Does Entering a Black Hole Speed Up Universal Aging?

[jimmycricket]

Would the universe end if you entered a black hole? What I mean by this is that due to time dilation would time elapse so fast for the universe outside the black hole relative to you inside it that all the stars would burn out and all that wouild be left would be other black holes?

 

[phinds]

Would the universe end if you entered a black hole? What I mean by this is that due to time dilation would time elapse so fast for the universe outside the black hole relative to you inside it that all the stars would burn out and all that wouild be left would be other black holes?

I do understand you question, and am not sure of the answer, but if you entered a BH, the universe would end for YOU, that's for sure

 

[jimmycricket]

I do understand you question, and am not sure of the answer, but if you entered a BH, the universe would end for YOU, that's for sure

Haha ok so assuming I can survive entering the black hole, is it possible that time would elapse so slowly there relative to the universe outside the black hole (apart from other similarly massive black holes) that the universe would essentially "run its course" during the time you spend in the black hole.

Edit: Oops misread your post I thought you said you didnt understand the question

 

[phinds]

I remember seeing other threads covering this so you might try a forum search.

The problems is not surviving since for a supermassive BH, there is nothing to survive out near the EH. Spaghetification doesn't occur w/ a SMBH until much closer to the singularity. You would not even notice crossing the EH. The problem is that you can't hover once inside, you HAVE to proceed to the singularity whether you like it or not. This does not, however, really answer your question, so keep looking.

 

[Chalnoth]

Haha ok so assuming I can survive entering the black hole, is it possible that time would elapse so slowly there relative to the universe outside the black hole (apart from other similarly massive black holes) that the universe would essentially "run its course" during the time you spend in the black hole.

Edit: Oops misread your post I thought you said you didnt understand the question

Unfortunately, once you've entered the black hole, you are drawn towards the center in a short, finite time (from your perspective). Even if you were (somehow) able to survive the trip towards the center, that trip wouldn't take enough time for you to observe much that goes on outside the black hole.

 

[timmdeeg]

Would the universe end if you entered a black hole? What I mean by this is that due to time dilation would time elapse so fast for the universe outside the black hole relative to you inside it that all the stars would burn out and all that wouild be left would be other black holes?

You would see the things far away extremely blue shifted due to gravitational time dilation (stars burning out) in case you would be hovering outside the black hole close to its event horizon. But it makes a big difference if you enter the black hole in free fall. Then (regardless if you are still outside or already inside the black hole), falling away from the stars outside causes redshift which is more or less canceled by the gravitational blueshift. I'm not sure about the net effect however.

 

[Chronos]

The external universe viewed by an infalling observer inside the event horizon is redshifted by a factor approaching 2. It is not blueshifted because you are falling away from the external universe, and you can only view photons approaching you from behind. There is no trajectory that permits photons from other directions to reach your eyes.

 

[rootone]

OK, so our indestructable observer has just fallen through what is the horizon as far as external observers are concerned.
He can observe only red shifted light coming from behind him, and that,s all.
If he could make sense of what he see behind him what does he see? - everything vastly sped up, though still looking very red.?
Galaxies and solar systems forming then dissipating in seconds?
Is there any idea of how much time passes from his own point of view until arrival at the dreaded singularity?

 

[Orodruin]

OK, so our indestructable observer has just fallen through what is the horizon as far as external observers are concerned.
He can observe red shifted light behind him, and that's all.
Is there any idea of how much time passes from his own point of view until arrival at the dreaded singularity?

This would depend on the initial conditions. There is a continuum of different coordinate velocities you can pass the horizon with. And there are of course also different black hole sizes.

 

[Nugatory]

Is there any idea of how much time passes from his own point of view until arrival at the dreaded singularity?

The proper time he experiences from horizon to singularity, assuming free fall, is something on the order of 15 microseconds for a sun-sized black hole.

It scales linearly with the mass of the black hole.

