Action for the relativistic point particleby bankcheggit6 Tags: action, particle, point, relativistic 

#1
Mar1210, 12:41 PM

P: 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 worldline. If we reparametrize the worldline 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 higherorder 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
Mar1210, 10:45 PM

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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