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9c. H. Bennett, "The thermodynamics of computation-a review," Int. J.

Theor. Phys. 21, 905-940 (1982). Reprinted in Ref. 4.

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110. Penrose, Foundations of Statistical Mechanics (Pergamon Press, Ox-

ford, 1970), pp. 221-238.

12R.Landauer, "Information is physical," Phys. Today 44 (5), 23-29 (May

1991).

131.D. Barrow, The World Within the World (OxfordUniversity Press, Ox-

ford, 1988), pp. 127-130.

14W.H. Zurek, "Maxwell's demon, Szilard 's engine and quantum measure-

ments," in G. T. Moore and M. O. Scully, Frontiers of Nonequilibrium

Statistical Physics (Plenum Press, New York, 1984), pp. 1 and 161. Re-

printed in Ref. 4.

15w. Ehrenberg, "Maxwell's demon," Sci. Am. 217 (5), 103-110 (Novem-

ber 1967).

l&rhis follows from the fact that the canonical partition function for classical

ideal gases has the form Z=[gV(27rmkT/h2)3/2qrot'lvibqeiY'/NI, where m

is the molecular mass, h is Planck's constant, g is the spin degeneracy, and

the q factors are the single-molecule partit ion functions for rotational,

vibrational, and electronic transitions, respectively. Each q depends only

on T and internal properties of the molecules ; see for example, D. S. Betts

and R. E. Turner, Introductory Statistical Mechanics (Addison-Wesley,

Reading, MA, 1992), pp. 125-129.

17Setting the derivative of the left side of Eq. (11) equal to zero gives

W2=WI exp([asl- as 21/k ) and wI=I![1+exp([as1- as 21/k ) . Substi-

tuting back into Eq. (11) yields, W1[asl/k+1n wil

X[1 +exp([as1- aS2]1k)1~0, which reduces to Eq. (12).

18p.A. Skordos and W. H. Zurek, "Maxwell's demon, rectifiers, and the

second law: Computer simulation of Smoluchowski's trapdoor," Am. 1.

Phys. 60, 876-882 (1992).

19A. F. Rex and R. Larsen, "Entropy and information for an automated

Maxwell's demon," in Workshop on Physics and Computation PhysComp

'92 (IEEE Computer Society Press, Los Alamitos, 1993), pp. 93-101.~. Landauer, "Information is physical," in Ref. 19, pp. 1-4.

21Arelated analysis is given by M. 1. Klein, "Note on a problem concerning

the Gibbs paradox," Am. 1. Phys. 26, 80-81 (1958). A gas with its nuclei

in an excited metastable state and another with the same type nuclei in the

ground state, are mixed. As one species decays into the other, the entropy

of mixing "disappears" but there is a compensating entropy increase in a

constant temperature reservoir. In contrast, our present example unmixes

(separates) the gases first and the subsequent equilibration induces new

mixing because the single species gases each become two-species gases.

On the resistance between two polnts on a grid

Giulio VenezianPhysics Department, Southeast Missouri State University, Cape Girardeau, Missouri 63701

(Received 8 November 1993; accepted 6 June 1994)

The resistance between two adjacent nodes on an infinite square grid of equal resistors can easily be

found by superposition. This paper addresses the corresponding problem for two arbitrary nodes. A

solution is found by exploiting the symmetry of the grid and using the method of superposition. Themathematical problem involves the solution of an infinite set of linear, inhomogeneous difference

equations which are solved by the method of separation of variables.

I. INTRODUCTION

A recently published general physics textbook! presents a

problem in which the student is asked to find the resistance

between two adjacent nodes in an infinite plane grid of iden-

tical resistors. As a follow-up question, the student is then

asked to find the resistance between two nodes which are an

arbitrary distance apart. From the statement of the problem,

it is apparent that the author of the problem assumed that the

second question could be answered as easily as the first. In a

private communication with the writer of this article, the

author of the problem confirmed that he had made that

assumption.' The second problem turns out to be consider-

ably more complicated than the first. It requires the applica-

tion of mathematical techniques which are beyond the reach

of the average general physics student, but would make it a

good problem to use in a mathematical physics class to show

an application of two-dimensional difference equations, a

topic that is often neglected.

IT.DESCRIPTION OF THE PROBLEM

The physical situation is illustrated in Fig. 1. An infinite

number of identical resistors of resistance R are connected to

1000 Am. J. Phys. 62 (11), November 1994

form a square grid. The problem is to find the resistance that

would be measured between two arbitrarily spaced nodes.

The basic approach used here is the same as that used by

Paul3 to treat the case of adjacent nodes. For this reason, the

problem of adjacent nodes will be discussed first.

