SSC/Structure/BiPolytropes/Analytic51

BiPolytrope with n_c = 5 and n_e = 1 (Pt 1)

Part I:  (nc,ne) = (5,1) BiPolytrope

Part II:  Example Models

Part III:  Limiting Mass

Part IV:  Free Energy

Eggleton, Faulkner & Cannon (1998) Analytic (nc, ne) = (5, 1)

Comment by J. E. Tohline on 30 March 2013: As far as I have been able to determine, this analytic structural model has not previously been published in a refereed, archival journal. Subsequent comment by J. E. Tohline on 23 June 2013: Last night I stumbled upon an article by Eagleton, Faulkner, and Cannon (1998) in which this identical analytically definable bipolytrope has been presented. Insight drawn from this article is presented in an additional subsection, below.

Here we construct a bipolytrope in which the core has an $n_c=5$ polytropic index and the envelope has an $n_e=1$ polytropic index. This system is particularly interesting because the entire structure can be described by closed-form, analytic expressions. In deriving the properties of this model, we will follow the general solution steps for constructing a bipolytrope that we have outlined elsewhere.  
 
 
 

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 ≤ ξ ≤ ξ_i)

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

Equations (A2) from 📚 P. P. Eggleton, J. Faulkner, & R. C. Cannon (1998, MNRAS, Vol. 298, issue 3, pp. 831 - 834) — hereafter, EFC98 — present the same relations but adopt the following notations:* $P$ $=$ $p_0{\theta }^6$         $\Rightarrow$ $K_c$ $=$ $p_0{\rho }_0^{-6/5};$ $\rho$ $=$ ${\rho }_0{\theta }^5$       where: $\theta$ $=$ $a_{\mathrm{e}\mathrm{f}\mathrm{c}}[a_{\mathrm{e}\mathrm{f}\mathrm{c}}^2+r^2{]}^{-1/2}=[1+(\frac{r}{a_{\mathrm{e}\mathrm{f}\mathrm{c}}}{)}^2{]}^{-1/2}$ $a_{\mathrm{e}\mathrm{f}\mathrm{c}}^2$ $\equiv$ $\frac{18p_0}{4\pi G{\rho }_0^2}$         $\Rightarrow$ $a_{\mathrm{e}\mathrm{f}\mathrm{c}}^2$ $=$ $\frac{18K_c{\rho }_0^{6/5}}{4\pi G{\rho }_0^2}=3[\frac{K_c}{G{\rho }_0^{4/5}}\cdot (\frac{3}{2\pi }\left)\right]$           $\Rightarrow$ $a_{\mathrm{e}\mathrm{f}\mathrm{c}}$ $=$ $\sqrt{3}\left[\frac{K_c}{G{\rho }_0^{4/5}}{]}^{1/2}\right(\frac{3}{2\pi }{)}^{1/2}=\sqrt{3}\left(\frac{r}{\xi }\right)$           $\Rightarrow$ $\frac{r}{a_{\mathrm{e}\mathrm{f}\mathrm{c}}}$ $=$ $\frac{\xi }{\sqrt{3}};$ Hence, $\theta$ $=$ $[1+\frac{{\xi }^2}{3}{]}^{-1/2},$ which matches our expression for the core's polytrope function, $\theta$. Now, look at the EFC98 expression for the core's integrated mass. $M_r$ $=$ $\frac{4\pi {\rho }_0{a}_{\mathrm{e}\mathrm{f}\mathrm{c}}^3}{3}\cdot \frac{r^3}{(a_{\mathrm{e}\mathrm{f}\mathrm{c}}^2+r^2{)}^{3/2}}$   $=$ $\frac{4\pi {\rho }_0}{3}\left[a_{\mathrm{e}\mathrm{f}\mathrm{c}}^3\right](r/a_{\mathrm{e}\mathrm{f}\mathrm{c}}{)}^3[1+(r/a_{\mathrm{e}\mathrm{f}\mathrm{c}}{)}^2{]}^{-3/2}$   $=$ $\frac{4\pi {\rho }_0}{3}[\frac{K_c}{G{\rho }_0^{4/5}}\cdot (\frac{3^2}{2\pi }\left){]}^{3/2}\frac{{\xi }^3}{3^{3/2}}\right[1+\frac{{\xi }^2}{3}{]}^{-3/2}$   $=$ $\frac{4\pi }{3}\left(\frac{3}{2\pi }{)}^{3/2}\right[\frac{K_c^3}{G^3}\cdot \frac{{\rho }_0^2}{{\rho }_0^{12/5}}{]}^{1/2}{\xi }^3[1+\frac{{\xi }^2}{3}{]}^{-3/2}$   $=$ $\left[\frac{K_c^3}{G^3{\rho }_0^{2/5}}{]}^{1/2}\right(\frac{2\cdot 3}{\pi }{)}^{1/2}{\xi }^3(1+\frac{{\xi }^2}{3}{)}^{-3/2}$ This expression matches ours.

