= Eddington inversion
{c}
{title2=$f(E)=\frac1{\sqrt8\pi^2}\frac d{dE}\int_0^E\frac{\rho'(\psi)}{\sqrt{E-\psi}}\,d\psi$}
= Eddington's formula
{c}
{synonym}
For a spherical isotropic bound distribution $f(E)$ with $E=\psi-v^2/2$, no unbound population and $\rho(0)=0$, <velocity> <integration> gives $\rho(\psi)=4\pi\sqrt2\int_0^\psi f(E)\sqrt{\psi-E}\,dE$. Differentiation followed by <Abel transform> inversion yields $f(E)=(\sqrt8\pi^2)^{-1}(d/dE)\int_0^E\rho'(\psi)/\sqrt{E-\psi}\,d\psi$. The factor is $1/(\sqrt8\,\pi^2)$, with $\pi^2$ outside the square root. Regularity sufficient for these operations and nonnegativity of the resulting distribution are separate requirements; an arbitrary density-potential pair need not give a physical distribution.
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