= Instantaneous quadrupole luminosity of a Kepler binary
{title2=$L=32G^3M^2\mu^2(v^2-11\dot r^2/12)/(5c^5r^4)$}
For a Newtonian <Kepler orbit>, let $M=M_1+M_2$ be the total <mass> and $\mu=M_1M_2/M$ the <reduced mass>; put $k=GM$, $\mathbf n=\mathbf r/r$, $u=\dot r$ and $\mathbf v_\perp=\dot{\mathbf r}-u\mathbf n$, with $v^2=\dot{\mathbf r}\cdot\dot{\mathbf r}$. In the convention $q_{ij}=\mu(3r_ir_j-r^2\delta_{ij})/2$, differentiating the acceleration $\ddot{\mathbf r}=-k\mathbf r/r^3$ gives
$$
\dddot q_{ij}=\frac{\mu k}{r^2}\left[u(\delta_{ij}-3n_in_j)-6(n_iv_{\perp j}+v_{\perp i}n_j)\right].
$$
The two terms have zero cross <tensor contraction>, with squared norms $6u^2$ and $72v_\perp^2$. Therefore $\dddot q_{ij}\dddot q_{ij}=72\mu^2k^2(v^2-11u^2/12)/r^4$. Substitution into the <quadrupole formula> in this convention, $L=4G\dddot q_{ij}\dddot q_{ij}/(45c^5)$, proves the luminosity. It is nonnegative since $v^2-11u^2/12=v_\perp^2+u^2/12$. The formula uses a weak gravitational field, slow motion and leading radiation order; orbital averaging gives a secular luminosity.
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