# Passing through event horizon

1. Jul 3, 2013

### ShamelessGit

Wikipedia says, "Likewise, any object approaching the horizon from the observer's side appears to slow down and never quite pass through the horizon, with its image becoming more and more redshifted as time elapses." So does this mean from the outside observer's perspective, that nothing ever actually falls into a black hole? Isn't this kind of strange? That would seem to mean that nothing (from our perspective) could ever actually fall into a black hole since its creation, it would just get stuck on the outside of the hole. But when things fall in they seem to disappear, don't they? Does it just redshift outside the range of any measurements? And from the perspective of the thing falling in, it just falls in and doesn't notice anything, right? But since it doesn't fall in from our perspective, does that mean that the inside of the event horizon has already experienced an infinite amount of time?

2. Jul 3, 2013

### ShamelessGit

If it is true that the thing never actually falls into the black hole, then wouldn't the correct model for a black hole that has eaten things since its creation, from an observer's perspective, be a singularity at the center plus a hollow shell at the event horizon of the things it has eaten?

3. Jul 3, 2013

### WannabeNewton

In order to "see" the observer falling in, light incident on the infalling observer would have to reach our eyes or the infalling observer would have to send some kind of signal but these signals/light will get infinitely redshifted as the observer approaches the EH so the observer would get dimmer and dimmer and dimmer. Needless to say, the infalling observer will still pass through the EH in finite proper time, keyword being proper. This is a frame invariant quantity along the worldline of the infalling observer and it will be a finite value when the observer's e.g. purely radial free fall path from infinity to the EH itself is integrated along. I don't understand your last question, the above is by definition for eternal Schwarzschild black holes.

Things certainly do fall into black holes.

Last edited: Jul 3, 2013
4. Jul 3, 2013

5. Jul 3, 2013

### ShamelessGit

That other topic is NOT the same as this one. It gets frustrating when it seems like 2/3 of the people who answer my question don't understand what I'm talking about.

Wikipedia says that matter falling into the event horizon approaches, but never crosses the event horizon. This directly implies that observers outside the event horizon can never see anything cross the event horizon, even if the light wasn't red shifted. What is there not to understand?

6. Jul 3, 2013

### Staff: Mentor

We understand what you're talking about; but after the four thousandth or so thread asking the same question, our answers can be a bit clipped sometimes.

Wikipedia is not a good starting point for actually learning about physics. Some Wiki pages are OK, but the only way to know which ones they are is to already know the subject matter well enough.

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

7. Jul 4, 2013

### Staff: Mentor

When more than half of the people who answer don't understand what you are talking about then it is a pretty good indication that you didn't ask your question clearly.

Here are some other related threads:

The basic idea is that the idea that it takes an infinite amount of time to reach the EH is simply an artifact of a particular choice of coordinates. The coordinates are not themselves physical.

Last edited: Jul 4, 2013
8. Jul 4, 2013

### Naty1

It took scientists some time to figure out what BH solutions to Einsteins equations meant....Einstein himself did not believe for some time that BH were 'physical'....

Two explanations of some mathematical complexities one doesn't normally see in short BH
descriptions: [To summarize a BH event horizon is a frame dependent measurement [i.e., relative]. Some see it, some don't.]

PAllen:

The following may be from Jambaugh of these forums:

Here is some detail on 'coordinates' described in summary by Dalespam:

From Kip Thorne in BLACK HOLES AND TIME WARPS

[Torne here describes the Eddington-Finklestein coordinate system which shows the apparent singularity at the Schwarzschild radius is only a coordinate singularity not a true physical singularity.]

Finkelsteinâ€™s Reference Frame

when the star forms a black hole:

http://www.jimhaldenwang.com/black_hole.htm
(separate file: Introduction to Mathematical Einstein)

9. Jul 4, 2013

### Markus Hanke

You appear to be making a tacit assumption that all observers agree on what happens here. However, that is not the case; a far-away observer will never see anything fall across the horizon, while an observer falling in together with the object will not notice anything special at all ( for him the singularity is reached in a finite, well-defined time ) when he crosses the EH. This is perfectly fine - both are right, but only in their own frames of reference. In GR, not all observers necessarily agree on physical outcomes as viewed from their own frames of reference.

10. Jul 4, 2013

### maxverywell

As have been said, objects certainly fall into black holes and they don't notice anything special while crossing the event horizon (in the case of massive black hole, so that the tidal forces are negligible at EH). But for an static observer outside the BH, the infalling objects don't cross the EH, he will see them approaching it asymptotically forever (but because of the redshift, the light from them will eventually diminish in finite time).

Think it differently, in terms of the gravitational time dilation. For you, as an observer far away from the black hole, a clock near the event horizon will tick in much much slower rate than your clock. And the things there become slower and slower as that clock moves towards the event horizon, and eventually at the event horizon the time dilation is infinite. So the signals from there, sent in a constant rate according to their clock, won't be received in a constant rate from us. Each signal will be received later and later according to our clock, and finally the last signal from the event horizon will be received in infinite time...

Last edited: Jul 4, 2013
11. Jul 4, 2013

### robphy

Last edited: Jul 4, 2013
12. Jul 4, 2013

### Staff: Mentor

This is not correct. Actual outcomes, meaning observables like the amount of proper time elapsed on a particular clock, are the same in all reference frames that cover the appropriate portion of spacetime.

