Black hole time dilation

  • #26
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.
 
  • #27
PAllen
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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).
 
  • #28
PeterDonis
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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.
 
  • #29
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.
 
  • #30
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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.
 
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  • #31
PAllen
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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.
 
  • #32
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?
 
  • #33
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@PAllen
OK, I got about 8/10 there, not too bad, but thanks for further clearing things up.
 
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  • #34
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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.
 
  • #35
PeterDonis
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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.
 
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  • #36
PAllen
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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.
 
  • #37
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I'm not sure if anyone has answered the question yet as I dont 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.
 
  • #38
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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.
 
  • #39
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-(rs/r) where rs 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?
 
  • #40
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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.
 
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  • #42
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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. :wink:
 
  • #43
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Well that's enough time to listen to a couple of pop songs, so it's not too bad.
 
  • #44
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Well that's enough time to listen to a couple of pop songs, so it's not too bad.
I suggest this one:
 

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