# Electrostatic Potential in a Hollow Cylinder

An electrostatic field $E$ (i.e. an electric field produced only by a static charge) is a conservative field, i.e. there exists a scalar potential $\psi$ such that $E=-\nabla\psi$. This is clear from Maxwell’s equations. Since there is no change of the magnetic field $B$ in time, $\nabla\times E=0$. If there is no charge present in a region, $\nabla\cdot E=0$. Together with $E=-\nabla\psi$, we obtain the Laplace equation $\nabla^2\psi=0$. Thus the Laplace equation can be used to find the electrostatic potential $\psi(\rho,\varphi,z)$ in a hollow cylinder with radius $a$ and height $l$ ($0\leq z\leq l$).

Using the separation of variables, we find the mode
\begin{align*}
\psi_{km}(\rho,\varphi,z)&=P_{km}(\rho)\Phi_m(\varphi)Z_k(z)\\
&=J_m(k\rho)[a_m\sin m\varphi+b_m\cos m\varphi][c_1e^{kz}+c_2e^{-kz}].
\end{align*}
The boundary conditions are:
$$\psi(\rho,\varphi,l)=\psi(\rho,\varphi),$$
where $\psi(\rho,\varphi)$ is a potential distribution. Elsewhere on the surface $\psi=0$. Now we find electrostatic potential
$$\psi(\rho,\varphi,z)=\sum_{k,m}\psi_{km}$$
inside the cylinder. From the boundary condition $\psi(\rho,\varphi,0)=0$, we find $c_1+c_2=1$. So we choose $c_1=-c_2=\frac{1}{2}$ and thereby $c_1e^{kz}+c_2e^{-kz}\sinh kz$. Since $\psi=0$ on the lateral surface of the cylinder, $\psi(a,\varphi,z)=0$. This implies that $J_m(ka)=0$. If we write the $n$-th Bessel zero as $a_{mn}$, then $k_{mn}a=a_{mn}$ or $k_{mn}=\frac{a_{mn}}{a}$. Hence,
$$\psi(\rho,\varphi,z)=\sum_{m=0}^\infty\sum_{n=1}^\infty J_m\left(\alpha_{mn}\frac{\rho}{a}\right)[a_m\sin m\varphi+b_m\cos m\varphi]\sinh\left(\alpha_{mn}\frac{z}{a}\right).$$
Finally using the boundary condition
$$\psi(\rho,\varphi)=\sum_{m=0}^\infty\sum_{n=1}^\infty J_m\left(\alpha_{mn}\frac{\rho}{a}\right)[a_m\sin m\varphi+b_m\cos m\varphi]\sinh\left(\alpha_{mn}\frac{1}{a}\right)$$ and the orthogonality of $\sin m\varphi$ and $\cos m\varphi$, we can determine the coefficients $a_m$ and $b_m$ as
\begin{align*}\left\{\begin{aligned}a_{mn}\\b_{mn}\end{aligned}\right\}=\frac{2}{\pi a^2\sinh\left(\alpha_{mn}\frac{1}{a}\right)J_{m+1}^2(\alpha_{mn})}\int_0^{2\pi}\int_0^a\psi(\rho,\varphi)&J_m\left(\alpha_{mn}\frac{\rho}{a}\right)\\
&\left\{\begin{aligned}
\sin m\varphi\\
\cos m\varphi
\end{aligned}\right\}\rho d\rho d\varphi.\end{align*}

# Cylindrical Resonant Cavity

In this lecture, we discuss cylindrical resonant cavity as an example of the applications of Bessel functions.

