a beautiful problem in mathematical physics francis wray 23 march 2011a beautiful problem in...

A Beautiful Problem in Mathematical Physics

Francis Wray

23 March 2011 A Beautiful Problem in Mathematical Physics 1

An Inspiring Cartoon


The ProblemFind the resistance between arbitrary nodes on

an infinite square lattice of unit resistorsFirst solved by Van der Pol in the 50s


Why is This a Beautiful Problem• It’s not completely specified

What happens at infinity?What happens before steady state?

• It’s got a trivial solution for adjacent nodes• The solution needs infinities to cancel• Easily extended to three dimensions• It’s got nice symmetries• Triangular and hexagonal geometries are also

interesting• It’s a whole course in applied mathematics• http://www.mathpages.com/home/kmath668/kmath668.htm


http://www.mathpages.com/home/kmath668/kmath668.htm

The One-Dimensional Case (1)


1 2 3 s n-2 n-1 n

Solution by superposition•Inject 1 amp at node s

•The current flowing to the right is (s-1)/(n-1)•The current flowing to the left is (n-s)/n-1

•Remove 1 amp from node t (t>s)•The current flowing to the right is –(t-1)/(n-1)•The current flowing to the left is –(n-t)/(n-1)

• The current flowing between nodes s and t is ((s-1)+(n-t))/(n-1)•The voltage between node t and node s is (t-s)((n-1)-(t-s))/(n-1) for R=1•The resistance between nodes s and t is (t-s){1-(t-s)/(n-1)} for R=1



1 2 3 s n-2 n-1 n

A more formal approach• Let the potential at node r be V(r)• Net zero current into a node implies {V(r)-V(r-1)}+{V(r)-V(r+1)}=0• V(r+1)-2V(r)+V(r-1)=0 for r≠s• V(r+1)-2V(r)+V(r-1)=-1 for r=s• Look for solutions of V(r+1)-2V(r)+V(r-1)=0 in the form V(r)=σr

• σ=1 is a repeated root of the resulting quadratic• Solutions take the form V(r)=αr+β for r≤s and for r≥s• The two solutions are continuous at s, but the gradients (differences) are discontinuous

The One-Dimensional Case (3)• The solution can be written

V(s,r)=(r-1)(n-s)/(n-1) for r≤s V(s,r)=(s-1)(n-r)/(n-1) for r≥s We note that V(s,s+1)-2V(s,s)+V(s,s-1)=-1

• Inject 1 amp at node s, remove 1 amp at node t where t>s• The voltage at s is:

(s-1)(n-s)/(n-1)-(s-1)(n-t)/(n-1) = (s-1)(t-s)/(n-1)• The voltage at t is

-(t-1)(n-t)/(n-1)+(s-1)(n-t)/(n-1) = -(n-t)(t-s)/(n-1)• The resistance between nodes s and t is

(t-s)(s-1+n-t)/(n-1) = (t-s){1-(t-s)/(n-1)}• This is exactly the result we got by considering currents



• What about the infinities?• Consider V(s,r)=(r-1)(n-s)/(n-1) for r≤s Put s=r=n/2 V(s,r)≈(n/2)(n/2)/n As n tends to infinity this diverges• The superposition of a source and sink cancels

the infinities as can be seen in (s-1)(t-s)/(n-1) and -(n-t)(t-s)/(n-1) for the voltages at

nodes s and t


The Two-Dimensional Case (1)

A Trivial Solution for Adjacent Nodes



Resistance to infinity is infinite


R ~ (1+1/3+1/5+1/7+...........)/4


Let V(m,n) be the voltage at node (m,n)As before, we inject current at (0,0)Away from sources of current we have:V(m+1,n)+V(m-1,n)+V(m,n+1)+V(m,n-1)-4V(m,n)=0Look for a general solution of the formV(m,n)= μ mνn

then μ+μ-1 + ν+ν -1 – 4 =0Let μ =exp(iα) and ν =exp(iβ),then cos(α)+cos(β)=2



Consider V(m,n,α, β)=exp(i(|m|α+|n|β))This satisfies the defining equation except wherem=0 or n=0. This also reflects the symmetry of

the problem V(±m,±n)=V(m,n). We need to combine such solutions so that

V(m+1,n)+V(m-1,n)+V(m,n+1)+V(m,n-1)-4V(m,n)=0

except where m=n=0, where we inject a unit current.


The Two-Dimensional Case (5)We look for a solution of the form

where

and m≥0 and n ≥0

We need to choose A(α, β) so that the current I(m,0)=0 for m≠0 and I(0,0)=1


π

-πα+ β ) ) dαV( m, n )= A( α, β ) exp( i( m n

cos( α )+cos( β )=2



π

-π[ 4-exp i -exp -i -2exp i ]dαm, 0 ) ( ) ( ) ( )I( )= A( α, β ) exp( i mα

π

-π[ 4- -2exp i ]dα) 2cos( ) ( )I( m, 0 )= A( α, β ) exp( i mα α

π

-π[ -2exp i ]dα) 2cos( ) ( )I( m, 0 )= A( α, β ) exp( i mα

π

-πdα)I( m, 0 )=-2i [A( α, β ) sin( ) ] exp( i mα

∞

mm=-∞

f( )= c exp( -im ) =m1

2π

π

-πdm )c f( ) exp( i

For m>0

Put ck=I(k,0)=0 for k≠0 and c0=I(0,0)=1→ -1

4πi sin( β )A( α, β )