 

[PAllen]

To answer the unanswered question, red shift correlates with observed clock rates. This must always be true, because a light emission process is effectively an oscillator clock. So, for the initial condition Chronos described, you would visually see the external universe going at half speed as you approached the singularity.

 

[timmdeeg]

The external universe viewed by an infalling observer inside the event horizon is redshifted by a factor approaching 2. It is not blueshifted because you are falling away from the external universe, and you can only view photons approaching you from behind.

If just falling away was to be considered one should expect that the infalling observer would see the stars outside infinitely red shifted while crossing the horizon at the speed of light. On the other side the light of the stars gains energy entering the gravity well of the black hole. But I'm not sure if this reasoning is correct. Anyhow, how do you calculate the factor 2?

 

[Chronos]

You need to chose the proper coordinate system to calculate the shift in frequency between emitted vs received photons. In Schwarzschild coordinates the infalling observer views the external universe at z=2 at the event horizon. Unfortunately, you cannot validly extend Schwarzschild coordinates beyond the event horizon. You need something like Panleve coordinates to do that - re: http://arxiv.org/abs/gr-qc/0001069. I believe this is what Hamilton used in his presentation at http://casa.colorado.edu/~ajsh/schw.shtml

 

[George Jones]

An earlier post of mine:

Actually, if the astronaut started falling from a great distance, the astronaut will see light red-shifted, i.e,, the distant outside universe will appear to run slow for the astronaut.

Suppose that observer A hovers at a great distance from a black hole, and that observer B hovers very close to the event horizon. The light that B receives from A is tremendously blueshifted. Now suppose that observer C falls freely from a great distance. C whizzes by B with great speed, and, just past B, light sent from B to C is tremendously Doppler redshifted. What about light from A to C? The gravitation blueshift from A to B is less that the Doppler redshift from B to C. As C crosses the event horizon, C sees light from distant stars redshifted, not blueshifted.

If observer A, who hovers at great distance from the black hole, radially emits light of wavelength \lambda, then observer C, who falls from rest freely and radially from A, receives light that has wavelength

\lambda' = \lambda \left( 1+\sqrt{\frac{2M}{R}}\right).
The event horizon is at R = 2M, and the formula is valid for all R, i.e., for 0 < R < \infty. In particular, it is valid outside, at, and inside the event horizon.

See posts 5 and 7 in

https://www.physicsforums.com/showthread.php?p=861282#post861282

I have since done the calculations using Painleve-Gullstrand coordinates that are vaild even on the event horizon.

 

[timmdeeg]

An earlier post of mine:
... The gravitation blueshift from A to B is less that the Doppler redshift from B to C. As C crosses the event horizon, C sees light from distant stars redshifted, not blueshifted.

If observer A, who hovers at great distance from the black hole, radially emits light of wavelength λ \lambda , then observer C, who falls from rest freely and radially from A, receives light that has wavelength

λ ′ =λ(1+2MR − − − − √ ).
\lambda' = \lambda \left( 1+\sqrt{\frac{2M}{R}}\right).

Great, that's what I have been looking for, thanks George Jones and also Chronos.
So, the factor 2 is clarified.

Unfortunately my intuition is still a bit behind.
A sees B (hovering) extremely redshifted and C (in free fall) even more redshifted, because in the latter case the redshift consists of a gravitational and a kinematic component additional.
Passing by, C sees B extremely redshifted (purely Doppler) but sees A at z = 2 according to the formula you have shown. Would it be correct to say that also this redshift (z = 2) consists of a kinematic (not purely Doppler, because Doppler depends on a well defined relative velocity which in curved space-time is possible only locally) and a gravitational shift. So, it seems that the kinematic effect prevails over the gravitational effect. But why?
Kindly correct my reasoning.

 

[PAllen]

Timmdeeg, I'm not sure there is a satisfying 'why', especially for what is observed by a freefaller past the horizon. Note that gravitational time dilation is only well defined for stationary spacetimes, and SR Doppler is well defined only over a spacetime region that may be considered flat (either due to small size or distance from any massive bodies). You are asking to extend these concepts beyond their applicability, especially to the interior case (the spacetime is not static or stationary, and curvature becomes unbounded as the singularity is approached).

