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The P-star approximation describes the sampling density of a regular maximum-likelihood estimator in sample coordinates consisting of the fitted parameter $\hat{\theta }$ and a suitable ancillary statistic $a$. For a $d$-dimensional parameter, write $\ell (\theta ;\hat{\theta },a)$ for the log-likelihood expressed in these coordinates, and $j(\hat{\theta };\hat{\theta },a)$ for the fitted observed information. The approximation isThe determinant factor supplies the local volume scale, while the exponential retains the full likelihood ratio, rather than replacing it by a quadratic. The normalizing constant is chosen by integrating over the fitted-parameter coordinates; its leading regular large-sample value is $(2\pi {)}^{-d/2}$. These ingredients arise from a statistical saddlepoint approximation. Their use requires an interior, locally unique fit, nonsingular information and suitable smoothness; an ancillary statistic specifies the conditioning surface when the fit alone does not exhaust the relevant sample coordinates. In a regular model the unnormalized leading expression has relative error of order $n^{-1}$ locally; normalization and conditioning matter for refined inference. It is not a universal exact density. Under a smooth one-to-one parameter transformation, the Hessian at the fit transforms without a score term, so its determinant supplies precisely the density Jacobian determinant. This proves the invariance of the construction under one-to-one changes of statistical parameters.

Here put $S={\sum }_i{X}_i$ and $T={\sum }_i{Y}_i$. The log-likelihood, apart from a data-only constant, isThe two score equations giveThe sample sums form a sufficient statistic, and there is no need to retain an additional ancillary in calculating their joint density. Inverting the fitted coordinates gives $S=n/\hat{\lambda }$ and $T=n/(\hat{\psi }\hat{\lambda })$. With parameter order $(\psi ,\lambda )$, the fitted observed information isAlso,Thus the required joint P-star approximation isTo integrate out $\hat{\lambda }$, substitute $u=b/\hat{\lambda }$ inTaking $b=n\lambda (1+\psi /\hat{\psi })$ and rearranging the powers yieldsThe beta function identity ${\int }_0^{\infty }{z}^{n-1}(1+z{)}^{-2n}dz=\Gamma (n{)}^2/\Gamma (2n)$ therefore givesBoth normalized approximations are actually exact. To check the joint assertion independently, $S$ and $T$ have independent gamma distributions with shape $n$ and rates $\lambda$ and $\psi \lambda$. Their joint density is ${\psi }^n{\lambda }^{2n}{S}^{n-1}{T}^{n-1}{e}^{-\lambda S-\psi \lambda T}/\Gamma (n{)}^2$. The inverse-coordinate Jacobian determinant iswhich reproduces the normalized joint expression. This is the P-star density for two exponential rates; its marginal gives the stated F-distribution. Stirling's approximation gives $c_n=(2\pi {)}^{-1}(1-1/(6n)+O(n^{-2}))$, so using only the leading normalizer retains the exact shape but not its exact total mass.

The original PDF's displayed density constant is missing a square on $\Gamma (n)$. Its printed expression integrates to $\Gamma (n)$, rather than one in general. The normalized density above is the one consistent with the stated F-distribution.

For an interest parameter $\psi$ and nuisance parameter $\lambda$, the profile likelihood is $L_p(\psi )=L(\psi ,{\tilde{\lambda }}_{\psi })$, where ${\tilde{\lambda }}_{\psi }$ maximizes the likelihood at fixed $\psi$. Its maximum occurs at the joint maximum-likelihood estimate of $\psi$. It preserves the optimized likelihood ratios for different interest values and is unchanged by one-to-one reparametrization of the nuisance at each fixed $\psi$. It generally is neither a marginal sampling density nor a conditional likelihood, and eliminating nuisance parameters by maximization can introduce appreciable small-sample distortion. Differentiating the constrained score equation gives the curvature formulaat a joint stationary point. The corresponding expected-information expression is the efficient information for the interest parameter. Under the regular conditions of Wilks theorem, twice the profile log-likelihood drop has an asymptotic chi-squared distribution with the dimension of the interest parameter as its degrees of freedom.

