Appendix/Ramblings/Interrelating51and00Bipolytropes/Organization

Interrelating (5, 1) and (0, 0) Bipolytropes

Structure of (n_c, n_e) = (0, 0) Bipolytropes

Here we draw heavily from an accompanying discussion to construct a bipolytrope in which both the core and the envelope have uniform densities, that is, the structure of both the core and the envelope will be modeled using an $n=0$ polytropic index.

Assuming that the central density, ${\rho }_0$, and central pressure, $P_0$, are specified, the natural dimensionless radius is given by the expression,

$\chi$

$\equiv$

$r[\frac{G{\rho }_0^2}{P_0}{]}^{1/2}.$

Throughout the core (0 ≤ χ ≤ χ_i)

In equilibrium, the radial profile of the density, pressure, and integrated mass are, respectively,

$\rho$	$=$	${\rho }_0$
$P$	$=$	$P_0(1-\frac{2\pi }{3}{\chi }^2)$
$M_r$	$=$	$\frac{4\pi }{3}[\frac{P_0^3}{G^3{\rho }_0^4}{]}^{1/2}{\chi }^3.$

Interface Conditions

In terms of the (as yet unspecified) total radius, $R$, we use $q$ to specify the fractional radial location of the core/envelope interface, that is,

$q$	$\equiv$	$\frac{r_i}{R}$
$\Rightarrow {\chi }_i$	$\equiv$	$q[\frac{G{\rho }_0^2{R}^2}{P_0}{]}^{1/2}.$

And whether viewed from the perspective of the core or the envelope, the pressure at the interface is given by the expression,

$P_i$

$=$

$P_0(1-\frac{2\pi }{3}{\chi }_i^2)$

Envelope Solution (χ ≥ χ_i)

After specifying the envelope-to-core density ratio, ${\rho }_e/{\rho }_0$, the envelope's equilibrium radial profile of the density, pressure, and integrated mass are, respectively,

$\rho$	$=$	${\rho }_e$
$\frac{P}{P_0}$	$=$	$1-\frac{2\pi }{3}{\chi }_i^2+\frac{2\pi }{3}\left(\frac{{\rho }_e}{{\rho }_0}\right){\chi }_i^2\left[2\right(1-\frac{{\rho }_e}{{\rho }_0}\left)\right(\frac{{\chi }_i}{\chi }-1)-\frac{{\rho }_e}{{\rho }_0}(\frac{{\chi }^2}{{\chi }_i^2}-1\left)\right]$
$M_r$	$=$	$\frac{4\pi }{3}\left[\frac{P_0^3}{G^3{\rho }_0^4}{]}^{1/2}{\chi }_i^3\right[1+\frac{{\rho }_e}{{\rho }_0}(\frac{{\chi }^3}{{\chi }_i^3}-1)]$

Related Discussions

• Analytic solution with $n_c=5$ and $n_e=1$.