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Do balck holes violate the constancy of the speed of light

  1. Feb 1, 2012 #1
    I read that the speed of light is constant, that everyone measures the same speed of light from their own reference frame.

    Yet I have also read that light can't escape from a black hole, because of the gravitational effects. How can you measure it's speed at 'c' and yet at the same time not escape?
  2. jcsd
  3. Feb 1, 2012 #2
    One could say that the space the photon is traveling through is "falling" inward faster than c.
  4. Feb 1, 2012 #3
    I think the explanation uses tilted light cones to hold the light inside the Schwarzschild radius...

    One could say that the space the photon is traveling through is "falling" inward faster than c.

    That would lead me think that maintaining an orbit around the BH might be a problem - that orbits would drift closer, and closer ones would drift faster... but isn't current thinking that orbits around BHs are just fine, outside the photon sphere?
  5. Feb 1, 2012 #4


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    The speed of light is always equal to 'c' , when measured using local clocks and rulers. This is true anywhere, even arbitrarily close to the event horizon of a black hole. It would be true even at the event horizon, except that it's a mistake to think that one can "hover" at the event horizon and construct a frame of reference there. No material object can hover at the event horizon, only light can.

    The difference is that the light that travels at 'c', measured using local clocks and rulers, doesn't actually go anywhere when measured in terms of global coordinates.

    In terms of global coordinates the rate of change of a global space coordinate with respect to a global time coordinate (sometimes called a speed, though that's not really strictly accurate and it's a different concept than the concept of speed measured using local clocks and local rulers which is much more physical) is NOT equal to 'c'.

    So - local measurements of speed using local clocks and rulers will always give you 'c'. Global measurements of "speed" do not always give 'c'.

    The difference can be attributed to the curvature of space-time. It's not possible to draw an accurate constant scale-factor map of space-time on a flat sheet of paper. One either has to introduce local scale factors to the map to relate map distances to real distances (the metric tensor), or draw the map of space-time on a surface that's not flat. The former is more useful mathematically, the later may be easier to visualize when one restricts oneself to 2 dimensional space-time diagrams (1 space + 1 time drawing).
  6. Feb 1, 2012 #5


    Staff: Mentor

    The speed of light being the same for everyone "from their own reference frame" is a local statement; to verify it, you just measure the speed of light that is going past you at your current location. By itself, it tells you nothing about how light moves globally in the spacetime you're in.

    Light being unable to escape from a black hole is a global statement; to verify it, you need to know the global configuration of the spacetime. So it's referring to something more than just the local speed of light; it's referring to how the different local movements of light "add up" to global paths that the light follows.

    This is a good way of looking at it, yes.

    This is another good way of looking at it; a good resource is this paper on the "river model" of black holes:


    This viewpoint can in fact be related to the first one. The amount of inward "tilting" of the light cones in the first viewpoint corresponds to the rate at which space is falling inward in the second viewpoint. At the black hole horizon, the rate of infall of space becomes equal to the speed of light, so outgoing light at the horizon just manages to stay in the same place (it moves outward just as fast as space falls inward). This corresponds to the light cones being tilted inward just enough for the "outgoing" side to be vertical in a spacetime diagram (i.e., the "outgoing" side of the light cone points purely in the "time" direction).

    Yes, "outside the photon sphere". But the photon sphere is at r = 3M, and the horizon (the surface from which light can't escape) is at r = 2M. (I'm using geometric units, where the speed of light c = 1 and Newton's gravitational constant G = 1. M is the mass of the black hole.) So there's a region between the photon sphere and the horizon where no closed orbits around the hole are possible.

    (Actually, technically there are no stable orbits possible for massive objects, like rocket ships, inside r = 6M; orbits for massive objects between r = 3M and r = 6M are unstable, small perturbations will cause the object to either fall into the hole or fly outward and escape. Photons can only orbit the hole exactly at r = 3M.)
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