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Action for the relativistic point particle

by bankcheggit6
Tags: action, particle, point, relativistic
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bankcheggit6
#1
Mar12-10, 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 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?
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Mentz114
#2
Mar12-10, 10:45 PM
PF Gold
P: 4,087
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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