Consider the Nambu-Goto action \begin{equation} S=\int d\sigma d\tau \sqrt{(\partial_\sigma X^\mu \partial_\tau X_\mu)^2-(\partial_\sigma X^\mu\partial_\sigma X_\mu)(\partial_\tau X^\mu\partial_\tau X_\mu)}. \end{equation} We have that $\partial_\tau X^\mu$ and $\partial_\sigma X^\mu$ are element of the tangent space to the string. Alternatively, $S$ can be written in terms of the induced metric \begin{equation} S=\int d\sigma d\tau \sqrt{-\text{det}{g}}. \end{equation} where \begin{equation} g=\left[\begin{array}{cc} \partial_\tau X^\mu\partial_\tau X_\mu & \partial_\sigma X^\mu \partial_\tau X_\mu \\ \partial_\sigma X^\mu \partial_\tau X_\mu & \partial_\sigma X^\mu\partial_\sigma X_\mu \end{array}\right]. \end{equation} Let us now work in the static gauge. We have seen in class that there is a specific reference frame in which the calculations are easier to perform, that is one constrained by, calling the new reference frame $(\sigma',\tau')$, \begin{align} \partial_{\sigma'} \vec{X}^\mu \partial_{\tau'} \vec{X}_\mu =0 \\ \partial_{\tau'} \vec{X}^\mu\partial_{\tau'} \vec{X}_\mu + \partial_{\sigma'} \vec{X}^\mu\partial_{\sigma'} \vec{X}_\mu=1. \end{align} These conditions can be obtained by requiring that the old equation of motion for the Nambu-Goto action take the form of a wave equation, $\partial_\tau^2 X^\mu-\partial_\sigma^2 X^\mu=0$. The first thing to note is that under the first constraint on the chosen parametrization the induced metric is diagonal, that is \begin{equation} g=\text{diag}(\partial_{\tau'} X^\mu\partial_{\tau'} X_\mu, \partial_{\sigma'} X^\mu\partial_{\sigma'} X_\mu)=\text{diag}(\partial_{\tau'} X^\mu\partial_{\tau'} X_\mu, \partial_{\sigma'} \vec{X}^\mu\partial_{\sigma'} \vec{X}_\mu). \end{equation} where the second equality was obtained by using that the zeroth component of $X$, in the static gauge, is independent of $\sigma$. Furthermore, according to the second condition: \begin{equation} \partial_{\sigma'} \vec{X}^\mu\partial_{\sigma'} \vec{X}_\mu=1-\partial_{\tau'} \vec{X}^\mu\partial_{\tau'} \vec{X}_\mu=-\partial_{\tau'} X^\mu\partial_{\tau'} X_\mu. \end{equation} It follows that in this frame \begin{equation} g=-\partial_{\tau'} X^\mu\partial_{\tau'} X_\mu \text{diag}(-1,1). \end{equation} My first question is, assuming everything I wrote is correct:
- Is it true that, up to rescaling of $\tau$, the new frame corresponds to the zweibein frame of the string? That is, if we further rescaled $\tau$, we would find $g=\text{diag}(-1,1)$, which seems to me the defining property of the orthonormal frame.
My second question is
- What does the action look like in this specific frame? I would expect it to be an action reproducing immediately the wave equation $\partial_\tau^2 X^\mu-\partial_\sigma^2 X^\mu=0$.
- At the same time, consider a frame in which $g=\text{diag}(-1,1)$. Then I would expect the action to be constant! So what would the equations of motion be?
I'm aware these questions most likely come from confusing objects and transformations, but I cannot wrap my head around them.
