= Solution
The trace-free <mass quadrupole moment> is diagonal:
$$
Q_{ij}=I_{zz}\,\operatorname{diag}(-1/3,-1/3,2/3).
$$
The same constant factors multiply its third derivatives, giving $\dddot Q_{ij}\dddot Q_{ij}=(2/3)(\dddot I_{zz})^2$. The <quadrupole formula> and the preceding infall result therefore give
$$
\boxed{P(z)=\frac{m^5}{15z^4}\left(\frac1z-\frac1{z_0}\right).}
$$
For infall from infinity, $P=m^5/(15z^5)$ and $|\dot z|=\sqrt{m/(2z)}$. To find the emitted energy, integrate power over time, using $dt=|dz|/|\dot z|$:
$$
\begin{aligned}
E_{\rm rad}
&=\int_{2m}^\infty\frac{m^5}{15z^5}\sqrt{\frac{2z}{m}}\,dz\\
&=\frac{2\sqrt2\,m^{9/2}}{105}(2m)^{-7/2}
=\boxed{\frac m{420}}.
\end{aligned}
$$
The total initial mass is $2m$, so \b[the radiated fraction is $1/840\simeq1.19\times10^{-3}$]. The <Sun> would emit that fraction of its <rest energy> in
$$
\boxed{t_\odot=\frac{t_{\rm all}}{840}\simeq1.76\times10^{10}\,\mathrm{yr},}
$$
under 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.
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