theory of resistor networks

7/21/2019 Theory of Resistor Networks

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arXiv:math-ph/0402038v2

19Feb2004

Theory of resistor networks: The two-point

resistance

F. Y. WuDepartment of Physics,Northeastern University

Boston, Massachusetts, 02115

Abstract

The resistance between arbitrary two nodes in a resistor network is

obtained in terms of the eigenvalues and eigenfunctions of the Lapla-

cian matrix associated with the network. Explicit formulas for two-

point resistances are deduced for regular lattices in one, two, and

three dimensions under various boundary conditions including that of

a Mobius strip and a Klein bottle. The emphasis is on lattices of finite

sizes. We also deduce summation and product identities which can be

used to analyze large-size expansions of two-and-higher dimensionallattices.

Key words: Resistor network, two-point resistance, finite lattices, summa-tion and product identities.

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1 Introduction

A classic problem in electric circuit theory studied by numerous authors overmany years is the computation of the resistance between two nodes in aresistor network (for a list of relevant references up to 2000 see, e.g., [1]).Besides being a central problem in electric circuit theory, the computation ofresistances is also relevant to a wide range of problems ranging from randomwalks (see [2, 3] and discussions below), the theory of harmonic functions [4],first-passage processes [5], to lattice Greens functions [6]. The connectionwith these problems originates from the fact that electrical potentials on agrid are governed by the same difference equations as those occurring in theother problems. For this reason, the resistance problem is often studied from

the point of view of solving the difference equations, which is most conve-niently carried out for infinite networks. Very little attention has been paidto finite networks, even though the latter are the ones occurring in real life.In this paper we take up this problem and present a general formulation forfinite networks. Particularly, known results for infinite networks are recov-ered by taking the infinite-size limit. In later papers we plan to study effectsof lattice defects and carry out finite-size analyses.

The study of electric networks was formulated by Kirchhoff[7] more than150 years ago as an instance of a linear analysis. Here, we analyze the prob-lem along the same line of approach. Our starting point is the consideration

of the Laplacian matrix associated with a network, which is a matrix whoseentries are the conductances connecting pairs of nodes. Just as in graphtheory that everything about a graph is described by its adjacency matrix(whose elements is 1 if two vertices are connected and 0 otherwise), we expecteverything about an electric network is described by its Laplacian. Indeed,in Section 2 below we shall derive an expression of the two-point resistancebetween arbitrary two nodes in terms of the eigenvalues and eigenvectorsof the Laplacian matrix [8]. In ensuing sections we apply our formulationto networks of one-dimensional and two-dimensional lattices under variousboundary conditions including those embedded on a Mobius strip and a Kleinbottle, and lattices in higher spatial dimensions. We also deduce summation

and product identities which can be used to reduce the computational la-bor as well as analyze large-size expansions in two-and-higher dimensions.In subsequent papers we shall consider large-size expansions and effects ofdefects in finite networks, the latter a problem that has been studied in the

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past only for infinite networks [9].

Let Lbe a resistor network consisting ofN nodes numbered by i =1, 2, ,N. Let rij =rji be the resistance of the resistor connecting nodes iandj . Hence, the conductance is

cij =r1ij =cji

so that cij = 0 (as in an adjacency matrix) if there is no resistor connectingiand j .

Denote the electric potential at the ith node by Vi and the net currentflowing intothe network at the ith node by Ii (which is zero if the ith nodeis not connected to the external world). Since there exist no sinks or sourcesof current including the external world, we have the constraint

Ni=1

Ii= 0. (1)

The Kirchhoff law states

Nj=1

cij(Vi Vj) =Ii, i= 1, 2, ,N, (2)

where the prime denotes the omission of the term j =i. Explicitly, (2) reads

L V =I (3)

where

L =

c1 c12 . . . c1Nc12 c2 . . . c2N

... ...

. . . ...

c1N c2N . . . cN

(4)

and

ci N

j=1

cij . (5)

Here V and I areN-vectors whose components are Vi and Ii respectively.We note in passing that if all nonzero resistances are equal to 1, then the

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off-diagonal elements of the matrixL are precisely those of the adjacencymatrix ofL.

The Laplacian matrix Lof the networkLis also known as the Kirchhoffmatrix, or simply the tree matrix; the latter name derived from the factthat all cofactors of L are equal, and equal to the spanning tree generatingfunction forL[10].

