Re: A partial black hole?

Erik Max Francis <max@[EMAIL PROTECTED]
> wrote:

> sigidunum@[EMAIL PROTECTED]
 wrote:
> 
> > On Feb 28, 11:19 am, Erik Max Francis <m...@[EMAIL PROTECTED]
> wrote:

> >> Freefall time is ~10^-5 s per solar mass of the black hole.
> > 
> > Really!  Huh.  Where does that come from?
> 
> _Gravitation_ by Misner, Thorne, and Wheeler, e.g., p. 863.
> 
> >> So that's a few days here.
> > 
> > I thought at first that this article
> > 
> > http://www.publish.csiro.au/?paper=AS07012
> > 
> > contradicted you, but it doesn't.  The freefall time from rest at the
> > event horizon to impact with the singularity is a fixed  maximum.  It
> > turns out that, if you don't start at rest -- if you cross the event
> > horizon moving in any direction -- then you can increase your survival
> > time by firing a reaction drive.  But only up to that fall-from-rest
> > maximum limit.
> 
> That's correct.  Remember the twin paradox; a geodesic is the path the
> maximizes proper time.

Eh? That doesn't make sense. I had a look at the abstract mentioned, and
it doesn't seem to say that. (Is it only an abstract?)

Perhaps I am misunderstanding you.

I can believe "T, the time from rest at the event horizon to impact with
the singularity is a fixed  maximum" and also "Tf, the freefall time
from rest at the event horizon to impact with the singularity is a fixed
maximum" but not that necessarily T < Tf. And I think the abstract is
saying the opposite - it is saying people assume T < Tf, but actually
this is not true. Actual quote from the abstract:

   In general, the use of such rockets can increase your remaining time,
   but only up to a maximum value; this is at odds with the 'more you
   struggle, the less time you have' statement that is sometimes
   discussed in relation to black holes.

Maybe I misread the abstract, too! [Sometimes I have bad brain days. <g>
But I'll risk posting anway.]

Suppose a very large black hole, where the acceleration due to gravity
is 1 g at the event horizon. Suppose you have a space****p capable of
thrusting at 1g for 1 minute, and you are just outside the horizon.
You are going to allow it to drift arbitrarily slowly across the
horizon.

Three cases:
(a) You drift across the horizon (from rest) and do nothing. Splat! T
seconds later.
(b) You fire your engines for 1 minute, then (when they cut out) drift
across the horizon and splat! T seconds later, or T+60 after the start
of the experiment.
(c) You drift a couple of metres, fire your engines 1 minute *just
inside* the horizon, and then T seconds later, splat! Again, almost
exactly T+60 seconds are you start.

You seem to be saying that (c) is wrong. What am I missing?

As I said, I have no problem with the idea of some maximum number, M, of
seconds you can survive once you cross, or that T+60 < M, and also I
note that the rocket exhaust in cases (b) and (c) will reach the
singularity quicker than the rocket.

Since the event horizon is not a physical discontinuity, cases (b) and
(c) ought to differ by an arbitrarily small amount - yet I'm sure you
would not deny that it is possible to hover *outside* the event horizon
(at least until you run out of fuel).

I get what you mean when you say "a geodesic is the path the maximizes
proper time" and what you say (ISTM) does follow from it -- but I can't
reconcile that with my thought experiment above.

My argument might be like the design for a perpetual motion machine: you
know it can't work, but you can't always easily see the flaw.

Maybe I need more coffee ... but an elucidation would be appreciated!

Jonathan Cunningham