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## Black Holes: Infalling Observers and BH Evaporation

An observer in free fall does not experience gravitational time dilation, as I recall. I am entirely open to correction on this point, if in error.

 Quote by phinds So the fact that you are traveling near c is irrelevant to the spaghettification? I guess that's the part I got confused on.
I'm pretty sure that no, traveling near c is unrelated to spaghettification.

I'm not 100%, ofc, but I don't see how a relativistic velocity can lead to tidal forces. If the traveler is accelerating to c... but then it would need to have different parts of his or her body accelerating at different rates, right? But since we're talking about a free-falling person falling into a supermassive blackhole, I don't think that that will happen, according to my understanding, at least.
 Recognitions: Gold Member So back to the original topic. How long would it take for an infalling observer to reach the event horizon of a black hole? According to an outside observer, they would never reach it. But they would also disappear and never reappear as the black hole would evaporate over time and leave nothing behind. Am I correct?

 Quote by Drakkith So back to the original topic. How long would it take for an infalling observer to reach the event horizon of a black hole?
Using whose clocks? They don't call it relativity for nothing..... :-) :-) :-)

 According to an outside observer, they would never reach it.
An outsider stationary observer would never see the object cross the event horizon.

 But they would also disappear and never reappear as the black hole would evaporate over time and leave nothing behind. Am I correct?
Something like that.

But....

There are some theories of quantum gravity in which the Hawking radiation would be affected by the object crossing the event horizon. This is one solution to the black hole information paradox.

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 Quote by twofish-quant Using whose clocks? They don't call it relativity for nothing..... :-) :-) :-)
Calculated in the reference frame of the stationary outside observer?

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 Quote by Drakkith Assuming that black holes do in fact evaporate via Hawking Radiation, how can an infalling observer ever get inside the event horizon if the black hole evaporates in a finite time from an outside observers frame?
Consider two observers, observer A that falls across the the event horizon and observer B that hovers at a finite "distance" above the event horizon, and two types of (uncharged) spherical black holes, a classical black hole that doesn't emit Hawking radiation and a semi-classical black hole that does.

For the classical black hole case, B "sees" A on the event horizon at infinite future time, and B never sees the singularity.

For the semi-classical black hole case, at some *finite* time B simultaneously "sees": A on the event horizon; the singularity. In other words, the singularity becomes naked, and A winks out of existence at some finite time in the future for B.

In both cases, A crosses the event horizon, remains inside the event horizon, and hits the singularity. In both cases, B, does not see (even at infinite future time) A inside the event horizon, as this view is blocked by the singularity.

These conclusions can be deduced from Penrose diagrams, FIGURE 5.17 and FIGURE 9.3 in Carroll's text, and Fig. 12.2 and Fig, 14.4 in Wald's text, or

http://motls.blogspot.ca/2008/11/why...nto-black.html.

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This is one of my favorite explanations fromearlier discussions in these forums. I'm guessing this is the 'classical' version as described by George above...
[unsure which 'expert' originally posted this]

 Won't it take forever for you to fall in? Won't it take forever for the black hole to even form? Not in any useful sense. The time I experience before I hit the event horizon, and even until I hit the singularity-- the "proper time" calculated by using Schwarzschild's metric on my worldline-- is finite. The same goes for the collapsing star; if I somehow stood on the surface of the star as it became a black hole, I would experience the star's demise in a finite time. On my worldline as I fall into the black hole, it turns out that the Schwarzschild coordinate called t goes to infinity when I go through the event horizon. That doesn't correspond to anyone's proper time, though; it's just a coordinate called t. In fact, inside the event horizon, t is actually a spatial direction, and the future corresponds instead to decreasing r. It's only outside the black hole that t even points in a direction of increasing time. In any case, this doesn't indicate that I take forever to fall in, since the proper time involved is actually finite.

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 Quote by Chronos An observer in free fall does not experience gravitational time dilation, as I recall. I am entirely open to correction on this point, if in error.
You can say that if you also note that they don't experience speed related time dilation relative to a distant observer in the 'normal' way. That is, their effective relative speed compared to a distant observer should not be considered the same as their speed relative to a local stationary observer. To compare relative speed to a distant observer, pick some (non-unique) path over which to parallel transport their 4-velocity. Then you find that the transport through the gravitational well [of distant observer's 4-velocity] gives them (free faller) much smaller relative velocity compared to distant observer than compared to a nearby stationary observer.

Alternatively, for a simple geometry like SC, you can factor effects on a free faller into speed time dilation (symmetric) relative to a local stationary observer, and gravitational redshift/blueshift relative to a distant observer.

Either way, the upshot is that a free faller crossing the event horizon has only modest time difference compared to infinity [at the moment of crossing the horizon, from free faller's point of view], with the exact result depending on the free fall trajectory (that is, whether the fall is 'from infinity' or from a closer start).

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 Quote by Naty1 [unsure which 'expert' originally posted this]
Matt McIrvin on John Baez's website,

http://math.ucr.edu/home/baez/physic...s/fall_in.html.

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Here is another perspective...but I don't like the word 'illusion' as it implies to me something faulty with that distant perspective when it is as valid as any other.