From the definition of a discrete Fourier transform, we have:



π

-π

-1 1 α+ β ) ) dα4πi sin( β)

V( m,n )= exp( i( m n

For m≥0 and n≥0

where cos( α )+cos( β )=2

Hence we deduce that V(m,n)=V(n,m)

However, from I(0,n)=0 for n>0, we have

π

-π

-1 1 α+ β ) ) dβ4πi sin( α)

V( m,n )= exp( i( m n

But, β and α(β) are dummy variables and we could write

π

-π

-1 1 α+mβ ) ) dα4πi sin( β)

V( m,n )= exp( i( n



π

-π

-1 1 α+ β ) ) dα4πi sin( β)

V( m, n )= exp( i( m n

| |2β =acos( 2-cos( α ) ) acos( 1+α /2) i α as α → 0

•The integral for V(m,n) has a logarithmic singularity•The infinity is no surprise for finite m and n•1 amp has been injected into an infinite impedance earthed at infinity

•As m and n →∞, V(m,n)→0•We note that R(m,n)=2[V(0,0)-V(m,n)] •Inject 1 amp at 0,0 and remove 1 amp at m,n



π

-π[ 1 + ) ] -1 1 - α β dα

2πi sin( β )R( m, n )= exp( i( m n

Hence we have:

π

-π

-1 1 [ 1 α ) ] dα2πi sin( β )

-R( 1, 0 )= exp( i

In particular,

π

0

-1 1 [ 1 α ) ] dαπi sin( β )

R( 1, 0 )= -cos(

1

-1

11 dsπ ( 3-s)( 1+s)R( 1, 0 )=

1

-1

1+sπ 3-s2R( 1, 0 )= [atan ] 1/2

Substitute s=cos(α), then

The Two-Dimensional Case (10) Some observations

• Our prototype solution is V(m,n)= μ mνn

μ =exp(iα) and ν =exp(iβ)

• The range of integration for α is [-π, π ]μ lies on the unit circle

• The range of integration for β is [acos(3),0]• ν lies in the interval [3-√8,1]• This all seems somewhat arbitrary!


The Two-Dimensional Case (11) Some observations

• The solution fits the boundary conditions for I(m,0)=0;

• The choice of β in slide 16 guarantees that |ν|≤ 1• This ensures that V(m,n)→0 as n→∞• Since V(m,n)=V(n,m) we satisfy the boundary

conditions I(0,n)=0 and V(m,n)→0 as m→∞• We have satisfied all the boundary conditions• The solution is unique


The Three-Dimensional Case

• It’s easy to see how to adapt the 2-D solution to 3-D

• We deal with a 2-D rather than 1-D delta function to join the solutions across quadrants

• A crucial difference is that the resistance to infinity is now finite

• R≈(1+1/32+1/52+1/72.....)/6 = π2/48


Other Cases

• Only rectanglar polytopes tesselate in 3 or more dimensions

• In two dimensions, only rectangles, equilateral triangles and regular hexagons tesselate

• As an example consider the case of tesselated equilateral triangles.


The Triangular Case (1)The highlighted lines show the axes of symmetry


0,0

0,1 1,0

0,2 1,1 2,0

0,3 1,2 2,1 3,0

0,4 1,3 2,2 3,1 4,0

0,5 1,4 2,3 3,2 4,1 5,0

1,1

2,1

3,1

4,1

3,2

2,2

1,2

The Triangular Case (2)

As before we have:V(m+1,n)+V(m,n+1)+V(m-1,n+1)+V(m-1,n)+V(m,n-1)+V(m+1,n-1)-6V(m,n)=0Look for a general solution of the formV(m,n)= μ mνn

then μ+μ-1 + ν+ν -1 + μ ν -1+ μ-1ν -6 =0Let μ =exp(iα) and ν =exp(iβ),then cos(α)+cos(β)+cos(α-β) =3


The Triangular Case (3)We look for a solution of the form

where

and m≥0 and n ≥0

We need to choose A(α, β) so that the current I(m,0)=0 for m≠0 and I(0,0)=1


π

-πα+ β ) ) dαV( m, n )= A( α, β ) exp( i( m n

)cos( α )+cos( β )+cos( α β =3

The Triangular Case (4)


As before we have

Hence

π

-π

-1 1 α+ β ) ) dα4πi [ sin( β-α )+sin( β ) ]

V( m, n )= exp( i( m n

]π

-πα+ β ) ) dαI( m, 0 )=-2i A( α, β ) [ sin( β-α )+sin( β ) exp( i( m n

We can substitute cos(α)=s and cos(β)=t into the characteristic equation to gives+t+st+√(1-s2)√(1-t2)=3 → s+t+st-3= -√(1-s2)√(1-t2)which gives a quadratic expression for t as a function of s. Hence we can determine all the terms in the above integrals as a function of α.WORK IN PROGRESS!!!

And

)π

-π[ + ) ] )-1 1 1- α β dα

2πi [sin( β α )+sin( β]R( m, n )= exp( i( m n

Conclusions• This is an interesting problem involving several

branches of mathematics with some subtle features.

• The closed forms involve tricky integrals many of which can be avoided by using recursion using the potential equation.

• It’s a good problem to teach mathematics because each of the stages in the solution require careful thought.

• It is related to practical problems in solid state physics and in crystallography.


a beautiful problem in mathematical physics francis wray 23 march 2011a beautiful problem in...

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