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Poincare invariant action of a point particle

  1. Nov 29, 2005 #1
    I am an MPhys graduate currently reading Joseph Polchinski’s, String Theory, Vol. 1. Unsurprisingly I’m stuck on the first real bit of maths… :p
    I quote from page 10, heh:
    “The simplest Poincaré invariant action that does not depend on the parametrization would be proportional to the proper time along the world line,
    S_pp = -m∫dτ(- δX¬μ/δτ δX_μ/δτ )^1/2 “
    Where X¬μ is a covariant tensor of time dependent equations describing the position of the particle in all space time dimensions and X_μ is the contravariant tensor. (I don't know how to write subscripts or superscripts in this btw).
    Now I understand basic tensor rules, and know that a Poincare invariant action is given by: dS = L dt. My questions are: why does the negative mass come into the equation? ; why does the negative in front of the derivative come into it?; why is the term in brackets square rooted?; and where does the pp subscript come from? Perhaps someone could point me to an article or web page that could help, or if you know the answers and would be so kind, perhaps explain this equation to me.
    Any help will be much appreciated. Thanks.
    Last edited: Nov 29, 2005
  2. jcsd
  3. Nov 29, 2005 #2

    Physics Monkey

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    The negative inside the square root simply comes from the choice of metric. If you choose the metric so that timelike vectors satisfy [tex] g(v,v) < 0 [/tex], then you have to have a negative inside to take the square root and avoid imaginary thingies.

    The m can be seen as arising from two places, first it insures the action is dimensionless. Second, the overall coeffecient is fixed by requiring that the classical limit of your Lagrangian reproduce the usual result, namely [tex] L = \frac{1}{2} m v^2 [/tex].

    Hope this helps.

    Edit: I noticed some other questions so let me answer them also. The 'pp' presumably stands for point particle. The square root comes from the fact that the proper time is [tex]
    d\tau = (-g_{\mu \nu} dx^\mu dx^\nu)^{1/2}.
    Last edited: Nov 29, 2005
  4. Nov 29, 2005 #3
    Thankyou very much for your speedy reply. I'm happy with the negative in the brackets now and the dimensionally necessary m. I'm looking into the other points you mentioned as I write, but I strongly suspect you have given me much needed pointers with the word "metric", the kinetic energy equation and the word "Lagrangian". :) Hopefully my physics brain will warm up once the cobwebs accumulated over long disuse have been brushed away, heh. Thanks again.

    P.S. should this thread be in the homework or in the classical physics section?
  5. Dec 4, 2005 #4
    A question about string theory seems oddly out of place in the classical physics section ;)
  6. Dec 9, 2005 #5


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    It's not about string theory at all and is about classical physics. A more elegant form for that action is the einbein formulation

    [tex] S^{L}\left[x^{\mu},e \right] =-\frac{1}{2}\int_{\tau_{1}}^{\tau_{2}} d\tau \left(\frac{-\dot{x}^{\mu}\dot{x}_{\mu}}{e}-m^{2}e\right) [/tex]

  7. Dec 9, 2005 #6
    thx dan =)

    for those tempted to post more in this thread, the question was resolved about a week ago, but thx all!
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