Question 3

Consider an infinite network of equal
resistors as in https://xkcd.com/356/ and two nodes separated by
m resistors in one direction and n resistors in the other direction.
The cartoon asks for the resistance between the two nodes when
m=2,n=1.  Our astrophysical-thinking question, however, is the
resistance if m^2 + n^2 >> 1.  Also the resistance in the analogous
three-dimensional problem.

We can simplify the problem a little, as follows.  Normally, we think
of applying a potential (say V) across the two nodes, and measure the
current I flowing into one node and out of the other node.  The
resistance is then V/I.  This can be considered as a superposition of
two scenarios.  Suppose we set the boundary (i.e., infinity) to some
constant potential, by grounding it.  In the first scenario, we inject
a current I into one node, and measure the voltage (say U) at the
_other_ node.  There is no net current at the second node, it all
flows to the boundary.  In the second scenario, we inject a current -I
into the second node, and measure the voltage at the first node.
Well, we don't really need to measure, because by symmetry the voltage
will be -U.  Superposing the scenarios, we have V = 2U.

The simplified version of the problem now reads: if we apply a current
source I at some node, which flows out to the boundary, what will be
the voltage at some distant node?

In one dimension, the answer will clearly by I times half the sum of
intervening resistors.  (Half because the current I splits at the node
into forward and backward flows.)  What about in two and three
dimensions?

Try writing down Kirchhof's laws for the circuit, and then going to
the limit where there are so many nodes, that it's like a continuum.
The continuum limit is pretty familiar to astrophysicists, at least in
three dimensions...