 Solution

The separation of the stars is $2R$. Newton's law of universal gravitation and circular acceleration giveTo leading nonrelativistic order, the mass density is the sum of the two translated point-mass Dirac delta distributions. Thuswith all components involving $z$ zero. The trace is $2M{R}^2$, a constant. Define the trace-free mass quadrupole moment $Q_{ij}=I_{ij}-{\delta }_{ij}{I}_{kk}/3$; its third derivatives equal those of $I_{ij}$. If $K=8M{R}^2{\Omega }^3$, thenBoth off-diagonal entries count in the contraction. Therefore ${\stackrel{...}{Q}}_{ij}{\stackrel{...}{Q}}_{ij}=2K^2=128M^2{R}^4{\Omega }^6$, independent of phase. The quadrupole formula gives the equal-mass circular-binary quadrupole luminosityRestoring units, ${\Omega }^2=GM/(4R^3)$ andHere $R$ is each star's radius about the centre of mass, not the separation. The rest-frame prescription supplies the leading mass density; a moving star does not still have zero momentum density and spatial stress. The calculation uses the assumed quadrupole formula and the Newtonian orbit rather than imposing those rest-frame zeros on the moving binary.