To compute the resistance Rbetween two nodes and , we connectand to the two terminals of an external battery and measure the currentIgoing through the battery while no other nodes are connected to externalsources. Let the potentials at the two nodes be, respectively, V and V.Then, the desired resistance is the ratio

R=V V

I . (6)

The computation of the two-point resistance Ris now reduced to solving(3) forV and Vwith the current given by

Ii = I(i i). (7)A probabilistic interpretation:

The two-point resistance has a probabilistic interpretation. Consider arandom walker walking on the networkLwith the probability

pij =cij/ci (8)

of hoping from node i to node j , where pij can be different frompji. LetP(; ) be the probability that the walker starting from node will reachnodebefore returning to, which is the probability of first passage. Thenone has the relation [2,5]

P(; ) = 1

cR(9)

where c is defined in (5). If all resistances are 1, then (9) becomes

P(; ) = 1

zR. (10)

where z is the coordination number of, or the number of nodes connectedto, the node .

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2 The two-point resistance: A theorem

Let i andli be the eigenvectors and eigenvalues ofL, namely,

L i = li i, i= 1, 2, ,N.Leti, = 1, 2, ,Nbe the components of i. Since L is Hermitian, theis can be taken to be orthonormal satisfying

(i , j) =

ij = ij .

Now the sum of all columns (or rows) of L is identically zero, so one ofthe eigenvalues of L is identically zero. It is readily verified that the zeroeigenvaluel1 = 0 has the eigenvector

1 = 1/N, = 1, 2, ,N.

We now state our main result as a theorem.

Theorem:

Consider a resistance network whose Laplacian has nonzero eigenvalues

li with corresponding eigenvectorsi = (11, 12, , 1N), i= 2, 3, ,N.Then the resistance between nodes and is given by

R=Ni=2

1li

i i2. (11)

Proof.

We proceed by solving Eq. (3) by introducing the inverse of the LaplacianL, or the Greens function[6]. However, since one of the eigenvalues ofL iszero, the inverse ofL must be considered with care.

We add a small term Ito the Laplacian, where Iis theN N identitymatrix, and set = 0 at the end. The modified Laplacian

L() =L + I

is of the same form ofL except that the diagonal elements ci are replaced byci+. It is clear that L() has eigenvalues li+ and is diagonalized by thesame unitary transformation which diagonalizes L.

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The inverse ofL() is now well-defined and we write

G() =L1().

Rewrite (3) as L()V() = I and multiply from the left by G() to obtainV() =G()I. Explicitly, this reads

Vi() =N

j=1

Gij()Ij , i= 1, 2, ,N, (12)

where Gij() is the ij -th elements of the matrix G().

We now compute the Greens functionGij ().

Let U be the unitary matrix which diagonalizes L() and L, namely,

ULU =

UL()U = (). (13)

It is readily verified that elements of U are Uij = ji , and and () are,respectively, diagonal matrices with elements li ij and (li+) ij .

The inverse of the second line of (13) is

UG()U= 1()

where 1() has elements (li+)1ij . It follows that we have

G() =U1()U

or, explicitly,

G() =Ni=1

U i

1

li+

U i

= 1

N

+ g() (14)

where

g() =Ni=2

iili+

. (15)

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The substitution of (14) into (12) now yields, after making use of the con-

straint (1),

Vi() =N

j=1

gij ()Ij .

It is now safe to take the 0 limit to obtain

Vi=N

j=1

gij(0)Ij. (16)

Finally, by combining (6), (7) with (16) we obtain

R=g(0) +g(0) g(0) g(0)which becomes (11) after introducing (15). QED

The usefulness of (11) is illustrated by the following examples.

Example 1: Consider the 4-node network shown in Fig. 1 with theLaplacian

L=

2c1 c1 0 c1c1 2c1+c2 c1 c2

0 c1 2c1 c1

c1

c2

c1 2c1+c2

,

wherec1= 1/r1, c2 = 1/r2. The nonzero eigenvalues ofL and their orthonor-mal eigenvectors are

l2 = 4c1, 2 =1

2(1, 1, 1, 1)

l3 = 2c1, 3 = 1

2(1, 0, 1, 0)

l4 = 2(c1+c2), 4= 1

2(0, 1, 0, 1).

Using (11), we obtain

R13 = 1

l2

21 23

2+

1

l3

31 33

2+

1

l4

41 43

2=r1,

R12 = 1

l2

21 22

2+

1

l3

31 32

2+

1

l4

41 42

2=

r1(3r1+ 2r2)

4(r1+r2) .

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Example 2: We considerN(N 1)/2 resistors of equal resistance r ona complete graph ofNnodes. A complete graph is a network in which everynode is connected to every other node. The Laplacian is therefore

L complete graph =r1

N 1 1 1 11 N 1 1 1

... ...

. . . ...