For scalar nuisance, the Barndorff-Nielsen modified profile likelihood may be writtenwhere the sample-coordinate derivative holds $\hat{\psi }$ and the ancillary statistic fixed. For vector nuisance the last two quantities are determinants. An equivalent form is ${\ell }_m={\ell }_p+\frac{1}{2}\log |j_{\lambda \lambda }|-\log |{\ell }_{\lambda ;\hat{\lambda }}|$, since differentiating the constrained nuisance score gives ${\ell }_{\lambda ;\hat{\lambda }}=j_{\lambda \lambda }\partial {\tilde{\lambda }}_{\psi }/\partial \hat{\lambda }$. The semicolon denotes differentiation in sample coordinates, not another ordinary parameter derivative. This Jacobian determinant adjustment is essential to the definition.

In the present model setThe log-likelihood is $\ell =-n\log \lambda -\psi {\sum }_i{x}_i-A(\psi )/\lambda$. ConsequentlyThe profile score is $nB/A-{\sum }_i{x}_i$, and its derivative is $-n(C/A-(B/A{)}^2)$. The expression in parentheses is the variance of the fixed design values under positive weights proportional to $Y_i{e}^{-\psi x_i}$. Thus the profile log-likelihood is strictly concave for a nonconstant design. Its score changes from $n{\max }_i{x}_i-{\sum }_i{x}_i>0$ to $n{\min }_i{x}_i-{\sum }_i{x}_i<0$ as $\psi$ goes from $-\infty$ to $+\infty$, giving a unique finite fit. If all $x_i$ coincide, only the mean $\lambda e^{\psi x_i}$ is identifiable and inference on $\psi$ alone is not defined; that exceptional design must be excluded.

The specified ancillary coordinates give $Y_i=\hat{\lambda }\exp (\hat{\psi }{x}_i+a_i)$. At fixed $\hat{\psi },a$, therefore,Meanwhile the constrained observed information isSubstitution provesEquivalently the sample derivative of the nuisance score is ${\ell }_{\lambda ;\hat{\lambda }}=A/({\lambda }^2\hat{\lambda })$, which equals $n/({\tilde{\lambda }}_{\psi }\hat{\lambda })$ at the constrained fit and gives the same cancellation. Thus the modified profile likelihood for exponential regression is proportional to the ordinary profile likelihood. This conclusion uses the given ancillary; no proof of its ancillarity is required.

A transformation model consists of a group $G$ acting on the sample space and on the set of statistical parameters, with the compatibility propertyIn words, transforming a sample from $P_{\theta }$ produces a sample from $P_{g\theta }$. A transitive transformation model is generated from one reference distribution by the group action. For densities, the compatibility property includes the Jacobian determinant of the sample transformation; it is not merely equality of density values at transformed points.

A maximal invariant $A$ is constant on orbits of a group action and separates them: $A(gx)=A(x)$, and $A(x)=A(y)$ implies $y=gx$ for some $g$. An equivariant estimator $T$ of the parameter satisfies $T(gx)=gT(x)$. Equivariance means the estimate follows the transformed parameter, whereas invariance means the reported value stays unchanged.

First suppose the set of statistical parameters can be identified with the group, as happens for a simply transitive action. An equivariant estimator then provides a group-valued fit $\hat{g}(x)$ with $\hat{g}(gx)=g\hat{g}(x)$. Normalize the sample by this fit:Indeed $A(gx)=\hat{g}(x{)}^{-1}{g}^{-1}gx=A(x)$. Conversely, if $A(x)=A(y)$, then $y=\hat{g}(y)\hat{g}(x{)}^{-1}x$, placing $x,y$ in the same orbit of a group action. This establishes maximality, rather than only invariance.

For a transitive action with a nontrivial stabilizer $H$ of a reference parameter ${\theta }_0$, a parameter-valued equivariant estimator does not specify a unique normalizing transformation. Choose a section $h_{\theta }$ with $h_{\theta }{\theta }_0=\theta$ and form $b(x)=h_{T(x)}^{-1}x$. Under $g$,Thus the $H$-orbit of $b(x)$ is invariant. It is maximal: if $b(y)=hb(x)$ with $h\in H$, then $y=h_{T(y)}h{h}_{T(x)}^{-1}x$. Appropriate measurable sections and orbit of a group action coordinates are assumed when treating this as a statistic. This qualification explains why an arbitrary equivariant statistic alone is not automatically a complete set of invariant coordinates.

