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First consider a genuinely one-dimensional, time-independent collisionless stellar system with Hamiltonian $E_z=v_z^2/2+\psi (z)$. Choose $\psi (0)=0$. For bound vertical oscillations, a phase-mixed population has equal occupations of the upward and downward branches of each orbit. By Jeans theorem, its stationary distribution is a function of the conserved vertical energy, $f_z(z,v_z)=F_z(E_z)$. Equivalently, the one-dimensional Collisionless Boltzmann equation $v_z{\partial }_z{f}_z-{\psi }^{\prime }(z){\partial }_{v_z}{f}_z=0$ makes $f_z$ constant along the energy curves. Branch-dependent incoming and outgoing populations or time-dependent waves would not satisfy this assumption.

Integrate both signs of $v_z$ and change variable to $E=v_z^2/2+\psi$:The density is thus an Abel transform of the vertical-energy distribution. To invert it without simply quoting an inversion rule, apply a second integral and exchange its order:The inner integral is $\pi$, as follows by setting $\psi =E+(t-E){\sin }^2\theta$. Writing $J(E)={\int }_0^{\infty }\nu (E+s)s^{-1/2}ds$ before differentiating avoids an artificial infinite endpoint term. Under sufficient differentiability and decay,The vertical-energy Abel inversion of a stellar distribution is thereforeThis proves the normalization, including the factor from the two velocity signs. The additive origin of $\psi$ merely changes the origin of energy.

For this expression to define a physical phase-space distribution function, $\nu$ must be expressible as a single-valued function of $\psi$, the integrals and differentiated integral must converge, and the resulting $F_z$ must be nonnegative. Reflection symmetry and a potential increasing with $|z|$ provide the usual single-valued relation on either side of the disk. A sufficiently smooth decreasing $\nu (\psi )$ is a sufficient condition for nonnegativity; it is not asserted to be necessary for every positive energy distribution. If there is a finite escape energy, replace the infinite limits by that energy and impose the appropriate cutoff and boundary conditions. The displayed infinite-limit formula presumes either a confining potential or a distribution extended by zero above its allowed energy, with no omitted boundary population.

A three-dimensional galactic disk does not automatically meet these hypotheses. Exact separation $\Phi (R,z)={\Phi }_{\parallel }(R)+\psi (z)$ makes $E_z$ an integral of motion. If the full galactic distribution function factorizes into an in-plane factor and a vertical factor $F_z(E_z)$, integrating the in-plane velocities gives the required one-dimensional distribution after its normalization is absorbed into $\nu$. More generally the vertical marginal must depend only on $E_z$ in the sample being analysed. Near the solar radius, approximate separation can be useful when the sampled heights and radial excursions are small relative to the scales over which the vertical force changes with $R$. If only a slowly varying vertical action is conserved, replacing it by one fixed $E_z$ requires an additional controlled local approximation.

The population must be sufficiently stationary and phase mixed, with negligible mean vertical flow, and the data must measure a well-defined tracer population with a known selection function. Radial coupling and the tilted velocity ellipsoid must be negligible or modelled. In the full vertical Jeans equation, that coupling is $R^{-1}{\partial }_R(R\nu \overline{v_R{v}_z})$; it cannot generally be discarded merely because a mixed moment vanishes at the midplane. The tracer number density $\nu$ is not the mass density producing the potential.

To infer the local mass, measure the tracer density as a function of height and its vertical velocity distribution near the midplane. With $\psi (0)=0$, the latter directly samples $F_z(v_z^2/2)$. Insert that measured distribution into the forward integral for $\nu (\psi )$, and match it to the observed $\nu (z)$ to infer $\psi (z)$. Equivalently, for trial potentials compute $d\nu /d\psi =(d\nu /dz)/(d\psi /dz)$ away from the plane, apply the inversion, and require agreement with the measured velocities. Density data alone do not determine both an unknown potential and an unknown energy distribution.

Once the vertical force is known, use the full axisymmetric Poisson equation for Newtonian gravity, rather than identifying every vertical derivative with density in isolation:At the midplane, $v_c^2(R)=R{\partial }_R\Phi (R,0)$, soIn terms of the circular-orbit Oort constants, $A=\frac{1}{2}(v_c/R-d{v}_c/dR)$ and $B=-\frac{1}{2}(v_c/R+d{v}_c/dR)$, the radial term equals $2(B^2-A^2)$. It vanishes for a locally flat rotation curve, not for an arbitrary rotating disk. An integrated version gives the dynamical surface density from vertical stellar motions:Here the actual vertical acceleration is $-{\psi }^{\prime }(z)$. The sign of the radial correction follows directly from the Poisson equation for Newtonian gravity.

