# Classical open string in Polchinski -- consistency of Neumann boundary conditions with gauge choice

In Section 1.3 of String Theory, Volume 1, Polchinski derives the open string spectrum from the Polyakov action with Neumann boundary conditions, by first considering the classical open string in lightcone coordinates.

After fixing the gauge via $$X^+ = \tau , \partial_{\sigma}\gamma_{\sigma\sigma} = 0 , \text{ and } \text{det}\gamma_{ab} = -1\tag{1.3.8}$$ (with $$\gamma_{ab}$$ the worldsheet metric), it is shown that $$\gamma_{\tau\sigma} = 0$$ everywhere and that $$\partial_{\sigma}X^i = 0\text{ at } \sigma = 0,l.\tag{1.3.15}$$ This seems to imply the following two points:

• $$\partial_{\sigma}X^{\mu} = 0$$ at $$\sigma = 0,l$$ in $$D \geq 4$$ (since a generic embedding of the worldsheet should locally be a function of any two spatial dimensions $$X^i$$).
• $$\gamma_{\sigma\sigma}$$ and $$\gamma_{\tau\tau}$$ are constant and nonzero on lines of constant $$\sigma$$.

The first point implies that the induced metric $$h_{ab} = \partial_a X^{\mu}\partial_b X_{\mu}$$ satisfies $$h_{\sigma\sigma} = 0$$ at $$\sigma = 0,l$$. But in the previous section, it is shown (from the equations of motion for the worldsheet metric) that $$h_{ab}$$ should be a constant multiple of $$\gamma_{ab}$$ at every point, and so $$h_{\sigma\sigma} \neq 0$$ at the worldsheet boundary, from the second point. What am I missing?

Let's use a trivial generalization of what you wrote which is \begin{align} \partial_\sigma X^\mu = 0 \; \text{at} \; \sigma = 0,l \Rightarrow h_{\sigma b} = 0 \; \text{at} \; \sigma = 0,l. \end{align} I.e. we set the $$a$$ index to $$\sigma$$ but not $$b$$. Now note that the equation showing $$h_{ab} \propto \gamma_{ab}$$ is \begin{align} h_{ab} (-h)^{-1/2} &= \gamma_{ab} (-\gamma)^{-1/2} \quad (1.2.17) \\ h_{ab} (-\gamma)^{1/2} &= \gamma_{ab} (-h)^{1/2}. \end{align} If we now write the determinant as $$$$h = h_{\sigma\sigma} h_{\tau\tau} - h_{\sigma\tau}^2,$$$$ we find that $$h = 0$$ at the boundary. This makes both sides of (1.2.17) consistent without the need for $$\gamma_{ab} = 0$$.
• Ah, so the induced metric vanishes completely at the boundary. And $h_{\tau\tau} = 0$ is exactly the statement that the ends of the string move at the speed of light, so this shouldn't be surprising.