For the location-scale family, write $X_i=\mu +\sigma Z_i$ with $\sigma >0$ and a fixed joint distribution of $Z$. The group consists of maps $x_i\mapsto a+b{x}_i$ with $b>0$, acting on parameters by $(\mu ,\sigma )\mapsto (a+b\mu ,b\sigma )$. For a nonconstant sample of size $n\ge 2$, take the equivariant estimatorIt gives the maximal invariantIts coordinates satisfy ${\sum }_i{A}_i=0$ and ${\sum }_i{A}_i^2=n$. They stay unchanged under location and positive-scale transformations. If two samples have the same standardized vector, thenso a single group transformation maps the first sample onto the second. The vector retains the observation labels: sorting it would discard information not removed by this location-scale group. The construction is algebraic and does not require the underlying distribution to have finite moments. Constant samples can be assigned a separate invariant value; they form a single exceptional orbit of a group action and have probability zero under an independent continuous base distribution.

Finally, in a transitive transformation model every invariant statistic is ancillary. Choose $g$ taking ${\theta }_0$ to $\theta$ and use $X_{\theta }\stackrel{d}{=}g{X}_{{\theta }_0}$; invariance gives $A(X_{\theta })\stackrel{d}{=}A(X_{{\theta }_0})$. Thus the standardized residual configuration provides a natural conditioning variable, while the equivariant fit contains the location and scale information.

A statistical functional is a map $T$ defined on a class of probability distributions. Its associated functional statistic is $T(F_n)$, where $F_n=n^{-1}{\sum }_i{\delta }_{X_i}$ is the empirical distribution and each ${\delta }_{X_i}$ is a Dirac measure. For example, $T(F)=\int xdF(x)$ yields the sample mean. The influence function records the derivative in the direction of point contamination:It is a local, infinitesimal robustness diagnostic, not a description of arbitrary large contamination.

Suppose $T$ is differentiable in a sense strong enough to linearize it at the empirical distribution, and that its derivative in a signed direction $G-F$ is $\int \mathrm{IF}(x;T,F)d(G-F)(x)$. Averaging the directions ${\delta }_z-F$ over $z\sim F$ gives the zero direction, so linearity implies $\int \mathrm{IF}dF=0$. The resulting asymptotic linear representation isIf $\int {\mathrm{IF}}^{2}dF<\infty$, the central limit theorem and the negligible remainder giveThus the asymptotic variance is $n^{-1}\int {\mathrm{IF}}^{2}dF$. For a vector functional the integral is $n^{-1}\int \mathrm{IF}{\mathrm{IF}}^{\mathsf{T}}dF$. Convergence of the actual finite-sample variance to this leading expression additionally requires control of second moments of the remainder, or an appropriate uniform integrability condition; convergence in distribution alone does not provide that. Existence of a contamination derivative by itself also does not guarantee the required empirical linearization.

Several complementary robustness measures follow. Gross-error sensitivity is ${\gamma }^{\ast }={\sup }_z|\mathrm{IF}(z)|$. Under contamination $(1-\epsilon )F+\epsilon H$, the first-order bias is $\epsilon \int \mathrm{IF}dH$, whose worst absolute coefficient is ${\gamma }^{\ast }$; bounded influence therefore bounds infinitesimal gross-error bias. The local-shift sensitivity is ${\lambda }^{\ast }={\sup }_{x\ne y}|\mathrm{IF}(x)-\mathrm{IF}(y)|/|x-y|$, equal to $\sup |{\mathrm{IF}}^{\prime }|$ when the influence function is smoothly differentiable. It measures the effect of moving a contaminant a small distance. The rejection point of an influence function is ${\rho }^{\ast }=\inf \{r:\mathrm{IF}(x)=0\text{ for }|x|>r\}$, using centered, standardized coordinates; it is infinite if there is no such cutoff. A finite cutoff completely rejects sufficiently extreme observations to first order. These measures address different aspects of robustness. The integral of squared influence controls asymptotic precision and efficiency, while breakdown point concerns finite amounts of contamination and is not determined by bounded influence alone.

