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Write $\mathcal{R}=k/(\mu m_H)$ for the specific gas constant, with the printed $m_H$ denoting the atomic mass unit, and $a_{\mathrm{r}}=4\sigma /c$ for the radiation constant. In the low-mass envelope the enclosed mass is approximately $M$ and the outward luminosity is approximately $L$. The stellar hydrostatic equation, stellar radiative diffusion, and ideal gas equation giveThe Kramers opacity law becomes $\kappa =({\kappa }_0/\mathcal{R})P{T}^{-9/2}$. Dividing the two differential equations eliminates radius and mass density:Integrating from a surface matching point $(P_s,T_s)$ givesDeep compared with the photosphere, the surface terms are negligible. The radiative-zero white-dwarf envelope therefore hasThis is the leading envelope approximation used in the later parts, rather than an exact atmospheric boundary condition at $T_{\mathrm{eff}}$.

At the base, temperature, mass density and pressure match the isothermal degenerate core. Let $K_{\mathrm{NR}}=K_1{\mu }_e^{-5/3}$, where ${\mu }_e$ is the mean molecular weight per electron. Equating the gas and electron degeneracy pressure givesHence $P_b={\mu }_e^{5/2}{\mathcal{R}}^{5/2}{K}_1^{-3/2}{T}_c^{5/2}$. Substituting this into the envelope relation at $T=T_c$ yieldsThe matching of a white-dwarf envelope to a degenerate core thus gives $L/M\propto T_c^{7/2}$, the envelope relation underlying Mestel's cooling law. It depends on pressure as well as mass density continuity at the interface; otherwise the supplied core equation would not determine the matching mass density.

Apply the preceding temperature profile at the envelope base. Introduce the dimensionless parameterThen $R/r_b-1=q$, so the geometrical thickness of a white-dwarf radiative envelope isThe smallness to be established is therefore $\mathcal{R}{T}_c\ll GM/R$, up to the numerical factor $17/4$.

That ordering follows from the cold, strongly electron-degenerate white dwarf model. Throughout the mass-bearing core, $k{T}_c\ll E_F$, where $E_F$ is the Fermi energy. For nonrelativistic electron degeneracy pressure,The stellar hydrostatic equation gives a characteristic core $P_{\mathrm{deg}}/\rho$ of order $GM/R$. Consequentlyfor fixed composition with order-one molecular-weight ratios, and $s\ll R$. Equivalently, the matched base mass density is much lower than a characteristic core mass density, since ${\rho }_b/{\rho }_{\mathrm{core}}\sim [\mathcal{R}{T}_c/(GM/R){]}^{3/2}$. The degeneracy ordering applies in the bulk core; at the matching point degeneracy is becoming marginal.

The inequalities $T_c\gg T_{\mathrm{eff}}$ and $k{T}_c\ll E_F$ express different facts. The first justifies ignoring surface-temperature terms, while the second makes the envelope geometrically thin. If the core were hot enough that $q$ were not small, the exact expression $s/R=q/(1+q)$ would not justify a thin-layer approximation.

For fixed composition, electron-scattering opacity is effectively independent of mass density and temperature, so $\lambda =\nu =0$. With the specified CNO cycle exponent $\eta =16$, the homology relations for radiative stars giveThe radiative homology with CNO burning and electron scattering is thereforeThe Stefan–Boltzmann law now givesEquivalently $T_{\mathrm{eff}}\propto L^{9/76}$. These slopes are illustrated with arbitrary separate reference-model normalizations:

The proton–proton/Kramers model applies to the lower-mass regime; the CNO/electron-scattering model applies to the higher-mass regime. Higher-mass cores are hotter, making the more temperature-sensitive CNO cycle efficient, and highly ionized material favors electron-scattering opacity. These are the stated idealized radiative, gas-pressure-supported sequences: a fully convective low-mass star or a radiation-pressure-dominated massive star needs a different structural model. The intersection of independently normalized lines in the figure does not specify a physical crossover mass.

Let $S$ denote specific entropy, reserving $s$ from the first question for envelope thickness. The stellar adiabatic exponents are defined at fixed composition byIt is useful to introduce ${\Gamma }_3-1=(\partial \log T/\partial \log \rho {)}_S$, so $({\Gamma }_2-1)/{\Gamma }_2=({\Gamma }_3-1)/{\Gamma }_1$.