That last part in italics is the key: the reference frame (a better term would be "coordinate chart") in which an observer outside the horizon is at rest *cannot* cover the horizon or anything inside it. But any reference frame that does cover the horizon and the region inside will agree on what happens there.

13. Jul 4, 2013

### Staff: Mentor

This isn't strictly correct, because the statements you are making, strictly speaking, only apply to clocks that are static--i.e., they stay at the same radius above the horizon. A clock can't be static at or inside the horizon. And the time dilation you are talking about, strictly speaking, can't be defined for clocks that are not at rest relative to each other, so it doesn't apply to clocks that are falling inward, even if they are outside the horizon.

14. Jul 4, 2013

### maxverywell

Can't we relate the time interval $\Delta \tau_{AB}=\tau_A-\tau_B$ between the events of emission of two signals A and B, as measured by a freely falling observer outside the EH of a Schwarzschild BH ($\tau$ is his proper time), to the time interval $\Delta t$of an static observer at infinity(his proper time is the Schwarzschild coordinate $t$)?

15. Jul 4, 2013

### Staff: Mentor

Sure - they're numbers, we can define a mapping between them. But what physical significance does this relationship have?

16. Jul 4, 2013

### WannabeNewton

You might be interested in this related exercise from Padmanabhan's text:

17. Jul 4, 2013

### maxverywell

Here's a nice diagram from Carroll's notes in GR.
It clearly demonstrates my point.

https://www.physicsforums.com/attachment.php?attachmentid=60078&d=1372967470

What I asked is if it's possible to find a relation between $\Delta \tau_1$ and the $\Delta \tau_2$. The problem here is that one observer is infalling. I know how to find the relation for the case when both observers are static.

18. Jul 4, 2013

### Naty1

Shamless...by now you should be pretty confused!! Every time I read one of these BH thingys I pick up a tad more....especially from thus guy....

PeterDonis posts:
That's a 'crazy' statement, but is an appropriate example of 'some mathematical complexities' I mentioned in a post above. It reminds me of one thing I have learned: Unless you are a local inertial observer, things get weird. Looking at things from an 'infinite distance' or accelerating outside an event horizon to remain 'stationary' can fortell odd observations.

Think as a crude example, about hearing thunder. Say you are two miles away and a buddy is one mile away from the lightning strike. You'll never resolve your differences in recording WHEN the event happened unless you adjust for the travel time of sound. In SR, GR and cosmology, each gets successively more detailed trying to describe distance, speed, even energy.

'At rest' in PeterDonis description above means accelerating to remain in place. Such an observer will generally see something very different from an adjacent [local] free falling observer.

19. Jul 4, 2013

### maxverywell

Actually I think that the light rays (dashed curves) in this diagram are drawn incorrectly.
I think that the $\Delta t$ between two consequent dashed curves should be constant.

Last edited: Jul 5, 2013
20. Jul 4, 2013

### pervect

Staff Emeritus
Red shfiting is NOT logically sepearate from the ratio of "time of arrival" to "time of emission". In fact, redshift is the ratio of the change time of emission to $\Delta \tau_{emission}$ to the change in time of arrival $\Delta \tau_{arrival}$, where the $\tau$ symbol represents proper time.

I haven't read the previous discussion, because after 5,000 times of going through this, I do get a bit tired of it. But I thought I'd point out how a discussion about redshfit actually WAS answering your question , even if you didn't quite realize it.

I am concerned that you might not know what I mean when I say proper time. Is the term familiar to you, and do you understand the difference between proper time and coordinate time? If not, it would be a good place to start understanding what people have been saying.

I'll say one more thing. If an observer is at rest a long way away from a black hole, the proper time (there's that word again!) before they reach the event horizion can be calculated, and is finite. The Schwarzschild time, which is a coordinate time assigned to the event of "falling into the black hole" is infinite. The proper time according to an infalling observer is not.

Saying that "the infalling observer never crosses the event horizon" is a bad idea, because it suggests that both times should be infinite. And, they're not. The proper time is finite.

Now that I've hopefully established some relevance of why you should care about the differene between proper time and cooridnate time, I'll try to give a (very brief) sketch of the difference between them.

Coordinate times simply assign a number, a label, to an event. In General Relativity, the laws by which this assignment made are very lax, because of this laxness the assignement of coordinate times dont have any real physical significance. Coordinate time can (and often does) advance at a different rate than the time which actual clocks keep. In particular Schwarzschild time is a coordinate time, and changes in Schwarzschild time aren't the same time as what actual clocks keep.

The time which actual clocks keep is called proper time. The clock must be directly present at the same (or nearly the same) spatial location of all events being measured in order for proper time to be applicable. If the clock was not present at all events, one is actually using a coordinate time, and not proper time.

21. Jul 5, 2013

### Markus Hanke

True, bad choice of words on my part. What I meant to say was that the observers don't agree in that the infalling one sees himself crossing the horizon, whereas the one stationary at infinity does not.

22. Jul 5, 2013

### maxverywell

The diagram I was talking about:

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23. Jul 5, 2013

### Staff: Mentor

No, it isn't, because the later light rays are emitted from a smaller radius (closer to the horizon) than the earlier ones. So it takes them longer to get out to the distant observer even according to the $t$ coordinate.