From: mcirvin@husc8.harvard.edu (Matt McIrvin)
Newsgroups: sci.physics
Subject: Re: Black hole insights
Date: 24 Jan 93 19:34:32 GMT

brahm@cco.caltech.edu (David E. Brahm) writes:

>bbowen@megatest.com (Bruce Bowen) writes re: an uncharged, non-rotating
>black hole,
>> The coordinate time for an infalling infinitesimal testpoint to reach
>> the horizon is infinite.  The proper time is finite.

>I've been trying to get this straight too, so someone let me know if I'm
>mistaken, but I think your first statement is only true in some coordinate
>systems (e.g. Schwarzschild coordinates), not in others (e.g. ingoing
>Eddington-Finkelstein coordinates).  See MTW pp. 828-829, 872-875.

That's correct; he's referring to the Schwarzschild t coordinate.
Ingoing Eddington-Finkelstein and Kruskal coordinates avoid this difficulty.

>What I want to know is, when people calculate how much "time" it takes for
>a black hole to evaporate (proportional to M^3 they say), what coordinate
>are they talking about?  I'm still confused by the fact that a test
>particle takes infinite Schwartzschild "t" to pass the horizon, but the
>hole evaporates in finite "t", and I'm not convinced that the "bulging"
>effect people have mentioned is the answer.  Perhaps I'm just being
>irresponsible using Schwarzschild coordinates when M is changing.

I think that's the problem-- the infinite-Schwarzschild-t effect applies
to a nonevaporating Schwarzschild hole.  Wald's GR textbook has the 
following conformal diagram which he claims applies to an evaporating
black hole:
               /|     |\ 
              / |     | \
             /\ |     | /\ 
            /  \|_____|/  \
           /    \     /    \ <---------future infinity 
          /      \   /      \
         /        \ /<------------ event horizon 
        <          X          >
         \        / \        / 
          \      /   \      /                / \ X  lightlike lines    
           \    /     \    /                 _____  spacelike singularity 
            \  /       \  / <-- past infinity
             \/_________\/

(I know someone in alt.fan.warlord is going to call this the worst ascii
map of Australia ever.) 

(Now that I think about it, the one in Wald might have just been the
right half of this diagram.  The left half isn't physical anyway for
a black hole that comes from a star.)  Imagine the diagonals are at 45-degree
angles; they're lightlike.  It's just like the Penrose diagram for a
Schwarzschild nonevaporating hole except that the future singularity occurs
sooner than future infinity.  Now you can see what an infalling object ought
to look like from the outside:  whereas in the case of the nonevaporating
hole the light from the object crossing the horizon takes forever to get to
someone outside the hole, in this case an exterior observer will cross a
future light cone of the infall event when the observer passes into the
little (really infinite) triangular "horn" at the top of the diagram.  
Those regions correspond to times after the black hole evaporates.  Now
at that same moment the observer will see the evaporation event, and 
presumably observe something singular happening, so GR is actually silent
about what the observer will observe, I suppose.  But moments arbitrarily
close to the object's infall will be observable by the observer at finite
observer proper times.  The observer (neglecting the usual dimmer-switch
effects, photon discreteness, finite mass of the object, etc.) will see
the object fall into the hole just as the hole evaporates, though that's
not what is actually going on-- it's a trick of the light cones.

Now, what *I'm* confused about is the situation in which the object has
a finite mass.  I agree that the horizon will come out to meet the object
and engulf it, but does this mean the observer will see it wink out
abruptly?  Forget "coordinate time"-- that's by definition dependent on an
arbitrary definition.  What do the future light cones do?  Do light cones
originating arbitrarily close to the engulfment event hug the horizon
arbitrarily closely, in which case the situation is really no different
except in detail from the massless one?  Or does something more complicated
happen?

 
t
-- 
Matt McIrvin