Electromagnetic waves > Properties of emission and reflection of electromagnetic waves > Problem 12.1.29

Statement

$12.1.29$ The frequency of a sinusoidal wave incident on a moving metal wall perpendicular
to its surface changes by $\Delta$ during reflection. The initial frequency of
the wave is $\nu_0$. Determine the wall speed.

Solution

The change in frequency of the electromagnetic wave is due to the relativistic Doppler effect. The frequency of the reflected wave is:

\begin{equation}
\nu = \nu_0 \frac{1+v/c}{1-v/c}
\end{equation}

We can obtain this result using the Lorentz transformation for energy and momentum:

\begin{equation}
h \nu = \frac{h \nu_0 - (-v) h \nu_0 / c}{\sqrt{1-v^2/c^2}} \rightarrow \nu = \nu_0 \sqrt{\frac{1+v/c}{1-v/c}}
\end{equation}

After reflection in the moving frame, the wave will have the same frequency; coming back to the Earth frame:

\begin{equation}
h \nu_1 = \frac{h \nu + v h \nu / c}{\sqrt{1-v^2/c^2}} \rightarrow \nu_1 = \nu_0 \frac{1+v/c}{1-v/c}
\end{equation}

So, the difference is:

\begin{equation}
\Delta = \nu_1 - \nu_0 = \nu_0 \left(\frac{1+v/c}{1-v/c} - 1\right) = \nu_0 \frac{2 v/c}{1-v/c}
\end{equation}

And finally:

\begin{equation}
v = \frac{c \Delta }{2 \nu_0 + \Delta}
\end{equation}

Answer

\begin{equation}
v = \frac{c \Delta }{2 \nu_0 + \Delta}
\end{equation}