Re: A partial black hole?

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

> Jonathan L Cunningham wrote:
> 
> > 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.
> 
> But this isn't what happens.  The event horizon is the surface at which
> one _cannot_ stay at constant radial coordinate.  In other words, the
> required acceleration to stay at constant r at the event horizon is 
> _infinite_.
> 
> That means that as you approach the horizon (from the outside, of 
> course), the required acceleration to hover diverges.  To stay 
> arbitrarily close to the horizon requires arbitrarily high
accelerations.

I thought the event horizon was where the escape velocity becomes equal
to c ? Presumably you are saying that, at that distance, the
gravitational force is infinite (= arbitrarily high acceleration). Since
it increases as you approach the singularity, sounds like you are saying
that it exceeds infinity as you get closer! :-)

Being serious for the moment:

I think we agree that the tidal forces are not infinite, but vary as the
inverse cube? And certainly the gravitational force varies as the
inverse square at large distances ... and with a very large mass black
hole, the event horizon *is* at a large distance. But is it large
enough?

So what's the formula for the gravity at a distance D from a black hole?

For a Newtonian "black hole" (i.e. its escape velocity is greater than
c, the speed of light) the potential energy is inversely pro****tional to
the distance, the gravitational force to the inverse square of the
distance, and the tidal forces the inverse cube.

And, for a Newtonian black hole, the distance at which escape velocity =
c is given by R = 2GM/(c^2).

The event horizon of a real (GR) black hole is given by R = 2GM/(c^2) --
which, remarkably, is the same formula.

Now, I'm not saying that the Newtonian and GR black holes are the same
thing, but I think the above entitles me to ask what the formula for the
gravitational acceleration is! The Newtonian acceleration is, of course,
GM/(D^2) at *all* distances. The relativistic formula must agree at
large enough distances, even if it *does* go to infinity at the event
horizon (which I'm still sceptical of). I can guess a formula ...
something like:
   GM/((D-R)^2)

would behave as you describe - but I see no basis for it. And I would
expect (if it's correct) that the formula for the event horizon radius
ought to differ from the Newtonian formula - but it doesn't.

Comments?

Jonathan