= Solution
{c}
Use <metric signature> $-+++$ and <transverse-traceless gauge>, $g_{ij}=\delta_{ij}+h_{ij}^{\rm TT}$. To resolve the source's origin ambiguity, put the detector centre at $z=0$ and the source centre at $z=-r_0$, so the <gravitational wave> travels in the positive $z$ direction. Reversing that direction replaces $z$ by $-z$. Work in the <weak-field approximation>, with $\omega l\ll c$, $r_0\gg l,c/\omega$, and detector size $a\ll c/\omega,r_0$. The <radiation reaction> is neglected, so the rod maintains the prescribed steady motion.
At the retarded source time $u=t-(r_0+z)/c$, take the source positions as $\mathbf x_\pm=\pm l(\cos(\omega u+\beta),\sin(\omega u+\beta),0)$. The arbitrary constant $\beta$ specifies the initial source phase. The <second mass moment tensor> $I_{ij}=\sum_A Mx_{Ai}x_{Aj}$ obeys
$$
I_{xx}=Ml^2[1+\cos\Phi],\quad I_{yy}=Ml^2[1-\cos\Phi],\quad
I_{xy}=I_{yx}=Ml^2\sin\Phi,\qquad
\Phi=2\omega\left(t-\frac{r_0+z}{c}\right)+2\beta.
$$
The <quadrupole formula> $h_{ij}^{\rm TT}=2G\ddot I_{ij}^{\rm TT}/(c^4r_0)$ projects away the transverse trace, which is already constant here. Differentiating twice therefore gives
$$
\boxed{h_{xx}=-H\cos\Phi,\qquad h_{yy}=H\cos\Phi,\qquad
h_{xy}=h_{yx}=-H\sin\Phi,\qquad
H=\frac{8GMl^2\omega^2}{c^4r_0}.}
$$
All time and longitudinal <metric perturbation> components vanish in <transverse-traceless gauge>. The <gravitational-wave frequency> is $2\omega$ in angular units; $M$ is each source <mass>, not their sum, and the detector <mass> $m$ does not enter the strain. The <plane wave> approximation holds the amplitude fixed at $r_0$; retaining spherical spreading replaces its denominator by $r_0+z$ near the axis. The retarded <quadrupole formula> and the distinction between source and wave <frequency> are derived in <General Relativity by David Tong>[Sections 5.3.1–5.3.2, retarded source solution and circular binary]{ref}.
Back to article page
