= Head-on quadrupole radiation from equal masses
{title2=$P=m^5(1/z-1/z_0)/(15z^4)$}
For two equal <point masses> at $\pm z$ falling from rest at $\pm z_0$ under <Newtonian gravity>, in <geometrized units> one has $\ddot z=-m/(4z^2)$ and $\dot z=-\sqrt{m(1/z-1/z_0)/2}$. The only nonzero component of the <second mass moment tensor> is $I_{zz}=2mz^2$, and its third derivative is $-m^2\dot z/z^2$. The trace-free <mass quadrupole moment> and <quadrupole formula> give the displayed power. From infinity, integrating down to $z=z_c$ gives $E=2\sqrt2\,m^{9/2}/(105z_c^{7/2})$. Relative to the total initial mass $2m$ the fraction is $\sqrt2(m/z_c)^{7/2}/105$. The weak-field slow-motion approximation must be distinguished from its extrapolation to a compact endpoint.
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