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

See Also

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