Bekenstein Bound

[Hal]

From: Jim choate <ravage at bga.com>
> The problem I see with this is that there is no connection between a black holes
> mass and surface area (it doesn't have one). In reference to the 'A' in the    
> above, is it the event horizon? A funny thing about black holes is that as the
> mass increases the event horizon gets larger not smaller (ie gravitational
> contraction). 

Actually black holes do have a defined surface area, which is basically, as
you suggest, the area of the event horizon.  And of course this is larger
for more massive black holes, as you say.

I believe the Bekenstein bound is based on reasoning that suggests that
if the state density of a region exceeds that bound, it will essentially
collapse into a black hole and be inaccessible to the rest of the universe.
The surface area in that context can be the conventionally defined area.

To bring this back to crypto a bit, the point of this discussion was that
there can be only a finite amount of processing done in finite time by
a finite-sized machine, even when QM is taken into consideration.  Note,
though, that this result appears to require bringing in quantum gravitation,
a very poorly understood theory at present.

Hal