Infinite Grid of
Resistors

Remain, remain thou here,

While
sense can keep it on. And, sweetest, fairest,

As
I my poor self did exchange for you,

To
your so infinite loss, so in our trifles

I
still win of you: for my sake wear this…

Shakespeare

There is a well-known puzzle based on the premise of an
infinite grid of resistors connecting adjacent nodes of a square lattice. A
small portion of such a grid is illustrated below.

Between every pair of adjacent nodes is a resistance R,
and were told that this grid of resistors extends to infinity in all
direction, and were asked to determine the effective resistance between two
adjacent nodes, or, more generally, between any two specified nodes of the
lattice.

For adjacent nodes, the usual solution of this puzzle is
to consider the current flow field as the sum of two components, one being
the flow field of a grid with current injected into a single node, and the
other being the flow field of a grid with current extracted from a single
(adjacent) node. The symmetry of the two individual cases then enables us to
infer the flow rates through the immediately adjacent resistors, and hence we
can conclude (as explained in more detail below) that the effective
resistance between two adjacent nodes is R/2. This solution has a certain
intuitive plausibility, since its similar to how the potential field of an
electric dipole can be expressed as the sum of the fields of a positive and a
negative charge, each of which is spherically symmetrical about its
respective charge. Just as the electric potential satisfies the Laplace equation, the voltages of the grid nodes satisfy the discrete from of the Laplace equation, which is to say, the voltage at each node is the average of the voltages
of the four surrounding nodes. Its also easy to see that solutions are
additive, in the sense that the sum of any two solutions for given boundary
conditions is a solution for the sum of the boundary conditions.

If we accept the premise of an infinite grid of resistors,
along with some tacit assumption about the behavior of the voltages and
currents at infinity, and if we accept the idea that we can treat the
current fields for the positive and negative nodes separately, and that
applying a voltage to a single node of the infinite grid will result in some
current flow into the grid, the puzzle is easily solved by simple symmetry
considerations. We assert (somewhat naively) that if we inject (say) four
Amperes of current into a given node, with no removal of current at any
finite point of the grid, the current will flow equally out through the four
resistors, so one Ampere will flow toward each of the four adjacent nodes.
This one Ampere must flow out through the three other lines emanating from
that adjacent node, as indicated in the left hand figure below.

The figure on the right shows the four nodes surrounding
the negative node, assuming we are extracting four Amperes from that node
(with no current injected at any finite node of the grid). Again, simple
symmetry dictates the distribution of currents indicated in the figure.
Adding the two current fields together, we see that the link between the
positive and negative nodes carries a total of 2 Amperes away from the
positive node, and the other three links emanating from the positive node
carry away a combined total of 1 + 1 + 1 –
(2α + β) = 2 Amperes. Thus the direct link carries
the same current as all the other paths, so the resistance of the direct link
equals the effective resistance of the entire grid excluding that link. The
direct link is in parallel with the remainder of the grid, so the combined
resistance is simply R/2.

This is an appealing argument, and it certainly gives the
right answer (as can be verified by other methods), but the premises are
somewhat questionable, and the reasoning involves some subtle issues that
need to be addressed before it can qualify as a rigorous proof. The
fundamental problem with the simple argument, as stated above, is that it
relies on the notion of forcing current into a node of an infinite grid,
without satisfactorily explaining where this current goes. One hand-waving
explanation is that we may regard the grid as being grounded at infinity,
but this isnt strictly valid, because the resistance from any given node to
infinity is infinite. This is easily seen from the fact that a given node
is surrounded by a sequence of concentric squares, and the number of
resistive links beginning from the central node and expanding outward to
successive concentric squares are 4, 12, 20, 28, , etc., which implies that
the total resistance to infinity is (at least roughly)

The odd harmonic series diverges, so this resistance is
infinite. Therefore, in order for current to enter the grid at a node and
exit at infinity, we would require infinite potential (i.e., voltage) between
the node and the grid points at infinity. To make the argument rigorous, we
could consider a large but finite grid, and convince ourselves that the
behavior approaches the expected result in the limit as the grid size
increases. This is not entirely trivial, because we must be sure the two
sequences of expanding grids, one concentric about the positive node and one
concentric about the negative node, approach mutually compatible boundary
conditions in the limit, giving zero net flow to infinity. To evaluate this
limit, the voltage at the central node (relative to the voltage at infinity)
must approach infinity to provide a fixed amount of current. This is
discussed further in another note, where we also discuss the
arbitrariness of the solution for a truly infinite grid. Strictly speaking,
for a truly infinite grid, the solution is indeterminate unless some
asymptotic boundary conditions are imposed (which are not specified in the
usual statements of the problem).

