Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 62 (Stellar Magnetohydrodynamics)

Choose the planar magnetic field convention $\mathbf{B}=\nabla \times (A{\mathbf{e}}_z)$, so $B_s=s^{-1}{A}_{\varphi }$ and $B_{\varphi }=-A_s$. The resistive induction equation, with constant magnetic diffusivity and a suitable time-dependent gauge constant removed, reduces to advection and diffusion of this Cartesian magnetic flux function:For the specified differential rotation, its mth azimuthal Fourier mode obeysThe physical field is the real part of the complex mode. Substitute $a=s^{m}g(t)\exp [-im{\Omega }_{0}t-i{s}^{2}f(t)]$. A direct radial differentiation givesOn the left, rigid rotation cancels, leaving $\dot{g}/g-i{s}^2\dot{f}+im\omega s^2$. Equating the constant and quadratic coefficients yieldsThe initial condition fixes $f(0)=0$ and $g(0)=1$, since the constant C was already factored out. Set ${\mu }^2=4i\eta m\omega$. Then the exact quadratic-shear magnetic flux solution follows by differentiation:Indeed $\dot{f}=m\omega {\mathrm{sech}}^2(\mu t)$ and $4i\eta f^2=m\omega {\tanh }^2(\mu t)$. Also $\dot{g}/g=-(m+1)\mu \tanh (\mu t)=-4i\eta (m+1)f$. For an unspecified initial normalization the second formula has a multiplicative constant $C^{\prime }$, but here $C^{\prime }=1$.

For $\eta >0$ and $\omega >0$, choose $\mu =(1+i)\sqrt{2\eta m\omega }$. The asymptotic condition is $\mathrm{Re}(\mu )t\gg 1$: then $f$ tends to the finite constant $m\omega /\mu$ and $g\sim 2^{m+1}\exp [-(m+1)\mu t]$. At any fixed s, differentiating A to obtain the magnetic field multiplies it by factors involving $m/s$ and $sf$, which have no additional exponential time dependence. Therefore its envelope decays asUsing the paper's magnetic Reynolds number definition gives the precise scalingThus the decay time scales as ${\mathrm{Rm}}^{1/2}/{\Omega }_0$ for positive ${\Omega }_0$ and fixed m. The mechanism is magnetic phase mixing under differential rotation: winding produces progressively fine radial structure on which magnetic diffusion acts efficiently. For negative $\omega$, choose the square root with positive real part, replacing $\omega$ by $|\omega |$ in the decay time. If $\omega =0$, the initial harmonic mode simply rotates, with $f=0$, $g=1$; there is no shear-enhanced decay. The printed positive-root and Reynolds-number expressions implicitly use $\omega >0$.

Write the poloidal magnetic flux function and toroidal amplitude explicitly in spherical components:Dotting the magnetostatic equilibrium equation with $\mathbf{B}$ gives $\mathbf{B}\cdot \nabla p=0$. Since all fields are axisymmetric, the poloidal magnetic field is tangent to the contours of $\chi$ in a meridional section. Thus, locally on regular connected flux surfaces, $p=p(\chi )$. This is a local flux-surface statement; disconnected surfaces with the same numerical value of $\chi$ need not have the same pressure without further global assumptions.

Taking the azimuthal component of the Lorentz force givesIts numerator is the meridional Jacobian determinant of $\psi$ and $\chi$. Vanishing therefore implies locally $\psi =\psi (\chi )$ on the same regular surfaces. DefineA direct curl calculation now givesSince ${\mathbf{e}}_{\varphi }\times {\mathbf{B}}_{\mathrm{pol}}=\nabla \chi /(r\sin \theta )$, the remaining force isConsequently the spherical magnetostatic pressure balance equation following from the stated physical force balance isThe PDF's displayed scalar equation has the wrong sign and omits ${\mu }_0$ for its stated pressure p. This is not corrected by setting ${\mu }_0=1$, which leaves the sign discrepancy. For example, $\chi =r^4{\sin }^2\theta$, $\psi =0$, $p=p_0-10\chi /{\mu }_0$ satisfies the stated vector force balance and the corrected scalar equation, because ${\Delta }_{\ast }\chi =10r^2{\sin }^2\theta$; it fails the printed scalar equation. One could obtain the printed form by replacing its pressure variable by $-{\mu }_0{p}_{\mathrm{physical}}$, but that is not the variable in the given force balance.

Now set $\psi =\alpha \chi$, $p=p_0+P\chi$, and $\chi =F(r){\sin }^2\theta$. The angular operator gives $-2F{\sin }^2\theta /r^2$, soFor the trigonometric form in the question, direct substitution yieldsFor $\alpha \ne 0$, this is a solution providedThe same coefficient relation makes the trigonometric pair regular at the origin: $\cos \alpha r-\sin \alpha r/(\alpha r)=-{\alpha }^2{r}^2/3+O(r^4)$. Thus all components of the magnetic field are finite there. With $p_0=0$, this explicitly constructs the requested proportional-pressure family; adding $p_0$ does not change force balance. The coefficient c is the one called C in the final sentence of the PDF.

For a confined linear-pressure spherical magnetic equilibrium, the condition that the entire magnetic field vanish on $r=a$ is precisely $F(a)=F^{\prime }(a)=0$. In components these conditions readThey eliminate both the radial and meridional components; the toroidal component vanishes too because $\psi =\alpha \chi$. For a nontrivial solution with $A\ne 0$, substituting $B=-A/\alpha$ givesThe last equation selects the allowed nonzero values of $\alpha a$. Equivalently, when division is legitimate, $\tan x=3x/(3-x^2)$; the undivided equation also handles vanishing denominators correctly. Since $\chi (a,\theta )=0$, $\boxed{p(a,\theta )=p_0}$ is independent of latitude. In the strictly proportional convention $p=P\chi$, this is zero. The zero-field solution is a trivial possibility; the displayed conditions describe the nontrivial confined family.

The truncated fields must be interpreted as a Galerkin method approximation: project each equation onto its retained spatial modes and discard orthogonal higher harmonics. They are not an exact pointwise solution of the full partial differential equations, because the nonlinear terms generate additional harmonics. Write $X=kx$, $Z=\pi z$, and abbreviate the printed normalizations byThen $\psi =Sa\sin X\sin Z$, $\theta =Tb\cos X\sin Z-c\sin 2Z/\pi$, and $\chi =Dd\sin X\cos Z+Ee\sin 2X$. All five amplitudes a,b,c,d,e depend on $\tau =pt$. Use the spatial Jacobian determinant $J(F,G)=F_x{G}_z-F_z{G}_x$ and the inner product over the stated rectangle. Orthogonality of the trigonometric modes makes projection equivalent to retaining their coefficients. In particular, ${\nabla }^2\psi =-p\psi$, so $J(\psi ,{\nabla }^2\psi )=0$.

For the temperature equation, direct products giveThe other temperature contribution is $-2kSac\cos X\sin Z\cos 2Z$. Since $\sin Z\cos 2Z=(\sin 3Z-\sin Z)/2$, its retained component is $kSac\cos X\sin Z$. Thus the projected temperature equation isHere the dot denotes $d/d\tau$. Since $kS/(pT)=1$ and $\phi =4{\pi }^2/p$, these equations give $\dot{b}=-b+a(1-c)$ and $\dot{c}=\phi (-c+ab)$.

Similarly, the two magnetic-advection products arewhile $J(\psi ,Ee\sin 2X)=-2\pi Sae\sin X\cos Z\cos 2X$ has retained component $\pi Sae\sin X\cos Z$. The projected induction equation is thereforeUsing $\pi S/(pD)=1$ and $4k^2/p=4-\phi$ gives $\dot{d}=-\zeta d+a(1-e)$ and $\dot{e}=-(4-\phi )\zeta e+\phi ad$.

It remains to project the Lorentz force in the vorticity equation. The linear term is $J(x,{\nabla }^2\chi )=p\pi Dd\sin X\sin Z$. The nonlinear cross term isThe Jacobian on the right is $2\pi kDEde\sin X\sin Z\cos 2X$. Its retained component is $-\pi kDEde\sin X\sin Z$, so the total retained magnetic term isMeanwhile the projected acceleration, viscous term and temperature forcing have coefficients $-p^{2}S\dot{a}$, $p^{2}Sa$, and $-RkTb$. Divide the projected vorticity equation by $-p^{2}S$ to obtain $\dot{a}=\sigma [-a+rb+\zeta qd\{(\phi -3)e-1\}]$, using $r=k^{2}R/p^3$ and $q={\pi }^{2}Q/p^2$. This derives the complete five-mode vertical-field magnetoconvection system:The Rayleigh number, Chandrasekhar number, Prandtl number and magnetic-to-thermal diffusivity ratio enter through r,q,$\sigma$ and $\zeta$ respectively.

Linearization at the zero-amplitude state leaves c and e with negative rates $-\phi$ and $-(4-\phi )\zeta$, while a,b,d have coefficient matrixA zero eigenvector satisfies $b=a$, $d=a/\zeta$ and $[-1+r-q]a=0$. Thus the stationary threshold is $\boxed{r_c=1+q}$. The reversal symmetry $(a,b,c,d,e)\mapsto (-a,-b,c,-d,e)$ makes the nontrivial steady states occur as a pair.

For these nonzero steady states put $A=a^2$. The temperature equations give $c=ab$ and $b=a(1-c)$, so $b=a/(1+A)$ and $c=A/(1+A)$. The induction equations give $\zeta d=a(1-e)$ and $(4-\phi )\zeta e=\phi ad$. With $\mu =(4-\phi ){\zeta }^2/\phi$, their solution isHere $k>0$ gives $0<\phi <4$, and positive diffusivity gives $\mu >0$. Substituting these amplitudes into the stationary a equation and dividing by a yields the stationary branch of five-mode magnetoconvection:At $a=0$ this meets $r=1+q$. Expansion in $A=a^2$ givesThe denominator of its initial slope is positive, and its numerator after multiplication by $(4-\phi ){\zeta }^2$ isTherefore $\mathcal{N}>0$ gives a supercritical pitchfork bifurcation direction, and $\mathcal{N}<0$ gives a subcritical pitchfork bifurcation direction. If $\mathcal{N}=0$, the leading slope vanishes and higher terms must be retained.