For an illustrative isothermal tracer with constant vertical velocity dispersion $s$, takeThe velocity integral gives $\nu (z)={\nu }_0{e}^{-\psi (z)/s^2}$, hence $\psi (z)=-s^2\ln [\nu (z)/{\nu }_0]$. For a locally flat rotation curve this yieldsThis example shows explicitly how a measured tracer scale and velocity scale constrain the gravitating density; using the tracer's own density as the mass source would instead impose an unjustified self-gravitating model.

The inferred density includes luminous and faint stars, gas and dark matter. Comparing it with an independently estimated baryonic census tests for extra gravitating mass. An excess need not be a dark disk: a dark matter halo also contributes to the local force. Conversely, agreement with the baryonic density locally does not exclude a halo at other radii or heights. Vertical disequilibrium, radial mixing, errors in selection or tracer populations, and neglected velocity tilt can mimic a mass discrepancy, so the one-dimensional equilibrium assumptions must be tested when interpreting such an inference.

A phase-space distribution function describes the mass in a small cell of positions and velocities: $F(\mathbf{x},\mathbf{v},t)d^{3}x{d}^{3}v$ is that mass. Its velocity integral is the spatial mass density, and $F/\rho$ at a fixed position is a normalized velocity probability density. A number-normalized galactic distribution function differs by the stellar mass convention.

Take $a>0$ and $v_0>0$. For the cusped logarithmic spherical galaxy model, differentiate the gravitational potential and use the spherical Poisson equation for Newtonian gravity:ThusA circular orbit requires $v_c^2/r={\Phi }^{\prime }$. Its galaxy rotation curve is thereforeIt rises as $v_0\sqrt{r/a}$ at small radius and approaches $v_0$ from below at large radius.

The outer flat galaxy rotation curve is qualitatively useful for galaxies. However, the density has an inner $1/r$ cusp and an outer $1/r^2$ tail, while $M(r)\sim v_0^{2}r/G$ diverges. The model therefore requires an outer truncation for a finite galaxy, and does not describe a central constant-density core. Its gravitational potential cannot be normalized to zero at infinity.

For a star, the specific energy is $E=v^2/2+\Phi$ and the specific angular momentum is $\mathbf{L}=\mathbf{x}\times \mathbf{v}$. Since the gravitational potential is stationary and spherical,Thus both the specific energy and the entire angular-momentum vector are conserved.

For Jeans theorem, follow a stellar trajectory in phase space. The Collisionless Boltzmann equation is preciselyIn a steady system ${\partial }_{t}F=0$, so the value of the galactic distribution function is constant along every trajectory: it is itself an integral of motion. Conversely, if $I_1,\dots ,I_m$ are time-independent integrals of motion, then $F=h(I_1,\dots ,I_m)$ satisfies $dF/dt=\sum h_{,a}{\dot{I}}_a=0$. Locally, a complete set of invariant orbit labels expresses any steady galactic distribution function this way. This is the weak form of Jeans theorem; a stronger assertion about a small set of isolating integrals needs additional orbital regularity or mixing assumptions. In particular, the theorem does not say that every steady galaxy must have a galactic distribution function depending on only $E$ and $L$.

For the proposed constant-anisotropy distribution function, write $v_T=(v_{\theta }^2+v_{\varphi }^2{)}^{1/2}$ and $L=r{v}_T$. Introduce a polar angle $\psi$ in the tangential velocity plane. Then $d^{3}v=d{v}_r{v}_{T}d{v}_{T}d\psi$, so the apparent $1/L$ singularity cancels against the velocity Jacobian. This inverse-angular-momentum velocity integration makes the density finite for every $r>0$.

Let $t=e^{-\Phi /v_0^2}=a/(a+r)$, $c_1=1$, $c_2=2$, and ${\alpha }_k=k/(2v_0^2)$. Each exponential component contributesHere the two Gaussian integrals and the angular integral have product ${\pi }^2/{\alpha }_k$. Consequently,The galactic distribution function is nonnegative, gives the required density, and depends on conserved $E,L$; by Jeans theorem it is a self-consistent steady solution.

Differentiating the Gaussian integral with respect to ${\alpha }_k$ shows that each component has $⟨v_r^2{⟩}_k=1/(2{\alpha }_k)=v_0^2/k$, and the same value for $⟨v_T^2{⟩}_k$. Angular averaging gives $⟨v_{\theta }^2{⟩}_k=⟨v_{\varphi }^2{⟩}_k=v_0^2/(2k)$. Weighting these by ${\rho }_1,{\rho }_2$ yields the velocity dispersionsThe radial numerator includes $2r$, as in the original PDF. All mean velocities vanish by parity. The same parity, or the zero angular average of $\sin \psi \cos \psi$, gives $⟨v_r{v}_{\theta }⟩=⟨v_r{v}_{\varphi }⟩=⟨v_{\theta }{v}_{\varphi }⟩=0$.

The velocity-anisotropy parameter isThe constant-anisotropy distribution function therefore favors radial motion at every radius: the radial variance is twice either individual tangential variance. It still has a substantial population with nonzero angular momentum, rather than exclusively radial orbits. The $L^{-1}$ factor enhances low-angular-momentum stars. The radial velocity dispersion approaches $3v_0^2/4$ at the centre and $v_0^2$ far out; each individual tangential variance is half as large. There is no mean rotation, despite the well-defined galaxy rotation curve for hypothetical circular orbits.

Use a local orthonormal tetrad and units in which the speed of light is one. A particle has four-momentum $p^{\hat{a}}=E(1,\mathbf{e})$ and moves with velocity $\mathbf{e}$. Thus its energy density contributes $E$, its momentum density contributes $E{e}^i$, and its momentum flux contributes $E{e}^i{e}^j$. Integrating the phase-space distribution function gives the kinetic stress-energy tensor. The measure $d^{3}p/E$ is the future mass-shell Lorentz-invariant phase-space measure, so the same expression transforms as a spacetime stress-energy tensor; fixed state-counting or polarization factors are understood to be included in $f$.

Define $⟨F{⟩}_{\Omega }=(4\pi {)}^{-1}\int Fd\Omega$. The photon momentum measure is $d^{3}p=E^{2}dEd\Omega$, so every component of the kinetic stress-energy tensor has a radial factor $E^{3}dE=a^{-4}{ϵ}^{3}dϵ$. An isotropic background has $⟨e^i{⟩}_{\Omega }=0$ and $⟨e^i{e}^j{⟩}_{\Omega }={\delta }^{ij}/3$. HenceThis radiation pressure equation of state requires isotropy and masslessness, not a thermal spectrum.

The distribution perturbation is $\delta f=-ϵ{\bar{f}}^{\prime }(ϵ)\Theta$. Assume finite energy density and endpoint behavior ${ϵ}^4\bar{f}(ϵ)\to 0$ at zero and infinity, as for the Planck photon distribution. Integration by parts then givesConsequently the perturbed energy density and momentum density areSince $q^i=(\bar{\rho }+\bar{P})v^i=(4\bar{\rho }/3)v^i$, the photon angular temperature moments give $\delta =4⟨\Theta {⟩}_{\Omega }$ and $v^i=3⟨\Theta e^i{⟩}_{\Omega }$.

The sign of anisotropic stress must be specified. Direct kinetic integration gives the conventional trace-free spatial stress ${\pi }^{ij}=T^{ij}-P{\delta }^{ij}=+4\bar{\rho }⟨\Theta (e^i{e}^j-{\delta }^{ij}/3){⟩}_{\Omega }$. The PDF uses the opposite sign, ${\Pi }^{ij}=-{\pi }^{ij}$, equivalently $T^{ij}=P{\delta }^{ij}-{\Pi }^{ij}$. With that convention, all requested moments areTaking the spatial trace also givesThe angular temperature description assumes $\Theta$ is independent of photon energy; independent spectral distortions would require additional energy-dependent moments.

Let $f(\mathbf{p})d^{3}p$ be particle number per unit volume in a momentum element, including the spin-state count in $f$. This is the local momentum part of a phase-space distribution function, normalized by the number density $n=\int f(\mathbf{p})d^{3}p$. For an isotropic distribution with particle energy $E(p)=\sqrt{m^2{c}^4+p^2{c}^2}$, the kinetic pressure of an isotropic gas is the average momentum flux:The kinetic energy density, explicitly excluding rest mass, isFor $p\ll mc$, $E-m{c}^2=p^2/(2m)$ and $\mathbf{v}=\mathbf{p}/m$, giving $P=2u_{\mathrm{kin}}/3$. For $p\gg mc$, $E\simeq pc$ and $v\simeq c$, giving $P\simeq u_{\mathrm{kin}}/3$. ThusNeither relation assumes a Maxwellian distribution. At intermediate momenta neither constant ratio is exact, and anisotropic distributions require a pressure tensor rather than this scalar pressure.

