SSC/Structure/BiPolytropes/Analytic51Renormalize

BiPolytrope with $n_c=5$ and $n_e=1$

This chapter very closely parallels our original analytic derivation — see also, 📚 P. P. Eggleton, J. Faulkner, & R. C. Cannon (1998, MNRAS, Vol. 298, issue 3, pp. 831 - 834) — of the structure of bipolytropes in which the core has an $n_c=5$ polytropic index and the envelope has an $n_e=1$ polytropic index. Our primary objective, here, is to renormalize the principal set of variables, replacing the central density with the configuration's total mass, so that the mass is held fixed along each model sequence.

From Table 1 of our original analytic derivation, we see that,

$(\frac{{\mu }_e}{{\mu }_c}{)}^2{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}$	$=$	${{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}(\frac{K_c}{G}{)}^{3/2}{\rho }_0^{-1/5}$
$\Rightarrow {\rho }_0$	$=$	$\left\{{{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}\right(\frac{K_c}{G}{)}^{3/2}(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{-1}{\}}^5,$

where,

${{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}$

$\equiv$

$\left(\frac{2}{\pi }{)}^{1/2}{\theta }_i^{-1}\right(-{\eta }^2\frac{d\varphi }{d\eta }{)}_s=(\frac{2}{\pi }{)}^{1/2}\frac{A{\eta }_s}{{\theta }_i}.$

Steps 2 & 3

Based on the discussion presented elsewhere of the structure of an isolated $n=5$ polytrope, the core of this bipolytrope will have the following properties:

$\theta (\xi )=[1+\frac{1}{3}{\xi }^2{]}^{-1/2}\Rightarrow {\theta }_i=[1+\frac{1}{3}{\xi }_i^2{]}^{-1/2};$

$\frac{d\theta }{d\xi }=-\frac{\xi }{3}[1+\frac{1}{3}{\xi }^2{]}^{-3/2}\Rightarrow (\frac{d\theta }{d\xi }{)}_i=-\frac{{\xi }_i}{3}[1+\frac{1}{3}{\xi }_i^2{]}^{-3/2}.$

The first zero of the function $\theta (\xi )$ and, hence, the surface of the corresponding isolated $n=5$ polytrope is located at ${\xi }_s=\mathrm{\infty }$. Hence, the interface between the core and the envelope can be positioned anywhere within the range, $0<{\xi }_i<\mathrm{\infty }$.

Step 4: Throughout the core ($0\le \xi \le {\xi }_i$)

Specify: $K_c$ and ${\rho }_0\Rightarrow$
$\rho$	$=$	${\rho }_0{\theta }^{n_c}$	$=$	${\rho }_0(1+\frac{1}{3}{\xi }^2{)}^{-5/2}$
$P$	$=$	$K_c{\rho }_0^{1+1/n_c}{\theta }^{n_c+1}$	$=$	$K_c{\rho }_0^{6/5}(1+\frac{1}{3}{\xi }^2{)}^{-3}$
$r$	$=$	$[\frac{(n_c+1)K_c}{4\pi G}{]}^{1/2}{\rho }_0^{(1-n_c)/(2n_c)}\xi$	$=$	$\left[\frac{K_c}{G{\rho }_0^{4/5}}{]}^{1/2}\right(\frac{3}{2\pi }{)}^{1/2}\xi$
$M_r$	$=$	$4\pi \left[\frac{(n_c+1)K_c}{4\pi G}{]}^{3/2}{\rho }_0^{(3-n_c)/(2n_c)}\right(-{\xi }^2\frac{d\theta }{d\xi })$	$=$	$\left[\frac{K_c^3}{G^3{\rho }_0^{2/5}}{]}^{1/2}\right(\frac{2\cdot 3}{\pi }{)}^{1/2}\left[{\xi }^3\right(1+\frac{1}{3}{\xi }^2{)}^{-3/2}]$