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- From: jbaez@riesz.mit.edu (John C. Baez)
- Subject: Re: 3-sided coin
- Message-ID: <1992Nov17.071017.13888@galois.mit.edu>
- Sender: news@galois.mit.edu
- Nntp-Posting-Host: riesz
- Organization: MIT Department of Mathematics, Cambridge, MA
- References: <1dsj4gINNeuh@agate.berkeley.edu> <1e98b8INN6s@manuel.anu.edu.au>
- Date: Tue, 17 Nov 92 07:10:17 GMT
- Lines: 25
-
- In article <1e98b8INN6s@manuel.anu.edu.au> butler@rschp2.anu.edu.au (Brent Butler) writes:
- >The solution to this problem is simple. When a coin is flipped it
- >will *come to rest* in a position that represents a minimum in its
- >(gravitational) potential energy. For a real coin this means one
- >of its two faces with equal probability - it will never *come to
- >rest* on its edge even though it has a finite area because this is
- >a metastable state.
- >
- >Our three sided coin, therefor, should be 1 coin diameter thick. This
- >is where the center of mass is at equal heights in each of the three
- >states and so each configuration represents an equally likely
- >stable state for the coin to *come to rest* after a toss.
-
- It seems that by *come to rest* you simply mean come to a
- lowest-energy state. While all metastable states should eventually
- tunnel to ground states quantum-mechanically at zero temperature,
- "eventually" can be a long time. (Also, near-ground states will have a
- by chance of being occupied at nonzero temperature.)
-
- A more major nitpick is the following: the chance of one of several
- macrostates being occupied (lying on one side vs. lying on an edge)
- depends on the volue of phase space occupied by the state, not just
- the energy. Someone already noted this and I believe it is a serious
- problem for your line of reasoning.
-
-