For a classical Maxwell-Boltzmann distribution, $f=n(2\pi m{k}_{B}T{)}^{-3/2}\exp [-p^2/(2m{k}_{B}T)]$. Three Gaussian component integrals give $⟨p^2⟩=3m{k}_{B}T$, so the ideal gas hasIn a fully ionized mixture, sum over independent species to obtain $P_{\mathrm{gas}}=\rho k_{B}T/({\mu }_{\mathrm{mol}}{m}_u)$, where the mean molecular weight counts ions and free electrons. This excludes partial-ionization and interaction corrections. A dilute classical ultrarelativistic gas still has $P=n{k}_{B}T$ but $u_{\mathrm{kin}}\simeq 3n{k}_{B}T$; its momentum distribution is proportional to $e^{-pc/(k_{B}T)}$ instead of the nonrelativistic Gaussian.

For a thermal photon gas, the two polarization states and zero photon chemical potential give the Planck photon distributionUse $x=pc/(k_{B}T)$ and ${\int }_0^{\infty }{x}^3/(e^x-1)dx={\pi }^4/15$. The radiation constant and photon equation of state are thereforeLikewise $n_{\gamma }=2\zeta (3)(k_{B}T/\hslash c{)}^3/{\pi }^2$. Unlike a gas with a fixed particle number, photons do not have pressure proportional to baryonic mass density.

For fully degenerate electrons, the Fermi-Dirac distribution becomes a filled momentum sphere with two spin states. Counting them gives the Fermi momentumHere ${\mu }_e$ is the mean molecular weight per electron. The equation of state of a cold electron gas follows by integrating momentum flux up to $p_F$:Its two limits areThe corresponding kinetic energy densities are $3n_e[p_F^2/(2m_e)]/5$ and $3n_e{p}_{F}c/4$, respectively. Thus in the high-density relativistic limit the electron pressure scales as ${\rho }^{4/3}$ and is nearly independent of temperature. This is the ideal noninteracting, fixed-composition electron result, not a universal equation of state at nuclear densities where captures, interactions and the composition change.

The boundaries on a stellar equation-of-state regime diagram concern the electron component. Define the electron relativistic density threshold and thermal electron relativistic threshold byA degenerate electron gas is nonrelativistic well below ${\rho }_{\ast }$ and ultrarelativistic well above it: this is an approximately vertical division on a log-density plot. A nondegenerate electron gas instead becomes thermally relativistic near $T_{\ast }$: this is an approximately horizontal division. The regimes have broad crossovers rather than a discontinuity at either line.

The exact electron Fermi temperature, subtracting electron rest energy, isFor electrons with negligible thermal pairs, strong degeneracy requires $T\ll T_F$, whereas $T\gg T_F$ gives a nondegenerate gas. In the pair-rich regime the actual electron and Positron distributions must instead be treated with their chemical potentials, as discussed below. The two limiting degeneracy boundaries areThese slopes $2/3$ and $1/3$ explain the bent degeneracy boundary on the logarithmic graph. A thermal-wavelength test $n_e[2\pi {\hslash }^2/(m_e{k}_{B}T){]}^{3/2}\sim 2$ gives the same nonrelativistic density/temperature scaling, with an order-one definition of the crossover.

The radiation-to-gas pressure boundary in the nondegenerate fully ionized regime isRadiation dominates above this line. Once electrons are strongly degenerate, compare radiation pressure with $P_e(\rho )$ instead of continuing the ideal-gas comparison into that regime. The radiation-to-degeneracy pressure boundary is $T=[3P_e(\rho )/a_{\mathrm{rad}}{]}^{1/4}$, with logarithmic slopes $5/12$ and $1/3$ in the nonrelativistic and ultrarelativistic limits. Degeneracy of the electrons and dominance of their pressure are distinct criteria.

