# Time inside a black hole?

1. May 21, 2014

### Sorthal

Is there time inside of a black hole?

2. May 21, 2014

Actually, if you could survive a trip into a black hole (which you cannot) you would not be aware of any slowing down of any clock you carried as you fell in. However, if you could compare the speed of your clock with that of a reference clock kept far away, then then the clock falling into the black hole would appear to slow down relative to the clock far from the hole. The place where the falling in clock would appear to 'stop' is the spherical surface called the event horizon. This is also regarded as the boundary of the black hole, since nothing crossing this surface can escape.

Last edited: May 21, 2014
3. May 21, 2014

### bapowell

Notions of time and space reverse inside the black hole. Observers inside the black hole move through space towards the singularity as inevitably as observers outside the black hole move through time towards the future.

4. May 21, 2014

### DrGreg

That's not really true. If you use Schwarzschild coordinates then inside the event horizon the time coordinate is called "r" and the space coordinate is called "t", but there are other coordinate systems (e.g. Kruskal coordinates) where there is no such "reversal".

Observers inside the black hole move through time towards the singularity (even though in Schwarzschild coordinates the time is labelled "r".)

5. May 21, 2014

### dauto

One important point is that the singularity is space-like

6. May 28, 2014

### ABunyip

...but never reach it (from the point of view of observers outside the BH). Once an object reaches the EH it continues to accelerate into the (our) future, but observed from outside, time starts slowing down until it pretty well stops altogether for objects at the singularity. Eventually everything that entered the BH will be evaporated out as Hawking Radiation zillions of years later. When the BH is so depleted that it finally allows light to escape it will most likely, IMHO, rapidly release the last remnants of its energy in a huge protracted pulse of radiation.

So the answer is yes, and no.

7. May 28, 2014

### bapowell

Almost: to an observer outside, time dilation tends to infinity at the event horizon. So to outside observers, objects falling towards the black hole appear to come to rest asymptotically at the event horizon. The inside of the black hole is of course invisible to outside observers.

It's not clear that things that fall into the black hole are reconstituted as Hawking radiation -- this is the founding problem of the information paradox.

8. May 28, 2014

### ABunyip

I think of the EH as an optical effect - like a rainbow. Its just the point at which everything beyond it becomes invisible. As objects approach the EH they are already being thrust into the future and will fade away as they go through it. As they get closer to the singularity over time, they are accelerated more and more into the future and less and less toward the singularity. So I beg to differ on this small point, bapowell (pun intended).

9. May 28, 2014

### bapowell

This is physics, so it's really not a matter of opinion. It can be demonstrated objectively that objects appear to freeze at the event horizon to observers sufficiently far outside the black hole. If you beg to differ, please substantiate your claim formally.

10. May 28, 2014

### Staff: Mentor

The singularity is in the future, so "accelerating more into the future", to the extent that makes sense, is also accelerating more toward the singularity.

11. May 28, 2014

### Staff: Mentor

You're confusing the singularity with the horizon. Observers far away from the hole see light from infalling objects more and more redshifted (which can also be interpreted as "time slowing down" for those objects) as the objects approach the EH. No light from the EH or anywhere inside it can ever reach any observer outside the EH; so to observers outside the EH, an object disappears when it reaches the EH, and its entire future trajectory inside the EH is invisible to observers outside the EH.

We don't actually know this; as bapowell pointed out, whether or not things that enter the hole get evaporated out is one of the key questions involved in the BH information paradox, which has not been fully resolved.

12. May 28, 2014

### pervect

Staff Emeritus
I beg to differ in that at minimum the statement needs further qualification. Specifically, it is true that objects appear to freeze at the event horizon for static observers using the conventional notion of simultaneity outside the black hole. In essence, no radar signal could return to a static observer from the event horizon.

It is also true that objects falling into a black hole take a finite proper time to reach the event horizion (and even beyond, to the singularity). This proper time could be measured, for instance, by an atomic clock they carry with them. If you need a reference for this, I can provide one, but most any GR textbook will mention this. The reference I'd provide would be looking up the specific page in MTW's Gravitation that tells you this, which may or may not be useful, in that you can find it easily yourself if you already have or can borrow the textbook, and if you don't you have to take my word for it anyway.)

Non-static observers who are themselves falling into a black hole will not see the freeze effect. Basically, when they reach the event horizon, they'll get their radar return signal.

13. May 28, 2014

### ABunyip

What?!!?

14. May 28, 2014

### skeptic2

Both statements need further qualification. To observers sufficiently far outside the black hole, objects do freeze, not just appear to freeze, at the EH and every experiment that could be done from that distance would support that observation.

The notion that objects only appear to freeze comes from the assumption that the viewpoint of the infalling observer is somehow more legitimate than that of a distant observer. Both observations are equally valid. One cannot pick one observer, e.g. the infalling one, and claim that his perspective is reality and what the distant observer sees and measures is not.

15. May 28, 2014

### ABunyip

Yes.

16. May 28, 2014

### PAllen

A more accurate description is as follows. This is readily derived by following light rays in a Kruskal chart with Schwarzschild r and t lines represented.

Imagine a long ship heading toward an isolated (no accretion disc) supermassive BH. Imagine an observer hovering a good distance from the horizon, off somewhat to the side of the rocket's radial infall trajectory. For any detection technology available, there is a surface (just) outside the horizon such that any light reaching the hovering observer from this surface is red shifted to the point where it is unobservable - effectively as black as possible (e.g. much blacker than CMB).

