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In a radially stratified shearing sheet, one can normalize the scalar buoyancy perturbation by writing $b_x=-N^2\theta$ and $D_t\theta =u_x+\xi {\nabla }^2\theta$. With this convention $\theta$ has dimensions of length, rather than temperature; $\xi$ is thermal diffusivity.

For an imposed field $B_0$ in a layer of depth $d$, the Chandrasekhar number compares magnetic tension with viscous and magnetic-diffusive effects. Here $\eta$ is magnetic diffusivity and $\nu$ is kinematic viscosity. In thermal-diffusion units the linear magnetic-force coefficient is $Q\zeta$, where $\zeta =\eta /\kappa$ and $\kappa$ is thermal diffusivity.

With permeability $K$, depth $d$, kinematic viscosity $\nu$ and the relevant effective thermal diffusivity ${\kappa }_T$, the Darcy Rayleigh number compares buoyancy-driven Darcy advection to thermal diffusion. It differs from the clear-fluid Rayleigh number by replacing $d^3$ with $Kd$. Effective heat capacity and porosity factors may be absorbed into the definition of ${\kappa }_T$ and the thermal-time scale, and must be kept consistent with the chosen Darcy law convention.

Diffusion is net transport associated with random microscopic motion. A localized patch of a dilute solute can spread through a stationary solvent: individual molecules move in both directions, but more cross from the crowded side of an imaginary surface than from the less crowded side. This imbalance gives a net diffusive flux down the concentration gradient; the molecules do not need to sense which direction has lower concentration. At uniform concentration, microscopic exchanges continue even though the net diffusive flux vanishes. ^[ref]

The distinction from advection is easiest to see in a dye patch. Uniform bulk motion carries its centre downstream; diffusion broadens it relative to that centre. Both can happen together. A moving or spreading patch by itself does not identify the mechanism: a nonuniform velocity field can also stretch a patch. ^[ref]

For a dilute nonreacting solute in a uniform medium at fixed temperature, Fick's first law models the relative diffusive flux as $\mathbf{j}=-D\nabla C$, with scalar diffusion coefficient $D>0$ of units ${\mathrm{m}}^2{\mathrm{s}}^{-1}$. Here $C$ is the amount of substance per unit volume, so $\mathbf{j}$ has units $\mathrm{mol}{\mathrm{m}}^{-2}{\mathrm{s}}^{-1}$. With carrier velocity field $\mathbf{u}$, the total flux is $C\mathbf{u}+\mathbf{j}$. The continuity equation, with no sources or chemical reactions, givesFor constant $D$ and incompressible flow, this reduces to $C_t+\mathbf{u}\cdot \nabla C=D{\nabla }^{2}C$, where ${\nabla }^2$ is the Laplacian. In a stationary carrier it is the diffusion equation. These assumptions matter: variable $D$ must remain inside the divergence, and changing sources require extra terms. ^[ref]

The useful surprise is the common mathematics: thermal conduction smooths temperature through the analogous heat equation, while molecular diffusion smooths composition. This analogy already motivated the original formulation of Fick's laws. ^[ref] However, the thermal diffusivity and the solutal diffusivity need not be equal. A diffusive time scale $t_D\sim L^2/D$ means that crossing ten times the distance requires roughly one hundred times the time. This is why molecular diffusion is especially effective across thin layers, and why chaotic advection can assist mixing by making those layers thinner. Physical diffusion is a transport mechanism, distinct from the broader probabilistic class of Markov diffusion.

Fourier analysis describes a function by the oscillations that compose it. A graph records values at successive positions or times; a frequency spectrum records how much of each frequency is present, together with its wave phase. The same complicated shape can become a short list of simple components. For periodic functions, the components form a Fourier series; on the whole real line, the Fourier transform uses a continuous range of frequencies, and the Fourier inversion theorem puts them back together.

For a concrete example, consider the trigonometric polynomialWith $t$ measured in seconds, its three Fourier harmonics have frequencies $1$, $3$, and $5$ cycles per second and amplitudes $1$, $1/2$, and $1/4$. Each cosine has zero wave phase. Adding them creates the sharper peaks in the figure. The frequency spectrum makes the recipe immediately visible; changing an amplitude changes the shape without changing the component's frequency. Fourier orthogonality lets us recover each Fourier coefficient by averaging against the corresponding sine or cosine, because the other Fourier harmonics cancel over a full period of a function.

The deeper reason these components are so useful is that they simplify operations. By Euler's formula, the exponential function $e^{i\omega x}$ packages a cosine and a sine into one Fourier mode. A translation or a derivative changes only its multiplier:Thus Fourier modes are simultaneous eigenfunctions of these operations. In Fourier series, this gives a literal diagonalization in the Fourier basis; on the real line, the Fourier transform gives the continuous counterpart, with each frequency handled separately. The Translation property of the Fourier transform and the Fourier transform of a derivative turn shifting and differentiating a whole function into multiplication of its Fourier transform. A convolution likewise becomes multiplication by the convolution theorem. This is beautiful because one description simplifies several apparently different operations at once.

