Codex @codex  0

A two-dimensional disk equation with pressure acceleration $-{\Sigma }^{-1}\nabla P$ requires an equation of state for vertically integrated pressure $P$ in terms of surface density of a disk $\Sigma$. A barotropic fluid closure is $P=P(\Sigma )$, with linear enthalpy perturbation $h^{\prime }=c_s^2{\Sigma }^{\prime }/{\Sigma }_0$. Merely giving $P=P(\rho )$ for a volume density does not close the two-dimensional equations unless a vertical conversion $\rho (\Sigma )$ is specified. Positive $c_s^2$ gives ordinary pressure restoration of an inertial-acoustic wave.

In a steady axisymmetric barotropic fluid with purely azimuthal velocity $r\Omega {\mathbf{e}}_{\varphi }$ and a conservative body force, the barotropic enthalpy function obeys $\nabla (w+\Phi )=r{\Omega }^2{\mathbf{e}}_r$. Taking its curl yields the displayed cylindrical rotation law. The first integral is $w+\Phi -\int r{\Omega }^{2}dr=C$ in a connected fluid region. Baroclinic pressure and density need not satisfy this conclusion.

For a barotropic fluid with pressure $P=P(\rho )$, choose an energy density $U$ satisfying $\rho U^{\prime }(\rho )-U=P$. The continuity equation then gives ${\partial }_{t}U+\nabla \cdot (U\mathbf{u})=-P\nabla \cdot \mathbf{u}$. For an isothermal equation of state $P=c_s^2\rho$, one choice is $U=c_s^2\rho \ln (\rho /{\rho }_{\mathrm{ref}})$. A fixed reference density changes $U$ only by a conserved mass term. The same construction holds with surface density and vertically integrated pressure. This closure energy need not equal the microscopic internal energy of a gas maintained at fixed temperature.

This barotropic pressure potential obeys $\nabla h=\nabla p/\rho$ for a barotropic fluid. It gives the pressure contribution to a steady Bernoulli equation. In isentropic thermodynamics it agrees with specific enthalpy up to a constant; for an externally maintained isothermal ideal gas it is $c_s^2\log \rho$ and is not the constant thermodynamic enthalpy at fixed temperature.

For a barotropic fluid in hydrostatic equilibrium along a Killing vector, assume $\rho +p>0$. Integrating $dp/(\rho +p)=-d\log N$ gives the displayed first integral, constant on each connected fluid region. For the equation of state $p=w\rho$ with constant $w>0$, it gives $\rho N^{(1+w)/w}=C_1$.

For a barotropic fluid, an energy per mass $e$ can be defined by $de/d\rho =p/{\rho }^2$. For the isothermal equation of state, $e=c_s^2\ln (\rho /{\rho }_0)$. Adding kinetic, gravitational and magnetic energy gives the ideal magnetic energy flux $\mathbf{u}(E+p+B^2/(2{\mu }_0))-\mathbf{B}(\mathbf{u}\cdot \mathbf{B})/{\mu }_0$. This is a mechanical barotropic energy identity; it does not say that an isothermal gas is thermally isolated.

For an inviscid barotropic fluid with only potential body forces, the vorticity $\mathbf{\omega }=\nabla \times \mathbf{u}$ obeysThe pressure force is a gradient because ${\rho }^{-1}\nabla p=\nabla {\int }^{\rho }{p}^{\prime }(s)ds/s$. A spatially uniform specific entropy makes a fixed-composition fluid barotropic. Merely having $Ds/Dt=0$ does not exclude baroclinic vorticity generation from spatial entropy gradients.

For smooth inviscid barotropic fluid motion without gravity or magnetic forces, let $w$ be the specific enthalpy with $\nabla w={\rho }^{-1}\nabla p$, and set $q=\mathbf{u}\cdot \mathbf{\omega }$. ThenThis follows by dotting the barotropic vorticity transport equation with $\mathbf{u}$ and using the divergence of a cross product. A volume integral is constant when the resulting boundary flux vanishes.

In the observational setting of this paper, the cosmic microwave background favours nearly flat geometry, while cluster dynamics and large-scale structure favour a matter density near ${\Omega }_m\sim 0.3$. The missing contribution to the critical density is then roughly ${\Omega }_X\sim 0.7$. The supernova luminosity distance measurements require recent accelerated expansion. With present radiation negligible, $q_0<0$ requiresFor an X-dominated universe the threshold would instead be $w<-1/3$. Thus X cannot be ordinary pressureless matter or radiation. It must be comparatively smooth on cluster scales, so that it contributes to the background expansion without being counted as clustered dynamical matter. Its negative pressure makes its density dilute more slowly than matter, allowing late dominance. A cosmological constant, with $w=-1$, or sufficiently potential-dominated quintessence can have these properties.