Take the stellar gas to be a fully ionized, nonrelativistic monatomic ideal gas, with fixed specific gas constant $\mathcal{R}$. Its mixture with thermal radiation hasHere $u$ is specific internal energy, and the stellar gas-pressure fraction is $\beta =\mathcal{R}\rho T/P$. The assumption of monatomic gas fixes its heat capacity; the perfect-gas pressure law alone would not determine it.

At fixed $T$ and fixed $\rho$, respectively, the pressure derivatives areFor an isentropic process, the first law of thermodynamics gives $du=Pd\rho /{\rho }^2$. Differentiate the internal energy rather than artificially holding $\beta$ fixed during the perturbation:The parenthesis is $(P/\rho T)[(3/2)\beta +12(1-\beta )]$, givingSince $d\log P={\chi }_{\rho }d\log \rho +{\chi }_{T}d\log T$,Using the relation between the exponents gives the adiabatic exponents of a monatomic gas-radiation mixtureThe requested values, including the adiabatic temperature gradient, areThe pure-radiation and pure-gas rows are understood as limits of the mixture.

For uniform composition, the Schwarzschild criterion for stellar convective instability isTo test whether a radiative configuration becomes unstable, use its required stellar radiative temperature gradient ${\nabla }_{\mathrm{rad}}$ for $\nabla$. Since $dP/dr<0$, the equivalent radial condition isAn outward-displaced parcel then cools less than its new surroundings, remains less dense at the same pressure, and is further accelerated outward.

The requested Eddington-model convective-core mass fraction follows from the usual global constant-$\beta$ closure. At the outer radiative surface, put $m=M$, $L_r=L$, and $P_{\mathrm{rad}}=(1-\beta )P$. Dividing the radiation-pressure gradient by the stellar hydrostatic equation givesThere is no energy generation outside the convective core, so use $L_r=L$ there. The local radiative-gradient formula isThe same representative $\beta$ has been used in the global luminosity closure and at the boundary. Setting this gradient equal to ${\nabla }_{\mathrm{ad}}$ at $m=M_c$ producesThis is $1$, $77/80$, and $5/8$ for $\beta =0,1/2,1$, respectively, with $\beta =0$ a marginal radiation-dominated limit.

There is a consistency qualification to this last model estimate. It cannot be an exact stellar solution if all the printed assumptions are enforced pointwise. An exactly uniform nonzero radiation fraction would give $d{P}_{\mathrm{rad}}/dP=1-\beta$ at every radiative point. The exact equations would then requireIf $L_r$ is constant throughout an envelope with positive mass density, $m$ increases with radius, so this equality cannot hold there. Equivalently uniform $\beta$ fixes the actual gradient to $1/4$, whereas ${\nabla }_{\mathrm{ad}}>1/4$ for $\beta >0$. The boxed core fraction is the intended Eddington closure and boundary estimate, which relaxes exact constancy of $\beta$ in the detailed envelope. The constant-beta radiative-envelope obstruction identifies the missing approximation; it is not legitimate to silently assert exact compatibility.

A spherically symmetric stellar model has no preferred direction in the tangent plane at a fixed radius. Rotations about the radial axis exchange its two transverse velocity directions, so $⟨v_{\theta }^2⟩=⟨v_{\varphi }^2⟩$. This uses spherical symmetry of the velocity distribution, not just a spherical mass density.

The augmented density $\rho (r,\psi )$ treats radius and relative potential as independent variables, agreeing with the physical density only along $\psi =\psi (r)$. For any regular bound spherical galactic distribution function $F(E,L)$, write $P_r=\int v_r^{2}F{d}^{3}v$ and $P_t=\int v_{\theta }^{2}F{d}^{3}v$. At fixed $r$, $F_{\psi }=F_E$ and ${\partial }_{v_r}F=-v_r{F}_E$. Integration by parts therefore givesFor $v_t^2=v_{\theta }^2+v_{\varphi }^2$, the velocity measure is $2\pi v_{t}d{v}_{t}d{v}_r$, and $r{F}_L={\partial }_{v_t}F+v_t{F}_E$. Integrating the transverse derivative gives $-2P_r$, while integrating the radial derivative in $\int v_r^2{v}_t^2{F}_E{d}^{3}v$ gives $2P_t$. Consequentlywhere $\psi$ is fixed. These identities show that the augmented representation holds for any such spherical distribution, not only a monomial ansatz. With radial pressure vanishing at the zero-binding boundary, set Along the physical curve, the chain rule gives $dA/dr={\int }_0^{\psi }{\partial }_r\rho (r,q)dq+\rho (r,\psi ){\psi }^{\prime }(r)$. The Spherical Jeans equation then requiresThus the Jeans moments from an augmented density areThe partial derivative holds $q$ fixed. Substitution cancels the integral of ${\partial }_r\rho$ and leaves precisely $\rho {\psi }^{\prime }$ in the Spherical Jeans equation. This is a representation by a chosen augmented density, not a unique solution determined by the one-variable density: different extensions off the physical curve encode different anisotropies, and not every formal extension necessarily gives a nonnegative galactic distribution function.

