Past exam of the mathematics course of the University of Cambridge / 2011 / ia / Paper 4 / 9B (Dynamics and Relativity)

Take the scalar product of Newton's second law with the velocity $\dot{\mathbf{r}}$. The cross product is perpendicular to its first factor, soConsequentlyThis is conservation of energy: the sum of kinetic energy and potential energy stays fixed. The magnetic-type Lorentz force $\dot{\mathbf{r}}\times \mathbf{B}$ does no work, because it is perpendicular to the instantaneous velocity. It may turn the trajectory but cannot directly change its speed, regardless of the spatial variation of $\mathbf{B}$.

For $\psi =\mu /r$, the radial gradient givesThe force $-\nabla \varphi$ is radial and has no torque about the origin. For the remaining force, use the vector triple product:Meanwhile the radial unit vector obeysSince $\psi \mathbf{r}=\mu \hat{\mathbf{r}}$, these derivatives cancel. Thus the generalized angular momentum in a radial monopole field isThe argument applies away from $r=0$, where the field is defined, for any radial $\varphi$. In contrast to ordinary angular momentum, this conserved vector has $\mathbf{L}\cdot \hat{\mathbf{r}}=-\mu$; it fixes a cone rather than generally fixing a plane.

With $\psi =0$, the angular momentum becomes $\mathbf{L}=m\mathbf{r}\times \dot{\mathbf{r}}$ and is constant. The identity $\mathbf{L}\cdot \mathbf{r}=0$ confines the trajectory to the plane through the origin perpendicular to $\mathbf{L}$ when $\mathbf{L}\ne 0$. If $\mathbf{L}=0$, the velocity is radial and a central force preserves the radial line; this degenerate motion is also planar.

For the proposed Laplace-Runge-Lenz vector convention, differentiate $\mathbf{K}=\mathbf{L}\times \dot{\mathbf{r}}-\varphi (r)\mathbf{r}$. Since $\ddot{\mathbf{r}}=-{\varphi }^{\prime }(r)\hat{\mathbf{r}}/m$ and $\mathbf{L}$ is constant, the vector triple product givesThereforeIt vanishes for general motion precisely when $(r\varphi {)}^{\prime }=0$, givingThis is an inverse-square potential, attractive when $C<0$ and repulsive when $C>0$; $C=0$ is the free-particle case. An additive constant in $\varphi$ leaves the force unchanged but would change this particular $\mathbf{K}$ by a constant multiple of $\mathbf{r}$, so its definition requires the displayed choice of energy zero.