**From:** Matt Mahoney (*matmahoney@yahoo.com*)

**Date:** Fri Mar 21 2008 - 13:27:16 MDT

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

*> On Fri, Mar 21, 2008 at 7:34 AM, Matt Mahoney <matmahoney@yahoo.com> wrote:
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*> > Now if you let R = cT be the "size" of the universe of age T, and set R =
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*> > 2Gm/c^2 to the Schwarzchild radius and solve for the mass of the universe
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*> m,
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*> > you get S ~ T^2 c^5/hG ~ 10^122, a unitless number.
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*>
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*> Ok, I think I see why you got the right answer. It's because of two things.
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*>
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*> First, you're using the formula for the entropy of a black hole. And
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*> this is where the Bekenstein bound comes from. It's just the Area in
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*> Planck units divided by 4. The reason they are the same is because a
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*> black hole is the maximum amount of stuff you can store in a region of
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*> space... it's the only thing that saturates the bound. So that part
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*> 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
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*> baryon.
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*> > S depends only on T, c, h, and G, not on the properties of baryons or any
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*> > other particles.
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*>
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*> Same thing. You're dividing the volume of the universe by S to get
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*> the volume of a baryon (or equivalently, dividing the volume of the
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*> universe by the volume of a baryon to get S). Either way, this is not
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*> meaningful since it's based on volume and not area. You can fit a
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*> whole lot of bits inside the volume of a baryon (about 10^30 bits I
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*> think). So that's a lot more than 1. It's just that you can't fit
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*> that number of bits in *every* volume that size at once, because they
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*> will have correlations between them due to their gravitational
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*> interactions.
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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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