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Speed of light in an inertial frame

  1. Jul 27, 2006 #1
    Does the speed of light for an observer falling into a black-hole remain the same?
  2. jcsd
  3. Jul 27, 2006 #2
    in which frame?
  4. Jul 27, 2006 #3


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    Using local clocks and local rulers, the speed of light stays constant. However, the rate of change of the Schwarzschild 'r' coordinate with respect to the Schwarzschild 't' coordinate for a light beam falling into a black hole does not remain the same.

    I think this has been discussed several times, you might want to look up the old threads.
  5. Aug 1, 2006 #4
    Yeah that was a great analogy you gave, pervect.
  6. Aug 3, 2006 #5
    That is a good question, MeJenn. The answer is yes, the speed of light does remain constant for an observer who is accelerating towards a siingularity.o:) However, there is a paradox to be dealt with.:bugeye: That is that the light never does reach the observer. :eek: Similarly, if another observer outside the Swarschild radius were to send a messaage toward the observer in the black hole, it would never be received.:surprised Furthermore, and slightly less astonishing, if the inner observer were to attempt to radio or signal the outer observer, the signal would never pass the Swarschild radius.:rolleyes: All this is due to the relativistic warping of space-time in the presence of a singularity.:confused:
  7. Aug 3, 2006 #6


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    The light does reach the infalling observer. For an observer free-falling into a Schwarzschild black hole from infinity, light from infinity will be visible to the observer crossing the event horizon and will be redshifted by a factor of 2:1.

    (This was worked out in some other thread, I could dig for the details if you're really interested, but you'd need some familiarity with GR to follow the calculation).
  8. Aug 4, 2006 #7
    Pervect, I don't want to get into an argument with you about this. However I strongly beleive that I said the same in my post that you said in yours before it, only using less-abstract terms.:confused:
    Here's my understanding of the subject; The light entering the black hole would never reach an observer already inside the black hole due to the infinite warping of space-time. Now the light would be red-shifted to obliviion but it still would never reach the observer. An outside observer would never see the light cross the Schwarschild radius.
    I am probably not as familiar with general relativiity as you so I wouldn't benefit from seeing those equations. But I would benefit from an explanation of how I'm wrong and the equations are right.
  9. Aug 4, 2006 #8


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    baryon, that's not what he said, he said the opposite
  10. Aug 4, 2006 #9
    Isn't that just a coordinate singularity? According to the Schwarzschild metric, nothing crosses the horizon in finite coordinate time, but it does cross in finite proper time.
  11. Aug 5, 2006 #10


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    I don't believe that this is correct. In Schwarzschild coordinates, an infalling observer gets stuck on the event horizon. In falling light also gets stuck on the event horizon, but if the light is sufficiently delayed with respect to the observer, it will not catch up with him.

    To see this, transform to Painleve coordinates where the observer reaches the singularity in finite coordinate time. If the infalling light begins travel in such a way that it does not reach the singularity at that time, then there is no way for the observer to have seen that light.

    If you transform this situation to Schwarzschild coordinates then the observer and the light can be made to approach the same point on the event horizon which seems to be a violation of the Painleve calculation which shows that the two never meet. But in the Schwarzschild coordinates the two never actually reach the event horizon so there is no contradiction. Instead, one must remember that one cannot make statements about events at points where there is a coordinate singularity.

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