ITI.ADJACENT RESISTORS

Following Paul or Purcell;' for instance, let a current 1

enter the grid at a node P and let it come out of the grid at a

distant point. If the return point is removed to infinity, the

problem is invariant under 90° rotations, so the current flow-

ing through each of the four resistors connected to node P

will be equal. Thus each of the four resistors carries a current

equal to 1/4. The resulting voltage drop between node P and

an adjacent node Q will be equal to (1/4)R, the product of

the resistance of the resistor and the current passing through

it.Now consider the problem of a current 1entering the grid

at a distant point and exiting at the adjacent node, Q. In thiscase, the current pattern at Q will be symmetric and the

current flowing from P to Q will be 1/4, and the voltage drop

from P to Q will again be given by (1/4)R.

© 1994 American Association of Physics Teachers 1000



Fig. 1. A portion of an array of equal resistors.

The superposition of the two problems results in the prob-

lem of a current 1 entering the grid at node P and exiting atthe adjacent node Q with a net voltage drop IR/2 between P

and Q. The distant point is eliminated since the net current

there is zero. The resistance between adjacent nodes is there-

fore R/2. It should be noted that because of the fourfold

symmetry of the current and voltage pattern in this configu-

ration, the voltage drop between any two points P and Q,adjacent or not, that results when the current enters at P and

leaves at infinity has to be equal to the voltage drop between

P and Q which results when the same current enters at in-

finity and leaves at Q. Thus the resistance between any two

points P and Q can be found as 2V/1 where V is the voltage

drop from P to Q which results when the current 1 enters at

P and leaves at infinity. This is the essential element in thesolution presented in this paper.

IV. ANALYSIS IN TERMS OF NODE VOLTAGES

Consider an infinite square array of resistors such as that

shown in Fig. 1. Let the nodes be numbered from minus

infinity to plus infinity in each direction, with the voltage at

the node located in the mth row and nth column denoted by

Vm ,n . It is convenient at this point to allow external currents

1m,n to enter the nodes, although in our problem only one of

these external currents will be nonzero.

Applying Kirchhoff's current law at node (m,n), the ex-

ternal current entering the node must be equal to the currentleaving the node through the four resistors which are at-

tached to it. Thus

Im ,n= (Vm ,n- Vm ,n+l) /R+ (Vm ,n- Vm,n-d/R

+ (Vm,n- V m +l,n)/R+ (Vm ,n- Vm-1 ,n)/R.

At a node where 1m,n is zero, this reduces to

4 V m,n= Vm,n+ l + Vm,n-l + Vm+1 ,n+ Vm- 1,n '

To define the problem completely, let a current 1enter node

(0,0) and leave the grid at infinity. Then

Im ,n=O unless m=O and n=O, and 10,0=1. (3 )

Equations (1) and (2) are the finite difference equivalent of

Poisson's and Laplace's equations, respectively. Indeed, the

1001 Am. J. Phys., Vol. 62, No. 11, November 1994

latter equations are the ones that need to be solved in prob-

lems related to current flow in thin films, of which the

present problem is a finite difference model. However, while

Laplace's equation in two dimensions can be solved el-

egantly by conformal mapping" techniques, this is not the

case for the finite difference equations considered here. Thus

it becomes necessary to use other methods. In preliminary

tests using a spreadsheet, the author found that satisfactory

answers could be found when a current was introduced at a

grid point and withdrawn at another. The node voltages

would quickly settle down to steady values after a few itera-tions, and if the grid size was sufficiently large compared to

the spacing between the two nodes, the voltage pattern was

reasonably independent of grid size. On the other hand, if the

current was simply introduced at one grid point and allowed

to exit at the boundaries of the grid which were held at zero

voltage, the node voltages tended to keep changing as the

grid size was increased, although the currents flowing

through the grid tended to settle to constant values.

The reason for this effect can be seen by referring to the

corresponding continuous medium problems. For a dipole,

the distant field dies off and the effect of the boundaries

becomes negligible, whereas for a monopole, the electric po-

tential is logarithmic, with an infinite voltage difference be-tween the source and the boundaries, but the electric field,

and hence the current, die off in inverse proportion to the

distance from the source.

After this preliminary numerical study, the author was

then led to look for a mathematical solution to the problem,

which is presented here.

