# Falling into a black hole

1. Aug 7, 2008

### phja

i was wondering what everyone thought they would see if they fell into a black hole. i know from relativity that from the perspective of me or you on earth, it would take an infinite amount of time for that person to fall in since as they approach the even horizon time dialation becomes very large.

But what about from the perspective of the person falling in (asuming they could servive the tidal forces on their bodies and all the other things trying to kill them). obviously they would see time pass at a normal rate in their referance frame but what would the rest of the universe seem like. i have always thought that as they cross the event horizon they would view all time instantly (i have always also thought that if one could travel at the speed of light they would also see all of time pass in an instant, and could be everywhere in the universe at once). this just seems to make sence to me. i suppose it could be that instead, the person falling in see's the rest of the universe slow down.

if anyone could clear this up for me it would be greatly appreciated.

2. Aug 7, 2008

### DaveC426913

All EM radiation from outside falling upon the observer would be blue-shifted, climbing up the spectrum until the observer was being bombarded entirely with huge doses of deadly gamma rays.

For however long he survives that, I do believe he would see the universe (or what small component of it he could see) start to speed up.

3. Aug 7, 2008

### phja

a bit off topic but i was just thinking. black holes have a large gravitational potential well (is it infinite?)and so anything close to them appears to slow down up to the point that objects never cross the event horizion. to me that signals that time has stopped inside the black hole. if this is true then could the black hole could exist for an infinitly small amount of time, yet still appear to us to exist forever.

4. Aug 7, 2008

5. Aug 7, 2008

### DaveC426913

I stand corrected. Feel a bit sheepish for falling into such an obvious pitfall. The quote above explains it very eloquently.

6. Aug 19, 2008

### aranoff

There are two solutions of the equations of general relativity. One from the point of view of the external observer. This solution is that the object takes forever to reach the event horizon. The other solution is from the point of view of the falling observer. This solution has the observer crossing the horizon in finite time; however, there is a singularity at the center.

What does singularity mean?

7. Aug 19, 2008

I do not believe that the object takes forever to cross the event horizon only that the objects clock stops as it crosses.

8. Aug 19, 2008

### aranoff

Wrong. It takes forever to reach the event horizon.

9. Aug 20, 2008

### stevebd1

Here's an extract from 'Exploring black holes' by Edwin Taylor and John Wheeler, page 3-19-

"No special event occurs as we fall through the Schwarzschild horizon. Even when we cross into a black hole at the event horizon r=2M, we experience no shudder, jolt, or jar. True, the tidal forces are ever increasing as we fall inward, and this increase continues smoothly at the horizon. But we are not suddenly torn apart at r=2M. True also, the curvature factor (1-2M/r) in the Schwarzschild metric goes to zero at this radius. But the resulting zero in the time term of the metric and the infinity in the radial term turn out to be singularities of the bookkeeper coordinates r and t, not singularities in spacetime geometry. They do not lead to discontinuites in our experience as we pass through the radius. There are other coordinate systems whose metrics do not have a discontinuity at the Schwarzschild radius. (See Project B, Inside the Black Hole, Section 4, page B-12 (metric for the rain frame)."

Rain frame- http://www.bun.kyoto-u.ac.jp/~suchii/inH.html

Last edited: Aug 20, 2008
10. Aug 20, 2008

### yuiop

I would like to discuss this webpage http://casa.colorado.edu/~ajsh/schwp.html about freefalling, Eddington-Finkelstein and Kruskal-Szekeres coordinate systems that are used to describe falling into a black hole. The webpage is not a refereed paper but it has frequently been recommended to newcomers by senior PF members so it is already in the “PF domain” and I think it reasonable to discuss it and as far as I know it represents the mainstream view.

The diagram I have attached to this post contains the first and last frames of this animation of Schwarzschild geometry transforming to freefalling coordinates.

In the diagram the white line is an infalling photon and the light green line is an infalling observer. On the left I have added dates to the Schwarzschild time lines. Event A1 is an infalling observer outside the event horizon at year 2007. Event A2 is year 2008 and he still outside the event horizon at that point. We can also see from event A3 that is also on the 2008 timeline that our observer is two places at once, simultaneously at the singularity and still outside the event horizon. At year 2009 things become clearer. Our observer is still falling towards the event horizon from outside but he also simultaneously moving away from the singularity back towards the event horizon. By interpreting events A3 and A2 as events belonging to a single observer in two places at one time the implication is that in Schwarzschild coordinates the infalling observer takes infinite time to arrive at the event horizon and then travels back in time to arrive at the singularity in finite time. The downward pointing arrow on the green line clearly shows this interpretation of free falling observer travelling backwards in time in order to arrive at the singularity. It then appears that by a suitable manipulation of the time coordinate to give "free falling" coordinates, the free falling observers falls smoothly and progressively in finite real time towards the singularity and everything is solved as indicated by the right hand diagram.

So the question I want to ask is why can't we consider that observers at A3 at the singularity and A2 outside the event horizon are not the same observer? To me, all the diagram is saying is that an observer at A2 would fall to the event horizon in an infinite amount of time and if a different observer happened to be at A3 they would fall forward in time outwards from the singularity towards the event horizon. Is it not possible that the Schwarzschild metric has been incorrectly interpreted?

