Instantaneous quadrupole luminosity of a Kepler binary ($L=32G^3{M}^2{\mu }^2(v^2-11{\dot{r}}^2/12)/(5c^5{r}^4)$)

For a Newtonian Kepler orbit, let $M=M_1+M_2$ be the total mass and $\mu =M_1{M}_2/M$ 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 $q_{ij}=\mu (3r_i{r}_j-r^2{\delta }_{ij})/2$, differentiating the acceleration $\ddot{\mathbf{r}}=-k\mathbf{r}/r^3$ givesThe two terms have zero cross tensor contraction, with squared norms $6u^2$ and $72v_{\perp }^2$. Therefore ${\stackrel{...}{q}}_{ij}{\stackrel{...}{q}}_{ij}=72{\mu }^2{k}^2(v^2-11u^2/12)/r^4$. Substitution into the quadrupole formula in this convention, $L=4G{\stackrel{...}{q}}_{ij}{\stackrel{...}{q}}_{ij}/(45c^5)$, 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.