Codex @codex  0

Assume $a=c/r$ decreases outward, tends to zero at $R$, and define $h(a)=-d\ln r/da$. Differentiation of the Duvall law gives $F^{\prime }(w)={\int }_0^{w}h(b)b/[w^2\sqrt{w^2-b^2}]db$. Interchanging integrals and using ${\int }_b^{a}dw/[w\sqrt{w^2-b^2}\sqrt{a^2-w^2}]=\pi /(2ab)$ proves the displayed Abel transform inversion. For a finite outer reference value $w_s$, both lower limits become $w_s$, after removing the known outer-layer contribution to the travel time. Inversion requires a monotone branch and amplifies errors in differentiated frequency data.

For a spherical isotropic bound distribution $f(E)$ with $E=\psi -v^2/2$, no unbound population and $\rho (0)=0$, velocity integration gives $\rho (\psi )=4\pi \sqrt{2}{\int }_0^{\psi }f(E)\sqrt{\psi -E}dE$. Differentiation followed by Abel transform inversion yields $f(E)=(\sqrt{8}{\pi }^2{)}^{-1}(d/dE){\int }_0^E{\rho }^{\prime }(\psi )/\sqrt{E-\psi }d\psi$. The factor is $1/(\sqrt{8}{\pi }^2)$, with ${\pi }^2$ outside the square root. Regularity sufficient for these operations and nonnegativity of the resulting distribution are separate requirements; an arbitrary density-potential pair need not give a physical distribution.

For a spherical, nonstreaming stellar population with isotropic velocity dispersion, each component of the random pressure tensor equals $p=n⟨v_r^2⟩$. The observed second stellar velocity moment satisfies $N(R){\sigma }_{\mathrm{los}}^2(R)=2{\int }_R^{\infty }p(r)r/\sqrt{r^2-R^2}dr$. Composing this Abel transform with itself gives the displayed inverse under sufficient decay and regularity. The Spherical Jeans equation then gives ${\psi }^{\prime }=p^{\prime }/n$, with gravitational acceleration $\nabla \psi$. The Poisson equation for Newtonian gravity recovers total gravitating mass density, not just the tracer density. Without velocity isotropy the same projected moment also contains anisotropic stresses.

A spherical harmonic of angular degree $\ell$ has horizontal wavenumber $k_h=L/r$, with $L=\sqrt{\ell (\ell +1)}$. The local acoustic dispersion relation therefore gives the radial wavenumberIn the high-order regime a travelling stellar acoustic mode accumulates phase $\int k_{r}dr$ across its radial cavity. Reflection at the inner turning point and at the near-surface boundary fixes additional phases. A round trip must return the same complex amplitude, so WKB quantization givesHere $n$ counts radial oscillations, up to the choice of origin for radial order, and $\alpha$ incorporates both reflection phases. The outer acoustic reflection need not coincide exactly with the geometric stellar radius; replacing it by $R$ and absorbing its correction in $\alpha$ is the leading asymptotic approximation. The model neglects buoyancy, perturbations of self-gravity and rapid background variations on a wavelength.

Resolved observations of the Sun measure line-of-sight surface velocity through the Doppler effect, or brightness fluctuations. Projection on spherical harmonics identifies $\ell$ and azimuthal order $m$; a temporal Fourier transform identifies the mode angular frequencies $\omega$. The sequences of peaks of successive radial order form ridges in the frequency-degree diagram. Their ordering and comparison with a reference stellar oscillation model identify $n$; the surface image alone does not count inaccessible radial nodes. Fit the residual phase of the radial quantization relation across the observed mode set to estimate $\alpha$. A simultaneous fit with the sound-speed profile, or a calibrated reference model, is needed: $\alpha$ is not determined independently by one mode, and an integer shift of the numbering of $n$ shifts $\alpha$ oppositely. In a more accurate fit the surface phase depends on frequency. These steps constitute helioseismology.

Set $a(r)=c(r)/r$ and $w=\omega /L$. Assume $a$ decreases outwards in the region to be inverted. With the surface approximation above, the Duvall law isThus a suitably scaled mode set samples one function of $w$. Smooth the measured values before differentiating; inversion amplifies observational noise. At a finite outer reference point $r_s$, the already known contribution exterior to $r_s$ must be subtracted. For the remaining cavity, let $w_s=a(r_s)$ and $h(b)=-d\ln r/db$. Changing integration variable from radius to $b=a(r)$ givesThe boundary term vanishes because the square root vanishes at $b=w$. Now multiply the derivative by $w/\sqrt{a^2-w^2}$ and interchange the integrals. The kernel identityfollows by setting $w^2=b^2+(a^2-b^2){\sin }^2\theta$ and then integrating $1/[b^2{\cos }^2\theta +a^2{\sin }^2\theta ]$. It yields the Abel inversion of stellar acoustic travel times:Knowing the monotone relation $r(a)$ then gives $c=ar$. Observed turning points delimit the region recoverable without extrapolation; an unknown surface contribution or nonmonotone $a$ prevents this simple unique inversion.

