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Use a steady axisymmetric flow governed by ideal magnetohydrodynamics and take the Newtonian gravitational potential to be axisymmetric. Work locally on connected regular magnetic flux surfaces with nonzero poloidal flow. As usual for a wind without an imposed toroidal loop voltage, $E_{\varphi }=0$; regularity on an included rotation axis guarantees this condition. It is needed to exclude steady cross-field flow in an annular domain.

Since $\nabla \varphi ={\mathbf{e}}_{\varphi }/R$, the poloidal magnetic flux function givesThe poloidal magnetic field is divergence-free. The zero toroidal electric field, with $\mathbf{E}=-\mathbf{u}\times \mathbf{B}$, gives ${\mathbf{u}}_p\parallel {\mathbf{B}}_p$. Define $k$ by $\rho {\mathbf{u}}_p=k{\mathbf{B}}_p$. Steady mass conservation then givesThus $k=k(\psi )$ is the magnetohydrodynamic mass loading, the mass flux per unit poloidal magnetic flux. Subtracting $(k/\rho )\mathbf{B}$ from the full velocity leaves only an azimuthal component. The cross product isThe steady ideal magnetohydrodynamic induction equation requires $0=\nabla \times (\omega \nabla \psi )=\nabla \omega \times \nabla \psi$. Consequently the field-line angular velocity is $\omega =\omega (\psi )$ and

Neither gas pressure nor axisymmetric gravity exerts an azimuthal torque. Multiplying the azimuthal momentum equation by $R$ combines the cylindrical curvature terms into angular-momentum transport:Insert $\rho {\mathbf{u}}_p=k{\mathbf{B}}_p$ and use ${\mathbf{B}}_p\cdot \nabla k=0$. It follows thatso the magnetohydrodynamic angular-momentum invariant isThe two terms account respectively for matter and Maxwell stress tensor transport of angular momentum.

Finally, subtracting $\gamma$ times the logarithmic mass density equation from the logarithmic pressure equation gives $D\ln (p/{\rho }^{\gamma })/Dt=0$. The specific entropy of a fixed-composition ideal gas is a function of $p/{\rho }^{\gamma }$. Steadiness and axisymmetry reduce its advection to ${\mathbf{u}}_p\cdot \nabla s=0$. HenceThese flux-label functions are local on connected regular surfaces; disconnected components with the same numerical label need not share constants without an additional matching condition.

To distinguish Eulerian and Lagrangian fluid perturbations, use $\mathbf{\xi }=({\xi }_x,{\xi }_y,{\xi }_z)$ for the fluid displacement, $\varphi ={\Phi }^{\prime }$ for the Eulerian potential perturbation, and $\delta \rho$ for the Eulerian mass density perturbation. The equilibrium mass density $\rho (z)$ is the profile from part (a); subscript $z$ below denotes differentiation. Linearized mass conservation and the isothermal equation of state giveThe velocity perturbation is $-i\omega \mathbf{\xi }$. Linearized momentum initially has the formSince $\nabla {\Phi }_0=-c_s^2\nabla \ln \rho$, the pressure and perturbed mass density gravity terms combine into $-\rho \nabla (c_s^2\delta \rho /\rho )$. Define $W=c_s^2\delta \rho /\rho +\varphi$. The closed linearized system isThe $y$ component is a neutral transverse sector; it vanishes for nonzero-frequency modes.

To prove reality rather than assume it, multiply momentum by $\rho {\mathbf{\xi }}^{\ast }$ and integrate over $z$. With $I=\int \rho |\mathbf{\xi }{|}^{2}dz>0$, integration by parts and the conjugate continuity equation giveUsing the conjugate Poisson equation for Newtonian gravity in the last integral givesTake disturbances with finite $I$, finite pressure and gravitational integrals, and vanishing pressure work and gravitational boundary terms. For $k\ne 0$, require the potential to decay at both infinities and the fluid displacement to make $\rho {\xi }_z^{\ast }W\to 0$ as well. In a source-free outer region the decaying potential is proportional to $e^{-|k||z|}$. The isothermal displacement energy of a self-gravitating slab then givesSince $I$ is strictly positive for a nonzero displacement, ${\omega }^2$ is real. Positive values describe oscillations; negative values give pure growth and decay. The result is the energy identity for the pressure-gravity displacement operator, a self-adjoint operator in the mass density-weighted inner product. The identity excludes an oscillatory growing eigenmode under these boundary conditions. At $k=0$ the same identity applies when its boundary terms vanish and its integrals remain finite, with the gravitational integral involving only $|{\varphi }_z{|}^2$.