From Kip Thorne in BLACK HOLES AND TIME WARPS

when the star forms a black hole:
 Finkelstein's reference frame was large enough to describe the star's implosion ...simultaneously from the viewpoint of far away static observers and from the viewpoint of observers who ride inward with the imploding star. The resulting description reconciled...the freezing of the implosion as observed from far away with (in contrast to) the continued implosion as observed from the stars surface....an imploding star really does shrink through the critical circumference without hesitation....That it appears to freeze as seen from far away is an illusion....General relativity insists that the star's matter will be crunched out of existence in the singularity at the center of the black...
and a related description [source unknown]

 One often sees people interested in the question "where is the infalling probe "now"". For instance, they want to know if the probe has crossed the horizon "now" yet, or not. The best answer to this question is the same as it was in special relativity - there is no universal notion of "now" - the question is ambiguous. It may be slightly annoying to attempt to think of everything in terms of the raw data that one will actually receive (such as curves of redshift vs time), but this is really the safest course. Thinking of things in terms of "where the probe is now" will inevitably lead to confusion, because there is no universal definition of what "now" means, different observers will regard different points as being simultaneous even in SR, and this does not change in GR.
Here is another perspective [source unknown] :

 ..... the Schwarzschild metric has a coordinate singularity at the event horizon, where the coordinate time becomes infinite. Recall that the coordinate time is approximately equal to the far away observer's proper time. However, a calculation using transformed coordinates shows that the infalling observer falls right through the event horizon in a finite amount of time (the infalling observer's proper time). How can we interpret solutions in which the proper time of one observer approaches infinity yet the proper time of another observer is finite? The best physical interpretation is that, although we can never actually see someone fall through the event horizon (due to the infinite redshift), he really does. As the free-falling observer passes across the event horizon, any inward directed photons emitted by him continue inward toward the center of the black hole. Any outward directed photons emitted by him at the instant he passes across the event horizon are forever frozen there. So, the outside observer cannot detect any of these photons, whether directed inward or outward. There's no coordinate-independent way to define the time dilation at various distances from the horizon—a clock is ticking relative to coordinate time, so even if that rate approaches zero in Schwarzschild coordinates which are the most common ones to use for a nonrotating black hole, in a different coordinate system like Kruskal-Szekeres coordinates it wouldn't approach zero at the horizon,

I can't find it, but "As the free-falling observer passes across the event horizon.." Leonard Susskind has explained the 'information' of the infalling object/observer gets 'smeared' across the horizon...so I continue to wonder if one could assume an image of the object remains on the horizon for that distant observer...while the actual infalling object/ observer continues inward, uninterrupted, in his own proper time.

Ah well, time to go and walk my Yorkies!!
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 Quote by Naty1 Here is another perspective...but I don't like the word 'illusion' as it implies to me something faulty with that distant perspective when it is as valid as any other.
Yes. That bothers me too.

Maybe "perspective" or "vantage point" would be better term,

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 This is because the infaller approaches the speed of light as the event horizon is approached making it increasingly difficult for external photons to 'catch up' with the infaller.
I believe Chronos explains this by noting that the horizon can be viewed as a light hypersurface....which is moving at lightspeed...I don't fully understand that perspective that but he's seem right about everything else.

edit: nope, its pervect:
"There's no such thing as a stationary clock at the event horizon..... any clock crossing the event horizon must be moving at the speed of light - or rather, since the event horizon can be thought of as trapped light, any physical infalling clock, which is stationary in its own frame, will see the event horizon approaching it at the speed of light."

One thing I do understand: Approaching a big BH from the exterior is no different than approaching a big dense planet...except, I guess, the BH is, well, black....the gravity itself [gravitational potential] is strong up close, but the gravitational potential gradient [the curvature of tidal force spaghettification] is nothing unusual. In other words, the gravitational gradient becomes extreme at the singularity not at the horizon; apparently the only 'unusual' thing at the horizon is a Schwarszchild coordinate ['fictitous'] singularity in time....so things appear to slow down from a stationary distant frame, but locally to a free falling observer things all seem 'normal' and no horizon can even be detected by such an soberver.

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 Quote by Naty1 I believe Chronos explains this by noting that the horizon can be viewed as a light hypersurface....which is moving at lightspeed...I don't fully understand that perspective that but he's seem right about everything else. One thing I do understand: Approaching a big BH from the exterior is no different than approaching a big dense planet...except, I guess, the BH is, well, black....the gravity itself [gravitational potential] is strong up close, but the gravitational potential gradient [the curvature of tidal force spaghettification] is nothing unusual. In other words, the gravitational gradient becomes extreme at the singularity not at the horizon; apparently the only 'unusual' thing at the horizon is a Schwarszchild coordinate ['fictitous'] singularity in time....so things appear to slow down from a stationary distant frame, but locally to a free falling observer things all seem 'normal' and no horizon can even be detected by such an soberver.
Well, I disputed this statement of Chronos, and stand by my disputation. From the point of view of the free faller, light from distant sources is not highly redshifted, and distant clocks do not appear to run very slow. On the other hand, the distant observer does see light from the infaller extremely redshifted and their clocks run slow then stop. I provided two different explanations of these facts.

The infaller continues to receive light from distant sources, with no difficulty, until catastrophe at the singularity.

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Hey PAllen....

 This is because the infaller approaches the speed of light as the event horizon is approached making it increasingly difficult for external photons to 'catch up' with the infaller. I believe Chronos explains this by noting that the horizon can be viewed as a light hypersurface....which is moving at lightspeed...I don't fully understand that perspective that but he's seem right about everything else.
PAllen
 Well, I disputed this statement of Chronos, and stand by my disputation.
Disputation!!! Cool [LOL]

Actually we agree. I was NOT trying to sneak in a 'last word' contrary view in the vain hope you would not catch me!!!

It took me a few moments to see my error: I should have quoted simply this from Chronos:

 This is because the infaller approaches the speed of light as the event horizon is approached....
because I thought he might be adopting a perspective relative to the event horizon....I was only wondering about looking inward toward the black hole...... I have never quite understood that perspective. I figure I am missing something if both he and pervect have adopted that 'frame' [bad word I know] for some reason I still do not get....

Anyway, your posted point that light from the distant cosmos is NOT radically redshifted I have read multiple times and even posted quotes supporting that view elsewhere from Kip Thorne and maybe Brian Greene. So you are in good company!! Cheers.