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Function: m_and_eta()

0 int m_and_eta(double tau0, double tau1, double *M, double *eta, double Mmin, double Mmax, double pf)
This function takes as inputs the coordinates $(\tau_0,\tau_1)$. If these correspond to individual masses $m_1$ and $m_2$ each lying in the range from $M_{\min}$ to $M_{\rm max}$ then the function sets the total mass $M=m_1+m_2$ and sets $\eta=m_1 m_2/(m_1+m_2)^2$ and returns the value 1. Otherwise, the function returns 0 and does not change the values of mass $M$ or $\eta$.

The arguments are:

tau0 Input. The value of $\tau_0$ (positive, sec).
tau1 Input. The value of $\tau_1$ (positive, sec).
M Output. The total mass $M$ (solar masses). Unaltered if no physical mass values are found in the desired range.
eta Output. The value of $\eta$ (dimensionless). Unaltered if no physical mass values are found in the desired range.
Mmin Input. Minimum mass of one object in the binary pair, in solar masses (positive).
Mmax Input. Maximum mass of one object in the binary pair, in solar masses (positive).
pf: Input. The value $\pi f_0$. Here $f_0$ is the frequency at which the chirp first enters the bandpass of the gravitational wave detector.
The algorithm followed by m_and_eta() is as follows. Eliminate $\eta$ from the equations defining $\tau_0$ ([*]) and $\tau_1$ ([*]) to obtain the following relation:
\begin{displaymath}
c_1 + c_2 \left( {M \over M_\odot} \right)^{5/3} - c_3 \left( {M \over M_\odot} \right) = 0,
\end{displaymath} (9.4.196)

with the constants given by:
$\displaystyle c_1$ $\textstyle =$ $\displaystyle 1155 \; T_\odot$  
$\displaystyle c_2$ $\textstyle =$ $\displaystyle 47552 \; (\pi f_0 T_\odot )^{8/3} \tau_0$ (9.4.197)
$\displaystyle c_3$ $\textstyle =$ $\displaystyle 16128 \; (\pi f_0 T_\odot )^2 \tau_1.$  

Given $(\tau_0,\tau_1)$ our goal is to find the roots of equation ([*]). It is easy to see that the function on the lhs of ([*]) has at most two roots. The function is positive at $M=0$ but decreasing for small positive $M$. However it is positive and increasing again as $M \rightarrow \infty$. Hence the function on the lhs of ([*]) has at most a single minimum for $M>0$. Setting the derivative equal to zero and solving, this minimum lies at a value of the total mass $M_{\rm crit}$ which satisfies
\begin{displaymath}
{M_{\rm crit} \over M_\odot} =
\left( {3 \over 5} \; {c_3 \over c_2} \right)^{3/2}
\end{displaymath} (9.4.198)

Hence the lhs of ([*]) has no roots if its value is positive at $M=M_{\rm crit}$ or it has two roots if that value is negative. (The ``set of measure zero" possibility is a single root at $M_{\rm crit}$.)

If $2 M_{\rm min} < M_{\rm crit} < 2 M_{\rm max}$ then m_and_eta() searches for roots $2 M_{\rm min} < M < M_{\rm crit} $ and $ M_{\rm crit} < M< 2 M_{\rm max}$ separately, else it looks for a root $M$ in the range $2 M_{\rm min} < M < 2 M_{\rm max}$. If the lhs of ([*]) changes sign at the upper and lower boundaries of the interval, then a double-precision routine, similar to the Numerical Recipes routine rtsafe(), is used to obtain the root with a combination of ``safe" bisection and ``rapid" Newton-Raphson.

If a root $M$ is found in the desired range, then $\eta$ is determined by ([*]) to be

\begin{displaymath}
\eta={5 \over 256} \left( {M \over M_\odot} \right)^{-5/3}
(\pi f_0 T_\odot)^{-8/3} {T_\odot \over \tau_0}
\end{displaymath} (9.4.199)

If $\eta \le
1/4$ then the smaller and larger masses are calculated from
\begin{displaymath}
m_1 = {M \over 2} \left(1-\sqrt{1-4\eta}\right) \quad
m_2 = {M \over 2} \left(1+\sqrt{1-4\eta}\right).
\end{displaymath} (9.4.200)

(If both roots for $M$ correspond to $\eta \le
1/4$ then an error message is generated and the routine aborts.) If both $m_1$ and $m_2$ are in the desired range $M_{\rm min} < m_1,m_2 < M_{\rm max}$ then m_and_eta() returns 1 and sets $M$ and $\eta$ appropriately, else it returns 0, leaving $M$ and $\eta$ unaffected.
Author: Bruce Allen, ballen@dirac.phys.uwm.edu
Comments: Although the arguments to this function are double precision floats, the values of $m_1$ and $m_2$ that may be inferred from them can generally only be determined to single precision, particulary in the neighborhood of $m_1=m_2$. The reason is that in the vicinity of $\eta
\sim 1/4$, a fractional error $\epsilon$ is the value of $\eta$ produces a fractional error $\sqrt{\eta}$ in the masses.


next up previous contents
Next: Function: template_area() Up: GRASP Routines: Template Bank Previous: Function: tau_of_mass()   Contents
Bruce Allen 2000-11-19