I was just thinking about this the other day. Given Einstein's Relativity etc. assigning the cosmic "speed limit" the value of $c=299792458\:\frac{m}{s}$, we know that the event horizon is the point at which not even light can escape. The acceleration due to the immense gravitational force is so great that light cannot escape. But what occurs to light on the other side of this boundary? In other words, what is the "speed limit" inside of a black hole?

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    $\begingroup$ Being on one side or the other makes no difference. Which side of the event horizon of the various black holes in our universe are you on? $\endgroup$
    – hft
    Feb 25, 2015 at 21:01
  • $\begingroup$ Light-like geodesics extend across the horizon; in-falling observers within the horizon measure the speed of light to be $c$. $\endgroup$ Feb 25, 2015 at 23:55

1 Answer 1


First, I highly suggest reading up on the concept of Locality. The issue is where you're measuring the speed from...

Remember that it isn't so much that light can't escape due to the escape velocity, as it is that space itself is being dragged into the black hole (and anything residing in it), which happens to be falling in at the speed of light where you observe the event horizon to be.

Light trying to escape has to travel along this in-falling space, and once the space itself exceeds light speed (this is permissible) even light traveling in the opposite direction will be moving 'backwards' relative to an observer in 'stationary' space far away.

Light will always move at light speed relative to the space it is traveling in (in a vacuum at least). If that space happens to be moving, a distant observer may 'see' a different speed but could never measure it as this requires a 'local' measurement, and every local measurement will agree with light speed, be it in space or in a black hole.


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