For the quantile, assume $0<p<1$, a unique solution $q=q_p(F)$, and a density continuous and positive at $q$. Define quantiles of the contaminated distribution by generalized inversion, since a distribution with a point mass need not attain $p$ as an equality. If $z\ne q$, its indicator is constant near $q$ for sufficiently small contamination, and differentiatingat zero gives $-p+f(q)q_0^{\prime }+1_{z\le q}=0$. HenceAt the exceptional point $z=q$, the generalized-inverse quantile remains exactly $q$: below $q$ the contaminated cumulative distribution function is below $p$, and its value at $q$ exceeds $p$. Its actual derivative there is zero. The usual step-function expression is an almost-everywhere version, which is all that the asymptotic variance needs. Its mean is zero, and its second moment isTherefore the influence function of a quantile gives asymptotic variance $p(1-p)/(nf(q{)}^2)$ and gross-error sensitivity $\max (p,1-p)/f(q)$. Its jump makes local-shift sensitivity infinite, and its nonzero constant tails make its rejection point infinite. If $f(q)=0$ or the quantile is not unique, this regular derivation does not apply.

An ancillary statistic has a sampling distribution independent of the unknown parameter. It describes a realized aspect of the sample configuration without supplying a marginal likelihood for the parameter. Its observed value can nevertheless be crucial to the precision of inference from the rest of the data. In a transitive transformation model, invariance gives a common distribution under every parameter value, so maximal invariants provide natural ancillaries, such as the standardized configuration in a location-scale family.

Conditional inference uses the sampling distribution of an estimator or sufficient statistic conditional on the observed ancillary. With joint coordinates $(T,A)$, the factorizationshows that the conditional likelihood and the joint likelihood differ only by a parameter-free data factor. Conditional reference distributions, confidence intervals and p-values can still differ markedly from unconditional ones, because they adapt to the realized configuration. For a continuous ancillary one uses a regular conditional distribution or conditional density; dividing by the zero probability of a single ancillary value is not a valid definition.

An exact example makes this adaptation visible. Let independent observations be uniform on $(\theta -1/2,\theta +1/2)$, with $n\ge 2$. Put $U=X_{(1)}$, $V=X_{(n)}$, $R=V-U$ and $C=(U+V)/2$. Specifying the minimum and maximum contributes $n(n-1)$ choices of observations, while each remaining observation must lie between them. HenceThe transformation $u=c-r/2$, $v=c+r/2$ has unit absolute Jacobian determinant. Its support is $0<r<1$, $|c-\theta |<(1-r)/2$. Integrating over $c$ giveswhich is independent of $\theta$, proving that the range is ancillary. Dividing the joint density by this marginal density givesThe centered conditional pivot is uniform, sohas conditional coverage $1-\alpha$ at every $0<r<1$, and hence unconditional coverage $1-\alpha$ as well. A large realized range leaves little room for translating the interval and gives more precise inference. This is range-conditioned uniform location inference. The range and center are not independent: the conditional spread depends on the range, even though the range is ancillary.

Completeness can give independence in other models. Suppose $S$ is a complete sufficient statistic and $A$ is ancillary. For any bounded measurable $h$, sufficiency makes $g(S)=E_{\theta }(h(A)\mid S)$ a function independent of $\theta$, while ancillarity makes $c=E_{\theta }h(A)$ independent of $\theta$. Since $E_{\theta }(g(S)-c)=0$ for every $\theta$, completeness implies $g(S)=c$ almost surely. Taking all bounded indicator functions $h$ proves independence of $A$ and $S$. This is the content of Basu's theorem, with the proof displaying precisely where sufficiency and completeness are used. Without completeness the conclusion fails: in the uniform location example the minimum and maximum are sufficient, and contain the nondegenerate ancillary range.

There are practical qualifications. Ancillaries need not be unique, and some models supply only approximate ancillary coordinates; one must specify the conditioning surface and the accuracy intended. Conditioning on a statistic sufficient for a nuisance parameter is a different device: it may remove that nuisance even though the statistic is not ancillary for the full parameter. Conditioning indiscriminately can discard useful information. In higher-order likelihood inference, the ancillary coordinates specify how the fitted data are varied, entering both the p* density and the sample derivatives in the modified profile likelihood. Their role is therefore substantive, rather than merely a label for an extra statistic.

