# Variation under infinitesimal reparametrization

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?

• 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 – Wakabaloola Apr 22 at 14:38
• @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 $\xi$? – Jxx Apr 22 at 16:18
• @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? – Jxx Apr 22 at 16:26