# Black Hole XVI

1. Apr 26, 2005

### Ich

I´m sure that I am not the first to ask this question, but I did not find the answer in recent threads-

There are three statements about black holes that I can´t bring together:
1. I will experience nothing special when I cross the event horizon of a massive black hole, and I will cross it in finite proper time.
2. For an observer in flat space far away from the black hole I will never reach the event horizon - the procedure takes infinite time.
3. An observer in flat space far away from the black hole will see it explode in finite time due to hawking radiation.

So will I suddenly get killed crossing the event horizon? Or will I reach he singularity?

Could someone please point out what I got wrong?

2. Apr 26, 2005

### pmb_phy

All seems correct accept #3. As I recall, since there is a 3k radiation background the black hole will absorb energy at a rate which is greater than it radiates energy. What you're speaking about is micro-black holes in the absence of 3k radiation.

Pete

3. Apr 27, 2005

### Ich

As I understood the temperature of a black Hole depends inversely on it´s mass.
So in 10^xyz years, when the background radiation is 10^-xy K the black hole will start to lose energy and finally explode.
And then the problem returns.

4. Apr 27, 2005

### JesseM

The last section of this page seems to say that the observer falling in will observe himself crossing the horizon long before the black hole evaporates, but an observer outside would see him reach 0 distance from the horizon at the exact moment the black hole evaporated to nothing:

5. Apr 27, 2005

### chronon

The commonly accepted answer is that you will fall through the horizon, whatever an external observer sees. However, I find that hard to accept too. See www.chronon.org/articles/blackholes.html

From the relativity faq in JesseM's post:
Note that this has a hidden assumption of an absolute time - not allowed!

6. Apr 27, 2005

### Ich

thanks JesseM, that gave me something to think about.
As I understood, the external observer will never see me crossing the horizon, but will see me dissolved together with the horizon.
And I will experience falling in the black hole the same way I expected to, i.e. the black hole has roughly the properties it had back then when I started falling in - I will not see the explosion at t~10^xyz.

There still remains something that bothers me: Does the external observer really merely see my image frozen at the event horizon while I have already been crushed?
Or, to avoid the simultaneity issue, is it physically possible to rescue me after, say, 1000 years flat time? After 10^xy years? When am I out of reach for the universe?
If I could be reached until the black hole vanishes there would really be a paradox.

7. Apr 27, 2005

### Ich

Could you explain that?

8. Apr 27, 2005

### chronon

The verb tenses used in the passage quoted imply that while the external observer goes on seeing your image, you have already fallen through the horizon. However, this implies a comparison of the times of two separated events, and so is not meaningful - it is only meaningful to say event A is before event B if B lies in the future light cone of A.

No after a certain time light from the external observer will not be able to catch you up. Since the external observer can't travel faster than light it will be impossible to rescue you. (Of course if you believe my arguments that the black hole will evaporate away before you reach it then that complicates things - but then you won't need rescuing).

9. Apr 27, 2005

### Ich

Ok, that´s the "simultaneity issue" I brought up. I read the original statement different: It sais that
1. I will fall through and
2. The observer will not see me fall through.
It doesn´t explicitly compare times. But still there´s the other point-->
So after a certain time nobody outside can interact with me. That´s sufficient for me to state that it´s not me there at the horizon but merely my image.
I´m trying to resolve the problem like that:
At the time I lost contact with the universe, I´m as good as through the event horizon - I can´t return (can I?) and nothing from the outside can bring me back. So I will be killed by singularity.
What remains after that time is an image which will fade away and finally be destroyed in the explosion. It´s merely an artifact of curved space time (something like trapped photons) and has nothing to do with me.

Does this make sense?
And how does one calculate that time?

10. Apr 27, 2005

### chronon

It doesn't really make sense to say that what an external observer sees is just an image rather than the real thing (unless you say the same thing about everything you see). In particular, the external observer may see you and your clock (running slower and slower) falling into the black hole. Suppose that he eventually gets tired of watching, when he sees your clock at some time T, thinking that you must have in reality fallen through the event horizon. However, you have a super-powerful rocket which you start when your clock says time T, (when you haven't yet crossed the event horizon) and so you escape from the black hole, much to his surprise. Hence at no time can the external observer claim that you have actually passed through the event horizon.

