Appendix/Ramblings/51BiPolytropeStability/RethinkEnvelope

Rethink Handling of n = 1 Envelope

Solution Steps

Drawing from an accompanying discussion …

• Step 1: Choose $n_c$ and $n_e$.
• Step 2: Adopt boundary conditions at the center of the core ($\theta =1$ and $d\theta /d\xi =0$ at $\xi =0$), then solve the Lane-Emden equation to obtain the solution, $\theta (\xi )$, and its first derivative, $d\theta /d\xi$ throughout the core; the radial location, $\xi ={\xi }_s$, at which $\theta (\xi )$ first goes to zero identifies the natural surface of an isolated polytrope that has a polytropic index $n_c$.
• Step 3 Choose the desired location, $0<{\xi }_i<{\xi }_s$, of the outer edge of the core.
• Step 4: Specify $K_c$ and ${\rho }_0$; the structural profile of, for example, $\rho (r)$, $P(r)$, and $M_r(r)$ is then obtained throughout the core — over the radial range, $0\le \xi \le {\xi }_i$ and $0\le r\le r_i$ — via the relations shown in the $2^{\mathrm{n}\mathrm{d}}$ column of Table 1.
• Step 5: Specify the ratio ${\mu }_e/{\mu }_c$ and adopt the boundary condition, ${\varphi }_i=1$; then use the interface conditions as rearranged and presented in Table 3 to determine, respectively:
  • The gas density at the base of the envelope, ${\rho }_e$;
  • The polytropic constant of the envelope, $K_e$, relative to the polytropic constant of the core, $K_c$;
  • The ratio of the two dimensionless radial parameters at the interface, ${\eta }_i/{\xi }_i$;
  • The radial derivative of the envelope solution at the interface, $(d\varphi /d\eta {)}_i$.
• Step 6: The last sub-step of solution step 5 provides the boundary condition that is needed — in addition to our earlier specification that ${\varphi }_i=1$ — to derive the desired particular solution, $\varphi (\eta )$, of the Lane-Emden equation that is relevant throughout the envelope; knowing $\varphi (\eta )$ also provides the relevant structural first derivative, $d\varphi /d\eta$, throughout the envelope.
• Step 7: The surface of the bipolytrope will be located at the radial location, $\eta ={\eta }_s$ and $r=R$, at which $\varphi (\eta )$ first drops to zero.
• Step 8: The structural profile of, for example, $\rho (r)$, $P(r)$, and $M_r(r)$ is then obtained throughout the envelope — over the radial range, ${\eta }_i\le \eta \le {\eta }_s$ and $r_i\le r\le R$ — via the relations provided in the $3^{\mathrm{r}\mathrm{d}}$ column of Table 1.

Setup

Drawing from the accompanying Table 1, we have …

Core	Envelope
$n_c=5$	$n_e=1$
$\frac{1}{{\xi }^2}\frac{d}{d\xi }\left({\xi }^2\frac{d\theta }{d\xi }\right)=-{\theta }^5$ sol'n: $\theta (\xi )$	$\frac{1}{{\eta }^2}\frac{d}{d\eta }\left({\eta }^2\frac{d\varphi }{d\eta }\right)=-\varphi$ sol'n: $\varphi (\eta )$
Specify: $K_c$ and ${\rho }_0\Rightarrow$ $\rho$ $=$ ${\rho }_0{\theta }^5$ $P$ $=$ $K_c{\rho }_0^{6/5}{\theta }^6$ $r$ $=$ $[\frac{3K_c}{2\pi G}{]}^{1/2}{\rho }_0^{-2/5}\xi$ $M_r$ $=$ $4\pi \left[\frac{3K_c}{2\pi G}{]}^{3/2}{\rho }_0^{-1/5}\right(-{\xi }^2\frac{d\theta }{d\xi })$	Knowing: $K_e$ and ${\rho }_e\Rightarrow$ $\rho$ $=$ ${\rho }_e\varphi$ $P$ $=$ $K_e{\rho }_e^2{\varphi }^2$ $r$ $=$ $[\frac{K_e}{2\pi G}{]}^{1/2}\eta$ $M_r$ $=$ $4\pi \left[\frac{K_e}{2\pi G}{]}^{3/2}{\rho }_e\right(-{\eta }^2\frac{d\varphi }{d\eta })$

From an accompanying discussion of $(n_c,n_e)=(5,1)$ bipolytropes, we know that the solution to the pair of Lane-Emden equations is …

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

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

and,

$\varphi =A\left[\frac{\sin (\eta -B)}{\eta }\right],$

$\frac{d\varphi }{d\eta }=\frac{A}{{\eta }^2}[\eta \cos (\eta -B)-\sin (\eta -B)].$

