I have a stupid question about Eq. (1.3.22) in Polchinski's string theory volume 1. In the equation of motion for an open string, Eq. (1.3.22),

$$X^i (\tau, \sigma) = x^i + \frac{ p^i}{p^+} \tau + i \bigl(2 \alpha'\bigr)^{1/2} \sum_{\substack{n= -\infty,\\n\neq0}}^{\infty} \frac{1}{n} \alpha_n^i \exp\biggl( -\frac{ \pi i n c \tau}{ l}\biggr) \cos \frac{ \pi n \sigma}{l} $$

How do I get the factor $ \frac{ p^i}{p^+} $?

  • $\begingroup$ The coefficient is called this way to agree with (1.3.20both) with the Hamiltopnian in (1.3.19), it's a result of the light cone gauge. $\endgroup$ – Luboš Motl Jul 14 '13 at 4:29
  • $\begingroup$ If I start from Eq. (1.3.20b), $$\partial_{\tau} X^i = \frac{ \delta H}{\delta \Pi^i}= \frac{l}{p^+} \Pi^i $$, still didn't get $$p^i/p^+$$ $\endgroup$ – user26143 Jul 14 '13 at 10:42

From $(1.3.18$), we have :

$$\Pi^i = \frac{p^+}{l} \partial_\tau X^i$$

The definition of the total momentum is $(1.3.23 b)$ :

$$p^i = \int_0^ld\sigma ~\Pi^i(\tau, \sigma)$$

So, by definition :

$$p^i = \frac{p^+}{l} \int_0^ld\sigma ~\partial_\tau X^i(\tau, \sigma))~~~~~~~~~~~~~~~~(1)$$

Now, considering equation $(1.3.22)$, and taking the $\tau$ derivate, we get :

$$ \partial_\tau X^i(\tau, \sigma) = \frac{p^i}{p^+} + \sum a_n (\tau) \cos (\frac{\pi n\sigma}{l})$$

The integral of space-periodic-excitations on the interval $[0,l]$ is zero, so we get : $$\int_0^ld\sigma ~\partial_\tau X^i(\tau, \sigma)) = \frac{lp^i}{p^+}~~~~~~~~~~~~~~(2)$$

Obviously, $(1)$ is the same thing as $(2)$, which explain the factor $\frac{p^i}{p^+}$ in $(1.3.22)$


Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.