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Geodesic quation coordinate time

  1. Nov 11, 2014 #1
    Hi guys

    So im having trouble reparameterizing the geodesic equation in terms of coordinate time.
    Normally you have:

    [tex] \frac{d^2 x^{\alpha}}{d \tau^2} + \Gamma_{nm}^{\alpha} \frac{d x^{n}}{d \tau}\frac{d x^{m}}{d \tau} = 0 [/tex]

    Where [itex] \tau [/itex] is the proper time. I class we were told to express the above in terms of coordinate time and so i reasoned that one would use the chain rule:

    [tex] \frac{d }{d \tau} = \frac{d t}{d \tau} \frac{d }{d t} [/tex]

    When i do so i get the following:

    [tex] \frac{d^2 x^{\alpha}}{d t^2} + \Gamma_{nm}^{\alpha} \frac{d x^{n}}{d t}\frac{d x^{m}}{d t} = - \frac{d^2t/d \tau^2}{dt/d\tau} \frac{d x^{\alpha}}{dt} [/tex]

    Which - i guess - has the form that one would expect because t is non-affine.

    However i dont know how to proceed from here. I´ve tried to use [itex] d\tau ^2 = g_{nm}dx^ndx^m [/itex] to find [itex]dt/d \tau [/itex], but i can´t seem to get anything meaningful.

    Thanks for your time!
     
  2. jcsd
  3. Nov 12, 2014 #2

    pervect

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    I did something related to this once upon a time, but it's been a while. Are you looking to find the form of the geodesic equations given that you have ##x^1##, ##x^2##, ##x^3## as functions of t? If so I might try to reconstruct what I did if you haven't already figured it out by the time I get to it., As I recall it was a matter of using the chain rule, plus some algebra involving the geodesic equation for ##x^0## (which is another name for "t") in it's original form as a function of ##\tau##, i.e. the standard geodesic equation for ##x^0(\tau)## or ##t(\tau)##.
     
  4. Nov 14, 2014 #3

    WannabeNewton

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    You cannot simplify it any further but you can express it in terms of meaningful quantities: ##\frac{dt}{d\tau} = u^0 = \gamma## where ##\gamma## is the "time dilation factor" of the particle in the coordinate system that it is moving through. In stationary space-times, that is those with a time-like Killing field, you can also express ##\frac{dt}{d\tau}## in terms of the conserved energy per unit mass ##e##. For example in Schwarzschild space-time we have ##e = (1 - \frac{2M}{r})\frac{dt}{d\tau}##. Furthermore you can use the geodesic equation for ##\alpha = t## to get rid of ##\frac{d^2 t}{d\tau^2}##.
     
    Last edited: Nov 14, 2014
  5. Nov 14, 2014 #4
    What did you mean by meaningful? This is the expression for geodesic parametrized by non-affine parameter and any actual physical manifestation might require information on the physical system. If you mean to eliminate the τ derivatives; then I think it is possible to set α to 0th component.

    In case you have never done so, physical interpretations usually rely on the Newtonian weak field limit, where the τ derivatives are eliminated by the original geodesic equation under that limit (again by setting α to 0 and making suitable assumptions about the Christoffel connection coefficients).
     
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