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In article <callahan.667353668@newton.cs.jhu.edu> callahan@cs.jhu.edu (Paul Callahan) writes:
>Is there a closed form solution for the effective resistance between two
>points on an infinite 2-D mesh whose edges are unit resistors?

I posted a solution to this problem which consisted of finding the
potential for a current sink of one amp at a vertex of the mesh and a
one amp current source "at infinity" and then using the superposition
principle.  As someone else also pointed out, you can immediately
deduce that the e.r. between two adjacent nodes is 1/2, and as I showed
in my previous post you can find all the effective resistances with
more work.  (In particular the e.r. between two diagonally separated
nodes is 2/pi.)

In that post I glibly wrote off the serious issue which Brendan McKay
brought up of "unphysical", i.e. mathematically fallacious, arguments:

In article <1991Feb28.203048.13891@agate.berkeley.edu> greg@math.berkeley.edu writes:
>As Brendan McKay pointed out, the voltages for the solution are unbounded
>as you go to infinity.  But it's a well-posed mathematical problem
>anyway.

I thought about this issue some more last night and came to a strange
conclusion:  The argument using the current source at infinity is valid
precisely because of and not in spite of the fact that the resistance
between a vertex and the periphery is infinite.

What does it mean to have an infinite 2-D mesh whose edges are unit
resistors?  I decided that there are not one but two natural
interpretations.  On the one hand, you could have an infinite 2-D grid
of disconnected electrical nodes, and you could proceed to wire in the
resistors one by one.  Each time you throw a resistor in the effective
resistance between any two given nodes goes down, so the e.r. must
necessarily settle to some limit.  Clearly, the limit does not depend
on the order that the resistors arrive.  On the other hand, you could
have an infinite 2-D grid of nodes all shorted together with
superconducters, and you could replace the shorts one by one with unit
resistors.  In this case, the resistance goes up with each new resistor
and therefore also must settle at a limit, and once again the order of
the resistors doesn't matter as long as every resistor eventually
arrives.

The first interpretation is probably the one that most people have in
mind when the hear the problem.  However, the idea of placing a current
source at infinity only jibes with the second interpretation.  Thus, it
is a crucial point that in 2 dimensions, the two answers that you get
are the same.  Roughly speaking, if the resistance between the periphery
and the center is large, it doesn't matter if the periphery is insulated
from itself or if it is shorted together.  I haven't completely worked
out a mathematical proof, but I'm confident that this reasoning
validates my derivation of all of the effective resistances between
two nodes in addition to supporting the claim that the e.r.
between adjacent nodes is 1/2.

It's a different story in 3 and higher dimensions, because there the
resistance between the center and infinity for an infinite mesh is
*not* infinite.  Let us call the limit obtained by replacing insulation
by resistors the upper limit of the e.r. and the limit obtained by
replacing superconducters by resistors the lower limit of the e.r.  I
know that for a 3D infinite mesh the lower limit of the e.r. between
two adjacent nodes is 1/3.  But what is the upper limit?

The best illustration of my point is the example of an infinite tree
in which all vertices have valence 3 with a unit resistor running along
each edge.  In this case, the e.r. between two adjacent nodes is
"obviously" 1, because the entire circuit is irrelevant except for the
resistor connecting the two nodes directly, but it is equally "obvious"
by symmetry and the superposition principle that the resistance is 2/3.
How can this be?  The symmetry argument implicitly assumes that the entire
periphery of the tree is shorted together.  Given that, the circuit reduces
by Kirchoff's laws to a single resistor with resistance 2/3.
----
Greg Kuperberg                 Reply only to postings you like.
greg@math.berkeley.edu         Ignore postings you dislike.