...1 1 N 1 11 1 1 N 1

. (17)

It is readily verified that the Laplacian (17) has eigenvalues 0 = 0 andn =

Nr1, n = 1, 2, ...,

N 1, with corresponding eigenvectors n having

components

n = 1N exp

i2n

N

, n, = 0, 1, ...,N 1.

It follows from (11) the resistance between any two nodes and is

R, = rN

n=1

n n2N =

2

N r. (18)

In subsequent sections we consider applications of (11) to regular lattices.

3 One-dimensional lattice

It is instructive to first consider the one-dimensional case of a linear array ofresistors. We consider free and periodic boundary conditions separately.

Free boundary condition:

ConsiderN 1 resistors of resistance r each connected in series forminga chain ofNnodes numbered from 0, 1, 2, , to N 1 as shown in Fig. 2,where each of the two end nodes connects to only one interior node. This is

the Neumann (or the free) boundary condition. The Laplacian (4) assumesthe form

L free{N1} = r1 T freeN

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where T freeN is theN N matrix

T freeN =

1 1 0 0 0 01 2 1 0 0 0

... ...

... . . .

... ...

...0 0 0 1 2 10 0 0 0 1 1

. (19)

The eigenvalues and eigenvectors of TNcan be readily computed (see, forexample, [11]), and are

n = 2 1 cos n, n= 0, 1, , N 1(N)nx =

1N

, n= 0, all x

=

2

Ncos

(x+ 1/2)n

, n= 1, 2, , N 1, all x, (20)

where n= n/N. Thus, using (11), the resistance between nodesx1 andx2is

R free{N1}(x1, x2) = r

N

N1n=1

cos(x1+

12 )n cos(x2+ 12 )n

21 cos n

= rFN(x1+x2+ 1) +FN(x1 x2)

12

FN(2x1+ 1) 12

FN(2x2+ 1), (21)

where

FN() = 1

N

N1n=1

1 cos(n)1 cos n . (22)

Note that without loss of generality we can take 0 < 2N.The function FN() can be evaluated by taking the limit ofl

0 of the

function I1(0) I1() evaluated in (61) below. It is however instructive toevaluateFN() directly. To do this we consider the real part of the summation

TN() 1N

N1n=1

1 ein1 ein .

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First, writing out the real part of the summand, we obtain

Re TN() = 1N

N1n=1

1 cos n cos n+ cos( 1)n2(1 cos n)

= 1

2

FN(1) +FN() FN( 1)

. (23)

Second, we expand the summand to obtain

TN() = 1

N

N1n=1

1=0

ein/N

and carry out the summation over n. The term = 0 yields FN

(1) andterms 1 can be explicitly summed, leading to

TN() = FN(1) + 1

N

1=1

1 (1)1 ei/N 1

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which is the expected expression.

Periodic boundary conditions:

Consider next periodic boundary conditions for which nodes 0 andN 1are also connected as shown in Fig. 3. The Laplacian (4) of the lattice isthereforeLper{N1} = r

1 TperN where

TperN =

2 1 0 0 0 11 2 1 0 0 0

... ...

... . . .

... ...

...0 0 0 1 2 1

1 0 0

0

1 2

. (29)

The eigenvalues and eigenfunctions of TperN are well-known and are, respec-tively,

n = 2

1 cos2n

,

pernx = 1

Nei2xn, n, x= 0, 1, , N 1. (30)

Substituting (30) into (11), we obtain the resistance between nodes x1 andx2

Rper{N1}(x1, x2) = r

N

N1n=1

ei2x1n ei2x2n

2

2(1 cos2n)= r GN(x1 x2) (31)

where

GN() = 1

N

N1n=1

1 cos(2n)1 cos2n .

The function GN() is again evaluated in a special case of the identity(62) below. Following analyses in section 9, one obtains the recursion relation

GN() GN( 1) = 1 1N

(2 1)

which can be solved to yield

GN() = || 2/N. (32)

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It follows that we have

Rper{N1}(x1, x2) =r |x1 x2|

1 |x1 x2|N

. (33)

The expression (33) is the expected resistance of two resistors |x1 x2| rand(N |x1 x2|) r connected in parallel as in a ring.

4 Two-dimensional network: Free boundaries

Consider a rectangular network of resistors connected in an array ofM Nnodes forming a network with free boundaries as shown in Fig. 4.

