initial positions set at $t\rightarrow-\infty$. In order to have the motion of the test particle in the case of the plane wave propagating along the unit vector ${{\boldsymbol \Omega}}=(-\cos{\phi}\sin{\theta},-\sin{\phi}\sin{\theta},-\cos{\theta})$ one rotates the frame $$\begin{aligned}
\label{eq:rot1}
\left(
\begin{array}{c}
x_{new} \\
y_{new} \\
z_{new}
\end{array}
\right) & = &
\left(
\begin{array}{ccc}
\cos{\theta}\cos{\phi} & -\sin{\phi} & \sin{\theta}\cos{\phi} \\
\cos{\theta}\sin{\phi} & \cos{\phi} & \sin{\theta}\sin{\phi} \\
-\sin{\theta} & 0 & \cos{\theta}
\end{array}
\right)
\left(
\begin{array}{c}
x_{} \\
y_{} \\
z_{}
\end{array}
\right).\end{aligned}$$ to get in the new coordinates (we omit the subindex ’$new$’): $$\begin{aligned}
\label{eq:solvec}
{\bf x}(t) & = & {\bf x}_{0} + \frac{{{\boldsymbol \Omega}}}{2 {\rm v}}\int\limits_{-\infty}^{t-\frac{{{\boldsymbol \Omega}}\cdot{\bf x}_{0}}{{\rm v}}}h_{s}(v)\,dv \\
\label{eq:solt}
\tau(t) & = & t + \frac12\int\limits_{-\infty}^{t-\frac{{{\boldsymbol \Omega}}\cdot{\bf x}_{0}}{{\rm v}}}h_{s}(v)\,dv,\end{aligned}$$ with ${\bf x}_{0}=(x_{0},y_{0},z_{0})$ and ${{\boldsymbol \Omega}}\cdot{\bf x}_{0} = -x_{0}\cos{\phi}\sin{\theta} - y_{0}\sin{\phi}\sin{\theta} - z_{0}\cos{\theta}$. To obtain the detector response we place the freely moving emitter and the detector initially at the points ${\bf x}_{E0}$ and ${\bf x}_{D0}={\bf x}_{E0} + L\,{\bf n}_{ED}$ respectively, where ${\bf n}_{ED}$ is the unit vector from the emitter to the detector and $L$ is the length of the detector arm both defined with respect to the $3$–dimensional Euclidean metric $\delta_{ij}$. We assume that clocks that measure the proper times along 