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- From: ted@physics3 (Emory F. Bunn)
- Newsgroups: sci.physics
- Subject: Re: Black hole insights
- Message-ID: <1k472o$p7o@agate.berkeley.edu>
- Date: 26 Jan 93 20:31:20 GMT
- References: <mcirvin.727904072@husc.harvard.edu> <C1FFtH.Gyq@megatest.com>
- Sender: ted@physics.berkeley.edu
- Followup-To: sci.physics
- Organization: Physics Department, U.C. Berkeley
- Lines: 51
- NNTP-Posting-Host: physics3.berkeley.edu
-
- In article <C1FFtH.Gyq@megatest.com> bbowen@megatest.com (Bruce Bowen) writes:
-
- > Say, we have a black hole of mass M. We drop a small test mass into
- >it. We wait long enough for it to get within a small distance epsilon
- >of the horizon. We then radially symmetrically dump another amount of
- >mass M in on top of it. Very soon in observer/coordinate time the
- >event horizon has moved far above the position of our small test mass,
- >so it is well within the event horizon in a finite amount of
- >schwartzchild "t". What now is it's coordinate time to reach the
- >central singularity?
-
- It seems to me that you can answer this question by using Birkhoff's theorem.
- The theorem says that if you have a spherically symmetric distribution of
- matter, you can always ignore the matter outside of a spherical
- shell if you want to determine the dynamics inside of that shell. So
- since the extra mass you've dumped in is always outside of the shell
- on which the test particle sits, the test particle just falls as if
- in a Schwarzschild geometry of mass M. Of course, it disappears from
- the view of a person sitting on the outside looking in, but that's because
- the photons that it would have to emit in order for the person to
- observe it have their motions altered by the extra mass.
-
- > Here's another question:
- >
- > Kruskal coordinates, etc. are different parameterizations of
- >spacetime that avoid the coordinate singularity at the horizon that
- >results in schwartzchild coordinates. There is of course a mapping
- >between the two coordinate systems. What does one get for "r" and "t"
- >when one maps back into schwartzchild coordinates after the particle
- >has passed the horizon, and does anyone give these values
- >significance? What is the final value of "t" when an infalling
- >particle reaches the central singularity?
-
- You can figure this out, and its not too hard. The path of the particle
- on an r-t diagram looks like this: It starts at r>R (R is the schwarzschild
- radius); then r decreases towards R as t goes to infinity. Then you
- see t coming "back from infinity" as r goes from R to zero.
- I don't imagine that the value of t at r=0 is of any physical significance,
- since you could alter it by just time-translating the whole coordinate
- system.
-
- One rather strange thing is that for r<R, r is a timelike coordinate and
- t is spacelike. So that means that the singularity r=0 is not a
- single point in space that exists at all times; rather, it is a single
- point in time that exists at all "distances" t. This makes it clear
- why you can't avoid the singularity once you've crossed the horizon.
- Since the singularity is a future point in time, you can't alter your
- world line to avoid hitting it, any more than you can alter your world
- line to avoid hitting next Thursday.
-
- -Ted
-