V. SOLUTION BY SEPARATION OF VARIABLES

Although the method of separation of variables is com-

monly applied to partial differential equations having suit-able boundaries, it is equally applicable to difference

equations'Consider

(4 )

Substituting this expression into Eq. (2), the result is

4e mu+ in .B= e (m+ l )u+ in .B+ e (m -l )u+ in .B

+ emu+ i( n+ l) .B+ emu+ i (n -l ).B ,

or

4e mu+in .B= em u+ in .B ( e" + e-U+ ei.B+ e -i.B),

so that Eq. (2) is satisfied provided that

cosh a+cos {3=2. (5)

(1)

Keeping in mind that our ultimate goal is to solve the

problem with a source at (0,0), and taking advantage of the

symmetries in the problem, which require that

Vm,n=vn,m = V-m,n= vm,-n=v -m,-n' (6 )

(2)take

(7)

The functions v m,n corresponding to a particular pair of

values (a,{3) will be denoted by vm,n({3). Because of the ab-solute value sign in the exponential terms, these functions do

not satisfy the source-free difference relation given by Eq.

Giulio Venezian 1001



(2) along the lines n =0 and m =0, so there will be residual

currents entering or leaving the grid at each node along those

lines.

Let us find the external currents im n which produce the

voltage pattern v m.n • '

From Eq. (1),

= 8 -4( cos {3+e-a),

which, using Eq. (5) simplifies to

Rio ,0=4 sinh a.

Similarly, for n :#=0,let

Rio,n=4v O,n - VO,n+1- VO,n-1 - VI,n - V -l,n'

=4 cos n{3+4e1n1a-cos(n+ 1){3-eln+lla

-cos(n-l){3-eln-Ilu-2e-a cos n{3-e1n1a cos {3,

=2 cos n{3(2-eU-cos {3)

+2e1n1a(2-cos {3-cosh a),

which, using Eq. (5) simplifies to

Rio ,n=2 cos n{ 3 sinh a .

Vi. CURRENT ENTERING AT (0,0)

Since the assumed node voltage pattern vm,n({3) requires

external currents im n ({ 3 ) not only at node (0,0) but at all

nodes for which either m =0 or n =0, it is necessary to form

a superposition of such voltages, with different values of {3,

to suppress all the external currents except the one at (0,0).

Accordingly, let

vm,n= t T F({3)vm,n({3)d{3, (10)

where the limits of the integral have been chosen to cover the

entire applicable range of values of {3.The function F({3) is

an amplitude function that must be chosen to make 10,0= Iand 1 0 n=O when n:#=O.

Substituting Eq. (10) into Eq. (1), the corresponding ex-

pression for the external currents is

R I m ,n= t T F({3)im ,n({3)d{3.

From Eqs. (8) and (9),

RIo 0= (1I"[F(f3)4 sinh a]d{3,, J o

(11)

(12)

and

RIo n= ( 1 1 " F({3)2 sinh a cos n{ 3 d{ 3,, J o (13)

where

1 0,n =lo ,-n= ln,0= 1 -n,O '

By inspection, the expression

F({3)=IR/(47T" sinh a)

(14)

(15)

satisfies the requirements of the present problem since it

makes 10,0=1 and 10,n=0 when n'*O.


Thus

V m n= (IR/47T") ( 1 1 " vm n({3)/sinh a d{3 ,, J o '

= ( IR/47T") f01l"(e-,m,a cos n{ 3

+e-1n1u cos m{3)/sinh a d{3 . (16)

(8 )

It must be remembered thatais a function of {3,where their

relationship is given by Eq. (5).

VII. RESISTANCE BETWEEN TWO POINTS IN A

LARGE GRID

(9)

It was mentioned earlier that the resistance between (0,0)

and (m,n), Rm,n, could be obtained directly from the solu-

tion of the problem in which the current Ienters at (0,0) and

leaves at infinity. In terms of the node voltages, the resistance

is Rm,n=2(Vo,0- Vm,n)/I, thus the final result is

Rm,n=(R! 27T ) f 01 l" (2 -e -,m, a cos n{ 3

_e-1n1a cos m{3)/sinh a d{3 , (17)

where a and {3are related through Eq. (5). It is shown in the

appendix that Rm,n can also be written as

Rm n=(R /7T") (1I"(I_e-1m1a cos n{3)/sinh a d{3 ,, J o

= (R/7T")f01l"(I-e-lnlu cos m{3)/sinh a d{3 . (18)

VIII. SPECIAL CASES

In general the integrals in Eqs. (17) or (18) have to be

evaluated numerically. Results for Ro,o through Rto, to are

presented in Table I.Some special cases that can be worked

out or approximated analytically are noted below:

(a) Ro,oSetting n =0, m =0 in Eq. (17), the numerator of the inte-

grand becomes zero, so that Ro,o=O, as it should.

