Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 69 (Astrophysical Fluid Dynamics)

For an axisymmetric magnetic field, the solenoidal condition becomes ${\partial }_R(R{B}_R)+{\partial }_z(R{B}_z)=0$. Locally introduce a poloidal magnetic flux function $\Psi$ byThe divergence condition is exactly the integrability condition for these two derivatives. The additive constant in $\Psi$ fixes the reference flux. For a field regular on the axis choose $\Psi (0,z,t)=0$. Then the magnetic flux through a horizontal disk of radius $R$ is $2\pi \Psi (R,z,t)$, and the flux between two such surfaces is $2\pi \Delta \Psi$.

Since ${\mathbf{B}}_p\cdot \nabla \Psi =0$ and ${\partial }_{\varphi }\Psi =0$, the level surfaces of $\Psi$ contain the complete magnetic field, including its toroidal magnetic field component. Thus regular levels are poloidal flux surfaces. Their meridional contours trace the projected field lines. The representation is local on regular regions; disconnected flux surfaces with the same label need not share additional physical data.

The azimuthal component of $\mathbf{u}\times \mathbf{B}$ is $u_z{B}_R-u_R{B}_z=-{\mathbf{u}}_p\cdot \nabla \Psi /R$. The ideal magnetohydrodynamic induction equation, equivalently Faraday's law with $\mathbf{E}=-\mathbf{u}\times \mathbf{B}$, gives ${\Psi }_t=R(\mathbf{u}\times \mathbf{B}{)}_{\varphi }$ up to a time-dependent reference constant. Fixing that reference gives the material advection of an axisymmetric magnetic flux function:Axisymmetry removes the azimuthal contribution to the material derivative. This is magnetic flux freezing expressed using one scalar flux label.

The azimuthal component itself is generally not an advected scalar. Including the cylindrical basis-vector derivatives in the induction equation givesUsing mass conservation, $D\rho /Dt=-\rho \nabla \cdot \mathbf{u}$, and $DR/Dt=u_R$, this becomes the toroidal magnetic-field winding equation:Thus $B_{\varphi }/(\rho R)$ is conserved following the flow if the angular velocity is constant along the poloidal magnetic field, including the case of no rotation. Otherwise differential rotation winds poloidal field into toroidal field.

There is also an explicit flux-conservation form,Integrating this over a meridional region converts its change of toroidal flux to boundary terms. Hence such a flux is conserved when the indicated boundary transport vanishes. More generally magnetic flux freezing conserves flux through every material surface, but a meridional coordinate surface need not remain material when the fluid rotates. Toroidal flux conservation therefore does not imply that $B_{\varphi }$ itself is materially constant.

Use the standard uniform-density incompressible flow version of ideal magnetohydrodynamics, in which $\nabla \cdot \mathbf{u}=0$ and pressure enforces that constraint. The finite-$\gamma$ compressible adiabatic pressure equation is replaced by this pressure constraint. The consequences of retaining that cover-sheet equation literally as an additional condition are given below.

Write $F=f(R,z+v(R)t)$, $G=g(R,z-v(R)t)$, $U=F+G$ and $b=F-G$, where $b$ is the toroidal Alfvén velocity. The azimuthal velocity is $U$, while the Alfvén velocity vector is $b{\mathbf{e}}_{\varphi }+v{\mathbf{e}}_z$. Both solenoidal constraints hold automatically. Since there is no meridional flow, the azimuthal magnetohydrodynamic momentum equation and ideal magnetohydrodynamic induction equation reduce toDifferentiating the traveling arguments verifies both equations for arbitrary smooth $f$, $g$ and $v$. In particular, no condition $v^{\prime }=0$ is needed.

The remaining force balance is essential. Define the total pressure per unit mass, including the gravitational potential, byThe radial acceleration is $-U^2/R$, and magnetic curvature tension contributes $-b^2/R$. The radial and axial momentum equations are consequentlyThey are compatible if and only ifThis is the pressure compatibility of nonlinear torsional Alfvén profiles. It is also sufficient locally: when $FG$ is independent of $z$, chooseA stationary axisymmetric $\Phi$ satisfying the Poisson equation can be absorbed in this pressure. Gravity need not vanish. If the axis is included, smoothness requires $U,b=O(R)$ and a regular axial field; $f,g=O(R)$ and smooth even $v$ give the usual regular extension. The pressure constant may be chosen to make $p$ positive on a bounded region where the other terms are bounded.

The compatible profiles have the following interpretation. At a cylinder with $v\ne 0$, the variables $a=z+vt$ and $c=z-vt$ are independent as $z,t$ vary. Compatibility reads $f_a(R,a)g(R,c)+f(R,a)g_c(R,c)=0$. If both profiles are nonzero, division gives $f_a/f=-g_c/g=\kappa (R)$, independent of both traveling arguments. Thereforelocally on such a region. Alternatively, one profile vanishes identically and the other is unrestricted. Choosing a point where one profile is nonzero shows that a nontrivial opposite profile obeys the same linear ordinary differential equation everywhere on a connected axial interval, so isolated zeros do not create additional overlapping branches.

If $v>0$, a pure $f$ profile travels towards negative $z$, and a pure $g$ profile towards positive $z$, each at the local Alfvén speed $v(R)$. Their velocity and toroidal-field perturbations have equal magnitudes, with respectively the same or opposite signs. They are finite-amplitude torsional Alfvén waves. For a single profile, centrifugal force and magnetic curvature tension cancel, leaving spatially constant $\Pi$; gas pressure compensates the varying magnetic pressure. The waveform is transported without axial steepening. Different cylinder speeds shear its phase radially, magnetohydrodynamic phase mixing.

At linear order both wave families may be superposed, since their interaction product is second order in amplitude. Two arbitrary opposing profiles cannot simply be superposed at finite amplitude: their product would generate an axial variation in the radial force imbalance, incompatible with purely azimuthal velocity. The exponential pair is a formal exception, but is unbounded in an axial direction when $\kappa \ne 0$. If both profiles must be bounded on the whole axial line, $\kappa =0$, so $f=A(R)$ and $g=C(R)$ describe a steady rotating, twisted-field equilibrium. Their pressure gradient balances the remaining centrifugal and magnetic curvature forces. At $v=0$ there is no propagation; the condition is just $f(R,z)g(R,z)=h(R)$, giving additional steady profiles. These cases exhaust the local possibilities under the stated ansatz and domain qualifications.

There is a closure distinction in the wording. Incompressible flow is the formal infinite-sound-speed limit, not the simultaneous imposition of finite-$\gamma$ adiabatic compression and an independently fixed density; see Ogilvie's discussion of simplified fluid models, section 2.8. If the cover-sheet pressure equation is nevertheless imposed in addition, then axisymmetry and purely azimuthal motion give $p_t=0$. In addition to $(FG{)}_z=0$, the necessary and sufficient local pressure-gradient conditions, with stationary $\Phi$, areThey follow by requiring the gradients $p_z/\rho =-{\Phi }_z-\frac{1}{2}(b^2{)}_z$ and $p_R/\rho =4FG/R-{\Phi }_R-\frac{1}{2}(b^2+v^2{)}_R$ to be time independent; the remaining spatially uniform pressure variation is removed by $C(t)$.

For a single traveling profile with $v\ne 0$, these extra conditions require its square to be affine in its traveling argument, $f^2=a(R)(z+vt)+b_0(R)$ or $g^2=a(R)(z-vt)+b_0(R)$, with $v(R)a(R)$ constant across a connected radial region. Smooth bounded profiles on the entire axial line then have $a=0$ and are steady. For two nonzero exponential profiles the extra conditions force $\kappa =0$. Thus arbitrary finite-amplitude traveling pulses are valid for the intended incompressible pressure model, but not generally for the overconstrained finite-$\gamma$ closure. The force compatibility derived above makes the distinction explicit.

Let $e$ be the specific internal energy, $s$ the specific entropy, $T$ the absolute temperature, and $w=e+p/\rho$ the specific enthalpy. The first law of thermodynamics for a simple fluid of fixed composition is $de=Tds-pd(1/\rho )$, soThe unmagnetized Euler equations for an inviscid fluid are ${\partial }_t\mathbf{u}+(\mathbf{u}\cdot \nabla )\mathbf{u}=-\nabla \Phi -\nabla p/\rho$. Define the vorticity $\mathbf{\omega }=\nabla \times \mathbf{u}$ and use the vorticity cross-product identity $(\mathbf{u}\cdot \nabla )\mathbf{u}=\mathbf{\omega }\times \mathbf{u}+\nabla (u^2/2)$. Substitution gives the Crocco form of the unsteady Euler equation:Adiabaticity means $Ds/Dt=0$ for smooth flow without dissipative heating. It does not require $s$ to be uniform between different fluid elements, so the entropy-gradient term generally remains. For a perfect gas with constant adiabatic exponent $\gamma$, one may write $w=\gamma p/[(\gamma -1)\rho ]$, but the thermodynamic derivation does not require that particular equation of state.

