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Breaking the speed of light?

  1. Nov 22, 2013 #1
    Ok I have a question regarding black holes and light. If a photon is travelling directly towards the centre of a black hole, when it comes within its gravitational pull, theoretically it is being given GPE therefore it must go faster than the speed of light because a black hole has enough energy to overcome a photon travelling in the opposite direction. IF this is true then surly it breaks some of the laws of physics? IF not why not?
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  3. Nov 22, 2013 #2


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    Giving energy to light doesn't change its speed, it changes its frequency/wavelength. Light is blue-shifted as it falls into a gravity well, red-shifted as it climbs out. You don't need a black hole to do this, any gravitating object works. The black hole case is interestingly different because the red shift (increase in wavelength, decrease in frequency) becomes infinite for light that is traveling outwards from the event horizon.

    Note that all this discussion of redshift and blueshift is relative to some observer, just as the speeds of ordinary objects would be. You don't need a black hole or gravitational time dilation to see why this must be: if two observers are traveling towards or away from a light source, they will measure different frequencies and wavelengths because of the Doppler effect.
  4. Nov 22, 2013 #3
    If giving energy to light does not change its speed, how does a photon travelling directly away from a black hole not detectable? While its wavelength is changing it will still be readable no? IF not then why not? and if the wavelength change is infinite surely it becomes either a flat line, not a wave of a solid block because the wave lengths are infinitely close together?
  5. Nov 22, 2013 #4


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    The outgoing light trapped at the horizon, its speed is c relative to any local inertial frame. Speed of at a distance is ill defined in GR; further speed of light in non-inertial coordinates is easily greater than c even in special relativity. A static observer outside the horizon is a non-inertial observer.

    (Note that speed of light can be greater than c in pure SR in Fermi-Normal coordinates (the most physical) for a uniformly accelerating observer).
  6. Nov 22, 2013 #5
    Thank you, this is teaching me a lot, however, it does not explain why the light would not change into either a block or a straight line. Also if the wavelength is being constantly shortened at an infinite rate how does the light go anywhere as the speed of light is not infinite
  7. Nov 22, 2013 #6
    Light emitted 'outward' exactly from the event horizon stays at the event horizon, so any light that actually leaves the event horizon must be less than infinitely red shifted. I guess there might be some oddball situations such as an evaporating black hole with a receding event horizon where a photon trapped at the event horizon might escape, but I am not sure what the redshift would be in that case.
  8. Nov 22, 2013 #7
    Why does the light get trapped in the event horizon, there must be enough energy to do so and therefore enough to break the speed of light
  9. Nov 22, 2013 #8
    In Kruskal-Szekeres coordinates it can be seen that the event horizon is (very loosely speaking) moving outwards at the speed of light, so outgoing light can only keep pace with it but not overtake it.

    In the above image the event horizon of the black hole is the diagonal red line going from bottom left to top right. The outgoing light rays (orange) are parallel to the event horizon, so if emitted at the event horizon, they cannot escape.

    P.S. To provide some context, the above coordinate system is the point of view of an observer moving towards the black hole and to such an observer, the event horizon appears to be coming towards him at the speed of light.
    Last edited: Nov 22, 2013
  10. Nov 22, 2013 #9

    Simon Bridge

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    When talking about relativity, be sure to always include the observer.
    Explanations for how things happen tend to depend on where the observer is and what the have been doing.
    Consider the explanations for the twins paradox for example.
  11. Nov 22, 2013 #10


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    What's breaking the speed of light? Nothing is. Light trapped at the event horizon still travels at ##c## relative to an observer falling into the event horizon at the very instant this observer passes by the event horizon. The event horizon is a null surface! It is not an object in space as you may be imagining in your head.

    What you should do is solve for the radially ingoing/outgoing radiation in the Schwarzschild geometry and sketch the light cones defined by the worldlines of the ingoing/outgoing radiation inside the event horizon in terms of Eddington-Finkelstein coordinates. Even though the local light cone structure is trivial, globally there is a big difference due to the highly non-trivial space-time structure inside of the event horizon. Recall that future-directed time-like curves and null geodesics must pass through the future lobes of light cones from event to event.
    Last edited: Nov 22, 2013
  12. Nov 22, 2013 #11


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    Giving energy to light doesn't change its speed, but light moving in curved spacetime will travel on paths that don't match your intuitions, which are based on flat spacetime. The light at the event horizon of a black hole is moving radially outward at c (relative to local observers): but spacetime is curved enough at the horizon that radially outgoing light stays at the horizon.

    One way of visualizing this is to think of light as traveling along the local light cones, and gravity as "tilting" the light cones inward towards the gravitating body. (WannabeNewton gave similar advice.) Far away from a black hole, spacetime is nearly flat, so the light cones go up and to the left, and up and to the right, at close to 45 degrees (in perfectly flat spacetime, they would be at exactly 45 degrees, viewed on a spacetime diagram). But as you get closer to the hole, the light cones tilt inward: more precisely, the outer sides of the light cones get closer and closer to vertical. At the event horizon, the outer sides of the light cones are exactly vertical, meaning that radially outgoing light stays at the same radius.

    This page has some diagrams that show what I described above:


    There are better diagrams in relativity textbooks, such as MTW, but unfortunately they aren't easily visible online.
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