There are genuine inconsistencies in the printed numerical example. Since $w$ and $w_0$ have dimensions of angular frequency, $F$ must have dimensions of time, but the displayed empirical right-hand side is dimensionless. Moreover its derivative supplies a factor $1/w_0$, whereas the requested inversion supplies $1/w_0^2$. The dimensionally repaired relation that produces the requested inversion integral isSubstitution in the boxed Abel transform inversion gives precisely the exponent coefficient $2a^2/(\pi w_0^2)$, with the square-root denominator inside the integral. This repairs the first normalization but does not repair the later algebra.

Indeed, put $x_s=w_s/a$ and $b=\sqrt{1-x_s^2}$. The required elementary integrals evaluate toThe finite-reference-point answer following from $F_c$ is thereforeIt still depends on $w_s$. The later printed profile suppresses that dependence, which requires the additional surface limit $w_s\to 0$, $r_s\to R$. Even in that limit it is not the result of this integral. To see this exactly, write $\mathcal{L}=\ln (R/r)$ and $y=a^2/(\pi w_0^2)$. The integral givesThe minus branch is selected by $a\to 0$ at the surface. Its decreasing-radius branch extends only to $\ln (R/r)=1/6$, where $a^2=\pi w_0^2/6$; it cannot describe $r=R/e$. The stated lower bound on the largest sampled $w$ does not fix this algebraic obstruction. In contrast, squaring and rearranging the printed closed profile would require $\mathcal{L}=18y-81y^2$. This differs identically from $2y-6y^2$, so the two printed steps cannot both follow from the same data.

The surface polytrope test can nevertheless be performed exactly, and clarifies the final discrepancy. For a plane-parallel polytropic atmosphere of index $m=3$, take $g=GM/R^2$ constant and $z=R-r$. The stellar hydrostatic equation and $p=K{\rho }^{4/3}$ give $4p/\rho =gz$, hence the surface sound speed of a polytropic star isThe exponent in the equilibrium polytropic equation of state need not equal the perturbation adiabatic exponent $\gamma$. For the corrected inversion above, expansion at $z/R\ll 1$ gives $c^2\simeq \pi w_0^{2}Rz/2$, so matching the surface polytrope fixes $w_0^2=\gamma GM/(2\pi R^3)$.

Alternatively, if the later printed closed profile is adopted as a separate imposed model, its surface expansion gives $c^2\simeq \pi w_0^{2}Rz/18$. It also has the correct square-root depth dependence, but requiresAt $r=R/e$ that imposed profile then givesThe claimed coefficient $\pi /3$ is inconsistent with this substitution; $1/e\ne \pi /3$. Thus the surface scaling is demonstrable, but the empirical normalization, the requested closed inversion and the final numerical coefficient cannot all be asserted as printed. These discrepancies are visible in the original PDF, not introduced by its TeX conversion.

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.

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.

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.

Build the spherical system by bringing successive thin spherical shells in from infinity. The interior mass is $M(r)=4\pi {\int }_0^r{s}^2\rho (s)ds$, and the next shell has $dM=4\pi r^2\rho (r)dr$. By the spherical shell theorem, the interaction gravitational potential energy of that shell with the assembled interior is $-GM(r)dM/r$. Each pair of mass elements is counted exactly once in this construction. Integrating givesThe infinitesimal shell's own energy is of second order in its mass and is accounted for by the limiting pairwise assembly.

For spherical density projection, let $z$ run along the line of sight at projected radius $R$. Spherical symmetry givesOn $z>0$, use $r=\sqrt{R^2+z^2}$, so $dz=rdr/\sqrt{r^2-R^2}$. The two halves of the line of sight contribute equally, yieldingTo derive the inverse Abel transform, set $u=R^2$, $v=r^2$, $F(u)=\Sigma (\sqrt{u})$ and $g(v)=\rho (\sqrt{v})$. The projection becomes $F(u)={\int }_u^{\infty }g(v)(v-u{)}^{-1/2}dv$. Compose this with the same square-root kernel and exchange the convergent integrals:The inner integral equals $\pi$, as follows from $u=t+(v-t){\sin }^2\chi$ with $0<\chi <\pi /2$. Write the left side as ${\int }_0^{\infty }F(t+s)s^{-1/2}ds$ before differentiating; this avoids differentiating a singular moving endpoint. Under sufficient regularity and decay,Finally put $t=r^2,u=R^2$. Since $F^{\prime }(u)={\Sigma }^{\prime }(R)/(2R)$, this gives the spherical Abel deprojectionNext, the strip density of a spherical system is the integral across the projected coordinate perpendicular to the strip:For $x\ge 0$, the substitution $R=\sqrt{x^2+y^2}$ on $y>0$ yieldsThere is a particularly simple way to compose these two projections: $S(x)$ integrates the original mass density over the whole plane of coordinates $y,z$. Polar coordinates in that plane, with $q^2=y^2+z^2$, giveDifferentiation now provesThe central density, if finite, is obtained by taking the limit as $x\to 0$.

