10 Nov 2024

circular disk cavity

From equation (7) of [Modes of Slab Waveguide] the scalar mode equation can be expressed by

$$ \begin{align} \nabla^2 \psi(x,y) + n^2 k^2 \psi(x,y) = 0 \tag{1} \end{align} $$

Since the system has rotational symmetry, the ansatz $\psi(r,\phi) = \Psi(r)e^{im\phi}$ is used. where the $2\pi$ periodicity implies $m \in \mathbb{Z}$ and can be connected with the slab waveguide propagtion constant as $\beta=\frac{m}{R}$. Then by assuming $k_z \approx 0$ that propagation in the z-direction is negligible,

$$ \begin{align} [r^2 \partial^2_r + r\partial_r + (n^2k^2r^2 - m^2)]\Psi = 0 \tag{2} \end{align} $$

is obtained. which is Bessel-like ODE that can be solved independently by Bessel, Hankel an Neumann functions.
With outgoing wave condition, the reasonable ansats is

$$ E_{m,l}(r,\phi)= \begin{cases} \dfrac{J_m(nkr)}{J_m(nkR)} e^{im\phi}, & r < R, \\[6pt] \dfrac{H_m(kr)}{H_m(kR)} e^{im\phi}, & r \ge R . \end{cases} $$

where the refractive index outside the cavity is assumed zero. Now applying the boudary condition for electromagnetic waves yields,

$$ S_m(x) := \frac{n}{\zeta}\,\frac{J_m'(nx)}{J_m(nx)} \;-\; \frac{H_m'(x)}{H_m(x)} = 0 . $$

where $x=kR, \zeta=1(n^2)$ for TM(TE) pol.

$kR$ values are complex with $\Re(kR)=\frac{2\pi R}{\lambda}, Q=\frac{\Re(kR)}{-2\Im(kR)}$

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