Theory of impedance networks: A new formulation F. Y. Wu FYW, J. Phys. A 37 (2004)

R Resistor network  

Ohm’s law R V I Combination of resistors

=  -Y transformation: (1899) Star-triangle relation: (1944) Ising model = =

 -Y relation (Star-triangle, Yang-Baxter relation) A.E. Kenelly, Elec. World & Eng. 34, 413 (1899)

r1r1 r1r1 r1r1 r1r1 r2r2

r1r1 r1r1 r1r1 r1r1 r2r

r1r1 r1r1 r1r1 r1r1 r2r I I/2 I r1r1 r1r1 r1r1 r1r1 r2r2

1 2 r r r r r r r r r r r r

1 2 r r r r r r r r r r r r I I/3 I 1 2 r r r r r r r r r r r r I/6

Infinite square network I/4 I

V 01 =(I/4+I/4)r I/4 I I

Infinite square network

Problems: Finite networks Tedious to use Y-  relation 1 2 (a) (b) Resistance between (0,0,0) & (3,3,3) on a 5×5×4 network is

I0I Kirchhoff’s law r 01 r 04 r 02 r 03 Generally, in a network of N nodes, Then set Solve for V i

2D grid, all r=1, I(0,0)=I 0, all I(m,n)=0 otherwise I0I0 (0,0) (0,1) (1,1) (1,0) Define Then Laplacian

Harmonic functions Random walks Lattice Green’s function First passage time Related to: Solution to Laplace equation is unique For infinite square net one finds For finite networks, the solution is not straightforward.

General I1I1 I2I2 I3I3 N nodes

Properties of the Laplacian matrix All cofactors are equal and equal to the spanning tree generating function G of the lattice (Kirchhoff). Example c3c3 c1c1 c2c2 G=c 1 c 2 +c 2 c 3 +c 3 c 1

I2I2 I1I1 ININ network Problem: L is singular so it can not be inverted. Day is saved: Kirchhoff’s law says Hence only N-1 equations are independent → no need to invert L

Solve V i for a given I Kirchhoff solution Since only N-1 equations are independent, we can set V N =0 & consider the first N-1 equations! The reduced (N-1)×(N-1) matrix, the tree matrix, now has an inverse and the equation can be solved.

II   We find Writing Where L  is the determinant of the Laplacian with the  -th row & column removed L  = the determinant of the Laplacian with the  -th and  -th rows & columns removed

Example c3c3 c1c1 c2c2 or The evaluation of L  & L  in general is not straightforward!

Spanning Trees: x x x x xx yy y y y y y y x G(1,1) = # of spanning trees Solved by Kirchhoff (1847) Brooks/Smith/Stone/Tutte (1940)

x x yy G(x,y)= + x x x xx x ++ yyyyyy =2xy 2 +2x 2 y N=4

Consider instead Solve V i (  ) for given I i and set  =0 at the end. This can be done by applying the arsenal of linear algebra and deriving at a very simple result for 2-point resistance.

Eigenvectors and eigenvalues of L 0 is an eigenvalue with eigenvector L is Hermitian L has real eigenvalues Eigenvectors are orthonormal

Consider where Let This gives

Let = orthonormal Theorem:

Example r1r1 r1r1 r1r1 r1r1 r2r2

Example: complete graphs N=3 N=2 N=4

1 23 N-1 r rrr N  

If nodes 1 & N are connected with r (periodic boundary condition)

New summation identities New product identity

M×N network N=6 M=5 r s s I N unit matrix s r rr

M, N →∞

Finite lattices Free boundary condition Cylindrical boundary condition Moebius strip boundary condition Klein bottle boundary condition

Klein bottle Moebius strip

Orientable surface Non-orientable surface: Moebius strip

Orientable surface Non-orientable surface: Moebius strip

Free Cylinder

Klein bottle Moebius strip

Klein bottle Moebius strip Free Cylinder Torus on a 5×4 network embedded as shown

Resistance between (0,0,0) and (3,3,3) in a 5×5 × 4 network with free boundary