# Action for the relativistic point particle

1. Mar 12, 2010

### bankcheggit6

I'm interested in the following action for a relativistic point particle of mass m:

$S = \int d\tau (e^{-1}\dot{x}^2 - em^2)$

where $e = e(\tau)$ is an einbein along the particle's world-line. If we reparametrize the world-line according to

$\tau \to \overline{\tau}(\tau) = \tau + \xi(\tau)$

then the einbein apparently changes according to

$e(\tau) \to e(\tau) + \frac{d}{d\tau}(e(\tau)\xi(\tau))$

However, I can't seem to understand where the term $e(\tau)(d/d\tau)\xi(\tau)$ comes from in this. A Taylor expansion of $e(\tau + \xi(\tau))$ would seem to give me only $e(\tau) + \xi(\tau)(d/d\tau)e(\tau)$ plus higher-order terms.

Can anyone explain to me where the extra term $e(\tau)(d/d\tau)\xi(\tau)$ comes from? Is there something particularly special about the einbein that gives rise to this term?

2. Mar 12, 2010

### Mentz114

As far as I can make out, that transformation is assumed because it keeps the action invariant under the reparameterization.