For the hypervirial density-potential family, differentiate the potential and use the spherical shell theorem:Differentiating $M(<r)$ and dividing by $4\pi r^2$ givesThe enclosed mass tends to $M$ at infinity and to zero at the centre for every $p>0$. The same result follows from ${\nabla }^2\psi =-4\pi G\rho$, with the stated sign convention.

From now on set $G=M=a=1$. The separable augmented density is $\rho (r,q)=(p+1)r^{p-2}{q}^{2p+1}/(4\pi )$. Integrating it givesand henceWith $\psi =(1+r^p{)}^{-1/p}$ these are the requested radial expressions. The velocity-anisotropy parameter is $\beta =1-p/2$, and the total mean square speed is $⟨v^2⟩=\psi /2$.

The local kinetic-energy density is $t=\rho ⟨v^2⟩/2=\rho \psi /4$. With the potential zero at infinity, the gravitational energy density is $w=\rho \Phi /2=-\rho \psi /2$, where $\Phi =-\psi$. ThereforeThis is the local virial relation of the hypervirial model, a special property stronger than the global virial theorem. To compute the global energies as well, set $I=\int \rho \psi d^{3}x$. The substitution $u=r^p$ yields the global binding integral of the hypervirial modelThe beta function integral converges for all $p>0$. ThusRestoring dimensions multiplies $I,T,W$ by $G{M}^2/a$.

To derive the hypervirial distribution function, let $E=\psi -v^2/2$ be the positive relative energy and $L=rv\sin \vartheta$ the magnitude of the specific angular momentum. Seek $F=C{L}^{p-2}{E}^s$ for $E>0$ and zero otherwise. Its density isThe angular integral is $\sqrt{\pi }\Gamma (p/2)/\Gamma ((p+1)/2)$. Substitution $z=v^2/(2\psi )$ makes the radial integralMatching the augmented-density power ${\psi }^{2p+1}$ gives $s=(3p+1)/2$. Matching its coefficient gives the normalization of the hypervirial distribution functionAll integrals are finite in velocity for $p>0$, and $C>0$. Because $E$ and $L$ are orbital integrals, this is a steady solution by the Jeans theorem. As checks, $p=1$ gives $C=3/(4{\pi }^3)$ for the Hernquist model, and $p=2$ gives $C=24\sqrt{2}/(7{\pi }^3)$ for the isotropic Plummer model.

The density and potential do not uniquely fix the stellar distribution. The preceding coefficient is fixed within the chosen separable power-law model, but nonuniqueness of spherical galactic distribution functions remains without that extra restriction. An explicit positive counterexample uses the same $p=2$ density. Define the Osipkov-Merritt distribution function variable $Q=E-L^2/2$ and takeRescale the transverse velocity by $\sqrt{1+r^2}$. Then $Q=\psi -[v_r^2+(1+r^2)v_t^2]/2$ and the velocity Jacobian contributes $(1+r^2{)}^{-1}$. Direct integration givesexactly the same Plummer model density. This anisotropic Plummer distribution with unit anisotropy radius has $\beta =r^2/(1+r^2)$, so it is distinct from the isotropic $E^{7/2}$ distribution while remaining positive and spherical. Its second moments differ, so it does not share the extra augmented-density and local-virial restrictions of the chosen hypervirial model.

