Consider the total spherically symmetric collapse of stellar matter comprised of a perfect fluid/gas with zero or negligible pressure. If the star is above the Oppenheimer-Volkoff limit and has exhausted its thermonuclear fuel, it is too heavy to reach a new stable collapsed state of extreme matter, such as a white dwarf or neutron star(refs.), or even a superdense ‘exotic star’. The star then collapses to a  point within a finite comoving, proper time interval, with the exterior metric matching the Schwarzchild (black hole) metric.  Since the particles comprising the star are acted upon by gravitational forces only, they can form the basis of a comoving coordinate system with metric (ref).

$ds^{2}$ = $dt^{2} - A(r,t) dt^{2}$$B(r,t)(d\theta^{2}+ sin^{2}\theta d \varphi^{2})$

The energy-momentum tensor equation  for perfect fluid with zero pressure is given by

$T_{\mu\nu}=\rho(r,t)U^{\mu}U^{\nu}$

where $\rho(r,t)$ is the proper energy density and $U^{\mu}$ is the velocity 4-vector with $U_{r}=U^{\theta}=U^{\varphi}=0$ and $U^{t}=1$ for a comoving system. Since the stellar matter has zero (or negligible) pressure, the equations of energy-momentum conservation are automatically satisfied so that $\nabla_{\mu} T^{\mu}_{i}=0$. The energy conservation equation is then

$\nabla_{\mu} T^{\mu}_{0}=-\frac{\partial \rho}{\partial t}-\rho\Gamma^{\lambda}_{\lambda 0}=-\partial_{t}\rho-\rho\left(\frac{\dot{A}}{2A} + \frac{\dot{B}}{B}\right)$

where $\nabla_{\mu}$ is the covariant derivative. This is equivalent to $\partial_{t} (\rho B A^{1/2})=0$. The Einstein field equations are

$\mathbf{R}_{\mu\nu}=-8\pi G\mathbf{\Theta}_{\mu\nu}$  $-8\pi G\rho[\tfrac{1}{2}g_{\mu\nu}+U_{\mu}U_{\nu}]$

with $\Theta_{rr}=\frac{1}{2}\rho A$, $\Theta_{\theta\theta}=\frac{1}{2} \rho B$,  $\Theta_{tt} = \frac{1} {2} \rho$,  $\Theta_{\theta \theta}$ =$\Theta_{\varphi \varphi} sin^{2} \theta$.

The components of the field equations are:

$\mathbf{R}_{tt}=-8\pi G \mathbf{\Theta}_{tt}= \frac{1} {2} \rho$

$\mathbf{R}_{rr}=8\pi G \mathbf{\Theta}_{rr}=\frac{1}{2}\rho A$

$\mathbf{R}_{\theta\theta}=8\pi G \mathbf{\Theta}_{\theta\theta}= \frac{1}{2}\rho B$

$\mathbf{R}_{\varphi\varphi}=8\pi G \mathbf{\Theta} _{\varphi\varphi}$

Using the metric (-) and the source tensor (-) gives the Einstein equations for the gravitational collapse of a supermassive star with zero pressure

$\frac{1}{A}\bigg[\frac{B^{\prime\prime}}{B}-\frac{B^{\prime 2}}{2B^{2}} -\frac{A^{\prime}B^{\prime}}{2AB}\bigg]- \frac{\ddot{A}}{2A} + \frac{\dot{A}^{2}}{ 4A^{2}} - \frac{\dot{A}\dot{B}}{2AB}=-4\pi G$

$-\frac{1}{B}+\frac{1}{A}\bigg[\frac{B^{\prime\prime}}{2B}-\frac{A^{\prime}B^{\prime}}{4AB}\bigg]- \frac{\ddot{B}}{2B}-\frac{\dot{A}\dot{B}}{4AB}=-4\pi G\rho$

$\frac{\ddot{A}}{2A}+\frac{\ddot{B}}{B}-\frac{\dot{A}^{2}}{4A^{2}}- \frac{\dot{B}}{2B^{2}}=-4\pi G\rho$

$\frac{\dot{B}^{\prime}}{B}-\frac{B^{\prime}\dot{B}}{2B^{2}}- \frac{\dot{A}B^{\prime}}{2AB}=0$

If $\rho(r,t)=\rho(t)$ then one seeks a separable solution of the form $A(r,t)=R^{2}(t)f(r)$ and $B(r,t)=R^{2}(t)r^{2}$. From (-) it follows that $\dot{S}/S=\dot{R}/R$ so $f$ and $g$ can be normalised and $\latex S(t)=R(t)$. The energy conservation condition requires $\rho(t) R^{3}(t)=const$. If $R(t)$ is normalised to $R(0)=1$ then $\rho(t)=\rho(0)R^{-3}$. The field equations are now reduced to a pair of ODEs of the form (ref.)

$-2k-\ddot{R}(t)R(t)-2\dot{R}^{2}(t)=-4\pi G\rho(0)R^{-1}(t)$

$\ddot{R}(t)R(t)=-\frac{4}{3}\pi G \rho(0)R^{-1}(t)$

Eliminating $\ddot{R}(t)$ gives

$\dot{R}^{2}(t)=-k+\frac{8 \pi G}{3} \rho(0) R^{-1}(t)$

Since the stellar matter is initially at rest, the initial data is $(R(0)=1,\dot{R}(0)=0)$ so that $k=8\pi G/3 \rho$, giving

${\dot{R}}^{2}(t)$  $=k (R^{-1} (t)-1)$

The final result reduces down to a simple nonlinear ODE for $R(t)$

$\dot{R}(t)=-k^{1/2} (R^{-1}(t)-1)^{1/2}$

The formal solution is then

$R(t)=R(0)-k^{1/2}{\displaystyle \int_{0}^{t}} (R^{-1}(v)-1)^{1/2} dv$

The actual solution is given by the parametric equations of a cycloid:

$R=\frac{1}{2}R(0)(1+\cos\chi)=\frac{1}{2}(1+\cos\chi)$

$t=\frac{1}{2k^{1/2}}(\chi+\sin(\chi))$

with $\chi\in[0,2\pi]$. The star collapses to zero size when $R(t)=0$ or for $\chi=\pi$ and hence at $t=t_{*}$ where

$t_{*}=\pi/2k^{1/2}=\frac{\pi}{2}(3/8\pi G\rho(0))^{1/2}$.

Hence a pressureless supermassive spherical star of initial density $\rho(0)$ collapses from rest to to a point within a finite proper time $t^{*}$. In other words, the collapse time to zero size from the perspective of a comoving observer falling with the stellar matter, is finite. The interior space-time geometry can be matched to the FRW metric with $R(t)$ the scale factor. For an exterior observer, the exterior of the star can be matched to the usual Schwarzchild metric by the Birkhoff theorem.

The differential equation (-) can also be integrated directly so that
${\displaystyle \int_{R(0)=1}^{R(t)}}d\bar{R}(t)(\bar{R}^{-1}(t)-1)^{-1/2}=- k^{1/2}{\displaystyle \int_{0}^{t}}d\bar{t}$

$t=\frac{1}{k^{1/2}}R(t)(R^{-1}(t)-1)^{1/2}$+$\frac{1}{k^{1/2}}\tan^{-1}[\frac{1}{2}(2R(t)-1)(R^{-1}(t)-1)^{1/2}(R(t)-1)^{-1/2}]$

$-\frac{1}{k^{1/2}}R(0)(R^{-1}(0)-1)^{1/2}$+$\frac{1}{k^{1/2}}\tan^{-1}[\frac{1}{2}(2R(0)-1)(R^{-1}(0)-1)^{1/2}(R(0)-1)^{-1/2}]$

The second term vanishes since $R(0)=1$. When the star has collapsed to zero size then $R(t_{*})=0$ and $\tan^{-1}(\infty)=\pi/2$ so that (-) gives $t_{*}=\pi/2k^{1/2}$ as before.

The differential equation (-) can be reformulated in terms of a matter density function. Let a pressureless perfect-fluid star have normalised initial radius $laetx R(0)=1$ at $t=0$ and initial density $\rho(0)$, and the collapse for all $t>0$ within the Oppenheimer model is given by the parametric cycloid differential equation

$\dot{R}(t)=-k^{1/2}(R^{-1}(t)-1)^{1/2}$

Since $\rho(t)=\rho(0)R^{-3}(t)$, define a ‘matter density function’

$D(t)=[\rho(t)/\rho(0)]^{1/3}=R^{-1}(t)$

for $D(t)\in[D(0),\infty)=[1,\infty)$ such that the total collapse of the star to a point is now equivalent to the blowup in the matter density function with $D(t_{*})=\infty$ at $t=t_{*}=\pi/2 k^{1/2}$. Taking the derivative and using (-)

$\dot{D}(t)=R^{-2}(t)\dot{R}(t)=k^{1/2}R^{-1}(t)(R^{-1}(t)-1)^{1/2}$

which is

$\dot{D}(t) = k^{1/2} D^{2}(t) (D(t)-1)^{1/2} \equiv \Psi[D(t),k]$

The collapse can be studied in terms if this non-linear differential equation with a singularity or blowup in the matter density function $D(t)$. The parametric solutions are now

$D(t)=2D(0)[1+\cos\chi]^{-1}= 2[1+\cos\chi]^{-1}$

$t=\frac{1}{2k^{1/2}}(\chi+\sin(\chi))$

since $D(0)=R^{-1}(0)=1$ . As a consistency check, this ODE is readily integrated so that

${\displaystyle \int_{D(0)}^{D(t)}}d\bar{D}(t)(\bar{D}(t))^{-2}(\bar{D}(t)-1)^{-1/2}= k^{1/2}{\displaystyle \int_{0}^{t}}d\bar{t}$

which gives

$k^{1/2}t=(D(t)-1))^{1/2}(D(t))^{-1}+ tan^{-1}(D(t)-1))^{1/2})$$(D(0)-1))^{1/2}(D(0))^{-1}+\tan^{-1}(D(0)-1))^{1/2}$

Since $D(0)=1$ and $tan^{-1}(0)=0$ then

$k^{1/2}t=(D(t)-1))^{1/2}(D(t))^{-1}+tan^{-1}(D(t)-1))^{1/2}$

When $D(t)=\infty$ then $tan^{-1}(\infty)=\pi/2$ and $(D(t)-1))^{1/2}D(t))^{-1}=0$ so that this blowup still occurs when $t=t_{*}=\pi/2k^{1/2}$ as required.