Black Holes: Infalling Observers and BH Evaporation

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?

twofish-quant

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.

Drakkith

Does an infalling observer ever enter the black hole?

twofish-quant

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.

Drakkith

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

phinds

The length of his body times his speed. It's just d=rt. The in-faller feller is not aware of the EH.

Naty1

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.

phinds

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.

Drakkith

Sorry, I mean according to an outside observer. If that even makes sense since we can't see him fall in.

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

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.

Yeah, I wasn't sure, thus my quibble.

Chronos

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.

Naty1

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:

found the answer here:
http://en.wikipedia.org/wiki/Hawking...le_evaporation

Evaporation of a black body is not dependent on surrounding temps:
[you can find some formulas there]
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: 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.

phinds

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

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

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

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

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.