# Where does $p^i/p^+$ come from in the EOM of an open string?

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 $\frac{ p^i}{p^+}$?

• 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. – Luboš Motl Jul 14 '13 at 4:29
• 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 $$p^i/p^+$$ – 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 $(1.3.22)$