# Action for the relativistic point particle

by bankcheggit6
Tags: action, particle, point, relativistic
 P: 1 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?
 PF Gold P: 4,081 As far as I can make out, that transformation is assumed because it keeps the action invariant under the reparameterization. See here http://www.physicsforums.com/showthread.php?t=127956

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