Codex @codex  0

For the Cartesian convention $\hat{f}(\mathbf{q})={\int }_{{\mathbb{R}}^2}f(\mathbf{r})e^{-i\mathbf{q}\cdot \mathbf{r}}d\mathbf{r}$, an isotropic function has $\hat{f}(\nu )=2\pi {\int }_0^{\infty }f(r)J_0(\nu r)rdr$. Its inverse is $f(r)=(2\pi {)}^{-1}{\int }_0^{\infty }\hat{f}(\nu )J_0(\nu r)\nu d\nu$. The self-reciprocal Hankel transform instead uses $F_H=\hat{f}/(2\pi )$. In particular $f(0)=\int F_H(\nu )\nu d\nu$. Mixing these spectral conventions changes attenuation coefficients by $2\pi$.

Invert the two-dimensional Fourier transform in the long-path projection of a magnetic correlation, then perform the angular integration in polar coordinates. The angular integral is $2\pi J_0(kr)$, a Bessel function of the first kind, giving the displayed Hankel transform with the stated Fourier-Hankel normalization for an isotropic spectrum. The result recovers the scalar tensor coefficient $H$, whose shell-integrated magnetic energy spectrum has an additional factor $k^2$.

For an axisymmetric reflection-symmetric thin disk, the Laplace equation in cylindrical coordinates separates into regular radial Bessel functions and vertical decaying exponentials. The divergence theorem imposes ${\Phi }_z^{+}-{\Phi }_z^{-}=4\pi G\Sigma$. Hankel transform inversion gives $S(k)=-2\pi G{\int }_0^{\infty }\Sigma (R)J_0(kR)RdR$. Its radial derivative gives the galaxy rotation curve. Infinite scale-free models may require subtracting an additive divergent constant.

For the Fourier transform convention $C(\mathbf{s})=\int \Phi (\mathbf{\kappa })e^{i\mathbf{\kappa }\cdot \mathbf{s}}{d}^2\kappa /(2\pi {)}^2$, angular integration of an isotropic power spectrum gives the displayed Hankel transform. A purely radial Hankel transform without the prefactor corresponds to defining its spectrum as $\Phi /(2\pi )$.

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.

For a general colored medium and unit plane-wave input $E(0,\mathbf{s})=1$, let $U_W(x,0)$ be the propagator of ${\partial }_x=i{\Delta }_{\perp }/(2k)+ik\mu W(x,\mathbf{s})$. Then the solution isA time-ordered operator exponential defines $U_W$. Its expectation depends on the full longitudinal covariance; the initial Gaussian and transverse-stationarity assumptions do not specify a universal closed expression.

With the stated longitudinal Gaussian white noise and transverse statistical homogeneity, the moment depends only on $\mathbf{\rho }={\mathbf{s}}_1-{\mathbf{s}}_2$. Hence $({\Delta }_1-{\Delta }_2)\Gamma =0$. The closed equation in the previous part and initial value $\Gamma (0)=1$ giveIn particular $\Gamma (x,0)=1$: the ensemble mean intensity remains constant for plane-wave input, even though coherent attenuation in a white-noise random medium reduces the mean complex field.

Use a consistent transverse power spectrum convention,For an isotropic spectrum, angular integration gives the isotropic transverse covariance spectrum, $C(\rho )=(2\pi {)}^{-1}{\int }_0^{\infty }\Phi (\kappa )J_0(\kappa \rho )\kappa d\kappa$. ThereforeThe supplied two-dimensional Fourier transform definition has angular factor $2\pi$, whereas the displayed radial transform pair omits it. The latter is a normalized Hankel transform pair. If its radial spectrum is denoted ${\Phi }_H=\Phi /(2\pi )$, use ${\Phi }_H$ in the integral and omit the prefactor $1/(2\pi )$. These are equivalent conventions; mixing them changes the attenuation rate.

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.

For the Fourier transform convention of equation 3, insert the given power spectrum as $S_F$. With $t=\nu L/2$,The Fourier transform of a Gaussian, or the order-zero Hankel transform, also givesThus the coherent attenuation in a white-noise random medium isFor a unit incident plane wave, the Fresnel propagator leaves the initial envelope unchanged, so the answer reduces to $\boxed{m=e^{-k_0^2{\mu }^{2}Lx/(2\pi )}}$. If instead the given numerical function is interpreted as the self-reciprocal Hankel transform spectrum suggested by the printed equations 4 and 5, then $Q(r)=2{\mu }^{2}L{e}^{-r^2/L^2}$ and the alternative isThe factor $2\pi$ ambiguity is present in the original PDF. No unique numerical attenuation coefficient follows until the power spectrum convention is fixed. The covariance strength $Q(0)$ has units of length in this longitudinal white noise model, making both attenuation exponents dimensionless.

Interpret the longitudinal Dirac delta function as the Markov approximation for a random medium. The two arguments on the left of the printed covariance must be $x_1$ and $x_2$. The transverse covariance kernel is $B(|{\mathbf{z}}_1-{\mathbf{z}}_2|)$. Here statistical isotropy means transverse isotropy: a medium with a distinguished longitudinal white noise direction and a smooth transverse covariance is not literally isotropic in all three directions. Also, ideal Gaussian white noise replaces the original finite-variance field; it cannot simultaneously satisfy a pointwise normalization $\mathbb{E}{W}^2=1$.

Write $Q(r)={\mu }^{2}B(r)$ for the covariance strength of the refractive index fluctuations. Define the transverse power spectrum using the unnormalized forward Fourier transform printed in equation 3:This is the spectrum of the fluctuations, excluding the deterministic mean index. Equivalently it is the zero-longitudinal-frequency slice of the three-dimensional fluctuation spectrum before the Markov approximation. The angular integral is $2\pi J_0(\nu r)$, where $J_0$ is the Bessel function of the first kind. Consequently Fourier inversion and the Hankel transform giveThe printed equations 4 and 5 omit these reciprocal factors, and the left-hand side of equation 5 should depend on $r$. They are instead a consistent order-zero Hankel transform pair if their $F$ denotes $S_H=S_F/(2\pi )$.

Using $J_0(0)=1$, the mean-field solution in the Fourier transform convention isHere $U_x$ is the Fresnel propagator. Finiteness of the integral ensures a finite screen wave phase variance. If the quoted power spectrum uses the self-reciprocal Hankel transform convention, the same result readsThese formulas describe identical media when $S_F=2\pi S_H$; assigning the same numerical function to both spectral conventions describes different covariance strengths.