The Infinite-dimensional Resistor Lattice Bruce Francis

∗†

November 21, 2013

Abstract A popular puzzle is to compute the equivalent resistance between two nodes in an infinite resistive lattice. This note is an expansion of Zemanian [2].

1

Introduction

A popular puzzle is depicted in Figure 1 The lattice, infinite in both directions, is a network of 1

b a

Figure 1: 2D resistor lattice. ohm resistors. The goal is to find the equivalent resistance between two specified nodes, for example nodes a and b. The answer is 2/π ohms. This is the unique answer when the problem is formulated in a suitable Hilbert space, as we shall see.

2

Infinite chain

The 1D-case is very easy, but we do it anyway in order to highlight the mathematical issue. Figure 2 shows a one-dimensional chain of 1 ohm resistors. The resistors are connected by wires of equal length. The midpoints of the wires are regarded as nodes of the network, a and b being two examples. Embed the chain into the real line so that the nodes are at the integers 0, ±1, ±2, . . . , with a at the origin and therefore b at the point 1. We wish to compute the effective resistance between a and b. Effective resistance is regarded as having the following meaning. As shown in ∗ †

This research was supported by NSERC (Canada). B. Francis is with the Department of Electrical and Computer Engineering, University of Toronto, Canada

1

b

a

Figure 2: An infinite chain of resistors.

Figure 3, insert a 1 A current source from node 1 to node 0 and calculate the voltage drop from 0 to 1. Then the equivalent resistance equals this voltage drop, by Ohm’s law.

0

1

1A Figure 3: A current source inserted. Let us temporarily imagine currents into every node. Label the currents and voltages as in Figure 4. Kirchhoff’s current law at node m is

v0

v1 i0

i1

Figure 4: Current sources into every node.

im = (vm − vm−1 ) + (vm − vm+1 ) = 2vm − vm−1 − vm+1 .

(1)

For the single current source in the second figure, the applied current is im = δm − δm−1 , where δm equals 1 for m = 0 and 0 otherwise, and the equivalent resistance, Req , equals v0 − v1 . This suggests defining wm = vm − vm+1 , writing the equations for wm , and solving for w0 . We have im = −wm−1 + wm .

2.1

(2)

Operator-theoretic formulation

We shall convert equation (2) into a single vector equation. Stack im and wm into vectors i and w:     .. ..  .   .   i−1   w−1        . i =  i0  , w =  w 0    i1   w1      .. .. . . 2

The horizontal lines matrix  .. .   ... 0   ... 1  U =  ... 0  ... 0   ... 0  .. .

merely distinguish between index m < 0 and index m ≥ 0. Also, form the .. . 0 0 1 0 0 .. .

.. . 0 0 0 1 0 .. .

.. . 0 0 0 0 1 .. .

.. . 0 0 0 0 0 .. .

 ... ... ... ... ...

The inverse matrix, that is, the one  .. .. .. .. .. . . . . .   ... 0 1 0 0 0   ... 0 0 1 0 0  −1 U =  ... 0 0 0 1 0  ... 0 0 0 0 1   ... 0 0 0 0 0  .. .. .. .. .. . . . . .

     .     

satisfying U U −1 = U −1 U = I, is  ... ... ... ... ...

     .     

Then (2) can be written as i = (I − U )w. Let δ denote the vector with components δm . Then (I − U )δ is the vector of current inputs to the chain when the 1 A current source is connected from node 1 to node 0. Substituting i = (I − U )δ into the preceding equation gives (I − U )δ = (I − U )w. The matrix I − U is as follows:  .. .. .. .. .. . . . . .   ... 1 0 0 0 0   . . . −1 1 0 0 0   ... 0 −1 1 0 0   ... 0 0 −1 1 0   ... 0 −1 1 0 0  .. .. .. .. .. . . . . .

(3)

 ... ... ... ... ...

     .     

It’s time to say what kind of solution we’re looking for. We take two possibilities: the Hilbert space `2 (Z) of square-summable sequences, with the inner product X hx, yi = x n yn ; n

and the Banach space `∞ (Z) of bounded sequences. Equation (3) is equivalent to the condition that w − δ belongs to the nullspace of I − U . If I − U acts on `2 (Z), its nullspace equals the zero 3

vector 0. In this case w = δ, so w0 = 1 and the equivalent resistance equals 1 ohm. On the other hand, if I − U acts on `∞ (Z), its nullspace equals the 1-dimensional subspace spanned by the vector 1 of all 1’s. Then any vector of the form w = δ + c1,

c∈R

is a possible solution. In this case the equivalent resistance can be any real number—it is not uniquely determined. Figure 5 shows the corresponding current flow for this observation, where c is arbitrary.

1 c

c

1

c 1+c c

c

Figure 5: A solution in `∞ (Z).

