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Adiabatic initial conditions perturb all relevant local physical quantities along the same background time trajectory. For quantities with nonzero background time derivatives, their perturbations satisfy $\delta X/{\bar{X}}^{\prime }=\delta Y/{\bar{Y}}^{\prime }$. For separately conserved barotropic fluids, cosmological energy conservation then givesFor example, pressureless matter and radiation in cosmology obey ${\delta }_m=3{\delta }_{\gamma }/4$. Their relative particle-number ratio is unperturbed at leading order.

The density condition tests relative entropy between components. In a general nonbarotropic component it does not by itself eliminate intrinsic entropy, measured by $\delta P_i-({\bar{P}}_i^{\prime }/{\bar{\rho }}_i^{\prime })\delta {\rho }_i$. The stronger statement that every physical variable follows the same time shift includes the pressure and any independent field-velocity data. A single scalar species alone does not make the density-ratio test sufficient, since that test is then vacuous.

For vanishing total non-adiabatic pressure perturbation and negligible regular gradient terms, uniform-density curvature is conserved on superhorizon scales.^[ref] Relating this to conserved comoving curvature perturbation also requires the appropriate regular adiabatic mode; a non-attractor phase such as ultra-slow-roll inflation needs separate analysis. The ordinary growing adiabatic photon mode selects coherent acoustic phases, explaining the adiabatic acoustic-peak spacing.

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.

A barotropic fluid has pressure determined by density alone, $P=P(\rho )$, so pressure and density surfaces coincide in hydrostatic equilibrium.

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 with potential body forces, barotropic vorticity transport and $\nabla \cdot \mathbf{\omega }=0$ give $D\mathbf{\omega }/Dt=(\mathbf{\omega }\cdot \nabla )\mathbf{u}-\mathbf{\omega }\nabla \cdot \mathbf{u}$. Combine this with material surface element evolution to obtain $D(\mathbf{\omega }\cdot d\mathbf{S})/Dt=0$. Integrating over a material surface proves the flux law. Its boundary must move with the fluid; flux through a fixed open surface is generally not conserved. By Stokes theorem the same statement is Kelvin's circulation theorem around the material boundary.

In an inviscid barotropic fluid subject to conservative body force, the circulation of a material loop is constant. For constant density, parameterize that loop by $\mathbf{r}(s,t)$ with ${\mathbf{r}}_t=\mathbf{u}$. Differentiating $\Gamma =\int \mathbf{u}\cdot {\mathbf{r}}_{s}ds$ gives an integral of $(D\mathbf{u}/Dt)\cdot d\mathbf{r}$ plus $\int {\partial }_s(|\mathbf{u}{|}^2/2)ds$. The latter vanishes on a closed loop, and the Euler equations make the former the integral of a gradient, also zero. For a barotropic density, $\nabla p/\rho$ is still a gradient of specific enthalpy. Constant circulation is consistent with changing local vorticity when a vortex tube changes its cross-sectional area.

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.

For a steady inviscid barotropic fluid in a conservative gravitational potential, define $h(\rho )=\int dp/\rho$. Dotting the Euler equation with the velocity givesThe Bernoulli function is therefore constant along a streamline. It balances kinetic energy, pressure work or enthalpy, and gravitational potential energy per unit mass. Different streamlines can have different constants; for an irrotational flow, the identity $\nabla H=\mathbf{u}\times (\nabla \times \mathbf{u})$ makes the constant common throughout a connected region.

For an incompressible fluid, $h=p/\rho$ up to a constant, giving $u^2/2+p/\rho +\varphi =C$. For a constant-entropy perfect-gas streamline, $h=\gamma p/[(\gamma -1)\rho ]$. Together with mass conservation, this explains conversion of enthalpy into kinetic energy in a nozzle and the need to include gravity in an escaping wind. Viscous losses, heating and entropy production require their corresponding energy balance. Across an adiabatic shock the total stagnation enthalpy can still be conserved, but the upstream isentropic pressure-density law cannot simply be continued through the entropy jump.

For a barotropic fluid, define $h(\rho )=\int dp/\rho$. Dotting the steady Euler equation with $\mathbf{u}$ givesThus the Bernoulli function $H=u^2/2+h+\psi$ is constant along each streamline. Different streamlines can have different constants; steady flow alone does not imply one spatially uniform $H$.

