= Solution
For each <Volvox> colony, the excess gravitational force is
$$
F=\frac{4\pi}{3}R^3(\rho-\rho_w)g
=\frac{4\pi}{3}\rho_w\epsilon gR^3.
$$
For a normal <Stokeslet> a distance $R$ below a flat <stress-free boundary condition>[stress-free interface], the image is an oppositely directed Stokeslet a distance $R$ above it. The real force creates no lateral velocity at the other colony because both centres have the same height. Their separation from the image is $(x,-2R)$, however, so each colony moves toward the other with speed
$$
U_x=\frac{FRx}{4\pi\mu(x^2+4R^2)^{3/2}}.
$$
Both colonies move, and therefore
$$
\boxed{\frac{dx}{dt}
=-\frac{FRx}{2\pi\mu(x^2+4R^2)^{3/2}}
=-\frac{2\rho_w\epsilon gR^4}{3\mu}
\frac{x}{(x^2+4R^2)^{3/2}}.}
$$
Define
$$
\xi=\frac{x}{2R},
\qquad
\tau=\frac{Ft}{16\pi\mu R^2}
=\frac{\rho_w\epsilon gR}{12\mu}t.
$$
The dimensionless dynamics are
$$
\frac{d\xi}{d\tau}
=-\frac{\xi}{(1+\xi^2)^{3/2}}
=-\frac{dV}{d\xi},
\qquad
\boxed{V(\xi)=-\frac1{\sqrt{1+\xi^2}}.}
$$
This <gradient flow> descends an even attractive potential with minimum $V(0)=-1$ and $V\to0^-$ as $|\xi|\to\infty$.
For $\xi\gg1$, $d\xi/d\tau\simeq-1/\xi^2$, so the approach from $\xi_0=x_0/(2R)$ takes $\tau_c\simeq\xi_0^3/3$. Restoring dimensions,
$$
\boxed{T\simeq\frac{\mu x_0^3}
{2\rho_w\epsilon gR^4}.}
$$
With water viscosity $\mu\simeq10^{-3}\,\mathrm{Pa\,s}$, $\rho_w\simeq10^3\,\mathrm{kg\,m^{-3}}$, $\epsilon=0.03$, $R=200\,\mu\mathrm m$, and $x_0=5R$, this gives
$$
\boxed{T\simeq1.1\,\mathrm s.}
$$
Because $x_0/(2R)=2.5$ is only moderately large, this is a far-field estimate rather than the exact collision time.
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