- #1
bankcheggit6
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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?
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?