Use radiative homology with proton-proton burning and inverse-fourth-power opacity, retaining the composition factors throughout. To distinguish the stellar radius from the stellar gas constant, write them as $R$ and $\mathcal{R}$. With gas pressure alone, introduce the scale factorsThe ideal gas equation gives $\rho ={\rho }_{\ast }b$, $P=P_{\ast }p$ and $T=T_{\ast }p/b$. Substitution into hydrostatic equilibrium and mass conservation givesso $p^{\prime }=-bq/x^2$ and $q^{\prime }=x^{2}b$.

Set $\zeta =Z(1+X)$. The radiative diffusion in a star equation, with the stipulated opacity, becomesHere ${\rho }^2{T}^{-7}={\rho }_{\ast }^2{T}_{\ast }^{-7}{b}^9{p}^{-7}$. ThusThe luminosity equation follows from the specific heating rate of the Proton–proton chain:ConsequentlyThese give all four dimensionless stellar structure equations with their numerical coefficients.

The stellar homology assumption means the dimensionless profiles are shared across the models, not merely that an individual model can be written in dimensionless coordinates. The displayed equations then require $D$ and $E$ to be fixed. Holding the universal constants and ${\kappa }_0,{ϵ}_0$ fixed givesEquating the two cancels $M^6$ and yields $R^6\propto X^2\zeta {\mu }^{-4}$. Therefore the radius and mass-luminosity relation areThe mass-independent radius is a consequence of the particular heating and opacity exponents in this model; it is not a general radius law for main-sequence stars.

Initially $X=X_0$ is the same for all stars. The adopted mean molecular weight $\mu =4/(5X+3)$ also has the same initial value, neglecting its small direct dependence on $Z$. Thus $R\propto Z^{1/6}$. At fixed luminosity, the Stefan–Boltzmann law $L=4\pi R^2\sigma T_e^4$ givesThe stars compared here can have different masses; the luminosity law adjusts the mass to keep $L$ fixed.

For the evolution, take the stellar mass as constant and maintain homogeneous composition. The available energy in the remaining hydrogen is $M{E}_{H}X$, so conservation of energy in the nuclear-powered model givesThis includes the entire mixed hydrogen reservoir, not just the mass in a central burning region.

To determine the turnoff mass of a homogeneously mixed proton-proton-burning population, write the luminosity law as $L=C{M}^6{Z}^{-7/6}H(X)$, whereSeparating variables, the time to evolve from $X_0$ to $f{X}_0$ isFor a common $X_0$ and $0\le f<1$, this integral is a positive finite constant independent of $M,Z,t$. It follows that the turning-off stars satisfyThe integration retains the evolving luminosity and mean molecular weight; replacing $L$ by its initial value is unnecessary.

Treat the stars as point masses for the orbital motion, neglecting their spin angular momenta. Let $M=M_1+M_2$. Their distances from the centre of mass are $a_1=a{M}_2/M$ and $a_2=a{M}_1/M$. The sum of their orbital angular momenta isThus the circular-binary orbital angular momentum is

For conservative binary mass transfer, both $M$ and $J$ are fixed. In terms of the binary mass ratio $q=M_1/M_2$, $M_1=Mq/(1+q)$ and $M_2=M/(1+q)$, henceMass conservation also gives $d{M}_2=-d{M}_1$, soDifferentiating $J\propto M_1{M}_2{a}^{1/2}$ at fixed $M,J$ gives $d\log a/d\log M_1=2(q-1)$. Therefore the stipulated Roche lobe law has the Roche-lobe radius response exponent

Define the overfill by $\Delta =\log (R_1/R_L)$. On a time scale short compared with the slow change of $K$, the donor's response is $d\log R_1=\beta d\log M_1$. ThusDuring mass loss $d\log M_1<0$. If ${\zeta }_L>\beta$, mass loss increases the overfill and therefore drives more mass loss: this is positive feedback, giving more rapid Roche-lobe overflow. If ${\zeta }_L<\beta$, mass loss decreases the overfill and can balance the slow expansion due to stellar evolution. More explicitly, with $k_{\mathrm{ev}}=d\log K/dt>0$, contact requiresA slow negative transfer rate can satisfy this only on the stable side $\beta >{\zeta }_L$. The rapid-transfer condition is thereforeIts time scale depends on the physical donor response represented by $\beta$; it need not be a dynamical time scale if $\beta$ is a thermal-equilibrium mass-radius exponent.

For Roche-lobe overflow feedback with a constant wind torque factor, retain mass conservation to the stated approximation but allow the prescribed angular-momentum loss. Since $d\log J=fd\log M_1$, differentiation givesThe new Roche-lobe radius response exponent isThe same overfill calculation givesA loss of angular momentum has $f>0$ because ${\dot{M}}_1<0$, and it lowers the critical ratio by making the orbit contract more strongly.

For the specified solar-mass donor exponent $\beta =1/2$, the answer isWithout the wind torque this becomes $q>22/23\simeq 0.957$, so donors slightly less massive than their companions can already lie on the rapid-transfer side of this model. For $0<f<11/10$ the lower threshold is $(22-20f)/23$, and for $f\ge 11/10$ every positive $q$ satisfies the strict inequality. There is no finite upper bound on $q$ from this criterion. Since the paper does not specify $f$, the wind case has a conditional range rather than a unique numerical cutoff.

The Batchelor entrainment hypothesis makes the inward entrainment velocity proportional to the local plume speed:where $\alpha >0$ is the dimensionless entrainment coefficient. For an upward plume $w>0$, this is $u_e=\alpha w$. The hypothesis represents turbulent ingestion of ambient fluid; it is a closure rather than a consequence of mass conservation.

Let $A=b^2$, suppressing the common factor $\pi$ from cross-sectional integrals. In the Boussinesq approximation, use a constant reference density ${\rho }_r$ in the leading-order mass conservation equation. Dividing that equation by ${\rho }_r$ gives volume conservation:The factor two is the ratio of the perimeter $2\pi b$ to the suppressed area factor $\pi$. Density differences are retained in buoyancy, even though they are neglected in the leading inertial and volume terms.

To derive the density-deficit balance, first leave the small density differences explicit in the supplied mass equation. Because the ambient density ${\rho }_0(z)$ is time independent, the product rule givesThus the ambient mass entrained through the perimeter cancels exactly against its contribution to the volume budget, yielding the required buoyancy balance. It is understood to the accuracy of the underlying Boussinesq approximation; the volume equation is not an additional exact constraint on a fully variable-density fluid.

For later use define reduced gravity $g^{\prime }=g({\rho }_0-\rho )/{\rho }_r$ and the squared buoyancy frequency $N^2=-g{\rho }_{0z}/{\rho }_r$. With $B=A{g}^{\prime }$ and $Q=Aw$, the balance isA stably stratified ambient has $N^2>0$, so upward motion consumes the plume's positive density deficit; an unstable ambient reverses this sign.

For lubrication theory, take layer thickness $H$, horizontal length $L$ and speed $U$. Require aspect ratio $\epsilon =H/L\ll 1$, small surface slopes, inertial ratio $U{H}^2/(\nu L)\ll 1$, and slow evolution $H^2/(\nu T)\ll 1$. The longitudinal viscous term then dominates the longitudinal derivative term, while transverse momentum is hydrostatic.

Here $\epsilon \sim \alpha$ and $U\sim g{H}^2\alpha /\nu$, so sufficient conditions are $\alpha \ll 1$ and $g{H}^3{\alpha }^2/{\nu }^2\ll 1$, together with the slow-time condition. With atmospheric pressure at the surface, $p=p_{\mathrm{atm}}+\rho g\cos \alpha (h-y)$. No slip at the wall and zero surface shear giveDepth-integrated mass conservation, $h_t+q_x=0$, therefore gives the stated equation withA steady linear profile with $h_x=A/B=\tan \alpha$ has zero flux: its free surface is horizontal in laboratory coordinates, so the hydrostatic gradient balances downslope gravity. A constant-depth profile also gives steady through-flow.

