From: Hal <hfinney@shell.portal.com>
To: cypherpunks@toad.com
Message Hash: 4f89fabd084c04888947b7180b2590cbeb21e12ab935dd3fdb80c42e6eac8521
Message ID: <199403311608.IAA05406@jobe.shell.portal.com>
Reply To: N/A
UTC Datetime: 1994-03-31 16:03:00 UTC
Raw Date: Thu, 31 Mar 94 08:03:00 PST
From: Hal <hfinney@shell.portal.com>
Date: Thu, 31 Mar 94 08:03:00 PST
To: cypherpunks@toad.com
Subject: Bekenstein Bound
Message-ID: <199403311608.IAA05406@jobe.shell.portal.com>
MIME-Version: 1.0
Content-Type: text/plain
From: Jim choate <ravage@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
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