Loss-less propagation

The boundary condition for the eigenfunction of the $n$th higher mode defined in equation (2.31) is:

$$\frac{\partial\psi_n}{\partial x} = 0, \mbox{\hspace{1cm}} x = -a,a$$ (2.63)

$$\frac{\partial\psi_n}{\partial y} = 0, \mbox{\hspace{1cm}} y = -b,b$$ (2.64)

The solution is most conveniently expressed by separating it into the $x$ dependent and $y$ dependent parts:

$$\psi_n = \phi_{n_x}\sigma_{n_y}$$ (2.65)

where

$$\phi_{n_x} = \left\{ \begin{array} {r@{\quad:\quad}l} 1 & ... ...\sqrt{2} \cos(n_x \pi x / a) & n_x>0. \\ \end{array} \right.$$ (2.66)

$$\sigma_{n_y} = \left\{ \begin{array} {r@{\quad:\quad}l} 1 ... ...\sqrt{2} \cos(n_y \pi y / b) & n_y>0. \\ \end{array} \right.$$ (2.67)

Performing the differentiation from equation (2.31) gives the corresponding eigenvalues as

$$\alpha_n=\sqrt{\left(\frac{n_x \pi}{a}\right)^2 + \left(\frac{n_y \pi}{b}\right)^2}.$$ (2.68)

As with circular cross-section it is possible to use a lossy boundary condition to give lossy versions of $\psi$ and $\alpha$ but the effect of losses will be noticeable in the $z$ direction only and will therefore be represented entirely by the choice of $z$ direction wavenumber, $k_n$.

$$k_n = \left\{ \begin{array} {r@{\quad:\quad}l} -\sqrt{k^2 ... ...right)^2 + \left(n_y \pi / b \right)^2. \end{array} \right.$$ (2.69)