It won't help your intuition necessarily, but there a couple of universal approaches to red shift that don't involve gravitational redshift at all (which is strictly a non-essential, derived concept in GR, that is useful only for the special case of stationary spacetimes).

1) Establish the 4-momentum of the emitted light pulse (it will be a null vector, whose time component will be the energy if the vector is expressed in a local frame of the emitter). Parallel transport this 4-momentum along the null geodesic path (must specify path, because, of course, parallel transport is path dependent) to the receiver. Compute the dot product with the receiver world line's tangent vector (4-velocity) at the event of reception. The ratio of the energy specified in emitter local frame and this dot product will be the red/blue shift.

2) Equivalently (not obvious at all), you can parallel transport the emitter 4 velocity along the null geodesic to the reception event, express the transported 4-velocity in the local frame of the receiver world line at that event, and apply the pure SR Doppler formua (as if the emitter were local, with transported 4-velocity).

Both of these methods are universally applicable in any GR situation, no matter how complex. They explain why there is really only one fundamental phenomenon, not two. So called pure gravitational redshift for static observers in a static spacetime falls right out of either of these computational methods. They also cover any combination of non-static observers, in any spacetime, as noted.

 

[bcrowell]

The external universe viewed by an infalling observer inside the event horizon is redshifted by a factor approaching 2. It is not blueshifted because you are falling away from the external universe, and you can only view photons approaching you from behind. There is no trajectory that permits photons from other directions to reach your eyes.

This doesn't sound right to me. There can't be a fixed, finite Doppler shift for a free-falling observer observing an external signal, because two different free-falling observers could have different states of motion.

 

[bcrowell]

Would the universe end if you entered a black hole? What I mean by this is that due to time dilation would time elapse so fast for the universe outside the black hole relative to you inside it that all the stars would burn out and all that wouild be left would be other black holes?

I think this is one of those questions that is very easy if you understand how to use Penrose diagrams, very hard otherwise. The question here is basically whether or not an event inside the event horizon has timelike infinity i+ in its past light cone. The answer is no if you look at the Penrose diagram.

By the way, it's a common misconception that the fate of all matter in the universe is that it will ultimately end up in black holes. Not true. In the distant future, due to the acceleration of cosmological expansion, there will be many cases where you have a single subatomic particle that is all alone inside its own cosmological horizon.

 

[PAllen]

This doesn't sound right to me. There can't be a fixed, finite Doppler shift for a free-falling observer observing an external signal, because two different free-falling observers could have different states of motion.

I assumed what was meant was the classic free faller from infinity. This is unique - the limit of free falling at zero speed from a static observer, as the static observer approaches infinite distance from the BH.

 

[timmdeeg]

1) Establish the 4-momentum of the emitted light pulse (it will be a null vector, whose time component will be the energy if the vector is expressed in a local frame of the emitter). Parallel transport this 4-momentum along the null geodesic path (must specify path, because, of course, parallel transport is path dependent) to the receiver. Compute the dot product with the receiver world line's tangent vector (4-velocity) at the event of reception. The ratio of the energy specified in emitter local frame and this dot product will be the red/blue shift.

2) Equivalently (not obvious at all), you can parallel transport the emitter 4 velocity along the null geodesic to the reception event, express the transported 4-velocity in the local frame of the receiver world line at that event, and apply the pure SR Doppler formua (as if the emitter were local, with transported 4-velocity).

Both of these methods are universally applicable in any GR situation, no matter how complex. They explain why there is really only one fundamental phenomenon, not two. So called pure gravitational redshift for static observers in a static spacetime falls right out of either of these computational methods. They also cover any combination of non-static observers, in any spacetime, as noted.

Thanks, PAllen, for your explanation. Yes, I think it is important to know that there is one phenomenon only, the observed frequency-shift, which is invariant.