Saddlepoint approximation methods retain the nonquadratic shape of a distribution through its cumulant-generating function. Suppose $X_1,\dots ,X_n$ are independent and identically distributed, $K(t)=\log E(e^{t{X}_1})$ is finite on an open interval containing zero, and $K^{\prime \prime }>0$. For an interior target mean $x$, solve the saddlepoint equation $K^{\prime }(t)=x$. Introduce exponential tilting byUnder the tilted distribution the mean and variance are $K^{\prime }(t)=x$ and $K^{\prime \prime }(t)$. For the product sample the Radon-Nikodym derivative is $e^{t{\sum }_i{X}_i-nK(t)}$. Consequently the exact density relation for the sample mean isA local central limit approximation to the tilted density at its own mean gives $f_{\overline{X},t}(x)\simeq \sqrt{n/(2\pi K^{\prime \prime }(t))}$. HenceFor the sum $S_n$, the equivalent expression is $f_{S_n}(s)\simeq e^{nK(t)-ts}/\sqrt{2\pi n{K}^{\prime \prime }(t)}$, with $K^{\prime }(t)=s/n$. Under the usual smooth nonlattice density conditions this approximation has relative error of order $n^{-1}$ for a fixed interior target. In the tilted Edgeworth expansion the order $n^{-1/2}$ odd term vanishes at the mean. The exponent retains the Legendre-transform rate $I(x)=tx-K(t)$, rather than a quadratic Taylor approximation to that rate, explaining the useful tail and skewness behavior. Normalization is not automatic, and accuracy statements need their regularity and range restrictions.

For a normal distribution, $K(t)=\mu t+{\sigma }^2{t}^2/2$ and $t=(x-\mu )/{\sigma }^2$. Substitution recovers exactly the normal density of the sample mean. For an exponential distribution of rate $\lambda$, $K(t)=-\log (1-t/\lambda )$, $t=\lambda -1/x$ and $K^{\prime \prime }(t)=x^2$. The approximation becomesIt has exactly the gamma density shape. Replacing its leading constant by the exact normalizing constant makes it exact, illustrating why exactness of shape and exactness of the unnormalized approximation are separate questions.

For a vector sample mean, tilt by $e^{t^{\mathsf{T}}y-K(t)}$, solve $\nabla K(t)=x$, and use the local multivariate normal density at the tilted mean. This givesJoint and marginal saddlepoint densities can also be divided to approximate conditional densities, for example after conditioning on a nuisance sufficient statistic. The p* approximation uses likelihood and fitted-information coordinates to obtain a closely related approximation to a maximum-likelihood estimator's density, usually conditional on an ancillary statistic.

Tail probabilities need a corresponding integration approximation. The Lugannani-Rice saddlepoint tail approximation usesand givesHere $\Phi ,\varphi$ are the standard normal distribution functions. The upper-tail correction has the opposite sign: $1-\Phi (w)+\varphi (w)(1/u-1/w)$. Inverting a cumulative distribution function introduces a pole in addition to the density's saddlepoint; treating both supplies the correction term, which is why simply integrating a normal approximation at the untilted mean is less accurate. At $x=K^{\prime }(0)$ the apparent singularities cancel. With $v=K^{\prime \prime }(0)$ and $k=K^{\prime \prime \prime }(0)$, expansions give $w=t\sqrt{nv}(1+kt/(3v)+O(t^2))$ and $u=t\sqrt{nv}(1+kt/(2v)+O(t^2))$, so $1/w-1/u\to k/(6\sqrt{n}{v}^{3/2})$. For the normal distribution the correction is zero everywhere.

These methods require care for lattice distributions, where summation and lattice corrections replace the density inversion, and near boundaries or singular covariance matrices. A distribution with no moment-generating function in a neighborhood of zero does not satisfy the starting assumptions. A successful saddlepoint approximation is therefore a structured refinement of regular asymptotic inference, not a formula guaranteed uniformly in every tail or for every model.

A density deficit in a flux tube is not by itself an instability calculation. For example, a straight magnetized tube in otherwise field-free gas at the same temperature has lateral balance $p_i+B_i^2/(2{\mu }_0)=p_e$. The perfect gas law then gives ${\rho }_i<{\rho }_e$, so the tube rises unless some other force balances its weight deficit. This configuration simply lacks magnetostatic equilibrium. A magnetic buoyancy instability, in contrast, starts from a force-balanced atmosphere and asks whether a small allowed displacement grows. A balanced horizontal magnetic layer can be stable or unstable; the presence of magnetic pressure alone does not decide this.

Assume a smooth equilibrium, with downward gravitational acceleration $g>0$, and write $\mathcal{R}$ for the gas constant per unit mass. The ambient total pressure and force balance areWe use ideal magnetic flux freezing on the displacement time scale and rapid lateral pressure equilibration. A bodily displacement perpendicular to a straight horizontal tube does not change its length or bend its field. If its area is $\mathcal{A}$, conservation of magnetic flux and mass per unit length gives $B\mathcal{A}=\mathrm{constant}$ and $\rho \mathcal{A}=\mathrm{constant}$. Thus the tube preservesEfficient heat exchange fixes its temperature to the ambient temperature at its new height, not to its original temperature. This is the thermal assumption that distinguishes the requested criterion from the adiabatic one.

Take a tube initially at $z_0$ and let $f_0=B_0(z_0)/\rho (z_0)$. After a small upward displacement $\xi$, let ${\rho }_i(z)$ be the density it would have in lateral balance at $z=z_0+\xi$. Its preserved $f_0$ and imposed local temperature giveAt the initial position ${\rho }_i=\rho$. Differentiate this relation at $z_0$:For the ambient atmosphere the same differentiation must also differentiate $f(z)=B_0(z)/\rho (z)$. It givesSubtracting isolates the actual displaced density contrast:The pressure-gradient force on the tube balances the weight of the displaced ambient gas, so its leading vertical buoyancy acceleration isThe coefficient multiplying $\xi$ is positive precisely when the tube becomes lighter on upward displacement. Thus the isothermal interchange criterion for magnetic buoyancy isWhere $B_0>0$, this is simply $d(B_0/\rho )/dz<0$, or $d\log B_0/dz<d\log \rho /dz$. The squared form is independent of the field's choice of direction. Equality is neutral at this order, and the reversed inequality is restoring for these displacements. A layer with an interval satisfying the strict inequality admits local growing interchange disturbances under the assumed pressure/thermal equilibration. The calculation does not address field-bending undular modes, finite heat-transfer times or magnetic diffusion; those can change the criterion. In particular, the localized interchange criterion for a magnetized atmosphere with an adiabatic gas response is not interchangeable with this fast-thermal-exchange result.

Magnetic flux concentration and flux expulsion describe redistribution of an existing field, not necessarily creation of net flux. For a weak field in a prescribed conducting flow, the resistive induction equation isIts ideal part combines transport, stretching and compression:Thus a converging flow can intensify a component, and stretching can increase magnetic energy even if the signed net flux is zero. At large magnetic Reynolds number $\mathrm{Rm}=UL/\eta$, diffusion is initially slow on the eddy scale, so magnetic flux freezing carries field lines with the flow. These statements do not imply that an ideal closed eddy can simply lose its initially enclosed flux.

The missing step is the production of small scales by differential circulation. In a two-dimensional incompressible eddy, the Cartesian magnetic flux function obeysDifferent streamlines have different circulation rates. An initially smooth field is wound into fine folds; diffusion and reconnection then smooth the variation of $A$ inside the eddy. As $A$ becomes nearly uniform there, its gradient and hence the interior magnetic field become small. The surviving flux lies mainly near the circulation boundary and adjacent separatrices, where stronger gradients remain. This is flux expulsion. With exactly zero diffusivity, field topology and the flux of material loops remain frozen, so a relaxed flux-free interior is not obtained by pure advection alone. Rigid rotation also fails to generate the differential winding responsible for this process.

The phase mixing in magnetic flux expulsion estimate makes the enhanced role of small diffusion explicit. If the angular-velocity gradient is of order $U/L^2$, a wound radial wavenumber grows as $k_r\sim Ut/L^2$. The accumulated damping exponent is of orderIt becomes order one atThis is much shorter than the large-scale diffusion time $(L/U)\mathrm{Rm}$ at large $\mathrm{Rm}$, but longer than one turnover time. It is a kinematic differential-shear estimate, not a universal time for every turbulent or rapidly evolving convection cell.

