Razor-thin disk Poisson kernel

The razor-thin disk Poisson kernel converts a horizontal Fourier transform of surface density of a disk into its gravitational potential. Write the razor-thin disk as $\rho (\mathbf{x},z)=\Sigma (\mathbf{x})\delta (z)$ and use the Poisson equation for Newtonian gravity ${\nabla }^2\Phi =4\pi G\rho$. For horizontal wavevector $\mathbf{k}$ of magnitude $k>0$, the transformed equation isA solution decaying away from the disk has the form $C(\mathbf{k})e^{-k|z|}$. Integrating across $z=0$ gives the derivative jump $-2kC=4\pi G\hat{\Sigma }$, hence the midplane relationThis multiplier turns the nonlocal horizontal gravity calculation into multiplication in Fourier space. It is useful in calculations of the self-gravity of an astrophysical disk, where longer wavelengths have a larger potential response at fixed density amplitude. The $k=0$ mode requires separate boundary information: an infinite uniform sheet has a gravitational potential proportional to $|z|$, which cannot decay at both vertical infinities.

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.