Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 314 / 1

In the inertial frame, rigid corotation gives $\mathbf{u}=\mathbf{\Omega }\times \mathbf{r}$. The ideal magnetohydrodynamics condition is the motional electric fieldApply Gauss's law in Gaussian units and use the divergence and curl of a cross product:The exterior field is produced by currents inside the star, so $\nabla \times {\mathbf{B}}_0=0$ there, while rotation with constant angular velocity has $\nabla \times \mathbf{u}=2\mathbf{\Omega }$. Hence the required Goldreich-Julian charge density isHere $n_q$ denotes electric charge per unit volume, as in the question. If each carrier has charge $q$, its signed number density is $n_q/q$.

For the aligned magnetic dipole field,Since ${\mathbf{e}}_z=\cos \theta {\mathbf{e}}_r-\sin \theta {\mathbf{e}}_{\theta }$,The corotation velocity is $\mathbf{u}=\Omega r\sin \theta {\mathbf{e}}_{\varphi }$. The current density is therefore the advected charge density,

In the midplane, the vacuum magnetic dipole field has scale $B_0\sim \mu /r^3$, while the result of part i gives $j\sim {\Omega }^2\mu /(c{r}^2)$. Ampère's law then gives the induced-field estimatewhere numerical factors are immaterial in this dimensional analysis. Thus $B_i\sim B_0$ atThis is the light cylinder radius: rigid corotation would have $u=\Omega R_e\sim c$. The nonrelativistic approximation and the assumed unmodified vacuum field therefore fail at the same scale, and a self-consistent pulsar magnetosphere must replace them.

The magnetic-field-line equation for the dipole isIntegration gives the dipole magnetic-field lineThe line crosses the midplane at $r=R_e$, so $C=R_e$. At the stellar surface,Consequentlyfor a small polar-cap opening angle.