Past exam of the options I course of the University of Oxford / 2009 / H / H1

With $\mathbf{d}={\mathbf{q}}_1-{\mathbf{q}}_2$ and $r=|\mathbf{d}|$, the Lagrangian isAt configurations with $r>0$, the Euler-Lagrange equations giveThese formulas apply on the differentiable part of the potential; collision configurations require any additional smooth extension of $f(|\mathbf{d}|)$ to be specified. Summing the equations givesso the total linear momentum is conserved. The total energy isDifferentiating, the two interacting forces do work $-f^{\prime }(r)\dot{r}$, while the potential derivative is $f^{\prime }(r)\dot{r}$ and particle three contributes no force term. Hence $\dot{E}=0$. Both total momentum and total energy are conserved.

For the first particle, the derivative of its angular momentum about the origin is its torque:This need not vanish. For example take $f(r)=r^2/2$ and instantaneous positions ${\mathbf{q}}_1=(1,0,0)$, ${\mathbf{q}}_2=(0,1,0)$; the torque is $(0,0,1)$. Therefore it is not conserved in general, although it is in special configurations or for a constant potential.

The third particle is force-free, so its torque about the origin vanishes:Thus this angular momentum is conserved.

Let $\mathbf{F}$ be the force on particle one, so the force on particle two is $-\mathbf{F}$, with $\mathbf{F}$ parallel to ${\mathbf{q}}_1-{\mathbf{q}}_2$. The combined torque isHence the combined angular momentum of particles one and two is conserved. Centrality of their interaction, in addition to equal and opposite forces, is what makes this torque sum zero.