Action for the relativistic point particle

  1. Mar 12, 2010 #1
    I'm interested in the following action for a relativistic point particle of mass m:

    [itex]S = \int d\tau (e^{-1}\dot{x}^2 - em^2)[/itex]

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

    [itex]\tau \to \overline{\tau}(\tau) = \tau + \xi(\tau)[/itex]

    then the einbein apparently changes according to

    [itex]e(\tau) \to e(\tau) + \frac{d}{d\tau}(e(\tau)\xi(\tau))[/itex]

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

    Can anyone explain to me where the extra term [itex]e(\tau)(d/d\tau)\xi(\tau)[/itex] comes from? Is there something particularly special about the einbein that gives rise to this term?
     
  2. jcsd
  3. Mar 12, 2010 #2

    Mentz114

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