Razor-thin disk approximation

The razor-thin disk approximation replaces a disk's vertical density profile by $\rho (x,y,z)=\Sigma (x,y)\delta (z)$. It is useful for horizontal dynamics on scales much longer than the disk thickness.

At an abruptly truncated razor-thin disk with nonzero edge surface density of a disk, the midplane radial force diverges logarithmically. Locally, mass occupies a half-plane and the inward force contains a nonzero angular factor multiplying $G{\Sigma }_{\mathrm{edge}}{\int }_{ϵ}^{\ell }ds/s$. The opposite exterior half-plane is absent, so its cancellation does not occur. Finite thickness or a taper taking the edge density to zero regularizes the force. Therefore a finite-edge galaxy rotation curve requires more than the ideal scale-free Mestel disk normalization.

A two-dimensional disk equation with pressure acceleration $-{\Sigma }^{-1}\nabla P$ requires an equation of state for vertically integrated pressure $P$ in terms of surface density of a disk $\Sigma$. A barotropic fluid closure is $P=P(\Sigma )$, with linear enthalpy perturbation $h^{\prime }=c_s^2{\Sigma }^{\prime }/{\Sigma }_0$. Merely giving $P=P(\rho )$ for a volume density does not close the two-dimensional equations unless a vertical conversion $\rho (\Sigma )$ is specified. Positive $c_s^2$ gives ordinary pressure restoration of an inertial-acoustic wave.

For a horizontal Fourier mode of wavenumber $k>0$, the midplane gravitational potential of a razor-thin disk is

For an isolated razor-thin disk, define the two-dimensional Fourier transform by $\tilde{g}(\mathbf{k})=\int e^{-i\mathbf{k}\cdot \mathbf{x}}g(\mathbf{x})d^{2}x$. At $k>0$, the razor-thin disk Poisson kernel gives the displayed inversion. Since the transform of $1/|\mathbf{x}|$ is $2\pi /k$, inversion also gives $\Sigma =(4{\pi }^{2}G{)}^{-1}(1/|\mathbf{x}|)\ast {\Delta }_2\Phi$. The forward relation is $\Phi =-G(1/|\mathbf{x}|)\ast \Sigma$. An additive potential constant is irrelevant; a uniform infinite sheet and scale-free divergent potentials require separate zero-mode or potential-difference treatment rather than an arbitrary decaying boundary condition.

For $k>0$, horizontal Fourier transform gives $({\partial }_z^2-k^2)\tilde{\Phi }=4\pi G\tilde{\Sigma }\delta (z)$. Decay and continuity give an exponential on each side; its derivative jump fixes the coefficient. The uniform mode needs a separate linear vertical solution and a potential reference.

The vertical field of a disk has kernel $Gz/(|\mathbf{R}-{\mathbf{R}}^{\prime }{|}^2+z^2{)}^{3/2}>0$. The triangle inequality therefore bounds a signed-surface density field by the field of its absolute surface density. If that surface density is everywhere smaller than a nonnegative comparison surface density, their vertical fields satisfy the displayed strict comparison, provided the integrals exist and the strict bound holds on positive measure. This is useful when comparing gravity-equivalent magnetic surface density with actual disk mass.