Frankly speaking, I was inspired by authors like Peacok and Chodorowski,

http://arxiv.org/PS_cache/arxiv/pdf/0809/0809.4573v1.pdf
http://arxiv.org/PS_cache/arxiv/pdf/0911/0911.3536v3.pdf

who think of the cosmological redshift as being combined of kinematic and gravitational shifts. In the expanding FRW-model co-moving galaxies fall away from each other. Since the same is true in the Schwarzschild case regarding objects in radial free fall, I started reasoning why not thinking of said decomposition as well. Just to get a better feeling or intuition of the situation that the observer at R = 2M sees the far away exterior redshifted at z = 2 only, though moving locally with c.
But, of course, it is questionable whether the decomposition of the redshift makes any sense, because there is no invariant definition of a gravitational and a kinematic shift, which in combination yield the measured redshift.

 

[Chronos]

Two papers that may be of interest relative to this subject: http://arxiv.org/abs/1201.4250, Exchange of signals around the event horizon in Schwarzschild space-time, http://link.springer.com/article/10.1007/s10714-012-1495-4/fulltext.html, On the nature of cosmological redshift and spectral shift in Schwarzschild-like and other spacetimes.

 

[timmdeeg]

Oh yes, very interesting papers, thanks.

 

[PAllen]

Two papers that may be of interest relative to this subject: http://arxiv.org/abs/1201.4250, Exchange of signals around the event horizon in Schwarzschild space-time, http://link.springer.com/article/10.1007/s10714-012-1495-4/fulltext.html, On the nature of cosmological redshift and spectral shift in Schwarzschild-like and other spacetimes.

For the second of these papers, the author tries to distinguish a classic result of Synge from a newer equivalent derivation by Narlikar, and seems to argue that Narlikar's result is not universally true. I find this to be a mis-representation.

The problem starts from the summary of Synge's result (I have his book; it is a favorite of mine). The author describes Synge's result as if there were two independent results of the parallel transport, a relative speed and recession speed. The authors do concede that both are derived locally (at receiver) from parallel transport of the emitter 4-velocity, so only one transported vector is involved. However, the key point is that after accounting for Synge's quite unusual notations, his resulting formula is nothing but the standard SR Doppler formula for arbitrary relative motion, whence you have gamma(total relative speed) expressing the time dilation contribution, and relative speed along the light path expressing the classical Doppler aspect. The key, that I just worked through myself, is that after adjusting for Synge's notations, his formulas are truly exactly the same as e.g. http://en.wikipedia.org/wiki/Relativistic_Doppler_effect#Motion_in_an_arbitrary_direction .

Thus, Synge's result, which the author admits is general, already establishes the universality of redshift in GR as 'pure Doppler' given acceptance of using source 4-velocity transported along the connecting null geodesic.

[Edit: To flesh this out a bit, Synge's equation (45) in Ch. III, section 7 is identical to equation (1) in the above wikipedia link once it is understood that in Synge's unusual notation:

1 + v² = γ²

and

vR = γ (v cos θ sub o), with v on the RHS having the normal meaning as wikipedia.

The key is that Synge's spatial speeds have scaling by γ built into them by the particular geometric definitions he is using.]

 

[jimmycricket]

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.

 

[PAllen]

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.

Ah, but starting from SR (let alone GR) any such question has no unique meaning. It all depends on the simultaneity convention adopted (equivalently, the foliation into time (per infaller) X space, chosen by the infaller). Only what you observe is devoid of convention - GR makes specific predictions for what you observe, and predicted observations don't depend on coordinate choices. But what is happening 'now' at a distance, is a choice of convention in all cases in GR and SR.

For a given moment on the infaller's world line after horizon crossing but before reaching the singularity, the only constraint on what distant event is considered 'now' is that it can't be in either the past or the future light cone of that event. Thus, you can choose, for the now time of (e.g. earth) anything after the last observed time on earth, up until infinite future (since none of the exterior is in the future light cone of an interior event). Thus, if at some moment you see 3PM today timestamped on signal from earth, you can say 'now' there is anything from 3PM + epsilon, up until infinite future. Pick whatever you want, all choices in this range are valid.

