Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 314 (Astrophysical Fluid Dynamics)

In a magnetostatic equilibrium, inertia is absent and all forces balance. A force-free magnetic field is the limiting case in which the Lorentz force density vanishes:Thus the current density is parallel to the magnetic field, so for some scalar force-free parameter $\alpha (\mathbf{r})$,Using Ampère's law in magnetostatics givesTaking the divergence and using both the divergence of a curl is zero and $\nabla \cdot \mathbf{B}=0$ yieldssoThe force-free parameter is therefore constant along every magnetic field line.

PutFor an axisymmetric vector field depending only on the cylindrical radius $R$, the solenoidal vector field condition isHence $R{B}_R$ is constant. The hypothesis $B_R\to 0$ on the axis forcesThe azimuthal and axial components of $\nabla \times \mathbf{B}=q\mathbf{B}$ are thenEliminating $B_{\varphi }=-B_z^{\prime }/q$ gives the closed ordinary differential equationOnce $B_z$ is known, $B_{\varphi }=-B_z^{\prime }/q$ and $B_R=0$ determine the full cylindrical force-free magnetic field.

Here $q=\kappa /R$, so the equation from part (b) becomes the Euler-Cauchy equationThe power-law ansatz $B_z=R^s$ gives the indicial equationFor $0<\kappa <1/2$, the general field is thereforeThe polynomial discriminant changes sign atFor $\kappa >{\kappa }_c$, define $\mu =\sqrt{{\kappa }^2-1/4}$. A real form of the solution isThus the field has a log-periodic oscillation: its phase is periodic in $\ln R$, so it oscillates as the radius changes geometrically.

At $\kappa ={\kappa }_c=1/2$, the indicial equation has the multiple root $s=-1/2$. The general axial component isSince $B_{\varphi }=-2R{B}_z^{\prime }$, the azimuthal component isConsequentlyas $R\to \infty$; the same ratio is identically one when $D=0$. A magnetic field line has tangent parallel to $\mathbf{B}$, so its asymptotic angle $\vartheta$ with the $z$-axis satisfiesHence the field lines become helices making the constant anglewith the axis.

Let $u(r)>0$ denote the inward radial velocity. Steady spherical symmetry and mass conservation giveThe radial Euler momentum equation for the globally isothermal equation of state $p=c_s^2\rho$ isThe logarithmic derivative of mass conservation is ${\rho }^{\prime }/\rho =-u^{\prime }/u-2/r$. Substitution gives the Isothermal Bondi equationAt a smooth sonic point, $u=c_s$ makes the coefficient of $u^{\prime }$ vanish, so the right-hand side must vanish too. ThereforeThe solution that crosses this critical point of the isothermal Bondi equation continuously is the transonic accretion solution.

The radial equation can be writtenThus the Bernoulli function is constant on each streamline:The boundary conditions $u\to 0$ and $\rho \to {\rho }_0$ as $r\to \infty$ fix $\mathcal{B}=c_s^2\ln {\rho }_0$.

At the Bondi sonic point, $u_s=c_s$ and $GM/r_s=2c_s^2$. Evaluating the Bernoulli function there givesand henceThe conserved mass accretion rate is thereforeThus

With the dimensionless variables $x=r/r_s$ and $y=u/c_s$, mass conservation and part (c) giveSubstitution into the Bernoulli function, together with $GM/r=2c_s^2/x$, eliminates every dimensional parameter and gives the transcendental equationEquivalently,orThe transonic branch passes through $(x,y)=(1,1)$.

Put $Y=y^2$. On the inner supersonic branch, the equation from part (d) isIts asymptotic expansion as $x\to 0^{+}$ beginsTaking the positive square root givesSince $x=r/r_s$ and $y=u/c_s$,The leading term is the free-fall speed $\sqrt{2GM/r}$; the logarithmic term is the first pressure correction to the inner Isothermal Bondi accretion flow.

In the wing frame the unperturbed plasma moves with velocity $-u_0{\mathbf{e}}_x$. The ideal magnetohydrodynamics condition $\mathbf{E}+\mathbf{u}\times \mathbf{B}/c=0$ therefore produces the motional electric fieldThe wing's surface conductivity then gives the surface current densityIts direction also follows directly from the Lorentz force on the wing's mobile charges.

The wing and its forcing are time independent in the co-moving frame, so after transients have propagated away the perturbation is stationary there. Write the background velocity and field asand take every perturbation to be independent of $y$. The cold-plasma linearized ideal magnetohydrodynamic equations areTheir relevant components areDifferentiate the momentum equation with respect to $x$ and use the two induction relations. With the Alfvén speedthe result is

Integrating the $y$ component of Ampère's law,through the current sheet gives the magnetic-field jump across a surface currentReflection in the wing plane reverses the tangential perturbation $\delta B_x$. Ifthe upper and lower boundary values on the wing are consequentlyOutside the wing, the corresponding boundary value is zero.

For $u_0>u_A$, define the Alfvénic Mach-cone slopeThe equation in part (b) is the hyperbolic partial differential equationwhose characteristic curves are $x\pm mz=\text{constant}$. Let $W(s)$ be one for $|s|<L/2$ and zero otherwise. Selecting the characteristics that trail downstream, toward negative $x$, and imposing part (c) givesThe stationary induction and mass conservation equations then givewith $\delta u_z=\delta u_y=\delta B_y=0$. In the strongly super-Alfvénic limit, $\delta u_x\simeq u_{A}b/B_0$ and $\delta B_z\simeq b{u}_A/u_0$ inside the disturbed region.

The perturbations have support only whereBecause the wing is infinite in $y$, this is the union of two inclined slabs bounded by the characteristic planes $x+m|z|=\pm L/2$. These two trailing slabs are the Alfvén wings generated by the conductor.

The given velocity is a homologous spherical flow,with the spatially uniform velocity divergenceEvery shocked fluid element therefore expands rather than undergoing continued compression, and the velocity fills the remnant smoothly instead of concentrating its mass in a thin cooling shell. In the absence of radiative losses, the entropy advection equation then makes each element follow an adiabatic process after its one entropy-producing passage through the shock. This broad expanding structure is the expected adiabatic phase of a supernova remnant.

The Strong-shock Rankine-Hugoniot conditions give the compression ratiofor $\gamma =5/3$. ThusIn the shock frame, mass conservation saysUsing ${\rho }_2=4{\rho }_0$ gives

Resolve the upstream uniform field ${\mathbf{B}}_0=B_0{\mathbf{e}}_z$ into components normal and tangential to the spherical shock:The ideal magnetohydrodynamic shock conditions preserve the normal magnetic component and multiply the tangential component of a weak passive field by the gas compression ratio. Part (b) therefore gives

For the homologous spherical flow $\mathbf{u}=q\mathbf{r}$, the ideal magnetohydrodynamic induction equation and $\nabla \cdot \mathbf{B}=0$ give the material derivativeLet $x=r/R(t)$ and use the self-similar ansatzPart (b) gives $q=3\dot{R}/(4R)$, and thereforeThe radial induction equation becomes $x{f}^{\prime }=6f$. The boundary value $f(1)=1$ gives $f=x^6$. The solenoidal vector field condition requireswhich also matches the tangential shock value in part (c). Hence the interior field is

Outside the shock, the uniform-field lines obey $r\sin \theta =\text{constant}$. Inside, the magnetic-field-line equation givessoA sketch therefore shows straight exterior lines refracting at the spherical shock into north-south symmetric curves that bow toward the equatorial interior before leaving through the opposite hemisphere. The field strength falls as $r^6$ toward the centre.