Number the nodes by coordinates {m, n}, 0 m M1, 0 n N1and denote the resistances along the two principal directions by r and s. TheLaplacian is therefore

L free{MN}= r1T freeM IN+ s1IM T freeN (34)

where denotes direct matrix products and T freeN is given by (19). TheLaplacian can be diagonalized in the two subspaces separately, yielding eigen-values

(m,n) = 2r1(1 cos m) + 2s1(1 cos n), (35)

and eigenvectors free(m,n);(x,y)=

(M)mx

(N)ny . (36)

It then follows from (11) that the resistance R free between two nodes r1 =(x1, y1) and r2= (x2, y2) is

R free{MN}(r1, r2) =M1m=0

N1n=0 (m,n)=(0,0)

free(m,n);(x1,y1) free(m,n);(x2,y2)2

(m,n)

= r

N

x1 x2

+ s

M

y1 y2

+ 2

MN

M1m=1

N1n=1

cos

x1+ 12

mcos

y1+ 12

n cos x2+ 12m cos y2+ 12n2

r1(1 cos m) +s1(1 cos n) ,(37)

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where

m= mM

, n= nN

.

Here, use has been made of (28) for summing over them= 0 andn= 0 terms.The resulting expression (37) now depends on the coordinates x1, y1, x2 andy2 individually.

The usefulness of (37) is best illustrated by applications. Several examplesare now given.

Example 3: For M = 5, N = 4, r = s, the resistance between nodes{0, 0, }and{3, 3}is computed from (37) as

R free{54}({0, 0, }, {3, 3}) = 34+3

5+ 9877231

27600540

r

= (1.70786...) r. (38)

Example 4: ForM=N= 4, we find

R free{44}({0, 0}, {3, 3}) =(r+s)(r2 + 5rs+s2)(3r2 + 7rs+ 3s2)

2(2r2 + 4rs+s2)(r2 + 4rs+ 2s2) . (39)

Example 5: We evaluate the resistance between two nodes in the interiorof a large lattice. Consider, for definiteness, both M, N = odd (for otherparities the result (40) below is the same) and compute the resistance between

two nodes in the center region,

r1 = (x1, y1) =M 1

2 +p1,

N 12

+q1

r2 = (x2, y2) =M 1

2 +p2,

N 12

+q2

,

where pi, qi


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and the summation in (37) breaks into four parts. In the M, N limitthe summations can be replaced by integrals. After some reduction we arriveat the expression

R(r1, r2) = 1

(2)2

20

d 2

0d

1 cos

x1 x2) cos(y1 y2)

r1(1 cos ) +s1(1 cos )

.

(40)Our expression (40) agrees with the known expression obtained previously[1]. It can be verified that the expression (40) holds between any two nodesin the lattice, provided that the two nodes are far from the boundaries.

5 Two-dimensional network: periodic bound-ary conditions

We next consider anMNnetwork with periodic boundary conditions. TheLaplacian in this case is

Lper{MN} = r

1TperM IN+ s1IM TperN (41)

where TperN is given by (29). The Laplacian (41) can again be diagonalizedin the two subspace separately, yielding eigenvalues and eigenvectors

(m,n) = 2r1(1 cos2m) + 2ss(1 cos2n)(m,n);(x,y) =

1MN

ei2xmei2yn . (42)

This leads to the resistance between nodes r1= (x1, y1) and r2= (x2, y2)

Rper{MN}(r1, r2) =M1m=0

N1n=0 (m,n)=(0,0)

(m,n);(x1,y1) (m,n);(x2,y2)2(m,n)

= r

N x1

x2(x1 x2)2

M +

s

M y1

y2 (y1 y2)2

N

+ 1

MN

Mm=1

Nn=1

1 cos2(x1 x2)m+ 2(y1 y2)n

r1(1 cos2m) +s1(1 cos2n) ,

(43)

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where the two terms in the second line are given by (33). It is clear that the

result depends only on the differences|x1 x2| and|y1 y2|, as it shouldunder periodic boundary conditions.Example 6: Using (43) the resistance between nodes{0, 0} and{3, 3}

on a 5 4 periodic lattice with r= s is

Rper{54}

{0, 0}, {3, 3}

=

3

10+

3

20+

1799

7790

r

= (0.680937...) r. (44)

This is to be compared to the value 1.707863... r for free boundary conditionsgiven in Example 3. It can also be verified that the resistance between nodes({0, 0}) and ({2, 1}) is also given by (44) as it must for a periodic lattice.

In the limit ofM, N with|r1 r2| finite, (43) becomes

R(r1, r2) = 1

(2)2

20

d 2

0d

1 cos(x1 x2)+ (y1 y2)

r1(1 cos ) +s1(1 cos )

= 1

(2)2

20

d 2

0d

1 cos(x1 x2) cos(y1 y2)r11 (1 cos ) +s1(1 cos )

,(45)

which agrees with (40).