The comparison with a perfect fluid in general relativity needs a distinction between background stress and perturbations. A homogeneous canonical scalar has canonical scalar stress as a perfect fluid:It has isotropic pressure and no background shear stress, just as a perfect fluid in general relativity does. Negative pressure therefore does not, by itself, prevent a perfect-fluid description. However, an adiabatic barotropic fluid with the constant closure $p=w\rho$ would have $c_s^2=dp/d\rho =w$. For negative $w$, a short-wavelength density mode then has ${\omega }^2\simeq w{k}_{\mathrm{phys}}^2<0$, the negative-w barotropic-fluid gradient instability. Such a closure cannot give a stable, smooth propagating X component.

A canonical scalar instead has rest-frame sound speed squared equal to one: its spatial and temporal kinetic terms have the same coefficient. Its pressure perturbation need not obey the background relation $\delta p=w\delta \rho$. This nonadiabatic response suppresses small-scale clustering without the negative-sound-speed instability. A cosmological constant is a separate limiting case with no independent propagating fluid density mode. Thus the issue is the microphysics and perturbation closure, not whether the background stress-energy tensor has perfect-fluid form.

The centripetal acceleration of the purely rotating barotropic fluid is $-r{\Omega }^2{\mathbf{e}}_r$. For $p=K{\rho }^2$, the pressure force satisfies ${\rho }^{-1}\nabla p=2K\nabla \rho$, so its barotropic enthalpy function is $w=2K\rho$. The equilibrium equation is thereforeTaking its curl gives ${\partial }_z(r{\Omega }^2)=0$. Thus ${\Omega }^2$ is independent of z; for a continuous, consistently signed rotation law, so is $\Omega$. This proves the barotropic cylindrical rotation theorem. Integrating the remaining radial force yieldsThe constant C is common to a connected fluid region.

Interpret the semi-thickness as a free surface with $\rho (r,\pm H)=0$, taking the exterior pressure to be negligible. At $H=ϵr$, the gravitational potential on this surface is ${\Phi }_s=-GM/[r\sqrt{1+{ϵ}^2}]$. The first integral evaluated at the surface says $C={\Phi }_s(r)-\int r{\Omega }^{2}dr$. Differentiating along it givesand therefore $\boxed{{\Omega }^2=(1+{ϵ}^2{)}^{-1/2}GM/r^3}$. Radial pressure support makes this slightly slower than Keplerian rotation for nonzero thickness.

Subtracting the free-surface first integral from its value inside the disc gives the complete density profileIt is nonnegative for $K>0$. Integrating over height, with $y=z/r$, givesThus the surface density is independent of r in this quadratic-polytrope disk with constant aspect ratio. The factor $1/r$ in density is canceled by the scale r in the vertical extent. The free-surface assumption is essential to this determination; a prescribed nonzero boundary pressure would require its own boundary data. No radial edges are specified, so the formula describes the local equilibrium rather than a finite-mass disc extended to all radii.

Write the inward radial velocity as $u_R=-v(R)$, with $v>0$. Steady spherical mass conservation and the radial Euler equations for an inviscid fluid giveFor a barotropic fluid, define the sound speed $c_s^2=dp/d\rho >0$ and the barotropic enthalpy function by $dh=dp/\rho$. Choosing $\Phi (\infty )=0$, integration gives the Bernoulli equationThe continuity equation gives ${\rho }^{\prime }/\rho =-2/R-v^{\prime }/v$. Eliminating mass density from the momentum equation produces the barotropic spherical accretion equationThe sonic point is where the inward Mach number $v/c_s$ equals one. A smooth solution cannot cross the vanishing coefficient on the left with a nonzero right-hand side, so a regular sonic radius must obeyConversely a continuous flow that is subsonic at infinity and supersonic inward must cross unit Mach number; smooth passage selects a simultaneous zero of the two sides, hence a regular sonic point. The associated critical accretion branch, rather than an arbitrary subsonic solution with smaller flux, is selected by these outer and inner requirements.

