Oppenheimer-Snyder model

The Oppenheimer-Snyder model describes a homogeneous ball of pressureless matter collapsing from rest, with a closed FLRW metric inside and a Schwarzschild metric outside. In geometric units its interior is $d{s}^2=-d{\tau }^2+a^2(d{\chi }^2+{\sin }^2\chi d{\Omega }^2)$, with $a=a_{\max }(1+\cos \eta )/2$ and $\tau =a_{\max }(\eta +\sin \eta )/2$. The boundary is a fixed comoving angle. Density conservation gives $\rho a^3$ constant, and the Friedmann equation at the initial turning point gives $8\pi {\rho }_0{a}_{\max }^2/3=1$.

The comoving dust ball $0\le \chi \le {\chi }_0<\pi /2$ in the Oppenheimer-Snyder model is an Oppenheimer-Snyder cloud. Its pressureless matter has uniform density on each constant-proper time slice and reaches a simultaneous final singularity in the ideal homogeneous model.

Radial outgoing null geodesics in the interior of Oppenheimer-Snyder collapse obey $d\chi /d\eta =1$. Trace the ray crossing the boundary at $({\eta }_H,{\chi }_0)$ backward: it starts at the centre at $\eta ={\eta }_H-{\chi }_0$. This ray is the event horizon; it can exist inside the cloud before the surface crosses $R=2M$. An event horizon is determined by global escape to future null infinity, rather than by a local stopwatch signal.

A boundary clock in Oppenheimer-Snyder collapse reads $\tau =a_{\max }(\eta +\sin \eta )/2$. Its event horizon crossing occurs at $R=2M$, giving the displayed angle, and the final curvature singularity occurs at $\eta =\pi$. The remaining proper time is $a_{\max }(\pi -{\eta }_H-\sin {\eta }_H)/2$. For $R_0\gg 2M$ this approaches $4M/3$, but it is not exactly that value for finite release radius.

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.