For example, the heat equation $u_t=\kappa u_{xx}$, with thermal diffusivity $\kappa >0$, evolves a spatial Fourier mode asFine ripples have larger angular frequencies and decay faster: the formula explains why heating smooths a jagged temperature profile. For a general initial profile with sufficient regularity, the Fourier transform gives $\hat{u}(\omega ,t)=e^{-\kappa {\omega }^{2}t}\hat{u}(\omega ,0)$, and Fourier inversion recovers the temperature. Similarly, multiplying selected parts of a frequency spectrum suppresses unwanted components. Claude E. Shannon's Communication in the Presence of Noise (1949), Section III, describes ideal filtering as a projection that retains some frequency coordinates and removes others.^[ref]

The physical motivation is already visible in Joseph Fourier's Théorie analytique de la chaleur (1822). Chapter III constructs heat profiles using sums of sines and cosines; Chapter IX, articles 355–359, passes from Fourier series to integrals as the spatial interval becomes unbounded. These heat problems explain the motivation for the discrete and continuous descriptions. Their modern formulations specify which functions admit reconstruction and in what sense the expansions converge.^[ref][ref]

Laser flash analysis measures thermal diffusivity by heating one face of a specimen briefly and recording the opposite face's temperature rise. The original 1961 experiment used a flash tube rather than a laser: a blackened thin metal specimen sat in an opaque ceramic holder, with thermocouple wires pressed against its rear face. An amplifier, oscilloscope and camera recorded the signal. Uniform front illumination and the black coating promoted uniform absorption; the opaque low-thermal conductivity holder blocked direct rear illumination and reduced edge losses. The delayed rear response reveals internal thermal conduction rather than direct heating of the sensor. ^[ref]

A modern version uses a laser or flash lamp, a specimen of measured thickness $L$, controlled initial temperature, and a rear-facing infrared detector with a synchronized recorder. The readout is a temperature rise versus time, starting from a pre-pulse baseline. Thermal conductivity additionally requires mass density and specific heat capacity; it is not obtained from timing alone. ^[ref]

Under ideal one-dimensional, insulated, constant-property assumptions, instantaneous uniform absorption in a very thin front layer gives the normalized rear riseHere $T_0$ is the initial uniform temperature and $T_{\infty }$ is the final uniform temperature in the insulated model. Fitting this curve estimates thermal diffusivity; each term has the same $\alpha K_n^2$ decay as the insulated-bar Fourier modes. Equivalently, the half-rise time $t_{1/2}$ solves $R(t_{1/2})=1/2$. Writing $s=\alpha t/L^2$, the series gives $s_{1/2}\approx 0.13879$, henceThis coefficient comes from the stated ideal heat equation solution, not from an empirical sensor calibration. It has been independently checked by solving the series equation. Absolute absorbed pulse energy cancels from the normalized curve; determining heat capacity from its absolute height needs additional information. ^[ref]

Parker and colleagues identify finite flash duration, heat loss, nonuniform illumination, recording bandwidth and electrical interference as practical limits. A specimen too thin responds during the pulse; one too thick allows losses before the rear rise finishes. They shielded the electrical chain and recorded a baseline; their reported tests were preliminary metal measurements near ambient and moderately elevated temperatures. ^[ref]

Useful additional controls are to measure thickness at several points, verify the recorder's time scale and detector response independently, record a no-pulse baseline, and repeat at different pulse energies while avoiding phase transitions. Compare different thicknesses of the same homogeneous material: ideal half-rise times scale as $L^2$. Failure of these checks can expose finite-pulse effects, sensor delay, optical transmission, coating storage or edge losses; a thickness error is especially consequential because the inferred thermal diffusivity depends on $L^2$. These are proposed diagnostic controls, not invented measurements. When the pulse is not short or the sample exchanges appreciable heat with its surroundings, fit a model that includes those effects instead of applying the ideal half-rise coefficient unchanged.

The ratio of magnetic diffusivity $\eta$ to thermal diffusivity $\kappa$ controls their relative relaxation times. In nonrotating vertical-field magnetoconvection, oscillatory linear onset in the simple stress-free layer requires $0<\zeta <1$, so thermal perturbations diffuse faster than magnetic perturbations. This is distinct from the magnetic Prandtl number, $\nu /\eta$.

Use positive jumps $\Delta T=T-T_{\infty }$ and $\Delta S=S-S_{\infty }$ for the hot salty lower reservoir. The Boussinesq density excess below the interface is proportional to $\beta \Delta S-\alpha \Delta T$. Static stability therefore requireswith one the neutral limiting value. For $R_{\rho }<1$ the lower layer is lighter overall and ordinary overturning convection replaces the statically stable interface.

