Function: template_area()

0 float template_area(struct Scope *Grid)
This function computes the area of the enclosed region of parameter space shown in Figure .

The arguments are:

Grid: Input. This function uses only the minimum mass, maximum mass and the cut-off frequency $f_0$ fields of Grid.

The function returns the numerical value of the area in units of ${\rm sec}^2$. See the example in the following subsection.

The function uses an analytic expression for the area obtained by integration of formulae (,) for $\tau_0$ and $\tau_1$ given earlier. For example, to obtain the area of the trapezoidal region bounded above by the maximum-mass curve and below by the $\tau_0$ axis, we integrate

$$\begin{eqnarray*} A_{1} &=& \int_{m_{\rm max}}^{m_{\rm min}} \tau_1 (m_{\rm min... ...2/3} } \right ] \biggr\}_{u=m_{\rm min}/m_{\rm max}}^{u=1} \; . \end{eqnarray*}$$

Here $a=924/743$ and $A_0$ is a quantity with dimensions ${\rm sec}^2$ given by

$$\begin{eqnarray*} A_0={18575 \over 49545216 }{ M_{\odot}^2 \over (\pi M_{\odot} f_0 )^{14/3} } \left ( {c^3 \over G } \right )^{8/3} \;. \end{eqnarray*}$$

The area $A_{2}$ under the minimum-mass curve can be obtained from the formula above by interchanging $m_{\rm min}$ and $m_{\rm max}$. (If you wish to use geomtrized units in which the solar mass is $4.92\times10^{-6} \> {\rm sec}$ simply set $G=c=1$.) The area under the equal-mass curve $A_{3}$ can be obtained by performing a similar integration along the equal-mass curve

$$\begin{eqnarray*} A_{3} &=& \int_{m_{\rm max}}^{m_{\rm min}} \tau_1 (m,m) {d \t... ...right )^{8/3} \right ] \left ( {c^3 \over G } \right )^{8/3} \;. \end{eqnarray*}$$

These three results can be combined to give the total area enclosed

$$A_{total} = A_{1} + A_{2} - A_{3} \;.$$ (9.5.201)

Equation () is the basis of template_area(); the next example shows an application of this function.

Author: Alan Wiseman, agw@tapir.caltech.edu

Comments: None.