= Solution
For <neutrino-electron elastic scattering>, work in the laboratory frame with a stationary <electron> and a massless incoming <neutrino> of <energy> $E$. If the <electron> receives <kinetic energy> $T$ and <momentum> $\mathbf p$, its <relativistic energy-momentum relation> gives
$$
p^2=(m+T)^2-m^2=T^2+2mT.
$$
<conservation of energy> fixes the outgoing <neutrino> <energy> to $E-T$, and <conservation of momentum> fixes its <momentum> to $\mathbf k-\mathbf p$. Its zero <mass> therefore requires
$$
(E-T)^2=E^2+p^2-2Ep\cos\theta.
$$
Substitution and cancellation give
$$
(E+m)T=E\sqrt{T(T+2m)}\cos\theta.
$$
For $T>0$, this requires $\cos\theta>0$; squaring must not introduce a backward-electron branch. Squaring, cancelling $T$ and solving gives
$$
\boxed{T=\frac{2mE^2\cos^2\theta}{(E+m)^2-E^2\cos^2\theta}.}
$$
The expression increases with $\cos^2\theta$ on the physical interval $0\leq\theta\leq\pi/2$. Maximum transfer occurs for a forward <electron> and a backward outgoing <neutrino>:
$$
T_{\max}=\frac{2E^2}{m+2E}
=\frac{2(0.86)^2}{0.511+2(0.86)}\,\mathrm{MeV}
=\boxed{0.663\,\mathrm{MeV}.}
$$
The remaining $0.197\,\mathrm{MeV}$ is carried by the <neutrino>. An elastic collision cannot transfer the entire incident <energy> to a free massive <electron>.
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