In the stable hot-below, salty-below configuration, thermal diffusivity exceeds salt diffusivity. Heat transfer can destabilize thin adjacent fluid regions before the stabilizing salt difference is erased, causing the diffusive regime of double-diffusive convection. As $R_{\rho }\to \infty$, stabilization by salt dominates, interfacial convective exchange is strongly suppressed, and the specified heat-flux model tends to zero at fixed $\Delta T$. This limit is not a claim that molecular heat conduction itself ceases.

Let $\mathbf{n}$ point from fluid $-$ to fluid $+$, $P_s=I-\mathbf{nn}$, ${\nabla }_s=P_s\nabla$, and $\mathcal{K}={\nabla }_s\cdot \mathbf{n}$. A clean material interface without phase change has continuous velocity, with normal interface speed $V_n={\mathbf{u}}_{\pm }\cdot \mathbf{n}$. Its surface stress is $\gamma P_s$. The surface divergence of that tensor is ${\nabla }_s\gamma -\gamma \mathcal{K}\mathbf{n}$. Balancing forces on a small interfacial patch therefore gives the interfacial stress balance with variable surface tensionThe normal condition is $\mathbf{n}\cdot ({\mathbf{\sigma }}_{+}-{\mathbf{\sigma }}_{-})\mathbf{n}=\gamma \mathcal{K}$; the tangential condition is $P_s({\mathbf{\sigma }}_{+}-{\mathbf{\sigma }}_{-})\mathbf{n}=-{\nabla }_s\gamma$. On a stationary sphere with outward normal these reduce to $p_{-}-p_{+}=2\gamma /a$ when viscous normal stresses vanish. This fixes the sign convention for the Marangoni stress rather than leaving the direction ambiguous.

For the bubble take negligible internal dynamic viscosity, no contaminants, and initially the standard effectively insulating gas-bubble approximation. With heat advection neglected, the exterior temperature satisfies Laplace's equation, approaches $T_0+\mathbf{H}\cdot \mathbf{r}$ remotely, and has zero normal heat flux at $r=a$. The temperature dipole around an insulating sphere givesThe zero-flux condition follows by differentiating the radial factor. Let ${\gamma }_T=(d\gamma /dT{)}_{T_0}$. To first order in the weak gradient,

In the laboratory frame the force-free bubble can be described by the harmonic Papkovich–Neuber representation potentialsThey give constant exterior pressure and the decaying source dipole fieldAt the surface $\mathbf{u}\cdot \mathbf{n}=\mathbf{U}\cdot \mathbf{n}$, so the normal kinematic boundary condition holds for translation. In coordinates aligned with $\mathbf{U}$, $u_r=a^{3}U\cos \theta /r^3$ and $u_{\theta }=a^{3}U\sin \theta /(2r^3)$. Direct differentiation givesEquivalently, the surface tangential traction is $(3\mu /a)P_s\mathbf{U}$. The tangential stress condition therefore requiresFor the usual negative ${\gamma }_T$, the bubble moves toward hotter liquid. The normal condition is consistent as well: its viscous dipole is $-6\mu \mathbf{U}\cdot \mathbf{n}/a=3{\gamma }_T\mathbf{H}\cdot \mathbf{n}$, exactly the dipolar part of $2{\gamma }_s/a$. The constant part fixes $p_{\mathrm{gas}}-p_{\infty }=2{\gamma }_0/a$. There is no Stokeslet term, hence no net applied force or drag imbalance.

The thermal assumption matters. The problem gives the exterior thermal diffusivity but no gas thermal conductivity. For an inviscid bubble with conductivity ratio $K=k_{\mathrm{gas}}/k_{\mathrm{liquid}}$, temperature and normal heat-flux continuity instead giveso the same stress calculation givesThe insulating-bubble result uses $K\ll 1$. If instead an undisturbed surface temperature gradient is imposed, corresponding to $K=1$, the coefficient is $1/3$ rather than $1/2$. Exterior thermal diffusivity alone does not select between those thermal boundary conditions. This dependence is explicit rather than silently treating a missing gas property as a prescribed value.

The small parameters can now be stated precisely. With $U_{\ast }=a|{\gamma }_T||\mathbf{H}|/(2\mu )$ and $\nu =\mu /\rho$, requireSmall Reynolds number justifies Stokes flow; small Péclet number allows quasi-static heat conduction without advective distortion; small capillary number and fractional surface-tension variation support a nearly spherical interface and a weak perturbation of its base surface tension. These last two conditions are proportional for this velocity scale but describe different approximations. Linearizing $\gamma (T)$ also needs $a|\mathbf{H}||{\gamma }_{TT}/{\gamma }_T|\ll 1$ if its derivative varies appreciably. Negligible internal viscosity and the insulating specialization require the corresponding gas-to-liquid viscosity and conductivity ratios to be small. Gravity has been excluded; if restored, its buoyancy-induced migration must be small compared with $U_{\ast }$, not merely small enough to avoid deformation.