For the travelling front of a gravity-driven thin film, set $h=F(\xi )$, $\xi =x-ct$. Integrating and using the dry tip gives $-cF=-A{F}^3+B{F}^3{F}^{\prime }$. The state $F\to h_0$, $F^{\prime }\to 0$ behind the front fixes $\boxed{c=A{h}_0^2}$. Consequently $F^{\prime }=(A/B)(1-h_0^2/F^2)$, and integration with $F(0)=0$ yieldsThis tends to $-\infty$ as $F\uparrow h_0$. Expanding at zero gives $\xi \sim -B{F}^3/(3A{h}_0^2)$, so $\boxed{F\sim (-3c\xi /B{)}^{1/3}.}$ The divergent slope at the formal tip violates the small-slope approximation sufficiently close to the contact line; the interior profile is the lubrication result.

Let $D=Q/V_0$ be the dilution rate. Population loss is $DN$, and nutrient loss by consumption is $NK(C)/V_0$, so mass conservation givesUse the dimensionless variables $\tau =Dt$, $c=C/K_0$, and $n=N{K}_{\max }/(Q{K}_0)$. Substitution into the Monod equation gives the stated system withFor a positive equilibrium, the first equation requires $\alpha c/(1+c)=1$. Hence $c_{\ast }=1/(\alpha -1)$ and $n_{\ast }=\alpha (\beta -c_{\ast })$, both positive exactly under the specified inequalities.

Writing $A=\alpha n_{\ast }/(1+c_{\ast }{)}^2>0$, the Jacobian matrix at this equilibrium isBoth eigenvalues are negative, so the state is locally asymptotically stable. The Monod chemostat equilibrium and relaxation also makes one decay rate transparent: $z=c+n/\alpha$ satisfies $z^{\prime }=\beta -z$, independently of the uptake kinetics.

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.

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 the gross star formation rate as $\psi =k{M}_g$, with $k>0$ constant. Use the instantaneous recycling approximation: a fraction $F$ returns promptly, while a fraction $1-F$ stays in long-lived stars and remnants. Assume $0\le F<1$, instantaneous mixing, and no galactic gas inflow or galactic outflow. If $M_s$ denotes the permanently locked stellar mass, mass conservation givesSet $\lambda =(1-F)k$. Integration, with $M_g(0)=M_0$, gives the exponential closed-box gas consumption lawThe locked mass is $M_s=M_0(1-e^{-\lambda t})$, so $M_g+M_s=M_0$ at every time. The decay time is $[(1-F)k{]}^{-1}$, longer than the gross gas-processing time $k^{-1}$ because returned gas can form subsequent generations. The formal $F=1$ limit retains all gas and locks no mass; a finite stellar yield defined per locked mass then needs separate treatment.

The entrainment coefficient $\alpha$ is the ratio of the inward ambient-fluid velocity normal to a buoyant plume edge to the characteristic buoyant plume velocity. For an upward buoyant plume, take $u_e=\alpha w>0$. There are two edges at half-width $b$, so the full-section volume source is $2u_e$. Dividing every integral balance by two gives consistent half-section equations.

For top-hat plume profiles, volume conservation and mass conservation areThe mass entering the buoyant plume is ambient mass with local mass density ${\rho }_0(z)$. The Boussinesq approximation uses a common reference mass density in inertia while retaining the mass density deficit in buoyancy. The mass equation may be kept to the first order in that deficit to derive its transport.

Multiply the volume equation by the time-independent ambient mass density. The product rule givesSubtract the mass equation and multiply by $g$. The entrainment sources cancel exactly, leavingwhereFor stable stratification, ${\rho }_0^{\prime }<0$ and the buoyancy frequency $N$ is real. Rising fluid moves into lighter ambient fluid and loses its mass density deficit relative to that local ambient. Thus environmental stratification reduces the buoyant plume's buoyancy flux, even though the ambient fluid entering by entrainment contributes no mass density deficit at its own height.

For a regular spherical star, mass conservation gives $m^{\prime }=4\pi r^2\rho$. Multiplying the hydrostatic equilibrium equation by $r^2/\rho$ and differentiating therefore yieldsAt the center, finite density implies $m(r)=O(r^3)$, so a regular pressure satisfies $P(0)=P_c<\infty$ and $P^{\prime }(0)=0$. At a free outer boundary in vacuum, $P(R)=0$; a surrounding medium would instead prescribe its external pressure.

