Does Entering a Black Hole Speed Up Universal Aging?

[PAllen]

I'm just as confused as the OP. The geometry of the black hole may allow for you to pass through the EH relatively unmolested (at least in the case of supermassive black holes), but the equation describing time dilation as you approach it looks pretty unambiguous itself - it appears that the universe, and the black hole you are trying to fall into, should both functionally evaporate as you attempt to cross the boundary.

What we are asking is, why ISN'T this the case, when the basic equation describing time dilation around a black hole seems so unambiguous? What part of the theoretical framework is preventing it from dilating you into a future where the BH no longer exists?

Your error is: "he basic equation describing time dilation around a black hole seems so unambiguous". This is false. Time dilation is coordinate dependent, NOT unambiguous. Redshift/blue shift measured by a particular observer is unambiguous, but simultaneity is not unambiguous. If you use coordinates that cover the horizon the interior and exterior of a collapsing body, you can model the competing processes and find that collapse and infall occur way before evaporation.

 

[Nugatory]

I'm not sure if anyone has answered the question yet as I don't have the background to even begin to understand some of the concepts presented here. I see a lot of people talking about what you would observe while inside the black hole. What I was trying to get at is not what one would observe but what actually happens to the universe outside of the black hole during the time that would elapse between crossing the event horizon and reaching the singularity.

Not much at all.

Depending on the size of the black hole, you'll fall through the horizon to the central singularity in anywhere between a few tens of microseconds to a few hours. That's not enough time for the universe to do anything interesting while you're falling in. (I've wasted that much time over a few nice cups of coffee, and the universe never does anything interesting while I'm drinking coffee).

Observers outside the event horizon won't be able to see you crossing the event horizon and reaching the singularity, so the phrase that I've bolded above isn't meaningful to those observers and the question makes no sense to them. But that's their problem, not yours.

 

[PeterDonis]

the equation describing time dilation as you approach it

Can you give the specific equation you are referring to? I suspect that you will find that it is an equation for the time dilation of static observers--i.e., observers who are hovering at a fixed radius above the horizon. You can't use that equation to draw conclusions about what happens to infalling observers.

 

[Jesse King]

Can you give the specific equation you are referring to? I suspect that you will find that it is an equation for the time dilation of static observers--i.e., observers who are hovering at a fixed radius above the horizon. You can't use that equation to draw conclusions about what happens to infalling observers.

t²=t²/1-(r_s/r) where r_s is the Schwarschilde radius.

So there you have what amounts to a pretty clear infinite progression as you attempt to reach the horizon. The naive assumption is that you'd be prevented from ever reaching it. This is clearly not what most astrophysicists believe happens under the current set of theories, but I haven't been able to dig deep enough to figure out why as yet. This equation is useful only for a 'static' observer? What would the correct equation for an infalling observer be?

 

[PeterDonis]

t2=t2/1-(rs/r) where rs is the Schwarschilde radius.

This formula, as I suspected, only applies to static observers. It does not apply to infalling observers. That is why you are getting incorrect conclusions from it.

What would the correct equation for an infalling observer be?

For an observer that falls from a radius $r$ that is above the horizon, the proper time for him to reach the horizon is given (at least to a very good approximation) by the formula I gave in post #35: $\tau = 2m \left[ ( r / 2m )^{3/2} - 1\right]$. If you take away the $-1$ inside the brackets, you get the formula for the proper time for the observer to reach the singularity at $r = 0$.

Note that I said "proper time". That is, the formula gives the time elapsed on the infalling observer's clock. That is the only invariant meaning of "time" that there is. But one can also use a coordinate chart, called Painleve coordinates, in which this time is also coordinate time, so you can use the simultaneity convention of these coordinates to assign "times" to spatially separated events. These coordinates cover the horizon and the region inside it, as well as the region outside, so it is a much better choice than Schwarzschild coordinates for analyzing scenarios like the one under discussion. The figure quoted earlier of $10^{68}$ years for a stellar-mass hole to evaporate by Hawking radiation is really given in Painleve coordinates, not Schwarzschild coordinates, so it clearly is much, much larger than the time for an observer to fall through the horizon, which is on the order of ten microseconds for a stellar-mass hole.

 

[Dale]

If you can tell me how long it takes to fall through the horizon from, say, r = 3m, and its not infinity, please do it.

It is not infinity, it is about 3.993 m.

see: http://mathpages.com/rr/s6-04/6-04.htm

 

[PeterDonis]

it is about 3.993 m.

Just a note, this is a more accurate answer than the one I gave in post #35. The formula I used there is for an object that starts from rest at infinity, so at any finite radius, it is falling inward with some nonzero velocity. That is a reasonable first approximation for an object that starts from rest at a finite radius, but the formulas in the mathpages article you linked to give the exact solution for that case. I was just too lazy to look it up, since the key point for this discussion is that the answer is finite.

 

[rootone]

Well that's enough time to listen to a couple of pop songs, so it's not too bad.

 

[Orodruin]

Well that's enough time to listen to a couple of pop songs, so it's not too bad.

I suggest this one: