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Work on a closed compact smooth Riemannian manifold $M$ without boundary, as in the usual discrete spectral setting here. Use the paper's $\Delta ={\mathrm{tr}}_g{\mathrm{Hess}}_g$, and put $P=-\Delta$. The Riemannian heat kernel $H(t,x,y)$ is the smooth kernel, for $t>0$, representing the solution ofThus $({\partial }_t-{\Delta }_x)H=0$ and its distributional initial value in $x$ is the Dirac delta distribution at $y$. The heat trace isThe heat invariants are the coefficients $a_r(M)$ in the short-time heat trace expansion specified below.

To derive its spectral form, take an orthonormal eigenbasis $P{\varphi }_j={\lambda }_j{\varphi }_j$, with $0={\lambda }_0\le {\lambda }_1\le \cdots$, repeating eigenvalues according to multiplicity. For the initial datum ${\varphi }_j$, the function $e^{-t{\lambda }_j}{\varphi }_j$ solves the heat equation. Uniqueness identifies it with the heat-kernel solution. Expanding arbitrary initial data in this basis therefore gives the spectral expansion of the Riemannian heat kernel:Standard elliptic estimates and eigenvalue growth imply smooth convergence for $t$ bounded away from zero: polynomial bounds on derivatives of eigenfunctions are dominated by the exponential factors. On the diagonal all summands are nonnegative, so monotone convergence also justifies integration without needing an interchange of conditionally convergent terms. Since each eigenfunction has unit $L^2$ norm,Equivalently, if the paper's Laplacian eigenvalues are written ${\nu }_j=-{\lambda }_j$, the formula is ${\sum }_j{e}^{t{\nu }_j}$; a growing exponential is not the heat semigroup.

Here are the needed heat kernel expansion facts, stated without proof. On a closed $d$-dimensional smooth Riemannian manifold,with the diagonal remainder uniform in $x$ after any finite truncation. The coefficients are smooth local geometric quantities; $u_1=R/6$. Compactness therefore allows integration term by term:These integrated coefficients are the heat invariants. The uniform leading remainder in particular gives $Z_M(t)=(4\pi t{)}^{-d/2}(\mathrm{Vol}(M)+O(t))$.

Isospectral manifolds have identical heat traces by the derived spectral formula. If their dimensions were $d<e$, multiplying the common trace by $t^{d/2}$ would give a finite positive limit on the $d$-dimensional manifold but divergence on the $e$-dimensional one. Thus the dimensions are equal, and equality of the leading coefficient gives equal Riemannian volume. Explicitly,The absence of a boundary fixes this particular integer-power expansion. With a boundary one must first specify boundary conditions and include the appropriate half-integer boundary terms; the leading dimension-and-volume conclusion remains valid in the usual compact smooth setting.

A spatially uniform pressure is a useful leading approximation when the sound speed makes the acoustic crossing time much shorter than the radiative cooling and thermal conduction times. The motions must remain at low Mach number, and gravity and other stresses must be negligible. Sound then redistributes pressure before appreciable thermal evolution occurs. Small pressure corrections can still drive the slow motion; imposing exactly zero pressure gradient in the full momentum equation at every order would be a stronger assumption.

The equation of state gives the specific internal energy of an ideal gas as $e=p/[(\gamma -1)\rho ]$. With $D/Dt={\partial }_t+v{\partial }_x$, the continuity equation is $D\rho /Dt=-\rho v_x$. Differentiating $e$ therefore yieldsInsert this into the internal energy equation, whose compressional term is $-p{v}_x$, and use spatially uniform pressure, so $Dp/Dt=p_t$. The result isFor temporally constant pressure put $A=\mu p/\mathcal{R}$, so the mass density is $\rho =A/T$. The continuity equation now gives $v_x=T^{-1}DT/Dt$. Multiplication of the last equation by $(\gamma -1)T/(\gamma p)$, with $\lambda ={\lambda }_0{T}^{\alpha }$, givesThis is the isobaric conductive cooling in mass coordinates transformation before changing coordinates.

There is a boundary assumption implicit in the proposed Lagrangian coordinate. For a mass integral from a fixed origin, integration of the continuity equation givesThus $m$ is material when the base mass flux vanishes, as appropriate for the stationary cold reservoir below. If the origin permits mass flux, replace $m$ by $m-{\int }^t\rho (0,s)v(0,s)ds$; this corrected label has zero material derivative. In a material mass coordinate, $D/Dt={\partial }_t{|}_m$ and ${\partial }_x=(A/T){\partial }_m$. In particularThe conduction coefficient therefore becomes $C{\lambda }_0$, with the same $C$ as the cooling coefficient. With $\tau =Ct$, the equation is

In the hot gas there is no volumetric radiative cooling. The initial temperature is uniform and the cold boundary introduces no finite mass length. The remaining nonlinear diffusion equation is invariant under $m\mapsto bm$, $\tau \mapsto b^2\tau$ with the same temperature scale. Its natural similarity solution is thereforeThe far-field condition also recovers the initial temperature at every fixed $m>0$ as $\tau \downarrow 0$. Substitution gives the ordinary differential equationFor $\alpha =1$, this reduces to $f^{\prime \prime }+\xi f^{\prime }/2=0$. Integrating first for $f^{\prime }$ gives $f^{\prime }=K\exp (-{\xi }^2/4)$. The two boundary conditions fix $K=1/\sqrt{\pi }$, hence the error-function isobaric cooling front isHere the error function has normalization $2/\sqrt{\pi }$, as printed in the original PDF; the converted TeX's $2/\pi$ is incorrect. The following original sketch shows the temperature profiles. The cooling layer broadens as $\sqrt{\tau }$, and at fixed positive mass label the temperature falls as time increases, while the remote hot gas remains at $T_0$.

For $\alpha =1$, the magnitude of the heat flux into the cold reservoir isAlthough the thermal conductivity tends to zero there, the physical temperature gradient diverges: their product has this finite limit. The stationary reservoir has no incoming advective energy flux and radiates the conducted energy immediately. Since $T_m(0,\tau )=T_0/\sqrt{\pi {\lambda }_0\tau }$, the radiated power per unit area isThus the requested exponent is $\boxed{k=1/2}$. The initial divergence is the usual idealization of an instantaneous temperature discontinuity in a diffusion equation.

Use the relative potential $\Psi ={\Phi }_{\ast }-\Phi$ and integrate the stated, untruncated Maxwell-Boltzmann velocity distribution over ${\mathbb{R}}^3$. The normalization cancels its three Gaussian integrals:The gravitational potential sign reverses under the relative-potential convention, so the spherical Poisson equation for Newtonian gravity isAlso ${\Psi }^{\prime }=-Gm(r)/r^2$, with $m^{\prime }=4\pi r^2\rho$. Differentiating the density relation yieldsAn ideal gas with constant temperature has $P=\rho k_{B}T/m_0$. Its hydrostatic equilibrium equation is precisely the same whenTogether with the same enclosed-mass equation, central density and boundary conditions, this proves the isothermal stellar and gas density equivalence. In dimensionless terms, $y=-[\Psi -\Psi (0)]/{\sigma }^2$ and $x=r\sqrt{4\pi G\rho (0)}/\sigma$ obey $x^{-2}(x^2{y}^{\prime }{)}^{\prime }=e^{-y}$ with $y(0)=y^{\prime }(0)=0$ in both systems.

This calculation uses the distribution as printed for all velocities. Imposing an additional cutoff at positive relative energy changes the integral to ${\rho }_1[e^W\mathrm{erf}\sqrt{W}-2\sqrt{W}/\sqrt{\pi }]$, where $W=\Psi /{\sigma }^2\ge 0$, and no longer gives an exactly isothermal gas profile. An untruncated self-gravitating isothermal sphere also needs an outer modification for finite total mass. The function $\mathrm{erf}$ here is the error function. Finally, the half-line Gaussian integral printed in the note lacks a factor: the correct value is ${\int }_0^{\infty }{e}^{-\alpha x^2}dx=\frac{1}{2}\sqrt{\pi /\alpha }$.

