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I Path of light in curved spacetime with metric

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  1. Feb 16, 2017 #1
    Suppose we have a curved spacetime with metric g, how can we find out the path of light throughout that space?
     
  2. jcsd
  3. Feb 16, 2017 #2

    stevendaryl

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    Light follows geodesics, so the path of a light pulse can be found by solving the geodesic equation:

    [itex]\frac{d^2 x^\mu}{ds^2} + \Gamma^\mu_{\nu \lambda} \frac{dx^\nu}{ds} \frac{dx^\lambda}{ds} = 0[/itex]

    If you haven't studied a little General Relativity, then this equation is going to be pretty opaque. But here are a few pointers that might help understand it a little better:
    1. [itex]s[/itex] is the path parameter, which is just a way of specifying the path. In classical physics, you always use time as the parameter, but in GR, time is one of the coordinates, so it's actually inconvenient to use it as a path parameter.
    2. [itex]x^\mu(s)[/itex] is the 4-D "position" of the light pulse as a function of [itex]s[/itex]. As [itex]s[/itex] increases, so does the time coordinate (often [itex]x^0[/itex]. The index [itex]\mu[/itex] ranges over the 4 dimensions of spacetime.
    3. [itex]\Gamma^\mu_{\nu \lambda}[/itex] is a quantity constructed from derivatives of the metric tensor, [itex]g[/itex]. The indices [itex]\mu, \nu, \lambda[/itex] again range over the 4 dimensions of spacetime, and the meaning of that equation is that the repeated indices, [itex]\lambda[/itex] and [itex]\nu[/itex], are summed over. In the absence of gravity, if you use inertial Cartesian coordinates, then [itex]\Gamma^\mu_{\nu \lambda} = 0[/itex].
    4. In the special case of a pulse of light, the 4-velocity [itex]\frac{dx^\mu}{ds}[/itex] is what's called a "null vector". What this means is that if you take the scalar product---[itex]\frac{dx}{ds} \cdot \frac{dx}{ds} \equiv \sum_{\mu \nu} \frac{dx^\mu}{ds} \frac{dx^\nu}{ds}[/itex]--the result is always zero. This means that out of the 4 components of the "4-velocity", you can solve for one of them in terms of the other 3.
    In the absence of gravity, using inertial Cartesian coordinates, and assuming only one spatial dimension, this all becomes extremely simple: [itex]\Gamma^\mu_{\nu \lambda} = 0[/itex], so the geodesic equation is just the pair of equations:

    [itex]\frac{dx^0}{ds} = 0[/itex]
    [itex]\frac{dx^1}{ds} = 0[/itex]

    which has the extremely simple solution: [itex]x^0 = a + bs[/itex], [itex]x^1 = d + es[/itex], where [itex]a, b, d, e[/itex] are constants of the motion. The additional constraint that it is a null vector implies that [itex](\frac{dx^0}{ds})^2 - (\frac{dx^1}{ds})^2 = 0[/itex] (the minus sign is from the metric tensor). This implies that [itex]e = \pm a[/itex]. So writing [itex]x^1 \equiv x[/itex] and [itex]x^0 \equiv ct[/itex], we can eliminate [itex]s[/itex] in terms of [itex]t[/itex] to get:

    [itex]x = \pm ct + d'[/itex]

    where [itex]d'[/itex] is a constant computed from the original constants [itex]a, b, d[/itex].
     
    Last edited by a moderator: Feb 16, 2017
  4. Feb 16, 2017 #3

    Ibix

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    I did this recently for the Schwarzschild metric (a nice simple case). A few examples are here: https://www.physicsforums.com/threads/null-geodesics-in-schwarzschild-spacetime.895174/

    Note that the pictures don't really represent the paths. Spacetime is curved and can't be drawn on a plane without distortion. So these are like the paths of airlines drawn on a Mercator projection map: sensible representations but not wholly accurate.

    Edit: @m4r35n357 has written some simulations as well.
     
  5. Feb 16, 2017 #4
    Thanks for reminding me ;) They are here for what it is worth.

    The quality is not great, I never got round to sorting out the issue with the video encoding (which worked fine for my twin paradox videos). So anyway you get to see the program running and displaying to visual python with me panning a zooming like an idiot.

    The results do fit with scientific papers on the subject, though, and the spherical light orbits in particular still fascinate me.
     
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