Special theory of relativity > The transformation of electric and magnetic fields > Problem 14.3.18

Statement

$14.3.18$ Solve problem 14.3.17 when an electromagnetic wave hits a moving wall at an
angle $\alpha$.

$14.3.17$ A plane electromagnetic wave is incident perpendicularly on a metal wall moving at the speed $\beta c$. How many times will the wave amplitude change during
reflection?

Solution

To solve this problem we can use the Lorentz transformation for momentum and energy in the horizontal direction.
The horizontal component of the momentum is:

\begin{equation}
p_x = \frac{h \nu }{c} \cos\alpha \qquad E = h \nu
\end{equation}

Everything else is the same as in problem $14.3.17$.

\begin{equation}
h \nu' = \frac{h \nu - (-\beta c) p_x}{\sqrt{1-\beta^2}} = \frac{h \nu + h \nu \beta \cos\alpha}{\sqrt{1-\beta^2}} \rightarrow \nu' = \nu \frac{1+\beta \cos \alpha}{\sqrt{1-\beta^2}}
\end{equation}

The negative sign appears because the velocity of the wall points in the negative direction.
The reflected wave will have this frequency; then transforming back to the Earth frame:

\begin{equation}
h \nu_{\text{reflected}} = \frac{h \nu' + \beta c \frac{h \nu'}{c} \cos\alpha}{\sqrt{1-\beta^2}} \rightarrow \nu_{\text{reflected}} = \nu \frac{(1+\beta\cos\alpha)^2}{1-\beta^2}
\end{equation}

Answer

\begin{equation}
\nu_{\text{reflected}} = \nu \frac{(1+\beta\cos\alpha)^2}{1-\beta^2}
\end{equation}