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- From: jbaez@riesz.mit.edu (John C. Baez)
- Subject: Re: ... an infinite mesh of 1ohm resistors ...
- Message-ID: <1992Jul28.141407.17816@galois.mit.edu>
- Sender: news@galois.mit.edu
- Nntp-Posting-Host: riesz
- Organization: MIT Department of Mathematics, Cambridge, MA
- References: <1992Jul25.210947.12316@cs.yale.edu> <1992Jul27.210947.5820@fs7.ece.cmu.edu> <1992Jul28.000844.27051@mixcom.com>
- Date: Tue, 28 Jul 92 14:14:07 GMT
- Lines: 34
-
- NOTE: John C. Baez is the author of this article. I am only using my
- account to post this article for John as a favor - Mark Corscadden
-
- In article <1992Jul28.000844.27051@mixcom.com> ttyytt@mixcom.com (Adam Costello) writes:
- >In article <1992Jul27.210947.5820@fs7.ece.cmu.edu> snyder@henry.ece.cmu.edu (John Snyder) writes:
- >>
- >>I believe that you are missing the point of the previous post here.
- >
- >But I think you are missing the point of the second post. The "solution"
- >involved imagining that you could inject an amp of current into a network
- >of resistors. No capacitors! Where is the charge supposed to go?
- >Regardless of whether this can be done in the laboratory, there's still
- >the question of whether it can be done in the mind. Certainly, if the
- >network were finite, it would make no sense. Does the infinitude of the
- >network allow a source with no sink? We can argue that it does this way:
-
- Yes, an infinite network is necessary and sufficient to have a source
- but no sink. The charge goes "off to infinity". This isn't as weird as
- one may think. Look at the equation of electrostatics div E = rho.
- In a closed universe (e.g. a three-sphere) this implies (with Gauss's
- theorem) that the total charge in the universe must be zero because
- there's "nowhere for the electric field lines to go". However in an
- infinite universe one can have nonzero total charge.
-
- That example may have made it seem more rather than less weird. But
- mathematically they are very analogous.
-
- The fact that one can only do this trick for an infinite network doesn't
- mena that the answer is very far off for a big finite network. One
- could grunge through the problem on a big finite network (hopefully with
- aid of a computer) and see, hopefully, that the answer provided by the
- slick trick was very close.
-
- John Baez
-