Adopting the same normalizations as before, we have,

Core	Envelope
${\rho }^{*}\equiv \frac{\rho }{{\rho }_0}$ $=$ ${\theta }^5$ $P^{*}\equiv \frac{P}{K_c{\rho }_0^{6/5}}$ $=$ ${\theta }^6$ $r^{*}\equiv r\left[\frac{G^{1/2}{\rho }_0^{2/5}}{K_c^{1/2}}\right]$ $=$ $[\frac{3}{2\pi }{]}^{1/2}\xi$ $M_r^{*}\equiv M_r\left[\frac{G^{3/2}{\rho }_0^{1/5}}{K_c^{3/2}}\right]$ $=$ $4\pi \left[\frac{3}{2\pi }{]}^{3/2}\right(-{\xi }^2\frac{d\theta }{d\xi })$	${\rho }^{*}$ $=$ $\left(\frac{{\rho }_e}{{\rho }_0}\right)\varphi$ $P^{*}$ $=$ $\left[\frac{K_e{\rho }_e^2}{K_c{\rho }_0^{6/5}}\right]{\varphi }^2$ $r^{*}$ $=$ ${\rho }_0^{2/5}[\frac{K_e}{2\pi K_c}{]}^{1/2}\eta$ $M_r^{*}$ $=$ $4\pi \left[{\rho }_e{\rho }_0^{1/5}\right]\left[\frac{K_e}{2\pi K_c}{]}^{3/2}\right(-{\eta }^2\frac{d\varphi }{d\eta })$

Interface Conditions

Now, at the core-envelope interface …

• ${\rho }^{*}{|}_c={\theta }_i^5$
• ${\rho }^{*}{|}_e=({\rho }_e/{\rho }_0){\varphi }_i$
• By choice, ${\varphi }_i=1$
• ${\rho }^{*}{|}_e{\mu }_c={\rho }^{*}{|}_c{\mu }_e$

Hence,

$\frac{{\rho }_e}{{\rho }_0}$

$=$

${\rho }^{*}{|}_e={\rho }^{*}{|}_c\left(\frac{{\mu }_e}{{\mu }_c}\right)=\left(\frac{{\mu }_e}{{\mu }_c}\right){\theta }_i^5.$

Also, setting the value of $P^{*}$ equal across the boundary gives us,

${\theta }_i^6$	$=$	$\left[\frac{K_e{\rho }_e^2}{K_c{\rho }_0^{6/5}}\right]{\varphi }_i^2={\rho }_0^{4/5}\left(\frac{K_e}{K_c}\right)(\frac{{\mu }_e}{{\mu }_c}{)}^2{\theta }_i^{10}$
$\Rightarrow {\rho }_0^{4/5}\left(\frac{K_e}{K_c}\right)$	$=$	$(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}{\theta }_i^{-4}$

As a result, throughout the envelope,

$P^{*}$	$=$	$\left(\frac{{\rho }_e}{{\rho }_0}{)}^2\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}{\theta }_i^{-4}{\varphi }^2={\theta }_i^6{\varphi }^2;$
$r^{*}$	$=$	$(2\pi {)}^{-1/2}[{\rho }_0^{4/5}\cdot \frac{K_e}{K_c}{]}^{1/2}\eta =(2\pi {)}^{-1/2}[\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-1}{\theta }_i^{-2}\right]\eta ;$
$M_r^{*}$	$=$	$2(2\pi {)}^{-1/2}[\frac{{\rho }_e}{{\rho }_0}\left]\right({\rho }_0^{4/5}\cdot \frac{K_e}{K_c}{)}^{3/2}(-{\eta }^2\frac{d\varphi }{d\eta })=2(2\pi {)}^{-1/2}[\left(\frac{{\mu }_e}{{\mu }_c}\right){\theta }_i^5\left]\right[\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}{\theta }_i^{-4}{]}^{3/2}\right(-{\eta }^2\frac{d\varphi }{d\eta })$
	$=$	$\left(\frac{2}{\pi }{)}^{1/2}\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}{\theta }_i^{-1}(-{\eta }^2\frac{d\varphi }{d\eta }).$

In summary, then,

Core

Envelope

${\rho }^{*}$ $=$ ${\theta }^5=[1+\frac{1}{3}{\xi }^2{]}^{-5/2}$ $P^{*}$ $=$ ${\theta }^6$ $r^{*}$ $=$ $[\frac{3}{2\pi }{]}^{1/2}\xi$ $M_r^{*}$ $=$ $4\pi \left[\frac{3}{2\pi }{]}^{3/2}\right(-{\xi }^2\frac{d\theta }{d\xi })$   $=$ $\left[\frac{2^4\cdot 3^3{\pi }^2}{2^3{\pi }^3}{]}^{1/2}\right\{\frac{{\xi }^3}{3}[1+\frac{1}{3}{\xi }^2{]}^{-3/2}\}$   $=$ $\left(\frac{6}{\pi }{)}^{1/2}{\xi }^3\right[1+\frac{1}{3}{\xi }^2{]}^{-3/2}$

