# light entering a black hole's singularity

I already understand that light cannot escape a black hole after passing the event horizon, so please do not explain that to me. What I would like to know is this: a well known fact about light (a photon specifically) is that it travels at the speed of light, and at no other speed, which means that it has no rest mass, as an example, as it does not stop. As a photon aproached a black hole, it would begin to spiral around it as it got ever closer to the singularity at the centre. The closer the photon got to the singularity, the shorter the amount of time it would take to go once around the singularity, as it remained at its constant speed. However, on reaching the perfect centre, it would stop moving completely relative to the blackhole, so would no longer be travelling at the speed of light. Can you explain why this happens or (more likely) where I have gone wrong?

You have misunderstood the nature of a singularity.

The singularity at the centre of a black hole is not a point in space. Instead it is a place where spacetime becomes infinity curved, and it isn't possible to describe what happens there. Well, it's not possible using General Relativity, but we hope some future theory of quantum gravity will explain what happens at the singularity.

If you jumped into a black hole, and timed your fall using your wrist watch, then you would hit the singularity in a finite (and short!) time, but what happens to you at the singularity it isn't possible to say. A light ray can't time its travel time to the singularity, but it would also hit the singularity (still traveling at $c$) and once again what happens afterwards we can't say.

If you're interested in finding out more about this then the phenomenon is called geodesic incompleteness. A geodesic is the path taken by a freely falling observer, like you and the light ray, and we describe it as incomplete because as far as we can tell the path stops abruptly at the centre of the black hole and there is no space or time beyond it.

• well, i guess that is the definition of 'singularity'. Thanks Apr 11, 2014 at 16:11
• yeah, the geodesic is the 'path of least resistance' of an object, that causes gravity Apr 11, 2014 at 16:14

The problem here is that you need to think of it from the point of view of an observer. If we, on Earth, (or anywhere else for that matter) try to watch a photon (which travels along a radial null geodesic) approach a black hole, we will never see it 'enter' the back hole. It takes an infinite amount of time for the photon to reach the black hole.

So indeed it does not 'stop', to an observer.

The other thing is that a Black Hole has a finite size, limited by the event horizion (for a Schwarzschild Black Hole. I believe it is more complicated for a Kerr Black Hole, for example, anyone?)

A (Schwarzschild) Black Hole with the mass of about 1 solar-mass will have a Schwarzschild raduis of about 3km, while one the mass of the Earth would have a radius of 0.9cm, plenty of space for a photon!

Also, I don't think that your model of the photon circling the black hole and falling in is correct?

Saying 'the shorter the amount of time it would take to go once around the singularity' is not true. To an observer, it takes more and more time to circle the black hole as it gets closer!

It imagine you mean to say something like 'it takes less affine parameter' to circle the black hole. These are two different things.

So to recap, the photon will indeed cross the event horizion of the black hole in finite proper time, but an observer will never see it cross the event horizion.

It is possible that the assumption about a photon "it does not stop" may not be totally true. Light certainly interacts with the environment such as in an eye or solar generator. It may not stop in the traditional sense but may be captured both in an eye and in a black hole. For more information on the possible properties of light http://youtu.be/B1DCP4C4MnY and the properties of a black hole http://youtu.be/Y5XzPOrItaI