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An augmented density $\rho (r,\psi )$ treats radius and relative potential as independent variables and reduces to the physical density on $\psi =\psi (r)$. Its extension away from that curve contains additional orbital-anisotropy information. Different extensions can give the same physical density; physical realizability additionally requires a nonnegative galactic distribution function.

The Poisson equation for Newtonian gravity givesFor $V>0$, nonnegative mass density away from the origin is equivalent to $q^2\ge 1/2$. Strict positivity requires $q^2>1/2$. This condition does not by itself guarantee a nonnegative galactic distribution function.

In a spherical Newtonian gravitational potential, use binding energy $E=\psi -v^2/2$ and specific angular momentum $L$. The displayed stationary galactic distribution function has locally Gaussian radial and transverse velocities: ${\sigma }_r^2={\sigma }^2$, ${\sigma }_{\theta }^2={\sigma }_{\varphi }^2={\sigma }^2/(1+r^2/r_a^2)$. Its mass density is $\rho ={\rho }_c{e}^{[\psi (r)-\psi (0)]/{\sigma }^2}/(1+r^2/r_a^2)$. The velocity-anisotropy parameter is ${\beta }_{\mathrm{an}}=r^2/(r^2+r_a^2)$, so the core is nearly isotropic and the halo nearly radial. These exact Gaussian integrals use the untruncated distribution over all velocities; imposing an escape-energy cutoff changes them.

In a spherical galactic distribution function with $F=L^{-1}{f}_E(E)$, write $L=r{v}_T$ and $d^{3}v=d{v}_r{v}_{T}d{v}_{T}d\psi$. The velocity Jacobian cancels the apparent singularity at $v_T=0$. Integration of Gaussian energy components then reduces to one full-line and one half-line Gaussian integral. The angular average makes each tangential velocity dispersion half the radial one, so the velocity-anisotropy parameter is $1/2$.

A steady galactic distribution function depending only on relative energy $E=\psi -v^2/2$ is isotropic and has zero mean velocity. If its one-dimensional velocity dispersion is constant, the Spherical Jeans equation reduces to ${\sigma }_0^2{\rho }^{\prime }=\rho {\psi }^{\prime }$, yielding the displayed density. This is a tracer relation in a specified gravitational potential, not a requirement that the tracer generate that potential. With no escape-energy cutoff its local velocity distribution is Maxwell-Boltzmann velocity distribution.

Without an escape-energy cutoff, a galactic distribution function proportional to $\exp [(\Psi -v^2/2)/{\sigma }^2]$ gives $\rho \propto e^{\Psi /{\sigma }^2}$ and a constant one-dimensional velocity dispersion. Since $\Psi ={\Phi }_0-\Phi$, differentiation gives the displayed hydrostatic relation. An ideal gas at constant temperature has pressure $P=\rho k_{B}T/m_0$, so its hydrostatic equilibrium equation is identical at the displayed dispersion-temperature identification. Coupling either relation to the same Poisson equation for Newtonian gravity, with matched central density and boundary data, therefore gives the same density structure. Truncating the stellar velocities at escape changes this equivalence.

The singular axisymmetric logarithmic gravitational potential admits this stationary galactic distribution function withThe zero of specific orbital energy is the printed logarithmic convention. Products of Gaussian integrals normalize its two spatial density terms. For $1/2\le q^2\le 1$ the two terms are separately nonnegative; mildly prolate models require a bound on their sum.

For a nonrotating axisymmetric galactic distribution function even in $L_z$, reverse every retrograde azimuthal velocity. The resulting distribution is $F_{+}(E,L_z)=2F(E,L_z)\Theta (L_z)$: it preserves density and even velocity moments but makes the streaming velocity $⟨|v_{\varphi }|⟩$. Dependence on conserved $E,L_z$ preserves stationarity. A centered Gaussian distribution of width $\sigma$ becomes a half-normal distribution, with mean $\sqrt{2/\pi }\sigma$.

For the logarithmic-potential two-integral distribution, $q^2>1$ makes $A<0$. Accessible phase space satisfies $L_z^2{e}^{-2E/V^2}\le V^2/e$, attained by equatorial circular motion. Hence the galactic distribution function is nonnegative precisely when $B+A{V}^2/e\ge 0$, giving the displayed upper bound. A larger prolate flattening parameter produces negative phase-space density even though the spatial mass density remains positive.

