# Falling into a black hole

1. Apr 11, 2005

### The Rev

Since the increasing gravity (as one falls) causes time to slow for those observing me fall into the black hole, would my observations of the universe speed up equally? In other words, as others saw my time slow to zero, would I see the Universe age more and more rapidly until, at the moment before I hit the event horizon, I saw it age and die before my eyes?

$$\phi$$

The Rev

2. Apr 11, 2005

### DaveC426913

Yup.

(In addition, you'd also see the radiation from the universe around you getting harder and harder (more energetic) until you were being pummeled by cosmic rays).

3. Apr 11, 2005

### lawtonfogle

would not the light reaching your eyes also slow

and of course you should already be dead by the time you reach the event horizon

4. Apr 11, 2005

### Billy T

Interesting point. Assume the REV were falling in feet first. How does he die? Would he be ripped appart of "cooked" first?

5. Apr 11, 2005

### The Rev

If you think tidal waves are bad, just try tidal forces.

The Rev

6. Apr 11, 2005

### pervect

Staff Emeritus
No. Not if you _fall_ into the black hole. Take a look at the Eddington-Finkelstein diagram in the

This is actually a diagram of a collapsing sphere, not a pre-existing black hole, but it illustrates the point.

E-F coordinates are good for this problem because they are reaonably well behaved (for infalling light rays, at least), and in addition the event horizon and the singularity itself are both represented by vertical lines. Kruskal coordinates are also well behaved, even better behaved than E-F coordinates because they handle outgoing light rays, but the singularity (r=0) and the event horizon (r=schwarzschild radius) are no longer simple vertical straight lines. The web-page has Kruskal diagrams as well, though, they are just harder to interpret.

The white line on the E-F diagram represents the infalling observer. The yellow diagonal lines represent incoming light rays. You can see that only a finite number of infalling light rays hit the infalling observer, both before he passes the horizon, and even after he passes beyond the event horizon. (When the white line hits the blue line at the left hand side of the screen, the observer has reached the singularity at r=0).

If you had a spaceship capable with unlimited acceleration capabilities, you could hover arbitrarily near the event horizon of a BH, and watch the universe go by in "fast motion". (Right at the event horizon you'd need infinite acceleration, but with unlimited acceleration you could hover as close to it as you liked). When you fall into a black hole (at least a non-rotating one), you will NOT see the entire universe go by.

Note that these diagrams apply directly only for a non-rotating black hole. Things get considerably more complex when the BH is rotating, especially after one passes the event horizon. Somewhere or other I posted some links for a recent paper about current state-of-the art simulations with respect to the collapse of a rotating BH.

7. Apr 11, 2005

### Janus

Staff Emeritus
It depends on the mass of the black hole. The more massive the black hole, the less intense the tidal forces at the event horizon. Remember, the radius of the event horizon increases directly with the mass, and the tidal force increases directly with the mass but inversely to the cube of the distance( in this case, the radius of the black hole.)
Double the mass of the black hole and you actually decrease the tidal force by a factor of 4.

For instance, with a super massive black hole of 1 million solar masses, the tidal force acting across the length of a person would be under 0.002g at the event horizon.

8. Apr 11, 2005

### Billy T

Thanks for reminding me of what I already knew. (See my post 46 of thread "creating a black hole" where I told SpaceTiger (not that he did not already know also):
"I suspect you know it well, but since you told me the better known r^-2 and r^-3 laws and did not mention it, I will just note that the gradient of a 1gm BH is greater than that of a 1kg one (at the event horizon). Thus, it really is a question of how close the atom must get before it is ripped up and ionized."

I was too quick to "kill the Rev," and only wondering how to do it (Like the wicked witch in Wizzard of Oz, when she is trying to decide how to kill Dorthy without damage to the magic power of the red slippers) and did not think.

To estimate the mass of the "Die-both-ways concurrently" black hole we will need to know details about the nature of the radiation being captured (near a star or not etc) and the stress the Rev can widthstand intact. A good physicist can tell the first (for various assumptions - see final note) and the Vatican's inquistion's specifications for converting heritics ("ropes for the rack" subsection) should tell something about the second.

As the Rev may be reluctant to let me confirm the "die-both-ways-concurrently" BH size experimentally and the lab store room is fresh out of small black holes, perhaps I should drop the "kill the Rev" idea and ask instead: What size BH makes the CBMR into visible light at the event horizon? I bet it is very tiny, perhaps much smaller than the photon so this too is all nonsense and just in fun (at least for me if not the Rev.)

Last edited: Apr 11, 2005
9. Apr 12, 2005

### pervect

Staff Emeritus
Well, as I said before, Rev won't fry. Another useful reference besides the page I previously quoted is the sci.physics.faq about falling into a black hole.

http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/fall_in.html

As far as the CMBR goes, while a stationary observer would see it blueshifted, the free-falling observer doesn't. There's a blue-shift due to the gravitatioanl potential difference, but there is also a red-shift from the doppler effect from the free-falling observers velocity. I'm fairly sure the two effects must cancel in the Schwarzschild black hole because E0 is a conserved quantity in the Schwarzschild metric.