2.2

Solution via Fourier transforms

As we just saw, the unique solution of (3) in the Hilbert space `2 (Z) is w = δ. It is on this Hilbert space where the Fourier transform provides a powerful tool. The z-transform of a sequence in `2 (Z) ss X ˆ X(z) = xn z −n . n

Setting z = ejα we get
X ˆ ejα = X xn e−jαn , n

which, if n represented a time variable, would be called the discrete-time Fourier transform. The
ˆ ejα lives in the Hilbert space L2 (−π, π], on which the inner product is function X Z π

1 ˆ ˆ ˆ ejα Yˆ e−jα dα. hX, Y i = X 2π −π ˆ in L2 (−π, π] is a Hilbert space isomorphism, which is to say, it The mapping from x in `2 (Z) to X is linear and continuous, has a continuous inverse, and it preserves inner products. Taking z-transforms in (3) gives ˆ (z) 1 − z −1 = (1 − z −1 )W

(4)

ˆ (z) = 1 and hence wm = δm . Notice that the factors I − U didn’t cancel in from which we get W (3), whereas the factors (1 − z −1 ) did cancel in (4) because we had already constrained the search for wm to the Hilbert space. Fourier transforms don’t work well on `∞ (Z). As a final comment on the 1D case we mention that another common, heuristic approach is to impose a symmetry to the problem. The method is first to apply the current im = δm and 4

1 1 2

1 2

1 2

1 2

1 2

1 2

1 2

1 2

1 1 2

1 2

1 2

1 2

1

1

1 Figure 6: A solution by invoking symmetry. The node voltages are not bounded.

find a symmetric voltage vm ; then apply the current im = δm−1 and find a symmetric voltage; then subtract the two voltages. This approach is depicted in Figure 6, where currents are shown with arrows: We arrive at the same value for Req . Notice that in the top figure |vm | increases as m → ∞ and also as m → −∞. Such a sequence does not have a z-transform and hence this solution is not obtainable by z-transform methods. We also remark that there is no justification for the symmetric current pattern. One cannot invoke a physical reason, because infinite chains do not exist physically.

3

Infinite grid

Now we turn to the 2D grid in Figure 1. We define the problem as follows. Embed the lattice in the plane with the nodes at the points (m, n), where m, n are integers. Place a 1 A current source from node (1, 1) to node (0, 0) as shown in Figure 7. Let vm,n denote the electric potential of node (m, n) with respect to ground. Then the potential difference v0,0 − v1,1 equals the equivalent resistance. With the current source as shown in the preceding figure, only two nodes have current inputs: Node (0, 0) has current input 1 A and node (1, 1) has current output 1 A. Then Kirchhoff’s current law (KCL) at node (m, n) gives this algebraic equation: δm δn − δm−1 δn−1 = 4vm,n − vm−1,n − vm+1,n − vm,n−1 − vm,n+1 .

(5)

Let us bring in matrix notation. First, let V denote the matrix with entries vm,n . We display

5

(1, 1) (0, 0)

(1, 0)

Figure 7: 2D resistor lattice.

V like this:       V =     

.. . ... ... ... ... ...

.. .

.. .

.. .

.. .

v−2,−2 v−2,−1 v−2,0 v−2,1 v−2,2 v−1,−2 v−1,−1 v−1,0 v−1,1 v−1,2 v0,−2 v0,−1 v0,0 v0,1 v0,2 v1,−2 v1,−1 v1,0 v1,1 v1,2 v2,−2 v2,−1 v2,0 v2,1 v2,2 .. .. .. .. .. . . . . .

Let ∆ denote the matrix with  .. .. .. .. . . . .   ... 0 0 0 0   ... 0 0 0 0  ∆=  ... 0 0 1 0  ... 0 0 0 0   ... 0 0 0 0  .. .. .. .. . . . .

 ... ... ... ... ...

     .     

elements δm δn :  .. .  0 ...   0 ...   0 ...  . 0 ...   0 ...   .. .

Notice that if M is a matrix, then the multiplication U M results in shifting all the rows down, U ∗ M all the rows up, M U all the columns left, and M U ∗ all the columns right. Equation (5) can be written succinctly as (2I − U − U ∗ )V + V (2I − U − U ∗ ) = ∆ − U ∆U ∗ , or, defining A = 2I − U − U ∗ , AV + V A = ∆ − U ∆U ∗ .

(6)

To recap, this equation encodes the KCL equations. The matrix V is the matrix of voltages at all the nodes, and ∆ − U ∆U ∗ is the matrix of current inputs to the nodes when a 1 A current source

6

is inserted from node (1, 1) to  .. .   ... 0   ... 0  ∗ ∆ − U ∆U =   ... 0  ... 0   ... 0  .. .

node (0, 0): .. . 0 0 0 0 0 .. .

.. .. .. . . . 0 0 0 0 0 0 1 0 0 0 −1 0 0 0 0 .. .. .. . . .

      .     

... ... ... ... ...

