# Black Holes: Infalling Observers and BH Evaporation

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## Main Question or Discussion Point

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?

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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?
Think of it as an optical illusion.

It **appears** from an outsider observation that the falling object never enters the black hole, that's because light from the infall takes an infinite amount of time to reach the outside observer.

Staff Emeritus
Think of it as an optical illusion.

It **appears** from an outsider observation that the falling object never enters the black hole, that's because light from the infall takes an infinite amount of time to reach the outside observer.
Does an infalling observer ever enter the black hole?

Does an infalling observer ever enter the black hole?
Yes, and they hit the singularity in finite time.

It's easiest to think of the apparent "freezing" of an object as it falls into a black hole as an "optical illusion". As something gets closer to the event horizon the time it takes for a light flash to make it to a distant observer increases. Once you cross the event horizon, the light never makes it out.

But if you try to flash the light to someone that is falling with you, then you'll see nothing unusual.

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Approximately how long would it take for an observer to cross the event horizon?

phinds
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Approximately how long would it take for an observer to cross the event horizon?
The length of his body times his speed. It's just d=rt. The in-faller feller is not aware of the EH.

A characteritic aspect of BH of any size is that they are not actually evaporating these days but rather accreting as they absorb matter and energy. They'll likely begin to decrease in size when their temperature exceeds the slowly cooling CMBR temperature....currently around 2.73 degrees K some billions of years from now. And even then big BH will evaporate very slowly as the Hawking radition in inversly proportional to their size and such evaporation will take billions upon billions of years additional time.

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phinds
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A characteritic aspect of BH of any size is that they are not actually evaporating these days but rather accreting as they absorb matter and energy. They'll likely begin to decrease in size when their temperature exceeds the CMBR temperature....around 2.73 degrees K some billions of years from now. And even then big BH will evaporate very slowly as the Hawking radition in inversly proportional to their size and such evaporation will take billions upon billions of years additional time.
While I agree w/ the end result of events as you describe them, I disagree that there is no Hawking radiation at present. What there IS is a whole lot LESS Hawking radiation than there is infalling stuff, including from the CMB so that the net result IS as though the BH is not radiating. I don't think this is just semantics.

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The length of his body times his speed. It's just d=rt. The in-faller feller is not aware of the EH.
Sorry, I mean according to an outside observer. If that even makes sense since we can't see him fall in.

While I agree w/ the end result of events as you describe them, I disagree that there is no Hawking radiation at present. What there IS is a whole lot LESS Hawking radiation than there is infalling stuff, including from the CMB so that the net result IS as though the BH is not radiating. I don't think this is just semantics.
I don't think Naty meant that there is no Hawking radiation present, only that the BH will not evaporate until billions of years from now.

phinds
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Sorry, I mean according to an outside observer. If that even makes sense since we can't see him fall in.
Ah ... I think the answer there is "forever" but that can't be quite right since it doesn't take forever for the BH to evaporate.

I don't think Naty meant that there is no Hawking radiation present, only that the BH will not evaporate until billions of years from now.
Yeah, I wasn't sure, thus my quibble.

Chronos
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Keep in mind the external universe is also time dilated from the infalling observers perspective. If the external observer ship sent a constant interval pulsed laser signal toward the hapless infalling volunteer, the time between pulses would increase as the infaller approached the event horizon. 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 don't think Naty meant that there is no Hawking radiation present, only that the BH will not evaporate until billions of years from now.
I edited my wording above for clarity reasons but DID mean a black hole doesn't radiate now. [edit: I am wrong see below.]

Does a black body cooler than it's surroundings radiate?? apparently so:

Evaporation of a black body is not dependent on surrounding temps:
[you can find some formulas there]
The power in the Hawking radiation from a solar mass () black hole turns out to be a minuscule 9 × 10−29 watts. It is indeed an extremely good approximation to call such an object 'black'.
I withdraw my assertion...apparently, it does radiate but we can't observe it because it's so small.

Obviously it has no effect on the original question.

Separately,

two fish:
Think of it as an optical illusion.
You can do that as a distant observer, but woe be to the free falling inbound observer who decides 'to stop and take a look' nearby outside the horizon....and accelerates there to remain stationary.....The horizon is 'VERY real' there and the observer will be fried almost instantaneously by high energy radiation.

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phinds
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... no one in these forums has provded a reason why they should radiate
There was a long thread about this some time back if you're interested (sorry, I don't have a link), which is why I happened to remember about it. There was a strong assertion that there is no Hawking radiation when the BH is below the CMB temp, but that assertion was shown to be incorrect.

phinds
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... because the infaller approaches the speed of light as the event horizon is approached
I don't dispute this, but it leads me to a quesiton:

For really large BH's the infaller does not experience sphagettification at the EH, but that seems inconsistent w/ travelling near c. It seems to me that such a high speed would imply a strong tidal force. I have no great sense that I'm right about this, and I'd appreciate your comment on what I'm not understanding.

Chronos
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The acceleration gradient is very small at the EH of a supermassive black hole, which means [assuming you jump in feet first] your feet are not being accelerated much faster than your head. You would not be 'spaghettifed' until the parts of you nearer the singularity were subject to much greater acceleration than your more distant parts.

phinds
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The acceleration gradient is very small at the EH of a supermassive black hole, which means [assuming you jump in feet first] your feet are not being accelerated much faster than your head. You would not be 'spaghettifed' until the parts of you nearer the singularity were subject to much greater acceleration than your more distant parts.
So the fact that you are traveling near c is irrelevant to the spaghettification? I guess that's the part I got confused on.

PAllen
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Keep in mind the external universe is also time dilated from the infalling observers perspective. If the external observer ship sent a constant interval pulsed laser signal toward the hapless infalling volunteer, the time between pulses would increase as the infaller approached the event horizon. 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.
The effect of the infaller's speed relative to distant observers competes with gravitational blueshift of light/sgnals from distant observers. Both, taken seperately, approach infinity. The ratio, is quite finite (I think about 2 for infaller's from infinity). The result is that infaller sees only modest shift and time distortion from distant observers. In particular, if an infaller counts pulses a millisecond apart as emmitted from a distant source, they receive a well defined finite number of such pulses before hitting the singularity, and they don't see the pulses becoming ever slower (as in the reverse case - the infaller sending millisecond pulses to the distant observer - these slow down by an unbounded factor as the infaller nears the horizon as perceived by the distant observer).

This doesn't contradict what you say, but I think it is important to emphasize the competing effects for the infaller, and the resulting asymmetry between infaller and distant observer.

Chronos
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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.

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.

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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?

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

George Jones
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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-can-anything-ever-fall-into-black.html.

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.

PAllen