Free-fall Time

Even if the pressure force can be neglected in the equation of motion (A.1), the gravitational force remains. Assuming the spherical symmetry, consider the gravity $g_r(r)$ at the position of radial distance from the center being equal to $r$. Using the Gauss' theorem, $g_r$ is related to the mass inside of $r$, which is expressed by the equation

$$M_r=\int_0 ^r \rho 4\pi r^2 dr,$$ (2.12)

and $g_r$ is written as

$$g_r(r)=-\frac{GM_r}{r^2}.$$ (2.13)

This leads to the equation motion for a cold gas under a control of the self-gravity is written

$$\frac{d^2 r}{d t^2}=-\frac{GM_r}{r^2}.$$ (2.14)

Analyzing the equation is straightforward, multiplying $d r/d t$ gives

$$\frac{d v^2/2}{d t}=+\frac{d }{d t}\left(\frac{GM_r}{r}\right),$$ (2.15)

which leads to the conservation of mechanical energy as

$$\frac{1}{2}\left(\frac{d r}{d t}\right)^2-\frac{GM_r}{r}=E,$$ (2.16)

in which $E$ represents the total energy of the pressureless gas element and it is obtained from the initial condition. If the gas is static initially at the distance $R$, the total energy is negative as

$$E=-\frac{GM_r(R)}{R},$$ (2.17)

because at $t=0$, $r=R$ and $d r/d t=0$.

The solutions of equation (2.16) are well known as follows:

1. the case of negative energy $E<0$. Considering the case that the gas sphere is inflowing $v<0$, equation (2.16) becomes
   

   $$\frac{d r}{d t}=-\left[2GM_r(R)\right]^{1/2}\left(\frac{1}{r}-\frac{1}{R}\right)^{1/2},$$ (2.18)

   
   
   where we assumed initially $d r/d t=0$ at $r=R$. Using a parameter $\eta(t)$, the radius of the gas element at some epoch $t$ is written
   

   $$r=R\cos^2 \eta.$$ (2.19)

   
   
   In this case, equation (2.18) reduces to
   

   $$\cos^2 \eta \frac{d \eta}{d t}=\left(\frac{GM_r(R)}{2R^3}\right)^{1/2}.$$ (2.20)

   
   
   This gives us the expression of $t$ as
   

   $$t=\left(\frac{R^3}{2GM_r(R)}\right)^{1/2}\left(\eta+\frac{\sin 2\eta}{2}\right).$$ (2.21)

   
   
   This corresponds to the closed universe in the cosmic expansion ($\Omega_0 > 1$).
2. if the energy is equal to zero, the solution of equation (2.16) is written as
   

   $$\left(r^{3/2}-R^{3/2}\right)^{2/3}= \left(\frac{9GM_r(R)}{2}\right)^{1/3}t^{2/3},$$ (2.22)

   
   
   where $R=r(t=0)$.