Photons hitting the Black Hole Singularity

Question

I was watching one of Dr. Susskind's Lectures on String theory and he was talking about black hole singularities (see the 40 minute mark of this lecture http://www.youtube.com/watch?v=SkEvsxg5Tu4&list=PL3E633552E58EB230). I'm curious, given that the singularity becomes a 'time' and not a 'place', if at the moment you cross the horizon, you emit a photon, according to the explanation given, does this mean that it would take an infinite amount of time for the photon to hit the singularity? And this would also mean that the faster you move after crossing the horizon, the longer it would take to hit the singularity (which seems kind of cool)? I'm just curious to know if I'm interpreting the explanation correctly.

Answer 1

I'm curious, given that the singularity becomes a 'time' and not a 'place', if at the moment you cross the horizon, you emit a photon, according to the explanation given, does this mean that it would take an infinite amount of time for the photon to hit the singularity?

Time is relative in general relativity, so questions like this do not automatically have well-defined answers.

If you want to know how much time a material object like a rock takes to hit the singularity, the most natural way of defining this is to ask how much time will elapse on a clock attached to the rock. The answer is then a finite number that depends on the size of the black hole. It isn't infinite.

In your example of a photon, we don't have this option. We can't duct-tape a clock to a photon, so it's not possible to define the amount of time experienced by a photon. Therefore the answer to your question is fundamentally undefined. It's not that we don't know the answer -- the question just isn't meaningful. It's like asking whether happiness is blue.

Answer 2

And this would also mean that the faster you move after crossing the horizon, the longer it would take to hit the singularity (which seems kind of cool)? I'm just curious to know if I'm interpreting the explanation correctly.

You have to solve numerical equations

If you begin with a free falling observer, being at rest at $r=R$ ($R>R_s=2MG$), one may use these parametric equations (Ref ; Susskind/Lindesay An introduction to Black Holes, Information, and the String Theory Revolution, World Scientific), with $0 \leq \eta \leq \pi$ :

$r = \frac{R}{2}(1+ \cos \eta) \tag{1}$

$\tau = \frac{R}{2} (\frac{R}{R_s})^{\frac{1}{2}} (\eta + \sin \eta) \tag{2}$

where $\tau$ is the proper time of the observer, and $r$ is the Schwarzschild radial coordinate.

At $\tau = 0$, that is $\eta=0$ the free falling observer is at rest at $r=R$.

The observer hits the singularity $r=0, \eta= \pi$, at : $\tau = \frac{\pi R}{2} (\frac{R}{R_s})^{\frac{1}{2}} \tag{3}$

Physically, more $R$ increases, more the speed at the horizon would be greater. Now, that you are interested in is :

$\Delta \tau(R) = \tau(r=0,R) - \tau(r=R_s,R)$

We have already the first term in $(3)$, but the second term is obtained only numerically, due to the parametric equations $(1),(2)$

Now, you have to check if $\Delta \tau(R)$ is increasing or decreasing with $R$. If this function increases when $R$ increases, this means that observers with greater speed take more time to hit the singularity , if the function decreases when $R$ increases, this means that observers with greater speed take less time to hit the singularity. Intuitively, the second case should be correct.