A small-angle gravitational encounter between a test star and a field star of mass $m$ gives a transverse kick $\Delta v_{\perp }\simeq 2Gm/(bu)$, where $b$ is the impact parameter and $u$ the relative speed. For number density $n$, encounters in $b,b+db$ occur at rate $2\pi bdbnu$. Independent kicks add in mean square, giving the random-walk derivation of stellar relaxationThe Coulomb logarithm in stellar dynamics has $b_{\max }$ of order system size $r_s$, and $b_{\min }$ of order the strong-deflection scale $Gm/u^2$. For a virialized $N$-star system, $u^2\sim GNm/r_s$, so $\Lambda$ is of order $N$. The stellar relaxation time is the time for accumulated velocity variance to become comparable with $u^2$:Using $n\sim 3N/(4\pi r_s^3)$ gives $t_{\mathrm{rel}}\sim [N/(6\ln N)]r_s/u$. With a diameter-crossing convention $T_{\mathrm{cross}}\sim 2r_s/u$, this is $[N/(12\ln N)]T_{\mathrm{cross}}$, conventionally rounded toThe coefficient is approximate: spatial profile, velocity averages and crossing-time convention change it by factors of order unity. The robust result is $N/\ln N$ crossing times.

For $N\sim 10^{10}$ and $T_{\mathrm{cross}}\sim 10^8$ years, the estimate is about $4\times 10^{15}$ years, enormously longer than a Hubble time. Most galaxies therefore behave as collisionless stellar systems. Dense nuclei and star clusters can relax more rapidly. Negligible stellar encounters do not mean negligible collective gravitational instabilities.

Define the mass-weighted galactic distribution function by $dM=F(\mathbf{x},\mathbf{v},t)d^{3}x{d}^{3}v$, so $\rho =\int F{d}^{3}v$. Hamiltonian gravitational motion preserves phase space volume by the Liouville theorem in Hamiltonian mechanics. In the collisionless regime no encounter term redistributes stars between neighbouring phase space trajectories, so $dF/dt=0$. With acceleration $\nabla \psi$, the Collisionless Boltzmann equation is

For a steadily rotating galactic bar, write the inertial polar angle as ${\varphi }_{\mathrm{I}}=\varphi +{\Omega }_{b}t$. The inertial radial and angular equations are $\ddot{R}-R{\dot{\varphi }}_{\mathrm{I}}^2={\psi }_R$ and $R{\ddot{\varphi }}_{\mathrm{I}}+2\dot{R}{\dot{\varphi }}_{\mathrm{I}}={\psi }_{\varphi }/R$. Substitution gives the equations of motion in a rotating frame:Here the velocities are measured in the bar frame. Set $v_R=\dot{R}$, $v_{\varphi }=R\dot{\varphi }$ and define the positive-force rotating-frame relative effective potentialThe centrifugal sign is positive in this relative-potential convention. The characteristics areConsequently the rotating-frame collisionless Boltzmann equation isFor these planar equations, integrate over $d{v}_{R}d{v}_{\varphi }$ and interpret $\rho$ as the planar or vertically integrated mass density. Define $\rho ⟨A⟩=\int AFd{v}_{R}d{v}_{\varphi }$. Boundary terms in velocity vanish for a sufficiently decaying distribution.

Let the velocity accelerations above be $A_R,A_{\varphi }$. Their velocity divergence is ${\partial }_{v_R}{A}_R+{\partial }_{v_{\varphi }}{A}_{\varphi }=-v_R/R$. Thus integration by parts in the zeroth moment contributes $\rho ⟨v_R⟩/R$, producing the cylindrical continuity equation for a rotating stellar systemThis is conservation of mass: change of density balances flux through the radial and azimuthal sides of a small cylindrical element. The factor $R$ is geometric; the physical phase space measure is $RdRd\varphi d{v}_{R}d{v}_{\varphi }$.

For the radial first moment, multiply the rotating-frame collisionless Boltzmann equation by $v_R$. Its velocity integrations are $\int v_R{A}_R{F}_{v_R}{d}^{2}v=-\rho ⟨A_R⟩$ and $\int v_R{A}_{\varphi }{F}_{v_{\varphi }}{d}^{2}v=\rho ⟨v_R^2⟩/R$. Hence the radial cylindrical Jeans equations in a rotating frame areFor the azimuthal first moment, the $A_R$ velocity integral vanishes, while ${\partial }_{v_{\varphi }}(v_{\varphi }{A}_{\varphi })={\psi }_{\mathrm{eff},\varphi }/R-2{\Omega }_b{v}_R-2v_R{v}_{\varphi }/R$. This givesThe Jeans equations are local momentum-balance equations. The second moments include streaming momentum flux and random-velocity stress, while gravity, centrifugal acceleration and Coriolis acceleration supply the frame-dependent forces. They do not by themselves close the full distribution: a stress prescription or a galactic distribution function is also needed.