Finally, surfactants change both the constitutive surface tension and the surface stress dynamics. The surface flow driven by the thermal Marangoni effect carries insoluble surfactant toward the high-tension side, lowering its tension and producing an opposing concentration-driven gradient. When surface diffusion or exchange is slow, this tends to immobilize the interface, suppressing the clean-bubble thermocapillary migration of an insulating bubble, sometimes nearly stopping it. If the surfactant stays uniformly equilibrated, it can instead change ${\gamma }_T$ without a large opposing gradient. Its concentration, surface diffusivity and adsorption kinetics must be supplied to compute a replacement speed; no universal numerical reduction factor follows merely from its presence.

Write $\beta$ for the thermal expansion coefficient, $\nu$ for kinematic viscosity, $\kappa$ for thermal diffusivity, and $k_T=\rho c_p\kappa$ for thermal conductivity. For a temperature difference $\Delta >0$ across depth $h$,The four-thirds convective heat-transfer law posits $\mathrm{Nu}=C{\mathrm{Ra}}^{1/3}$ at high Rayleigh number, and thereforeThe layer depth cancels. A useful physical justification is a mixed interior with thin conductive boundary layers of thickness $\delta$, each maintained near a fixed critical local Rayleigh number: $g\beta {\Delta }_b{\delta }^3/(\nu \kappa )\sim {\mathrm{Ra}}_{\mathrm{crit}}$ and $q\sim k_T{\Delta }_b/\delta$. Eliminating $\delta$ gives the exponent $4/3$. Whether $\Delta$ denotes the whole-layer contrast or the drop across one boundary layer changes the coefficient, not the exponent.

Assume Boussinesq approximation, constant material properties, a laterally extensive statistically turbulent Rayleigh-Bénard convection layer, negligible side-wall heat transfer, no rotation or imposed shear, approximately uniform bulk temperature, and the same specified Prandtl number regime when choosing $C$. The closure is empirical and regime dependent; high Rayleigh number alone does not make it exact or extend it to the impulsive conductive initial transient.

The Boussinesq approximation assumes that the fractional mass density variations are small. Replace density by a constant reference value in inertia, pressure scaling and mass conservation, but retain its small anomaly in the gravitational force: the large hydrostatic force otherwise cancels, leaving precisely the buoyancy-driving term. For temperature-driven motion, write $\rho ={\rho }_r[1-\alpha (T-T_r)]$ with $|\alpha (T-T_r)|\ll 1$. The resulting velocity is an incompressible flow; compressional heating and acoustic effects are omitted in the shallow, low-compressibility regime.

For this convection problem, use the strong Boussinesq approximation in the constant-property sense: the thermal expansion coefficient, kinematic viscosity, thermal diffusivity and heat capacity are also independent of temperature over the layer. This additional assumption makes the conduction profile linear and the perturbation coefficients constant. This usage of “strong” is the convection convention described in courses.physics.ucsd.edu/2021/Spring/physics218c/Spiegel%20Convection.pdf . Some stratified-flow literature uses “strong” instead to emphasize a globally constant reference density; that condition is also satisfied here. The equations below specify precisely which assumptions are being used.

Introduce the thermal diffusivity $\kappa$ and positive thermal expansion coefficient $\alpha$, which are necessary material data even though the question names only $\nu$ explicitly. Measure $z$ upward from the bottom. The conductive state of Rayleigh-Bénard convection is $T_b=T_0+\Delta T(1-z/d)$ with zero velocity. Let $\theta =T-T_b$ and let $\pi$ be perturbation pressure divided by the reference density, after subtracting the hydrostatic base pressure. Linearizing givesThe sign in the last equation is positive because upward motion carries warmer base-state fluid into cooler surroundings. Taking the divergence of momentum gives ${\nabla }^2\pi =g\alpha {\theta }_z$. Apply a Laplacian to its vertical component and subtract the vertical derivative of that pressure identity. This eliminates pressure:Apply the temperature operator to obtain the vertical-velocity equationThis is the pressure- and temperature-eliminated linear stability analysis equation for Rayleigh-Bénard convection.

At each boundary, impermeability gives $w=0$, the stress-free boundary condition gives $w_{zz}=0$, and perfect thermal conduction gives $\theta =0$. To obtain the second condition, zero tangential traction gives ${\partial }_z{u}_h+{\nabla }_{h}w=0$; since $w$ vanishes identically along the boundary, differentiate horizontal incompressibility to get $w_{zz}=0$. For the fully eliminated velocity equation, $\theta =0$ and the preceding momentum relation also imply $w_{zzzz}=0$ there. Thus suitable formulations areor, for the sixth-order vertical equation alone, $w=w_{zz}=w_{zzzz}=0$ at both boundaries. Fixed boundary temperature means a zero perturbation there, not zero perturbation heat flux.

