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

From: Matt Mahoney (matmahoney@yahoo.com)
Date: Thu Mar 20 2008 - 19:16:09 MDT

--- Mike Dougherty <msd001@gmail.com> wrote:

> On Wed, Mar 19, 2008 at 4:10 PM, Matt Mahoney <matmahoney@yahoo.com> wrote:
> > The model assumes that the set of states is isomorphic to N. Any real
> > implementation with finite memory must have finite subjective experience.
> All
> > implementations must have finite memory because the universe has a
> Bekenstein
> > bound of about 10^122 bits.
>
> I looked up Bekenstein bound on wikipedia (it's a start)
>
> "S=A/4 where A is the two-dimensional area of the black hole's event
> horizon in units of the Planck area, \hbar G/c^3."
>
> Can someone explain how a two dimensional area is used to measure the
> amount of information that can be stored in a 3 (or more) dimensional
> universe? (Or if this is too basic a question, simply email me
> off-list with some educational URLs)

I had independently derived an order of magnitude estimate of the information
content of the universe a few years ago using a different approach. I
estimated the number of possible standing wave patterns in a volume of size R
as a function of the mass-energy E = mc^2 in that space, which depends on
Planck's constant h. Then I set R = Tc where T = 4.3 x 10^17 s is the age of
universe and c is the speed of light. Then I estimated the largest mass that
could be contained in a sphere of radius R before the escape velocity reached
c (the Schwartzchild radius). I came up with S ~ T^2 c^5/hG ~ 10^122 ignoring
small constants because the radius of the universe is not well defined in
expanding, curved spacetime, the exact mass of the universe isn't known, and I
don't have the physics background to compute the exact value anyway.

I only recently came across the Bekenstein bound, which gives S = 2.02 x
10^122 if you let the surface of the universe be 4*pi*(Tc)^2. Actually this
is 2.91 x 10^122 bits because thermodynamic entropy uses natural logs.
Calculating with Bousso's figure of 1.4 x 10^69 bits/m^2 gives the same
result.

It is rather interesting that the volume of a bit is about that of a proton or
neutron, but S does not depend on the properties of any particles. Either
this is a coincidence or the size or number of baryons is changing because S
increases with T^2 while the volume increases with T^3. There are about
S^(2/3) ~ 10^80 baryons in the universe, enough to make a coating one particle
thick.

It has always puzzled me why the universe didn't collapse into a black hole
shortly after the big bang if it had the same mass as today. It also puzzles
me that a black hole is not symmetric with respect to time (stuff goes in but
not out) even though Einstein's general field equations are. Shouldn't there
be another solution with the opposite sign, one where stuff goes out but not
in, like an expanding universe? In that case, dark energy is just gravity?

-- Matt Mahoney, matmahoney@yahoo.com

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