11. Apr 27, 2005

### Mortimer

For what it's worth, here are my own weird ideas about what happens to objects falling into a black hole. See http://www.rfjvanlinden171.freeler.nl/idea/index.htm at sections 2 and 3. The assumptions behind it are: a black hole (and gravity) is 5D; everything always moves at c (also in 5D). The clue is that the velocity vector of the falling object rotates towards $$x_5$$ during the fall.
So the effect of the frozen image at the event horizon is merely the linear projection to 3D space of the moving image in $$x_5$$.
I'm not completely alone in my assumption that black holes may be 5D objects. Just Google with http://www.google.com/search?hl=en&q=5+dimensional+black+holes to see that.
As already stated in the beginning of my page, don't take it too serious.

BTW have a look at Andrew Hamilton's homepage where you can watch yourself falling into a black hole! http://casa.colorado.edu/~ajsh/home.html

Last edited by a moderator: Apr 21, 2017
12. Apr 27, 2005

### robphy

This spacetime diagram might help (adapted from Geroch's General Relativity from A to B).

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13. Apr 27, 2005

### pervect

Staff Emeritus

The person who is falling into the black hole can still receive signals from the outside world. But, once he passes the event horizon, he cannot send signals back.

The question arises - if you have an observer sitting at infinity, sending out pulses once a second, how many of them does the infalling observer see? He can see these pulses even after he crosses the event horizon.

It takes math to really calculate the answer - the only really interesting part of the answer is that he sees a finite number of pulses from the external observer, not an infinite number, before he reaches the black hole.

There is a fairly simple argument to show that the infalling observer can't see an infinte number of pulses, without a lot of math. You have to belive that photons are redshifted when they go "uphill", and blueshifted when they go "downhill". You also have to believe that the tidal forces near a black hole at a given point don't act weirdly as a function of velocity. You also have to believe that the observer hits the singularity in a finite amount of his own proper time.

Given all these assumptions, you can say that from the infalling observers POV, the stretching tidal forces mean that the photons from infinity have to climb "uphill" to reach him. This means that the photons are redshifted.

The point of this is that your mental model must give the following results to be correct:

Photons emitted by the infalling observer don't make it back out
Infalling photons tracing a radial path (ones that fall directly in to the black hole) are redshifted when they reach the infalling observer.
The observer reaches both the event horizon (and even the singularity!) in a finite amount of proper time (umm, we're doing a non-rotating black hole here to keep life simple).

Because the infalling photons are redshifted, the observer sees only a finite number of pulses from the outside world before he reaches the event horizon (or before he hits the singularity for that matter). (Not only photons are redshifted, but any time interval is lengthend. If you send pulses out at once a second with a redshift factor of 2, pulses willl arive once every two seconds).

A model that the infalling ovserver never reaches the event horizon usually fails on this point (the finite number of pulses) - because if you think of him never reaching the point, you think that he must receive an unlimited number of pulses. This is wrong.

To actually calculate the paths of light and objects, one needs to use a well behaved coordinate system like the Finklestein ingoing coordinates, or Kruskal coordiantes. Of the two, Finklestein coordinates are the easiest to understand. The Finklestein coordinate 'v' is basically just the number of pulses an observer has seen while falling from infinity. Another way of saying this - by consturction, radially infalling light rays have a constant Finklestein coordinate 'v'. This coordinate 'v' is neither a space coordinate, nor a time coordinate, but a 'null' coordinate, becuase it desccribes a light-like path with a constant number.

There's a graph at http://casa.colorado.edu/~ajsh/collapse.html#kruskal which may be of some help if you study it enough of an infalling black hole in Finklestein coordinates (and in Kruskal coordinates, too).

Given the right coordinates, the job of calculating the trajectories is basically a matter of satisfying the geodesic deviation equations, which are a system of second order partial differential equations.

For specificity we say that the particle follows some parameterized path

$$x^i(\tau)$$, i.e. t(tau), x(tau), y(tau), z(tau), in the usual cartesian coordinates. In Finklestein coordinates it's actually r(tau), v(tau), theta(tau), phi(tau), where theta and phi are two angles (zero for someone falling straight in, so we can deal only with r(tau) and v(tau) for an observer with no angular momentum.

Then we write down the geodesic deviation equations

$$\frac{d^2 x^c}{d \tau^2} + \Gamma^a{}_{bc} \frac {d x^b}{d \tau} \frac {d x^c} {d \tau} = 0$$

Then we solve these equations.

The $$\Gamma$$ in the above expression are the Christoffel symbols for the given metric - they can be calculated by hand (or much more easily by programs like GRTensor II) from the metric coefficients. They are basically partial derivatives of the metric coefficients.