6 Cylindrical boundary conditions

Consider an M Nresistor network embedded on a cylinder with periodicboundary in the direction ofMand free boundaries in the direction ofN.

The Laplacian is

Lcyl{MN}= r

1 TperM IN+s1 IM T freeN

which can again be diagonalized in the two subspaces separately. This givesthe eigenvalues and eigenvectors

(m,n) = 2r1(1 cos2m) + 2s1(1 cos n),

cyl(m,n);(x,y) = 1

Mei2xm(N)ny .

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It follows that the resistance R free between nodes r1 = (x1, y1) and r2 =

(x2, y2) is

Rcyl{MN}(r1, r2) =M1m=0

N1n=0 (m,n)=(0,0)

cyl(m,n);(x1,y1) cyl(m,n);(x2,y2)2

(m,n)

= r

N

x1 x2(x1 x2)2M

+

s

M

y1 y2+

2

MN

M1m=1

N1n=1

C1

2 +C22 2C1C2cos 2(x1 x2)m

r1(1 cos2m) +s1(1 cos n)

,

where

C1= cos

y1+1

2

n, C2 = cos

y2+

1

2

n. (46)

It can be verified that in theM, N limit (46) leads to the same expres-sion (40) for two interior nodes in an infinite lattice.

Example 7: The resistance between nodes{0, 0} and{3, 3} on a 5 4cylindrical lattice with r= s is computed to be

Rcyl{54}

{0, 0}, {3, 3}

=

3

10+

3

5+

5023

8835

r

= (1.46853...) r. (47)

This is compared to the values of (1.70786...) r for free boundary conditionsand (0.680937...) r for periodic boundary conditions.

7 Mobius strip

We next consider an M N resistor lattice embedded on a Mobius strip ofwidthNand length M, which is a rectangular strip connected at two endsafter a 180o twist of one of the two ends of the strip. The schematic figure of

a Mobius strip is shown in Fig. 5(a). The Laplacian for this lattice assumesthe form

LMob{MN}=r1

HM IN KM JN

+s1IM T freeN (48)

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where

KN =

0 0 0 10 0 0 0...

... . . .

... ...

0 0 0 01 0 0 0

, JN=

0 0 0 10 0 1 0...

... . . .

... ...

0 1 0 01 0 0 0

,

HN = TperN + KN=

2 1 0 01 2 0 0

... ...

. . . ...

...0 0 2 10 0 1 2

are N N matrices. Now T freeN and JN commute so they can be replacedby their respective eigenvalues 2(1 cos n) and (1)n and we need only todiagonalize an M Mmatrix. This leads to the following eigenvalues andeigenvectors of the Laplacian (48) [11,12]:

(m,n) = 2r1 cos

4m+ 1 (1)n

2M

+ 2s1

1 cosn

N

,

Mob(m,n);(x,y) = 1

Mexp

i

4m+ 1 (1)nx

2M

(N)ny (49)

where(N)ny is given in (20). Substituting these expressions into (11) and aftera little reduction, we obtain

RMob{MN}(r1, r2) =M1m=0

N1n=0 (m,n)=(0,0)

Mob(m,n);(x1,y1) Mob(m,n);(x2,y2)2

(m,n)

= r

N

x1 x2(x1 x2)2M

+

1

MN

M1

m=0

N1

n=1

C12 +C2

2 2C1C2cos

(x1 x2)(4m+ 1 (1)n) 2M

r11 cos (4m+ 1 (1)n) 2M

+s1(1 cos n) ,

(50)

where C1 andC2 have been given in (46).

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Example 8: The 22 Mobius strip is a complete graph ofN= 4 nodes.For r =s the expression (50) gives a resistance r/2 between any two nodeswhich agrees with (18).

Example 9: The resistance between nodes (0, 0) and (3, 3) on a 5 4Mobius strip with r= s is computed from (50) as

RMob{54}{0, 0}, {3, 3}

=

3

10+

1609

2698

r

= (0.896367...) r. (51)

This is to be compared to the corresponding values for the same networkunder other boundary conditions in Examples 3, 6, and 7.

8 Klein bottle

A Klein bottle is a Mobius strip with a periodic boundary condition imposedin the other direction. We consider anM Nresistor grid embedded on aKlein bottle, a schematic figure of which is shown in Fig. 5(b).