 

[Asher Weinerman]

Would the universe end if you entered a black hole? What I mean by this is that due to time dilation would time elapse so fast for the universe outside the black hole relative to you inside it that all the stars would burn out and all that wouild be left would be other black holes?

Jimmycricket, I had the same question. My view is that not only would all the stars burn out, but all the black holes would evaporate by Hawking Radiation before you crossed the horizon, including the one you are falling into.

As for separating the time dilation effect into kinematic and gravitational, that's easy. The gravitational effect blue shifts the outside universe without bound as you approach the event horizon, but your speed approaches the speed of light as you approach the event horizon thus red-shifting the universe without bound as you approach the event horizon (all stationary people (dr = 0) that you pass on your way to the horizon will be waving at you very slowly because of their speed relative to you, which approaches that of light as you near the horizon). Taking the limit of these two "infinities" as R approaches 2m yields 1+√2m/R as stated by George Jones.

 

[PAllen]

Jimmycricket, I had the same question. My view is that not only would all the stars burn out, but all the black holes would evaporate by Hawking Radiation before you crossed the horizon, including the one you are falling into.

As for separating the time dilation effect into kinematic and gravitational, that's easy. The gravitational effect blue shifts the outside universe without bound as you approach the event horizon, but your speed approaches the speed of light as you approach the event horizon thus red-shifting the universe without bound as you approach the event horizon (all stationary people (dr = 0) that you pass on your way to the horizon will be waving at you very slowly because of their speed relative to you, which approaches that of light as you near the horizon). Taking the limit of these two "infinities" as R approaches 2m yields 1+√2m/R as stated by George Jones.

But that limiting process tells you nothing about the interior, which is not stationary and for which a potential (and therefore gravitational time dilation) are undefined. However, GR has an unambiguous prediction for this shift (given chosen initial conditions) as George Jones noted. This interior result can be derived many different ways, all producing the same result (but none involving gravitational blueshift).

Your first point (on BH evaporation) is wholly irrelevant to a classical discussion. Separately, a large majority of experts in the field says your position is wrong (that is wrong as a semiclassical prediction of QFT + GR, which is the only framework to talk about such things).

 

[PeterDonis]

My view is that not only would all the stars burn out, but all the black holes would evaporate by Hawking Radiation before you crossed the horizon, including the one you are falling into.

But this would mean that you would never actually fall into the hole--that is, you would observe it evaporating as you fell towards it, and it would disappear before you reached its horizon. As PAllen noted, the unambigous prediction of semiclassical QFT + GR is that this is not correct. Do you have a different theoretical framework that makes a different prediction? If not, your statement is a personal theory and is off limits for discussion here.

 

[Asher Weinerman]

But this would mean that you would never actually fall into the hole--that is, you would observe it evaporating as you fell towards it, and it would disappear before you reached its horizon. As PAllen noted, the unambigous prediction of semiclassical QFT + GR is that this is not correct. Do you have a different theoretical framework that makes a different prediction? If not, your statement is a personal theory and is off limits for discussion here.

PeterDonis, I'm no expert as I've stated before, but Hawking Radiation evaporates black holes in finite time (approx. 10^68 years or something) - this is not my personal theory - many sources quote this estimate or 'round about. 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.

And don't threaten me again about being off-limits - most of us here are searching for information and trying to understand things through conversation, while others just act as know-it-all bullies and they are the ones that should be off-limits.

 

[rootone]

Evaporation due to Hawing radiation is proportional to the size (mass) of the black hole.
It could be nearly instantaneously or it could be trillions of years.