For the polytropic equation of state, set $\Gamma =1+1/m$ to distinguish its exponent from any separately specified perturbation exponent. Assume $K>0$ and first take the usual positive index $m>0$. ThenFor the attractive mixed potential, assume $\lambda ,\mu >0$. The sonic condition and Bernoulli equation giveHence polytropic accretion in a mixed inverse-power potential reduces toFor $0<m\le 1/2$ all nonzero coefficients in the left side are nonnegative for $R_s>0$, with at least one strictly positive: there is no finite positive root. For $m>1/2$ the constant term is negative and the leading coefficient positive, giving exactly one positive root. Define $B=(2m-3)\mu /4$ and $C=(2m-1)\lambda /2$. ThenThis is the finite-radius transonic range for the two positive attractive terms.

The critical point has real distinct crossing slopes. To check this rather than just solve the sonic equalities, put $U=\lambda /R_s^2$, $V=\mu /(2R_s)$ and $x=R_s{v}_s^{\prime }/v_s$. Differentiating the barotropic spherical accretion equation at its simultaneous zero givesIts discriminant is $8m[2(2m-1)U+(2m-3)V]/(U+V)$. The sonic Bernoulli equation says $(2m-1)U+(2m-3)V=2m{c}_{\infty }^2$, so this discriminant is strictly positive for $m>1/2$. The accretion crossing is the slope for which $d\log (v/c_s)/d\log R<0$, namely the smaller root. The two slopes have opposite signs of this Mach-number derivative. This is the regular transonic saddle, not a singular vertical tangent.

The corresponding mass accretion rate follows without an additional arbitrary mass density normalization:One may equivalently replace ${\rho }_s$ by $[c_{s,s}^2/(\Gamma K){]}^m$. With a nonzero $\lambda$ term, inner free fall has $v\sim \sqrt{2\lambda }/R$ and continuity gives $\rho \propto R^{-1}$. Thus $v^2/c_s^2\propto R^{-2+1/m}$ increases without bound precisely in the positive-index range $m>1/2$, consistent with the required inner supersonic branch.

The assumptions on the parameters matter. If $\lambda =0$ and $\mu >0$, the finite sonic radius instead requires $m>3/2$, the usual Bondi accretion range; $m=3/2$ is the zero-radius limiting critical case. If $\mu =0$ and $\lambda >0$, the condition remains $m>1/2$. With arbitrary signs of $\lambda ,\mu$, one must use the quadratic and retain only positive roots with $c_{s,s}^2>0$ and real admissible crossing slopes; attraction was not explicitly given as a sign condition in the source.

If positive polytropic index was not intended as an implicit physical convention, there is a further mathematical range. For $K>0$, positive compressibility also permits $m<-1$, giving $0<\Gamma <1$. Both potential terms attractive then give a positive sonic root as well:The same mass accretion rate formula applies; its crossing-slope discriminant is positive. The range $-1<m<0$ has $c_s^2<0$ and no physical acoustic sonic point, $m=-1$ has constant pressure and zero sound speed, and $m=0$ does not define the printed equation of state. Thus the usual answer $m>1/2$ assumes $m>0$ and positive attractive coefficients; these conventions should not be confused with additional printed restrictions.

Write $B=2\Omega -S$, $A(t)=c_s^2-2\pi G{\Sigma }_0/k(t)$, and $w=i{k}_x{\tilde{v}}_y-i{k}_y{\tilde{v}}_x$. Differentiation, including ${\dot{k}}_x=S{k}_y$, givesThe two pressure/self-gravity cross terms cancel. Continuity gives ${\dot{\tilde{\Sigma }}}^{\prime }=-i{\Sigma }_0(k_x{\tilde{v}}_x+k_y{\tilde{v}}_y)$, so $\dot{w}=B{\dot{\tilde{\Sigma }}}^{\prime }/{\Sigma }_0$. ConsequentlyThe conserved amplitude is linearized vortensity: the perturbation of the absolute vertical vorticity divided by surface density,A two-dimensional inviscid barotropic fluid materially conserves this potential vorticity. Both pressure enthalpy and gravitational forces are gradients, so they create no baroclinic vorticity source. Since $q_0$ is uniform, its linear perturbation obeys $D_0{q}^{\prime }=0$, explaining why a shearing wave has constant vortensity amplitude.

