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The aspect ratio of an astrophysical disk is its vertical thickness scale divided by cylindrical radius. A thin disk has $H/r\ll 1$. In gas-supported vertical hydrostatic equilibrium with harmonic external gravity $-{\Omega }_z^{2}z$, $H/r\sim c_s/(r{\Omega }_z)$, where ${\Omega }_z$ is the vertical epicyclic frequency and $c_s$ is the sound speed. In a centrally dominated Keplerian disk, ${\Omega }_z=\Omega$, so this compares sound speed with orbital speed. If disk self-gravity supplies the vertical restoring force, the relation $H\sim c_s/\Omega$ requires replacement. A polytropic free-surface height and an isothermal scale height differ by model-dependent constants even though their aspect ratios have the same basic role.

A Kuzmin disk is an infinitesimally thin axisymmetric astrophysical disk whose relative potential is $\psi =Gm/\sqrt{R^2+(|z|+b{)}^2}$, with $m,b>0$. Away from the disk this is a point-mass vacuum field. The jump of its vertical derivative is $-2Gmb/(R^2+b^2{)}^{3/2}$, so the Poisson equation for Newtonian gravity, written ${\nabla }^2\psi =-4\pi G\rho$, gives the displayed surface density. Integrating $2\pi R\Sigma$ gives total mass $m$.

For a thin axisymmetric astrophysical disk, angular integration of the Newtonian kernel eliminates odd Legendre polynomials. Define $I_n={\int }_0^R\Sigma (s)s^{n+1}ds$ and $J_n={\int }_R^{\infty }\Sigma (s)s^{-n}ds$. The even coefficient is ${\alpha }_n=\pi [n!/(2^n((n/2)!{)}^2){]}^2$. In differentiating the paired inner and outer contributions, their moving-boundary terms cancel. The resulting galaxy rotation curve includes an inward inner-ring contribution and an outward exterior-ring contribution. For smooth surface density, the force at the disk is interpreted with a symmetric Cauchy principal value or a vanishing-thickness regularization.

A Mestel disk is a thin axisymmetric astrophysical disk with surface density inversely proportional to radius. Its galaxy rotation curve is exactly flat, $v_c=v_M$, and $v_c^2=GM(R)/R$ despite its nonspherical geometry. In the Legendre expansion of thin-disk gravity, all positive-degree interior and exterior terms cancel. Its total mass is infinite, so only potential differences, rather than a potential zero at infinity, are defined.

A passive astrophysical disk receives most of its heat from external irradiation rather than internal viscous dissipation. For a stellar disk, a rough criterion is $GM\dot{M}/R_{\ast }\ll L_{\ast }$. The actual local requirement compares absorbed stellar flux with viscous flux; incidence angle, shadowing and absorption can invalidate a luminosity comparison alone.

The exponent of the point-mass weight is $\beta$, with $0<\beta <1$. At the origin the unreferenced relative potential has integrand $GB{b}^{\beta -1}$, whose integral diverges at its upper endpoint. In the origin-referenced potential the two terms cancel pointwise there. More generally, first form their difference at finite cutoffs and then pass to the limit. For any fixed nonzero $(R,z)$, the small-$b$ singular term is $-b^{\beta -1}$, which is integrable because $\beta >0$. At large $b$,so the weighted difference is integrable because $\beta <1$. Hence the referenced potential is well defined, and its value at the origin is zero. It is a potential difference for an infinite astrophysical disk, not a potential with zero at infinity. In fact rescaling all lengths shows ${\psi }_{II}(\lambda R,\lambda z)={\lambda }^{\beta }{\psi }_{II}(R,z)$; its finite values approach zero at the origin even though its force is singular there.

For $R>0$, differentiation under the convergent force integral in the plane givesUse $b=R\sqrt{t/(1-t)}$, for which $db=(R/2)t^{-1/2}(1-t{)}^{-3/2}dt$. The powers of $R$, $t$, and $1-t$ then givewhere the Euler beta function evaluates the constant asThe first Gamma function argument is $(1+\beta )/2$: it follows from adding one to the exponent of $t$. Both Euler beta function arguments are positive in the stated range.

