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I'm reading Polchinski and am confused about equation (1.3.13), $$\gamma_{\tau\sigma}\partial_\tau X^\mu-\gamma_{\tau\tau}\partial_\sigma X^\mu=0~~~~~\text{at}~~~~~\sigma=0,l.\tag{1.3.13}$$
It says that this comes from $$\partial^\sigma X^\mu(\tau,0)=\partial^\sigma X^\mu(\tau,l)=0\tag{1.3.15}$$ along with the gauge conditions $$\partial_\sigma\gamma_{\sigma\sigma}=0;~~~~~\det\gamma_{ab}=-1,\tag{1.3.8bc}$$ but I really don't see it. How does the metric come into play?
Starting from the boundary condition $$ \partial^{\sigma}X^{\mu}(\tau,0)=0 $$ and lowering the index on the derivative using the metric gives $$ \gamma^{\sigma\tau}\partial_{\tau}X^{\mu}(\tau,0)+\gamma^{\sigma\sigma}\partial_{\sigma}X^{\mu}(\tau,0) =0. $$ Apparently Polchinski wants to express this in terms of the metric $\gamma_{ab}$ with its indices lowered, instead of in terms of $\gamma^{ab}$ (the inverse of $\gamma_{ab}$). We can express $\gamma^{ab}$ in terms of $\gamma_{ab}$ using the formula for the inverse of a $2\times 2$ matrix, where you flip the diagonal corners and put minus signs on the off-diagonal corners and then divide by the determinant. Conveniently, the determinant is $-1$ so dividing by it is not much work. The formula gives $$ \gamma^{\sigma\sigma} = -\gamma_{\tau\tau} \\ \gamma^{\sigma\tau} = \gamma_{\tau\sigma} $$ And so $$ \gamma_{\tau\sigma}\partial_{\tau}X^{\mu}(\tau,0) -\gamma_{\tau\tau}\partial_{\sigma}X^{\mu}(\tau,0)=0. $$ Likewise at $\sigma=l$.