The density and self-consistent potential of a spherical galaxy need not determine its galactic distribution function. Different angular-momentum dependence can reproduce the same density with different velocity-anisotropy parameters. An isotropic Plummer model and its positive anisotropic Plummer distribution with unit anisotropy radius provide an explicit example. A chosen full augmented density is additional information.

The pressure tensor has Cartesian components $p_{ab}=\int f(v_a-u_a)(v_b-u_b)d^{3}v$. With zero mean velocity and an isotropic galactic distribution function, off-diagonal components vanish by reflection symmetry and diagonal components are equal by rotational symmetry. ThereforeHere $p$ is a tracer number-weighted random-motion stress, rather than thermodynamic pressure per volume. Its tensor divergence is $(\mathrm{div}\mathbf{p}{)}_a={\partial }_b(p{\delta }_{ab})={\partial }_{a}p$. The time-independent, nonstreaming Jeans equation consequently reduces toThe gravitational acceleration convention is $\nabla \psi$: this $\psi$ differs by a sign from the usual Newtonian gravitational potential $\Phi$.

Let $z$ measure distance along the line of sight. Spherical density projection givesBecause isotropy makes the local second stellar velocity moment along every line of sight equal to $p/n$, and because there is no streaming motion, the projected moment isThis is the forward relation for isotropic stellar pressure deprojection.

Both inversions follow from one Abel transform calculation. Write $F(u)=N(\sqrt{u})$ and $a(v)=n(\sqrt{v})$, so $F(u)={\int }_u^{\infty }a(v)(v-u{)}^{-1/2}dv$. Compose the transforms, using Tonelli's theorem for nonnegative densities, or absolute convergence for signed profiles:Assume sufficient decay to make these integrals finite and sufficient regularity for the last derivative. Then $J^{\prime }(x)=-\pi a(x)$. Setting $x=r^2$ gives the spherical Abel deprojectionReplacing $F(u)$ by $N(\sqrt{u}){\sigma }^2(\sqrt{u})$ and $a(v)$ by $p(\sqrt{v})$ provesThese are the requested expressions, with $u=R^2$. The composed-transform argument avoids an invalid separate differentiation of a divergent lower-end kernel.

The separated disk modes. Away from the plane of a razor-thin disk, there is no volume source, so the Newtonian gravitational potential obeys the Laplace equation in cylindrical coordinates. Separation of variables givesHere $k>0$ is a radial wavenumber, with units of inverse length. Regularity on the axis selects the Bessel function of the first kind $J_0(kR)$ rather than the singular second solution. Decay away from the disk, continuity across it and reflection symmetry select $Z=e^{-k|z|}$. These conditions apply to each nonzero-wavenumber mode; a uniform sheet is a separate zero-mode problem.

Integrate the Poisson equation for Newtonian gravity across the disk to obtain the jump conditionFor the unit-amplitude separated mode, the jump is $-2k{J}_0(kR)$. ThusAn individual basis density is signed; positive physical surface densities are built by superposition.

An arbitrary axisymmetric surface density. Define its order-zero Hankel transform byMatching the coefficients of the mode densities gives the Hankel representation of thin-disk gravity,The boundary conditions exclude added external fields. For sufficiently decaying surface density, this is the isolated-disk potential; for infinite scale-free disks, only potential differences or regularized forces need be finite.

The Mestel disk. Put $C={\Sigma }_0{R}_0$. The Bessel function integral ${\int }_0^{\infty }{J}_0(k{R}^{\prime })d{R}^{\prime }=1/k$, understood with a decaying regulator, gives $\hat{\Sigma }=C/k$. Differentiate the potential before removing the regulator and use $J_0^{\prime }=-J_1$:The last integral is $1/R$, since ${\int }_0^{\infty }{J}_1(u)du=[-J_0(u){]}_0^{\infty }=1$. Therefore the circular speed and enclosed mass areConsequently $v_c^2=GM(R)/R$. This equality follows from the specific Mestel disk calculation; the spherical shell theorem cannot be applied to general disks. Its total mass and absolute potential referenced to infinity diverge. A convenient finite midplane reference is $\Phi (R,0)=v_0^2\log (R/R_0)$.

The rotating distribution function. Use specific energy $E=(v_R^2+v_{\varphi }^2)/2+\Phi$ and $L_z=R{v}_{\varphi }$. For the printed positive exponential, take $\epsilon =-E$ and $s:=\sigma >0$; $s$ has units of squared velocity. If instead $\epsilon$ means ordinary total energy and $\sigma >0$, the velocity integral diverges, so that literal convention cannot define the requested galactic distribution function. Equivalently one can retain ordinary energy and choose the printed $\sigma$ negative.