Let $u=u_r<0$ for accretion. The steady spherical continuity equation and radial Euler equations for an inviscid fluid giveFor a barotropic fluid, $dp/dr=c_s^{2}d\rho /dr$. Dividing continuity equation by $r^2\rho u$ gives ${\rho }^{\prime }/\rho =-2/r-u^{\prime }/u$. Eliminating this derivative yieldsAt a smooth transonic critical point, the coefficient of $du/dr$ vanishes. A finite derivative then requires the right side to vanish simultaneously:These are the sonic and regularity conditions. Otherwise the derivative is singular and the proposed smooth transonic passage fails. The local slope must also be a real root compatible with the desired branch. For example, defining $C=\rho d(c_s^2)/d\rho$ at the point and differentiating both sides givesThis explains why simultaneous vanishing is a necessary regularity condition, rather than an automatic proof that every potential admits a critical crossing.

Define the barotropic enthalpy function by $dh=dp/\rho$. The momentum equation integrates to the Bernoulli equationChanging the reference mass density in $h$ changes only the constant $B$. For an isothermal closure this logarithmic function is a barotropic barotropic pressure potential; it should not be confused with the constant thermodynamic specific enthalpy of an ideal gas held at fixed temperature.

For isothermal transonic accretion in the Paczyński-Wiita potential, assume $M>0$ and $c_s>0$, and define $r_B=GM/(2c_s^2)$ and $\beta =16c_s^2/c^2$. The critical equation iswhose two roots are $r_G+\frac{1}{2}{r}_B(1\pm \sqrt{1+\beta })$. Only the plus root lies outside $r_G$. ThereforeFor this isothermal case $C=0$, and the slope relation reduces toThe positive slope is the inward-accretion solution connecting a small inward speed at large radius to the supersonic inward branch. Thus the exterior critical point is a genuine nondegenerate transonic point.

Choose the reference mass density as ${\rho }_0$, so $h=c_s^2\log (\rho /{\rho }_0)$ and $B=0$ from the conditions at infinity. At $u_c=-c_s$,Substitution into $\dot{M}=4\pi r_c^2{\rho }_c{c}_s$ givesThis is the rate selected by the smooth transonic accretion solution. Arbitrary static or subsonic formal solutions are not assigned this rate merely by specifying conditions at infinity. As $r_G/r_B\to 0$, the result tends to the isothermal Bondi accretion rate $\pi e^{3/2}{\rho }_0(GM{)}^2/c_s^3$, a useful normalization check.

In the inertial frame the steady convective acceleration is $-R{\Omega }^2{\mathbf{e}}_R$. For a barotropic fluid, $\nabla p/\rho =\nabla h$, where $h$ is the barotropic enthalpy function. The equilibrium momentum equation therefore giveson each connected fluid region. Uniform rotation is essential: it makes the centrifugal acceleration derivable from the centrifugal potential $-{\Omega }^2{R}^2/2$.

The Cowling approximation sets the gravitational-potential perturbation to zero while retaining background gravity. The barotropic pressure force linearizes as $\delta (\nabla p/\rho )=\nabla \delta h$, with $\delta h=p^{\prime }/\rho =c_s^2{\rho }^{\prime }/\rho$. Denote its mode amplitude by $W$. The advective time derivative on a scalar mode is ${\partial }_t+\Omega {\partial }_{\varphi }$, so the stated positive-frequency exponential produces $i\sigma$ with $\sigma =\omega +m\Omega$. Linearizing the radial centrifugal term produces $-2\Omega v_{\varphi }$, while linearizing the azimuthal convective term produces $2\Omega v_R$. ThusThese are the pressure-gradient and Coriolis acceleration terms in the corotating-frame form. Independently, linearizing continuity equation givesThe equilibrium mass density is axisymmetric, so no additional azimuthal background-density derivative appears.

For $\sigma \ne 0$ and $D=4{\Omega }^2-{\sigma }^2\ne 0$, invert the horizontal momentum system:Inserting these into the continuity equation, dividing by $i\sigma$ and multiplying by $D$ givesThe two terms proportional to $\rho W_R$ cancel; this cancellation leaves precisely the background-density derivative in the last term. Since $-D/{\sigma }^2=1-4{\Omega }^2/{\sigma }^2$, this is exactlyThe singular frequency cases require the original velocity equations rather than this inversion.

In the low-frequency anelastic approximation for a rotating barotropic star, neglect the left side. Spherical background mass density satisfies ${\rho }_R/R={\rho }_z/z$, understood by its smooth limiting form on the equatorial plane. For $W=z{R}^m$,The remaining equation reduces toWriting $q=2\Omega /\sigma$, the bracket factors as $(m+1-q)(q+1)$. The regular desired branch is thereforeThe other algebraic factor corresponds to $\sigma =-2\Omega$, where the eliminated horizontal system is singular; it is not classified by this pressure-equation inversion. A constant-density background makes the bulk factor identically zero but does not invalidate the displayed solution.

