_{1}x_0 \partial_{2}x_0 + B\partial_{1}r\partial_{2}r.

The determinant of the induced metric will be denoted by:

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

We see that since we have:

\left(\partial_{1}C \cdot \partial_{2}C\right)^2 = A^2 (\partial_1 x_0)^2 (\partial_2 x_0)^2 + B^2 (\partial_1 r)^2 (\partial_2 r)^2 - 2AB \partial_1 x_0 \partial_2 x_0 \partial_1 r \partial_2 r ,

and

\left(\partial_{1}C\right)^{2} \left(\partial_{2}C\right)^{2} = A^{2}\left[\left(\partial_{1}x_{0}\right)^2 - 1\right] \left[\left(\partial_{2}x_{0}\right)^2 - 1\right] + B^{2} (\partial_1 r)^2 (\partial_2 r)^2 - AB\left[\left(\partial_{1}r\right)^{2}\left[\left(\partial_{2}x_{0}\right)^{2}-1\right]+\left(\partial_{2}r\right)^{2}\left[\left(\partial_{1}x_{0}\right)^{2}-1\right]\right].

we get some cancellations in -G. In the end we can write

-G = -A^{2} - 2AB\partial_{1}x_{0} \partial_{2}x_{0} \partial_{1}r \partial_{2}r + A^{2}\left[\left(\partial_{1}x_{0}\right)^{2} + \left(\partial_{2}x_{0}\right)^{2} \right] + AB\left[\left(\partial_{1}r\right)^{2}\left[\left(\partial_{2}x_{0}\right)^{2}-1\right]+\left(\partial_{2}r\right)^{2}\left[\left(\partial_{1}x_{0}\right)^{2}-1\right]\right].

We can write the lagrangian as:

L_{NG} = \sqrt{AJ},

with

J = A\left[\left(\partial_{1}x_{0}\right)^{2} + \left(\partial_{2}x_{0}\right)^{2} - 1 \right] + B \left[\right]^{2}

Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: