# External view of a black hole

1. Jul 25, 2011

### atomthick

Hi all,

Let's say we can witness a star going inside a black hole. From our perspective the star will approach the black hole but we will never see it reach the event horizon (due to the time contraction) instead it will remain motionless somewhere near it.

The questions are:

Can we really see the star for ever trying to approach the black hole? If so how can it emit light for that long, where does the energy come from?
Also, if we can see the star for ever trying to approach the black hole then why isn't the black hole visible due to all the stars that are trying to enter it?

Thank you.

2. Jul 25, 2011

### Ben Niehoff

The light from objects falling into a black hole gets more and more red-shifted as they fall (as seen by a distant observer). Eventually the redshift becomes infinite as the object approaches the horizon. So all of the radiation becomes undetectable (because any real detection device has some lower limit on the frequency it can detect).

3. Jul 25, 2011

### atomthick

By red-shifting the amount of energy decreases therefore the star can send light for a very long time without violating the energy conservation law.

Thanks!

4. Jul 25, 2011

### Staff: Mentor

What do you mean Atomthick? Are you saying that the red shift of light allows the star to "Exist" at the horizon for a near-infinite time without breaking conservation laws?
Remember that unless this is an absolutely astoundingly massive BH the the star will be ripped apart long before it reaches the horizon and will not exist as a star anymore.

5. Jul 25, 2011

### atomthick

Drakkith, you are right the star will get pooled inside the BH piece by piece but I was imagining this will take a while if viewed from outside due to time distortions.

6. Jul 25, 2011

### Naty1

By that logic we would be observing stars not black holes....yet that is not what is observed.
THAT is a question I have NOT figured out yet...It may be that we are not in an appropriate frame....

Incorrect....there is NOT infinite gravity at the black hole horizon...The horizon is NOT the singularity despite the incorrect name "Schwarschild singularity" which existed for many years.....for a massive black hole gravity is not all that strong....time dilation IS infinite....an observer, say a probe, can easily exist and continue to send information back to the outside universe until it passes thru the horizon.

The following description covers the formation of a black hole by a collapsing star.

The key to understanding this is relativity...different frames see different things...analogous to length contraction and as mentioned above and time dilation. This was NOT well understood until David Finkelstein unraveled the proper frame (observational prspective) to explain Schwarzschild's spacetime geometry.

The best description I have come across is Kip Thorne's in BLACK HOLES AND TIME WARPS

How can this be?? It seems crazy!!!! (It is!!!)

Example# 1:

Think about a near light speed observer looking at a clock tick off time back on earth,,,it seems almost stopped.....but not to the person on earth....nothing is unusual in the appropriate local frame.

Example #2:

If you go too fast do you become a black hole??

misconception: "When an object approaches the speed of light, its mass increases without limit, and its length contracts towards zero. Thus its density increases without limit."

But that is in the frame of a static observer ...in the frame of the "fast" moving individual everything appears normal in the local frame....

Last edited: Jul 25, 2011
7. Jul 25, 2011

### Staff: Mentor

You misunderstand me. The pull from a stellar mass black hole would rip material from the star very quickly as it approached. This material would be pulled down into the black hole or would be made part of its accretion disk. Hence, the STAR wouldn't exist anymore since its composite material is now spread out everywhere or falling into the BH.

8. Jul 25, 2011

### CJames

Another way to think about this is to realize that when time is slowing down, EVERYTHING slows down. If every peak or trough in an EM wave corresponds to the tick of a clock, then the fact that time slows down also means that the image HAS to be redshifted. Assuming that there aren't any frame dragging effects, you can actually say that the redshift tells you exactly how much slower their clock is running. You could even say that gravitational redshift is the REASON time is slowing down.

9. Jul 25, 2011

### pervect

Staff Emeritus
Remember (assuming you've learned it - if you haven't learned it yet, learn it first, then remember it):

Simultaneity is relative.

The "near infinite amount of time" that an object "exists" (I think is present at would be a better, less philosophical term) at the event horizon is only coordinate time.

In terms of proper time, the time a clock on the object itself measures, the time at the event horizon is finite.

In addition, the tidal forces measued by the object at the event horizon are finite, so an object is not necessarily ripped apart before it crosses the horizon. (The object will eventually be ripped apart later, at the central singularity).

Event horizons exist in General relativity in a context other than black holes, and they are perhaps easier to understand there. For instance, an accelerating rocket has a "Rindler horizon", which exhibits much the same properties. The Rindler horizon occurs when a rocket gets far enough away that light is not able to catch up to it. The rocket is always moving slower than the light, but due to the nature of it's headstart, light will never catch up to it unless the rocket stops accelerating.