The unphysical aspect of the problem can also be seen in
the fact that the flow field is assumed to be fully developed, to infinity, a
situation that could not have been established by any realistic physical
process in any finite amount of time. Of course, the postulated grid consists
purely of ideal resistances, with no capacitances or inductances, so there
are no dynamics to consider, and hence one could argue that the entire
current field is established instantaneously to infinity – but this merely
illustrates that the postulated grid is idealized to the point of violating
the laws of physics. All real circuits have capacitance and inductance, which
is why the propagation speed cannot be infinite. One might think that such
idealizations are harmless for this problem, but they actually render the
problem totally indeterminate if we apply them rigorously. Our intuitive
sense that there is a unique answer comes precisely from our unconscious
imposition of physically reasonable asymptotic behavior emanating from a
localized source, based on the asymptotic behavior of a finite grid as the
size increases a conception that arises from our physical notions of
locality and finite propagation of effects, notions which are not justified
in the idealized setting.

Setting aside these issues, and just naively adopting the
usual tacit assumptions about the asymptotic conditions of the grid, we can
consider the more general problem of determining the resistance between any
two nodes. The most common method is based on superimposing solutions of the
basic difference equation. Again this method tacitly imposes plausible
boundary conditions to force a unique answer, essentially by requiring that
the grid behaves like the limit of a large finite grid. Consider first the
trivial example of a one-dimensional grid of unit resistors, in which the
net current emanating from the nth node is given by

where we stipulate that I₀ = –1 Amp and I_n = 0 for all n
≠ 0. Notice that for all n ≠ 0 would could negate the signs of
the indices on the right hand side without affecting the equation, because
the net current is zero at those nodes, and the equations above and below the
origin are symmetrical. However, the case n = 0 is different if we stipulate
that V₁ = V_–1,
and if we stipulate that I₀ = 1, which implies

It follows that, for unit resistors, if we stipulate V₀
= 0, we must have V₁ = 1/2. Therefore, if we imagine current being
extracted from just the node at the origin, while all the other nodes have
zero net current flow, then for all n ≠ 0 we have the relation

The characteristic polynomial is

For any value of μ that satisfies this equation, its
clear that one solution of the preceding difference equation is V_n
= Aμⁿ for any constant A. However, the characteristic
equation has the repeated roots μ₁ = μ₂ = 1,
so we have a resonance term, and the general form of the solution of the
discrete difference equation is

for some constants A and B, chosen to make the solution
consistent with the specified boundary conditions. We want V₀ = 0,
so we must put A = 0. Also, since the recurrence relation doesnt apply at n
= 0, we can chose B equal to +1/2 for positive n, and 1/2 for negative n,
which amounts to taking the absolute value of n, giving the result V_n
= |n|/2. This implies that the effective resistance between nodes separated
by k resistors is (as expected) simply kR, where R is the resistance of an
individual resistor.

We can similarly consider the difference equation for
grids of higher dimensions. For a two-dimensional grid, the current emanating
from node (m,n) for all m,n > 0 is

We stipulate that I_m,n = 0 for all m,n > 0,
so the characteristic equation for this two-dimensional difference equation
is

This shows that there are infinitely many eigenvalues.
Indeed, for any value of μ we can solve for the corresponding value
ν, and vice versa. For any such pair of values μ,ν satisfying
this equation its clear that a solution of the preceding difference equation
is given by V_m,n = A(μ,ν) μ^mνⁿ
for any constant A(μ,ν). If we define parameters α,β such
that iα = ln(μ) and iβ = ln(ν), then these solutions can
be written as

In terms of α and β the characteristic equation
can be written as

Now, any combination of these solutions will satisfy the
original difference equation for nodes with zero net current, but we want I_0,0
= –1 Amp, and to achieve this we
(again) take the absolute values of the indices, i.e., we define

For values of m and n different from zero, taking the absolute
values has no effect on the difference equation, i.e., it still gives zero
net current. However, for m = n = 0 we get

This shows that if we put V_0,0 = 0 then for 1
Amp of current we must have V_1,0 = 1/4 volts, which is consistent
with the fact that the resistance between two adjacent nodes of 1/2 ohms,
because we superimpose this solution with an equal and opposite solution
centered on the adjacent node with +1 Amp current.

