NetNews Usenet Archive 1992 #27/NN_1992_27.iso/spool/sci/math/15084

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From: [e-mail] (Blair P. Houghton)
Newsgroups: sci.math,sci.physics
Subject: Re: Three-sided coin
Date: 16 Nov 1992 22:02:20 GMT
Organization: Intel Corp., Chandler, Arizona
Lines: 64
Message-ID: <[e-mail]>
References: <[e-mail]> <[e-mail]> <jim.721702281@st-michael>
NNTP-Posting-Host: alfalfa.intel.com
Summary: probability is distributed according to entropy, not energy

In article <jim.721702281@st-michael> [e-mail] (Jim Easton) writes:
>Given that it is possible to bounce from one state to another it
>is more probable to bounce out of a high energy state to a lower
>energy state than it is the reverse.  I claim that the lowest energy
>state is the most probable and therefore conclude that the three states
>should be equal in energy.

This is fallacious.

Imagine a three-sided "coin" made by bending a very thin
sheet (of something very low-density like titanium) in the
middle at about a 170-degree angle, then adding a thick
bead of gold or lead inside the crease.  The low-energy
states of this object are the two states that occur when
the coin lands with the convex side down (resting flat on
one leaf or the other).  The high-energy state occurs when
the coin lands with the concave side down (like a tent).

The concave and convex sides have nearly the same
probabilities, but there are two states on the convex side,
which splits the probability there; say it's something like
30, 30, 40 for convex/A, convex/B, and concave, respectively.

This is a direct contradiction of the idea that "the lowest
energy state is the most probable."

It's even worse if you make low pyramids with heavy apices
(say a styrofoam body but a unnilquadium capstone, in a
vacuum, to eliminate aerodynamics).

Like 11, 11, 11, 11, 11, 45 for a pentagonal pyramid, for
instance.

The key isn't energy; it's entropy.  The pointy end gets a
larger total of the probability (55%) because it has a
higher entropy because it has a lower energy (this is what
causes the confusion; the low-energy idea works perfectly
for two-state systems).  But the entropies of all of the
five faces of the pentagonal pyramid are equal, just as
their energies are all equal, so the probability for the
pointy end has to be shared among them (11% each).

Conversely, landing on the base has a lower probability
(45%) than landing on the point because it has a lower
entropy because it has a higher energy, but it doesn't have
to share the entropy, so it keeps all of the probability.

To solve the fat-cylinder problem (height H, radius R),
therefore, one considers the faces as similar, then one has
a two-state problem, choosing between a face-landing and an
edge-landing.  Set the probability (entropy) of a
face-landing to 2/3 and an edge landing to 1/3; thus the
energy of the edge-landing is (1 - 1/3)E = 2/3 E and that
of a face-landing is (1 - 2/3)E = 1/3 E.  The edge-landing
therefore has twice the energy of a face-landing.

Since the energy is simply mgh, and m and g are constants,
it follows that the h for an edge-landing is twice the h
for a face-landing, so R=H.

                --Blair
                  "If you catch an edge you
                   will land on your face."
                   -the Bunny-slope paradox