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Affine parameterization of a light ray

  1. Apr 28, 2015 #1
    Hello,

    Is this parameterization correct? -

    ##r(t) = R = \mbox{const}##
    ##\theta(t) = 0 = \mbox{const}##
    ##z(t) = ct##
    ##t = t##

    This is supposed to be the null geodesic curve in the case of a light ray, emitted at point {##r=R,\theta=0,z=0,t=0##} parallel to the ##z-##axis in flat spacetime. I'm posting because I am unsure if the time ##t## can serve as an affine parameter.

    Are there better parameterization choices here?
     
  2. jcsd
  3. Apr 28, 2015 #2

    PeterDonis

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    It can.

    Not really "better", but another equally good choice in this case would be ##z##; the only difference is the factor of ##c##, and if you use "natural" units in which ##c = 1##, that difference goes away.
     
  4. Apr 28, 2015 #3
    It seemed to me that the "coordinate time" ##t## would appear to be a reasonable parameter, because I understand it to be the reading of a stationary clock; while the "proper time" ##\tau## would be the reading of a clock traveling with the ray, and it would not be convenient to use ##\tau## as a parameter because it would show the same reading at any point of the curve.
     
  5. Apr 28, 2015 #4

    Matterwave

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    There can be no clock which travels along with the light ray. There is no ##\tau## for a light ray, only an interval ##s=0## along the whole light ray. And you are right, this is definitely not a good affine (or really any other kind as well) parameter. Null curves can not be parametrized by their arc length.
     
  6. Apr 29, 2015 #5

    stevendaryl

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    Something that always struck me as mysterious is that you can derive the geodesic equation

    [itex]\frac{dU^\mu}{ds} + \Gamma^\mu_{\nu \lambda} U^\nu U^\lambda = 0[/itex]

    by extremizing proper time: [itex]\tau = \int \sqrt{g_{\mu \nu} U^\mu U^\nu} ds[/itex] (where [itex]U^\mu = \frac{dx^\mu}{ds}[/itex])*

    but the equation works unchanged even for lightlike geodesics. The original integral is identically zero for lightlike geodesics, so extremizing it doesn't give you any equations of motion.

    * Actually, extremizing proper time gives something like:

    [itex]\frac{dU^\mu}{ds} + \Gamma^\mu_{\nu \lambda} U^\nu U^\lambda = U^\mu \frac{d}{ds} log(\mathcal{L})[/itex], where [itex]\mathcal{L} = \frac{d \tau}{ds}[/itex]. By choosing [itex]s[/itex] to be affine, you can make [itex]\mathcal{L} = [/itex] a constant, so the right-hand side vanishes.
     
  7. Apr 29, 2015 #6

    bcrowell

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    We don't consider the set of all geodesics from A to B and then extremize proper time to pick out one geodesic. We consider the set of all curves from A to B and extremize proper time, and that gives us one curve (or in some cases more than one), which is a geodesic. Did you mean to say "The original integral is identically zero for lightlike *curves*?"
     
  8. Apr 29, 2015 #7

    stevendaryl

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    Yes, that's what I meant.
     
  9. Apr 29, 2015 #8

    bcrowell

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    A couple of points that may or may not be relevant:

    - You can always consider a lightlike geodesic from A to B as the limiting case of a timelike geodesic, in the case where, say, the final point is brought closer and closer to B.

    - Suppose you define timelike geodesics by extremization among timelike curves, and similarly for spacelike ones. The lightlike case is qualitatively different, because the dimensionality of the set is smaller. In particular, in 1+1 dimensions there is at most one smooth, lightlike curve from A to B, so extremization isn't even relevant.
     
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