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The divergence theorem states ${\int }_V\nabla \cdot \mathbf{F}dV={\oint }_{\partial V}\mathbf{F}\cdot \mathbf{n}dS$. With the Poisson equation for Newtonian gravity, this givesor gravitational-acceleration flux $\oint \mathbf{g}\cdot \mathbf{n}dS=-4\pi GM(V)$.

Use cylindrical radius $R=r\sin \theta$, so $r(1+|\cos \theta |)=r+|z|$. An arbitrary reference length in the logarithm only adds a constant potential. Off the equatorial plane, the radial part of ${\nabla }^2\log [r(1+|\cos \theta |)]$ is $1/r^2$. Writing $c=\cos \theta$, its angular part is $r^{-2}d[(1-c^2)d\log (1+|c|)/dc]/dc=-1/r^2$ on either open hemisphere. The two cancel: there is no volume density away from $z=0$.

At the plane, ${\partial }_z\varphi {|}_{0^{+}}=v_0^2/R$ and ${\partial }_z\varphi {|}_{0^{-}}=-v_0^2/R$. Applying the gravitational flux law to a thin pillbox gives $4\pi G\Sigma ={\partial }_z\varphi {|}_{0^{+}}-{\partial }_z\varphi {|}_{0^{-}}$. ConsequentlyThis is the Mestel disk potential-density pair. There is no additional point mass at the origin: the enclosed mass tends to zero linearly as the enclosing radius shrinks.

At $z=0$, $\varphi =v_0^2\log R$ up to a constant, so radial circular balance givesOn a sphere of radius $r$, ${\partial }_r\varphi =v_0^2/r$ at every nonsingular angular point. Its flux is $4\pi v_0^{2}r$, and henceDirect integration of the surface density confirms $2\pi {\int }_0^r\Sigma (R)RdR=v_0^{2}r/G$. Thus $v_{\mathrm{circ}}^2=GM(<r)/r$ even though the source is a razor-thin disk. This equality is a special scale-free property of the Mestel disk, not evidence that the source is spherical. In a general disk, matter outside the orbit and nonspherical interior forces prevent this enclosed-mass formula for the local circular speed. The model has infinite total mass, so its potential is defined by differences rather than a zero at infinity.

Adopt the planar Fourier transform conventionExtend the disk potential to three dimensions. The Poisson equation for Newtonian gravity becomesFor $k>0$, decay away from an isolated disk and continuity across it give $\tilde{\varphi }=B(\mathbf{k})e^{-k|z|}$. Integration through the plane yields $-2kB=4\pi G\tilde{\Sigma }$, hence the razor-thin disk Poisson kernelThe given transform of $K(\mathbf{x})=1/|\mathbf{x}|$ is $\tilde{K}=2\pi /k$. The convolution theorem therefore gives the forward potential relationFor the reverse relation, $\widetilde{{\Delta }_2\varphi }=-k^2\tilde{\varphi }$, soInvert the transform to obtain the thin-disk potential-density inversionAn additive constant in $\varphi$ is annihilated by ${\Delta }_2$. These formulas hold for suitable isolated finite disks, and in a distributional or regularized sense for less regular profiles. The uniform zero-wavenumber sheet needs a separate vertical solution; the infinite Mestel disk requires potential differences or an infrared regularization, since its absolute forward integral diverges.

For an isolated, virialized three-dimensional Newtonian system with negligible boundary-pressure terms, the virial theorem gives $2K+W=0$. Hence its total energy is $E=K+W=-K$. Defining the kinetic temperature by $K=3NkT/2$ givesThus losing energy makes the system hotter, while adding energy makes it cooler. This negative heat capacity of a virialized gravitational system uses the inverse-distance potential and the stated boundary assumptions; it is not a dimension-independent consequence of self-gravity.

A gravothermal catastrophe occurs when a stellar core loses heat to its halo through two-body relaxation, contracts and becomes hotter because of its negative heat capacity. The larger temperature gradient then drives more heat outward and accelerates contraction. In the confined three-dimensional isothermal sequence this runaway is associated with loss of a regular thermodynamic equilibrium at a finite total energy. It is different from a fast collisionless instability.