We dont yet have a definite solution satisfying all the
conditions, because with the absolute values of the indices we generally get
non-zero currents for I_m,0 and I_0,n. To solve this
problem, we will simply superimpose several of these solutions, and impose
the requirement that the net currents of the form I_m,0 and I_0,n
all vanish. To do this, we integrate the solutions of the form (4) over
α ranging from π to π. Thus we have

with the understanding that β is given as a function
of α by equation (3). We will find that this determines the function
A(α,β). Consider first the requirement I_m,0 = 0.
Inserting the expression for V_m,0 from equation (4) into equation
(1), we have, for all m > 0,

Making use of the characteristic equation (3), this can be
written as

Now we seek to superimpose many of these solutions such
that the net current for these expressions is zero. To do this, we will
integrate this expression over α ranging from π to π. Thus we
have

At this point we recall that the exponential Fourier
series for an arbitrary function f(x) is

Therefore we have

where weve made use of the fact that I_0,0 = –1 and I_m,0 = 0 for all m
≠ 0. From this we infer that

Inserting this into equation (5), we get

Again, its understood that β is given as a function
of α by equation (3). It might seem as if we cannot now force the
currents I_0,n to equal zero. If we had imposed that requirement
first, instead of I_m,0 = 0, by symmetry we would have found

which seems superficially different from (7). However,
notice that the limits of integration are reversed, because α and β
progress in opposite directions. Furthermore, if we take the differential of
the characteristic equation (3) we get

which proves that (7) and (8) are equivalent. As discussed
previously, the resistance between the origin and the node (m,n) is twice the
value of V_m,n – V_0,0,
so we have the formula

This can be split into two integrals, one ranging from
π to 0, and the other ranging from 0 to +π, as follows:

Reversing the sign of α in the first integral, and
noting that cos(α) = cos(α), this can be written as

and hence we have

For example, the resistance between two diagonal corners
of a lattice square is given by

If we define the variable s = cos(α) we have α =
acos(s) and

As α ranges from 0 to π, the parameter s ranges
from 1 to 1, so the preceding integral can be written as

We now make use of the identity

to re-write the integral as

Also, from the identity

we have

so the integral can be written as

(This same result can be derived purely algebraically,
without the use of Fourier series, as described in another note.)
Knowing this resistance value for diagonal nodes, and the resistance value
1/2 for adjacent nodes, we can immediately compute the resistances to several
other nodes by simple application of the basic difference equation. Thus we
have the resistances relative to the origin shown below.

Returning to equation (9), we see that the same
substitutions and identities that we used to simplify R_1,1 enable
us to write the general expression as

where h_m(s) denote the trigonometric
polynomials giving cos(mα) as a function of s = cos(α). The first
several of these polynomials are

The coefficients of these polynomials are given by a
simple recurrence relation, and they also have a simple trigonometric
expression. However, we dont actually need to deal with these polynomials,
because it is sufficient to determine the values of R_0,n by the
above integral, and then all the remaining resistances are easily determined
by simple algebra. Hence we can focus on just the integrals

To simplify this still further, we can define the
parameter

in terms of which s and ds are given by

Re-writing the expression for R_0,n in terms of
the parameter σ, we get

Making use of the indefinite integrals

where

we can determine the values

and so on.

Another way of simplifying the integrals involved in the
infinite grid solution is to return to equation (7), focusing on the case of
positive m and n with m > n, and recalling that the resistance from the
origin to the node (m,n) is twice the voltage given by (7), so we have

where, for convenience, weve shifted the limits of
integration from the range [π,+π] to the range [0,2π]. Now
suppose we define new variables r,s such that α = r + s and β = r
s. Substituting for α and β in equation (3) gives the relation
cos(r)cos(s) = 1. In terms of these parameters equation (10) can be written
as

Obviously we have dα = dr + ds, and differentiating
the relation cos(r)cos(s) = 1 gives

so we have

Making this substitution into (11) gives

Now, since cos(r)cos(s)=1 where cos(z)=(e^iz+e^−iz)/2,
and in view of the identity

we see that r and s can be expressed parametrically in
terms of a single parameter t such that

From this we get

With these substitutions, equation (12) becomes

To this point we have continued to specify the limits of
integration in terms of α, but now we note as α ranges over the
real values from 0 to 2π we have t = (1i)τ where τ is a
real-valued parameter ranging from infinity to 0. Hence we make this change
of variable to express the result as

where weve made use of the fact that the variable τ
can be multiplied by an arbitrary factor inside the curved parentheses
without affecting the integral. For the first diagonal node we have m = n = 1
and the integral is simply

For the resistances to the other nodes along the diagonal
of the lattice, notice that for any m we have

Consequently we have the well-known result

from which all the other resistances are easily computed
using the basic recurrence relation (1). In another note we consider the same problem
from a more algebraic standpoint.

Return to
MathPages Main Menu