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.
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asked Oct 14, 2013 at 20:57
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1$\begingroup$ For the second question, it would take the longest(proper time) to hit the singularity if you do nothing. If you try to "run away" from it, it will come sooner. $\endgroup$– MBNCommented Oct 15, 2013 at 10:59
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2 Answers
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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.
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$\begingroup$ Ok, but is it true that if you (not a photon) fall past the horizon with a clock, the greater your speed, more times your clock will tick before hitting the singularity (given a certain size black hole)? $\endgroup$ Commented Oct 14, 2013 at 21:33
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$\begingroup$ Again, note that you're dealing with relativity here. Speed relative to what exactly? $\endgroup$– DanuCommented Oct 14, 2013 at 21:36
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1$\begingroup$ To add to and reinforce the photon part of this answer, since a photon moves at the speed of light and because of special relativity, photons do not "experience" time. This is why it is impossible to define the amount of time in the reference frame of a photon, or at least the way that I think of it. $\endgroup$ Commented Oct 14, 2013 at 21:41
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$\begingroup$ Ahhh..Now I see your point. Let's say two people fall in together at the same instant (t=0 on both clocks) and one is moving twice as fast as the other, I was thinking the faster moving person should have more ticks on their clock relative to the other, but I suppose that doesn't make sense because either person can say they are 'moving faster' than the other. Hahaha, physics is great. $\endgroup$ Commented Oct 14, 2013 at 22:09
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$\begingroup$ You got it backwards--they both see the other's clocks ticking slower than their own due to there being relative velocity between them. $\endgroup$– gregsanCommented Oct 14, 2013 at 22:36
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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.
