The Algebra of an Infinite Grid of Resistors
In a previous note we discussed the well-known problem of determining the resistance between two nodes of an “infinite” square lattice of resistors. The most common approach is to superimpose two “monopole” solutions, one representing the field for one amp of current entering a given node and flowing “to infinity”, and the other representing the field for one amp of current being withdrawn from a given node flowing in from infinity. If the two nodes are adjacent, the solution is fairly unambiguous, but the resistance between two arbitrary nodes of an “infinite” resistor grid is actually indeterminate unless we impose restrictions on the voltage and current levels “at infinity” (such as stipulating that we seek the solution for a finite grid in the limit as the size of the grid increases to infinity).
According to the naïve approach, we imagine injecting 1 amp of current into a single node at the origin, and allow it to flow outward symmetrically to infinity. Let V m,n denote the voltage drop from the source node with coordinates (0,0) to any given node with coordinates (m,n). Likewise we could imagine extracting one amp from the node at (m,n), flowing in symmetrically from infinity, and the voltage drop from the origin to (m,n) would also be V m,n . Superimposing these two solutions, we have one amp entering node (0,0) and exiting node (m,n), and the voltage drop is 2V m,n , so the net effective resistance between (0,0) and (m,n) is R m,n = 2V m,n /I, where I = 1 amp. Of course, as mentioned in the previous note, the voltage “at infinity” must go to infinity relative to the voltage at the source, because the resistance to infinity is infinite. Nevertheless, it might seem that with suitable care we can evaluate the limit to arrive at an unambiguous answer.
However, the stated problem has a unique solution only if we stipulate some restriction on the boundary conditions. This is a delicate proposition, since the boundary is “at infinity” and the voltage approaches infinity. The ambiguity can be seen immediately by examining the figure below, which shows a node from which two amp of current is being extracted, surrounded by nodes of higher voltage necessary to supply this current. The figure shows just the lower quadrant of the grid, with the understanding that that other three quadrants are symmetrical. The voltages along the diagonals are denoted by α 1 , α 2 , α 3 , …, and all the other voltages are expressed in terms of these, in such a way that the difference equation governing the system is satisfied.
For a grid with 1 ohm resistors between each pair of adjacent nodes, the effective resistance between the origin and any given node equals half the value listed for the node in the table below. Obviously this construction can be continued to infinity, since each succeeding row is computable from the prior rows, noting that the outer-most cells adjacent to the diagonals are determined by symmetry with the neighboring quadrants. Thus for any arbitrary choice of the diagonal parameters α 1 , α 2 , α 3 , …, we can construct an entire infinite grid. For example, if we set all the diagonals to zero, we get
We will refer to the value in the ith row and jth column of this array as ρ i,j . The sum of the terms of the nth row is n. The terms are generated by the basic two dimensional recurrence (representing the discrete form of the Laplace equation)
This table implies that the resistance between the origin and any node on either diagonal is zero. Does this mean that an infinite grid of resistors acts as a superconductor with zero resistance between two arbitrarily distant nodes? (It’s interesting to compare this with the pseudo-metric of Minkowski spacetime, in which the “diagonals” represent light-like intervals of null magnitude.) Notice that we achieved this “superconductivity” along the diagonals at the expense of introducing large voltages and currents at the off-diagonal nodes. The fact that these voltages increase to infinity is not really objectionable, because any monopole solution must have the voltages increasing to infinity as we move away from the origin. On the other hand, we might object to the fact that the voltages alternate between positive and negative values on adjacent nodes in the transverse direction, and hence the transverse currents between adjacent nodes go to infinity. However, nothing in the original problem statement restricts the behavior of the voltages and currents at arbitrarily large distances from the origin. The only requirement is to specify a set of voltages and currents on the infinite grid that everywhere satisfy Ohm’s Law across each resistor, with a certain amount of current entering one particular node and exiting one other particular node. One might argue that the superposition of the two monopole solutions yields finite behavior at infinite distances from the two monopoles, because the infinities of the individual monopole solutions cancel each other, but it’s conceivable that similar cancellation could occur for superpositions of diagonally null grids. Notice that the signs alternate by columns in the “past” and “future” regions, and by rows in the “present” regions, so if the two poles are separated by an even number of nodes in both directions, they will tend to cancel (i.e., destructively interfere with each other) in all regions except along the stripes between their null diagonals, as shown below.
In the “light-like” regions between the two interacting poles, the voltages and currents are alternately re-enforcing and canceling, as depicted below.
This could be regarded as an analog to polarized vacuum solutions of the Laplace equation. Of course, rather than setting the diagonal voltages to zero, we can set them to any values we choose. The result is given by superimposing the “null” solution (whose values were given in the preceding table) with the product of each diagonal value times the corresponding array that satisfies the usual Laplace recurrence relation (noting that the recurrence for the nodes adjacent to the diagonal actually involve the symmetrical nodes in the neighboring quadrant). For example, if α 1 is not zero, we must superimpose α 1 times the array shown below.
Similarly, if α 2 is not zero, we must superimpose α 2 times the array shown below.
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