Working the problem out in Finklestein coordinates, I get the following equations for what it's worth (I've represented differentiaton with respect to tau by a prime mark. You can see these are second order equations in r'', r', r, v'', v', v, where r and v are the Finklestein coordinates. (r is equivalent to the Schwarzschild radial coordinates).

r'' - 2m/r^2 r'v' +(1/r^2-2m^2/r^3) v' v' = 0
v'' + 1/r^2 v' v' = 0

Oh yeah - you need the boundary conditions to solve the equations - specifically, what are r, r', and v (the radial position, radial velocity, and you can safely set v=0 at tau=0, but you need to specify that that's what you're doing).

Last edited: Apr 27, 2005
14. Apr 27, 2005

### Ich

There is no time T when I am beyond any hope to return? I really have to cross the event horizon? Isn´t there some calculation that shows that it has to be decided in a finite time whether I will cross the horizon or not?
My problem with it is the following:
An observer will already have collected part of my remains as Hawking radiation at the time when he sees me cross the border. How could I come back then and say "April Fool! I never went down there!".
How would one resolve this?

15. Apr 27, 2005

### JesseM

If I am maintaining a fixed distance from the horizon, I think there should be a point beyond which I can't rescue you. The reason I think this is because I know that, despite the fact that an observer on the outside sees you take an infinite amount of time to cross the horizon, you do not see the entire infinite future history of the universe pass before your eyes as you cross it, you only see a finite amount of time pass on my clock (pervect discussed this above); this must mean that there is a time beyond which a light signal emitted from me will not have time to catch up with you before you cross the horizon. So beyond this point, nothing I do should be able to have any causal effect on you before you cross it.

Of course, this doesn't mean that you may not independently choose to fire your rockets right before you reach the horizon, so perhaps there is no time beyond which I can't be sure you won't avoid crossing the horizon, even if there is a point beyond which it's out of my hands to save you.

Last edited: Apr 27, 2005
16. Apr 27, 2005

### Ich

JesseM,

so my argument was flawed. I could still escape the black hole. But then, if I decided so just before entering the horizon, how could I return undamaged to the observer when he collected already a considerable amount of me as Hawking radiation?

17. Apr 27, 2005

### JesseM

I don't know. The problem is even worse if you assume that the Hawking radiation somehow encodes information about what objects fell into the black hole, as a few of the proposed resolutions to the black hole information loss paradox suggest. If I was able to decode the information in the Hawking radiation, could I determine that you did indeed cross the event horizon at a time when I still see you almost frozen outside of it? If this happened, would this allow me to predict with 100% certainty that I would not later see you decide to fire your rockets and move away from the event horizon? I guess as long as I didn't receive this Hawking radiation until after the point where it's no longer in my power to interfere with what happens to you, then this wouldn't lead to any time-travel-type paradoxes, but it still seems a bit weird. Maybe there's some way to avoid this conclusion, but I have no idea what it would be.

18. Apr 27, 2005

### Ich

Thanks pervect. I really appreciate the time you spent answering, and I will think about it further.
But right now, I don´t think I understand how the puzzle is resolved.
OK, I only see a finite portion of the universe during my fall. I interpret this fact similar to our universe: as it´s expansion is accelerated, the visible universe shrinks.
When I cross the event horizon, my past light cone does NOT extend to infinity.
Still, if I decided to accelerate and escape, is there a last flat time to do so? Is there a time when it´s safe to say "ok, all information about me is lost to the universe, now I can come back as Hawking radiation"?

19. Apr 27, 2005

### pervect

Staff Emeritus
There is a last time to do so in the infalling observers coordinate system. This is the time when he (the infalling observer) reaches the event horizon. Once he / you passes the event horizon, there is no return.

However, from the outside perspective, light being emitted close to the event horizon will be delayed, and as you get closer and closer to the horizon, the delay approaches infinity. So the outside observer will never actually see you reach the point of no return. He can see you get arbitrarily close, and if he knows the capabilities of your rocket, he can predict when that particular rocket doesn't have the capability to escape, but the outside observer will never actually see you pass through the critical point where you actually reach the event horizon, no matter how long he waits.

20. Apr 27, 2005

### JesseM

So no matter how long you wait, isn't it still possible that just before the astronaut crossed the horizon he fired his rockets and escaped the black hole, but you just haven't seen it happen yet because the light from the event of his beginning to fire the rocket hasn't reached you yet?