# How does the speed of light vary throughout a black hole?

I surmise that there is no equation defining the speed of light, as there is with the speed of sound. Presumably, we assume that, at the event horizon, it falls to zero. What is its value at the centre of the hole?

• We don't have a good understanding of what goes on beyond the event horizon. General relativity tries to explain the curvature of space time due to mass. This mathematical modelling could estimate the behavior of light due to its physical properties. But the key thing is, that space-time is curved so much beyond the event horizon of a black hole that light is unable to escape the gravitational acceleration towards the singularity. So as far as velocity is concerned; there are ways to estimate, but physically, we have no idea what really happens "at the center" of a BH. Jan 1, 2017 at 18:30
• Possible duplicates: physics.stackexchange.com/q/145110/2451 and links therein. Jan 1, 2017 at 19:00

Although interesting and very important questions have been raised about exotic behavior at the event horizon using quantum mechanical arguments, at least classically, i.e., in General Relativity, nothing dramatic happens at the event horizon and nothing catastrophic happens to Physics unless at the center. It is well-established that in a purely classical perspective (which we know is not enough for a comprehensive study of BHs) we can certainly calculate the coordinate speed of light.

In a spacetime with a Schwarzschild BH, the famous Schwarzschild metric is written as

$d\tau^2$ $=$ $\bigg(1-\dfrac{2M}{r}\bigg)dt^2 - \dfrac{1}{\bigg(1-\dfrac{2M} {r}\bigg)}dr^2 - r^2 d\Omega^2$

where symbols have the usual meaning and the non-rotating uncharged Schwarzschild BH is at $r=0$.

Now, we characterize a photon's motion by $d\tau^2=0$ assuming the validity of the Principle of Equivalence everywhere (at least everywhere except at the Singularity at $r=0$). Thus, the equation of motion of a photon becomes

$0$ $=$ $\bigg(1-\dfrac{2M}{r}\bigg)dt^2 - \dfrac{1}{\bigg(1-\dfrac{2M} {r}\bigg)}dr^2 - r^2 d\Omega^2$

Therefore, the coordinate speed (the only meaningful concept of speed for a photon anyway) can be easily found using the above equation of motion. In the special case of radial motion (i.e., $d\Omega=0$), the coordinate speed can be expressed as

$\bigg|\dfrac{dr}{dt}\bigg| = \bigg|1-\dfrac{2M}{r}\bigg|$

This expression is well-behaved an equally valid both inside (except the Singularity) and outside the event horizon.

One peculiarity that I don't know the full resolution to (and thus, actually questions the credibility of my answer. In the case of my answer being beyond repair, feel free to delete it.) is that although mathematically there is no trouble in calculating this coordinate speed inside the event horizon, since the time coordinate is actually the time as displayed on a clock far apart from the BH, it is impossible for the person measuring the radial displacement inside the event horizon to keep track of the $t$ coordinate owing to his inability to send back signals to that far apart clock.

• So speed (dr/dt) is zero at r=2M in plank units. I thought it might be interesting to footnote in SI. I think it is $\bigg|\dfrac{dr}{dt}\bigg| = c\bigg|1-\dfrac{2GM}{r}\bigg|$ so r=2GMc Jan 1, 2017 at 20:32

Well, the best answers to this question have pretty much been covered in the answers to the question at How does light behave within a black hole's event horizon?

The straight answer to the question about the speed of light inside the horizon is simple: the coordinate velocity of light depends on the coordinate system. In the Kruskal-Szekeres coordinate system shown below (from the reference above) the speed is c, or 1 when we use units where c=1. And yes, you can get other answers for the coordinate speed of light, they are just coordinate-dependent and not invariant in any sense. In a local coordinate frame it is always c - locally light always travels at c. As Rennie shows mathematically in his answer, even as a free falling observer gets close to the horizon from outside, the speed of light for him/her is still c.

The description below also tells you what happens to the light inside.

From the reference, the answer by Rennie is instructive. The one by Motl discusses how it looks different in different coordinate frames, with Penrose type coordinates better to understand the causal structure of the spacetime -- which means in essence that you can see the light geodesics explicitely. But they both emphasize the coordinate speed of light. The easiest to understand it, IMO, is @Alfred-Centauri's description and the diagram of the spacetime using Kruskal-Szekeres coordinates, which are a particularly nice set of coordinates to see the light geodesics, and thus the causal structure, of the spacetime, both INSIDE and OUTSIDE the horizon. I copied that diagram below from Alfred-Centauri's answer, where light geodesics are simply lines at 45 degrees, like an a Minkowski spacetime diagram. You can see that for light rays (45 degrees cones Inside the Black Hole) from inside the horizon, they never get closer to the horizon, they actually get closer to r= 0, the singularity, and if somehow they were to start at the horizon they remain there. The wiki article for those coordinates are at https://en.wikipedia.org/wiki/Kruskal%E2%80%93Szekeres_coordinates

And as you can see from the diagram, light (nor anything else) can ever escape the Black Hole from inside, it'll always go towards the singularity. Outside it also always equals c in those coordinates, but keep in mind that the horizon is always at t= infinity in Schwarszchild coordinates, so from the outside, for an observer at spatial infinity which is what the Schwartszchild coordinates represent) light never actually gets to the horizon either, though always traveling at c.