We could have the Earth fall behind the Rindler horizon of such an accelerating rocket. We wouldn't notice a thing, but the rocket would see us apparently frozen in space, approaching but never reaching its Rindler horizon. We would notice when we crossed the rocket's "Rindler horizon, but only in the sense that we'd compute that we couldn't send signals to the rocket anymore. Nothing dramatic would happen to our clocks, or anything else, life would go on as normal except for not being able to send signals to the rocet anymore.

We wouldn't have any doubt that we actually had actually reached the Rindler horizon, and passed it, either, and we would assign a finite time to the event (say noon, in the year 2200), even though the rocket assigned an infinite coordinate time to the event.

10. Jul 25, 2011

### CJames

I've got a related question: Does a black hole form before it evaporates?

From the perspective of an external reference frame, the star should take an infinite amount of time to collapse. And yet, according to Stephen Hawking, it should evaporate in a finite amount of time.

Is it possible for these two events to be causally reversed from the perspective of the proper time reference frame? The star collapses before it evaporates?

These two events (the formation of the black hole and its evaporation) aren't spacelike separated, are they?

11. Jul 25, 2011

### George Jones

Staff Emeritus
We have been through this before.

The standard opinion:

12. Jul 25, 2011

### CJames

Last edited: Jul 25, 2011
13. Jul 26, 2011

### Naty1

No, That post is simply wrong.

I guess some surfaces gases could get pulled off depending on the trajectory of the star, but all different masses are attracted with equal acceleration....so there is nothing to "rip apart" much of anything OUTSIDE the horizon, very distant from the central singularity. What rips apart things are tidal forces and curvature as the singularity is approached. A free falling obsever, for example, has no way of even knowing she has passed the event horizon...nothing happens to her...

Pervect says:
which is correct...and also describes the attractive gravitational forces outside the horizon. For one thing, the bigger the black hole, the weaker the gravity just outside it's (much larger) horizon.

Last edited: Jul 26, 2011
14. Jul 26, 2011

### Staff: Mentor

Only if the black hole is sufficiently massive enough. Otherwise the difference in attraction between points in the object is enough to "rip" it apart. I don't see how this is even debateable. A similar situation is easily seen by material being leeched from a stellar companion to a White Dwarf star, only the effect can be magnified since a black hole has more mass and has no "surface" for the material to impact on.

I assume you mean the gradiant is less extreme for a larger black hole?

15. Jul 26, 2011

### WannabeNewton

The surface gravity of the black hole $\kappa = \frac{1}{4M}$ decreases for larger black holes.

EDIT: This expression for surface gravity is only for schwarzchild black holes but the decrease in surface gravity (ill - defined as the term may be for black holes) for larger black holes also applies to rotating ones.

16. Jul 26, 2011

### Staff: Mentor

I am not disagreeing with that.

17. Jul 27, 2011

### stevebd1

I may be stating the obvious but $\kappa$ is normally referred to as the Killing surface gravity and is the surface gravity as observed from infinity for a static black hole (it's basically derived from Newtonian gravity $g=Gm/r^2$ where $r=r_s=2Gm/c^2$ and expressed in geometric units). Taking into account relativistic effects, the local gravity would be $g=Gm/r^2\cdot 1/\sqrt(1-r/r_s)$. Interestingly, the equation for tidal forces $dg=L \cdot 2Gm/r^3$ remains unchanged. The equation for the Killing surface gravity taking into account spin and charge is-

$$\kappa_\pm=\frac{r_\pm-r_\mp}{2(r_\pm^2+a^2)}$$

where $r_\pm=M \pm \sqrt(M^2-a^2-Q^2)$ where M=Gm/c2, a=Jm/c and $Q=C\sqrt(G k_e)/c^2$, $\kappa_\pm$ reducing to the static solution when a=Q=0.

Last edited: Jul 27, 2011
18. Jul 27, 2011

### WannabeNewton

Could you give the derivation from Newtonian Gravity per chance? I always thought it was simply the result of $\xi ^{\beta }\triangledown _{\beta }\xi ^{\alpha } = \kappa \xi ^{\alpha }$ for the time - like killing field $\boldsymbol{\xi }$.

19. Jul 27, 2011

### stevebd1

Newtonian equation for gravity-

$$g=\frac{Gm}{r^2}$$

substitute r with the Schwarzschild radius (the 'surface' of the black hole)-

$$g=\frac{Gm}{\left(\frac{2Gm}{c^2}\right)^2}=\frac{c^4}{4Gm}$$

multiply by the geometric units for force- G/c4 and you get 1/(4M). Alternatively, you can multiply by the geometric units for mass- G/c2 to get c2/(4M) then multiply by the geometric units for acceleration- 1/c2.

Last edited: Jul 27, 2011
20. Jul 31, 2011

### Zentrails

Remind me of a black hole that "we have observed" in any frame you would consider appropriate.

On the other hand we have observed PLENTY of stars that appear to be orbiting about the common center of gravity with a black hole and the intensity of the light from the star changes as one would expect if it were going "behind" a black hole.