With the binding-energy convention, write the amplitude as $A$ to distinguish it from the function:Integration over both radial velocities and positive azimuthal velocities givesThe half-line integral follows by setting $u=v_{\varphi }^2/(2s)$ in the defining Gamma function integral and requires $q>-1$. Matching its radial power and normalization to the Mestel disk yields the rotating Mestel disk distribution function parametersSince $v_0^2,s>0$, the required integrability condition is automatic. The amplitude uses the stated potential reference; changing the additive potential constant changes $A$ without changing the physical distribution.

First velocity moment. Assume a stationary, spherical, nonstreaming collisionless stellar system with constant mass-to-light ratio ${\rm Y}$. Use a luminosity-weighted galactic distribution function, with luminosity density $\lambda$. Multiplying the Collisionless Boltzmann equation by $v_i$ and integrating over velocities, with its velocity-space boundary terms vanishing, givesSpherical symmetry makes the stress tensorHere ${\sigma }_t^2$ is one tangential component, as in the paper; the total tangential variance is $2{\sigma }_t^2$. Its radial divergence is $q^{\prime }+2(q-q_t)/r$, since the divergence of the radial unit vector is $2/r$. With the velocity-anisotropy parameter $\beta =1-{\sigma }_t^2/{\sigma }_r^2$, and enclosed mass $M(r)={\rm Y}L(r)$, this gives the Spherical Jeans equationThis uses a constant ratio of total gravitating mass to luminosity, rather than an arbitrary independent dark halo.

Projection and elimination of anisotropy. Write $R$ for projected radius and $r$ for spherical radius. The Abel transform of the luminosity density giveswhose inverse is the displayed density inversion in the paper. At a point on this line of sight, ${\sin }^2\theta =R^2/r^2$. The line-of-sight second stellar velocity moment is ${\sigma }_r^2{\cos }^2\theta +{\sigma }_t^2{\sin }^2\theta ={\sigma }_r^2(1-\beta R^2/r^2)$. Therefore the measured quantity $J(R)=\mu (R){\sigma }_p^2(R)$ satisfiesThe Spherical Jeans equation gives $\beta q=-r{q}^{\prime }/2-\gamma \lambda L/(2r)$. Substitution yieldsThe inner $\gamma L$ printed in the PDF's definition of $p$ must therefore be $\lambda L$: the literal expression has an extra gravitational factor and lacks the density. Both this derivation and dimensional consistency require the correction. Define the projected spherical Jeans residual byTo express it entirely in terms of the radial stress, putDifferentiating the nonsingular $z$ representation gives $A^{\prime }(R)=R{\int }_R^{\infty }{q}^{\prime }(r)/\sqrt{r^2-R^2}dr$. HenceIf $R^{2}A$ vanishes at zero and infinity, then ${\int }_0^{\infty }6\pi Rp(R)dR=6\pi [R^{2}A{]}_0^{\infty }=0$. These are the absence-of-boundary-stress conditions needed for the global result.

The virial relation, term by term. Interchange the order in the residual integral and useIt follows thatThe first term is $2K/{\rm Y}$: spherical averaging assigns one third of the integrated velocity-square to any fixed viewing direction, giving $K=3\pi {\rm Y}\int RJdR$. The gravitational potential energy isThus the integrated residual is exactly $(2K+W)/{\rm Y}$, and its vanishing is the virial theorem. Solving for $\gamma$ gives the global projected virial estimate of a mass-to-light ratio,It is independent of the radial anisotropy profile when the full aperture and the stated boundary conditions are used. A finite aperture retains a boundary term.

Both models obtain their density by integrating a galactic distribution function over velocity. The planar model integrates one vertical velocity, whereas the spherical model integrates all three components. In each case the resulting density must also generate its own gravitational potential through the Poisson equation for Newtonian gravity. The two printed numerical subparts are treated separately below.

Take $V=|\mathbf{V}|$ to be the test body's instantaneous speed relative to the nonstreaming background, rather than an ensemble root mean square speed. Use a local isotropic galactic distribution function $f(v)$ normalized as a number distribution, and assume independent encounters and a large, approximately constant Coulomb logarithm in stellar dynamics.

