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For Keplerian rotation, initial $B_R\propto R^{-1}$ and thermal pressure $p\propto R^{-2}$, axisymmetric magnetic winding gives $B_{\varphi }=-(3/2)\Omega t{B}_R$. If the initial plasma beta is ${\beta }_p>1$, the formal equality of magnetic pressure and gas pressure occurs atThe magnetic-to-thermal pressure ratio decreases with radius, so magnetic pressure dominates inside this radius. In a disk with a finite radial range, an equality radius exists in the disk only when this formal radius lies within that range.

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.

In the Gaussian units of the paper, the total magnetic pressure is $p_B=(B_R^2+B_{\varphi }^2)/(8\pi )$. Define the initial plasma beta ${\beta }_p=8\pi p_0/B_0^2\gg 1$, assuming a nonzero initial field. The solution from part (b) givesFor any $t>0$ this ratio decreases strictly with $R$, tends to infinity as $R\downarrow 0$ and tends to $1/{\beta }_p<1$ as $R\to \infty$. The magnetic-pressure equality radius under Keplerian winding is therefore unique in the formal profile on $R>0$:The same expression proves $p_B>p$ for $R<R_{\mathrm{eq}}$ and $p_B<p$ for $R>R_{\mathrm{eq}}$. At equality, $|B_{\varphi }|/|B_R|=\sqrt{{\beta }_p-1}\gg 1$, justifying the toroidal-dominated approximationThe source's existence claim refers to the indefinitely extended radial profile and the stated neglect of back-reaction. For a disk with a finite inner and outer edge, an equality radius lies in the disk only while the displayed value lies between those edges. If the initial field were zero, no magnetic winding or equality radius would occur.

Here $\beta =p_{\mathrm{rad}}/p_{\mathrm{gas}}$ is the radiation-to-gas pressure ratio, not plasma beta. Write the opacity as ${\kappa }_{\mathrm{op}}$ to distinguish it from the radial epicyclic frequency. The radiation pressure is $p_{\mathrm{rad}}=4\sigma T^4/(3c)$, so radiative diffusion can be written asSince $\beta$ is independent of height, $p_{\mathrm{rad}}=\beta p/(1+\beta )$. Substitution of hydrostatic equilibrium, $p^{\prime }=-\rho {\Omega }^{2}z$, givesDifferentiating and using the viscous heating equation, together with $r{\Omega }^{\prime }=-q\Omega$, yieldsThus the effective dynamic viscosity $\rho \nu$ is independent of height. If the surfaces are at $z=\pm H$, the vertically integrated kinematic viscosity satisfiesUse the previous result and the specified Eddington accretion rate convention,to obtain the full thicknessThe constant-pressure-ratio vertical disk model therefore has thickness proportional to the inward accretion rate. This is a full surface-to-surface thickness, rather than a density disk scale height. Its use as a thin disk requires $2H\ll r$.

Let $c_s^2\sim P_g/\rho$. The assumed magnetic pressure givesso the plasma beta obeys $\boxed{\beta =P_g/P_m\sim {\mathrm{Ma}}^{-1}}$. In the magnetically dominated regime, vertical hydrostatic equilibrium gives $P_m/\rho \sim {\Omega }^2{H}^2$. Combining this with $P_m/\rho \sim r\Omega c_s$ yields

The radiation-to-gas pressure ratio compares radiation pressure with gas pressure. It is distinct from plasma beta, which compares gas pressure with magnetic pressure. Both are sometimes denoted by $\beta$, so the convention must be stated.