Differentiate the auxiliary function, using both equations:where enclosed mass is nonzero. Since $m^2/r^4\to 0$ at the regular center, $F(0)=P_c$, whereas $F(R)=G{M}^2/(8\pi R^4)$. Integrating the strictly negative derivative givesWith nonzero external pressure, the corresponding bound is $P_c>P(R)+G{M}^2/(8\pi R^4)$.

Use stellar homology: dimensionless density, temperature and luminosity profiles are taken to have the same shape, so the dimensional scale relations follow from the structure equations. Mass conservation and hydrostatic support giveThe ideal gas equation then gives $T_c\propto \mu M/R$, with the dimensional constants held fixed. The specific stellar energy-generation rate has the prescribed hydrogen factor and temperature exponent, soIndependently, radiative diffusion in a star gives $T_c/R\propto \kappa {\rho }_{c}L/(R^2{T}_c^3)$, henceEquating the nuclear supply and radiative transport establishes composition-dependent CNO homology:At one fixed composition, these reduce to $R\propto M^{3/4}$ and $L\propto M^3$ as required.

The Stefan–Boltzmann law gives $T_e^4\propto L/R^2\propto M^{3/2}$, so $T_e\propto M^{3/8}$. Thus the logarithmic Hertzsprung-Russell diagram slope along this zero-age main sequence isIncreasing mass moves upward and leftward on the conventional diagram, whose temperature axis is reversed.

For composition changes, neglect the small heavy-element contribution to the mean molecular weight and set $Y\simeq 1-X$. Then ${\mu }^{-1}=(3+5X)/4$. The fixed trace catalyst abundance is absorbed into ${ϵ}_0$; this is not an assertion that an exactly metal-free star can undergo CNO burning. Substitution into the homology laws givesFor a retained, nonnegligible fixed metallicity $Z$, the corresponding mean-molecular-weight factor is $3+5X-Z$ instead. The displayed requested formula uses the trace-metal approximation.

At fixed mass, differentiating with respect to $X$ at the formal reference $X=1$ givesConsequentlyand the initial fully mixed hydrogen-burning evolutionary track has slopeHydrogen depletion has $dX<0$, so both luminosity and effective temperature initially increase. The star therefore moves upward and leftward, on a shallower logarithmic track than the mass sequence. The figure uses dimensionless luminosity and temperature relative to its initial reference star; the track is computed for declining $X$ at constant mass.

Ordinary cluster main-sequence turn-offs instead evolve toward brighter, cooler, more extended configurations and eventually the giant region. Normal stars are therefore not mixed throughout their entire mass during hydrogen burning: the core composition changes while the envelope retains hydrogen, and composition and entropy gradients develop. Core exhaustion and subsequent shell burning then produce very different structural evolution. Complete mixing is a special evolutionary scenario, not the standard explanation of ordinary giant formation.

These hydrogen-burning scalings must not be extrapolated literally to $X=0$. Their vanishing nuclear coefficient would make the inferred radius tend to zero and the effective temperature singular; in reality hydrogen exhaustion requires a different structural and energy-generation model. A homogeneous remnant may be helium-rich, but its later structure is not described by the same CNO law. Also, the cover-sheet radiation-pressure expression has a temperature-power typo: the physical term is $a{T}^4/3$. Radiation pressure was neglected in this derivation, so that typo does not enter the calculation.

Put $D_t={\partial }_t+\mathbf{u}\cdot \nabla$. The continuity equation and the expanded ideal magnetohydrodynamic induction equation areso $D_t(B_i/\rho )=(B_j/\rho ){\partial }_j{u}_i$. Because $\mathbf{B}=\nabla \times \mathbf{A}$, the induction equation permits the magnetic vector potential equation ${\partial }_t\mathbf{A}=\mathbf{u}\times \mathbf{B}-\nabla \phi$, with an arbitrary scalar electric potential $\phi$ carrying the gauge choice. Using $(\mathbf{u}\times \nabla \times \mathbf{A}{)}_i=u_j{\partial }_i{A}_j-u_j{\partial }_j{A}_i$ yieldsContracting with $B_i/\rho$ and differentiating that factor givesThe last two terms cancel by exchanging their dummy indices. Thus the requested material derivative isThis is material transport of magnetic helicity density.