I have a question that's bugging me for some time now:
Consider the typical doomed astronaut falling into a black hole. For a
distant observer, he does not cross the event horizon in a finite amount
of time. But the black hole is evaporating in a finite amount of time due
to Hawking radiation.
The astronaut on the other hand would just see the black hole
evaporating faster and faster the closer he gets and - assuming he's
shielded from the gamma rays emitted in the end - just ends up in empty
space and lives happily ever after. Some zillions of years later, that is.
So what's wrong with this picture? If nothing ever crosses the event
horizon, what's all this fuss about "information paradox" etc.? I know
next to nothing about QFT in curved spacetimes, so perhaps something's
wrong with my concept of Hawking radiation? Or is it that I can't join the two
processes of the astronaut getting closer to the event horizon and the
black hole evaporating in such a simplistic manner?


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Matthias Plaue
Homepage: http://www.math.tu-berlin.de/~plaue/

Dear Matthias Plaue:

"Matthias Plaue" wrote in message
news
I have a question that's bugging me for some time now:
Consider the typical doomed astronaut falling into a
black hole. For a distant observer, he does not cross the
event horizon in a finite amount of time.

This is because light has difficulty from getting from just above
the event horizon to the distant observer.

But the black hole is evaporating in a finite amount of
time due to Hawking radiation.

For stellar mass black holes, this is millions to billions of
years...

The astronaut on the other hand would just see the
black hole evaporating faster and faster the closer he
gets and

No. The astronaut "crosses over" the event horizon in a very
short time.... minutes to months depending on the size of the BH
and his / her initial speed.

David A. Smith

I finally found a web page that explains the case:
http://math.ucr.edu/home/baez/physic...s/fall_in.html

Quote:
"What about Hawking radiation? Won't the black hole evaporate before you
get there? (First, a caveat: Not a lot is really understood about
evaporating black holes. The following is largely deduced from information
in Wald's GR text, but what really happens-- especially when the black
hole gets very small-- is unclear. So take the following with a grain of
salt.)

Short answer: No, it won't. This demands some elaboration.
From thermodynamic arguments Stephen Hawking realized that a black hole
should have a nonzero temperature, and ought therefore to emit blackbody
radiation. He eventually figured out a quantum- mechanical mechanism for
this. Suffice it to say that black holes should very, very slowly lose
mass through radiation, a loss which accelerates as the hole gets smaller
and eventually evaporates completely in a burst of radiation. This
happens in a finite time according to an outside observer.

But I just said that an outside observer would never observe an object actually
entering the horizon! If I jump in, will you see the black hole evaporate
out from under me, leaving me intact but marooned in the very distant
future from gravitational time dilation?

You won't, and the reason is that the discussion above only applies to a
black holethat is not shrinking to nil from evaporation. Remember that
the apparent slowing of my fall is due to the paths of outgoing light rays
near the event horizon. If the black hole does evaporate, the delay in
escaping light caused by proximity to the event horizon can only last as
long as the event horizon does! Consider your external view of me as I
fall in.

If the black hole is eternal, events happening to me (by my watch) closer and
closer to the time I fall through happen divergingly later according to
you (supposing that your vision is somehow not limited by the discreteness
of photons, or the redshift).

If the black hole is mortal, you'll instead see those events happen closer
and closer to the time the black hole evaporates. Extrapolating, you
would calculate my time of passage through the event horizon as the exact
moment the hole disappears! (Of course, even if you could see me, the
image would be drowned out by all the radiation from the evaporating
hole.) I won't experience that cataclysm myself, though; I'll be through
the horizon, leaving only my light behind. As far as I'm concerned, my
grisly fate is unaffected by the evaporation.

All of this assumes you can see me at all, of course. In practice the time
of the last photon would have long been past. Besides, there's the
brilliant background of Hawking radiation to see through as the hole
shrinks to nothing. (Due to considerations I won't go into here, some
physicists think that the black hole won't disappear completely, that a
remnant hole will be left behind. Current physics can't really decide the
question, any more than it can decide what really happens at the
singularity. If someone ever figures out quantum gravity, maybe that will
provide an answer.)"


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Matthias Plaue
Homepage: http://www.math.tu-berlin.de/~plaue/

Dear Matthias Plaue:

Matthias Plaue wrote:
I finally found a web page that explains the case:
http://math.ucr.edu/home/baez/physic...s/fall_in.html

Quote:
....
If the black hole is mortal, you'll instead see those events
happen closer and closer to the time the black hole
evaporates. Extrapolating, you would calculate my time
of passage through the event horizon as the exact moment
the hole disappears! (Of course, ...

I have issue that your image will only appear to cross the event
horizon as the black hole completely disappears. Since the event
horizon has this "halo" of outbound light "trapped" just above it on
the light's way outwards, and the event horizon changes based on the
mass it contains, evaporation will move the EH inwards... not the halo
of outbound light. So evaporation should release the infaller's last
image long before the event horizon completely disappears. No infaller
information gets added to the external halo as the event horizon moves
inwards, so the infaller's "grizzly demise" is never available...
"Hawking flare" or not.