I think the problem lies in how the velocity equation v=dr/dt is interpreted. The Schwarzschild metric indicates dr/dt becomes negative for a falling photon below the event horizon. The equation is dr/dt = (1-2m/R) which is negative for R<2m. Now if we insist dr can not be negative and and motion always proceeds in the positive spatial direction towards the singularity, then negative dr/dt implies dt becomes negative and time starts running backwards. If we take an alternative interpretation and insist that dt always remains positive and that motion always take place in the positive temporal direction then negative dr/dt simply suggests the motion has changed direction which to me is more plausible. It all depends upon whether you want the to interpret the universe as always going forward in the time direction or the spatial direction. By not allowing negative dt (time reversal) then observers do not cross the event horizon.

My analysis of the coordinate acceleration in Schwarzschild coordinates shows the acceleration also becomes negative (but still real) below the event horizon which also suggests to me that objects below the event horizon are accelerated towards the event horizon and not towards the singularity.

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11. Aug 20, 2008

### aranoff

"...observers at A3 at the singularity..." A singularity of f(x)=1/x is at x=0. The equation which is a solution of general relativity is not valid at the singularity!

12. Aug 20, 2008

### stevebd1

I'm going to go out on a limb here and say that the singularity might be composed of (or replaced by) matter compressed to Planck density.

Technically, if something has density then it's not a singularity but in theory, space and matter combine at Planck dimensions, creating a quantum foam supported by either loop quantum gravity (LQG) or strings so even though it has dimensions, it might still be described as a singularity as it is a combination of both matter and space (and probably time) rather than just collapsed matter in space.

There's also a theory that the Planck length is the invariant quantum length-scale that replaces zero and that beyond this, the spacetime sheet collapses on itself, sustaining energy at Planck density (which I think is an aspect of LQG) which I find more convincing than the singularity being represented by r=0, ρ=∞ which I always thought was a geometric curiosity rather than a reality.

The link below looks like a good thread regarding the meaning of a singularity-

13. Aug 20, 2008

### aranoff

tan(x) has a singularity at x = 90°. The tan function is not defined here. It is not infinity, but not defined. A singularity is simply a point where the function or equation is not defined.

Let us not discuss Planck length when discussing gravity, for these two theories contradict each other. Physicists are working on a theory that will unify these. Until that happens, we must deal with what we have.

14. Aug 20, 2008

### DrGreg

I haven't had time to look at the site you mentioned or think about this in any depth, but a quick comment (with my usual caveat I don't claim to be an expert in this subject).

I think in previous threads I've told you (unless my memory is incorrect and it was someone else, sorry) that inside the event horizon the r coordinate is timelike and the t coordinate is spacelike. In view of our recent discussion in another thread about the meaning of "ds2", "spacelike" and "timelike", maybe I can now convince you of this.

Just consider the curve given by $dr = d\phi = d\theta = 0$ and consider the sign of ds2 within the event horizon -- this shows the curve is spacelike and cannot be worldline of any particle. t is a spacelike coordinate, a warped distance.

And consider the curve given by $dt = d\phi = d\theta = 0$ and consider the sign of ds2 within the event horizon -- this shows the curve is timelike, the worldline of a massive particle. r is a timelike coordinate, warped time.

Coordinate velocity and acceleration below the event horizon would therefore be $dt/dr$ and $d^2t/dr^2$, although these are not the same as velocity and acceleration measured in local inertial coordinates.

Note: it probably doesn't help to think of the outside t and inside t as representing the same coordinate. They just happen to use the same symbol. For a falling object the outside t goes to infinity as it approaches the event horizon. Once inside, the inside t (which is warped distance) decreases from infinity. I have great difficulty in picturing what all this really means, but you can't argue with the maths.

15. Aug 20, 2008

### [Matrix]

This may have been answered in one of the quoted resources in this thread, but I have never been able to get my head around the following conundrum:

An observer external to the black hole will never see something cross the event horizon, yet someone falling in should be able to cross the horizon with no problems. So, let's say you're falling in feet-first in your super-adamantium spacesuit that protects you from tidal forces and gamma rays and so forth; you should never see your feet cross the horizon, correct?

If this is the case, does that mean that you will carry on falling until you actually hit the singularity, never having any evidence your feet crossed the horizon?

16. Aug 20, 2008

### aranoff

It does not make sense, says you. I say nothing makes sense when we discuss division by zero or other illegal operations. A singularity is division by zero.

17. Aug 20, 2008

### DaveC426913

Incorrect. The freely-falling observer will not experience the same effect. His frame of reference is accelerating.

18. Aug 20, 2008

### George Jones

Staff Emeritus
If you fall feet-first into a black hole, you will see your feet right up unil the (proper) time that your head "hits" the singularity.

There are two possible points of confusion that are both wrong:

1) since my feet cross the event horizon before my head, I lose sight of my feet;

2) since my feet "hit" the singularity before my head, I lose sight of my feet.

See

19. Aug 21, 2008

### Ich

No, you will fall quite naturally into the light your feet sent off. You observe nothing unusual. Problem is, both you and the light fall into the singularity, which is why an outside observer never sees your feet after they crossed the horizon.

20. Aug 21, 2008

### [Matrix]

OK that clears it up quite a bit, thanks for the help - I guess it really is true that a lot of these problems can be fixed just by thinking about what happens in the frame of the light you want to observe. Thanks! :)