Re: A partial black hole?

Jonathan L Cunningham wrote:

> 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! :-)
	...
> 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.

The general problem you're having here is that general relativity is not 
Newtonian mechanics with some minor tweaks.  It is a fundamentally 
different approach to things, so you have to be very careful about what 
you mean.

In general relativity, velocities of and distances to other objects in a 
curved spacetime is actually a difficult concept to nail down.

Here's one way of looking at what happens at the event horizon that 
might help:  Think of the event horizon not as the point where the 
escape speed is c, but rather the point at which spacetime is ru****ng 
inward toward the black hole at c.  (This isn't really rigorous, but 
it's not a terribly bad model; a photon moving radially outward will be 
"suck" at r = 2 M; the event horizon is a null surface.)

For a massive particle, you'd have to reach c to be able to fight 
against the tendency for spacetime to rush in to the black hole at the 
event horizon.  You can't do that, since that would take an infinite 
amount of force.

Remember, in general relativity, gravity isn't even modeled as a force! 
  It's just the tendency of an object to naturally move along the 
shortest path in spacetime.  Gravity causes distortions in that 
spacetime, which causes those paths to change from what they'd be in a 
flat spacetime.  That's what gravity is.  Tidal forces, thus, are the 
tendency for two nearby geodesics to deviate from each other over an 
extended trip.

So you really need to be careful when you ask "What is the gravity?" -- 
how do you want to measure it, and relative to what?  Travelling along a 
geodesic is actually the natural state of matter, and doesn't require 
any forces in general relativity!  (This is the basis of the equivalence 
principle.)  The closest thing to the "gravitational field" would 
probably be the local curvature, but that's probably not really going to 
help you much.  (And it's finite at the horizon.)

If the question is, "What force is required to maintain constant 
Schwarzschild coordinate r at the event horizon?" then the answer is 
infinite -- the "weight" diverges.  But remember that r here is just a 
convenient coordinate; it doesn't even correspond to proper distances. 
(As should be evident because of the coordinate singularity at r = 2 M 
-- you have to switch to another coordinate system to truly verify that 
the event horizon is not singular.)

Similarly, the 1/r^3 tidal approximation is strictly Newtonian, not 
general relativistic.  (It's also an idealized case where you're dealing 
with an object of negligible size and/or very far from the gravitating 
body.)

It may sound like I'm being a jerk, but I'm really not.  You can't just 
take Newtonian mechanics, tweak it, and expect to get something 
meaningful from general relativity.

-- 
Erik Max Francis && max@[EMAIL PROTECTED]
 && http://www.alcyone.com/max/
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