(b) Ro ISetting n =0, m = 1 in Eq. (17), the numerator becomes

2-cos {3-e-u=cosh a-e-a=sinh a,

so that RO,1= (R!2), which is in agreement with the special

case discussed at the beginning of this paper.

(c) Asymptotic form for large m or n

When either m or n are large, the exponential terms in one

or both of the alternative expressions given by Eq. (18) be-

come negligible except when a is very small. When a is very

small,

cos {3=2-cosh a=l-a2/2 ,

so that a=sinh a={3 .Suppose, for example, that m is large. The first expression

in Eq. (18) can be rewritten as the sum of three integrals




Table I. Resistances Rm•n between node (0,0) and node (m,n) in units of R.

n\m 0 1 2 3 4 5 6 7 8 9 10

0 0.000 0.500 0.727 0.860 0.954 1.026 1.084 1.133 1.176 1.214 1.247

1 0.500 0.636 0.773 0.880 0.964 1.032 1.088 1.136 1.178 1.215 1.248

2 0.727 0.773 0.849 0.924 0.992 1.050 1.101 1.146 1.186 1.221 1.253

3 0.860 0.880 0.924 0.976 1.028 1.076 1.120 1.160 1.197 1.230 1.261

4 0.954 0.964 0.992 1.028 1.067 1.106 1.144 1.179 1.212 1.242 1.271

5 1.026 1.032 1.050 1.076 1.106 1.317 1.169 1.200 1.229 1.257 1.283

6 1.084 1.088 1.101 1.120 1.144 1.169 1.195 1.222 1.247 1.272 1.296

7 1.133 1.136 1.146 1.160 1.179 1.200 1.222 1.244 1.267 1.289 1.3118 1.176 1.178 1.186 1.197 1.212 1.229 1.247 1.267 1.287 1.306 1.326

9 1.214 1.215 1.221 1.230 1.242 1.257 1.272 1.289 1.306 1.324 1.342

10 1.247 1.248 1.253 1.261 1.271 1.283 1.296 1.311 1.326 1.342 1.358

e; n/R=(I/7r) (1T{(I_e-mIPI cos n{3)/{3}d{3, J o

+(1/17') Jo1T{1/sinh a-1/{3}d{3

+(1/11") Jo1T{(e-lmIPcos n{3)/{3

_(e-1m1a cos n{3)/sinh a}d{3

=11+12+13,

The first integral can be expressed in terms of the expo-

nential integral Ein( z), 6

Ein(z)= J:[(1-e-t)/t]dt,

(19)

so that

1= (l/7r)Re{Ein[ 1T(n+ im)]).

The second integral can be integrated numerically, and its

value is found to be 12=-0.033 425.Since the exponentials in the third integral are negligible

except for small values of a and {3, and for those values

a=sinh a= {3,the third integral can be neglected.

Thus

Rm.n/R=(1/1T)Re{Ein[ 1T(n+im)]}-0.033 425.

(20)

For large values of its argument, Ein(z)=ln z+0.577 21.

Thus

Rm n/R=(1/21T)[ln(n2+m2)+2(.577 21)+2 In ' 1 7 ' ]

-0.033 425,

=(1/21T)[ln(n2+m2)+0.51469]. (21)

To verify the validity of this asymptotic expression, con-

sider the values of R1,o, R1,1' R3,4' R5,o, and RlO,10' By direct

calculation their values, in units of R, are found to be 0.5,

0.636, 1.028, 1.026, and 1.358, respectively. Using the as-

ymptotic expression, the corresponding values are 0.515,


0.625, 1.027, 1.027, and 1.358.Thus for reasonable values of

m and n, the asymptotic form gives an excellent approxima-

tion.

IX . D ISCUSS ION

The array of resistors can be regarded as a lumped-

parameter model for a conducting sheet of conductivity a

and thickness d. Thus it is of interest to compare the results

obtained here to the corresponding results for an infinite con-

ducting sheet.

To establish the connection between the resistance R of

the resistors in the infinite grid and the parameters of impor-

tance in an infinite sheet of conductivity a and thickness d,

consider a square array N resistors long and N resistors wide

such as that illustrated in Fig. 2. The equivalent resistance

between two parallel edges is the same as that of N parallel

branches each having N resistors in series, so that the total

resistance is R. The resistors in the branches parallel to the

two edges play no part in this situation as they would carry

no current.

Similarly, if a square piece of side L is cut from a con-

ducting sheet, the resistance between two parallel sides is

given by R = L/uA, where A is the cross-sectional area, L d.Thus R = 1/ ad, independent of the size of the square sheet,

and, for this reason, this property of the sheet is termed the

Fig. 2. A square cut from the infinite array. Only the highlighted resistors

carry a current.




resistance per square. Thus the lumped-parameter model of a

sheet with a resistance per square equal to R is a grid of

equal resistors of resistance R.