According to relativity, the infalling observer notices nothing particularly strange locally while crossing the event horizon.
Locally, (his own frame of referance), other material would be infalling too,and not at unreasonable velocities, and his clock still ticks at one second per second.
He would certainly notice the effect of extreme gravity (spaghettification), in the case of a stellar sized black hole,
but for a supermassive black hole that would be lest drastic.

In any event though he becomes eventually causally disconnected from whatever happens outside of the event horizon.
It is no longer for possible for information to be received by him from 'outside' , and he no longer transmit information to 'outside' either.

 

[PAllen]

Evaporation due to Hawing radiation is proportional to the size (mass) of the black hole.
It could be nearly instantaneously or it could be trillions of years.

Actually, it is inversely proportional to the square of the mass.

According to relativity, the infalling observer notices nothing particularly strange locally while crossing the event horizon.
Locally, (his own frame of referance), other material would be infalling too,and not at unreasonable velocities, and his clock still ticks at one second per second.
He would ceetainly notice the effect of extreme gravity (spaghettification), in the case of a stellar sized black hole,
but for a supermassive black hole that would be lest drastic.

This is all correct.

In any event though he becomes eventually causally disconnected from whatever happens outside of the event horizon.
It is no longer for possible for information to be received by him from 'outside' , and he no longer transmit information to 'outside' either.

This is not correct. The infaller can receive information from the outside at a 'normal' rate until they reach the singularity. It is true that they can't transmit to the outside.

 

[Jesse King]

But this would mean that you would never actually fall into the hole--that is, you would observe it evaporating as you fell towards it, and it would disappear before you reached its horizon. As PAllen noted, the unambigous prediction of semiclassical QFT + GR is that this is not correct

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?

 

[rootone]

@PAllen
OK, I got about 8/10 there, not too bad, but thanks for further clearing things up.

 

[PAllen]

PeterDonis, I'm no expert as I've stated before, but Hawking Radiation evaporates black holes in finite time (approx. 10^68 years or something) - this is not my personal theory - many sources quote this estimate or 'round about. 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.

And don't threaten me again about being off-limits - most of us here are searching for information and trying to understand things through conversation, while others just act as know-it-all bullies and they are the ones that should be off-limits.

You made a definite statement (not asked a question), and that statement is considered simply wrong by experts.

In another recent thread, many references and explanations were given to you. There is no point in repeating them all here.

As for your question, you should know that it fundamentally ambiguous in even special relativity. How long does it take a muon to reach the ground on creation in the upper atmosphere? It depends on who is measuring and how. However, invariant is that it reaches the ground. How long an infaller takes to reach the horizon and then the singularity has one type of answer that is invariant: the time on the infaller's watch, which is quite short. Otherwise, time for an external observer depends only on coordinates chosen and can be short and finite or infinite depending on coordinates. The infinite value has the same content (or lack of it) as the statement that a ball dropped from an accelerating rocket never gets beyond the Rocket's Rindler horizon. This is just a matter of the choice of coordinates by the rocket. They can choose different coordinates and get different answer. However, the fact that, say, a dropped ball collides with some planet is invariant irrespective of coordinates (though ill chosen coordinates may not include that event).

The competition between evaporation and collapse can be modeled quantitatively, and relevant papers were provided to you in the other thread that establish that the horizon forms, and object fall through it, before evaporation occurs.

 

[PeterDonis]

Hawking Radiation evaporates black holes in finite time

And objects also fall into the hole in finite time. See below.

this is not my personal theory

The part about black hole evaporation taking a finite time isn't, true, but you are stating it in a way that does not give me confidence that you understand what it means.

However, your statement that the hole will evaporate before anything can fall into it, since you have continued to repeat it in the face of repeated demonstrations as to why it is incorrect according to mainstream science, is a personal theory. That's what I was responding to.

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.

Sure. It will take a time $\tau = 2m \left[ (r / 2m)^{3/2} - 1 \right] \approx 2.67m$. (Note: edited to correct the formula.)

don't threaten me again about being off-limits

I wasn't making a threat. I was reminding you of PF policy about personal theories. You will not get another reminder.

 

[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: