Jim Choate writes:
Jim Choate writes:
The Deutsch paper I quoted before was where I first heard of the Bekenstein Bound which Eric Hughes mentioned. According to Deutsch:
"If the theory of the thermodynamics of black holes is trustworthy, no system enclosed by a surface with an appropriately defined area A can have more than a finite number ...
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).
If I read the quote correctly, the surface area of the black hole itself is not under discussion. Rather, whether it can be contained in a surface with some area, which it can be.
Of course a singularity can be contained in a volume (not shure what you mean by surface), it is in the universe after all.
I fail to see how this solves anything.
When I read the quote being discussed, it seems to say that no system which can be contained in a surface with an appropriate area A can have more than a finite number of states. I don't think that volume is discussed at all, just a surface. If you are happy to contain the singularity in an imaginary cube with a million light years on each side, I'm happy to call the surface the sides of that cube. This may seem pointless, because, as you point out, everything in the universe can be contained in a surface (or volume). It is not pointless if we can imagine systems which cannot be contained in a surface. I'm guessing that a very large system, say everything in the universe, might not be containable in a surface. If the quote is correct that would imply that the universe may not have a finite number states. The cryptography tie in: if the quote is correct, then any computer we build is going to have a finite number of states which implies that the number of computrons is theoretically limited. And this implies that there may be RSA keys of sufficient size that they cannot be broken with brute force, which doesn't seem that surprising. Peter