Solution

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\begin{center}
\Large \textbf{Statement}
\end{center}

At what angle to the horizon is a luminous object seen moving horizontally at
a speed $\beta c$ at the moment when it is above the observer?

\begin{center}
\Large \textbf{Solution}
\end{center}

The key idea of this problem is that light has a finite velocity $c$. To understand the solution, you may use the figure below:

\begin{figure}[H]
\centering
\includegraphics[width=0.5\linewidth]{14.1.3_Observer.png}
\caption{The point $C$ is at the observer's head. The red vectors are the velocity vectors of the object at each position.}
\end{figure}

The light needs a time $t_1 = \frac{\overline{AC}}{c}$ to travel from $A$ to $C$. In this time, the object also moves. When the object reaches point $B$, the light reaches point $C$ at the same time. The time taken by the object is $t_2 = \frac{\overline{AB}}{\beta c}$. Since these times must be equal, we have:

\begin{equation} t_1 = t_2 \rightarrow \frac{\overline{AC}}{c} = \frac{\overline{AB}}{\beta c} \rightarrow \frac{\overline{AC}}{\overline{AB}} = \frac{1}{\beta} \end{equation}

The angle between $\overline{AB}$ and $\overline{AC}$ can be calculated as:

\begin{equation} \cos\alpha = \frac{\overline{AB}}{\overline{AC}} \rightarrow \alpha = \cos^{-1} \left(\frac{\overline{AB}}{\overline{AC}}\right) \end{equation}

Using equation usingequation1 we obtain:

\begin{equation} \alpha = \cos^{-1} \left(\frac{\overline{AB}}{\overline{AC}}\right) = \cos^{-1} (\beta) \end{equation}

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