In a quasi-one-dimensional flow through area $A(z)$ with negligible gravity, the axial momentum equation and $dp=c_s^{2}d\rho$ giveMeanwhile the mass flow rate $\dot{M}=\rho uA$ is constant, so $d\ln \rho =-d\ln u-d\ln A$. Substitution proves the nozzle area-velocity relationA subsonic flow accelerates where $A$ decreases, whereas a supersonic flow accelerates where $A$ increases. A smooth accelerating transonic branch in a de Laval nozzle therefore crosses $u=c_s$ at a minimum-area throat. Suitable reservoir and downstream pressures are also required: a converging-diverging shape by itself does not exclude a wholly subsonic solution or a downstream shock.

For the stipulated constant-$K$ polytropic flow, $c_s^2=\gamma K{\rho }^{\gamma -1}$. At the throat $u_m=c_{s,m}$, soThe specific enthalpy is $h=\gamma K{\rho }^{\gamma -1}/(\gamma -1)=c_s^2/(\gamma -1)$. Evaluating $H$ at the throat and substituting $\rho =\dot{M}/(Au)$ elsewhere givesOn the outward supersonic branch, $A\sim C{r}^2$ lets the gas expand and its specific enthalpy tend to zero for $\gamma >1$. ConsequentlyFor $\gamma =5/3$, $u_{\infty }=2c_{s,m}$. Also $T\propto {\rho }^{\gamma -1}\propto r^{-2(\gamma -1)}$. These are the adiabatic nozzle terminal speed and expansion laws for a lossless escaping branch; a subsonic branch could instead approach finite pressure with $u\propto A^{-1}$. A pressure-shaped astrophysical jet is being approximated by a nozzle here: the assumed geometry does not by itself explain the confinement or account for magnetic forces or relativistic motion in an actual active galactic nucleus.

For steady inviscid motion of a barotropic fluid in a conservative gravitational potential $\Phi$, the Euler equations for an inviscid fluid imply the Bernoulli equationalong each streamline. The barotropic enthalpy function $h$ has units of energy per unit mass. For an isentropic ideal gas, $p=K{\rho }^{\gamma }$ gives $h=\gamma p/[(\gamma -1)\rho ]$ up to an additive constant. For constant density, the corresponding term is $p/\rho$. Pressure work converts between kinetic, enthalpy and gravitational energy.

A converging subsonic nozzle can accelerate a fluid at the expense of its enthalpy; in the incompressible horizontal approximation, higher speed accompanies lower static pressure. Together with conservation of mass, this explains a Venturi effect measurement. Compressible nozzle flow additionally needs its equation of state and sonic conditions. In a stellar wind, the potential term is $-GM/r$, so enthalpy or mechanical energy can be spent escaping gravity. A pressure-negligible marginal-escape wind has $v^2/2-GM/r=0$.

The constant need not be the same on different streamlines. Irrotational barotropic flow in a connected region gives a common value; a general rotational flow does not. A pump, viscosity, a radiating shock, or heat addition changes the relevant energy balance. In particular, adiabaticity alone does not allow one to drop the enthalpy term.

Write $\mathbf{\omega }=\nabla \times \mathbf{u}$ and use the vector identity $(\mathbf{u}\cdot \nabla )\mathbf{u}=\nabla (|\mathbf{u}{|}^2/2)-\mathbf{u}\times \mathbf{\omega }$. Taking the curl of the momentum equation eliminates all pure gradients, including gravity and the $\nabla (\nabla \cdot \mathbf{u})$ viscous contribution. Before invoking barotropy, the pressure term givesThis is baroclinic vorticity generation. For a barotropic fluid, $p=p(\rho )$, so $\nabla p$ is parallel to $\nabla \rho$ and this source vanishes. Constant $\nu$ permits the curl and Laplacian to commute. The result isNo incompressibility assumption was used. Since $\nabla \cdot \mathbf{\omega }=0$, an equivalent form isThe terms represent vortex stretching/tilting, compression/dilatation and viscous diffusion. In particular, removing the baroclinic source does not make the vorticity of a three-dimensional parcel constant. A spatially varying entropy-dependent $K$ in $p=K{\rho }^{\gamma }$ need not define a single barotropic law; the common-pressure-law assumption matters.

Assume an inviscid barotropic fluid acted on by a conservative body force $-\nabla \psi$. Define its barotropic enthalpy function byThe Euler equations for an inviscid fluid are then ${\partial }_t\mathbf{v}+(\mathbf{v}\cdot \nabla )\mathbf{v}=-\nabla (H+\psi )$. Using the vorticity $\mathbf{\omega }=\nabla \times \mathbf{v}$ and the vector identitygives ${\partial }_t\mathbf{v}+\nabla B=\mathbf{v}\times \mathbf{\omega }$, where $B=v^2/2+H+\psi$.

