A long look back, an OOPS!, and a couple of questions.
Sixty years ago when I was a post-graduate student an old professor of
Electrical Engineering asked me this same question about the resistance
between two points in an infinite grid of resistors. In his question the
resistance was across the diagonal of one small square and the resistors
were each 2 ohms. He added in his crusty manner, "...and if you write
anything down, Park, you are wrong!". I assumed then that the expected
answer was one ohm which I was never able to demonstrate. I was rather
gruntled therefore to see the elegant proof posted by Kelly. After mulling
over that result for a few days I came to the conclusion that it was wrong.
At first I attributed the failure of the proof due to the presence
of infinities in the argument, but there I too was wrong. Consider the
simpler case of the resistance between two adjacent points on the grid.
Call these points A and B. There are four resistors connect to each of A
and B and there is one resistor, call it Rab which is common to both
points. We first inject a current into point A. Because the resistance of
the grid to infinity is infinite there must be infinite voltage at point A
under this condition. However this infinity does not cause any problem
because we only use this step to determine the current in the resistor Rab.
The current must be 0.25 amperes because the one-ampere current must
divide equally amongst the four resistors connected to point A. We then
remove the current from point A and pull one ampere out of point B. Again
there must be 0.25 amperes flowing in resistor Rab. We now apply the two
currents simultaneously. The principle of superposition allows us to
determine the current in Rab by simply summing the results obtained by
applying each current separately. There must then be 0.5 amperes flowing
in Rab which will produce a voltage of 0.5 volts. The resistance from A to
B is therefore 0.5 ohms. Rab is one ohm so the rest of the infinite grid
behaves as if it were a one-ohm resistor parallel to it.
Now consider the case where the two points are on opposite corners
of a square. As before call these points A and B. (If I am belabouring the
argument it is because there seemed to be so much confusion the last time
that this subject was discussed.) Call the four resistors connected to
point A a0,a1,a2 and a3. Similarly the four resistors connected to point
B are b0,b1,b2 and b3. The names are arbitrary so we may call the four
resistors that form the square a0,a1,b0 and b1 with a0 connected to b0 and
a1 connected b1. We now inject a current of one ampere into point A with
the current flowing out at infinity. The current flowing in a0 and a1 must
be 0.25 amperes. With the current removed from A we now pull one ampere
out of point B. Again it is certainly true that there must be 0.25 amperes
flowing in each of b0 and b1. By superposition we may find the current in
each resistor when both currents are applied by adding the results obtained
from applying each current separately . That is certainly true. So the
current in a0 and b0 must be 0.25 amperes with both currents applied, but
this is false, of course.
When we apply a current of one ampere to point A there is 0.25
amperes flowing in a0 and a1, but there is also a current flowing in b0 and
b1. Call this current OOPS! because we overlooked it. When we pull one
ampere out of point B there will be, of course, the same current OOPS!
flowing in a0 and a1. So under superposition the current in all four
resistors will be 0.25 + OOPS! amperes, so we may conclude that the
resistance between A and B is 0.5 + 2*OOPS! ohms. The value OOPS! is not
easy to calculate so superposition is of no help to solve this problem.
Don Kelly provided a reference to a solution to the problem at:
http://www.mathpages.com/home/kmath668/kmath668.htm
The unnamed author of the report derives an answer of 2%pi ohms for the
resistance across the diagonal. I do not understand the method or the math
involved. The answer is almost certainly true.
I have used J to build the admittance matrix for small square grids.
The resistance calculated for a small grid is too high, but it approaches
the correct value as the size of the grid is increased. So the result can
be considered as an upper bound for the true value. If one considers a
small grid for which the outer resistors are all short-circuited together
then one obtains a resistance that is too small but rises as the grid size
is increased. These results are then a lower bound for the true
resistance. For the largest grids that my computer could handle the upper
and lower bounds are 0.636945 and 0.636305 respectively. The mean of these
two bounds differs from 2 divided by pi by less than 0.001 percent. This
result may be more of a happy accident than reality as I do not know how to
properly compare the results for the "open" and the short-circuited grids.
If Roger will ever get around to implementing %. for sparse matrices I
could then perhaps squeeze out a couple more decimal places.
It would be sort of fun if someone would write an explanation of the
method used in the mathpages reference. Please keep it simple....something
even an old engineer could understand.
A couple of questions:
0) Why does pi appear in a calculation for a SQUARE grid!
1) Matrix division for this special matrix is therefore a way of
computing pi, albeit a very inefficient one. Is this a new way of
computing pi or is it somehow related to a known sequence?
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