For the pair curve, distinguish baryonic net electrons from thermally created electrons and Positrons. Chemical equilibrium with photons requires opposite electron/Positron chemical potentials when the one-particle energy includes rest energy, as in the distributions below. At low temperature and low degeneracy the zero-chemical-potential density per charge species isCharge neutrality gives $n_{-}-n_{+}=n_b=\rho /({\mu }_e{m}_u)$, while the nondegenerate equilibrium product is $n_{-}{n}_{+}=n_0^2$. HencePairs become important when $n_0$ is comparable with or exceeds $n_b$, approximately the electron-positron thermal pair abundance curve ${\rho }_{\mathrm{pair}}(T)={\mu }_e{m}_u{n}_0(T)$. Below its density at a fixed temperature, pairs dominate over the net charge electrons. The exponential makes the low-temperature portion steep in a log-log plot. For ${\mu }_e=2$, the full zero-potential integral gives approximate pair-marker densities $1.80\times 10^3\mathrm{g}\mathrm{c}{\mathrm{m}}^{-3}$ at $10^9\mathrm{K}$ and $7.52\times 10^5\mathrm{g}\mathrm{c}{\mathrm{m}}^{-3}$ at $3\times 10^9\mathrm{K}$. At relativistic temperature use the full zero-chemical-potential Fermi-Dirac distribution instead; it gives $n_0\to [3\zeta (3)/(2{\pi }^2)](k_{B}T/\hslash c{)}^3$, so the high-temperature pair curve approaches slope three in log-density versus log-temperature. Strong net-electron degeneracy suppresses Positrons and requires the full chemical-potential-dependent distribution.

The original diagram uses fully ionized helium, ${\mu }_e=2$ and ${\mu }_{\mathrm{mol}}=4/3$, only to set numerical locations. It plots the full $T_F$ curve and the zero-chemical-potential pair integral, rather than extending their asymptotes into the crossover. The pair boundary is deliberately an approximate abundance marker, not a phase transition; at high temperature the nonrelativistic ion approximation and fixed-composition model also have limits. For $T\gg T_{\ast }$ with abundant nondegenerate pairs, the combined electron/Positron energy density is $7a_{\mathrm{rad}}{T}^4/4$, so photons plus pairs have $u=11a_{\mathrm{rad}}{T}^4/4$ and $P=u/3$. Real stellar matter adds Coulomb effects, partial ionization, nuclear reactions and, at sufficiently high density, nuclear-matter physics beyond the ideal regime map.

Write the total derivative of the phase-space distribution function asSince $a\bar{T}$ is constant for the background photon gas, expand the Bose-Einstein distribution asAt first order the four terms are respectivelyThe angular-deflection velocity is already first order and multiplies the first-order angular dependence of $f$, so its contribution is second order. Thus the collisionless left-hand side of the Free-streaming photon Boltzmann equation is

The energy density from a phase-space distribution function with $g=2$ internal states isIn the ultrarelativistic limit $T_{\nu }\gg m_{\nu }$, set $x=p/T_{\nu }$. The Fermi-Dirac distribution then givesSince $\zeta (4)={\pi }^4/90$,

In natural units, the equilibrium phase-space distribution function iswhere the minus sign gives the Bose-Einstein distribution for bosons and the plus sign gives the Fermi-Dirac distribution for fermions.

The equal normal stresses in the two directions perpendicular to the local magnetic field for a gyrotropic phase-space distribution function. The factor one half averages the two perpendicular velocity components. With first adiabatic invariant $\mu =v_{\perp }^2/(2B)$, the integrand is $m\mu Bf$.

In the local rest frame of a gas, the pressure tensor is the directional momentum flux $P_{ij}=\int p_i{v}_{j}f(\mathbf{p})d^{3}p$, where $f$ is the phase-space distribution function per unit momentum volume and $\mathbf{v}$ is particle velocity. For an isotropic distribution, $P_{ij}=P{\delta }_{ij}$, recovering the kinetic pressure of an isotropic gas. An anisotropic distribution can exert different normal pressures along different directions.

A weak-fluctuation closure retaining the averaged quadratic fluctuation flux while linearizing the fluctuating kinetic equation about a slowly evolving mean phase-space distribution function. One solves the fast linear response with the mean frozen, then inserts its covariance into the slow averaged equation. A short-memory stationary limit can yield quasilinear diffusion, but finite-time responses retain memory. Self-consistent collective fields enter through the dielectric function in homogeneous electrostatic problems.

