# Infinite Time Dilation at the Surface of a Black Hole?

1. Oct 24, 2013

I've never fully understood how anything can actually fall into a black hole without the black hole evaporating first. Since time dilates exponentially as I fall into a black hole, a point will come where a few seconds for me will be millions of years in the outside world...trillions in fact. Perhaps long enough for the black hole to evaporate?

Let me set up an experiment as precisely as a I can to clarify my question.

Assumptions:

1) I am indestructible. Nothing short of the utter dismantling of space itself can destroy me
2) I have an almost infinitely powerful rocket capable of generating insane amounts of thrust and with a practically unlimited energy supply.

Here's the scenario:

I first set up a clock at a safe distance away from the black hole. I engineer it so that it has just one hand and each complete circle of the hand measures what I experience as one year. For each rotation, a counter increases and there is no limit to the number this counter can reach as long as the clock keeps working. And the clock has an inexhaustible energy supply.

Now, I turn around and start heading towards the black hole. I go closer...closer...closer. The gravitational pull keeps increasing. But because of (1), I don't die. I come closer to the event horizon never actually crossing it. In fact, I come as close to it as theoretically possible. Because of (2), my rocket is capable of counteracting the gravitational force since it's still finite no matter how great.

I hold my position for 5 years of my local time. After 5 years have expired (for me), I give my rocket an extra boost and escape the clutches of the black hole (technically I suppose I was never "in" it) at all.

I now go back to my clock which is still ticking away happily at a safe distance.

Question: Does the clock counter show that millions of years have passed? Or billions? Or is the above experiment moot because the black hole itself would have evaporated out from under me but I still survive because of (1)?

Note: I'm not proposing any explanation or putting forward a theory. I freely admit that my knowledge of general relativity has a lot to be desired. I'm merely asking the question - what does the counter on my clock show that I set up before my trip. Of course, this is dependent on the size of the black hole etc, but I want to know whether the counter can be in the millions or billions.

2. Oct 24, 2013

### Staff: Mentor

If you were going to just fall into the hole, no. For the experiment you describe, perhaps, but see below.

I assume you mean, what you experience as one year while you are co-located with the clock.

Classically, there's no limit to how close you could come, because classically, spacetime is a continuum, so there is no limit to how small a distance there can be.

When we include quantum mechanics, there should be a limit on how small a distance there can be, because spacetime should no longer act like a continuum on small enough scales. Typically the scale on which these effects become important is taken to be the Planck length, which is about 10^-35 meters. So you could not come within a Planck length or so of the horizon without falling inside.

If the hole is a quantum hole, so that it emits Hawking radiation, then you will also have to withstand the radiation, which becomes arbitrarily high frequency as you get close to the hole's horizon. You assumed you were indestructible, so you can withstand it.

That's right; you never crossed the horizon, so you never got to see what was "inside" the hole.

If the black hole was classical, it could show an arbitrarily large amount of elapsed time, depending on how close you got to the horizon. If it was a quantum hole, so that it was emitting Hawking radiation, it depends on the mass of the hole; see below.

If you got close enough to the hole's horizon and stayed long enough, you would find that the mass of the hole was gradually decreasing during the time you spent close to the horizon; you would see this as a reduction in the rocket thrust you had to keep up to maintain altitude. It's possible, if the mass of the hole were small enough, that it could indeed evaporate completely; but as it did so, the relationship between your local clock time and the time on the clock you left behind at a safe distance would change. I haven't run the numbers to see if there is a maximum elapsed time you could see on the clock you left behind (which would depend on the mass the hole started with), but I think there could be one.

3. Oct 24, 2013

### Bill_K

Don't get this. Hawking radiation does not even come from the horizon.

4. Oct 24, 2013

### Staff: Mentor

The time dilation will behave as you expect: a clock that is moved very close to the event horizon, kept there for a time, and then accelerated back out to where you left the safe-distance clock will show less time having passed than the safe-distance one. Other time-dependent processes will be consistent with these readings as well; for example, if we have two identical twins, one of them stays with the safe-distance clock, and the other takes the journey almost to the event horizon and back, the traveler will have aged less when they get back together.

However, this doesn't tell us anything about falling into a black hole, because the traveler isn't falling - he has a powerful rocket that carries him back out of the gravity well before he crosses the event horizon. Indeed, this situation is just the general relativistic version of the well-known "twin paradox" of special relativity; you will want to be sure that that you completely understand the SR version before you take on the GR version.

But you started your post with a different question: Given the gravitational time dilation effect, would it be possible for something to actually fall into the black hole if it didn't have that powerful rocket to rescue it before it crossed the event horizon?

There are a bunch of threads on this question already, and the short answer is "yes, you fall in".