For a background particle with relative incoming velocity $\mathbf{w}=\mathbf{V}-\mathbf{v}$, a gravitational scattering angle $\chi$ obeysIn the centre-of-momentum frame, the test body's share of the relative velocity is $m/(M+m)$. The azimuthal average cancels transverse changes, leaving a change parallel to the incoming relative velocity,The number of encounters per unit time in $d^{3}v$ and impact interval $db$ is $f(v)d^{3}vw2\pi bdb$. Integrating the longitudinal change from $b=0$ to $b_m$ givesAt leading logarithmic order replace the logarithm by $2\ln \Lambda$, evaluated at a representative speed. For isotropic $f$, the velocity-shell identity for dynamical friction isOne proof takes the divergence with respect to $\mathbf{V}$: the inverse-square vector field has divergence $4\pi {\delta }^{(3)}(\mathbf{V}-\mathbf{v})$, whose velocity integral is $4\pi f(V)$, and rotational symmetry plus the divergence theorem determines the radial field. Exterior velocity shells give zero field. Consequently the systematic Chandrasekhar dynamical friction is antiparallel to motion, withIn the massive-test-body approximation $M\gg m$, this becomesThe finite-test-mass dynamical friction coefficient shows the needed qualification: $M>m$ alone does not make $M+m=M$. The quoted expression also neglects diffusion and the velocity dependence of the logarithmic cutoff. For the lowered Maxwellian velocity distribution, set $f=0$ for $v>v_e$; continuing the printed subtraction beyond escape would give a negative distribution. If $V>v_e$, the integral stops at $v_e$.

Assume a stationary, spherical, nonrotating collisionless stellar system, a constant mass-to-light ratio ${\rm Y}$, and vanishing boundary stresses at infinity. Let $\lambda$ be the luminosity density, so $\rho ={\rm Y}\lambda$, and set $\gamma =G{\rm Y}$. Distinguish spatial radius $r$ from projected radius $R$. The enclosed luminosity and radial force areHere ${\sigma }_t^2$ is the velocity dispersion of either one of the two tangential components; thus $\beta =1-{\sigma }_t^2/{\sigma }_r^2$. If a combined tangential variance is used instead, its contribution must be divided by two in this definition.

To derive the Jeans equation, start with the steady Collisionless Boltzmann equation $v_j{\partial }_{j}f-({\partial }_j\Phi ){\partial }_{v_j}f=0$. Multiply by $v_i$ and integrate over all velocities. Integration by parts in velocity, with $f$ decaying fast enough to eliminate boundary terms, givesThe same equation holds with luminosity weighting because the tracer's luminosity is conserved along its orbit. Spherical symmetry makes the local stellar velocity moment tensor diagonal, with entries $q=\lambda {\sigma }_r^2$, $q_t=\lambda {\sigma }_t^2$, $q_t$. Its radial divergence is $q^{\prime }+(2q-2q_t)/r$: the second term accounts for the changing directions of the radial and tangential unit vectors. ThereforeThis is the Spherical Jeans equation; it follows from a stellar velocity moment rather than requiring a particular form of the galactic distribution function.

Integrating the luminosity density along a line of sight gives its surface brightness:The second identity is Spherical Abel deprojection. For the second stellar velocity moment, the angle between the radial direction and the line of sight has ${\sin }^2\theta =R^2/r^2$. The line-of-sight velocity dispersion is therefore ${\sigma }_r^2{\cos }^2\theta +{\sigma }_t^2{\sin }^2\theta ={\sigma }_r^2(1-\beta R^2/r^2)$, andFrom the Spherical Jeans equation, $\beta q=-r{q}^{\prime }/2-\gamma \lambda L/(2r)$. Substitution producesThe primary PDF has a genuine symbol error in its projected gravity term: it prints a second $\gamma$ multiplying $L$ inside the integral, where $\lambda$ is required. The corrected projected spherical Jeans residual isTo justify the derivative at the singular-looking lower limit, write $A(R)={\int }_0^{\infty }q(\sqrt{R^2+z^2})dz$. Differentiation then gives $A^{\prime }(R)=R{\int }_R^{\infty }{q}^{\prime }(r)(r^2-R^2{)}^{-1/2}dr$. In particular, $Rp=(R^{2}A{)}^{\prime }$. If $R^{2}A$ tends to zero both at zero and infinity, as for a finite isolated equilibrium with sufficiently decaying stresses,If a boundary stress survives, its contribution must be retained; the zero-integral statement is not a finite-aperture identity.

