# Using a Transfer Matrix

K.E. Schmidt
Department of Physics and Astronomy
Arizona State University
Tempe, AZ U.S.A.

Jackson problems 4.8, 5.8, and many similar problems can be solved using a transfer matrix. These problems are the shielding effect of a circular cross section dielectric or of a permeable sheath. For the magnetic case, there are no currents, so the curl of and the divergence of are both zero, and

 = - = . (1)
Except at the boundaries between media with different ,

 = 0, (2)

and tangential and normal are continuous at boundaries. Continuity of tangential implies that the potentials on either side of a surface can differ by at most a constant. We choose the constant to be zero so that that boundary condition is replaced by continuity of the potential. The dielectric case is identical to the magnetic case with the replacements
 (3)

For the two-dimensional case, where the fields go to a constant along far from the shield, only solutions with this cos() dependence will be produced. That is the solution in region i bounded by two circular cross section surfaces will have the form

 (,) = () (4)

The boundary conditions between and at = ai are then
 Aiai + = Ai + 1ai + = (5)
with the solution
 (6)
We can now use this transfer matrix to solve for a multilayered sheath. At each boundary we apply Eq. 6 to get the coefficients of the potential in the next region. For the simple case of a sheath of permeability from = a to = b , we have the additional boundary conditions, C1 = 0 and A3 = - B0 , and two boundaries. Identifying A1 = - Bin , we can write
 (7)
which gives immediately the result

 B0 = Bin (8)

or

 Bin = B0 (9)

The coeffients of the potential inside the sheath are
 A2 = - Bin = - B0 C2 = - Bin = - B0 (10)
and outside

 C3 = Bin = B0 (11)

Jackson's example in three dimensions can also be written in terms of a transfer matrix. The potential within each spherical shell is

 (r,) = (). (12)

The boundary conditions are
 Aiai + = Ai + 1ai + = (13)
Solving for the transfer matrix as in the two-dimensional case gives
 (14)
For the single shell from a to b ,
 (15)
with the immediate result

 B0 = Bin (16)

as given in Jackson.

These same techniques can be applied to higher l values too. For the spherical case, the equations are

 (r,) = l(cos()) (17)

The boundary conditions are
 Aiail + = Ai + 1ail + = (18)
or
 (19)
For the single shell from a to b ,
 (20)
with the result

 B0 = Bin. (21)

For ,

 Bin 1 (1-) B0 (22)