Let the network have a twisted boundary condition in the direction of thelengthMand a periodic boundary condition in the direction of the widthN.Then, in analogous to (48), the Laplacian of the network assumes the form

LKlein{MN} = r1HM

IN

KM

JN +s1IMT

perN . (52)

Now the matricesJNandTperN commute so they can be replaced by their

respective eigenvalues1 and 2(1 cos2n) in (52) and one needs only todiagonalize an M Mmatrix. This leads to the following eigenvalues andeigenvectors for LKlein{MN} [11, 12]:

(m,n)() = 2r1

1 cos

(2m+)

M

+ 2s1

1 cos2n

N

,

Klein(m,n);(x,y) = 1

Mexp

i

2m+x

M

(N)ny , (53)

where

=n = 0, n= 0, 1, ,

N 12

,

= 1, n=

N+ 1

2

, , N 1,

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and

(N)ny = 1N

, n= 0,

=

2

Ncos

(2y+ 1)

n

N

, n= 1, 2, ,

N 1

2

,

= 1

N(1)y n= N

2, for even N only,

=

2

Nsin

(2y+ 1)

n

N

, n=

N

2

+ 1, , N 1.

Substituting these expressions in (11), separating out the summation for

n= 0, and making use of the identity

sin

(2y+ 1)N

2 +n

N

= (1)y cos

(2y+ 1)

n

N

,

we obtain after some reduction

RKlein{MN}(r1, r2) =M1m=0

N1n=0 (m,n)=(0,0)

Klein(m,n);(x1,y1) Klein(m,n);(x2,y2)2

(m,n)(n)

= r

N

x1 x2

(x1 x2)

2

M

+M

1m=0

N

1n=1

1l(m,n)(n)

Klein(m,n);(x1,y1) Klein(m,n);(x2,y2)2

= r

N

x1 x2(x1 x2)2M

+ N

+ 2

MN

1=0

M1m=0

[N12

]n=1

C21 +C22 2(1)(y1y2)C1C2 cos[2(x1 x2)m() ]

l(m,n)() ,

(54)

where

m() =

m+

2

M ,

N = 2

MN

M1m=0

1 (1)y1y2 cos[2(x1 x2)m(1)](m,N/2)(1)

, N= even

= 0, N= odd, (55)

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andCi= cos[(yi+ 1/2)n/N], i= 1, 2,as defined in (46).

Example 10: The resistance between nodes (0, 0) and (3, 3) on a 5 4(N= even) Klein bottle with r = s is computed from (54) as

RKlein{54}{0, 0}, {3, 3}

=

3

10+

5

58+

56

209

r

= (0.654149...) r, (56)

where the three terms in the first line are from the evaluation of correspondingterms in (54). The result is to be compared to the corresponding value forthe same 5 4 network under the Mobius boundary condition considered inExample 9, which is the Klein bottle without periodic boundary connections.

9 Higher-dimensional lattices

The two-point resistance can be computed using (11) for lattices in anyspatial dimensionality under various boundary conditions. To illustrate, wegive the result for anMNLcubic lattice with free boundary conditions.

Number the nodes by{m, n.}, 0 m M 1, 0 n N 1, 0 L 1, and let the resistances along the principal axes be, respectively, r, s,andt. The Laplacian then assumes the form

L free

{MNL}= r1T free

M I

NI

L+s1I

MT free

N I

L+t1I

MI

NT free

L

where T freeN is given by (19). The Laplacian can be diagonalized in the threesubspaces separately, yielding eigenvalues

(m,n,)= 2 r1(1 cos m) + 2 s1(1 cos n) + 2 t1(1 cos ), (57)

and eigenvectors

free(m,n,);(x,y,z)= (M)mx

(N)ny

(L)z

where (M)mx is given by (20) and = /L. It then follows from (11) thatthe resistance R free between two nodes r1 = (x1, y1, z1) and r2 = (x2, y2, z2)

is

R free{MNL}(r1, r2) =M1m=0

N1n=0

L1=0 (m,n,)=(0,0,0)

1(m,n,)

free(m,n,);(x1,y1,z1) free(m,n,);(x2,y2,z2)

2 (58)20


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The summation can be broken down as

R free{MNL}(r1, r2) =M1m=1

N1n=1

L1=1

free(m,n,);(x1,y1,z1) free(m,n,);(x2,y2,z2)2

(m,n,)

+1

LR free{MN}

{x1, y1}, {x2, y2}

+

1

MR free{NL}

{y1, z1}, {y2, z2}

+1

NR free{LM}

{z1, x1}, {z2, x2}

1MN

R free{L1}(x1, x2) 1

NLR free{M1}(y1, y2)

1

LMR free{N1}(z1, z2). (59)

All terms in (59) have previously been computed except the summation in

the first line.Example 11: The resistance between the nodes (0, 0, 0) and (3, 3, 3) in

a 5 5 4 lattice with free boundaries and r = s = t is computed from (59)as

R free{554}({0, 0, 0}; {3, 3, 3}) =

327687658482872

352468567489225

r

= (0.929693...) r. (60)

Example 12: The resistance between two interior nodes r1 and r2 can

be worked out as in Example 4. The result is

R(r1, r2) = 1

(2)3

20

d 2

0d 2

0d

1 cos

x1 x2) cos(y1 y2) cos(z1 z2)r1(1 cos ) +s1(1 cos ) +t1(1 cos )

,

which is the known result [1].