Now identify each integral with a term in the virial theorem. Let $K$ be the total kinetic energy, including all three random components, and $W$ the gravitational potential energy. Averaging over spatial directions in a spherical system gives one third of the total stellar velocity moment along any chosen line of sight, even if the velocities are locally anisotropic. ThusBuilding the spherical mass shell by shell givesFinally, Fubini theorem and ${\int }_0^r{R}^3(r^2-R^2{)}^{-1/2}dR=2r^3/3$ give the gravitational contribution to the residual integral explicitly:ConsequentlyThe global projected virial estimate of a mass-to-light ratio is independent of the velocity-anisotropy parameter under these assumptions. It does not remove the mass--anisotropy--density degeneracy for arbitrary radial mass profiles. In particular, if an independent dark matter halo makes mass fail to follow light, a single constant ${\rm Y}$ is no longer the physical enclosed mass-to-light ratio at every radius.

In an axisymmetric cylindrical coordinate system, the steady Collisionless Boltzmann equation isLet $\nu =\int f{d}^{3}v$ and $\overline{A}={\nu }^{-1}\int Af{d}^{3}v$. Multiply by $v_R$ and integrate over velocities. The spatial terms give ${\partial }_R(\nu \overline{v_R^2})+{\partial }_z(\nu \overline{v_R{v}_z})$. Integration by parts in $v_R$ supplies $\nu {\Phi }_R-\nu \overline{v_{\varphi }^2}/R$, and the $v_{\varphi }$ derivative supplies $\nu \overline{v_R^2}/R$. The $v_z$ force term vanishes because $v_R$ does not depend on $v_z$. This assumes sufficient decay of the galactic distribution function to discard all velocity boundary terms. The radial Jeans equation is consequentlyReflection symmetry about $z=0$ makes $\nu$ even and $\overline{v_R{v}_z}$ odd. At the midplane the latter vanishes, and ${\nu }^{-1}{\partial }_z(\nu \overline{v_R{v}_z})={\partial }_z\overline{v_R{v}_z}$. Multiplying by $R/\nu$ provesAll quadratic velocities here are population moments, not individual stellar velocities. In particular, symmetry does not make the midplane tilt contribution to asymmetric drift disappear: an odd mixed moment can have a nonzero derivative at zero.

For application to the Milky Way, assume negligible mean radial and vertical flows, and define the circular speed by $v_c^2=R{\Phi }_R(R,0)$. Write $\overline{v_R^2}={\sigma }_R^2$ and $\overline{v_{\varphi }^2}={\overline{v}}_{\varphi }^2+{\sigma }_{\varphi }^2$. Rearranging gives the exact stress-support relationThus the stellar asymmetric drift $v_a=v_c-{\overline{v}}_{\varphi }$ is the right-hand side divided by $v_c+{\overline{v}}_{\varphi }$. For $v_a\ll v_c$, this denominator is approximately $2v_c$. If $\nu \propto e^{-R/h_{\nu }}$ and ${\sigma }_R^2\propto e^{-R/h_{\sigma }}$ locally, the bracket becomes $R/h_{\nu }+R/h_{\sigma }-1+{\sigma }_{\varphi }^2/{\sigma }_R^2$, making the density and dispersion-gradient contributions explicit.

A radially declining random-motion stress supplies some of the force needed to support a galactic disk, so its stars need less ordered azimuthal motion than a population on circular orbits. Hotter, often older, stellar populations therefore generally rotate more slowly than cold populations in the same gravitational potential; this is the observed stellar asymmetric drift. It is a stress-support effect, not a frictional slowing of each orbit. For a velocity ellipsoid aligned approximately with spherical coordinates, $\overline{v_R{v}_z}\simeq ({\sigma }_R^2-{\sigma }_z^2)z/R$, and the midplane tilt term contributes $-({\sigma }_R^2-{\sigma }_z^2)$ rather than zero.

Local measurements of mean rotation and all components of the velocity dispersion, combined with tracer-density and dispersion gradients, can therefore constrain $v_c$ and the radial Newtonian gravitational field. A hotter sample's mean rotation must not be identified directly with the circular speed. The tracer number density need not be the gravitating mass density. Nonaxisymmetric streaming, a nonstationary disk, poorly measured radial gradients, or a neglected tilt term can bias this inference; these are assumptions of the derivation rather than extra free corrections to the circular force.

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.