For the rods, use energies and masses per unit axial length, consistently with their logarithmic interaction. DefineIntegrating the two-dimensional Maxwell-Boltzmann velocity distribution gives $\rho =A{e}^{\beta \psi }$, with a positive normalization $A$. The proposed potential has $\psi ={\psi }_e-2\log F/\beta$, so this requires $\rho \propto F^{-2}$. Direct differentiation givesConsequently the Poisson equation for Newtonian gravity is satisfied precisely byThe free normalization of the velocity distribution can be chosen to give this density. Its enclosed line mass verifies total-mass normalization:At the wall $F(1)=1$, so the required potential value is also satisfied. This establishes the two-dimensional self-gravitating rod equilibrium, including its field equation, distribution law, mass and boundary condition.

A regular positive solution needs $F(0)=1-q>0$. Since $\beta =m/(kT)$,The minimum is an infimum of regular equilibrium temperatures. At equality the central logarithm is singular; for $T<T_{\min }$, $F$ vanishes inside the box. Thus equality is not a regular isothermal rod configuration.

Each of the two velocity components has second moment $1/\beta$, so the kinetic energy is $K=M/\beta =NkT$. With the given reference potential, the gravitational potential energy is $W=-\frac{1}{2}\int \rho \psi d^{2}R$. Using $dM=M(1-q)dx/F^2$ givesThe remaining integral can be evaluated explicitly:Adding $K$ therefore yieldsNow $Nk{T}_{\min }=G{M}^2/2$, $\Theta =T/T_{\min }=1/q$ and $M/\beta =Nk{T}_{\min }\Theta$. The energy-temperature curve of a two-dimensional self-gravitating rod gas is consequentlyAt fixed box area, differentiation givesTo establish its sign rather than infer it from a plot, expand the logarithm for $\Theta >1$:The series and its derivative converge locally uniformly in $\Theta >1$. Moreover $E\to -\infty$ as $\Theta \downarrow 1$ and $E\to +\infty$ as $\Theta \to \infty$. Thus every finite energy on this mean-field isothermal sequence has a unique regular equilibrium temperature, and there is no finite-energy termination or turning point. The central density $M/[V(1-q)]$ diverges only in the infinite-negative-energy limit.

There is no finite-energy gravothermal catastrophe in this two-dimensional rod model. Its logarithmic interaction gives a different caloric curve from the three-dimensional inverse-distance problem. This microcanonical conclusion does not prohibit a distinct fixed-temperature collapse: in the canonical ensemble, a bath at or below $T_{\min }$ admits no regular equilibrium of this family.

The strong equivalence principle asserts universality of free fall even for bodies with gravitational self-binding, together with independence of local experimental outcomes, including gravitational experiments, from the location and velocity of a freely falling laboratory. “Local” means that external tidal variation across the apparatus can be neglected. It does not mean that all curvature, or a laboratory's own gravitational field, can be removed. The weak principle concerns freely falling test bodies without appreciable self-gravity; laboratory composition tests alone therefore do not establish the strong version.

Lunar laser ranging supplies evidence specifically sensitive to the strong version. The Earth and Moon have different fractions of gravitational self-energy. If those fractions affected gravitational and inertial mass differently, their accelerations toward the Sun would differ and the lunar orbit would acquire a periodically varying polarization, the Nordtvedt effect. Combining lunar ranging with laboratory bounds on composition-dependent differential acceleration separates this self-energy effect. A determination available before this examination found the self-energy contribution to the Earth–Moon difference in $m_G/m_I$ to be $(-2.0\pm 2.0)\times 10^{-13}$, consistent with zero; see the 2004 experimental analysis. Universality-of-free-fall experiments, gravitational redshift measurements and tests of local Lorentz invariance support the weaker constituent principles, but should not be confused with a direct self-gravity test.

In general relativity, all matter couples to one metric tensor, and freely falling test particles follow its geodesics. At any event, Riemann normal coordinates give $g_{ab}={\eta }_{ab}$ and ${\Gamma }^a{}_{bc}=0$. Nongravitational local equations reduce to their special-relativistic forms in that frame. Gravitational binding is governed by the same metric field equations rather than a separate composition-dependent force, and the local gravitational coupling is universal. These features implement the strong equivalence principle. Curvature still governs geodesic deviation, so this construction removes uniform acceleration, not tidal gravity.