Then, the hovering observe sees the following, in a time span very short compared to the life time of the universe:

The front of the ship vanishes at this surface, then as each part of the ship advances to this surface, it vanishes. There is no perception of the ship being pancaked or frozen. Instead, it would appear as if the ship were progressively vanishing as it went through a surface of invisibility. In vernacular, it would appear as if the ship smoothly and slowly passed into a hole in space.

17. May 29, 2014

### .Scott

[But that's not what happens at the event horizon, that's what happens when light become trapped within the photon sphere.][STRIKE]No. That is a complete misunderstanding of an event horizon.[/STRIKE]
There is an event horizon shrouding the black hole from the observer hovering far above that BH. If a clock on the ship passed through noon just as it passed through that black hole, the observer would see the ship approach the event horizon but never cross it. He would also see the clock approach closer and closer to noon, but never reach it.

Event horizons are not specific to black holes. If you place yourself in a continuously accelerating spacecraft, you would observe an event horizon develop behind you [and you have objects "collecting" on its surface].

In that sense, the event horizon is like a rainbow (as described above). If you try to visit the end of the rainbow - it moves back from you or disappears altogether.

http://en.wikipedia.org/wiki/Event_horizon

Last edited: May 29, 2014
18. May 29, 2014

### PAllen

No, what I said is correct. The event horizon itself is undetectable. The surface of last visibility is outside the event horizon. The same description would apply to a Rindler horizon. An extended object would be seen to finally disappear as each part reached the surface of last visibility. The process of vanishing would be slow, but finite for any given sensitivity of imaging device.

Last edited: May 29, 2014
19. May 29, 2014

### Staff: Mentor

Careful.... PAllen isn't describing the event horizon, he is describing the behavior of light emitted from a free-falling object very near to but still outside the horizon. That doesn't mean his description is correct, but if it's wrong it's wrong for some reason deeper than "complete misunderstanding".
(I suspect that you and he may be working with a different threshold for "effectively as black as possible")

20. May 29, 2014

### .Scott

Yes. That is what's happening. He introduced his remark as a "more accurate description", but then what he described was not what happens at the event horizon.

He put the observer "off somewhat to the side" so that the observer would not be able to track the traveler all the way to the horizon. Instead, at a certain point in the descent light from the traveler headed towards the observer becomes trapped in the photon sphere and the traveler appears to simply vanish.

Last edited: May 29, 2014
21. May 29, 2014

### Staff: Mentor

He can't be seen passing through the event horizon. But here PAllen is suggesting that there is a surface of last visibility outside the event horizon. Because it is outside, it can be reached in a finite amount of the observer's proper time; the infaller passes through this surface before time is "completely frozen".

22. May 29, 2014

### .Scott

Another way to look at that is that the event horizon is undetectable, but you can set things up to make an observation arbitrarily close to it.
I'm going to drop a matter-antimatter bomb into the black hole, set to detonate a meter above the event horizon. Assuming I can still detect the explosion, the extreme time dilation should be very apparent.

23. May 29, 2014

### .Scott

Yes, the photon sphere - or rather a sphere within the photon sphere at a distance based on how far "aside" the observer is.
I already edited the post you quoted - as well as the earlier one.

24. May 29, 2014

### PAllen

The extreme time dilation is certainly apparent. There is an easy way to ball park some quantities I only gave qualitatively. Imagine a hypothetical 1km rocket falling radially into supermassive BH, with the long dimension oriented radially. You are looking at this from far enough away to hover safely, and on a different radial line from the rocket so you can see its whole length. Let's say the rocket is luminescent, emitting bright blue light. Now, make some assumption about what is the longest wavelength you can possibly image. For example, pretend you can still image blue light red shifted by a factor of a million. There is a specific surface outside the horizon that corresponds to this redshift factor. A rocket in free fall starting from reasonably far away will experience (but not detect - there is no plausible way to do so) this surface cross them at near c. Thus, rocket clocks will experience crossing this surface of last visibility in about 10-5 seconds. The red shift factor of a million means a distant observer will see this crossing time as appx. 10 seconds.

Thus, the distant observer sees the 1km ship vanish as each portion crosses this surface, the whole process (from front crossing to back crossing the same surface) lasting about 10 seconds.

Make different assumptions, and you get different numbers, but the general picture does not change. You see the long ship crossing into invisibility, with each part becoming invisible per some declared device sensitivity as it reaches this surface (there will be progressive dimming as each part of the rocket approaches this surface) (the location of the surface is function of assumed imaging technology). It will look exactly like the rocket crossed into a 'void' in 10 seconds.

Note, that what I was arguing against is the frequent statement that a distant observer sees objects frozen at the horizon. This is simply a false statement, in any plausible sense of the term 'see'. What I describe is what would actually be seen (for an isolated supermassive BH).

Last edited: May 29, 2014
25. May 30, 2014

### skeptic2

Let's plug some numbers into PAllen's example above.

For a 10 solar mass BH the radius at which there would be a dilation factor (gamma) of 1 million would be about 2.95x10^-8 meters above the event horizon. Both the time and length of the space ship would be diminished by gamma as seen from a distance. So at about 30nm above the horizon the space ship would be 1mm long traveling at a velocity of approximately 300m/s (as seen from a distance). It would disappear rather quickly. In fact, the space ship will disappear in about 3.333uS.

If the space ship were luminescent in gamma rays we might actually be able to see it freeze just above the horizon.