Subsections


Derivations


Free-Fall Derivation

Consider a uniform density spherical distribution of matter initially at rest, with mass $M$ and radius $R$. At time $t=0$ it begins to collapse under self gravity. Since the original distribution is spherical and homogenous, it remains so during the collapse. The free-fall equations can be derived by considering the energies present:

A small portion of surface material with mass $\mu$, has gravitational potential energy equal to $-\frac{GM}{R}\mu$. Since the body is initially static, this is the total energy associated with the small mass. At subsequent times the body will have both potential energy and kinetic energy, and this will be equal to the starting energy just calculated:

\begin{displaymath} \frac{1}{2}\bigg(\frac{dr}{dt}\bigg)^{2}\mu-\frac{GM}{r}\mu=-\frac{GM}{R}\mu \end{displaymath} (B.1)

where $r$ is the subsequent radial distance of the surface. Rearranging in terms of the potential energies gives:
\begin{displaymath} \frac{dr}{dt}=-\sqrt{2GM\bigg(\frac{1}{r}-\frac{1}{R}\bigg)} \end{displaymath} (B.2)

where the velocity is defined as negative, since it is in the opposite direction to the positively defined $r$. Substituting $r=R\sin^{2}(\theta)$, gives $dr=2R\sin(\theta)\cos(\theta) d\theta$ and:
\begin{displaymath} dt=-\frac{2R\sin(\theta)\cos(\theta) d\theta}{\sqrt{2GM\Big(\frac{1}{R\sin^{2}(\theta)}-\frac{1}{R}\Big)}} \end{displaymath} (B.3)

which conveniently simplifies to:
\begin{displaymath} dt=-\sqrt{\frac{2R^{3}}{GM}}\sin^{2}(\theta) d\theta \end{displaymath} (B.4)

In calculating the free-fall time, the integration is performed from $r=R$ to $r=0$ ( $\theta=\frac{\pi}{2}$ to $\theta=0$):
\begin{displaymath} t_{ff}=\Bigg[-\sqrt{\frac{2R^{3}}{GM}}\Bigg\{\frac{\theta}{2... ...cos(\theta)}{2}\Bigg\}\Bigg]_{\theta=\frac{\pi}{2}}^{\theta=0} \end{displaymath} (B.5)

The free-fall time is thus:
\begin{displaymath} t_{ff}=\frac{\pi}{2}\Bigg(\frac{R^{3}}{2GM}\Bigg)^{\frac{1}{2}} \end{displaymath} (B.6)

The equation for following the position of the surface can be obtained by integrating Equation B.4 from $r=R$ to $r=r$ ( $\theta=\frac{\pi}{2}$ to $\theta=\sin^{-1}\Big[\Big(\frac{r}{R}\Big)^\frac{1}{2}\Big]$)

\begin{displaymath} t=-\sqrt{\frac{R^{3}}{2GM}}\Bigg[\theta-\sin(\theta)\cos(\th... ...^{\theta=\sin^{-1}\big[\big(\frac{r}{R}\big)^\frac{1}{2}\big]} \end{displaymath} (B.7)

this leads to:
\begin{displaymath} t=-\sqrt{\frac{R^{3}}{2GM}}\Bigg\{\Bigg[\sin^{-1}\bigg[\bigg... ...R}\bigg]^{\frac{1}{2}}\Bigg]- \Bigg[\frac{\pi}{2}\Bigg]\Bigg\} \end{displaymath} (B.8)

Substituting for the free-fall time calculated in Equation B.6 gives:
\begin{displaymath} \frac{t}{t_{ff}}=1-\frac{2}{\pi}\Bigg\{\sin^{-1}\Bigg[\bigg(... ...g)^{\frac{1}{2}}\bigg[1-\frac{r}{R}\bigg]^{\frac{1}{2}}\Bigg\} \end{displaymath} (B.9)

