Falling into BH question

skeptic2

Naty1

I raised the question of how an object is able to cross the event horizon because I don't understand how it happens and what seems logical to me is at odds with the widely accepted interpretation of black hole geometries. Certainly I accept that observers in different reference frames don't always agree on observed time,distance,mass,etc. but I also believe that events in one reference frame can be mathematically transformed into any other reference frame to explain what those observers see. To suggest that simply because observers in different reference frames disagree about time, distance or mass, is sufficient reason to accept an ad hoc instance of differing observations without providing some sort of transformation between the reference frames is less than scientific.

I also raised the question hoping that someone here could point out the errors in my logic. The references I've seen, like xantox's posts, don't address the issue of the evaporation of the black hole during the extremely dilated time an object experiences as it falls towards the event horizon. Briefly put, is time infinitely dilated at the event horizon and if so, how does an object cross that event horizon in finite time? If it crosses in finite time in one frame of reference but not in another, what is the transformation between those reference frames that permits that?

Chalnoth

It doesn't. Basically in the case of a static black hole without any Hawking radiation, this means that it is impossible to do a transformation between the observer that has already passed the event horizon and an observer outside the event horizon. This is, in fact, what is meant by an event horizon in the first place: observers on different sides of an event horizon are causally disconnected, and it is therefore no longer possible to translate between their reference frames.

However, I'm beginning to think that with Hawking radiation, an infalling observer won't actually ever observe the interior of the black hole, but will instead just see the black hole evaporate to nothing, until the observer itself exits the black hole as Hawking radiation.

skeptic2

Of course not. I guess I should have been clearer. I meant a transformation between an observer very close to the EH where time is extremely dilated and flat space.

Exactly, and this is the crux of my problem. If nothing can pass through the horizon, how can there be large black holes? (Small black holes may still be possible.) If the weight of all the matter can be supported by the EH, why is there required to be a singularity at the center?

xantox

The infalling body falls only in one place, not two. His path in that one place is finite, as it may be calculated by integrating the proper time on his worldline. So, the infalling body reaches the horizon, and enters the black hole, and fast. However, when this same path is measured by a distant observer, and since he is not in the same curved spacetime, he will use as a result coordinates having a radically different meaning. So he may well obtain an infinite result since his clock is not measuring the same thing that the infalling body calls "time".

Maybe this metaphor can help get a very rough and partial image: if you can only measure your shadow, then when the sun approaches noon exactly above you, the shadow will approach zero length, while at sunset it will become longer and longer, until going ideally to infinity. This does not mean your height is infinite, but just that it has been projected in some way onto different coordinates.

Chalnoth

Well, ultimately, I think that describing what a black hole actually is would require an understanding of quantum gravity. And we know that there isn't going to be a real singularity at the center: that's a feature of General Relativity, and a signature that General Relativity is wrong.

Chronos

GR may not be complete, but, nobody has proven it wrong. That hits my hot button. Show me the math before prophetizing.

Chalnoth

Let me be clear with what I mean. We know that GR must break down at some point, because it provides nonsensical predictions (singularities). But we don't yet know where it breaks down, because so far all experiments and observations are exactly in line with the theory's predictions.

atyy

What exactly is so bad about a singularity, since it doesn't seem to be preventing predictions?

Chalnoth

Beyond the difficulties of having an actual infinity in the theory, General Relativity has a fundamental energy scale (the planck scale). Singularities necessarily go far beyond that energy scale. And the reason why it's a problem for General Relativity is because GR predicts such singularities under very general conditions.

To put it another way, even with this fundamental energy scale sitting in the theory, if there was no reason to believe that the energy density could ever get high enough to contest this energy scale we might well think that the theory could potentially be absolutely correct.

And, of course, there are the incompatibilities with quantum mechanics to consider.

Phrak

I don't know what you all are talking about, but the horizon can't be crossed in finite time in the refererence frame of the exterior universe. So is it crossed, or is the question a nonsequiter?

