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The Action[s] of relativistic particle

  1. Oct 11, 2014 #1

    ChrisVer

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    In case of a relativistic particle, one can try to minimize the length of the worldline of the particle, thus write the action as:
    [itex] S = -m \int_{s_i}^{s_f} ds = - m \int_{\tau_i}^{\tau_f} d \tau ~ \sqrt{\dot{x}^{\mu}(\tau) \dot{x}^{\nu} (\tau) \eta_{\mu \nu}}[/itex]
    Where the minus is to ensure minima and [itex]m[/itex] is the mass, chosen for dimensional reasons [also for the eq. of motion].

    However I heard that this action is problematic, in the case of [itex]m=0[/itex] (massless particles) and also due to the [itex]\sqrt{.}[/itex] it gets problematic at quantization.
    So to overcome those two problems, one can define another action:

    [itex]S= \int d \tau e [ e^{-2} \dot{x}^2 + m^2 ] [/itex]
    With [itex]e[/itex] now an auxilary field, transforming under reparametrizations as a vielbein. The Equation of Motions for the [itex]x[/itex] field for both actions are the same,so they are equivalent. The second however doesn't seem to have the same problems with the square root, neither with the massless case (due to the freedom of fixing [itex]e[/itex]).
    My main question is, however, how can someone build the second action? I mean did people find it by pure luck or are there physical reasons to write it down? eg. for the first action as I mentioned, the idea is to minimize the worldline length.
     
    Last edited: Oct 11, 2014
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  3. Oct 11, 2014 #2

    dextercioby

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    Educated guess would be the right answer: the particle is a constrained system and the square-root (as in the case of the Klein-Gordon equation without squaring) is tempting to make you look for a linear alternative. To my shame (for I love the history of physics a lot), I don't know who coined this einbein formulation.

    http://physics.stackexchange.com/questions/4188/whats-the-point-of-having-an-einbein-in-your-action

    And a little research on the history side, maybe this article (which for me is behind a paywall) is a start:

    http://www.sciencedirect.com/science/article/pii/0370269376901155
     
    Last edited: Oct 11, 2014
  4. Oct 12, 2014 #3

    ChrisVer

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    yes sorry I meant einbein.
     
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