Since the collapse is homogeneous, the initial surface radius $R$ in Equation B.9, can be replaced with any starting radius $r_{0}$, in which case $r$ then refers to the subsequent radial distance of that point:

\begin{displaymath} \frac{t}{t_{ff}}=1-\frac{2}{\pi}\Bigg\{\sin^{-1}\Bigg[\bigg(... ...\frac{1}{2}}\bigg[1-\frac{r}{r_{0}}\bigg]^{\frac{1}{2}}\Bigg\} \end{displaymath} (B.10)


Point Source Absorption Derivation

Figure B.1: Point source radiation test scenario.
\begin{figure}\centering {\epsfig{figure=/home/steve/tmp/hold/thesis/graphs/absorba1.eps,angle=0,width=1.0\linewidth}} \end{figure}

Figure B.1 shows a point source radiator, denoted by the small black circle on the right, and a uniform density spherical distribution of gas of known optical depth $\tau$, radius $R$, and distance $d$ from the point source. The radiation absorbed in the gas can be calculated by consideration of the radiation absorbed between two truncated cones of semi-angle $\theta$ and $\theta+d\theta$ centred on the point source:
\begin{displaymath} \frac{dE_{d\theta}}{dt}=\frac{L}{4\pi}2\pi\sin(\theta)\Bigg\{1-exp\bigg(-\frac{x}{2R}\tau\bigg)\Bigg\}d\theta \end{displaymath} (B.11)

where $L$ is the luminosity of the point source, and $x$ and $\theta$ are the quantities shown in Figure B.1. The total rate of energy absorbed in the sphere is thus:
\begin{displaymath} \frac{dE}{dt}=\frac{L}{2}\int_{0}^{\theta_{M}}\sin(\theta)\Bigg\{1-exp\bigg(-\frac{x}{2R}\tau\bigg)\Bigg\}d\theta \end{displaymath} (B.12)

where $\theta_{M}$ is the maximum value of $\theta$ coinciding with the surface of the sphere:
\begin{displaymath} \theta_{M}=\sin^{-1}\Bigg(\frac{1}{K}\Bigg) \end{displaymath} (B.13)

where $K$ is the dimensionless ratio of distances:
\begin{displaymath} K=\frac{d}{R} \end{displaymath} (B.14)

When $\tau$ is such that all radiation incident on the sphere is absorbed, Equation B.12 can be simplified:
\begin{displaymath} \frac{dE_{BB}}{dt}=\frac{L}{2}\int_{0}^{\theta_{M}}\sin(\theta)d\theta \end{displaymath} (B.15)

which integrates to:
\begin{displaymath} \frac{dE_{BB}}{dt}=\frac{L}{2}\Bigg\{1-\bigg(1-\frac{1}{K^{2}}\bigg)^\frac{1}{2}\Bigg\} \end{displaymath} (B.16)

When the simplification cannot be made, $x$ is required to be expressed as a function of $\theta$ and $K$. Using the law of cosines to solve for $D$:
\begin{displaymath} R^{2}=D^{2}+d^{2}-2Dd\cos(\theta) \end{displaymath} (B.17)

gives:
\begin{displaymath} D=d\cos(\theta)\pm\sqrt{R^{2}-d^{2}\sin^{2}(\theta)} \end{displaymath} (B.18)

One of the solutions of Equation B.18 corresponds to the distance $x+D$. The difference between the solutions is therefore $x$:
\begin{displaymath} x=2\sqrt{R^{2}-d^{2}\sin^{2}(\theta)} \end{displaymath} (B.19)

or more conveniently:
\begin{displaymath} x=2R\sqrt{1-K^{2}\sin^{2}(\theta)} \end{displaymath} (B.20)

Substituting for $x$ in Equation B.12 results in:
\begin{displaymath} \frac{dE}{dt}=\frac{L}{2}\int_{0}^{\theta_{M}}\sin(\theta)\B... ...p\bigg(-\tau\sqrt{1-K^{2}\sin^{2}(\theta)}\bigg)\Bigg\}d\theta \end{displaymath} (B.21)

This could not be solved analytically. Instead, numerical Gauss-quadrature was used to solve it for a given value of $K$ and $\tau$ whilst performing the tests in Section 4.3.2.
Front Page
Stephen Oxley 2002-01-19