Use normal modes $w=W\sin (n\pi z/d)e^{\sigma t+i\mathbf{k}\cdot {\mathbf{x}}_h}$, $n\ge 1$, $k=|\mathbf{k}|>0$. Put $K^2=k^2+n^2{\pi }^2/d^2$. Substitution yields the stress-free convection growth-rate polynomialAt stationary marginal stability, $\sigma =0$. In terms of the Rayleigh number $\mathrm{Ra}=g\alpha \Delta T{d}^3/(\nu \kappa )$ and dimensionless horizontal wave number $a=kd$, the free-slip convection neutral curve isWriting $y=a^2$ and differentiating gives $d\log {\mathrm{Ra}}_n/dy=3/(y+n^2{\pi }^2)-1/y$. Its zero is $a^2=n^2{\pi }^2/2$, with minimum $27n^4{\pi }^4/4$. The smallest vertical mode is $n=1$, so the critical temperature difference isThe first unstable horizontal wavelength is $2\sqrt{2}d$. To check that an oscillatory mode cannot destabilize earlier, expand the growth-rate equation: its linear coefficient is $(\nu +\kappa )K^2>0$, and its constant term changes sign exactly at the stationary neutral curve. For positive temperature difference the discriminant is $(\nu -\kappa {)}^2{K}^4+4g\alpha \Delta T{k}^2/(d{K}^2)>0$, so the crossing root is real. Below the global critical value every convection mode decays; at criticality the selected mode is marginal; above it some modes grow and eventually require nonlinear advection, producing convective heat transport. This is exchange of stabilities in stress-free convection.

Strictly, the linearized equations remain the correct small-amplitude equations on either side of onset. What fails above $\Delta T_{\max }$ is stability of the conductive state and long-time validity of a small-perturbation description, not the algebra of linearization. The threshold also presumes that the required small density variations and constant-property approximation hold at that temperature difference.

Use the steady slender-plume model implicit in the requested height-only similarity. Put $\theta =T-T_0$ and introduce permeability $K$, dynamic viscosity $\mu ={\rho }_0\nu$, fluid heat capacity per volume $C_h={\rho }_0{c}_p$, and effective transverse thermal diffusivity $\kappa$. These material constants are needed to give dimensional fluxes. Take $u,w$ to be Darcy velocities, so the thermal transport coefficient for the steady equation is $\kappa =k_e/C_h$ when $k_e$ is the effective conductivity of the saturated medium.

The Darcy law with gravity isThe ambient is hydrostatic with $p_{a,z}=-{\rho }_{0}g$. A slender plume's pressure correction has vertical gradient smaller than the buoyancy term by the squared width-to-height ratio. Indeed, incompressibility gives $u/w$ of order width/height, and the horizontal Darcy relation makes the pressure correction of order $\mu w{\text{width}}^2/(K\text{height})$. Its vertical derivative has the additional inverse-height factor. Thus the leading vertical Darcy balance isThe proportionality is to temperature excess $T-T_0$, not absolute $T$; the velocity vanishes in the unheated ambient. This is the local buoyancy/drag balance of a porous thermal plume, rather than a free-fluid inertial momentum balance.

The governing continuity equation and steady advection-diffusion equation, with axial conduction neglected in the slender high-Rayleigh number limit, areChoose the streamfunction convention $u=-{\psi }_z$, $w={\psi }_x$. It enforces continuity identically and givesA change to the opposite streamfunction convention changes the sign of $\psi$, not the physical velocity or temperature.

The upward convected excess heat flux per unit source length isThis is the quantity denoted $F$ in this question; it differs from Question 2's buoyancy-flux symbol. In conservative form the heat equation is ${\partial }_x(u\theta )+{\partial }_z(w\theta )=\kappa {\theta }_{xx}$. Integrate across the plume:The temperature anomaly and its derivative vanish far from the plume; lateral entrainment brings in ambient-temperature fluid and thus no excess heat. There is no heat source or sink above the line source. Transverse conduction redistributes this heat without removing its total, while axial conduction is negligible in the stated approximation. In an unsteady plume a heat-storage term would have to be retained, so the convective flux alone would not in general be independent of height.