This classification describes which side of the stationary threshold contains the steady branch. It does not alone establish dynamical stability or rule out an earlier oscillatory instability. For example, at $r_c$ the characteristic polynomial factors asThe zero eigenvalue is simple unless the final coefficient vanishes; the other two eigenvalues need not both be stable for every positive parameter choice. The stationary branch formula and its direction criterion remain the requested algebraic conclusions.

A solar-type main sequence star has a radiative interior and an outer convection zone. Rotating convection and differential rotation continually regenerate its magnetic field by dynamo action. The field is therefore not simply an initially stored field undergoing passive magnetic diffusion. In an alpha-Omega dynamo, the Omega effect winds a poloidal magnetic field into a toroidal magnetic field, while the alpha effect associated with helical motions can regenerate poloidal flux. In a solar dynamo, the Babcock-Leighton mechanism provides another poloidal source through the emergence and dispersal of tilted active regions. Shear near the tachocline and transport through the convection zone help organize the cycle; no single dynamo model is needed to claim that rotation and convection are essential ingredients.

Rotation acts on convective motions through the Coriolis force, altering their correlations and helicity. A useful activity measure is the stellar activity Rossby number, ${\mathrm{Ro}}_{\ast }=P_{\mathrm{rot}}/{\tau }_c$, where ${\tau }_c$ is a convective turnover time. At a given stellar structure, faster rotation ordinarily increases rotational influence and produces greater magnetic activity in the unsaturated regime. Young solar-type main sequence stars commonly rotate rapidly and show strong sunspots, flares, and chromospheric and coronal emission, although their initial rotation rates have substantial scatter. Very rapid rotators enter a saturated regime, so activity does not increase without limit as rotation increases. The saturation mechanism is not uniquely fixed by the correlation itself. The observed unsaturated correlation and saturated regime are documented in primary rotation–activity measurements.

The reverse interaction is wind-driven magnetic braking of a solar-type star. An ionized stellar wind is coupled to the star's open magnetic field. Matter carries angular momentum, and the Maxwell stress tensor also transmits a torque. Magnetic coupling provides an effective lever arm comparable to the Alfvén radius, rather than only the stellar radius. Writing ${\dot{M}}_w>0$ for the outward mass-loss rate, a schematic angular-momentum equation iswhere a geometry factor can be absorbed into an effective $R_A$. This lever-arm interpretation follows from the angular-momentum invariant of a magnetized wind; it is the central result of the original magnetized solar-wind calculation. Magnetic braking can be effective despite a small mass-loss rate, because $R_A$ can greatly exceed the stellar radius.

One can derive the familiar rotation-age scaling in a simple unsaturated model. For a radial open field, take $B_r(R_A)\simeq B_{\ast }(R_{\ast }/R_A{)}^2$, ${\dot{M}}_w\simeq 4\pi R_A^2{\rho }_A{v}_w$, and $v_w^2\simeq B_r(R_A{)}^2/({\mu }_0{\rho }_A)$. Eliminating ${\rho }_A$ givesIf the unsaturated large-scale field satisfies $B_{\ast }\propto \Omega$ and the other factors vary slowly, this gives $\dot{J}=-K{\Omega }^3$ with positive K. Approximating I as constant on this part of the main sequence, integration yieldsThus $\boxed{\Omega \propto t^{-1/2},P_{\mathrm{rot}}\propto t^{1/2}}$ in the late-time part of this approximation. This Skumanich rotation law captures the classic observed secular decline of rotation and magnetic activity; its observational basis is the original rotation and chromospheric-emission analysis. It is a scaling model, not an exact trajectory for every star. With a saturated field, the same simplified wind model instead gives a torque approximately proportional to $-\Omega$, so initially rapid rotation can decay more nearly exponentially until the unsaturated regime is reached.

Over the main sequence lifetime, wind loss therefore generally makes the star rotate more slowly, increases its stellar activity Rossby number, and reduces its unsaturated mean activity. Later changes of moment of inertia, wind properties and open-field topology alter the simple power law. Internal transport of angular momentum can also feed the braked envelope from the radiative interior, so surface rotation is not automatically a measure of the entire star's angular momentum. In a star with weak braking, this secular history differs from the ideal cubic-torque model; an age law alone is not a fundamental dynamo equation.

The stellar rotation-activity feedback also acts on shorter time scales. A growing magnetic field exerts a Lorentz force that changes differential rotation and convective motions, extracting kinetic energy and limiting further growth through dynamo quenching. Conversely, varying shear and poloidal regeneration change the magnetic cycle. The present solar-like regime can sustain repeated solar cycles, with polarity reversals and activity modulation superposed on the much slower rotational evolution. Rotation promotes dynamo activity; dynamo-generated fields brake rotation through the wind and react back on the motions that generate them. This coupled feedback explains the broad progression from young, rapidly rotating, strongly active stars toward older, more slowly rotating, less active solar-type stars, while allowing cycles, irregular modulation and different initial histories.