Write down Maxwell's equations in a region with no charges and no currents. Show that if $\mathbf{E}(\mathbf{x}, t)$ and $\mathbf{B}(\mathbf{x}, t)$ is a solution then so is $\widetilde{\mathbf{E}}(\mathbf{x}, t)=c \mathbf{B}(\mathbf{x}, t)$ and $\widetilde{\mathbf{B}}(\mathbf{x}, t)=-\mathbf{E}(\mathbf{x}, t) / c$. Write down the boundary conditions on $\mathbf{E}$ and $\mathbf{B}$ at the boundary with unit normal $\mathbf{n}$ between a perfect conductor and a vacuum.

Electromagnetic waves propagate inside a tube of perfectly conducting material. The tube's axis is in the $z$-direction, and it is surrounded by a vacuum. The fields may be taken to be the real parts of

$$\mathbf{E}(\mathbf{x}, t)=\mathbf{e}(x, y) e^{i(k z-\omega t)}, \quad \mathbf{B}(\mathbf{x}, t)=\mathbf{b}(x, y) e^{i(k z-\omega t)}$$

Write down Maxwell's equations in terms of $\mathbf{e}, \mathbf{b}, k$ and $\omega$.

Suppose first that $b_{z}(x, y)=0$. Show that the solution is determined by

$$\mathbf{e}=\left(\frac{\partial \psi}{\partial x}, \frac{\partial \psi}{\partial y}, i k\left[1-\frac{\omega^{2}}{k^{2} c^{2}}\right] \psi\right)$$

where the function $\psi(x, y)$ satisfies

$$\frac{\partial^{2} \psi}{\partial x^{2}}+\frac{\partial^{2} \psi}{\partial y^{2}}+\gamma^{2} \psi=0$$

and $\psi$ vanishes on the boundary of the tube. Here $\gamma^{2}$ is a constant whose value should be determined.

Obtain a similar condition for the case where $e_{z}(x, y)=0$. [You may find it useful to use a result from the first paragraph.] What is the corresponding boundary condition?