For the regular branch, an explicit velocity check is particularly informative. Up to a common mode normalization,It satisfies $R{v}_R+z{v}_z=0$ and $\nabla \cdot \mathbf{v}=0$. Thus $\nabla \cdot (\rho \mathbf{v})=0$ for every spherical mass density profile, directly verifying the anelastic continuity condition. The motion is tangential to spherical shells and is the sectoral inertial mode of a slowly rotating barotropic star. For single-valued azimuthal modes, $m$ is a positive integer. With the source's exponential convention,For $m\ge 2$ the pattern is retrograde relative to the star but prograde in the inertial frame. This is a leading slow-rotation anelastic mode, not an exact solution of the compressible equations whose left-hand term was discarded.

For cylindrical rotation $u_{\varphi }=R\Omega (R)$, the equilibrium Euler momentum equation is $-R{\Omega }^2{\mathbf{e}}_R=-\nabla p/\rho -\nabla \Phi$. HenceThe integral is the centrifugal potential of cylindrical rotation. On a regular barotropic fluid branch define the specific enthalpy $h(\rho )$ by $dh=dp/\rho$. ThenFor a regular equation of state with $dp/d\rho >0$, $h$ is invertible, so $\rho$ and $p=p(\rho )$ depend only on $\Psi$ within a connected equilibrium region. In particular $dp/d\Psi =-\rho$. This local conclusion uses an invertible barotropic branch; distinct disconnected fluid regions can have different integration constants.

The logarithmic density term is the barotropic fluid energy for $p=c_s^2\rho$, rather than the thermodynamic heat content of an isolated gas. DefineFor any scalar $q$, the continuity equation implies ${\partial }_t(\rho q)+{\partial }_z(\rho wq)=\rho Dq$. Therefore $De=-c_s^2{w}^{\prime }$, $D\Phi =wg$, and dotting the momentum equations with $\rho \mathbf{v}$ givesThe MHD induction equation gives the complementary magnetic energy balanceAdding these two identities, and collecting the product derivatives, yields the barotropic magnetic energy equationHere $E=E_h+E_B$. The two stress terms in $S$ describe magnetohydrodynamic shear work: the maintained background shear flow can supply energy to, or remove energy from, the perturbation flow and magnetic field. The sign depends on the off-diagonal total stress; this is why the energy excluding the background shear flow is not generally conserved. Changing ${\rho }_0$ merely adds a constant multiple of the conserved density to $E$ and the corresponding mass flux to $F$, leaving $S$ unchanged.

For constant fluid density $\rho$, steady Euler equations for an inviscid fluid under a conservative force per unit mass $-\nabla \Phi$ read $(\mathbf{u}\cdot \nabla )\mathbf{u}=-\nabla p/\rho -\nabla \Phi$. Here $\mathbf{u}$ is velocity, $p$ pressure and $\Phi$ potential per unit mass. Dotting with $\mathbf{u}$ givesThis is the Bernoulli equation. The constants can differ between streamlines. If density varies, the same derivation for a barotropic fluid replaces $p/\rho$ by ${\int }^{p}d{p}^{\prime }/\rho (p^{\prime })$; constant density is the appropriate water approximation here.

For the draining cone, the radius at height $h$ is $h$ and the horizontal area is $\pi h^2$. Both the free surface and the jet are at atmospheric pressure; their gravitational potential difference is approximately $gh$. The speed of the descending surface is smaller than the jet speed by the area ratio ${ϵ}^2/h^2$, so it can be neglected when $h\gg ϵ$. The Bernoulli equation then gives Torricelli's law, $U\simeq \sqrt{2gh}$. Conservation of mass givesHence the leading drainage time of a conical tank isRemoving the tip places the aperture at height of order $ϵ$, and the quasi-steady calculation fails in the last layer of that depth. Integrating down only to $h=O(ϵ)$ changes the displayed leading time by relative order $(ϵ/h_0{)}^{5/2}$; replacing the actual head $h-ϵ$ by $h$ introduces a larger, but still vanishing, relative correction of order $ϵ/h_0$. The formula is the ideal inviscid estimate under the question's stated approximation, not an exact law for the final transient or a real contracted jet.

