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Escape of light perfectly orthogonal to black hole

  1. Dec 28, 2011 #1
    I have a fairly decent understanding of black holes, but have always had one curiosity that I haven't found a distinct answer to:

    If light, through whatever reaction, is emitted inside the event horizon of a black hole such that it is directed in a path exactly orthogonal to the black hole, i.e., 180 degrees relative to the direction of the gravitational force, is it able to escape since the photon must travel at c?

    The reason I specify its direction is because I understand that if there is even the slightest angle, the fact that the escape velocity is greater than c would cause it to spiral back into the singularity.

    Obviously the photon cannot escape (or can it?), I am simply looking for the explanation. The only theory I have been able to formulate that would dictate the photon to return is that at the singularity, spacetime is distorted to such a point that any direction pointing away from the singularity loops back inward.

    Any insight would be greatly appreciated, thanks in advance!
     
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  3. Dec 28, 2011 #2

    tom.stoer

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    The problem kis that there is no "direction of the gravitational field" in GR; there is only a set of geodesics through a certain point. In additioon the event horizion itself is a light-like surface.
     
  4. Dec 28, 2011 #3

    Bill_K

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    For a point inside the black hole, all null geodesics lead inward and strike the singularity. Any photon emitted from such a point is doomed, regardless of the direction it may be headed.

    For a point on the event horizon, the 'outward going' photon will remain stationary on the horizon, others will fall inward.
     
  5. Dec 28, 2011 #4

    tom.stoer

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    The horizon itself can be understood as a light-like surface consistsing of all light rays neither escaping to infinity (region I = outside) nor hitting the singularity (region II = inside) i.e. separating region I from II
     
  6. Dec 28, 2011 #5
    Here is an analogy that perhaps will make it clear:

    Imagine a moving walkway that has a fixed velocity of x, then one can escape this walkway by going faster than x in the opposite direction.

    Now imagine a very special moving walkway which does not have a fixed velocity but instead has a zero velocity at the starting point and a velocity that increases the farther out one goes. At the 'event horizon' of this walkway the velocity is c. Then for light at this point a velocity of c in the opposite direction will only make light stand still on this moving walkway.

    Lookup "Lemaitre observers" to get the idea of a "vari-speed moving walkway" in mathematical terms.
     
  7. Dec 28, 2011 #6
    Okay thank you all, I believe I understand now. My the flaw in my understanding was that I believed light HAD to move at c in any reference frame and therefore I thought it would be impossible for it to remain stationary on the horizon. Thank you again!
     
  8. Dec 28, 2011 #7

    tom.stoer

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    No flaw!

    Light does indeed move at c, even at the horizon. The horizon is no 'physical' reference frame, i.e. no physical observer can be at rest at the horizon. The whole horizon itself is moving at c w.r.t. to any physical observer. That means that a physical observer will always cross the horizon at c, regardless what he/she is doing (free fall, acceleration in a rocket, ...).
     
  9. Dec 28, 2011 #8
    Oh I see! That is much easier to digest than questioning my understanding of reference frames haha.
    Thank you so much
     
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