From: Lee Corbin (lcorbin@tsoft.com)
Date: Tue Aug 05 2003 - 18:59:07 MDT
Anders writes
> On Mon, Aug 04, 2003 at 02:52:47PM -0700, Lee Corbin wrote:
> >
> > I maintain only that the event horizons of this object and
> > other black holes have not *yet* formed, and there are as
> > yet *no* singularities over which to worry.
> >
> > It's just a frozen star, as all the Russian scientists in
> > the 1950's would have told you.
>
> And how do you distinguish between frozen stars and black holes? Your
> complaint seems to be against the existence (in whatever tempus) of
> singularities, and to a lesser degree the existence of event horizons.
Yes, observationally they're the same to outside observers.
And perhaps yes, my complaint may come down to the tense
of sentences people use to describe black holes. But even
if that's all it is, then people have still said a lot of
very incorrect things, or, (of course) I do not understand
what is going on.
> But the singularity is not necessary to have a black hole, e.g. some
> quantum gravity theories propose...
Oh yes. In fact, Scientific American just had an article about
that. But here I am talking about the GR theory itself.
> Event horizons appear naturally even in classical physics if you view
> them as the effect of superluminal escape velocities.
I didn't know that superluminal velocities were a part of
mainstream theories at all. Yes, for sure, event horizons
appear even in very ordinary physics. You can outrun a
photon if you start about a light year in front, and maintain
about a 1 gee acceleration. It never catches you, even though
you do not yourself (of course) ever quite get to lightspeed.
> Even in GR there are no "real" objects but rather the spacetime
> hypersurfaces with limit light rays.
Yes, so I am told. And it does make sense that event horizons
capture outgoing photons (and, as I contend, incoming ones too!).
> The Schwarzschild solution is rather inescapable for any spherically
> symmetric static object, and perturbations don't seem to alter the
> essential characteristics. Do you disagree with GR altogether, or assume
> that some other effect causes the interior to behave wildly different (a
> la the baby universe example above or some weird force preventing
> collapse)?
No, I'm following standard GR so far as I understand it.
> http://antwrp.gsfc.nasa.gov/htmltest/gifcity/bh_pub_faq.html#evaporate
>
> (seems a bit too popular to me. Maybe the best way of finding out is
> simply to make a model of a classical black hole with a slowly
> decreasing mass and follow where the geodesics go)
Well, I read that FAQ many years ago in John Baez's stuff. But thanks
for directing me to it again. There are many passages in it that
almost but (to me) do not quite address my problem. Consider the
paragraph in section 5 that reads
If the black hole is mortal, you'll instead see those events
[infalling astronaut] 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!
Note the use of the word calculate. Often in the FAQ, the author
insists that we consider only the *appearance* of the black hole---
namely the photons that are reaching us from the events taking
(or which have taken) place near the event horizon. He calls
this merely an "optical effect", and I too am completely uninterested
in merely what we see.
(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.
So here it is said, unmistakably, that even though the hole is
only mortal, the infalling astronaut breezes through the event
horizon before that occurs. Yet if a tiny mirror (supported by
very powerful rockets) is suspended just above the event horizon,
and the hole is not going to evaporate before, say, 10^1000 years,
then you may calculate that if the mirror is close enough, then
a photon we emit may take 10^300 years to get to the mirror, and
10^300 years to get back (in our reference frame, a round trip
time of 2x10^300 years). So my confusion stems from the fact
that if we send a photon into the black hole just ahead of the
astronaut, then in *some* sense by our watch it strikes the
mirror only after a long time. The astronaut therefore does
not pass the mirror and does not go through the horizon also
until a *very* long time, that is, as we reckon it, not before
10^300 years. (Naturally, his watch is nearly frozen, and to
him it's no time at all.) The trajectories of the infalling
and outgoing photons are identical according to what I read.
There are many puzzles such as I have just outlined. I don't
see an answer to them. Any help is appreciated.
Lee
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