To geometrize Newtonian mechanics, use a four-dimensional manifold with coordinates $(t,x^1,x^2,x^3)$, absolute time ${\tau }_a=(dt{)}_a$ and a degenerate spatial inverse metric $h^{ij}={\delta }^{ij}$, $h^{0a}=0$. For a gravitational potential $\Phi (t,x)$ define the torsion-free Newton–Cartan gravitational connection byIt satisfies ${\nabla }_a{\tau }_b=0$ and ${\nabla }_a{h}^{bc}=0$. An affine geodesic has $d^{2}t/d{\lambda }^2=0$; choosing $\lambda =t$ makes its spatial equationsThis is Newton's free-fall law. The geometry is Newton–Cartan theory, with an affine connection and degenerate metric data, rather than a nondegenerate four-dimensional Lorentzian metric. The spatial slices themselves remain Euclidean.

Use the curvature conventionIn these coordinates the quadratic connection terms vanish, and the only nonzero curvature components, apart from antisymmetry in $c,d$, areIn particular $R_{00}={\nabla }^2\Phi$ and all other components of the Ricci tensor vanish. This Newtonian tidal curvature gives relative acceleration $\delta {\ddot{x}}^i=-{\partial }_i{\partial }_j\Phi \delta x^j$. Spatially constant gravitational acceleration can be removed by an accelerating frame because its Hessian vanishes. The Poisson equation for Newtonian gravity becomes $R_{ab}=4\pi G\rho {\tau }_a{\tau }_b$. Fluid mechanics is compatible with the same connection: the covariant continuity equation is the usual mass-conservation equation, and the spatial Euler equation is $u^b{\nabla }_b{u}^i=-{\rho }^{-1}{\partial }_{i}p$, which includes the gravitational force through the connection.

Now replace this absolute-time geometry by the Lorentzian metric of general relativity, with signature $(-,+,+,+)$. A perfect fluid haswhere $\epsilon$ is rest-frame energy density. In units $c=1$, $\epsilon \simeq \rho$ for a nonrelativistic fluid. The Newtonian equation suggests a covariant curvature equation sourced by $T_{ab}$. However $R_{ab}=\kappa T_{ab}$ alone conflicts with generic stress-energy conservation, because the contracted Bianchi identity gives ${\nabla }^a{R}_{ab}=\frac{1}{2}{\nabla }_{b}R$. Among expressions linear in Ricci curvature, conservation therefore selectsA constant multiple of the metric is also divergence-free, giving the possible cosmological constant term. This is a heuristic selection using locality, universal metric coupling and second-order metric equations; Newtonian matching by itself does not prove uniqueness among every possible gravity theory.

To fix the coefficient, first set the cosmological constant to zero. The trace-reversed equation would read $R_{ab}=\kappa (T_{ab}-\frac{1}{2}T{g}_{ab})$. In the fluid rest frame, $T=-\epsilon +3p$, so $R_{00}=\frac{\kappa }{2}(\epsilon +3p)$. In the weak static Newtonian limit, $g_{00}=-(1+2\Phi )$ and $R_{00}\simeq {\nabla }^2\Phi$. For negligible pressure the Poisson equation for Newtonian gravity then requires $\kappa /2=4\pi G$. Restoring $c$ uses $x^0=ct$, $R_{00}\simeq {\nabla }^2\Phi /c^2$ and $T_{00}\simeq \rho c^2$, giving $\kappa =8\pi G/c^4$. ThusPressure contributes to relativistic active gravity, while the low-pressure weak-field limit recovers Newton's Poisson equation. The comparison fixes the Newtonian normalization but does not fix $\Lambda$, which is an independent parameter.

Use a steady, isotropic, self-gravitating dark matter halo, with $V$ its constant one-dimensional velocity dispersion. The Spherical Jeans equation and spherical mass conservation readFor the scale-free outer region, put $\rho =A{r}^{-p}$. Its enclosed mass scales as $r^{3-p}$, so $GM(r)/r$ is constant only if $p=2$; the Spherical Jeans equation then gives $2V^2=GM(r)/r=4\pi GA$. Thus the singular isothermal sphere hasIts circular speed is $v_c=\sqrt{2}V$. If a speed denoted $V$ were instead the circular speed, the coefficient would be $V^2/(4\pi G)$, but the printed variable is a velocity dispersion.