${\rho }^{*}$ $=$ $\left(\frac{{\mu }_e}{{\mu }_c}\right){\theta }_i^5\varphi =A\left(\frac{{\mu }_e}{{\mu }_c}\right){\theta }_i^5\left[\frac{\sin (\eta -B)}{\eta }\right]$ $P^{*}$ $=$ ${\theta }_i^6{\varphi }^2$ $r^{*}$ $=$ $(2\pi {)}^{-1/2}[\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-1}{\theta }_i^{-2}\right]\eta$ $M_r^{*}$ $=$ $\left(\frac{2}{\pi }{)}^{1/2}\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}{\theta }_i^{-1}(-{\eta }^2\frac{d\varphi }{d\eta })$   $=$ $\left(\frac{2}{\pi }{)}^{1/2}\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}{\theta }_i^{-1}\left\{A\right[\sin (\eta -B)-\eta \cos (\eta -B)\left]\right\}$

This matches our earlier derivation. Remember, as well, that ${\varphi }_i=1$, that is to say,

$A$

$=$

$\left[\frac{{\eta }_i}{\sin ({\eta }_i-B)}\right].$

Suppose we use $\eta$ as the primary abscissa. Throughout the envelope, for various values of $\eta$, we set

${\rho }^{*}=Q_{\rho }\left[\frac{\sin (\eta -B)}{\eta }\right],M^{*}=Q_m[\sin (\eta -B)-\eta \cos (\eta -B)],\xi =Q_r\eta$

where,

$Q_{\rho }$	$\equiv$	$A\left(\frac{{\mu }_e}{{\mu }_c}\right){\theta }_i^5$
$Q_m$	$\equiv$	$A\left(\frac{2}{\pi }{)}^{1/2}\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}{\theta }_i^{-1}$
$Q_r$	$\equiv$	$3^{-1/2}\left[\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-1}{\theta }_i^{-2}]$

Earlier Examples

In our earlier analysis, we determined that the following relations hold in an equilibrium bipolytrope.

Keep in mind that, once ${\mu }_e/{\mu }_c$ and ${\xi }_i$ have been specified, other parameter values at the interface are: ${\theta }_i$ $=$ $(1+\frac{1}{3}{\xi }_i^2{)}^{-1/2},$ ${\eta }_i$ $=$ $\left(\frac{{\mu }_e}{{\mu }_c}\right)\sqrt{3}{\theta }_i^2{\xi }_i,$ ${\mathrm{\Lambda }}_i$ $=$ $\frac{1}{{\eta }_i}+(\frac{d\varphi }{d\eta }{)}_i=(\frac{{\mu }_e}{{\mu }_c}{)}^{-1}\frac{1}{\sqrt{3}{\xi }_i{\theta }_i^2}-\frac{{\xi }_i}{\sqrt{3}},$ $A$ $=$ ${\eta }_i(1+{\mathrm{\Lambda }}_i^2{)}^{1/2},$ $B$ $=$ ${\eta }_i-\frac{\pi }{2}+{\tan }^{-1}({\mathrm{\Lambda }}_i),$ ${\eta }_s$ $=$ $B+\pi .$

As a test case, let's draw from the accompanying B2 model for which, ${\mu }_e/{\mu }_c=0.25$ and ${\xi }_i=2.4782510$ and …

${\theta }_i$	${\eta }_i$	${\mathrm{\Lambda }}_i$	$A$	$B$	${\eta }_s$		$Q_{\rho }$	$Q_m$	$Q_r$
0.572857	0.352159	1.408807	0.608404	-0.265127	$2.876465$		0.00938349	13.558308	7.0373055

Profiles of Physical Variables

Let's begin by choosing the value of ${\xi }_i$ at which the core-envelope interface will occur. For example, setting ${\xi }_i=3^{-1/2}$ means that $r^{*}=(2\pi {)}^{-1/2}$ and that,

${\eta }_i$

$=$

$\left(\frac{{\mu }_e}{{\mu }_c}\right){\theta }_i^2=\left(\frac{{\mu }_e}{{\mu }_c}\right)[1+\frac{{\xi }_i^2}{3}{]}^{-1}=\frac{9}{10}(\frac{{\mu }_e}{{\mu }_c}).$

This means that we will be outside the core — and, hopefully, inside the envelope — for all values of $\eta >{\eta }_i$, which means for all values of $r^{*}>(2\pi {)}^{-1/2}$.

See Also

• Rappaport, Verbunt, & Joss (1983, ApJ, 275, 713) — A New Technique for Calculations of Binary Stellar Evolution, with Application to Magnetic Braking.