1) If you consider the eventual evaporation of the black hole : The safe-distance observer will eventually see you falling into the black hole, and then at some later time will see the final flash from the evaporation of the black hole. (The evaporation time of an astronomical black hole is easily fifty orders of magnitude greater than the age of the universe, so "eventually" is a long time).
2) If you don't consider the evaporation of the black hole, safe-distance observer will never see you falling into the black hole, because light from that event will never make it out to his eyes - but that doesn't mean that it didn't happen.

5. Oct 24, 2013

Thank you for that. So if I understand you correctly, it's possible for my clock to have measured trillions of years if the black hole is reasonably sized is that correct?

6. Oct 24, 2013

### Staff: Mentor

But it does get redshifted as it goes out to infinity, yes? So if you are close to the horizon, you will see the Hawking radiation as much higher frequency than an observer at infinity does.

7. Oct 24, 2013

### Staff: Mentor

For the clock you left at a safe distance, yes.

8. Oct 24, 2013

### Bill_K

Not if it originates above you. Remember that the predominant wavelength of the Hawking radiation is comparable to the size of the hole. In fact I believe that the amplitude of the radiation goes to zero as you approach the hole.

9. Oct 24, 2013

### Staff: Mentor

But this is the wavelength as seen by an observer at infinity, correct? The wavelength as seen by an observer close to the horizon would be blueshifted. At least, that's what various physicists who talk about a hot "membrane" close to the horizon (or "stretched horizon", which is what Susskind, for example, calls it) seem to be saying.

10. Oct 24, 2013

### Staff: Mentor

Sure, but you don't need a black hole to get a clock to measure trillions of years - you just have to wait trillions of years, and that's what the safe-distance observer is doing. The black hole is just one way of creating a situation in which some other clock following a different path through space-time (in this case, passing very close to the event horizon so experiencing extreme time dilation) will correctly measure much less time on its path through space-time.

I said above that you really want to nail down the special relativity version of the twin paradox before you take on this general relativistic version... There's a pretty decent summary here: http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html

11. Oct 24, 2013

### yuiop

For Unruh radiation, the temperature is is proportional to the proper acceleration. The closer you get to a black hole the greater the proper acceleration required to hold station, so it would seem by the equivalence principle, that the temperature increases the nearer you are to a black hole. The temperature of a black hole due to Hawking radiation is usually quoted for the measurement at infinity.

While thinking about this I came up with this simple thought experiment that might prove interesting. Consider a 12V car battery with a 120 amphour rating that is connected to a 5 amp lamp rated at 60 watts. Anne is lowered with the battery and lamp to a level in a Schwarzschild metric where the gravitational time dilation is 10 time greater than that experienced by Bob higher up. Anne runs the lamp for 24 hours and runs the battery flat and so uses up approximately 1440 Watt Hours of energy.

The light is focused up to Bob in a tight beam so that he collects all the transmitted photons. Due to time dilation, Bob sees the energy arrive over a period of 240 hours at a rate of 6 watts per hour. The total energy received by Bob is 1440 Watt Hours. The interesting aspect is that photons are said to lose energy as they climb out of a gravitational well, but the number of photons received by Bob is identical to the number of photons sent by Anne and the energy sent and received is identical, so the energy carried by each photon is unchanged.

<EDIT>Ooops, shooting from the hip again. Upon reflection, the energy carried by each photon must be reduced by the redshift factor, so the total energy E received by Bob is E'/gamma where E' is the energy sent by Anne. The power P or energy received per unit time by Bob is P'/gamma2 where P' is the power as seen by Anne. This implies the power of the Hawking radiation increases by a factor of gamma2 where gamma = √(1-2m/r), as you get closer to the black hole.

Last edited: Oct 25, 2013
12. Nov 10, 2013

### nickb123

This comment/question does not directly relate to the current topic. Someone I was talking to mentioned that 95% of the matter in the universe is unaccounted for. If this was the leading matter that was expelled in the big bang, it could explain why the universe is expanding at an accelerating rate (due to the gravitational force of this expanding spherical mass encompassing the universe).
If that is the case, I was wondering if the acceleration rate is increasing or decreasing (of course we can only determine if it was way back in time that we can currently see). Like a black hole, light could not escape inward (towards the visible universe).
If it is increasing, the expansion will eventually reach the point that the black hole mass dilutes to the point that can no longer sustain increased expansion acceleration.
On the other hand, if the acceleration rate is already decreasing, we may already be headed back to a singularity, once the acceleration reaches zero and contraction begins.

Last edited: Nov 11, 2013
13. Nov 11, 2013

### dauto

There is no leading matter. Big Bang happened everywhere. There is no center for it to expand from and there is no edge for it to expand into.