Note that net evaporation of the black hole will occur when its
temperature (based on its mass) is higher than background (not less
than the CMBR at that age).

I wonder how long it would take a 0.01 solar mass black hole to
evaporate completely... seems like that would be good for me to do a
Google search on...

David A. Smith

Matthias Plaue wrote:
I finally found a web page that explains the case:
http://math.ucr.edu/home/baez/physic...s/fall_in.html

But, as the page warns, this subject isn't really understood very well, so I
don't think you should take what it says there as authoritative.

As far as I know, the answer on that page is based on this Penrose diagram,
which is supposed to represent a black hole forming and then evaporating:

|\
| \
____| \
| . /
| . /
|. /
| /
| /
| /
|/

The vertical lines are the origin (center of symmetry), the horizontal line
is the singularity, the dotted line is the event horizon, and the other
diagonal lines are null infinity. On this diagram you can definitely cross
the event horizon before the hole evaporates. But I don't entirely trust
this diagram, and in fact I just started a thread about this on s.p.research:

http://groups.google.com/group/sci.p...147c2f662c387f

I think the following (from the page you linked) is wrong:

Extrapolating, you would calculate my time of passage through
the event horizon as the exact moment the hole disappears!

You would *see* the passage and the evaporation at the same time, but they
aren't the same spacetime event. In any case this isn't very interesting.
There's no empirical content to it -- it's basically just a restatement of
the definition of the event horizon.

-- Ben

dlzc wrote:
I have issue that your image will only appear to cross the event
horizon as the black hole completely disappears. Since the event
horizon has this "halo" of outbound light "trapped" just above it on
the light's way outwards, and the event horizon changes based on the
mass it contains, evaporation will move the EH inwards... not the halo
of outbound light. So evaporation should release the infaller's last
image long before the event horizon completely disappears.

This can't happen; the event horizon always moves at the speed of light. If
it didn't, light could cross from inside the event horizon to outside, which
contradicts the definition of event horizon.

There may be a sensible notion of an apparent horizon which does move slower
than light as the hole shrinks, and you may see the crossing of the apparent
horizon well before evaporation. But not the crossing of the real event horizon.

-- Ben

Dear Ben Rudiak-Gould:

Ben Rudiak-Gould wrote:
dlzc wrote:
I have issue that your image will only appear to cross the
event horizon as the black hole completely disappears.
Since the event horizon has this "halo" of outbound light
"trapped" just above it on the light's way outwards, and
the event horizon changes based on the mass it contains,
evaporation will move the EH inwards... not the halo of
outbound light. So evaporation should release the infaller's
last image long before the event horizon completely
disappears.

This can't happen; the event horizon always moves at the
speed of light.

.... at that "location" in the spacetime that contains the BH.

If it didn't, light could cross from inside the event horizon to
outside, which contradicts the definition of event horizon.

It could simply be that there are no spatial vectors / paths that can
point across the horizon from the inside, and this would not require
any surface simultaneously "moving at c" (local to the "surface") and
stationary (to the remote observer). And it could be that your
statement and mine don't differ.

There may be a sensible notion of an apparent horizon which
does move slower than light as the hole shrinks, and you
may see the crossing of the apparent horizon well before
evaporation. But not the crossing of the real event horizon.

OK, perhaps the invocation of "real" and "apparent" is the problem. I
find it inconceivable that the last "image" of an infaller waits at a
surface above / just-outside an event horizon until the EH completely
disappears. How does this "spherical surface of otherwise outbound
light" know to move inwards to maintain this relationship, or know when
the now-distant EH is finally gone so that it can then rush outwards?
A ship trapped in "orbit" around a vortex suddenly "shoots way" when
the driver for the vortex is removed and it starts to shrink.

And outbound light just above the event horizon is in no kind of orbit.
The only "stable" orbit for photons occurs at the photon sphere at
r=3M (in case any lurkers might get confused by this discussion).

Any discussion of light "trapped forever" at the event horizon (not
made by Baez) is also frequently untrue, since a BH immersed in a
Universe with a warmer CMBR temperature than the BH's Hawking
temperature will be increasing its event horizon... however slightly.
This will overtake some outbound light, consuming "instants" before
crossing. And since CMBR temperature is monotonically decreasing...
the instant when those two temperatures match is fleeting in a
universal moment sense.

Let me say I feel the last light you will ever get from an infaller,
you will get long before you get the information that the EH has
reduced to half the size it was when the infaller crossed (distinct
from when it is seen to cross by the distant observer). Correlating
distant observer time with infaller crossing time is non-trivial, at
least for me. But no improvement in imaging techniques will return
"outbound" information from the infaller, because that information has
passed the observer.

Sorry, Ben. I didn't mean to pirate the thread, I hope I stayed close
to the thread's "intent". I guess the OP's question is answered.

David A. Smith