 Solution

The trace-free mass quadrupole moment is diagonal:The same constant factors multiply its third derivatives, giving ${\stackrel{...}{Q}}_{ij}{\stackrel{...}{Q}}_{ij}=(2/3)({\stackrel{...}{I}}_{zz}{)}^2$. The quadrupole formula and the preceding infall result therefore giveFor infall from infinity, $P=m^5/(15z^5)$ and $|\dot{z}|=\sqrt{m/(2z)}$. To find the emitted energy, integrate power over time, using $dt=|dz|/|\dot{z}|$:The total initial mass is $2m$, so the radiated fraction is $1/840\simeq 1.19\times 10^{-3}$. The Sun would emit that fraction of its rest energy inunder the stipulated constant luminosity. This is the head-on quadrupole radiation from equal masses prediction with the prescribed stopping rule. At the endpoint $|\dot{z}|=1/2$ and $m/z=1/2$, so extending the slow-motion weak-field formula that far is an extrapolation, not a controlled strong-field prediction.