Recall that electromagnetic waves in vacuum space can be described by the following four equations, called Maxwell’s equations (in vacuum)
\begin{align*}
\nabla\cdot B=0,\\
\nabla\cdot E=0,\\
\nabla\times B=\epsilon_0\mu_o\frac{\partial E}{\partial t},\\
\nabla\times E=-\frac{\partial B}{\partial t},
\end{align*}
where $E$ is the magnetic field and $B$ the magnetic induction, $\epsilon_0$ the electric permittivity, and $\mu_o$ the magnetic permeability.
Now,
\begin{align*}
\nabla\times(\nabla\times E)&=-\frac{\partial}{\partial t}(\nabla\times B)\\
&=-\epsilon_0\mu_0\frac{\partial^2E}{\partial t^2}.
\end{align*}

Resonant cavity is an electromaginetic resonator in which waves oscillate inside a hollow space (device). For more details see Wikipedia entry for Resonator, in particular for Cavity Resonator.

Here we consider a cylindrical resonant cavity. In the interior of a resonant cavity, electromagnetic waves oscillate with a time dependence $e^{-i\omega t}$, i.e. $E(t,x,y,z)$ can be written as $E=e^{-i\omega t}P(x,y,z),$ where $P(x,y,z)$ is a vector-valued function in $\mathbb R^3$. One can easily show that $\frac{\partial^2E}{\partial t^2}=-\omega^2E$ or
$$\nabla\times(\nabla\times E)=\alpha^2E,$$
where $\alpha^2=\epsilon_0\mu_0\omega^2$. On the other hand,
\begin{align*}
\nabla\times(\nabla\times E)&=\nabla\nabla\cdot E-\nabla\cdot\nabla E\\
&=-\nabla^2E.
\end{align*}
Thus, the electric field $E$ satisfies the Helmholtz equation
$$\nabla^2E+\alpha^2E=0.$$
Suppose that the cavity is a cylinder with radius $a$ and height $l$. Without loss of generality we may assume that the end surfaces are at $z=0$ and $z=l$. Let $E=E(\rho,\varphi,z)$. Using separation of variables in cylindrical coordinate system, we find that the $z$-component $E_z(\rho,\varphi,z)$ satisfies the scalar Helmholtz equation
$$\nabla^2E_z+\alpha^2E_z=0,$$
where $\alpha^2=\omega^2\epsilon_0\mu_0=\frac{\omega^2}{c^2}$. The mode of $E_z$ is obtained as

$$(E_z)_{mnk}=\sum_{m,n}J_m(\gamma_{mn}\rho)e^{\pm im\varphi}[a_{mn}\sin kz+b_{mn}\cos kz].\ \ \ \ \ \mbox{(1)}$$ Here $k$ is a separation constant. Consider the boundary conditions: $\frac{\partial E_z}{\partial z}(z=0)=\frac{\partial E_z}{\partial z}(z=l)=0$ and $E_z(\rho=a)=0$. The boundary conditions $\frac{\partial E_z}{\partial z}(z=0)=\frac{\partial E_z}{\partial z}(z=l)=0$ result that $a_{mn}=0$ and $$k=\frac{p\pi}{l},\ p=0,1,2,\cdots.$$ The boundary condition $E_z(\rho=a)=0$ results $$\gamma_{mn}=\frac{\alpha_{mn}}{a},$$ where $\alpha_{mn}$ is the $n$th zero of $J_m$. Thus the mode (1) is written as
$$(E_z)_{mnp}=\sum_{m,n}b_{mn}J_m\left(\frac{\alpha_{mn}}{a}\rho\right)e^{\pm im\varphi}\cos\frac{p\pi}{l}z,\ \ \ \ \ \mbox{(2)}$$ where $p=0,1,2,\cdots$. In physics, the mode (2) is called the transverse magnetic mode or shortly TM mode of oscillation.

We have \begin{align*}\gamma^2&=\alpha^2-k^2\\&=\frac{\omega^2}{c^2}-\frac{p^2\pi^2}{l^2}.\end{align*} Hence the TM mode has resonant frequencies
\omega_{mnp}=c\sqrt{\frac{\alpha_{mn}^2}{a^2}+\frac{p^2\pi^2}{l^2}},\ \left\{\begin{aligned} m&=0,1,2,\cdots\\ n&=1,2,3,\cdots\\ p&=0,1,2,\cdots.\end{aligned} \right.