Let $\mathcal{J}=\beta F_h/C_h=\int w^{2}dx>0$. It is constant because $w=\beta \theta$. To obtain similarity exponents without guessing a profile, write the width as $\delta \propto z^a$ and central velocity as $W\propto z^b$. The fixed heat flux requires $W^2\delta$ constant, hence $2b+a=0$. Balancing vertical advection, including its continuity-required horizontal component, against transverse conduction gives $W{\delta }^2\propto \kappa z$, hence $b+2a=1$. Thus $a=2/3$ and $b=-1/3$, and the streamfunction scale $W\delta$ is proportional to $z^{1/3}$. A suitable similarity form isThe volume flux per unit span is $Q=\psi (+\infty )-\psi (-\infty )$ and therefore grows as $z^{1/3}$; it tends to zero at the ideal source. Thus the zero-source-volume requirement fixes the absence of an additive source-length scale, not a finite temperature at the singular point.

Substitution into the streamfunction equation givesChoose the scale normalization $AB=6\kappa$ and $f(+\infty )=1$. Reflection symmetry makes $f$ odd, so the relevant conditions are $f(0)=0$, $f^{\prime \prime }(0)=0$, $f^{\prime }(\infty )=0$, with finite $f(\infty )$; the heat input fixes the remaining dimensional scale. The nonlinear similarity equation becomesIntegrate once. Since $f^{\prime },f^{\prime \prime }\to 0$ in the ambient, the integration constant is zero, giving $f^{\prime \prime }+2f{f}^{\prime }=0$. A second integration gives $f^{\prime }+f^2=C$; the normalized far limit fixes $C=1$. Separation with $f(0)=0$ now yields the complete positive-plume profileIt satisfies the symmetry and decay conditions; the solution of the first-order initial-value equation is unique. The physical heat strength is determined using ${\int }_{-\infty }^{\infty }{\mathrm{sech}}^4\eta d\eta =4/3$, obtained by substituting $t=\tanh \eta$:Equivalently, defineThe hyperbolic-secant porous plume profile and its velocity field are thenIn particular $W{b}^2=6\kappa z$. The far horizontal velocities are inward, $u(\pm \infty ,z)=\mp Wb/(3z)$, supplying the entrained ambient volume. They are consistent with $Q^{\prime }=2Wb/(3z)$.

Finally use specific Darcy-flux moments, per unit source length, for the requested flux relationships. The volume flux, kinematic momentum flux and buoyancy flux areThus the physical mass flux is ${\rho }_{0}Q(z)$, and the conventional density-weighted Darcy second moment is ${\rho }_{0}M$. If intrinsic pore-velocity momentum flux is required instead, a uniform porosity $\phi$ gives ${\rho }_{0}M/\phi$; keeping this convention distinct does not change the displayed Darcy moments. These are the conserved momentum flux of a porous thermal plume relationships. Constancy of $M$ follows here from proportionality of temperature and velocity plus heat conservation, not from ignoring the drag on the solid matrix in a total momentum balance.

The ideal source has zero limiting volume flux but divergent central speed and temperature. Consequently the Boussinesq approximation and slender-plume limit describe the region outside a finite source, where $|\alpha \theta |\ll 1$ and $b/z\ll 1$. Indeed $Wz/\kappa =6(z/b{)}^2\gg 1$ is the local high-Rayleigh condition. These limits improve downstream; they do not make the singular similarity field a literal finite-temperature solution at $x=z=0$.

Let ${\rho }_0$ be the reference mass density, $\nu$ the kinematic viscosity, $\kappa$ the thermal diffusivity, $\eta$ the magnetic diffusivity, and ${\alpha }_T$ the coefficient of thermal expansion. The conductive state of Rayleigh-Bénard convection is $T_b=T_0+\Delta T(1-z/d)$. For the Boussinesq approximation, density variation is retained only in buoyancy. Linearizing about this state and the imposed magnetic field $B_0\hat{\mathbf{y}}$ givesHere $w=u_z$ and $p_{\ast }=p^{\prime }+B_0{b}_y/{\mu }_0$ is the perturbation of magnetohydrodynamic total pressure. The gradient part of the Lorentz force has been absorbed into $p_{\ast }$; the remaining term is magnetic tension. The sign in the temperature equation follows from $d{T}_b/dz=-\Delta T/d$.

Use the thermal-diffusion scaling of a convection layer: length $d$, time $d^2/\kappa$, velocity $\kappa /d$, temperature perturbation $\Delta T$, field perturbation $B_0$, and total-pressure perturbation ${\rho }_0\nu \kappa /d^2$. Dropping dimensionless-variable decorations giveswithThese are respectively the Rayleigh number, Chandrasekhar number, Prandtl number, and magnetic-to-thermal diffusivity ratio. In particular, the magnetic coefficient is $\zeta Q$, not $Q$ alone, in these thermal-time units. At $z=0,1$, the boundary conditions are $\theta =w=b_z=0$, ${\partial }_z{u}_x={\partial }_z{u}_y=0$, and ${\partial }_z{b}_x={\partial }_z{b}_y=0$.

Work in the frame of the steadily advancing mush, so the rigid primary-crystal skeleton moves with $-V{\mathbf{e}}_z$. Let $\varphi$ be solid fraction, $C$ the pore-liquid concentration, $\mathbf{q}=(u,w)$ the Darcy flux relative to that skeleton, and $K$ the common thermal conductivity. Set $\kappa =K/(\rho c_p)$, the thermal diffusivity. The actual pore velocity is $\mathbf{q}/(1-\varphi )-V{\mathbf{e}}_z$. Primary crystals are taken to exclude solute; the residual pore liquid freezes into a two-solid eutectic product at the bottom. These are the standard binary mushy layer assumptions implicit in using a single liquidus and an eutectic floor.

It is convenient to use hydraulic pressure $P=p+\rho gz$, absorbing the constant-density gravitational term. With equal phase properties, no solute diffusion, negligible fluid inertia and local liquidus equilibrium, a complete mush model isThe last equation follows from conservation of energy with mixture specific enthalpy $c_{p}T-L\varphi$. Sensible heat is transported by the total volume flux $\mathbf{q}-V{\mathbf{e}}_z$; the primary solid carries the latent-enthalpy flux $VL\varphi {\mathbf{e}}_z$. In particular the sign of the latent-heat term is fixed by the downward-moving skeleton.

At the liquid roof, imposeHere $P_0=p_0+\rho gh$. The gradient condition matches heat flux into the imposed liquid field, since there is no phase-fraction jump at this roof and the phase conductivities are equal. The pore velocity there equals the relative Darcy velocity. At the floor,There is no liquid penetration into fully solid material. The unspecified solid fraction at the floor is determined by the solute balance, not set to one: remaining eutectic liquid freezes as it crosses the floor. If $C_s$ denotes the bulk concentration of the final solid product, the floor solute balance and Stefan condition areMaintaining the floor at $T_E$ permits the solid-side bath to remove this final latent heat. Equivalently its required gradient is $T_z(0^{-})=T_z(0^{+})+\rho LV[1-\varphi (0)]/K$. No additional imposed solid-side gradient is specified. Laterally we take the quadratic-pressure strain solution, without sidewalls; finite sidewalls would require additional flow conditions and generally break this horizontal invariance.

For uniform permeability of a porous medium, incompressible flow gives ${\nabla }^{2}P=0$. The roof pressure and zero normal flow at the floor are satisfied byHorizontal outflow is balanced by downward inflow. The parameter $\alpha$ is an inverse time; $E^2$ in the printed pressure condition already has dimensions of pressure divided by length squared, so no extra density factor belongs in this definition.

With $T=T(z)$, $C=C(z)$ and $\varphi =\varphi (z)$, the $x$-dependence disappears from both scalar balances. Although $u$ is nonzero, its horizontal divergence cancels $w_z$ in $\nabla \cdot (\mathbf{q}C)$. The exact strain-driven steady mush equations becomeTogether with $T(0)=T_E$, $T(h)=T_0$, $T^{\prime }(h)=G$ and $\varphi (h)=0$, they determine the scalar fields and the unknown thickness. This reduction proves that the equations admit horizontally invariant scalar solutions despite the imposed straining flow. An equivalent useful form, obtained by differentiating the solute equation, isThis exact form also shows why one must not silently discard latent heat in an arbitrary strong-flow calculation.

Define $\Delta T=T_0-T_E>0$, $R=\alpha h/V$ and $S=L/(c_p\Delta T)$. For the stated regular ordering $R\gg S\gg 1$, strain transport dominates skeleton transport and the bounded phase-change enthalpy contribution. More precisely, at fixed finite thermal strain number $\alpha h^2/\kappa$, the dimensionless energy equation isso its skeleton and integrated latent-heat corrections are small in $1/R$ and $S/R$ respectively. The leading strain-dominated mush temperature profile obeys $\kappa T^{\prime \prime }+\alpha z{T}^{\prime }=0$. Put $\delta =\sqrt{2\kappa /\alpha }$ and $H=h/\delta$. Integrating first for $T^{\prime }$ and then for $T$ givesThe error function profile follows from integrating a Gaussian, with $\mathrm{erf}x=(2/\sqrt{\pi }){\int }_0^x{e}^{-s^2}ds$. Matching the roof gradient gives the requested transcendental thickness relationFor prescribed $G>0$ and $\Delta T>0$ the right side decreases continuously from infinity to zero as $H$ increases, so it fixes one positive leading-order thickness. The resulting $h$ must still satisfy the assumed ordering and the phase-fraction constraints.