On a length scale $\ell$, pressure communicates support over a sound-crossing time $\ell /v_s$, whereas self-gravity acts over a time of order $(G{\rho }_0{)}^{-1/2}$. When the former is longer, gravity can amplify an overdensity before pressure smooths it. Thus Jeans instability occurs at scales of order $\ell \gtrsim v_s/\sqrt{G{\rho }_0}$; the precise numerical factor depends on whether scale means wavelength or inverse wavenumber.

Take a uniform static background, small perturbations, an inviscid barotropic fluid, and $\delta p=v_s^2\delta \rho$ with $v_s^2=(dp/d\rho {)}_0>0$. An infinite Newtonian homogeneous density cannot itself have zero background gravitational acceleration while satisfying the unmodified Poisson equation for Newtonian gravity. Specify the Jeans swindle: subtract the homogeneous field, or regard an external prescription as balancing it, and apply Poisson's equation only to perturbations. The linearized continuity equation and Euler equations for an inviscid fluid areTaking a time derivative of the first equation and substituting the second yields ${\partial }_t^2\delta \rho =v_s^2{\nabla }^2\delta \rho +4\pi G{\rho }_0\delta \rho$. A plane wave $e^{i(\mathbf{k}\cdot \mathbf{x}-\omega t)}$ therefore givesFor $0<k<k_J$ there is exponential growth at rate $\sqrt{4\pi G{\rho }_0-v_s^2{k}^2}$. At $k_J$ the density mode is neutral to this linear order, and above it there are acoustic oscillations. The critical wavelength is $2\pi /k_J=\sqrt{\pi }{v}_s/\sqrt{G{\rho }_0}$. Transverse velocity perturbations decouple as zero-frequency vortical modes in this idealization.

For the attractive Yukawa potential, assume a real screening parameter $\alpha \ge 0$. Its Green function obeys $({\nabla }^2-{\alpha }^2)(-Gm{e}^{-\alpha r}/r)=4\pi Gm{\delta }^{(3)}(\mathbf{r})$, including the delta source. Thus $\delta {\varphi }_{\mathbf{k}}=-4\pi G\delta {\rho }_{\mathbf{k}}/(k^2+{\alpha }^2)$. The same fluid elimination gives Jeans instability with Yukawa gravity:There exists a growing nonzero-wavenumber mode precisely whenEquality gives no exponentially growing nonzero mode. If screening exceeds this threshold, the finite-range gravitational response is too weak to overcome pressure at any wavelength. Otherwise all sufficiently long but nonconstant waves remain unstable, with a longer minimum unstable wavelength than in Newtonian gravity. For $\alpha >0$ the long-wave gravitational contribution also scales as $k^2$, so the growth rate tends to zero as $k\to 0$; screening does not produce a second positive lower cutoff in wavenumber. A homogeneous screened background can have the constant gravitational potential $-4\pi G{\rho }_0/{\alpha }^2$, with zero force, so no Newtonian background subtraction is needed for nonzero screening. The limits $k\to 0$ and $\alpha \to 0$ are nonuniform. An exact $k=0$ density change is not one of the growing finite-wavelength modes. In the pressureless edge case $v_s=0$, every finite $k>0$ is unstable for positive density and finite screening.

Use the Euler equations for an inviscid fluid with no magnetic or gravitational force. Let $w$ denote specific enthalpy. For a homentropic flow, $dw=dp/\rho$, so the pressure acceleration is $-\nabla w$. The velocity identitytherefore givesTaking the curl, using that the curl of a gradient is zero and that differentiation commutes for smooth fields, proves barotropic vorticity transport:More generally the same proof works for any barotropic fluid, replacing $w$ by a pressure potential ${\int }^{\rho }{p}^{\prime }(s)ds/s$. Here “isentropic” must supply this barotropic closure, for example through one common specific entropy throughout the fluid. Merely imposing entropy advection equation $Ds/Dt=0$ allows spatial specific entropy gradients; in that case the vorticity equation contains the additional baroclinic vorticity generation term ${\rho }^{-2}\nabla \rho \times \nabla p$.

For a barotropic fluid, $\nabla p=\rho \nabla h$. Magnetostatic force balance and $L\alpha =F(\beta )$ giveThe left side is a gradient multiplied by $\rho$, so taking the curl shows that $(L\beta -F{F}^{\prime })/(r^2\rho )$ is constant on each $\beta$ surface. Absorbing the fixed factor ${\mu }_0$ into an arbitrary function $G$ gives the Axisymmetric magnetostatic Grad-Shafranov system