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I am reading Tong's notes about string theory, the second chapter, and I encountered this part that I don't know how is derived. We are considering the worldsheet $(\tau,\sigma)$ whose gauge we set to the flat metric $\eta_{\mu\nu}$. Then, Tong introduces lightcone coordinates $$\sigma^\pm = \tau\pm\sigma,\tag{2.9}$$ under which the metric is $$ds^2 = -d\sigma^+d\sigma^-.\tag{p.32}$$ Then, Tong states that any transformation of the form $$\sigma^+\to\tilde{\sigma}^-(\sigma^-),\sigma^-\to\tilde{\sigma}^-(\sigma^-),\tag{2.10}$$ which is equation (2.10), can be undone by a corresponding Weyl transformation. So this is a residual symmetry of the worldsheet metric.

What I have trouble understanding is this statement above eq. (2.12):

... we claimed that we could go to static gauge $X^0=R\tau$ for some dimensionful parameter $R$. It is easy to check that this is simple to do using reparameterizations of the form (2.10).

I guess I don't really understand what the reparametrization invariance has to do with it, or how these transformations would explicitly play out. I'm not sure how to write arbitrary $\tau,\sigma$ in terms of $t$ and the spatial coordinates $x^i$. I guess we would have arbitrary $\tau(t,x^i)$ and $\sigma(t,x^i)$ and then find some transformation $f(\tau+\sigma)$,$g(\tau-\sigma)$ to reparametrize $\sigma^\pm$, but I can't see how this is guaranteed to make $\tau$ proportional to $t$.

What would be these reparametrizations of the form (2.10) to get from an arbitrary $(\sigma,\tau)$ to static gauge?

The question seems very simple and probably has a simple solution I'm not getting.

My attempt to solve: We have the worldsheet metric $\eta_{ab}$ embedded inside a higher-dimensional metric $g_{ab}$, and we set $\eta$ to be flat.

$$\eta_{ab} = \partial_aX^\mu\partial_bX^\nu g_{ab}$$

so with $(\sigma_-,\sigma_+)$ coordinates on the worldsheet metric,

$$\partial_-X^\mu\partial_+X^\nu g_{\mu\nu} = -1/2$$ and $$\partial_-X^\mu\partial_-X^\nu g_{\mu\nu} = \partial_+X^\mu\partial_+X^\nu g_{\mu\nu} = 0.$$

We would like $\sigma_\pm \to \tilde{\sigma}_\pm$ such that $\tau = Rt$, which means $\tilde{\sigma}_+ + \tilde{\sigma}_- \propto X^0$. Here I'm not sure how to proceed.

Is the relevant factor/explanation that both the $\sigma_\pm$ and the $\tilde{\sigma}_\pm$ parametrizations have the same off-diagonal metric, so $\partial\tilde{\sigma}^\pm/\partial_\sigma^\pm = 0$?

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  • $\begingroup$ Edited to add: What would be these reparametrizations of the form (2.10) to get from an arbitrary (σ,τ) to static gauge? $\endgroup$ Commented Jun 28 at 18:48

1 Answer 1

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Let's start from the equations of motion under the conformal gauge (1.30) - the equation (1.31): $$\partial_\alpha\partial^\alpha X^\mu = 0,$$ or $$(\partial_\tau^2 - \partial_\sigma^2)X^\mu=0.$$ For $X^0$, in the lightcone coordinates, a general solution to this (linear) equation can be decoupled as $$ X^0(\tau, \sigma) = f(\sigma^+) + g(\sigma^-),$$ where $f,g$ are arbitrary functions (rather than $f(\sigma^+,\sigma^- )$, etc. Why? I will show this later. See also Eq. 7.33 in "A First Course in String Theory", by Barton Zwiebach).

Now, if we had $$f(\sigma^+)=\frac{R}{2}\sigma^+,g(\sigma^-)=\frac{R}{2}\sigma^-,$$ we would get the desired result $$X^0 = \frac{R}{2}(\sigma^+ +\sigma^-)=R(\sigma^+ +\sigma^-)=R\tau.$$ The residual transformations $\sigma^+ \rightarrow \tilde(\sigma^+)$ (and same for $\sigma^-$) should help us bring the functions $f,g$ to the desired linear form: just declare $\tilde(\sigma)^+=\frac{2}{R}f(\sigma^+)$, and then $X^0(\sigma^+)=f(\sigma^+)=\frac{R}{2}\sigma^+$ is automatically satisfied.

Now, regarding the decoupling: in lightcone coordinates, the d'Alembertian operator becomes $$\partial_\tau^2 - \partial_\sigma^2 = 4\partial_+\partial_-.$$ So the motion eq. for $X^0$ becomes $$ \partial_+\partial_-X^0=0.$$ $$\partial_-f(\sigma^+,\sigma^-)=0 \rightarrow f=f(\sigma^+)$$ and similarly for $\partial_+$. This implies $f(\sigma^+)+g(\sigma^-)$ is the most general solution of the wave equation we have.

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