For a circular orbit in the plane, $V^2/R=-{\partial }_R{\psi }_{II}$. The prescribed circular speed therefore fixes $C=K^2$ andIndependently, the Kuzmin reflection method givesHere the power of $b$ is one larger than in the force integral, so its Euler beta function arguments differ. Substituting the normalization fixed by the circular speed yieldsThis scale-free Kuzmin superposition has nonnegative surface density, finite central enclosed mass, and infinite total mass. Its force and surface density approach the Mestel disk values as $\beta \downarrow 0$; that limiting disk requires a different potential reference at the origin.

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.

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.

Integrating over the two-dimensional velocity plane gives the planar isotropic distribution inversion:On the astrophysical disk $R=R_{\ast }{e}^{\Phi /v_0^2}$, so $S(\Phi )=v_0^2{e}^{-\Phi /v_0^2}/(2\pi G{R}_{\ast })$. Differentiate the integral with respect to its lower limit:The boundary value $S\to 0$ as $\Phi \to \infty$ verifies the integrated equation as well as its derivative. Choosing the implicit length unit $R_{\ast }=1$ recovers the printed normalization. Changing the additive energy zero changes this prefactor accordingly.

At a fixed radius, the normalized velocity density isIt is a product of centered Gaussian distributions. Differentiating the supplied Gaussian integral with respect to its coefficient gives the second moments, and odd moments vanish. Thus the in-plane velocity dispersions areAn exactly planar astrophysical disk has $v_z=0$ and ${\sigma }_z=0$. The nonzero circular speed is a property of the force field, not a statement that this hot stellar distribution has net rotation.

Reversing every retrograde star folds the azimuthal Gaussian to a half-normal distribution. Equivalently the new steady galactic distribution function is $F_{+}(E,L_z)=2F(E)\Theta (L_z)$, because $E$ and $L_z=R{v}_{\varphi }$ are integrals of the motion. Its density and even velocity moments are unchanged. Its streaming velocity isThis maximally prograde stellar distribution still has radial motion and a spread of azimuthal speeds; it does not place every star on a circular orbit. In particular $⟨v_{\varphi }^2{⟩}_{+}=v_0^2$ while ${\sigma }_{\varphi ,+}^2=v_0^2(1-2/\pi )$. The mean speed is smaller than the root-mean-square speed $v_0$, which explains why it is not the circular speed.

For a spherical system, use the spherical shell theorem with shell mass $dm=4\pi \rho (s)s^{2}ds$. Inner shells contribute $-Gdm/r$ to the gravitational potential, while an outer shell contributes its constant interior potential $-Gdm/s$. With $\Phi (\infty )=0$, assuming the required integrals converge,On differentiation, the two terms containing $\rho (r)r$ cancel. Hence ${\Phi }^{\prime }=GM(r)/r^2$, where $M(r)=4\pi {\int }_0^r\rho (s)s^{2}ds$. Radial balance for a circular orbit gives $v_c^2=r{\Phi }^{\prime }=GM(r)/r$.

For a razor-thin axisymmetric astrophysical disk, the element of mass is $dm=\Sigma (s)sdsd{\varphi }^{\prime }$, where $\Sigma$ is the surface density. Superposing Newtonian gravitational potentials therefore givesAxisymmetry permits $\varphi =0$. Unlike the spherical case, an exterior annulus generally exerts a radial force: enclosed mass does not determine a disc rotation curve.

For the Legendre expansion of thin-disk gravity, split the radial integral at $s=R$. In the inner part the kernel expands in $s/R$, while in the outer part it expands in $R/s$. The angular integrals of odd Legendre polynomials vanish, and those of even degree $n$ equal $2{\alpha }_n$, whereIntroduce $I_n(R)={\int }_0^R\Sigma (s)s^{n+1}ds$ and $J_n(R)={\int }_R^{\infty }\Sigma (s)s^{-n}ds$. The gravitational potential becomesWhen differentiating each bracket, the moving-limit terms cancel: $R^{-(n+1)}{I}_n^{\prime }=\Sigma (R)$ and $R^n{J}_n^{\prime }=-\Sigma (R)$. Since $v_c^2=R{\Phi }^{\prime }$, this givesThe zeroth term has ${\alpha }_0=\pi$ and $M(R)=2\pi I_0$. Separating it provesThe inner correction is inward, while the exterior correction is outward. For a smooth surface density, the original potential singularity at coincident points is integrable. Its radial force is understood through a symmetric Cauchy principal value or a vanishing-thickness regularization; the paired interior and exterior terms above retain the cancellation at $s=R$. This avoids treating the two singular local force contributions separately.