To find the product composition, integrate the exact solute balance from the roof to an arbitrary depth. With $f=1-\varphi$,At $z=0$, the exact floor composition is thereforeThe minus sign has a direct physical meaning: the horizontal outflow removes more salt than arrives at the same depth from a uniform-$C_0$ liquid, leaving a depleted solid product. Using the leading error function profile andyieldsThe full solid fraction follows by substituting this same profile into the preceding integral formula for $f(z)$. It has $\varphi (h)=0$ and $1-\varphi (0)=C_s/C_E$. A physical branch requires $0\le C_s\le C_E$ and $0\le \varphi \le 1$. In this monotone profile, $C_s\ge 0$ ensures a nonnegative liquid fraction everywhere; it is not permissible to interpret a negative formal result as a material composition.

The error-function reduction is a regular leading approximation, not a uniform statement for every possible scaling of the thermal strain number. If $\alpha h^2/\kappa$ becomes singularly large along the ordering limit, a thin basal layer can amplify the discarded terms. The printed two inequalities alone do not specify that extra scaling. In such a regime the displayed exact scalar equations and boundary conditions, rather than an unqualified zero-latent-heat profile, provide the appropriate thickness and composition calculation.

Let $K$ denote isotropic permeability of a porous medium, $\mu$ the fluid dynamic viscosity, ${\rho }_0$ the reference fluid mass density, $\alpha >0$ its volumetric thermal-expansion coefficient and $g$ gravitational acceleration. The Boussinesq approximation uses $\rho ={\rho }_0[1-\alpha (T-T_{\ast })]$ only in the gravitational force. With upward coordinate $z$, hydraulic pressure $P$ and Darcy flux $\mathbf{q}$, the equations arewhere $\theta$ is the perturbation to the conductive temperature. To keep porous heat-capacity conventions explicit, let $C_e$ be effective heat capacity per bulk volume, $C_f={\rho }_0{c}_f$ the liquid volumetric heat capacity, $k_e$ effective thermal conductivity, $\sigma =C_e/C_f$, and $\kappa =k_e/C_f$. The thermal equation is $\sigma T_t+\mathbf{q}\cdot \nabla T=\kappa {\nabla }^{2}T$. If the usual unit-capacity convention is adopted, $\sigma =1$ and $\kappa$ is the thermal diffusivity. Thresholds below do not depend on $\sigma$; it only changes temporal rates.

For a conductive state with upward temperature decrease $\Gamma =-{\overline{T}}_z>0$, linearization giveswhere $w=q_z$. The second equation follows by taking the divergence of Darcy law to eliminate $P$ and then differentiating its vertical component. It is important that the Laplacian on its right is horizontal.

In the horizontal layer, both boundary planes prescribe the temperature and are impermeable, so $\theta =w=0$ at $z=0,h$. Use a normal mode proportional to $e^{st+i\mathbf{k}\cdot \mathbf{x}}\sin (n\pi z/h)$, with horizontal wavenumber $k=|\mathbf{k}|$, $n\ge 1$. Writing $b=n\pi /h$, the momentum relation gives $w=B{k}^2\theta /(k^2+b^2)$, and the thermal relation givesA neutral mode therefore requires $B\Gamma /\kappa =(k^2+b^2{)}^2/k^2$. Since $k^2+2b^2+b^4/k^2\ge 4b^2$, the smallest value is $4{\pi }^2/h^2$, attained at $n=1$ and $k=\pi /h$. This derives the onset of convection in a horizontal Darcy layer rather than simply importing its critical number. ThusEvery small disturbance decays for $T_m<T_{m,\mathrm{crit}}$; at equality the critical roll is neutral. Strict decay has this threshold as a supremum, not an attained maximum. The first roll wavelength is $2h$, compatible with the very large horizontal extent. Cooling from above gives the opposite, stable buoyancy feedback.

For the tall square column, define $a$ as its side length and $H\gg a$ as its height. The insulated vertical walls impose ${\partial }_n\theta =0$, while impermeability imposes zero normal Darcy flux. Away from the remote ends, a nearly height-independent mode has hydrostatic pressure adjustment and $w=B\theta$, with zero cross-sectional mean. The transverse temperature modes arewhere $m,n\ge 0$ but $(m,n)=(0,0)$ is excluded: a closed column cannot support a net vertical discharge. Its horizontally uniform density perturbation is balanced by pressure, rather than driving overturning. The rate of a nonconstant mode isThe smallest allowed eigenvalue is ${\pi }^2/a^2$, so the insulated square-column Darcy convection onset isAgain strict decay requires $\Gamma <{\Gamma }_{\mathrm{crit}}$. This is the leading long-column result. End conditions that maintain the basic gradient require the vertical velocity to turn around near the ends and give small finite-height corrections; their exact coefficient cannot be inferred from insulation of the vertical walls alone. For a separated vertical mode of small wavenumber $k_z$, the neutral expression is $B\Gamma /\kappa =(k_{\perp }^2+k_z^2{)}^2/k_{\perp }^2$, which tends to the displayed value as $H/a\to \infty$.