Why does reparameterisation invariance lead to gauge-fixing?

In Becker, Becker and Schwarz, the point particle action is given in terms of an auxiliary field $e(\tau)$ as: \begin{align} \tilde{S}_0 = \frac{1}{2}\int \,d\tau \left(e^{-1}\dot{X}^2 - m^2e\right) \end{align}

It is then shown that under infinitesimal reparametrizations of $\tau$, the action is unchanged. This allows us to pick a gauge, in particular $e(\tau) = 1$.

I'm not sure I'm understanding this right, but I have a few issues with this.

1. Doesn't this assume that $e$ takes the value $1$ somewhere?

2. Although $\tilde{S}_0$ (sorry, not $e(\tau)$) may be reparametrization invariant, I don't see how you could pick a reparametrization that can leave $e$ constant. Such a reparametrization $\tau'(\tau)$ would need to map all $\tau$ to a constant, but then $\frac{d\tau'}{d\tau} = 0$, which can't be the case.

3. Invariance is only under infinitesimal transformations. This is related to 2: how do we know that an infinitesimal reparametrisation could make $e$ constant?

• Is $e$ a gauge field? Or just some field transforming non-trivially under gauge? (If the latter, this is not gauge-fixing) Which gauge are we talking about? – ACuriousMind Aug 13 '14 at 18:19
• Sorry, $e$ transforms non-trivially, but the action is invariant under infinitesimal reparameterisations of $\tau$. Where I'm confused is why this allows us to pick $e = 1$. – user2520385 Aug 14 '14 at 5:21
• A while ago I found myself on a situation where a teacher recommended to use the concept of "rep. invariance" for deriving the eq. of motion of a rel. particle under a E-M field. I got the result but never quite understood why I could choose the gauge. Investigating a bit I found some information on a pdf file (that I can't seem to find today) and on the book Methods of Quantization by Latal & Schweiger, p.58, sec 2.1 (you can find it on Google Books). Maybe you can get something out of it, I couldn't get much. – Drarp Oct 14 '14 at 0:22

1. It seems that the resolution to OP's question essentially is that the einbein $e$ is not invariant under world-line reparametrizations $$\tag{1} \tau\longrightarrow \tau^{\prime}=f(\tau),$$ but transforms as a world-line co-vector/one-form, $$\tag{2} e~\mathrm{d}\tau~=~e^{\prime}~ \mathrm{d}\tau^{\prime},$$ cf. e.g. this Phys.SE post. In particular, if we are given an arbitrary einbein $e$ with antiderivative $f$, i.e. $$\tag{3} \frac{\mathrm{d}f}{\mathrm{d}\tau}~=~e,$$ then the world-line reparametrization (1) leads to a new einbein $$\tag{4} e^{\prime}~=~ 1$$ that is equal to one everywhere. [Since we work in units where speed of light in vacuum $c=1$ is one; where the parameter $\tau$ has dimension of time; and the vielbein $e$ has dimension of inverse mass, then eqs. (2) and (4) would literally imply that the new parameter $\tau^{\prime}$ has dimension of time over mass, cf. this Phys.SE post. A less radical approach would probably be to work in a gauge where $e^{\prime}$ is a constant with dimension of inverse mass.]
2. Let us add for completeness that the einbein $e$ is invariant (a scalar) under target-space reparametrizations, although this is not relevant for the issue at hand.
• Sorry, I mis-wrote. I know that $e$ is not invariant, but that the total action $\tilde{S}_0$ is invariant. The part I don't understand is how $e$ can be chosen to be equal to 1 because of this. – user2520385 Aug 14 '14 at 5:23