For $\mathbf{u}=\alpha y{\mathbf{e}}_x-\beta x{\mathbf{e}}_y$, the velocity divergence is zero and $\mathbf{u}\cdot \nabla \mathbf{u}=-\alpha \beta (x{\mathbf{e}}_x+y{\mathbf{e}}_y)$. Substituting into the steady Euler equations yieldsIntegration givesThe steady enthalpy equation also requires $\mathbf{u}\cdot \nabla Q=0$. Its $xy$ coefficient is $\alpha (3{\Omega }^2+\alpha \beta -{\beta }^2)$, so, since $\alpha >0$,For a bounded patch with positive mass density and a vacuum fluid free boundary, choose $Q_0>0$ and require both quadratic coefficients to confine the fluid. With $\Omega >0$ and $\alpha ,\beta >0$, this meansThe lower bound is imposed by $\alpha >0$; the upper bound by bounded positive $Q$. At $\beta =2\Omega$, $Q$ is constant and has no nontrivial finite vacuum boundary; $\beta \le \sqrt{3}\Omega$ violates or degenerates the stipulated elliptical flow.

For a polytropic elliptical patch in a shearing sheet, the physical fluid free boundary is $p=\rho =Q=0$. It is the ellipsewhere $a_x^2=2Q_0/[\beta (2\Omega -\beta )]$ and $a_y^2=2Q_0/[\alpha (2\Omega -\beta )]$. In the interior $\rho =[Q/((n+1)K){]}^n$. The boundary is a material streamline because $\mathbf{u}\cdot \nabla (\beta x^2+\alpha y^2)=0$. Since $\alpha <\beta$, it is elongated in the azimuthal $y$ direction.

This is a pressure-supported fluid patch in the supplied tidal model; no additional planetary self-gravity potential was assumed.

For the polytropic equation of state $p=K{\rho }^{1+1/n}$, take $K>0$ and $n>0$ for the physical model. The specific enthalpy relative to vacuum isIts gradient obeys $\nabla Q={\rho }^{-1}\nabla p$, so the Euler equations becomeFor the material derivative $D_t={\partial }_t+\mathbf{u}\cdot \nabla$, $D_{t}Q=(Q/n\rho )D_t\rho$. The continuity equation thus becomesThe choice $Q=0$ at zero mass density fixes the additive convention in this equation. Although the momentum equation alone is unchanged by adding a constant to $Q$, the polytropic relation and enthalpy evolution must use the same vacuum reference.

For constant component masses, the center of mass frame has ${\mathbf{r}}_1=(M_2/M)\mathbf{r}$ and ${\mathbf{r}}_2=-(M_1/M)\mathbf{r}$. Newton's law of universal gravitation gives the relative equation $\ddot{\mathbf{r}}=-GM\mathbf{r}/r^3$. Substitution in the component angular momenta yieldsHere $\mu =M_1{M}_2/M$ is the reduced mass, while $\mathbf{h}$ is specific angular momentum.

The printed total-energy expression lacks a factor of $\mu$ in its kinetic term. With the stated physical separation coordinate, the dimensionally consistent total two-body orbital energy isThe second expression is specific orbital energy; these two conventions must not be mixed. Its conservation follows directly:For the eccentricity vector, differentiate $GM\mathbf{e}=\dot{\mathbf{r}}\times \mathbf{h}-GM\hat{\mathbf{r}}$. Since $\mathbf{h}$ is constant,The cancellation is the vector triple product identity. Thus all three corrected Kepler integrals are conserved. Dotting the eccentricity relation with $\mathbf{r}$ also gives the useful orbit identity

