Wednesday, 29 January 2020

action - 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?






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