The assumptions matter. A constant velocity dispersion and spherical shape alone do not specify a unique central mass density or orbital distribution. Even with isotropy, the Spherical Jeans equation integrates to $\rho ={\rho }_0\exp [-(\Phi -{\Phi }_0)/V^2]$, and the self-gravitating Poisson equation for Newtonian gravity givesThere are finite-central-density isothermal solutions as well as the singular one; the $r^{-2}$ expression is the intended outer scale-free halo approximation. A finite halo must also end or steepen at large radius, since the untruncated singular isothermal sphere has infinite total mass.

For a spherically symmetric potential, the Poisson equation for Newtonian gravity becomesIntegrate from the center, excluding any additional point mass there. The boundary condition is $r^2{\varphi }^{\prime }(r)\to 0$ as $r\to 0$, soThis derives the enclosed-mass form of the spherical shell theorem from the differential equation. Outside the mass, integrating ${\varphi }^{\prime }=GM/r^2$ gives $\varphi =-GM/r+C$; the conventional zero at infinity sets $C=0$.

For the dyadic shells, multiply their constant mass density by their spherical volumes. The $n$th shell contributesSumming the geometric series yieldsAt radius $r=2^{-N}$, the enclosed shells are exactly those with $n\ge N+1$; a boundary itself contributes no mass. ThusDividing by $r^2=4^{-N}$ givesThese dyadic spherical shells with inverse-radius density scaling have the same field magnitude at each shell boundary, although the field varies inside an individual shell. Finally, outside the planet,with zero potential at infinity; without that convention an arbitrary additive constant remains.

Assume a steady, locally plane-symmetric galactic disk, with negligible radial tilt term in the vertical Jeans equation. Write $\rho (z)$ for the stellar mass density, and $K_z={\partial }_z\Phi$ for the upward gradient of the Newtonian gravitational potential; the actual vertical gravitational acceleration is $-K_z$. Constant vertical velocity dispersion givesIntegrating the plane-parallel Poisson equation for Newtonian gravity from the symmetry plane gives $K_z(z)=4\pi G{\int }_0^z\rho (z^{\prime })d{z}^{\prime }$. Outside most of the mass, this tends to $2\pi G\Sigma$ for $z>0$, where the surface density includes both sides of the galactic disk. Treating this field as constant and integrating the vertical Jeans equation yieldsThis exponential is the thin-sheet approximation or the outer tail, not the exact profile at every height. To see the distinction without an additional assumption, set $\rho ={\rho }_0{e}^{-u}$ with $u=(\Phi -\Phi (0))/{\sigma }^2$. The Poisson equation for Newtonian gravity becomes $u^{\prime \prime }=(4\pi G{\rho }_0/{\sigma }^2)e^{-u}$, with $u(0)=u^{\prime }(0)=0$, and is solved bySince $\Sigma =2{\rho }_0{z}_0$, the exact self-gravitating isothermal slab has $z_0={\sigma }^2/(\pi G\Sigma )$. At large $|z|/z_0$, ${\mathrm{sech}}^2(z/z_0)\sim 4e^{-2|z|/z_0}$, recovering precisely the boxed scale height; near the plane its slope vanishes, whereas an exponential has a cusp.

Measure the number-density profile $n(z)$ and vertical velocity dispersion ${\sigma }_z(z)$ of a tracer stellar population. The vertical Jeans equation estimates the restoring field fromThe second term accounts for the tilt of the velocity ellipsoid; an exponential tracer with constant velocity dispersion and negligible tilt gives $K_z={\sigma }_z^2/h$ above the plane. The tracer need not supply all the gravitating mass. To infer the dynamical surface density from vertical stellar motions, integrate the axisymmetric Poisson equation for Newtonian gravity:For a nearly flat galaxy rotation curve with a small radial correction, this reduces to $K_z/(2\pi G)$. Estimate stellar surface density independently from counts, the initial mass function and faint objects, and add the gas surface density and stellar remnants. The residuallimits the allowed unseen material within the same height interval. A fitted dark matter halo contributes too: measurements at several heights help distinguish its extended density from a thin dark component in the galactic disk. Under a specified decomposition, the disc dark-matter fraction is the inferred dark-disc surface density divided by the total disc surface density, with uncertainty or an upper limit if the residual is consistent with zero. Selection effects, disequilibrium, the tilt term and uncertain baryonic counts must enter the error budget; an excess alone does not establish that all of it is in the galactic disk.