For an exponential galactic disk, write $\Sigma (s)={\Sigma }_0{e}^{-s/R_d}$, so $M_{\infty }=2\pi {\Sigma }_0{R}_d^2$ and $M(R)=M_{\infty }[1-(1+R/R_d)e^{-R/R_d}]$. At large $R$, the missing mass and exterior-ring terms are exponentially small. The leading nonspherical interior term is $k=1$, with ${\alpha }_2=\pi /4$ and $I_2(\infty )=6{\Sigma }_0{R}_d^4$. Consequently the exponential-disk Keplerian asymptotic isThe positive leading correction shows that the rotation curve approaches the Keplerian limit from above. The finite-order large-radius expansion is sufficient here; an infinite moment expansion need not converge for a disk extending to infinity.

A useful special example is the Mestel disk, with surface density $\Sigma (s)=C/s$ for $s>0$, $C>0$. It has $M(R)=2\pi CR$. For every positive even $n$,All the nonspherical corrections cancel, givingThus the Mestel disk has a perfectly flat galaxy rotation curve.

This example has infinite total mass and a singular central surface density. The absolute gravitational potential cannot be set to zero at infinity, but the radial force exists as a limit of disks with increasing outer cutoff. Potential differences are $\Phi (R)-\Phi (R_0)=v_M^2\log (R/R_0)$. The earlier potential integral is therefore interpreted up to a radius-independent divergent constant for this example; the force calculation remains valid. Truncating the Mestel disk gives a more physical finite system but changes the exact flat curve near its edges.

The axisymmetric razor-thin astrophysical disk equations areFor the stationary background, mass and azimuthal momentum conservation are automatic, while radial force balance and the Poisson equation giveHere $d{p}_0/dR=0$ under the stated constant-pressure assumption.

Retaining first-order perturbations and using $c_s^2=dp/d\Sigma$ giveswhereis the squared radial epicyclic frequency. In the local $|k|R\gg 1$ limit, a Fourier mode $e^{i(kR-\omega t)}$ obeysEliminating ${\Sigma }^{\prime }$, $u_{\varphi }^{\prime }$, and ${\Phi }^{\prime }$ yields the local dispersion relation

For a Keplerian orbit, $\Omega \propto R^{-3/2}$, soInstability requires ${\omega }^2<0$, equivalently a mode with positive imaginary frequency. Treating the right-hand side as a quadratic in $x=|k|$, its two roots areReal distinct roots, and hence an unstable interval $x_{-}<|k|<x_{+}$, exist exactly when the Toomre stability criterion has $Q<1$.

With $h=c_s/\Omega$,The constant positive term is epicyclic restoration by rotation and stabilizes long wavelengths. The negative term is the razor-thin disk's self-gravity and drives collapse. The positive $k^2$ term is gas-pressure restoration and stabilizes short wavelengths. Gravitational instability can therefore survive only on an intermediate band of scales.

A protoplanetary disk is a gas-and-solid astrophysical disk around a young star in which planet formation occurs. Its evolution combines accretion, radiation, gas dynamics and the growth of solid material. A minimum-mass solar nebula is one reconstruction of the material needed to form the Solar System.

The radial buoyancy frequency describes restoration of a radially displaced parcel in a stratified astrophysical disk. With a dimensionless specific entropy $S$, its square is proportional to $-{\rho }^{-1}(dP/dr)(dS/dr)$. Negative $N_r^2$ means an adverse entropy stratification, but epicyclic motion can still stabilize the parcel. For smooth thin disk profiles, $|N_r^2|$ is usually of order $(H/r{)}^2{\Omega }^2$.

In a reflection-symmetric astrophysical disk, the vertical mass coordinate measures mass per unit area above the midplane. It satisfies $dm=\rho dz$, so ${\partial }_m={\rho }^{-1}{\partial }_z$ inside the material. If the total surface density is $\Sigma$, the upper face is $m=\Sigma /2$; the normalized coordinate $\zeta =2m/\Sigma$ ranges from zero to one. Integrals weighted by mass density become ordinary integrals in $m$, which is especially useful for radiative diffusion and the density-weighted mean kinematic viscosity.