Re: Black hole questions
- From: sal <pragmatist@xxxxxxxxxx>
- Date: Thu, 23 Nov 2006 21:32:43 -0500
On Thu, 23 Nov 2006 14:25:35 -0800, Eric Gisse wrote:
sal wrote:
On Thu, 23 Nov 2006 10:02:10 -0700, N:dlzc D:aol T:com (dlzc) wrote:
Dear sal:
"sal" <pragmatist@xxxxxxxxxx> wrote in message
news:pan.2006.11.23.03.01.59.538035@xxxxxxxxxxxxx
On Wed, 22 Nov 2006 17:12:19 -0700, N:dlzc D:aol T:com (dlzc) wrote:...
Dear sal:
"sal" <pragmatist@xxxxxxxxxx> wrote in message
news:pan.2006.11.22.20.50.15.364154@xxxxxxxxxxxxx
No it does not. It takes infinite (Schwarzschild) time for
anything, including light, to reach the event horizon.
Nonsense.
http://hermes.physics.adelaide.edu.au/~dkoks/Faq/Relativity/BlackHoles/fall_in.html
... "finite proper time"
Not nonsense, and your reference doesn't contradict me. I said
"infinite (Schwarzschild) time". I didn't say "infinite proper time".
The proper time *is* evaluated in Schwarzchild coordinates.
The proper time can be evaluated in _any_ coordinates: It's the
integral of
sqrt(|ds^2|) = sqrt(|g_00 dt^2 + g_11 dr^2 [ + cross terms ]|)
...and your path intersects a coordinate singularity.
If you are going to use Schwarzschild coordinates, stop being surprised
when you get a nonsense answer if you try to anything at or beyond the
event horizon.
First, you don't need to integrate _to_ the horizon. You just need to
integrate to points _close_ _to_ the horizon and take the limit of the
integral as the horizon is approached, which is legitimate, and which
gives a perfectly good (and finite!) value for the integral. This is no
different from evaluating any other "improper integral" -- if you play by
the rules it just works. The value obtained for the elapsed proper time
of the infalling object will be the same whether obtained this way, or any
other way.
Second, it's not a nonsensical answer, it's just surprising. For a
distant observer Schwarzschild coordinate time matches wall clock time,
and the elapsed wall clock time to bounce photons off a sequence of
(fixed) mirrors which are placed ever closer to the horizon approaches
infinity as the location of the mirror approaches the horizon. That's a
statement about elapsed proper time for a distant observer between actual
events in a real (hypothetical) situation, and it doesn't matter what
coordinates we use: we'd better get the same answer. (Using Schwarzschild
coordinates throughout just happens to make that particular calculation
easy.) And as I already observed, infinity/2=infinity, so by the isotropy
of the speed of light, if the round trip time approaches infinity, then it
must, in some sense, approach infinite time for the photon to reach the
mirror as the mirror approaches the horizon.
Make sense? (Or is there something wrong with what I just said? AFAIK
it's correct, and in fact is just a rehash of well-known facts.)
[...]
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.
- References:
- Black hole questions
- From: Wai Yu Wong
- Re: Black hole questions
- From: N:dlzc D:aol T:com \(dlzc\)
- Re: Black hole questions
- From: sal
- Re: Black hole questions
- From: N:dlzc D:aol T:com \(dlzc\)
- Re: Black hole questions
- From: sal
- Re: Black hole questions
- From: N:dlzc D:aol T:com \(dlzc\)
- Re: Black hole questions
- From: sal
- Re: Black hole questions
- From: Eric Gisse
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