Reflection of an Electromagnetic Wave from a Moving Mirror

An electromagnetic wave is reflected by a perfect conductor at $x=0$. The electric field has the form \[\mathbf{E}(t,\mathbf{x}) = \hat{\mathbf{y}}\left[f(t_-) - f(t_+)\right], \tag{1.1}\] where $f$ is an arbitrary function and $ct_\pm = ct\pm x$. Show that this satisfies the relevant boundary condition at the conductor. Find the corresponding magnetic field $\mathbf{B}$.

Show that under a Lorentz transformation to a frame moving with speed $v$ in the $x$-direction, the electric field is transformed to \[\mathbf{E}'(t',\mathbf{x}') = \hat{\mathbf{y}}\left[ \rho f(\rho t'_-) - \frac{1}{\rho}f\!\left(\frac{t'_+}{\rho}\right) \right], \qquad\text{where}\quad \rho = \sqrt{\frac{c-v}{c+v}}\,.\tag{1.2}\]

Hence for an incident wave $\mathbf{E}(t,\mathbf{x}) = \hat{\mathbf{y}}F(t_-)$, find the wave that is reflected after it hits a perfectly conducting mirror moving with speed $v$ in the $x$-direction.

It is clear to see that the electric field of this form has an ingoing part, $f(t_-)$, and an outgoing part, $f(t_+)$. The boundary condition at a perfect conductor is given by \[\hat{\mathbf{x}}\cdot (\mathbf{E}_+- \mathbf{E}_-) = \frac{\sigma}{\varepsilon_0} = 0, \quad \text{and}\quad \hat{\mathbf{x}}\times(\mathbf{E}_+- \mathbf{E}_-) =0.\tag{1.3}\] It is clear to see that (1.1) satisfies the boundary condition. To find the magnetic field, let's first take the curl of the electric field, which gives \[\begin{aligned}\nabla \times \mathbf{E} &= -\hat{\mathbf{y}} \times[\nabla f(t_-) - \nabla f(t_+)] = \frac{\hat{\mathbf{y}}\times \hat{\mathbf{x}}}{c}[\dot{f}(t_-) + \dot{f}(t_+)]= -\frac{\hat{\mathbf{z}}}{c}[\dot{f}(t_-) + \dot{f}(t_+)].\end{aligned}\tag{1.4}\] Hence, the magnetic field is given by \[\mathbf{B}(t,\mathbf{x}) = \frac{\hat{\mathbf{z}}}{c} [f(t_-) + f(t_+)].\tag{1.5}\]

For a boost in the $x$-direction, the electric field transforms like \[E_x'= E_x, \quad E_y' = \gamma(E_y-vB_z),\quad \text{and}\quad E_z'=\gamma(E_z+vB_y).\tag{1.6}\] Therefore, we have \[\mathbf{E}'(t',\mathbf{x}') = \gamma[f(t_-)-f(t_+)]\hat{\mathbf{y}}-\frac{v}{c}\gamma[f(t_-)+f(t_+)]\hat{\mathbf{y}} = \hat{\mathbf{y}}[\rho f(t_-) -\frac{1}{\rho}f(t_+)].\tag{1.7}\] Now, we express the coordinates in the primed coordinates, which is \[ct_\pm = \gamma\left(ct'+\frac{vx'}{c^2}\pm x' \pm vt'\right) = \frac{\gamma}{c}\left[(c\pm v)t' \pm(c\pm v)x'\right]=\begin{cases} \rho t_-',\\ \dfrac{1}{\rho}t_+'. \end{cases}\tag{1.8}\] Thus, we find \[\mathbf{E}'(t',\mathbf{x}') = \hat{\mathbf{y}}\left[ \rho f(\rho t'_-) - \frac{1}{\rho}f\!\left(\frac{t'_+}{\rho}\right) \right].\tag{1.9}\]

In the frame where the mirror is stationary, the incident electric field is given by $\hat{\mathbf{y}}\rho F(\rho t_-')$. Since the mirror is stationary in this frame, the reflected electric field must be $-\hat{\mathbf{y}}\rho F(\rho t_+')$. Given the transformation law we just derived, the reflected electric field in the rest frame is $-\hat{\mathbf{y}}\rho^2 F(\rho^2 t_+)$. The total electric field is given by \[\mathbf{E}(t,\mathbf{x}) = \hat{\mathbf{y}}\left[F(t_-) -\rho^2 F(\rho^2 t_+) \right].\tag{1.10}\]