Put $V=|v_0|>0$, take $q>0$ without loss of generality, and write $s=R^2+z^2/q^2$. The Newtonian gravitational potential is singular at the origin; all pointwise claims below concern $s>0$. The zero of the Newtonian gravitational potential is the one in the paper, with an implicit fixed length unit inside the logarithm.

The Poisson equation for Newtonian gravity in cylindrical coordinates givesThe two contributions to the Laplacian areConsequently the mass density of the axisymmetric logarithmic gravitational potential isThe coefficient of $R^2$ is positive. The coefficient of $z^2$ determines the density positivity for an axisymmetric logarithmic potential: strict positivity at every noncentral point requires $q^2>1/2$ and $v_0\ne 0$. Nonnegativity permits $q^2=1/2$, where the mass density vanishes on the noncentral symmetry axis. For $q^2<1/2$ it is negative near that axis. Either sign of a nonzero real $q$ gives the same model, and $q=0$ is undefined. Setting $v_0=0$ yields vacuum away from the singular origin, not strictly positive mass density. The $r^{-2}$ central cusp is locally integrable and contains no hidden point mass: the Newtonian gravitational field flux through a shrinking sphere is of order its radius.

An equatorial circular orbit requires inward gravitational acceleration equal to $v_c^2/R$. Thus the circular speed isThis is a flat galaxy rotation curve. The singular center does not define an additional circular orbit at $R=0$.

The specific orbital energy $E=\frac{1}{2}(v_R^2+v_{\varphi }^2+v_z^2)+\Phi$ is conserved because the Newtonian gravitational potential is time independent. The axial specific angular momentum $L_z=R{v}_{\varphi }$ is conserved because the Newtonian gravitational potential is axisymmetric. Hence the Jeans theorem, or directly $dF(E,L_z^2)/dt=0$ in the Collisionless Boltzmann equation, permits a stationary two-integral galactic distribution function. Using $L_z^2$ makes this galactic distribution function even under reversal of the azimuthal velocity, so it describes a nonstreaming population. A general axisymmetric population can have an odd part in $L_z$, or dependence on a third integral of motion; neither is required for this construction.

At each position define $\rho ⟨g⟩=\int gF{d}^{3}v$. The two-integral galactic distribution function is invariant under swapping $v_R$ and $v_z$, since only their sum of squares occurs in $E$. It is also even in each velocity component. Swapping integration variables proves meridional velocity isotropy of a two-integral distribution, while reflecting one variable makes each mixed integrand odd. ThereforeAll mean velocities vanish, so these are also the corresponding statements for the velocity ellipsoid and its centered covariance tensor. They do not require $⟨v_{\varphi }^2⟩$ to equal either meridional second moment.

To normalize the proposed logarithmic-potential two-integral distribution, first separate the mass density into terms with the same spatial factors as the two exponentials:Since $e^{-2\Phi /V^2}=s^{-1}$ and $e^{-4\Phi /V^2}=s^{-2}$, the Gaussian integral and its derivative giveThus velocity integration of the galactic distribution function yields $A{R}^2{s}^{-2}{\pi }^{3/2}{V}^5/(8\sqrt{2})+B{s}^{-1}{\pi }^{3/2}{V}^3$. Matching the independent spatial factors determinesThis verifies both the stationary Collisionless Boltzmann equation and the required Poisson equation for Newtonian gravity. A shift $\Phi \mapsto \Phi +C$ changes the normalizations to $A{e}^{4C/V^2}$ and $B{e}^{2C/V^2}$; the physical mass density is unchanged.

