Mode Expansion of Closed String with Compact Dimensions

[benbenny]

Im working through derivations of string equations of motion from the Nambu-Goto Action and I'm stuck on something that I think must be trivial, just a math step that I can't really see how to work through.
At this point I've derived the equation of motion for the closed string from the wave equation in a general \tau gauge, and now I am trying to do the same but for a string in space with compactified dimensions. So I am trying to implement

X^i (\tau, \sigma) = X^i (\tau , \sigma +2\pi) +2\pi R_i W^i

Where W is wrapping number of ther string around the dimension.

Now, without the compactified dimension it was simple to see the periodicity of the left moving wave and the right moving wave separately, and to expand using Fourier expansion and combine the 2 to get the equation of motion. But now I am stuck on how to use the constraint to again rederive the equation of motion from the general solution to the wave equation. how do you do the mode expansion now that you have this extra term of

2\pi R_i W^i ?

Thanks for any help.

B

 

[torquil]

Maybe you'll find your answer in one of these:

http://arxiv.org/abs/hep-th/9611050
http://arxiv.org/abs/hep-th/0007170

Torquil

 

[benbenny]

Thanks, but I am afraid I wasn't able to find what I'm looking for there. I'm sure that because this is such a trivial step, it is not explicitly referred to in the literature. I've also looked in Zwiebach. Its a trivial step that eludes me however.

It seems from what I have read, that what arises from the implementation of the closed string in compact space constraint, is a constraint on the values of the constants a_0 ^i,
where a_n ^\mu are the coefficients of the oscillatory modes in the equation of motion. ( \mu is regualr space coordinates and i denotes compact space coordinates).
I can't see how this comes about mathematically though.

Is what I'm asking clear?

Thanks again.

 

[xepma]

You need a zero-mode contribution that takes care of this. If I recall correctly from conformal field theory, which has a similar calculation, you need to implement something like

X(\tau,\sigma) = \ldots + R_iW^i \sigma +\ldots

The terms in front are \sigma independent. The latter terms are exponentials in \sigma and do not change.

However, there is also a term \tau times some coefficient. This coefficients also changes, but I don't know how.

Hope this helps a bit.

 

[torquil]

Thanks, but I am afraid I wasn't able to find what I'm looking for there. I'm sure that because this is such a trivial step, it is not explicitly referred to in the literature. I've also looked in Zwiebach. Its a trivial step that eludes me however.

It seems from what I have read, that what arises from the implementation of the closed string in compact space constraint, is a constraint on the values of the constants a_0 ^i,
where a_n ^\mu are the coefficients of the oscillatory modes in the equation of motion. ( \mu is regualr space coordinates and i denotes compact space coordinates).
I can't see how this comes about mathematically though.

Is what I'm asking clear?

Thanks again.

Since a circular dimension can be represented as \mathbb{R} where one identifies all points separated by 2*pi*R. So you can use the mode expansion for the non-compactified case, and just impose the restriction that comes from the space-identifications. Are you not doing this?

The periodicity condition only affect the components of X in the i-directions. In addition, the higher order terms a_n^{\mu} for n>1 won't be accected by the circular dimension, because their accompanying sin() function in the mode expansion will automatically have the same value both at \sigma=0 and at \sigma=2\pi (beware, some only use \sigma\in[0,\pi]).

So in the constraint equation in the i-direction reduces from

X(sigma=0) = X(sigma=2pi) + 2*pi*R*w

to something only involving the a^i_0 because all the other terms disappear automatically. I.e. to the usual formulas for the a0's in terms of the winding number w and the momentum number n.

Torquil

 

[benbenny]

Thank you both for your replies.
Torquil, spelling it out for me helped thanks! I just wasn't looking at it correctly. its amazing how much time one can spend on these small math steps sometimes.

Cheers!