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Say that under the infinitesimal reparametrization $\sigma \rightarrow \sigma' = \sigma-\xi(\sigma)$, $x^\mu$ transforms as a scalar, i.e. $x'^\mu(\sigma')=x^\mu(\sigma)$. I would like to show the following:

$$x'^\mu(\sigma)-x^\mu(\sigma)=\xi(\sigma) \dot{x}^\mu(\sigma),$$

where the dot refers to differentiation with respect to $\sigma$.

Here is how I would proceed:

\begin{align} x'^\mu (\sigma' = \sigma - \xi) &\approx x'^\mu(\sigma) - \xi \frac{d x'^\mu(\sigma')}{d\sigma'}\Big\lvert_{\sigma'=\sigma} \\ &\overset{!}{=} x^\mu(\sigma) \end{align}

This gives me:

\begin{align} x'^\mu(\sigma) - x^\mu(\sigma) = \xi \frac{d x'^\mu(\sigma')}{d\sigma'}\Big\lvert_{\sigma'=\sigma} \end{align}

I am almost there, however I have trouble to prove that:

$$\frac{d x'^\mu(\sigma')}{d\sigma'}\Big\lvert_{\sigma'=\sigma} = \frac{d x^\mu(\sigma)}{d\sigma}$$

Here is how far I could go:

\begin{align} \frac{d x'^\mu(\sigma')}{d\sigma'}\Big\lvert_{\sigma'=\sigma} &= \left( \frac{d\sigma}{d\sigma'}\frac{d}{d\sigma}x^\mu(\sigma) \right)\Big\lvert_{\sigma'=\sigma} \\ & = \left( \frac{1}{1-\dot{\xi}} \right)\Big\lvert_{\sigma'=\sigma}\cdot \dot{x}^\mu(\sigma), \end{align}

in which I use the fact that $x^\mu$ transforms as a scalar in the first equality. I am left with this annoying factor, which I don't manage to make disappear, unless I assume $\dot{\xi}$ is infinitesimal (but then I wouldn't know why). I could find this derivation in some scripts about string theory, but they never explain that one step that seems crucial for the whole proof to work. Or am I missing something simple?

Thank you very much in advance for your ideas and suggestions.

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  • $\begingroup$ the equation you’re trying to derive is only true to leading order in $\xi$, so neglecting higher order terms in your derivation leads to the correct result $\endgroup$ – Wakabaloola Apr 22 at 14:38
  • $\begingroup$ the point is that if a given equation is invariant under infinitesimal reparameteisations then it is also invariant under finite reparametrisations, since you can compose infinitesimal ones to reach any finite reparametrisation. All this is true for reparameteisations that are continuously connected to the identity; keep in mind that there are also reparametrisations that are not connected to the identity and requiring invariance under these is the requirement of modular invariance. $\endgroup$ – Wakabaloola Apr 22 at 14:42
  • $\begingroup$ @Wakabaloola Thank you for your comment! You mean that in the last line of my post I can expand and drop all the higher order terms? Isn't it a problem anyway that I have $\dot{\xi}$ instead of just $\xi$? $\endgroup$ – Jxx Apr 22 at 16:18
  • $\begingroup$ it is not a problem because dotxi always appears multiplied by xi so it doesn’t contribute to leading order in xi $\endgroup$ – Wakabaloola Apr 22 at 16:21
  • $\begingroup$ @Wakabaloola I see. Just to make sure that I understand you right, do you suggest to do the following? $\xi \frac{dx'^\mu(\sigma')}{d\sigma'} = \xi \frac{1}{1-\dot{\xi}} \dot{x}^\mu(\sigma) \approx \xi (1 + \dot{\xi}) \dot{x}^\mu(\sigma) \approx \xi \dot{x}^\mu(\sigma)$ (by dropping the $\xi \dot{\xi}$ terms in the last equality). It still seems kind of questionable to me. Or did I misunderstand you? $\endgroup$ – Jxx Apr 22 at 16:26

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