Steady axisymmetric mass conservation is $\nabla \cdot (\rho {\mathbf{u}}_p)=0$, permitting the axisymmetric hydrodynamic mass flux function representationThus stream surfaces are regular level surfaces of $\psi$. Adiabaticity gives $\mathbf{u}\cdot \nabla s=0$, and axisymmetry removes the azimuthal derivative, so the specific entropy is constant along their meridional streamlines. The azimuthal Euler equations for an inviscid fluid has no pressure or gravitational torque:The conserved specific angular momentum is $\ell =R{u}_{\varphi }$. Taking the scalar product of the steady Crocco form of the unsteady Euler equation with $\mathbf{u}$ also givesConsequently, on each connected regular family of stream surfaces,These functions express entropy, angular momentum and the Bernoulli function carried by a stellar wind. Stagnation regions and disconnected branches with identical flux labels require their own matching information; the argument establishes the functions locally where the poloidal flow is nonzero.

To derive the hydrodynamic transfield equation for an axisymmetric wind, decompose the vorticity. The poloidal part comes from the toroidal velocity:The azimuthal part comes from the poloidal velocity:The final divergence is the full axisymmetric cylindrical divergence, including its radial $1/R$ factor.

Since ${\mathbf{u}}_p=\nabla \psi \times {\mathbf{e}}_{\varphi }/(\rho R)$, the meridional part of $\mathbf{\omega }\times \mathbf{u}$ isThe remaining steady Euler equations for an inviscid fluid is thereforeCanceling $\nabla \psi$ on a regular stream surface and inserting ${\omega }_{\varphi }$ givesThe sign of the angular-momentum term follows from ${\mathbf{u}}_p\times {\mathbf{e}}_{\varphi }=-\nabla \psi /(\rho R)$. To determine the wind, this transverse force equation is supplemented by its Bernoulli function relationand the equation of state, with the gravitational potential supplied externally or by the Poisson equation.

Write $u=u_x$, $w=u_y$ and $B=B_y$. The isothermal equation of state replaces the adiabatic thermal equation by $p=c_s^2\rho$. Since the fields depend only on $x,t$, the solenoidal constraint gives ${\partial }_x{B}_x=0$ and the $x$ component of the ideal magnetohydrodynamic induction equation gives ${\partial }_t{B}_x=0$. Hence $B_x$ is constant in both space and time.

The continuity, two momentum and transverse induction equations areHere $B$ is the transverse magnetic component and the constant $B_x$ is the longitudinal component. The longitudinal magnetic force is the gradient of transverse magnetic pressure; the transverse force is magnetic tension. In primitive variables the isothermal coplanar magnetohydrodynamic characteristic matrix isThe imposed $z$ components stay zero because their corresponding equations have zero right-hand sides under this reduction.

Let $c=v-u$ be a characteristic speed relative to the fluid and define the Alfvén velocity ${\mathbf{v}}_a=\mathbf{B}/\sqrt{{\mu }_0\rho }$. To derive the characteristic polynomial, an eigenvector with components $(\delta \rho ,\delta u,\delta w,\delta B)$ obeysEliminating $\delta u$ and $\delta w$ on a nondegenerate branch givesTheir consistency yields $(c^2-c_s^2)(c^2-v_{ax}^2)=v_{ay}^2{c}^2$. Equivalently, direct expansion of $\det (\mathbf{A}-vI)$ gives the same polynomial without dividing by $c$ or a possibly vanishing field component:ThusThe discriminant is $(c_s^2-v_a^2{)}^2+4c_s^2{v}_{ay}^2\ge 0$, and both squared speeds are nonnegative. These are the fast magnetosonic wave and slow magnetosonic wave families, transported with the mean flow. The equations are hyperbolic; on a generic state the four distinct real eigenvalues give four characteristic directions. Special field alignments can make speeds coincide or the slow speed vanish. There is no separate entropy characteristic because the thermal model is isothermal, and the independent out-of-plane Alfvén wave polarization is excluded by $u_z=B_z=0$.

To construct a simple wave in magnetohydrodynamics, take $\mathbf{U}=\mathbf{U}(a)$ along a state-space curve whose tangent is a right eigenvector of one selected branch:Substitution into the partial differential equations then gives the scalar transport equationSuch curves exist locally by integrating the eigenvector field away from degeneracies. For example, parameterizing by $a=\rho$ where its eigenvector density component is nonzero gives the explicit state-curve equationswith signed $c$ equal to one of the four relative wave speeds. These formulas are used only where their denominators are nonzero.

