n}_{ED}\right)^2}{2} \int\limits_{t-L}^{t}h_{s}\left[v(\lambda)\right]\,d\lambda \nonumber \\
\label{eq:DT3c}
& = &
L + \frac{{\bf n}_{ED}\otimes{\bf n}_{ED}:\left[{{\boldsymbol \epsilon}}^{st}  + (1 - \frac{1}{{\rm v}^2}){{\boldsymbol \epsilon}}^{sl}\right] }{2} \int\limits_{t-L}^{t}h_{s}\left[v(\lambda)\right]\,d\lambda, \nonumber\end{aligned}$$ where the $3\times3$ matrices, ${{\boldsymbol \epsilon}}^{st}$ and ${{\boldsymbol \epsilon}}^{sl}$, are defined in the Appendix \[A1\]. They play the role of the two (spatial) polarization tensors of the scalar transversal and scalar longitudinal modes whose components in the synchronous gauge and in the source frame read $$\begin{aligned}
\label{eq:polarizations}
\epsilon^{st} & = &
\left(
\begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 0
\end{array}
\right),
\qquad
\epsilon^{sl} = \left(
\begin{array}{ccc}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 1
\end{array}
\right).\end{aligned}$$

Detector response in the synchronous gauge {#s:3}
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In the synchronous gauge the metric is given by $$\begin{aligned}
\label{eq:b:synch}
{\rm \bf g} = -dt^2 + \left(\delta_{ij} + {\rm h}_{ij}\right)dx^i\,dx^j.\end{aligned}$$ The condition ${\rm h}_{\mu0}=0$ implies that the motion of test bodies is trivial: one can assume that the emitter and detector stay at fixed coordinates, $x^{i}(t)=const.$ and the coordinate time $t$ is equal to the proper times along their trajectories, $\tau_{E}(t)=\tau_{D}(t)=t$. The time of flight is then given by the difference in 