14. Nov 11, 2013

### pervect

Staff Emeritus
If we take the limit where the black hole horizon approaches a Rindler horizon (I'd call it the flat space limit, as it corresponds to a very large black hole with no appreciable tidal force), shouldn't the temperature of the local vacuum be the Unruh temperature associated with the very large acceleration the observer needs to hold station?

I'm going to snip some stuff from Wiki, I think it maybe is too distracting in this thread and I'm not sure if it's accurate.

Last edited: Nov 11, 2013
15. Nov 12, 2013

### nickb123

Is it known (or can it even be measured) if the universe expansion acceleration rate is increasing or decreasing? I understand that even if known, it would reflect a distant past condition.

Last edited: Nov 12, 2013
16. Nov 12, 2013

### dauto

We know the universe is expanding.
We know the expansion rate is accelerating

That's all that's known for sure for now, based on direct observations

Models can be built in order to try to explain those facts
Within those models it is possible to calculate other variables and try to answer questions such as: "Is the acceleration of the expansion increasing or decreasing over time?". But the answers you get are model dependent and should be taken with a grain of salt. Until we have a better theoretical understanding about what causes the expansion acceleration, I wouldn't bet the farm on any of those models. Scientific progress is often slower than what we wish it would be. Gotta have patience.

17. Dec 29, 2013

### DKS

I don't understand how the safe-distance observer can observe you falling through the event horizon (assuming that's what you meant), with or without Hawking radiation. At what time (say some fraction of the evaporation time) will the safe-distance observer observe this happening?
As you mention it is not observed classically, how does the Hawking radiation cause you to observe the falling in?
Not sure what "did (not) happen" means exactly here. Should I interpret it as for the outside observer it never happens but for the falling observer it does? Since they can never compare notes there is no contradiction?

Kees

18. Dec 29, 2013

### PAllen

An observer never seeing something is a statement about light not 'reality', whatever that is. In the complete description of the universe in classical infall, crossing the horizon and approaching the singularity are part of the description. These are invariant facts of the complete description, not observer dependent.

That light = reality is a false concept is seen readily in SR. Consider a rocket uniformly accelerating away from earth, watching a someone drop a rock on earth. They would see the earth fade to black, and the rock never reach the ground. Is this a feature of reality or a feature of light not being able to catch the rocket? A scientist on the rocket can readily model dropping a rock on earth and conclude that it fell like any other rock, irrespective of light being able to catch the rocket.

The BH horizon behavior is completely equivalent.

Last edited: Dec 29, 2013
19. Dec 29, 2013

### DKS

I don't get that. The rocket would see the rock falling and hitting the ground at some finite time, perhaps substantially redshifted, but you'd see it.

20. Dec 30, 2013

### Staff: Mentor

Google for "Rindler horizon" - as long as the rocket is uniformly accelerating, there is a region of spacetime from which a light signal will never catch up with the rocket.

(Of course if the rocket stops accelerating, starts coasting at a constant velocity, then the light will eventually reach it)

21. Dec 30, 2013

### DKS

Thank you, now I understand what I didn't understand. Subtle stuff!

22. Dec 30, 2013

### Staff: Mentor

The Hawking radiation causes the black hole to eventually evaporate. When it does, the light from me falling through the horizon (this light is moving radially outwards all along but doesn't escape as long as the black hole exists) finally escapes, makes it to the safe-distance observer's eyes.

Conversely, if the black hole never evaporates then the light from me falling through the horizon never gets out, never makes it to the safe-distance observer.

It's much more straightforward than that. I do fall through the horizon and light from that event does not reach the safe-distance observer's eyes as long as the black hole is there. There's no contradiction between these two statements.

23. Dec 30, 2013

### DKS

What I still don't understand is that for the far-away observer you fall through the horizon at a time $t =∞ > T_e$ where $t$ is the Schwarzschild coordinate, which coincides (almost) with proper time for the distant observer, and $T_e$ is the evaporation time (in Schwarzschild coordinates). So it seems to me that no light from you falling through the horizon will ever reach me, as you will not have fallen through the horizon yet at $t = T_e$. If you survive the explosion you'd tell the distant observer that you saw the black hole explode before you could fall in.

What am I doing wrong?

24. Dec 30, 2013

### Staff: Mentor

No, this is not correct. You only fall through the horizon at $t = \infty$ if the black hole never evaporates. If the hole evaporates, you fall through the horizon at $t = T_e$, the evaporation time, according to the far-away observer. This is basically another way of saying what Nugatory was saying.

25. Dec 30, 2013

### DKS

__________________
But at $t = T_e$ there is no longer an event horizon and Hawking's calculation breaks down shortly before $T_e$ (say at time $t_q=T_e - \Delta T$ with $\Delta T$ something like the Planck time) as quantum gravity will kick in.

Would it be correct to say you don't fall through the horizon till time $t_q$ and we don't know what happens after that?