A related concentration mechanism occurs at converging flow boundaries. For a locally vertical bundle, conservation of flux gives $B\mathcal{A}\simeq \mathrm{constant}$, so reducing its cross-section increases $B$. In a simple converging boundary region with strain rate $U/L$, balancing strain and transverse diffusion givesThis is the thickness estimate for that local strain-diffusion model; it need not be the thickness or energy scaling of every closed-eddy flux-expulsion problem. Stronger field eventually exerts Lorentz force feedback and changes the flow. The equipartition magnetic field $B_{\mathrm{eq}}=\sqrt{{\mu }_0\rho }U$ is a useful indication that the kinematic assumption can fail, not a universal upper bound on concentrated fields.

At the Sun's photosphere, hot upflows form solar granules, and their spreading horizontal motion transports weak vertical field toward cooler intergranular lanes and downflows. A thin-layer illustration, neglecting vertical transport terms, is ${\partial }_t{B}_z+{\nabla }_h\cdot ({\mathbf{u}}_h{B}_z)=0$. It gives $D_h{B}_z/Dt=-B_z{\nabla }_h\cdot {\mathbf{u}}_h$: horizontal convergence intensifies the vertical flux density. This surface transport is related to the redistribution by convective eddies, but is not identical to the closed-streamline mechanism in the preceding planar example. Larger-scale supergranulation also gathers magnetic elements toward its boundaries, helping organize the solar magnetic network.

Concentration can be further intensified by compressible thermal dynamics. In the thin-tube picture of convective collapse, an unstable downward flow drains gas, reduces internal pressure and allows contraction at approximately conserved flux. Lateral balance $p_i+B^2/(2{\mu }_0)\simeq p_e$ then permits field stronger than the local kinetic equipartition estimate. This evacuation mechanism is excluded by a strict Boussinesq approximation. It is not a description of every evolving surface concentration: gas pressure, dynamic stresses and magnetic curvature can all matter when thin-tube equilibrium fails. The resulting strong, intermittent magnetic elements are associated with intergranular structure; their radiative appearance also depends on thermal structure and opacity. The organized fields of sunspots require a broader account of flux supply and magnetic feedback, rather than attribution to weak-field expulsion alone.

Thus convection redistributes and concentrates flux, finite diffusion enables classical expulsion, and nonlinear magnetic and thermal feedback determines the strong-field outcome. An expulsion argument alone neither supplies the original solar field nor establishes a self-sustaining solar dynamo.

Choose a reference circular orbit at radius $r_0$, rotate the frame with $\mathbf{\Omega }=\Omega (r_0){\mathbf{e}}_z$, and introduce local Cartesian coordinates $x=r-r_0$, $y=r_0(\varphi -\Omega t)$ and vertical coordinate $z$. On scales small compared with $r_0$, neglect curvature of the coordinate directions but retain the leading differential rotation. The background velocity relative to the rotating frame isThus $A=3\Omega /4$ for a Keplerian disk. The shearing sheet retains Coriolis acceleration and the linear tidal acceleration. The radial shearing-sheet tidal potential is $-2\Omega A{x}^2$, whose force $4\Omega Ax{\mathbf{e}}_x$ balances the background Coriolis acceleration. If vertical tidal gravity is included, its equilibrium contribution is balanced by a background pressure $p_0(z)$ and disappears when that equilibrium is subtracted.

Write the total velocity as $\mathbf{u}={\mathbf{u}}_0+\mathbf{v}$. In the rotating-frame magnetohydrodynamic momentum equation, background advection is $({\mathbf{u}}_0\cdot \nabla )\mathbf{v}=-2Ax{\partial }_y\mathbf{v}$ and advection of the background by the perturbation is $(\mathbf{v}\cdot \nabla ){\mathbf{u}}_0=-2A{v}_x{\mathbf{e}}_y$. The uniform-viscosity diffusion of the linear background shear is zero. For a solenoidal magnetic field the Lorentz force per unit mass isAt constant density, absorb the magnetic-pressure gradient into $\psi =(p-p_0)/\rho +B^2/(2{\mu }_0\rho )$. Subtracting the background force balance then gives the stated momentum equation, with the remaining nonlinear advection $\mathbf{v}\cdot \nabla \mathbf{v}$, Coriolis acceleration $2\mathbf{\Omega }\times \mathbf{v}$, magnetic tension and viscous diffusion.