10 Summation and product identities

The reduction of the two-point resistances for one-dimensional lattices tothe simple and familiar expressions of (31) and (33) is facilitated by theuse of the summation identities (27) and(32). In this section we extend theconsideration and generalize these identities which can be used to reduce the

21


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computational labor for lattice sums as well as analyze large-size expansions

in two-and-higher dimensions.We state two new lattice sum identities as a Proposition.

Proposition: Define

I() = 1

N

N1n=0

cos( nN)

cosh l cos( nN), = 1, 2.

Then the following identities hold forl 0, N= 1, 2, ,

I1() = cosh(N )l

(sinh l) sinh(Nl)

+ 1

N

1

sinh

2

l

+ 1 (1)

4 cosh

2

(l/2), 0

< 2N,

(61)

I2() = cosh ( N

2 )l

(sinh l) sinh(Nl/2), 0 < N. (62)

Remarks:

1. It is clear that without the loss of generality we can restrict to theranges indicated.

2. For = 0 and l 0, I1(0) leads to (27) and I2(0) leads to (32).3. In the N

limit both (61) and (62) become the integral

1

0

cos()

cosh l cos d = el

sinh l 0. (63)

4. Set = 0 in (61), multiply by sinh l and integrate over l, we obtainthe product identity

N1n=0

cosh l cosn

N

= (sinh Nl) tanh(l/2). (64)

5. Set = 0 in (62), multiply by sinh l and integrate over l. We obtainthe product identity

N1n=0

cosh l cos2n

N

= sinh2(Nl/2). (65)

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Proof of the proposition:

It is convenient to introduce the notation

S() = 1

N

N1n=0

cos( n)

1 +a2 2a cos n , a


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after making use of (25). So the summation over = odd terms yields

N1

m=0a2m+1

=a/N(1 a2

), and we have

ReN1n=0

1

1 a ein = 1

1 a2N + a

N(1 a2) (70)

Equating (69) with (70) we obtain

S1(0) = 1

1 a2

1 +a2N

1 a2N

+ 2a

N(1 a2)

. (71)

To evaluate S1() for general , we consider the summation

Re 1N

N1n=0

1 (a ein)1 a ein = Re

1N

N1n=0

(1 a ein)(1 a ein)|1 a ein |2

= S1(0) aS1(1) aS1() +a+1S1( 1),(72)where the second line is obtained by writing out the real part of the summandas in (69). On the other hand, by expanding the summand we have

Re 1N

N1n=0

1 (a ein)1 a ein = Re

1

N

N1n=0

1m=0

ameimn/N

= 1 +R

e 1

N

1

m=1

am1 (

1)m

1 eim/N= 1 +

a(1 a)N(1 a2) , = even< 2N

= 1 +a(1 a1)

N(1 a2), = odd


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The recursion relation (74) can be solved by standard means. Define the

generating function

G(t) =

=0

S() t, = 1, 2. (76)

Multiply (74) byt and sum over . We obtain

(1 at)G1(t) S1(0) = A a1t

1 a1t+ t+at2

N(1 a2)(1 t2) . (77)

This leads to

G1(t) = 11 at

S1(0) + A a

1

t1 a1t + t+at

2

N(1 a2)(1 t2)

= 1

(1 a2)(1 a2N)

1

1 at+ a2N

1 a1t

+ 1

2N(1 a)2(1 t) 1

2N(1 +a)2(1 +t),

from which one obtains

S1() = a +a2N

(1

a2)(1

a2N)

+ 1

2N(1

a)2

(1)

2N(1 +a)2

= a +a2N

(1 a2)(1 a2N)+ 12N

4a

(1 a2)2 +1 (1)

(1 +a2)2

. (78)

It follows that using I1() = 2 a S1() we obtain (61) after setting a = el.

