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Up: PHY531 Classical Electrodynamics Links

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 $\vec{H}$ and the divergence of $\vec{B}$ are both zero, and

$\displaystyle\vec{H}$ = - $\displaystyle\vec{\nabla}$$\displaystyle\Phi$   
$\displaystyle\vec{B}$ = $\displaystyle\mu$$\displaystyle\vec{H}$. (1)
Except at the boundaries between media with different $\mu$ ,

$\displaystyle\nabla^{2}_{}$$\displaystyle\Phi$ = 0, (2)

and tangential $\vec{H}$ and normal $\vec{B}$ are continuous at boundaries. Continuity of tangential $\vec{H}$ 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
$\displaystyle\mu$ $\textstyle\rightarrow$ $\displaystyle\epsilon$   
$\displaystyle\vec{H}$ $\textstyle\rightarrow$ $\displaystyle\vec{E}$   
$\displaystyle\vec{B}$ $\textstyle\rightarrow$ $\displaystyle\vec{D}$ (3)

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

$\displaystyle\Phi_{i}^{}$($\displaystyle\rho$,$\displaystyle\phi$) = $\displaystyle\left ( A_i \rho + \frac{C_i}{\rho} \right)\cos$($\displaystyle\phi$) (4)

The boundary conditions between $\Phi_{i}^{}$ and $\Phi_{i+1}^{}$ at $\rho$ = ai are then
Aiai + $\displaystyle{\frac{C_i}{a_i}}$ = Ai + 1ai + $\displaystyle{\frac{C_{i+1}}{a_i}}$   
$\displaystyle\mu_{i}^{}$$\displaystyle\left ( A_i -\frac{C_i}{a_i^2} \right)$ = $\displaystyle\mu_{i+1}^{}$$\displaystyle\left ( A_{i+1} - \frac{C_{i+1}}{a_i^2} \right)$ (5)
with the solution  
 \begin{equation}
\left (
\begin{array}
{c}
A_{i+1} \\ C_{i+1}\end{array} \right ...
 ...ght )
\left (
\begin{array}
{c}
A_{i} \\ C_{i}\end{array} \right )\end{equation}(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 $\mu$ from $\rho$ = a to $\rho$ = b , we have the additional boundary conditions, C1 = 0 and A3 = - B0 , and two boundaries. Identifying A1 = - Bin , we can write
\begin{equation}
\left (
\begin{array}
{c}
-B_0 \\  C_3 \end{array} \right ) =
\...
 ...ight )
\left (
\begin{array}
{c}
-B_{in} \\  0\end{array} \right )\end{equation}(7)
which gives immediately the result

B0 = $\displaystyle{\frac{(\mu+1)^2b^2-(\mu-1)^2 a^2}{4 \mu b^2}}$Bin (8)

or

Bin = $\displaystyle{\frac{4 \mu b^2}{(\mu+1)^2b^2-(\mu-1)^2 a^2}}$B0 (9)

The coeffients of the potential inside the sheath are
A2 = - $\displaystyle{\frac{\mu+1}{2\mu}}$Bin = - $\displaystyle{\frac{2 (\mu+1)b^2}{(\mu+1)^2b^2-(\mu-1)^2 a^2}}$B0        
C2 = - $\displaystyle{\frac{a^2(\mu-1)}{2\mu}}$Bin = - $\displaystyle{\frac{2 a^2 b^2 (\mu-1)}{(\mu+1)^2b^2-(\mu-1)^2 a^2}}$B0      (10)
and outside

C3 = $\displaystyle{\frac{(b^2-a^2)(\mu^2-1)}{4\mu}}$Bin = $\displaystyle{\frac{b^2 (b^2-a^2)(\mu^2-1)}{(\mu+1)^2b^2-(\mu-1)^2 a^2}}$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

$\displaystyle\Phi_{i}^{}$(r,$\displaystyle\theta$) = $\displaystyle\left ( A_i r + \frac{C_i}{r^2} \right)\cos$($\displaystyle\theta$). (12)

The boundary conditions are
Aiai + $\displaystyle{\frac{C_i}{a_i^2}}$ = Ai + 1ai + $\displaystyle{\frac{C_{i+1}}{a_i^2}}$   
$\displaystyle\mu_{i}^{}$$\displaystyle\left ( A_i -\frac{2 C_i}{a_i^3} \right)$ = $\displaystyle\mu_{i+1}^{}$$\displaystyle\left ( A_{i+1} - \frac{2 C_{i+1}}{a_i^3} \right)$ (13)
Solving for the transfer matrix as in the two-dimensional case gives  
 \begin{equation}
\left (
\begin{array}
{c}
A_{i+1} \\ C_{i+1}\end{array} \right ...
 ...ght )
\left (
\begin{array}
{c}
A_{i} \\ C_{i}\end{array} \right )\end{equation}(14)
For the single shell from a to b ,
\begin{equation}
\left (
\begin{array}
{c}
-B_0 \\  C_3 \end{array} \right ) =
\...
 ...ight )
\left (
\begin{array}
{c}
-B_{in} \\  0\end{array} \right )\end{equation}(15)
with the immediate result

B0 = $\displaystyle{\frac{(\mu+2)(2\mu+1)b^3-2(\mu-1)^2a^3}{9 \mu b^3}}$Bin (16)

as given in Jackson.

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

$\displaystyle\Phi_{i}^{}$(r,$\displaystyle\theta$) = $\displaystyle\left ( A_i r^l + \frac{C_i}{r^{l+1}} \right)P$l(cos($\displaystyle\theta$)) (17)

The boundary conditions are
Aiail + $\displaystyle{\frac{C_i}{a_i^{l+1}}}$ = Ai + 1ail + $\displaystyle{\frac{C_{i+1}}{a_i^{l+1}}}$   
$\displaystyle\mu_{i}^{}$$\displaystyle\left ( l A_i a_i^{l-1} -\frac{2 (l+1) C_i}{a_i^{l+2}} \right)$ = $\displaystyle\mu_{i+1}^{}$$\displaystyle\left ( A_{i+1} a_i^{l+1} - \frac{2 (l+1) C_{i+1}}{a_i^{l+2}}
\right)$ (18)
or
\begin{equation}
\left (
\begin{array}
{c}
A_{i+1} \\ C_{i+1}\end{array} \right ...
 ...ght )
\left (
\begin{array}
{c}
A_{i} \\ C_{i}\end{array} \right )\end{equation}(19)
For the single shell from a to b ,
\begin{equation}
\left (
\begin{array}
{c}
-B_0 \\  C_3 \end{array} \right ) =
\...
 ...ight )
\left (
\begin{array}
{c}
-B_{in} \\  0\end{array} \right )\end{equation}(20)
with the result

B0 = $\displaystyle{\frac{(l+1+l\mu)(l+(l+1)\mu)b^{2l+1}-l(l+1)(\mu-1)^2a^{2l+1}}{(2l+1)^2 \mu b^{2l+1}}}$Bin. (21)

For $\mu$ $\rightarrow$ $\infty$ ,

Bin $\displaystyle\rightarrow$ $\displaystyle\left ( 4+\frac{1}{l(l+1)} \right)\frac$1 $\displaystyle\mu$(1-$\displaystyle{\frac{a^{2l+1}}{b^{2l+1}}}$) B0 (22)


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