For steady flow, dotting this equation with $\mathbf{v}$ yields $\mathbf{v}\cdot \nabla B=0$. Thus Bernoulli's equation isThe constant need not agree between different streamlines. If the flow is also irrotational, the vector equation gives $\nabla B=0$, so one common constant applies throughout a connected flow region.

For an isentropic process with $p=K{\rho }^{\gamma }$, $H=\gamma p/[(\gamma -1)\rho ]$ up to an additive constant, equal to the thermodynamic specific enthalpy. For an externally maintained isothermal gas, $H=a^2\ln (\rho /{\rho }_{\mathrm{ref}})$; this pressure potential is not the constant thermodynamic enthalpy of a fixed-temperature ideal gas.

If steady flow is not assumed, a useful additional case is unsteady potential flow $\mathbf{v}=\nabla \varphi$. The same vector equation integrates spatially toGeneral unsteady barotropic flow does not possess the steady streamline constant without further conditions.

Following a parcel. Let $\mathbf{X}(t)$ be a fluid pathline, with $\dot{\mathbf{X}}=\mathbf{u}(\mathbf{X},t)$. The multivariable chain rule gives the material derivative,The first term measures change at fixed position; the advection term accounts for the parcel moving through spatial gradients. The derivative is also called the Lagrangian derivative.

The forces on a parcel. A material volume $dV$ has mass $\rho dV$. The pressure forces on its boundary sum to $-\nabla pdV$, while the gravitational acceleration is $-\nabla \psi$. Newton's second law therefore gives the Euler equations for an inviscid fluid,This neglects viscosity and other body forces. The left side is the parcel's acceleration, not just the time derivative at a fixed spatial location.

Bernoulli conservation. The claimed streamline result needs steady flow and a barotropic fluid, $p=p(\rho )$, or an equivalent enthalpy relation along each isentropic streamline. These hypotheses are necessary; the unsteady Euler equation alone does not imply the stated conservation law. Define the barotropic enthalpy functionDot the steady Euler equation with $\mathbf{u}$, using $\mathbf{u}\cdot [(\mathbf{u}\cdot \nabla )\mathbf{u}]=\mathbf{u}\cdot \nabla (u^2/2)$. This givesThus the Bernoulli function isEvery term has units of energy per mass, ${\mathrm{m}}^2{\mathrm{s}}^{-2}$. Different streamlines can have different constants.

The spherical accretion equation. Write the inward speed as $u(r)>0$, so the velocity vector is $-u(r)\hat{\mathbf{r}}$ and the positive mass accretion rate isAssume a prescribed central potential $\psi =-GM/r$ and negligible gas self-gravity. The steady radial Euler equation isFor a barotropic equation of state, $dp/dr=c_s^{2}d\rho /dr$, with $c_s^2=dp/d\rho$. Eliminate the density derivative to obtainA smooth sonic point requires the right side to vanish where $u=c_s$, henceFor a nonisothermal gas this is an implicit relation involving the local sound speed. In the isothermal case $c_s$ is constant and this is the Bondi sonic point.

The isothermal integral. For $p=c_s^2\rho$, choose the enthalpy reference so that $h_b=c_s^2\ln (\rho /{\rho }_{\infty })$. The reservoir is at rest at infinity and $\psi (\infty )=0$, making its Bernoulli constant zero. ThusorThe logarithm has a dimensionless argument. Constant temperature during compression generally requires heat removal, so this is Isothermal Bondi accretion, rather than adiabatic accretion with $\gamma >1$.

The critical accretion rate. Select the smooth reservoir-connected transonic branch. Evaluate the last relation at $u_s=c_s$ and $GM/r_s=2c_s^2$:Mass conservation then gives the isothermal Bondi accretion rate,The units are $\mathrm{kg}{\mathrm{s}}^{-1}$. Equivalently $\dot{M}=4\pi \lambda G^2{M}^2{\rho }_{\infty }/c_s^3$, with $\lambda =e^{3/2}/4\simeq 1.12$.

The inner-boundary selection of isothermal Bondi flow is important: a standard absorbing accretor inside the sonic radius selects this critical solution. The steady equations and ${\rho }_{\infty }$ alone also admit subcritical subsonic branches supported by a different inner boundary; they do not by themselves fix a unique rate. The usual answer presumes transonic capture.

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.