There is an important distinction between positive mass density and a nonnegative galactic distribution function. For $1/2\le q^2\le 1$, both coefficients above are nonnegative. For a prolate model $q^2>1$, $A<0$ and one must test the sum rather than reject the model just because one term is negative. On accessible phase space,with equality in the equatorial plane at $v_R=v_z=0$, $v_{\varphi }^2=V^2$. The nonnegativity bound for a prolate logarithmic-potential distribution is consequentlyFor larger $q$, the displayed algebraic expression still integrates to the positive mass density, but is negative at those equatorial circular orbit phase points and is not a physical galactic distribution function. This extra restriction is distinct from the earlier density-only answer.

For an axisymmetric Newtonian gravitational potential, the vacuum Laplace equation in cylindrical coordinates is $R^{-1}{\partial }_R(R{\partial }_R\Phi )+{\partial }_z^2\Phi =0$. Use separation of variables, $\Phi =X(R)Z(z)$. Dividing by $XZ$ and setting the separated constant to $k^2$, with $k>0$, givesThe first equation has solutions $e^{\pm kz}$; with $u=kR$, the second is the order-zero Bessel differential equation. Regularity at the axis selects the Bessel function of the first kind $J_0(kR)$, rather than the singular Bessel function of the second kind $Y_0$. ThusThere are also zero-separation-constant modes, and other boundary conditions can select other separated families. For a localized reflection-symmetric thin disk, decay away from the disk selects $e^{-k|z|}$ on the respective half-spaces.

Apply the divergence theorem to the Poisson equation for Newtonian gravity across a narrow pillbox enclosing disk area. Writing the volume mass density as $\Sigma (R)\delta (z)$ givesReflection symmetry makes ${\Phi }_z(R,0^{+})=2\pi G\Sigma (R)$. A superposition of the regular vacuum modes therefore satisfiesThe order-zero Hankel transform and its inversion now determine the Hankel representation of thin-disk gravity:The sign is fixed by the positive derivative of an attractive Newtonian gravitational potential above the disk. For ordinary finite disks the representation uses the isolated boundary condition, fixing any extra vacuum field or additive constant. Infinite scale-free disks require suitable regularization of the potential itself.

Equatorial circular orbit balance is $v_c^2=R{\Phi }_R(R,0)$. Using $J_0^{\prime }(u)=-J_1(u)$ in the Hankel representation of thin-disk gravity givesDifferentiation under the integral is justified by convergence for regular models, or by first working off the disk and taking the limiting radial field.

Write $C={\Sigma }_0{R}_0$. For the Mestel disk, the supplied Bessel function integral gives $S(k)=-2\pi GC/k$ for $k>0$. The circular speed is thenMeanwhile the interior mass is $M(R)=2\pi {\int }_0^R(C/s)sds=2\pi CR$. ConsequentlyThe Mestel disk has a flat galaxy rotation curve, despite its nonspherical geometry.

Its infinite total mass prevents setting the Newtonian gravitational potential to zero at infinity. Indeed the integral for $\Phi$ diverges at $k=0$ because $S(k)\sim 1/k$. Subtracting the potential at a reference point removes this additive divergence. As an explicit check, the Mestel disk potential-density pair isIt is harmonic for $z\ne 0$, its vertical derivative jump is $4\pi GC/R$, and $R{\Phi }_R(R,0)=2\pi GC$. Equivalently, an Abel regularization of an oscillatory integral with damping factor $e^{-ak}$ in the radial-field integral gives $R{\int }_0^{\infty }{e}^{-ak}{J}_1(kR)dk=1-a/\sqrt{a^2+R^2}$, which tends to one as $a\downarrow 0$. The force is well-defined although the unreferenced potential is not.

The equality with $GM(R)/R$ is special to the Mestel disk, not a thin-disk shell theorem. The spherical shell theorem normally makes that formula possible, but does not apply to an astrophysical disk. Exterior annuli exert outward radial gravitational acceleration inside their radii, while interior annuli do not act exactly as point masses at the origin. For this particular scale-free surface density, their combined contributions happen to produce the same result as the interior mass expression. Changing the radial surface density or truncating the disk generally destroys the equality: enclosed mass does not determine a disc rotation curve.

