 Homogeneous dust-ball matching

At the boundary of an Oppenheimer-Snyder cloud, its areal radius is $R=a\sin {\chi }_0$. Matching to a Schwarzschild metric gives $M=4\pi \rho R^3/3=a_{\max }{\sin }^3{\chi }_0/2$. Thus ${\sin }^2{\chi }_0=2M/R_0$ and $a_{\max }=\sqrt{R_0^3/(2M)}$. The boundary is a radial timelike geodesic with specific Killing energy $\sqrt{1-2M/R_0}=\cos {\chi }_0$. Its radial equation is ${\dot{R}}^2=2M/R-2M/R_0$. The induced metric and the angular extrinsic curvature agree on both sides, so no surface layer is required.