For ballistic streamline focusing by a moving star, work in the star's rest frame and take the incoming gas to have velocity $v{\mathbf{e}}_x$, with downstream behind the star at $x>0$. A streamline with impact parameter $d>0$ approaches from $(-\infty ,d,0)$. Its upstream specific angular momentum and eccentricity vector areAt its downstream axis crossing, $\mathbf{r}=b{\mathbf{e}}_x$. The orbit identity gives $b=d^2{v}^2/(GM)-b$, hence the collision distance isAt that point, $h_z=b{u}_y=-dv$, while the $y$ component of $GM\mathbf{e}=\mathbf{u}\times \mathbf{h}-GM\hat{\mathbf{r}}$ gives $dv{u}_x=d{v}^2$. ThereforeA symmetric streamline with opposite impact parameter has the opposite transverse velocity. Their shock wave removes the opposing transverse motion while preserving the common downstream component. Thus the remaining velocity is $v{\mathbf{e}}_x$ in the star frame. In the frame where the undisturbed medium is stationary, adding the star's velocity gives zero velocity immediately after this idealized collision.

After transverse kinetic energy is dissipated, the remaining specific orbital energy is ${\epsilon }_{\mathrm{after}}=v^2/2-GM/b$. It is negative if $b<2GM/v^2$, or equivalently if the upstream impact parameter satisfies $d<d_{\mathrm{cap}}=2GM/v^2$. This is the post-shock capture criterion in ballistic accretion; bound axial streams can return to the star. The captured incident mass flux through the corresponding disk gives the ballistic Bondi--Hoyle--Lyttleton accretion rate:This is the pressureless, dissipative capture model, with the shock's transverse energy unavailable to unbind the wake.

For wind mass transfer in a binary star, steady isotropic donor mass loss gives the local wind mass density at the companion:A fast wind has $v_w\gg 2\pi a/P$ and a capture scale small compared with $a$, so its relative incident speed is $v_w$ to leading order. Apply the same Bondi--Hoyle--Lyttleton accretion rate with accretor mass $M_2$ to obtainFinally, Kepler's third law eliminates the unquoted separation, $a=[G(M_1+M_2)P^2/(4{\pi }^2){]}^{1/3}$, giving the rate in terms of the stated orbital period:Equivalently, the fast-wind accretion fraction in a circular binary is $[M_2/(M_1+M_2){]}^2(v_{\mathrm{orb}}/v_w{)}^4$, with $v_{\mathrm{orb}}=2\pi a/P$ the relative circular orbital speed. Its smallness is consistent with using an almost undisturbed isotropic stellar wind.

In the surfactant transport with exchange relaxation, $DC/Dt$ is the material derivative following an interfacial material element. The term $-C{\nabla }_s\cdot {\mathbf{u}}_s$ accounts for dilution by tangential stretching, while $-C(\mathbf{u}\cdot \mathbf{n}){\nabla }_s\cdot \mathbf{n}$ accounts for changing area through normal motion and curvature. The surface diffusion term redistributes surfactant down concentration gradients. The exchange term relaxes its concentration towards $C_0$, through adsorption or desorption from a surrounding reservoir, on time scale $k^{-1}$.

In the bubble frame the spherical interface is stationary and impermeable. A steady material derivative is ${\mathbf{u}}_s\cdot {\nabla }_s{C}^{\prime }$. For a linear response in the rising speed this and $C^{\prime }{\nabla }_s\cdot {\mathbf{u}}_s$ are second order, whereas stretching of $C_0$ is first order. ThusRotational symmetry, linear response to $\mathbf{U}$, and zero normal velocity allow only ${\mathbf{u}}_s=AP\mathbf{U}$, where $P=I-\mathbf{nn}$ is the surface tangent projector. This is the dipolar, axisymmetric sector; axisymmetry alone, without the linear-response assumption, would allow other angular dependence.

On a sphere $\mathbf{n}=\mathbf{x}/r$ and ${\partial }_j{n}_i=({\delta }_{ij}-n_i{n}_j)/r$. Applying the surface tangent projector to the derivative gives the sphere surface derivative identitiesFor example, taking a second surface divergence of the first identity yields the third. Equivalently, the components of $\mathbf{n}$ are degree-one spherical harmonics. Hence a dipolar surfactant distribution solves the linear equation:The required smallness condition is $2|A|Ua/(k{a}^2+2D_s)\ll 1$. Surface diffusion or exchange must smooth the concentration faster than it is redistributed by the actual interfacial velocity. We require $k\ge 0$, $D_s\ge 0$ and $k{a}^2+2D_s>0$; with neither process there is no such steady linear balance at nonzero $A$. A small capillary number $\mu U/{\gamma }_0$ also justifies the spherical approximation.