Work at one radius and write $\mathcal{R}=k/({\mu }_m{m}_p)$. Use the surface density to fix the mass density normalization and introduce the height scaleChoose the dimensionless variablesThe pressure and temperature scales make the hydrostatic and ideal gas equations parameter-free. The flux scale fixes the heating coefficient at $9/4$, and the choice of $h$ makes the diffusion coefficient unity. Direct substitution gives the universal ordinary differential equationsFor example, the dimensionless coefficient in the temperature equation is $3\alpha \kappa {\Sigma }^2{\mathcal{R}}^4/(16\sigma {\Omega }^5{h}^6)=1$.

Let the upper free surface be at $\zeta ={\zeta }_s$, so its physical semithickness is $H=h{\zeta }_s$. Midplane symmetry and the radiative-zero disk surface conditions giveThe last condition is precisely $\Sigma ={\int }_{-H}^H\rho dz$. The unknown surface position and the two positive midplane values $P(0),\Theta (0)$ are determined as part of the normalized boundary value problem. Symmetry makes $P,D,\Theta$ even and $Q$ odd; $P^{\prime }(0)={\Theta }^{\prime }(0)=0$ already follow from the equations and are not additional independent boundary conditions.

The zero conditions do not mean zero surface flux. Integrating the heating equation gives $Q({\zeta }_s)=(9/4){\int }_0^{{\zeta }_s}Pd\zeta >0$. Near a regular surface with finite outgoing flux, dividing the hydrostatic equation by the temperature equation yields $dP/d\Theta \simeq {\zeta }_s{\Theta }^3/Q_s$. Thus $P\propto {\Theta }^4$, $D\propto {\Theta }^3$ and $\Theta \propto {\zeta }_s-\zeta$, so mass density also vanishes. The apparent singular quotient in the diffusion equation has a finite regular limit. All boundary conditions and equations are independent of the physical parameters: this is homologous vertical structure of a constant-opacity alpha disk.

Let the perturbation phase be $\vartheta =\mathbf{k}(t)\cdot \mathbf{r}$ and write $\psi ={\psi }_0+\mathrm{Re}[\Pi (t)e^{i\vartheta }]$. Incompressible flow imposes $\mathbf{k}\cdot \mathbf{v}=0$. Applying the background advective derivative to the phase givesCancel this position-dependent phase by choosingThe perturbation acting on the basic flow gives $({\mathbf{u}}^{\prime }\cdot \nabla ){\mathbf{u}}_0=-2A{u}_x^{\prime }{\mathbf{e}}_y$. Its self-advection vanishes exactly: the physical velocity ${\mathbf{u}}^{\prime }$ is perpendicular to $\mathbf{k}$, while every spatial derivative of ${\mathbf{u}}^{\prime }$ is proportional to a component of $\mathbf{k}$. Thus ${\mathbf{u}}^{\prime }\cdot \nabla {\mathbf{u}}^{\prime }=0$, including the products between the complex mode and its conjugate. Substituting into the full Navier-Stokes equation therefore gives an exact incompressible shearing wave, with no small-amplitude approximation:In components these equations areDifferentiating the transversality constraint closes the pressure:The coefficient $4A$ includes both the shear term in the momentum equation and $\dot{\mathbf{k}}\cdot \mathbf{v}$; treating the wavevector as fixed would miss one contribution. Initial data with $\mathbf{k}(t_0)\cdot \mathbf{v}(t_0)=0$ remain transverse under this closed evolution. These shearing waves are exact in an unbounded local shear flow or with compatible shearing-periodic boundaries; arbitrary rigid boundary conditions need not admit a single such wave.

Suppress the common Fourier mode factor and define the transverse and full divergences of the fluid Lagrangian displacement byThe perturbed magnetic field is $\mathbf{Q}=({\mathbf{B}}_0\cdot \nabla )\mathbf{\xi }-(\mathbf{\xi }\cdot \nabla ){\mathbf{B}}_0-{\mathbf{B}}_{0}D$. Its components areThus the magnetic part of the energy integrand isUse $p_0^{\prime }=-B_0{B}_0^{\prime }/(4\pi )$ from force balance. The pressure-gradient cross term isThe current term has the azimuthal cross-product component ${\xi }_z^{\ast }{Q}_r-{\xi }_r^{\ast }{Q}_z$, givingEvery term involving $B_0^{\prime }$ cancels in the sum, including the possibly complex ${\xi }_z^{\ast }{\xi }_r$ terms. The magnetohydrodynamic energy principle therefore reduces exactly toFor physical $p_0\ge 0$ and positive adiabatic index $\gamma$, every surviving term is nonnegative. On the admissible displacement domain for the supplied functional, squared ideal-mode frequencies cannot be negative. This is the positive ideal energy of a theta pinch, proving absence of exponentially growing ideal modes. The qualification is stable or marginal, not strictly positive energy for every motion: at $k=0$, suitable incompressible displacements can have zero energy. As usual, the energy principle presumes regularity and boundary conditions eliminating unwanted surface terms; additional external systems are not part of the given functional.

For $k>0$, put $q=\sqrt{k^2+l^2}$ and define the Squire transformationThe two-dimensional Orr-Sommerfeld equation then has the same transverse operator because $k^{\prime 2}=q^2$, and the same viscous coefficient because $k^{\prime }{R}^{\prime }=kR$. Its boundary conditions on $v$ and $v^{\prime }$ are unchanged. Thus it has the same eigenfunction and the same phase-speed eigenvalue.

If the original disturbance grows, $\mathrm{Im}c>0$, so the transformed disturbance also grows. Since $R^{\prime }\le R$, every unstable three-dimensional mode maps to an unstable two-dimensional mode at a no larger Reynolds number. It follows that the minimum critical Reynolds number is attained among two-dimensional disturbances. This is the Squire theorem for temporal normal-mode onset, not a statement that three-dimensional transient amplification is impossible. The case $k=0$ is treated separately without dividing by $k$; its streamwise-uniform viscous normal modes decay, although coupling can produce transient growth.

Let $q^2=k^2+l^2$ and $c=c_r+i{c}_i$, with $k>0$ chosen for the temporal mode convention. In the inviscid limit, the Orr-Sommerfeld equation reduces to the Rayleigh equation for inviscid shear flow:The impermeability boundary condition is $v(\pm 1)=0$; the extra viscous derivative condition is not an independent condition for this reduced second-order equation. For an unstable mode $c_i>0$, the denominator has no real zero. Multiply by $\bar{v}$ and apply integration by parts to obtainIts imaginary part givesFor a nontrivial mode, the weight is nonnegative and is positive except at isolated mode zeros. Thus $U^{\prime \prime }$ must take both signs, unless it vanishes identically. The identically zero case also has no unstable nontrivial mode, by the positive first integral in the displayed identity. Therefore the smooth real base-flow profile must have an interior inflection point. This proves Rayleigh's inflection-point theorem; it is a necessary condition, not a sufficiency test.

Let $m=\dim X$, $n=\dim Y$, and $x\in f^{-1}(y)$. In coordinate charts centered at $x$ and $y$, write the map as $F:{\mathbb{R}}^m\to {\mathbb{R}}^n$. Surjectivity of its derivative gives an invertible $n\times n$ minor; reorder the domain coordinates so it is the first one. The maphas invertible derivative at the point. By the inverse function theorem, it is a local coordinate change. In these coordinates the fiber is simply the set whose first $n$ coordinates are zero, so it is locally a coordinate plane of dimension $m-n$. These charts are inherited from $X$, provingIf the fiber is empty, it is the empty submanifold under the usual convention. The no-boundary assumptions avoid any extra boundary conditions in this local model.

For a normalized trial state in the quadratic-form domain of a self-adjoint Hamiltonian operator bounded below, its expectation of $H$ is at least the ground energy $E_0$. Minimizing this Rayleigh quotient over a tractable trial family therefore supplies an upper bound.

For the two states, set $a_{\pm }=(a_1\pm a_2)/\sqrt{2}$. Their quotient isThe strict assumption $ϵ<sE$ excludes $|s|=1$, since equality in the overlap Cauchy-Schwarz bound would then force the two states to be proportional and $ϵ=sE$. Thus $|s|<1$. The two generalized eigenvalues are $(E\pm ϵ)/(1\pm s)$, and their difference has the sign of $ϵ-sE$. The symmetric trial is the lower one, so

For the single attractive delta well, the supplied wavefunction has norm one because $\int \lambda e^{-2\lambda |x|}dx=1$. Its kinetic energy in quadratic-form notation is ${\hslash }^2\int |{\psi }_1^{\prime }{|}^2/(2m)={\hslash }^2{\lambda }^2/(2m)$, and its delta potential energy is $-K|{\psi }_1(0){|}^2=-K\lambda$. With $\lambda =mK/{\hslash }^2$ this givesIts derivative jump is ${\psi }_1^{\prime }(0+)-{\psi }_1^{\prime }(0-)=-2\lambda {\psi }_1(0)$, exactly the delta boundary condition, so it is an eigenstate. A decaying bound state on the two half-lines must have a common decay rate fixed by this same jump; there is only this one negative-energy state, confirming it is the ground state.

For $R>0$, integrate the overlap on the three regions $x<0$, $0<x<R$, $x>R$. Put $a=\lambda R$. The result isThe second equality is the single-well energy plus the extra delta expectation; the third follows by applying either single-well eigenvalue equation in the cross matrix element. Moreover $ϵ-sE=K\lambda e^{-a}[(1+a)e^{-2a}-1]<0$, so the preceding bound applies. Substitution givesFor widely separated wells, each localized orbital is close to an isolated-well ground state and the low-energy subspace is accurately their span; the symmetric combination captures the leading tunneling energy lowering. In the limit both exact and trial energies approach $E_B$. This illustrates electronic molecular binding: sharing an orbital between attractive centers can lower its energy below the separated-center value. A full equilibrium bond length would also require the repulsion of the positive centers.

Take the nonnegative scalar Laplace-Beltrami operator $\Delta =-\mathrm{div}\nabla$ on a closed compact Riemannian manifold, and include every eigenvalue with its multiplicity. The spectrum contains substantial global geometric information, but generally does not determine the Riemannian metric up to isometry. Specifying the operator and its domain matters: the spectrum on functions is different information from the spectra of the Hodge Laplacian on differential forms, and a boundary requires fixed boundary conditions.

The Weyl law makes dimension and volume audible. If $N(\lambda )$ counts eigenvalues at most $\lambda$, thenwhere ${\omega }_d$ is the volume of the Euclidean unit ball. Thus the growth exponent recovers $d$, and its leading coefficient recovers $\mathrm{Vol}(M)$. The multiplicity of zero gives the number of connected components: a function in the kernel has $\int |\nabla f{|}^2=0$ and hence is constant on each component.

More detailed invariants come from the heat trace, which is determined directly by the spectrum:The first heat invariants for a closed manifold areHere $R$ is scalar curvature, $\mathrm{Ric}$ the Ricci curvature, and $\mathrm{Rm}$ the Riemann curvature tensor. A divergence term in the local coefficient integrates to zero on a closed manifold. The expansion supplies further integrated curvature expressions, rather than a pointwise curvature map.

On a closed surface, $R=2K$, $|\mathrm{Ric}{|}^2=2K^2$ and $|\mathrm{Rm}{|}^2=4K^2$, so the same coefficients becomeThe Gauss-Bonnet theorem gives $\int KdA=2\pi \chi (M)$; hence the constant term of the heat trace is $\chi (M)/6$. In particular the Euler characteristic, and for a connected orientable surface its genus, are spectrally determined. There is even spectral determination of hyperbolic curvature on a surface: if this spectrum agrees with that of a curvature-$-1$ surface, the coefficients give $\int K=-A$ and $\int K^2=A$, whenceThus $K=-1$ everywhere. This determines local curvature, while leaving room for globally nonisometric hyperbolic surfaces.

Geodesic information is also accessible in important settings. For closed hyperbolic surfaces, the Selberg trace formula makes the scalar spectrum equivalent to the unmarked length spectrum with its appropriate multiplicities. This does not supply a marked length spectrum: it does not identify which group element or intersecting curve has a given length. For general manifolds, one must not simply assert this equivalence; the singularities of the wave trace can be affected by degeneracies and cancellations.

The historic development separates these positive results from the inverse problem. Weyl's 1911 eigenvalue asymptotics established the leading geometric information. The Minakshisundaram-Pleijel heat expansion of 1949 brought local curvature into the short-time analysis. In 1964 Milnor exhibited nonisometric flat tori in dimension sixteen with identical scalar spectra, showing that all eigenvalues together need not recover the metric. Kac's 1966 question about a vibrating drum popularized the corresponding inverse problem for planar domains. Vignéras's 1980 hyperbolic examples and the Sunada theorem of 1985 extended the supply of counterexamples; the latter turns Gassmann equivalence of finite subgroups into a geometric construction. The four-dimensional tetracode length-isospectral lattice construction and the planar examples of Gordon, Webb and Wolpert in 1992 made nonuniqueness explicit in smaller dimensions and in the original drum setting.

Nor is scalar isospectrality a universal topological invariant strong enough to determine the manifold. For example, Tetra and Didi are scalar-isospectral closed orientable flat three-manifolds with first Betti numbers one and zero. Their topology therefore differs. The spectra of all Hodge Laplacians would detect these Betti numbers through the dimensions of their zero eigenspaces, illustrating the extra information obtained by changing the operator. Dimension, volume and many integrated invariants are determined; the full metric and, in general, the topology are not.

An order parameter distinguishes phases by a macroscopic quantity that changes at the transition. For an Ising model ferromagnet it is the magnetization per Ising spin, $M=N^{-1}{\sum }_n⟨{\sigma }_n⟩$. At zero magnetic field the statistical Hamiltonian is invariant under spin inversion symmetry. The disordered phase preserves this symmetry, whereas a selected ordered phase has $M\ne 0$ and a spin-reversed partner with $-M$. At finite volume with symmetric boundary conditions the expected value is zero even below the transition; spontaneous magnetization means taking the thermodynamic limit before sending a symmetry-breaking field to zero.

Landau-Ginzburg theory introduces a coarse-grained local order parameter $\varphi (x)$ and a symmetry-constrained free-energy functional. For a short-range system with scalar spin inversion symmetry, a useful expansion isThe coefficients vary smoothly with microscopic control parameters at the level of the Landau approximation; normally $r=A(T-T_c)$ with $A>0$. The gradient term penalizes spatial variation. The magnetic field is conjugate to the order parameter, and odd powers at $h=0$ are excluded by spin inversion symmetry. With $u>0$, the sextic term may be omitted locally; with $u<0$, a stabilizing positive $v$ is essential. The equilibrium uniform order parameter minimizes the local potential $V(M)$, givingThis equation of state follows from minimization, rather than being imposed as an independent assumption. Vector or tensor order parameters and their allowed invariants describe other symmetry classes.

Take $L=-d^2/d{x}^2+x$. The Airy equation is $Lu=0$. To discuss positivity of this linear operator, use its homogeneous variation domain $D(L)=\{v\in H^2(0,1):v(0)=0,v^{\prime }(1)=0\}$. Integration by parts givesThe boundary term is zero at both ends. The same calculation with two different functions proves that the associated form is a symmetric bilinear form. Positivity here concerns the homogeneous domain; the condition $u(0)=1$ instead defines an affine set of admissible solutions.

For the variational problem defineand minimizeThe Sobolev trace at zero is meaningful in $H^1$. No derivative boundary value is imposed on this trial space: the Neumann boundary condition will emerge naturally. For $v\in V$, Cauchy-Schwarz inequality and integration giveso $a(v,v)\ge \frac{2}{3}\Vert v{\Vert }_{H^1}^2$. Thus $a$ is a bounded, symmetric, coercive bilinear form on the Hilbert space $V$. The Lax-Milgram theorem says that a bounded coercive bilinear form and a bounded linear functional determine a unique weak solution. Apply it toand set $u=1+v$. This is precisely the first variation condition $a(u,w)=0$ for $J$.

Compactly supported test functions first yield $u^{\prime \prime }=xu$ in the sense of distributional derivatives. Since $xu\in L^2$, the solution is in $H^2$. Integration by parts then leaves $u^{\prime }(1)w(1)=0$ for every $w\in V$, hence $u^{\prime }(1)=0$. The other boundary condition is built into $1+V$. Conversely a solution of this boundary value problem satisfies the first variation condition. Finally,This proves existence, uniqueness and the global minimum characterization, including the natural boundary condition. It is the mixed-boundary Airy energy principle.

Put $a_h^2=k^2+m^2>0$ and $\Lambda =a_h^2+{\pi }^2$. The plane-layer poloidal-toroidal decomposition supplies the assumed poloidal fields asSince $w=-{\Delta }_h\varphi$ and $b_z=-{\Delta }_h\chi$, potentials proportional to $e^{st+ikx+imy}\sin \pi z$ have the required vertical components. Their horizontal components are proportional to $\cos \pi z$, satisfying the stress-free boundary conditions and the horizontal-field boundary conditions. The double-curl representation automatically enforces incompressibility and the solenoidal magnetic-field constraint.

Let $W,\Theta ,C$ be the amplitudes of $w,\theta ,b_z$. The temperature and resistive induction equations immediately giveEliminate the total-pressure perturbation by applying $\Delta$ to the vertical momentum equation and subtracting ${\partial }_z$ times its divergence. The resulting vertical equation isIn particular, $imC=-m^{2}W/(s+\zeta \Lambda )$ shows explicitly that magnetic tension opposes the displacement.

Taking the determinant of these three amplitude equations gives the horizontal-field magnetoconvection dispersion relation:This cubic polynomial includes all roots of the assumed poloidal coupled system; using its determinant form avoids discarding possible roots by dividing by $s+\Lambda$ or $s+\zeta \Lambda$. As a check, $Q{m}^2=0$ factors off magnetic diffusion and leaves the standard stress-free Rayleigh-Bénard convection growth equation.

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 ${\partial }_z{b}_x={\partial }_z{b}_y=0$.

Let ${\rho }_0$ be a constant reference density and define total buoyancy acceleration by $\sigma =-g(\rho -{\rho }_0)/{\rho }_0$. Absorb the reference hydrostatic pressure into a kinematic pressure $p$. The ideal nonrotating Boussinesq equations areFor a resting density profile ${\rho }_b(z)$, the background satisfies $p_b^{\prime }(z)={\sigma }_b(z)$ and the squared buoyancy frequency isStable stratification has $N^2>0$. The Boussinesq approximation requires fractional density variations to be small over the domain and parcel excursions, so density can be replaced by ${\rho }_0$ in inertia while its small variation is retained in the gravitational force. For an approximately uniform density gradient over height $H$, this requires $N^{2}H/g\ll 1$. The stipulated incompressibility alone does not justify neglecting density variation in inertia.

Put $b=\sigma -{\sigma }_b$ and $p^{\prime }=p-p_b$. To linear order the two-dimensional internal gravity wave equations areThe exact impermeability condition on $z=\eta (x,t)$ is the kinematic boundary conditionTaylor expansion about $z=0$ permits $w(x,0,t)={\eta }_t$ at leading order if the boundary slope and its displacement relative to the vertical wave scale are small: $|kϵ|\ll 1$ and $|mϵ|\ll 1$. The second condition controls both the omitted evaluation correction $\eta w_z$ and the product $u{\eta }_x$, since continuity gives $u\sim (m/k)w$. Hence for the harmonic forcing,The propagating calculation assumes $k\ne 0$ and $0<|\omega |<N$; a time-independent boundary does no oscillatory work on fluid initially at rest. If $k=0$ but $\omega \ne 0$, incompressibility forces the vertical velocity to be independent of height, so the unbounded column has no finite-energy radiating internal-wave solution. The propagating wave-generation problem therefore needs the stated nonzero horizontal wavenumber.