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Free-fall Time

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,
\end{displaymath} (2.12)

and $g_r$ is written as
\end{displaymath} (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}.
\end{displaymath} (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),
\end{displaymath} (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,
\end{displaymath} (2.16)

in which $E$ represents the total energy of the pressureless gas element and it is fixed from the initial condition. If the gas is static initially at the distance $R$, the total energy is negative as
\end{displaymath} (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},
\end{displaymath} (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.
\end{displaymath} (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}.
\end{displaymath} (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).
\end{displaymath} (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},
\end{displaymath} (2.22)

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

next up previous contents
Next: Problem Up: Physical Background Previous: Problem   Contents
Kohji Tomisaka 2007-07-08