Past exam of the mathematics course of the University of Cambridge / 2026 / ia / Paper 4 / 10B (Dynamics and Relativity)

$m\ddot{x}=-\nabla V$. Dotting with $\dot{x}$ gives $d(\frac{1}{2}m|\dot{x}{|}^2+V)/dt=0$.

A central potential is $V(r)$. Then $\dot{L}=x\times (-\nabla V)=0$. Since $L\cdot x=0$, the trajectory lies in the fixed plane perpendicular to $L$.

The equation is $m\ddot{x}=-\nabla V+q\dot{x}\times B$; the magnetic force does no work. For $B=\beta \hat{r}/r^2$, the triple-product identity gives $\dot{L}=q\beta d\hat{r}/dt$, so $J=L-q\beta \hat{r}$ is conserved. Since $J\cdot \hat{r}=-q\beta$, the trajectory lies on a cone with $\cos \alpha =-q\beta /|J|$.

$L^2=J^2-(q\beta {)}^2=J^2{\sin }^2\alpha$. Splitting kinetic energy into radial and angular parts therefore gives $E=\frac{1}{2}m{\dot{r}}^2+J^2{\sin }^2\alpha /(2m{r}^2)+V(r)$, proving the stated effective potential.