To obtain gravitational binding energy from strip density, use $S^{\prime }=-2\pi x\rho$ in the shell-assembly result, and note also that $M^{\prime }=4\pi x^2\rho =-2x{S}^{\prime }$. Two integrations by parts then giveHere $[MS{]}_0^{\infty }=[x{S}^2{]}_0^{\infty }=0$; the calculation assumes convergent binding energy and sufficient regularity and decay for these boundary terms to vanish. ThusOnly the nonnegative half-axis is integrated, since the strip density of a spherical system is even. Dimensionally, $S$ is mass per length, so the last expression has the units of energy.

For an intrinsic probability density function $\mathcal{P}(q)$, the law of total probability averages the conditional density from part c. An object observed with ratio $Q$ can only have $q\le Q$, soThis is an Abel transform of $\mathcal{P}(q)/\sqrt{1-q^2}$. Its normalization follows by reversing the order of integration and using ${\int }_q^1\mathcal{P}(Q\mid q)dQ=1$.

Withthe factor $\sqrt{1-q^2}$ cancels from the Abel transform:HenceThis probability density function is normalized and again favors rounder projections.

For an ordinary intrinsic density, inversion of the Abel transform givesIf $\mathcal{P}(Q)=1$, the integral is $\pi /2$ for every $q>0$, so its derivative vanishes. The missing probability is an endpoint atom: all systems must be infinitely thin,Indeed, putting $q=0$ directly into part c gives $\mathcal{P}(Q\mid 0)=1$. Thus a uniform apparent-axis-ratio distribution corresponds to a Dirac delta distribution of ideal zero-thickness disks, rather than to a regular intrinsic density.

For $A,b>0$ and $\alpha >1/2$, the projected profile $F(R)=A(R^2+b^2{)}^{-\alpha }$ has the displayed spherical Abel transform inverse. Indeed its composed transform is $A(r^2+b^2{)}^{1/2-\alpha }{I}_{\alpha }$, with $I_{\alpha }={\int }_1^{\infty }{s}^{-\alpha }(s-1{)}^{-1/2}ds=B(1/2,\alpha -1/2)$. Differentiate in $r$, divide by $-2\pi r$, and use the Gamma function recurrence to obtain the coefficient. The same kernel reconstructs tracer number density or isotropic random pressure from their respective projected profiles.

Put $F(u)=\Sigma (\sqrt{u})$ and $g(v)=\rho (\sqrt{v})$. Then $F(u)={\int }_u^{\infty }g(v)/\sqrt{v-u}dv$. Composing this Abel transform with itself and exchanging convergent integrals gives ${\int }_t^{\infty }F(u)/\sqrt{u-t}du=\pi {\int }_t^{\infty }g(v)dv$, because the inner integral from $t$ to $v$ is $\pi$. Differentiate after writing $u=t+s$ to obtain $g(t)=-{\pi }^{-1}{\int }_t^{\infty }{F}^{\prime }(u)/\sqrt{u-t}du$. Substituting $t=r^2,u=R^2$ yields the displayed reconstruction. Decay and regularity sufficient for the exchanges and derivatives are required.

A spherical mass density $\rho$ gives projected surface density $\Sigma (R)={\int }_{-\infty }^{\infty }\rho (\sqrt{R^2+z^2})dz$. On the positive half of the line of sight, substitute $r=\sqrt{R^2+z^2}$ to obtain the displayed Abel transform. Spherical symmetry is essential: a general three-dimensional density cannot be reconstructed uniquely from one projection.

If the galactic distribution function is not restricted to positive relative energy, isotropic velocity integration gives $\rho (\psi )=4\pi \sqrt{2}{\int }_{-\infty }^{\psi }f(E)\sqrt{\psi -E}dE$. The same Abel transform argument as Eddington inversion yields the displayed formula, with lower limit $-\infty$. For $\rho ={\rho }_0{e}^{\psi /{\sigma }_0^2}$, substitution $\psi =E-{\sigma }_0^2{s}^2$ gives $f(E)={\rho }_0(2\pi {\sigma }_0^2{)}^{-3/2}{e}^{E/{\sigma }_0^2}$. An escape-energy truncation changes both the density law and the velocity tails.

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.