0
$\begingroup$

In the process of finding the equations of motion for a string from the Polyakov action (say in the conformal gauge), we have to implement some boundary conditions on the target spacetime coordinates $X^M(\xi)$ so that some surface terms vanish.

In particular, we imply that the time parameter $\tau$ runs from $-\infty$ to $+\infty$ and thus we suppose that the string is "static" at these times, i.e. $$\delta X^M(-\infty,\sigma)\equiv\delta X^M(+\infty,\sigma)\equiv 0.\tag{1}$$ This vanishes one of the surface terms that appear, while the other term vanishes if we imply periodic boundary conditions (closed strings) or Neumann/Dirichlet boundary conditions (open strings).

For the closed strings, we usually take the space variable $\sigma$ to run from $0$ to $\pi$ and then take the periodic b.c. $$X^M(\tau,\sigma)\equiv X^M(\tau,\sigma+\pi),\tag{2}$$ $$\partial_{\sigma}X^M(\tau,\sigma)\equiv \partial_{\sigma}X^M(\tau,\sigma+\pi).\tag{3}$$ My question is the following:

  1. Is it really harmless to restrict $\sigma$ so that $\sigma\in[0,\pi]$?

  2. If $\sigma\in[0,\pi]$, shouldn't we have the following periodic b.c. $$X^M(\tau,\sigma)\equiv X^M(\tau,\sigma+\pi/2),\tag{4}$$ $$\partial_{\sigma}X^M(\tau,\sigma)\equiv \partial_{\sigma}X^M(\tau,\sigma+\pi/2),\tag{5}$$ $$\forall \sigma\in[0,\pi/2]?$$

$\endgroup$
  • $\begingroup$ Why do you impose eq. (3) and (5)? They follow from eq. (2) and (4), respectively. $\endgroup$ – Qmechanic Aug 25 '18 at 15:20
  • $\begingroup$ Concerning question 1: Yes, because of world sheet reparametrization invariance. $\endgroup$ – Qmechanic Aug 25 '18 at 15:23
  • $\begingroup$ @Qmechanic Thanks for the answers. The main reason I asked the second question is that I think eq. (4) is the correct one, since we have the $\forall \sigma\in[0,\pi/2]$ condition. That is, I see eq.(2) in almost every textbook but everyone implies that $\sigma\in[0,\pi]$. If that was true, then we would have something like $X^M(\tau,\pi)=X^M(\tau,2\pi)$, but $X^M(\tau,2\pi)$ wouldn't have any meaning since $\sigma\in[0,\pi]$. $\endgroup$ – G K Aug 25 '18 at 16:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.