QED 2. Proof of (62):

Again, we first evaluateS2(0) by carrying out the summation

Re 1N

N1

n=01

1

a ei2n

, a


26/35

whereS2() is defined in (66). Secondly by expanding the summand we have

1

N

N1n=0

1

1 a ei2n = 1

N

N1n=0

=0

aei2n/N = 1

1 aN (81)

where by carrying out the summation over n for fixed all terms in (70)vanish except those with = mN, m= 0, 1, 2,...Equating (81) with (80) weobtain

S2(0) = 1

1 a2

1 +aN

1 aN

(82)

and from (68)

S2(1) = 1

1 aN.

We consider next the summation

Re 1N

N1n=0

1 (a ei2n)1 a ei2n a


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The recursion relation (86) can be solved as in the above. Define the

generating function G2(t) by (76). We find

G2(t) = 1

1 at

S2(0) + aN1t

(1 aN)(1 a1t)

= 1

(1 a2)(1 a2N)

1

1 at+ aN

1 a1t

, (87)

from which one reads off

S2() = a +aN

(1 a2)(1 a2N) . (88)

Using the relation I2() = 2 a S2() with a = el, we obtain (62). QED

Acknowledgments

I would like to thank D. H. Lee for discussions and the hospitality at Berkeleywhere this work was initiated. I am grateful to W.-J. Tzeng for a criticalreading of the manuscript and helps in clarifying the Mobius strip and Kleinbottle analyses, and to W. T. Lu for assistance in preparing the graphs. Workis supported in part by NSF Grant DMR-9980440.

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References

[1] J. Cserti, Application of the lattice Greens function for calculating theresistance of an infinite network of resistors, Am. J. Phys. 68 896-906(2002).

[2] P. G. Doyle and J. L. Snell, Random walks and electric net-works, The Carus Mathematical Monograph, Series 22 (The Math-ematical Association of America, USA, 1984), pp. 83-149; Also inarXiv:math.PR/0001057.

[3] L. Lovasz, Random walks on graphs: A survey, in Combinatorics, PaulErdos is Eighty, Vol. 2, Eds. D. Miklos, V. T. Sos, and T. Szonyi(Janos Bolyai Mathematical Society, Budepest, 1996) pp. 353-398. Alsoat http://research.microsoft.comlovasz/ as a survey paper. I am in-debted to S. Redner for calling my attention to this reference.

[4] B. van der Pol, The finite-difference analogy of the periodic wave equa-tion and the potential equation, in Probability and Related Topics inPhysical Sciences, Lectures in Applied Mathematics, Vol. 1, Ed. M. Kac(Interscience Publ. London, 1959) pp. 237-257.

[5] S. Redner, A Guide to First-passage Processes (Cambridge Press, Cam-bridge, UK 2001).

[6] S. Katsura, T. Morita, S. Inawashiro, T. Horiguchi, and Y. Abe, LatticeGreens function: Introduction, J. Math. Phys. 12 892-895 (1971).

[7] G. Kirchhoff, Uber die Auflosung der Gleichungen, auf welche man beider Untersuchung der linearen Verteilung galvanischer Strome gefuhrtwird, Ann. Phys. und Chemie, 72497-508 (1847).

[8] A study of random walks using matrices can be found in[3].

[9] J. Cserti, G. David, and Attila Piroth, Perturbation of infinitenetworks of resistors, Am. J. Phys. 70 153-159 (2002). Also in

arXiv:cond-mat/0107362.

[10] See, for example, F. Harary, Graph theory (Addison-Wesley, Reading,MA 1969).

28
http://arxiv.org/abs/math/0001057http://arxiv.org/abs/cond-mat/0107362http://arxiv.org/abs/cond-mat/0107362http://arxiv.org/abs/math/0001057


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[11] W.-J. Tzeng and F. Y. Wu, Spanning Trees on Hypercubic Lattices and

Non-orientable Surfaces, Appl. Math. Lett. 13 (7) 19-25 (2000).[12] I am indebted to W.-J. Tzeng for working out (49) and (53).

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Figure Captions

Fig. 1. A network of 4 nodes.

Fig. 2. A one-dimensional network ofNnodes with free ends.

Fig. 3. A one-dimensional network ofN nodes with periodic boundaryconditions

Fig. 4. A 5 4 rectangular network.

Fig. 5 (a) The schematic plot of an Mobius strip. (b) The schematic plotof a Klein bottle.

30


31/35

Fig.1

1

2 3

4

r1 r1

r1

r1

r2


32/35

Fig.2

0 1 2 N-1


33/35

Fig.3

0

1

2

3

N-1


34/35

Fig.4

(0,0) (5,0)

(0,1)

(0,2)

(0,3)

(5,1)

(5,2)

(5,3)

(1,0) (2,0) (3,0) (4,0)


35/35

Fig.5

(a) (b)

theory of resistor networks

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