A particularly transparent nonlinear example is an isothermal perpendicular magnetosonic simple wave. Set the allowed constant $B_x=0$, choose $B=K\rho$ and keep $w$ constant. For the fast branch,These relations satisfy the eigenvector equations and therefore all four original equations wheneverThe amplitude-dependent characteristic speed has derivativeThis already proves the existence of genuinely nonlinear waves in the stated system.

More generally, the method of characteristics applied to initial data $a(x,0)=a_0(x)$ givesIf the initial speed decreases towards increasing $x$ somewhere, the denominator reaches zero in finite time. When the negative minimum is attained, the first steepening time isFaster trailing states catch slower leading states, giving shock formation by characteristic intersection. For the right-moving perpendicular fast example, a negative density gradient provides such a compression. Beyond $t_{\ast }$ a smooth simple wave is no longer sufficient and an appropriate shock solution is required. Rarefactive profiles instead spread. A linearly degenerate characteristic field has $v^{\prime }=0$ along its state curve and avoids this mechanism; the conclusion is that compressive nonlinear magnetosonic simple waves steepen, not that every possible MHD waveform must do so.

Let $m(r)$ be the background enclosed mass, so $g=Gm/r^2$ and $m^{\prime }=4\pi r^2\rho$. The background hydrostatic equilibrium is $p^{\prime }=-\rho g$. Use a Lagrangian fluid displacement $\xi ={\xi }_r$ of each spherical mass shell; $\Delta$ denotes its Lagrangian perturbation. Conservation of shell mass givesand the adiabatic pressure law gives the Lagrangian pressure perturbationThis retains perturbed self-gravity: a shell keeps its enclosed mass while its gravitational acceleration changes with its radius.

In mass coordinates, the exact radial shell equation is $\ddot{r}=-4\pi r^2\partial p/\partial m-Gm/r^2$. Its first-order perturbation isHydrostatic equilibrium gives ${\partial }_{m}p=-g/(4\pi r^2)$, and ${\partial }_m\Delta p=(\Delta p{)}^{\prime }/(4\pi r^2\rho )$. Thus the perturbation of the pressure-force geometry contributes another $2g\xi /r$, in addition to the gravity term:Put $\eta =\xi /r$, so ${\xi }^{\prime }+2\xi /r=r{\eta }^{\prime }+3\eta$. Expanding the derivative, using constant $\gamma$ and $p^{\prime }=-\rho g$, yields the radial stellar pulsation equation:The coefficient $4$ includes both geometric pressure and self-gravity effects; dropping the gravitational perturbation would not give this equation.

For a normal mode $\xi (r,t)=\xi (r)e^{-i\omega t}$, multiply by $\rho r^2{\xi }^{\ast }$ and integrate. Since ${\xi }^{\ast }/r={\eta }^{\ast }$, integration by parts givesRegularity at the center requires $\xi =O(r)$ with finite $\eta$, so the lower endpoint vanishes. For an isolated star take $p(R)=0$ and the free-surface condition $\Delta p=0$, with a regular finite-energy displacement; these eliminate the upper boundary term. HenceThese boundary conditions give the self-adjoint radial Sturm-Liouville problem behind the weighted stellar pulsation Rayleigh quotient.

Use the homologous trial fluid displacement ${\xi }_r=r$, or $\eta =1$. It is regular at the center, and $\Delta p=-3\gamma p$ vanishes at the free surface. The derivative term in the Rayleigh quotient is zero, leavingThe ratio is a positive weighted average of $g/r$. Since the central mass density is its maximum,For $\gamma >4/3$ multiplication preserves this inequality, proving the central-density upper bound for a radial stellar frequency:The full positive quadratic form also shows radial stability for constant $\gamma >4/3$ under these assumptions. A uniform-density equilibrium has constant $g/r$, making the homologous displacement an exact mode and saturating the bound.

If $\gamma <4/3$, the same trial value is strictly negative. Therefore ${\omega }_{\min }^2<0$: the star has an exponentially growing radial instability, with growth rate $\sqrt{-{\omega }_{\min }^2}$. One must use this negative trial quotient, rather than carrying the central-density inequality across the sign change unchanged. At the limiting value $\gamma =4/3$, the homologous displacement has zero restoring force and is a neutral radial mode in this constant-exponent model.