Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 314 (Astrophysical Fluid Dynamics)

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.

The collapse is perpendicular to the initial field, so magnetic flux freezing preserves the mass-to-flux ratio of each material flux tube:With no toroidal field, radial magnetostatic equilibrium isDefine the effective polytropic constantThen gas and magnetic pressure combine as $P+B_z^2/(8\pi )=K_{\mathrm{eff}}{\rho }^2$. Dividing equilibrium by $\rho$, differentiating, and using the cylindrical Poisson equationgivesPut $a^2=K_{\mathrm{eff}}/(2\pi G)$. This is the order-zero Bessel differential equation, and regularity on the axis together with $\rho (0)={\rho }_1$ yields

The physical filament ends when the mass density first reaches zero. If $x_1\simeq 2.4$ is the first positive zero of the Bessel function $J_0$, thenThe first zero must be used because continuing to the next lobe would make the density negative.

The line mass isUsing $d[x{J}_1(x)]/dx=x{J}_0(x)$ and $J_1(x)=-J_0^{\prime }(x)$ givesSince $x_1\simeq 2.4$ and $J_0^{\prime }(x_1)\simeq -0.52$,

Write $Y=B_{\varphi }^2$, $x=R^2/L^2$, and retain the frozen axial field $B_z=(B_0/{\rho }_0)\rho$. The radial magnetostatic equilibrium equation now includes both the gradient of the toroidal magnetic pressure and the inward magnetic tension:For $\rho ={\rho }_1{e}^{-x}$, cylindrical gravity givesMultiplication by the integrating factor $R^2$ therefore producesThe integration constant must vanish for regularity on the axis. Direct integration givesIts Taylor expansion at the axis isso the regular field behaves as $B_{\varphi }=O(R)$. At large radius,Nonnegativity at infinity requiresThis condition is also sufficient: at the limiting value, the braces divided by $K_{\mathrm{eff}}{L}^2$ reduce to $2e^{-x}[1-(1+x)e^{-x}]\ge 0$, and decreasing $L^2$ only increases them. Thus the toroidal magnetic field is real and regular at every radius exactly in the stated range.

Steady spherical mass conservation gives $4\pi r^2\rho u=\dot{M}$. For the polytropic equation of state, $c_s^2=dp/d\rho$, soSubstitution into the radial Euler equations for an inviscid fluid gives the Parker wind equationAt a sonic point, the coefficient of $du/dr$ vanishes. A smooth transonic branch can pass it only if the right-hand side vanishes simultaneously. Hence

The steady Bernoulli equation isAt the regular sonic point, part a givesDefineThe isentropic flow relation $c_s^2\propto {\rho }^{\gamma -1}$ and mass conservation between the stellar surface and the sonic point giveEquating the surface and sonic values of the Bernoulli function therefore yields the required relationAn outflow reaching infinity with positive terminal kinetic energy requires $E>0$, henceAt $\gamma =5/3$ the terminal state is marginal with $E=0$; larger $\gamma$ cannot support the stipulated wind to infinity.

At $\alpha =1/2$, setting ${\mathcal{M}}_0=1$ gives $x=1$ in the mass relation, and both sides of the boxed relation in part b equal $\nu$. Moreover,so the stellar surface itself is the sonic point. This conclusion is independent of $\gamma$ throughout $1<\gamma <5/3$.

There is no other value of $\alpha$ giving a smooth solution with ${\mathcal{M}}_0=1$. Indeed, eliminating $\alpha$ in favour of $x$ reduces the Bernoulli relation toThe left side has its unique minimum at $x=1$, where it equals the right side. Thus $x=1$ and $\alpha =1/2$ are forced.

For fixed $0<{\mathcal{M}}_0<1$, eliminate $\alpha$ byThe remaining equation isIts left side diverges at both ends and has one minimum, atFor $0<{\mathcal{M}}_0<1$ this minimum lies strictly below the right side, so there are exactly two positive values of $x$, and hence exactly two values of $\alpha$. The branches merge at the marginal point $(\alpha ,{\mathcal{M}}_0)=(1/2,1)$.

On the very hot branch, $\alpha \to \infty$ and ${\mathcal{M}}_0=O({\alpha }^{-2})$. Dropping $1/\alpha$ and ${\mathcal{M}}_0^2$ in the relation from part b gives

The shock speed isApply the Strong-shock Rankine-Hugoniot conditions in the shock frame and transform the downstream velocity back to the ambient-medium frame. With the shock compression ratiothe post-shock values are

For $\gamma =5/3$, the shock data give ${\rho }_{\mathrm{ps}}=4{\rho }_0$ and $u_{\mathrm{ps}}=3V_s/4$. Since $V_s=R/(3t)$, the prescribed homologous spherical flow isPut $\xi =r/R(t)$ and $\rho ={\rho }_{0}D(\xi )$. The spherical continuity equation becomesso $D\propto {\xi }^9$. Matching the post-shock density givesThe material acceleration isThe radial Euler equations for an inviscid fluid thus give $dp/dr=3\rho r/(16t^2)$. Integrating from the centre to the shock,Since $p_{\mathrm{ps}}={\rho }_0{R}^2/(12t^2)$,

Resolve the uniform upstream field into components normal and tangential to the spherical shock:The ideal magnetohydrodynamic shock conditions leave the normal field continuous and multiply the tangential field by the gas compression ratio $\chi$. Therefore

The post-shock density is $\chi {\rho }_0$, while part c givesUsing the Alfvén speed $v_A=B/\sqrt{4\pi \rho }$ in Gaussian units,The ratio is largest in the equatorial plane, where the field is tangential to the shock: