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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.

Use complex Fourier amplitudes and let primes denote $z$ derivatives. Put $C=B^2/{\mu }_0$, $T=\gamma p+C$ and $d=i{k}_x{\xi }_x+i{k}_y{\xi }_y+{\xi }_z^{\prime }$. Static balance gives $(p+B^2/(2{\mu }_0){)}^{\prime }=-\rho g$. The perturbations from part (a) then implyFor the last identity, combine $\delta p$ with $B\delta B_y/{\mu }_0$ and use static balance; this is where the background magnetic-pressure gradient cancels.

Multiply the Fourier momentum equation by $-{\xi }^{\ast }$ and integrate from $a$ to $b$. Integrating the pressure-gradient term by parts gives $[{\xi }_z^{\ast }\delta \Pi {]}_a^b-\int d^{\ast }\delta \Pi dz$. The gravity term is $\int [-g{\rho }^{\prime }|{\xi }_z{|}^2-\rho g{\xi }_z^{\ast }d]dz$. The magnetic terms reduce toIndeed the term from $\delta B_z{B}^{\prime }{e}_y$ cancels exactly the $B^{\prime }$ term arising from $i{k}_{y}B\delta B$. This establishes, before completing the square,Now $\delta \Pi =A-Td$, soThus the magnetohydrodynamic energy principle has the requested quadratic formAll squares are squared complex moduli. With the prescribed vanishing surface term the quadratic form is real, as required for the self-adjoint stability problem.

Self-adjointness gives $⟨\mathbf{\xi },F\mathbf{\xi }⟩=⟨F\mathbf{\xi },\mathbf{\xi }⟩=⟨\mathbf{\xi },F\mathbf{\xi }{⟩}^{\ast }$, soThe integrated expression from part (a) also writes it explicitly asThis is the quadratic change in potential energy associated with the perturbation, including pressure, magnetic and imposed-gravity restoring effects. Its sign need not be positive term by term.

The perturbation kinetic energy is $T=\frac{1}{2}⟨\dot{\mathbf{\xi }},\dot{\mathbf{\xi }}⟩$, and the total quadratic perturbation energy isSince the equilibrium and $F$ are time independent, differentiating and using the self-adjoint operator identity givesThus $\boxed{dE/dt=0}$ along the linearized motion. This conserved kinetic-plus-restoring energy is the basis of the magnetohydrodynamic energy principle.

First suppose $W[\mathbf{\xi }]>0$ for every nonzero admissible displacement. The conserved energy from part (b) then bounds both $T$ and $W$ by $E$. In particular,where $\Vert \mathbf{\xi }{\Vert }_{\rho }^2=⟨\mathbf{\xi },\mathbf{\xi }⟩$. The positive quadratic form $\Vert \dot{\mathbf{\xi }}{\Vert }_{\rho }^2+2W[\mathbf{\xi }]$ defines the perturbation energy norm and is conserved. Small perturbations in this norm remain small, proving energetic linear stability. Also $\Vert \mathbf{\xi }(t){\Vert }_{\rho }\le \Vert \mathbf{\xi }(0){\Vert }_{\rho }+\sqrt{2E}t$, excluding exponential displacement growth. For any admissible normal mode $\mathbf{\xi }=\mathbf{\zeta }{e}^{-i\omega t}$, multiplication of $-{\omega }^2\mathbf{\zeta }=F\mathbf{\zeta }$ by ${\mathbf{\zeta }}^{\ast }$ gives directlyso all such frequencies are real. No variational principle has been assumed.

