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Equation for path of light in GR

  1. Oct 24, 2012 #1
    In several places I've seen a derivation of the radius of a black hole by simply determining when the Newtonian escape velocity will reach c. Similar analogies are commonly used to explain light being bent by a gravitational field.

    So will treating light as a particle traveling at speed c in a Newtonian universe actually give correct geodesics or is this just a +/- order of magnitude approximation only useful for explaining the concepts without going into tensor calculus?

  2. jcsd
  3. Oct 24, 2012 #2


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  4. Oct 24, 2012 #3

    But I assume that due to the equivalence of gravitational and linear acceleration, you CAN obtain correct geodesics if you use the equations for apparent bending of light inside a spaceship whose acceleration is the same as that in the local gravitational field.
  5. Oct 24, 2012 #4


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    No. The full theory predicts twice the light bending you'd get from this calculation. To get the path of light, you need to solve the geodesics. Which isn't terribly hard for the case of light, just remember that the Lorentz interval is zero,.
  6. Oct 24, 2012 #5


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    AFAIK, this is for the Schwarzschild spacetime, I think that he is correct for a rocket accelerating in flat spacetime.
  7. Oct 24, 2012 #6
    i was always taught that you cannot say anything about a photon in between the time it is emitted and the time it is absorbed. a photon takes "all possible paths". in GR, light travels at C - where there is no definition of either time or distance, ie, for the photon, there is no distance between the place where it is emitted and the place where it is absorbed. i do not think you can make any kind of statement about the path of a single photon.
  8. Oct 24, 2012 #7
    I meant the example to be only over short distances (so direction of gravitational "force" doesn't change much) in a case where tidal forces are minimal.
    According to the "weak equivalence principle", any experiments performed locally should be indistinguishable from the accelerated spacecraft, right?

    Also: does it actually come to exactly twice the light bending, or is this just an approximation?
  9. Oct 25, 2012 #8


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    People have fallen into the unfortunate habit of saying "photon" when what they really mean is light ray. When we talk about null geodesics in a gravitational field, we mean the propagation of a classical beam of light - nothing quantum mechanical involved.

    Quantum mechanics is appropriate in limited situations. E.g. when the light intensity is very low, a continuous light ray is resolved into individual photons. Experiments that fire lasers at retroreflectors on the surface of the moon receive back only a few photons. But did you notice something about this? even photons travel in a straight line. The photons don't zigzag around "all possible paths", they always come back right from the direction of the retroreflector! That's because in addition to the low intensity you have to ask about the wavelength. "All possible paths" is relevant for interference effects, and this matters only over short distances, the same distance scale as a wavelength.

    For the path of a light ray in the gravitational field of a black hole, you'd need to worry about this only if the wavelength of the light was very long, comparable to the size of the hole.
  10. Oct 25, 2012 #9
    I guess so... and It is exactly twice:


    (but I'm sure that there was a page with a graph... )
  11. Oct 25, 2012 #10
    thanks for the link!
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