The magnetic field satisfies $\nabla \cdot \mathbf{B}=0$. Expanding the divergence of its Maxwell stress tensor givesFor the Newtonian gravitational field $\mathbf{g}=-\nabla \Phi$, Poisson equation for Newtonian gravity gives ${\partial }_j{g}_j=-4\pi G\rho$, and the gradient representation gives ${\partial }_j{g}_i={\partial }_i{g}_j$. ConsequentlyThe negative of the Newtonian gravitational stress tensor, in this force-stress convention, therefore supplies $\rho g_i$. Adding the pressure stress supplies $-{\partial }_{i}p$, soEach summand is symmetric. Absence of external gravitational sources is needed to represent the full gravitational force by this self-gravitating stress.

Integrate Poisson equation for Newtonian gravity through a narrow slab around the disk. The horizontal derivative contributions vanish as its thickness tends to zero, leaving the normal-derivative jumpThe even function symmetry of $\Phi$ makes the derivatives opposite, soIn the current-free simply connected upper half-space, Ampère's circuital law gives $\nabla \times \mathbf{B}=0$ and permits a magnetic scalar potential. Rescale it so that $\sqrt{4\pi G/{\mu }_0}\mathbf{B}=-\nabla {\Phi }_M$. The divergence-free condition makes ${\Phi }_M$ satisfy Laplace's equation, withCompare with the gravitational jump condition. Subject to the same isolated-field boundary condition at infinity, ${\Phi }_M$ is the harmonic potential of the effective surface densityThis gravity-equivalent magnetic surface density can have either sign; it is a mathematical representation of the exterior magnetic field, not physical negative mass. An imposed nondecaying field would require additional boundary data and would not be fixed by the disk surface density alone.

Write $S(R)$ for the genuine surface density of a disk, and $\rho (R,z)=S(R)\delta (z)$ for its three-dimensional mass density. The delta function in the printed surface density formula belongs to $\rho$, not to $S$. Introduce a reference length $R_{\ast }$ to make logarithms dimensionless. The intended model is the infinite, self-gravitating, scale-free Mestel disk, with no extra source or imposed gravitational field.

The circular speed satisfies $v_c^2=R{\partial }_R\Phi (R,0)$. Thus the flat galaxy rotation curve givesA reflection-symmetric harmonic function with this midplane boundary value isVerify this continuation using the Poisson equation for Newtonian gravity. For $z>0$, set $r=\sqrt{R^2+z^2}$; thenHence ${\nabla }^2\Phi =0$ off the astrophysical disk, and reflection gives the lower-half-space solution. Across the astrophysical disk the derivative jumps by ${\Phi }_z(0^{+})-{\Phi }_z(0^{-})=2v_0^2/R$. The distributional Poisson equation for Newtonian gravity therefore givesAt the origin the enclosed disk mass tends to zero linearly with radius, so there is no additional central point mass. This verifies the Mestel disk potential-density pair. The astrophysical disk has infinite total mass and a logarithmic Newtonian gravitational potential, so it is not an isolated finite-mass model with Newtonian gravitational potential zero at infinity. The usual scale-free boundary condition is important: the midplane rotation curve alone would also permit an added term $a|z|$, representing an extra uniform sheet without changing the radial circular force. That contribution is excluded in the intended Mestel disk model.

Use a mass-weighted planar galactic distribution function, so $F(\mathbf{x},\mathbf{v})d^{2}x{d}^{2}v$ is the stellar mass in a small planar phase space element. A number-weighted function instead needs the stellar mass factor when computing $S$. Assume a steady collisionless stellar system and isotropy in the two in-plane velocity components. Put $u=|\mathbf{v}{|}^2/2$ and write $F=f(\mathbf{x},u)$. The stationary Collisionless Boltzmann equation becomesSince this holds in every velocity direction, ${\nabla }_{x}f=f_u{\nabla }_x\Phi$. In coordinates $(\mathbf{x},E)$ with $E=u+\Phi$, this is exactly ${\nabla }_{x}f{|}_E=0$. Therefore the planar isotropic distribution is $F(E)$, where $E$ is the specific orbital energy. Stationarity is essential; instantaneous isotropy alone would not imply this result.