For completeness, the matrix A is  .. .. .. .. .. . . . . .   ... 2 −1 0 0 0 ...   . . . −1 2 −1 0 0 ...  A= . . . 0 −1 2 −1 0 ...   ... 0 0 −1 2 −1 . . .   ... 0 −1 2 ... 0 0  .. .. .. .. .. . . . . .

      .     

Since the equivalent resistance equals v0,0 − v1,1 , it is convenient to introduce voltage drops wm,n = vm,n − vm+1,n+1 and the corresponding matrix W . Now we would like the equation satisfied by W that is analogous to (6). The definition of W is W = V − U ∗ V U.

(7)

Premultiply equation (6) by U ∗ and postmultiply by U and use the fact that A commutes with U and U ∗ : A(U ∗ V U ) + (U ∗ V U )A = U ∗ ∆U − ∆.

(8)

Subtract (8) from (6) and use (7): AW + W A = 2∆ − U ∗ ∆U − U ∆U ∗ .

(9)

Element w0,0 of W is exactly what we’re looking for, namely, the equivalent resistance, v0,0 − v1,1 . Assume for the moment that we know that a solution W to (9) exists. It certainly is not unique. For example, let E denote the infinite matrix of all 10 s. Since AE = 0 and EA = 0, W + E is also a solution to (9). So is W + cE for every constant c. We will show that the solution to (9) is unique if we require it to have a certain property, namely, XX 2 wm,n < ∞. m

n

Such a matrix is called a Hilbert-Schmidt matrix. Theorem 1 There exists a unique Hilbert-Schmidt matrix satisfying equation (9). w0,0 = 2/π.

7

Moreover,

Proof The set of Hilbert-Schmidt matrices forms a Hilbert space under the inner product hX, Y i = trace (X ∗ Y ). Denote this Hilbert space by HS. The (z, µ)-transform of such a matrix is XX ˆ µ) = X(z, xm,n z −m µ−n . m

n

Setting z = ejα and µ = ejβ , we get   XX ˆ ejα , ejβ = xm,n e−jαm e−jβn . X m

n

This function belongs to the Hilbert space L2 ((−π, π] × (−π, π]), whose inner product is Z πZ π     1 ˆ Yˆ i = ˆ ejα , ejβ Yˆ e−jα , e−jβ dα dβ. hX, X (2π)2 −π −π ˆ in L2 ((−π, π] × (−π, π]) is a Hilbert space isomorphism. The mapping from X in HS to X From equation (9) we get   1 1 ˆ 1 2−z− +2−µ− W (z, µ) = 2 − zµ − . z µ zµ Setting z = ejα and µ = ejβ , we get   ˆ ejα , ejβ = 1 − cos(α + β) (1 − cos(α) + 1 − cos(β)) W and so we arrive at   ˆ ejα , ejβ = W

1 − cos(α + β) . 2 − cos(α) − cos(β)

It remains to compute w0,0 = Req by taking the inverse mapping: Z πZ π 1 1 − cos(α + β) dα dβ w0,0 = 2 (2π) −π −π 2 − cos(α) − cos(β) 2 = by Mathematica. π 

4

Discussion

I had an interesting discussion with Professor David Atkinson of Groningen University. He has also worked on the problem [1]. I drew to his attention my claims in this note. Here was his reply: I completely agree with you that the infinite network of resistors yields a system of equations that possess an infinity of solutions. As we noted in the introduction of our paper, (p. 486), we tacitly obtain uniqueness by requiring that the currents at infinity vanish, ... 8

In your note you object that an infinite network is not physical, so appeal to physics is disallowed. Indeed, if you stick to the mathematics, the equations do admit multiple solutions, unless you add the requirement at infinity, in which case there is only one solution, namely the one we published. If you further press the point as to why we should be interested in this particular solution, then physics can be brought in via the Aristotelian distinction between actual and virtual infinity. In an actually infinite network (without a boundary condition at infinity), there is an infinity of solutions of the equations. In a virtually infinite network, which means that you consider a finite network of size N, and then let N tend to infinity, there is only one solution, and moreover it is one in which the currents at the periphery do indeed vanish in the limit. This is the physical solution, according to the Aristotelian canon. If I understand Professor Atkinson correctly, he is saying that an “actually infinite network” is imaginary, while the network that is a mathematical limit is real. These two networks have the same number (cardinality) of resistors, which happens to be more than the number of atoms in the universe. I would therefore claim that they are both imaginary, in the sense of not being physically real. The virtue of the second network is that it gives the answer Professor Atkinson thinks is the right one. Acknowledgement It is a pleasure to acknowledge that Professor Avraham Feintuch steered me to Hilbert Schmidt matrices.

References [1] D. Atkinson and F. J. van Steenwijk. Infinite resistive lattices. Am. J. Physics, 67(6):486 – 492, 1999. [2] A. H. Zemanian. A classical puzzle: the driving-point resistance of infinite grids. IEEE Circuits and Systems Magazine, pages 7 – 9, March 1984.

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