The diffusive evolution of an averaged phase-space distribution function caused by weak fluctuating fields in the quasilinear approximation. In a stationary homogeneous one-dimensional electrostatic problem, if the potential's temporal power spectral density is $S_{\phi ,k}(\omega )$ with covariance $\int S_{\phi ,k}(\omega )e^{-i\omega \tau }d\omega /(2\pi )$, thenThis follows from the real part of the one-sided time correlation along an unperturbed particle trajectory. Both signs of the spatial wavenumber are included. The action-space counterpart for regular stellar orbits is quasilinear diffusion in action space.

Relativistic kinetic theory describes particles by a scalar phase-space distribution function on the future mass shell. Its Boltzmann equation uses geodesic streaming and a collision operator. Integrating its momentum moments supplies a kinetic stress-energy tensor and particle-number currents. An orthonormal tetrad makes the local momentum integrals identical in form to those of special relativity.

For unit particle mass, this symmetric third-order tensor transports the pressure tensor in a kinetic fluid description. Its contraction gives the heat flux $q_i=Q_{ijj}/2$. It vanishes for a local Maxwellian, whose odd central velocity moments vanish, but it need not vanish for a general anisotropic or skewed phase-space distribution function.

For $N$ identical particles with symmetric, unit-normalized phase-space distribution function $D_N$, the number-normalized two-particle marginal is $f_2(z_1,z_2)=N(N-1)\int D_{N}d{z}_3\cdots d{z}_N$. Its integral over $z_2$ is $(N-1)f(z_1)$ and its full integral is $N(N-1)$. The factorial convention counts ordered distinct pairs and absorbs the combinatorial coefficient in the first BBGKY hierarchy equation. It does not imply statistical independence of the pair.

Integrating a phase-space distribution function against products of particle velocity components produces its velocity moments. The zeroth gives number density, the first gives number flux, and, after multiplying by particle mass, the second gives momentum flux. Replacing $\mathbf{v}$ by $\mathbf{v}-\mathbf{u}$, where $\mathbf{u}$ is the mean velocity, defines central velocity moments; the second is the pressure tensor divided by particle mass.

An interval of particle velocities on which an equilibrium phase-space distribution function is constant. Its zero derivative removes the resonant imaginary part of the linear dielectric function for waves with phase velocity inside the interval. This permits a completely undamped electron acoustic wave in suitable single-population, nonthermal electron distributions. Zero slope alone is necessary but does not guarantee that the real dispersion relation has a root.

For an isotropic, integrable background phase-space distribution function, rotational symmetry makes the integral radial in velocity space. Its divergence with respect to $\mathbf{V}$ is $4\pi f(V)$ by the Dirac delta identity for the divergence of the inverse-square vector field. Integrating over a velocity-space sphere and using the divergence theorem proves the displayed identity. Exterior velocity shells have zero contribution, just as in the spherical shell theorem. This explains why the leading-log Chandrasekhar dynamical friction integral contains only particles slower than the test body.

For a gyrophase-independent phase-space distribution function, $d^{3}v=2\pi v_{\perp }d{v}_{\perp }d{v}_{\parallel }$. With first adiabatic invariant $\mu =v_{\perp }^2/(2B)$, this gives the measure in the title. When the magnetic field changes, both the measure and any explicit factor $\mu B$ in a moment must be perturbed.

For a phase-mixed one-dimensional collisionless stellar system, $E=v_z^2/2+\psi (z)$ is an integral of motion and $f_z=F(E)$. Integrating both signs of velocity gives $\nu (\psi )=\sqrt{2}{\int }_{\psi }^{\infty }F(E)(E-\psi {)}^{-1/2}dE$. Applying the Abel transform a second time gives ${\int }_E^{\infty }\nu (\psi )(\psi -E{)}^{-1/2}d\psi =\pi \sqrt{2}{\int }_E^{\infty }F(t)dt$; differentiating proves the inverse. Sufficient convergence, regularity, and correct escape-energy boundary conditions are essential. An arbitrary tracer-density/potential pair is physical only if the reconstructed phase-space distribution function is nonnegative. Applying this inversion to a galactic disk requires a locally separable vertical dynamics or a justified approximation to it, and control of radial mixing and the tilted velocity ellipsoid.

A phase-space distribution function constant inside a prescribed region and zero outside. The one-dimensional example in the title has a velocity plateau and distributional derivatives at its edges. Its simple kinetic response can support undamped collective modes, illustrating why absence of growing modes does not imply Landau damping. For electrons with fixed ions, $ϵ(\omega ,k)=1-{\omega }_p^2/({\omega }^2-k^2{V}^2)$, so collective roots lie outside the populated velocity interval.