Are we talking physics here, or UFOs?

Chalnoth

In basic General Relativity with no Hawking radiation, the answer is a definitive yes, because the proper reference frame to consider is not the reference frame of an external observer, but rather the observer falling into the black hole.

I think that the definitive answer to how this works in reality may potentially require an understanding of quantum gravity, which we don't yet have.

xantox

The answer is a definitive yes -also- with Hawking radiation for any macroscopic black hole. For a free falling body starting at rest in the distant flat spacetime, the time it takes on the body's own stopwatch to go from say 3x(horizon radius) to the horizon is around 100 microseconds – rather small compared to the evaporation timescale into a surrounding vacuum of such black hole of around 10^68 years, so that such effect may be ignored.

stevebd1

These lecture notes by Kim Griest might be of interest-
http://physics.ucsd.edu/students/cou...161.8feb07.pdf

In all cases, [itex]r_s=2Gm/c^2[/itex]


'..Suppose we take the case where someone starts from rest at ∞ and falls into the hole.. as viewed from far away the person’s time slows down and then stops as it enters the Schwarzschild radius. The calculation is done starting from radial and time equations: [itex]dr/d\tau=\pm\sqrt{2Gm/r}=\pm\sqrt{r_s/r}[/itex], and [itex]dt/d\tau=\left(1-r_s/r\right)^{-1}[/itex], where we used conserved energy E=m which is valid starting at rest from infinity. Dividing these equation we find the relation between r and t, that is the speed as seen from from away:

[tex]v_{far\ away}=\frac{dr}{dt}=-\left(1-\frac{r_s}{r}\right)\sqrt{\frac{r_s}{r}}[/tex]

We see again that as r→r_s, v→0. The far away observer never sees the person fall in..'


The following is based on [itex]dr_{shell}=dr\left(1-r_s/r\right)^{-1/2}[/itex] and [itex]dt_{shell}=dt\left(1-r_s/r\right)^{1/2}[/itex]


'..We can also find the speed measured in the shell frame:

[tex]v_{shell}=\frac{dr_{shell}}{dt_{shell}}=\left(1-\frac{r_s}{r}\right)^{-1}\frac{dr}{dt}[/tex]

For the person falling in from far away, we put in the result for dr/dt above to find:

[tex]v_{shell}=\frac{dr_{shell}}{dt_{shell}}=-\sqrt{\frac{r_s}{r}}[/tex]

This gives the result that to a shell observer, sitting at r=r_s, the falling objects goes by at v_shell=1, the speed of light! Isn’t it strange that the same object doing the same thing can be moving at 0 speed or c from different vantage points..'

Steve

Chalnoth

Remember that the important metric for determining whether or not the observer sees itself passing the event horizon is the lifetime of the black hole in the infalling observer's frame, as opposed to an outside observer's frame. Because any reference frame is a valid reference frame for computing the results, provided that you're not attempting to talk about behavior beyond an event horizon from your reference frame, and because from the outside observer's point of view an object never falls beyond a black hole's event horizon, it's beginning to look to me that the black hole will always evaporate before the observer passes the event horizon, when considered by an outside observer.

Now, I'm not certain on this. The idea just occurred to me in reading this thread, and I haven't heard any GR experts' take on it. But it seems to make sense to me. Perhaps I'll send an e-mail to my old GR professor...

xantox

I fail to understand how you can consider that while without Hawking radiation the observer of the previous example can cross the final gap to the horizon in 100 microseconds, with Hawking radiation he would need to wait 10^68 more years. Or maybe you think the black hole will evaporate for him in 100 microseconds?

Chalnoth

That's precisely it. Remember that those 100 microseconds to the infalling observer are beyond positive infinity for the outside observer. So yes, I'm suggesting that perhaps the time dilation really is that extreme.

Phrak

So the answer is, "No, it can't be crosses, as the horizon retreats under Hawking radiation in finite time."