For constant finite electrical conductivity ${\sigma }_e$, put $\eta =({\mu }_0{\sigma }_e{)}^{-1}$. The resistive induction equation with uniform magnetic diffusivity is ${\partial }_t\mathbf{B}+\mathbf{u}\cdot \nabla \mathbf{B}=\mathbf{B}\cdot \nabla \mathbf{u}+\eta {\nabla }^2\mathbf{B}$. Substituting the background shear supplies the term $-2A{B}_x{\mathbf{e}}_y$ and the same background-advection operator as in the momentum equation. The two further conditions areThey express incompressibility and the absence of magnetic monopoles, respectively.

Now take real plane waves with common phase $\vartheta =\mathbf{k}(t)\cdot \mathbf{x}$ and real wavevector. The solenoidal conditions becomeEvery field depends on position only through $\vartheta$. Hence for either transverse vector field $\mathbf{P}$ and either wave field $\mathbf{Q}$, $(\mathbf{P}\cdot \nabla )\mathbf{Q}=(\mathbf{P}\cdot \mathbf{k}){\partial }_{\vartheta }\mathbf{Q}=0$. This proves that all the displayed quadratic advection and tension terms vanish pointwise, including products involving conjugate amplitudes. These waves are therefore exact solutions at finite amplitude, not merely a linearization.

Acting with $D_0={\partial }_t-2Ax{\partial }_y$ on the phase yieldsCanceling its position-dependent part requiresThe remaining amplitude equations areDifferentiating the velocity constraint determines the pressure amplitude:The resistive induction equation preserves its constraint because $\dot{\mathbf{k}}\cdot \tilde{\mathbf{B}}=2A{k}_y{\tilde{B}}_x$ cancels the corresponding shear term in $\mathbf{k}\cdot \dot{\tilde{\mathbf{B}}}$. The specified magnetic field has no spatially uniform part. Thus there is no background magnetic field producing an additional linear magnetic-tension coupling. The scalar $\psi$ is the modified pressure; the physical pressure can include a constant and a second spatial harmonic from $-B^2/(2{\mu }_0)$ while $\psi$ retains the stipulated plane-wave form. This is an exact zero-mean magnetic shearing wave.

For decay, take the real part of the scalar product of the velocity-amplitude equation with its complex conjugate. Pressure does no work because $\mathbf{k}\cdot \tilde{\mathbf{v}}=0$, and the real part of Coriolis work is zero. ThereforeThe resistive induction equation similarly givesusing $2|u_x{u}_y|\le |\mathbf{u}{|}^2$. Integrating these differential inequalities, defineThenA non-axisymmetric wave has $k_y\ne 0$. For nonzero differential shear and positive $\nu ,\eta$, the negative cubic terms dominate the linear upper bound on shear work. ConsequentlyThis cubic-exponent decay of a nonaxisymmetric shearing wave allows transient energy growth before the winding to short radial wavelengths makes diffusion dominant. The magnetic amplitudes can also be written explicitly aswhich confirms the damping independently. Positivity of the dissipative coefficients is necessary: with $k_z=0$, an inviscid vertical velocity amplitude can remain constant, and an ideal vertical magnetic amplitude can remain constant when $\eta =0$. The asserted decay applies to the viscous, finitely conducting model; taking an ideal limit before the long-time limit changes the conclusion.

In the specified dust model $H_0=2/(3t_0)$, so ${\rho }_{\mathrm{crit}}=m_{\mathrm{pl}}^2/(6\pi t_0^2)$ and $n_B={\Omega }_{B0}{\rho }_{\mathrm{crit}}/m_B$. Photons and the six neutrino/antineutrino helicity states giveUsing the supplied GeV estimates literally givesThe rounded time/temperature conversions are unusually crude: converting $10$ Gyr to $4.8\times 10^{41}{\mathrm{GeV}}^{-1}$ and $3$ K to $2.6\times 10^{-13}$ GeV instead gives $Y_B\sim 10^{-10}$. Either set gives the same order-of-magnitude mass conclusion below, but the numerical abundance should identify which supplied units are used.