10. Apr 13, 2005

### pervect

Staff Emeritus
For what it's worth, I worked the problem out in more detail and I'm currently getting the result that the "falling in" observer, assumed to be "at rest" at infinity, will see the CMBR redshift as he approaches the black hole, so I think my previous answer was incorrect.

Unfortunately, the problem is rather messy - the equations aren't horrendous, but there's lots of opportunity for sign errors. (It's even giving the symbolic algebra problem fits).

To describe the solution in brief, in schwarzschild coordinates, the 4-velocity is easily found as a funciton of radius, since E=1 and (dr/dtau) = E^2 - Veff^2, with Veff^2 = (1-2*m/r).

(We are assuming that there is no angular momentum, the fall is straight inwards).

This gives dr/dtau = sqrt(2*m/r), which gives a power law formula with an exponent of 2/3 for r as a function of tau It's convenient to take tau=0 as the time the object hits the singularity, and to have tau < 0.

dt/dtau is given by E/(1-2*m/r) (this from a textbook) with E=1 again - this allows solving for the coordinate time as function of proper time, but it's a rather messy function and this is the place where most of the opportunity for error occurs. It's also the place where the blowup occurs - the coordinate time goes to infinity at the event horizon.

Converting to Finklestein coordinates solves the problem, it subtracts out the terms which "blow up".

The finkelstein coordinate v is

v := t + r + 2*m*ln|(r/(2*m)-1|);

and it's well behaved. v as a function of tau gives the answer to the redshift/blueshift question directly - if it were a linear relationship, then there would be no redshift/blueshift. If one envisions a transmitter at infinity transmitting pulses once a second, every unit increase in 'v' represents the receipt of a pulse. I find that v is less than tau, hence there is redshift, though the effect appears to be on the order of a 2:1 redshift at the Schwarzschild radius (dv/dtau = .5 there - unless I've made an error, which is unfortunately all too possible).

While the detailed equations (which is the best way to tackle the problem IMO) are messy, looking at it from the POV of the infalling observer it's fairly easy to see that there should be a redshift. The infalling observer will not feel gravity, only tidal forces. These tidal forces will cause light falling from infinity to be red-shfited.

Last edited: Apr 13, 2005
11. Apr 14, 2005

### Dragongod

I am obviously out of my league on this topic, at least to certain extent I believe, but isn't the event horizon, as Stephen Hawking defines it, the point at which matter gets infinitly condensed. If thats true, which I am currently pondering, then how could you pass the event horizon. Wouldn't you stay inside it until the black hole died out.

12. Apr 14, 2005

### Phobos

Staff Emeritus
Nope...it's the distance around the singularity where the escape velocity is equal to the speed of light. The singularity is the point with infinite density.

13. Apr 14, 2005

### DaveC426913

Yep. You're corect. I was oversimplifying.

14. Apr 14, 2005

### Dragongod

Wait phobos I don't understand. You are agreeing with me when you say the singularity is the point of infinite density. If thats true then how can matter, once its reaches the event horizon and becomes part of that infinite density, go past it???

15. Apr 14, 2005

### chroot

Staff Emeritus
Dragongod,

Read Phobos' post more carefully. The "singularity" and the "event horizon" are not the same thing.

- Warren

16. Apr 14, 2005

### Dragongod

OHHHHHHHH. ok thanx. so what you are telling me is that the event horizon is the distance around the singularity where the escape velocity is equal to the speed of light. and the singularity is exactly how i defined it. Ok cool thanx.

There is still a problem however. U guys say its possible to pass the event horizon if you have infinite speed but the speed of light, according to GR and Einstein, is already infinite and therefore you would need something faster than infinity, which doesn't make sense, to break free and go past the event horizon.

There is something else I don't understand, how can people call the speed of light infinite if we can quantize it (give it a numerical value) such as it goes around the world 7 times in 1second. If it were infinite wouldn't we not be able to say how fast it is and just say it's as fast as present time, meaning instantaneous.

17. Apr 14, 2005

### chroot

Staff Emeritus
No one said that. It's not possible to have "infinite speed" in the first place.
The speed of light is not infinite. It is approximately 300,000 km per second.
No one says it's infinite.

- Warren

18. Apr 14, 2005

### Dragongod

wow i have heard some alot of people call the speed of light infinite but if what you say is correct thats great I thought society was behind me on this issue but i guess not. I guess everyone i heard was wrong. Thanx alot that clears a lot of stuff up :)

19. Apr 14, 2005

### chroot

Staff Emeritus
If you're not sure what an "event horizon" is, you can rest assured you're not ahead of society. Stay in school, you might learn something.

- Warren

20. Apr 14, 2005

### Dragongod

That makes no sense buddy. Intelligence isn't measured by how much LABELS you memorize in school or wherever. It's measured by how well you can reason and think logically. If I don't know the LABEL "event horizon" that doesn't mean I can't reason or think better than someone who does. All it means is that either that i heard wrong, misinterpreted someone else, just didn't hear about it, etc.