# Re: Bekenstein bound (Re: A model of consciousness)

From: Matt Mahoney (matmahoney@yahoo.com)
Date: Fri Mar 21 2008 - 13:27:16 MDT

--- Jeff L Jones <jeff@spoonless.net> wrote:

> On Fri, Mar 21, 2008 at 7:34 AM, Matt Mahoney <matmahoney@yahoo.com> wrote:
> > Now if you let R = cT be the "size" of the universe of age T, and set R =
> > 2Gm/c^2 to the Schwarzchild radius and solve for the mass of the universe
> m,
> > you get S ~ T^2 c^5/hG ~ 10^122, a unitless number.
>
> Ok, I think I see why you got the right answer. It's because of two things.
>
> First, you're using the formula for the entropy of a black hole. And
> this is where the Bekenstein bound comes from. It's just the Area in
> Planck units divided by 4. The reason they are the same is because a
> black hole is the maximum amount of stuff you can store in a region of
> space... it's the only thing that saturates the bound. So that part
> makes sense.

The Bekenstein bound is based on thermodynamics, an entirely different
approach. Of course there are only so many ways you can combine T, c, h, and
G to get a unitless number. It has to have the form (T^2 c^5/hG)^n times some
small constant. If n = 1 you have the entropy of the universe. If n = -1 you
have the cosmological constant. If n = 2/3 you have the number of baryons in
the universe.

> > No, I divide the volume of the universe by S and get the volume of a
> baryon.
> > S depends only on T, c, h, and G, not on the properties of baryons or any
> > other particles.
>
> Same thing. You're dividing the volume of the universe by S to get
> the volume of a baryon (or equivalently, dividing the volume of the
> universe by the volume of a baryon to get S). Either way, this is not
> meaningful since it's based on volume and not area. You can fit a
> whole lot of bits inside the volume of a baryon (about 10^30 bits I
> think). So that's a lot more than 1. It's just that you can't fit
> that number of bits in *every* volume that size at once, because they
> will have correlations between them due to their gravitational
> interactions.

That is not the same thing. My question is how do you derive the mass of a
proton from c, h, and G if it is not a coincidence? Why do you need the age
of the universe?

-- Matt Mahoney, matmahoney@yahoo.com

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