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TM Waves


$\displaystyle B_z$ $\textstyle =$ $\displaystyle 0$ (10.58)
$\displaystyle E_z\vert _S$ $\textstyle =$ $\displaystyle 0$ (10.59)

The magnetic field is strictly transverse, but the electric field in the $z$ direction only has to vanish at the boundary - elsewhere it can have a $z$ component.

Thus:

$\displaystyle \mbox{\boldmath$E$}_\perp$ $\textstyle =$ $\displaystyle \frac{i}{\mu\epsilon\omega^2 - k^2}\left( k\mbox{\boldmath$\nabla...
...mega (\hat{\mbox{\boldmath$z$}}\times\mbox{\boldmath$\nabla$}_\perp B_z)\right)$  
$\displaystyle (\mu\epsilon\omega^2 - k^2)\mbox{\boldmath$E$}_\perp$ $\textstyle =$ $\displaystyle i k\mbox{\boldmath$\nabla$}_\perp
E_z$  
$\displaystyle \frac{1}{ik}(\mu\epsilon\omega^2 - k^2)\mbox{\boldmath$E$}_\perp$ $\textstyle =$ $\displaystyle \mbox{\boldmath$\nabla$}_\perp E_z$  

which looks just perfect to substitute into:
$\displaystyle \mbox{\boldmath$B$}_\perp$ $\textstyle =$ $\displaystyle \frac{i}{\mu\epsilon\omega^2 - k^2}\left( k\mbox{\boldmath$\nabla...
...mega (\hat{\mbox{\boldmath$z$}}\times\mbox{\boldmath$\nabla$}_\perp E_z)\right)$  
$\displaystyle (\mu\epsilon\omega^2 - k^2)\mbox{\boldmath$B$}_\perp$ $\textstyle =$ $\displaystyle i\mu\epsilon\omega(\hat{\mbox{\boldmath$z$}}\times\mbox{\boldmath$\nabla$}_\perp E_z)$  
$\displaystyle (\mu\epsilon\omega^2 - k^2)\mbox{\boldmath$B$}_\perp$ $\textstyle =$ $\displaystyle \frac{\mu\epsilon\omega}{k}(\mu\epsilon\omega^2 - k^2)(\hat{\mbox{\boldmath$z$}}\times \mbox{\boldmath$E$}_\perp )$  

giving us:
\begin{displaymath}
\mbox{\boldmath$B$}_\perp =
\pm \frac{\mu\epsilon\omega}{k}(\hat{\mbox{\boldmath$z$}}\times \mbox{\boldmath$E$}_\perp )
\end{displaymath} (10.60)

or (as the book would have it):
\begin{displaymath}
\mbox{\boldmath$H$}_\perp =
\pm \frac{\epsilon\omega}{k}(\hat{\mbox{\boldmath$z$}}\times \mbox{\boldmath$E$}_\perp )
\end{displaymath} (10.61)

(where as usual the two signs indicate the direction of wave propagation).

Of course, we still have to find at least one of the two fields for this to do us any good. Or do we? Looking above we see:

$\displaystyle (\mu\epsilon\omega^2 - k^2)\mbox{\boldmath$E$}_\perp$ $\textstyle =$ $\displaystyle i k\mbox{\boldmath$\nabla$}_\perp
\psi$  
$\displaystyle \mbox{\boldmath$E$}_\perp$ $\textstyle =$ $\displaystyle \frac{\pm ik}{(\mu\epsilon\omega^2 - k^2)}\mbox{\boldmath$\nabla$}_\perp
\psi$  

Where $\psi(x,y) e^{ikz} = E_z$. This must satisfy the transverse wave function:
\begin{displaymath}
\left(\nabla_\perp^2 + (\mu\epsilon\omega^2 - k^2)\right)\psi = 0
\end{displaymath} (10.62)

and the boundary conditions for a TM wave:
\begin{displaymath}
\psi\vert _S = 0
\end{displaymath} (10.63)



Subsections
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Next: TE Waves Up: TE and TM Waves Previous: TE and TM Waves   Contents
Robert G. Brown 2014-08-19