Following the same general steps as before, let us investi-

gate the resistance between two points in a conducting sheet

by introducing a current Iat a point in the sheet. Let the

current be distributed as a line source across the thickness of

the sheet. Then at a distance r from this point the radial

current density will be j= 1/2 ttrd and the radial electric

field will be

E= j /u=I/2 'TTurd=IR/2 'TTr.

The corresponding electric potential is

V= - (/R/2 'TT )In r+ Vo ,

where V0 is a constant.

This expression is infinite both at r =0 and at infinity. The

singularity at r= O is due to the assumption that the current

enters as a line source and it can be removed by changing the

current to a current density over finite dimensions.

The singularity at infinity cannot be removed except by

providing an exit for the current at another point on the

sheet. The same arguments that were used before can be used

to find the resistance between two points, giving

R (r )= (R /'T T)ln r +c onsta nt

which agrees with Eq. (21).

ACKNOWLEDGMENTS

Dr. Jai Dahiya brought this problem to the author's atten-

tion. The author would like to thank him and Dr. Lawrence

Pinsky for some useful discussions.

APPENDIX

Instead of using the symmetric functions

v =e- m 1 a cos nf3+e-1n1a cos mf3m, n

defined in Eq. (7), let us use the functions

w =e-1m1a cos nf3 (A1).n ,

which are source free everywhere except along the line

m =0. The external currents io,n which correspond to this

voltage pattern are found to be

Ri o ,n=2 cos nf3 sinh a.

Now let

(A2)

(A3)

The corresponding expression for the external currents is

RIo n= ( 1 r F(f3)2 sinh a cos nf3 df3 ., J o

(A4)

Equation (A4) defines the coefficients of a Fourier cosine

series, so that

2 'TTF (f 3) si nh a=R{Io ,0+2IIo ,n cos nf3}. (AS)

For the case considered here, 2'TTF(f3) sinh a=RI; or

F (f3 ) =R I/ (2 'TT sinh a) , which leads directly to Eq. (18).

1Frederick 1. Keller, W. Edward Gettys, and Malcolm 1. Skove, Physics

(McGraw-Hill, New York, 1993), 2nd ed., problem 24.13, p. 622. The

corresponding 3D problem is given as 24.22.

2Lawrence Pinsky (private communication).

3Clayton R. Paul, Analysis of Linear Circuits (McGraw-Hill, New York,

1989), problem 2.12, p. 68; the answer is provided in the solut ions manual.

Edward M. Purcell, Electricity and Magnetism (McGraw-Hill , New York,

1965), 1st ed., p. 424.

4L . 1. van der Pauw, "A method of measuring the resistivity and Hall

coefficient on lamellae of arbitrary shape," Philips Tech. Rev. 20, 220-224

(1959).

5Phi lip M. Morse and Herman Feshback, Methods of Theoretical Physics

(McGraw-Hill, New York, 1953), pp. 695 and 700.6Milton Abramowitz and Irene A. Stegun, Handbook of Mathematical

Functions, NBS, AMS 55 (U.S. Government Printing Office, Washington,

D.C., 1964), p. 228.

INAN IMATE MACH INES ?

Indeed, progress in the field of the natural sciences has fundamentally transformed man's whole

mode of thinking and feeling. Whereas the earlier humanist age saw everything as animate and

sensitive, we are regretably becoming more and more inclined to view everything under the aspect

of a machine. In the past the rambler roamed singly through woods and fields while in a coach you

could do nothing better than dream and indulge poetic fancy, unless vexation happened to out-

weigh boredom; today we work and calculate even in express train or ocean liner. In the past the

coachman used to direct his horse's will by encouraging it in human language, today we operate

electric motors or crank up our motor car in silence.

And yet we cannot rid ourselves of the idea that nature is animate. Are not today's big machines

working like conscious beings? They snort and puff and howl and moan, uttering sounds of

complaint, fear and warning and whistling shrilly under excess power. They absorb from their

surroundings the materials needed to maintain their power, eliminating what is unusable, subject to

exactly the same laws as our own bodies.

Ludwig Boltzmann, "On the Principles of Mechanics," 1902, reprinted in Theoretical Physics and Philosophical Prob-

lems, edited by Brian McGuiness (Reidel, Boston, 1974), pp. 147-148.

1004 Am. 1. Phys., Vol. 62, No. 11, November 1994 Giulio Venezian 1004

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Infinite Resistor Grid