To prove instability, suppose some admissible ${\mathbf{\xi }}_0$ has $W_0=W[{\mathbf{\xi }}_0]<0$, and put $N_0=\Vert {\mathbf{\xi }}_0{\Vert }_{\rho }^2>0$. Choose the initial velocityIts kinetic energy exactly cancels $W_0$, so $E=0$ for the entire subsequent motion. Write $N(t)=\Vert \mathbf{\xi }(t){\Vert }_{\rho }^2$. Since $F$ is self-adjoint,Zero energy gives $⟨\mathbf{\xi },F\mathbf{\xi }⟩=\Vert \dot{\mathbf{\xi }}{\Vert }_{\rho }^2$, and hence $N^{\prime \prime }=4\Vert \dot{\mathbf{\xi }}{\Vert }_{\rho }^2\ge 0$. Initially $N^{\prime }(0)=2\alpha N_0>0$, so $N$ stays positive and its logarithm is defined. The Cauchy-Schwarz inequality now givesTherefore $(\ln N{)}^{\prime }(t)\ge (\ln N{)}^{\prime }(0)=2\alpha$, and integration provesThis is an exponentially growing perturbation. Scaling the initial displacement and velocity down leaves $\alpha$ unchanged, so arbitrarily small initial perturbations display the same linear growth mechanism. It proves that a negative-energy direction gives exponential growth in a self-adjoint wave equation, without assuming a discrete spectrum or quoting a trial-function variational theorem.

Positive $W$ on nonzero displacements gives energetic stability; a negative direction gives instability. For mathematical precision, a uniform bound on the mass-weighted displacement norm follows if $W\ge \frac{1}{2}{\omega }_{\min }^2\Vert \mathbf{\xi }{\Vert }_{\rho }^2$ for some ${\omega }_{\min }>0$, giving $\Vert \mathbf{\xi }{\Vert }_{\rho }^2\le 2E/{\omega }_{\min }^2$. Strict positivity alone in an infinite-dimensional space need not supply such a uniform spectral gap; the standard magnetohydrodynamic energy principle asserts energetic stability and absence of growing modes. Cases with zero directions are marginal and require separate consideration.

Write $c_s^2=p/\rho$ for the constant isothermal sound speed squared and $v_a^2=B^2/({\mu }_0\rho )$ for the constant Alfvén speed squared. The horizontal magnetic field has no vertical magnetic tension; magnetostatic equilibrium isConsequently the magnetized isothermal atmosphere hasThe sign of $B_0$ specifies the orientation of the magnetic field. The adiabatic sound speed is $v_s^2=\gamma p/\rho =\gamma c_s^2$, also constant.

For the proposed normal mode, let $A=\exp (i\mathbf{k}\cdot \mathbf{x}+z/(2H)-i\omega t)$, $a=1/(2H)$, and temporarily write $\mathbf{\xi }$ for its constant complex amplitude. The divergence amplitude is $\Delta =i{k}_x{\xi }_x+i{k}_y{\xi }_y+(i{k}_z+a){\xi }_z$. Using ${\rho }^{\prime }/\rho =-1/H$ and $B^{\prime }/B=-a$, the Eulerian and Lagrangian fluid perturbations giveSince ${\Pi }^{\prime }/\rho =-g$, the magnetohydrodynamic total pressure amplitude isThe vertical derivative of $\delta \Pi =\rho A\psi$ is $\rho A(i{k}_z-a)\psi$. For the magnetic tension perturbation, the terms proportional to $ai{k}_y{\xi }_z{\mathbf{e}}_y$ from $\delta \mathbf{B}\cdot \nabla \mathbf{B}$ and $\mathbf{B}\cdot \nabla \delta \mathbf{B}$ cancel, leavingThus the linearization really reduces to the constant-coefficient eigenvalue problem

For a static ideal magnetohydrodynamics equilibrium with conservative gravitational acceleration, the magnetohydrodynamic energy principle gives a Hermitian operator for the restoring force in the mass density weighted inner product, provided the surface terms vanish. The factor $e^{z/(2H)}$ removes the exponential mass density weight. This can be checked directly, without assuming stability: set $s=v_s^2$, $b=v_a^2$, and $q=s+b$. Then ${\omega }^2\mathbf{\xi }=\mathsf{K}\mathbf{\xi }$, whereThis is a Hermitian matrix for real $\mathbf{k}$ and real positive $s,b$. Therefore every squared frequency ${\omega }^2$ is real; negative values give exponential growth, rather than an oscillation with a complex squared frequency. In an infinite atmosphere these are generalized plane waves, not finite-total-energy global modes.

At ${\omega }^2=0$ and $k_y\ne 0$, the $y$ equation, after cancelling its magnetic terms, gives $s\Delta =g{\xi }_z$. For nonzero magnetic field, the $x$ equation also gives $i{k}_x{\xi }_x=k_x^2\psi /(k_y^{2}b)$. Substituting $\psi =-b[\Delta -i{k}_y{\xi }_y]+g{\xi }_z-s\Delta$ and then $s\Delta =g{\xi }_z$ yieldsFor a nontrivial mode ${\xi }_z\ne 0$; otherwise these equations and $k_{y}b\ne 0$ force every component to vanish. Substitution in the vertical equation givesFor example, the determinant provides an independent check:

To prove the claimed “if and only if”, rather than merely locating a zero eigenvalue, use the magnetic buoyancy energy criterionIf $gH\le s$, every term is nonnegative, so all eigenvalues are nonnegative for every wavevector, including $k_y=0$. If $gH>s$, choose $k_z=0$, any fixed $k_x\ne 0$, and ${\xi }_z=1$, ${\xi }_x=-a/(i{k}_x)$, ${\xi }_y=g/(si{k}_y)$. The first two squares vanish; the remaining value iswhich is negative for sufficiently small nonzero $k_y$. The Rayleigh quotient of this displacement is therefore negative, proving a negative eigenvalue and an unstable normal mode. This uses precisely the assumption that boundary conditions do not exclude the chosen wavelengths. Equality is a neutral threshold, not exponential growth.

Finally the plasma beta is $\beta =p/[B^2/(2{\mu }_0)]=2c_s^2/v_a^2$. HenceThe magnetic field is assumed nonzero in the marginal-mode elimination and in the finite $\beta$ expression. The positive-sound-speed, nonmagnetic limit remains covered by the energy identity; for the usual $\gamma >1$ it is stable. Parker instability occurs because magnetic pressure supports part of the atmosphere's weight while the gas can drain along gently bent magnetic field lines. The resulting buoyant displacement can release Newtonian gravitational potential energy; choosing long wavelengths along the magnetic field makes the stabilizing magnetic tension small. The completed-square calculation quantifies this competition, including the adiabatic process restoring force.

For an admissible Fourier fluid Lagrangian displacement $\mathbf{\xi }(r)e^{im\theta +ikz}$, put $D_{\perp }=(r{\xi }_r{)}^{\prime }/r+im{\xi }_{\theta }/r$ and $D=D_{\perp }+ik{\xi }_z$. Force balance cancels every magnetic-field-gradient cross term in the magnetohydrodynamic energy principle, leaving an integrand $\gamma p|D{|}^2+B^2[|D_{\perp }{|}^2+k^2(|{\xi }_r{|}^2+|{\xi }_{\theta }{|}^2)]/(4\pi )$. For $p\ge 0$, $\gamma >0$, this is nonnegative. Neutral displacements can exist, so stability here means absence of negative-energy exponentially growing ideal modes, not strictly positive energy for every displacement.

For a horizontal field, set $C=B^2/{\mu }_0$. Static force balance gives $\delta \Pi =\rho g{\xi }_z-(\gamma p+C)\nabla \cdot \xi +Ci{k}_y{\xi }_y$. Given smooth compactly supported vertical and along-field displacements, the displayed choice makes total-pressure perturbation zero while ${\xi }_x=O(k_x^{-1})$. The nonnegative pressure and transverse-tension terms vanish as $|k_x|\to \infty$. This constructs the infimum needed for the magnetohydrodynamic energy principle; a strictly negative limit is already negative at some finite wavenumber.