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- From: mcirvin@husc8.harvard.edu (Matt McIrvin)
- Newsgroups: sci.physics
- Subject: Re: Black hole insights
- Message-ID: <mcirvin.727904072@husc.harvard.edu>
- Date: 24 Jan 93 19:34:32 GMT
- References: <C18EqF.86x@megatest.com> <1jtf4sINNrcd@gap.caltech.edu>
- Lines: 79
- Nntp-Posting-Host: husc8.harvard.edu
-
- 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
-