Step 5: Interface Conditions

			Setting $n_c=5$, $n_e=1$, and ${\varphi }_i=1\Rightarrow$
$\frac{{\rho }_e}{{\rho }_0}$	$=$	$\left(\frac{{\mu }_e}{{\mu }_c}\right){\theta }_i^{n_c}{\varphi }_i^{-n_e}$	$=$	$\left(\frac{{\mu }_e}{{\mu }_c}\right){\theta }_i^5$
$\left(\frac{K_e}{K_c}\right)$	$=$	${\rho }_0^{1/n_c-1/n_e}(\frac{{\mu }_e}{{\mu }_c}{)}^{-(1+1/n_e)}{\theta }_i^{1-n_c/n_e}$	$=$	${\rho }_0^{-4/5}(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}{\theta }_i^{-4}$
$\frac{{\eta }_i}{{\xi }_i}$	$=$	$\left[\frac{n_c+1}{n_e+1}{]}^{1/2}\right(\frac{{\mu }_e}{{\mu }_c}){\theta }_i^{(n_c-1)/2}{\varphi }_i^{(1-n_e)/2}$	$=$	$3^{1/2}\left(\frac{{\mu }_e}{{\mu }_c}\right){\theta }_i^2$
$(\frac{d\varphi }{d\eta }{)}_i$	$=$	$\left[\frac{n_c+1}{n_e+1}{]}^{1/2}{\theta }_i^{-(n_c+1)/2}{\varphi }_i^{(n_e+1)/2}\right(\frac{d\theta }{d\xi }{)}_i$	$=$	$3^{1/2}{\theta }_i^{-3}(\frac{d\theta }{d\xi }{)}_i$

Step 6: Envelope Solution

Adopting equation (8) of 📚 M. Beech (1988, Astrophysics and Space Science, Vol. 147, issue 2, pp. 219 - 227), the most general solution to the $n=1$ Lane-Emden equation can be written in the form,

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

where $A$ and $B$ are constants. The first derivative of this function is,

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

Equations (A2) from 📚 Eggleton, Faulkner, & Cannon (1998) present the same relations but adopt the following notations:* $P$ $=$ $p_0{\theta }^6$         $\Rightarrow$ $K_c$ $=$ $p_0{\rho }_0^{-6/5};$

From Step 5, above, we know the value of the function, $\varphi$ and its first derivative at the interface; specifically,

${\varphi }_i=1\mathrm{a}\mathrm{n}\mathrm{d}(\frac{d\varphi }{d\eta }{)}_i=3^{1/2}{\theta }_i^{-3}(\frac{d\theta }{d\xi }{)}_i\mathrm{a}\mathrm{t}{\eta }_i=3^{1/2}{\xi }_i\left(\frac{{\mu }_e}{{\mu }_c}\right){\theta }_i^2.$

From this information we can determine the constants $A$ and $B$; specifically,

${\eta }_i-B={\tan }^{-1}({\mathrm{\Lambda }}_i^{-1})=\frac{\pi }{2}-{\tan }^{-1}({\mathrm{\Lambda }}_i),$

$A=\frac{{\varphi }_i{\eta }_i}{\sin ({\eta }_i-B)}={\varphi }_i{\eta }_i(1+{\mathrm{\Lambda }}_i^2{)}^{1/2},$

where,

${\mathrm{\Lambda }}_i=\frac{1}{{\eta }_i}+\frac{1}{{\varphi }_i}(\frac{d\varphi }{d\eta }{)}_i.$

Step 7

The surface will be defined by the location, ${\eta }_s$, at which the function $\varphi (\eta )$ first goes to zero, that is,

${\eta }_s=\pi +B=\frac{\pi }{2}+{\eta }_i+{\tan }^{-1}({\mathrm{\Lambda }}_i).$

Equations (A2) from 📚 Eggleton, Faulkner, & Cannon (1998) present the same relations but adopt the following notations:* $P$ $=$ $p_0{\theta }^6$         $\Rightarrow$ $K_c$ $=$ $p_0{\rho }_0^{-6/5};$

Step 8: Throughout the envelope (η_i ≤ η ≤ η_s)

					Knowing: $K_e/K_c$ and ${\rho }_e/{\rho }_0$ from Step 5   $\Rightarrow$
$\rho$	$=$	${\rho }_e{\varphi }^{n_e}$	$=$	${\rho }_0\left(\frac{{\rho }_e}{{\rho }_0}\right)\varphi$	$=$	${\rho }_0\left(\frac{{\mu }_e}{{\mu }_c}\right){\theta }_i^5\varphi$
$P$	$=$	$K_e{\rho }_e^{1+1/n_e}{\varphi }^{n_e+1}$	$=$	$K_c{\rho }_0^{6/5}\left(\frac{K_e{\rho }_0^{4/5}}{K_c}\right)(\frac{{\rho }_e}{{\rho }_0}{)}^2{\varphi }^2$	$=$	$K_c{\rho }_0^{6/5}{\theta }_i^6{\varphi }^2$
$r$	$=$	$[\frac{(n_e+1)K_e}{4\pi G}{]}^{1/2}{\rho }_e^{(1-n_e)/(2n_e)}\eta$	$=$	$\left[\frac{K_c}{G{\rho }_0^{4/5}}{]}^{1/2}\right(\frac{K_e{\rho }_0^{4/5}}{K_c}{)}^{1/2}(2\pi {)}^{-1/2}\eta$	$=$	$\left[\frac{K_c}{G{\rho }_0^{4/5}}{]}^{1/2}\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-1}{\theta }_i^{-2}(2\pi {)}^{-1/2}\eta$
$M_r$	$=$	$4\pi \left[\frac{(n_e+1)K_e}{4\pi G}{]}^{3/2}{\rho }_e^{(3-n_e)/(2n_e)}\right(-{\eta }^2\frac{d\varphi }{d\eta })$	$=$	$\left[\frac{K_c^3}{G^3{\rho }_0^{2/5}}{]}^{1/2}\right(\frac{K_e{\rho }_0^{4/5}}{K_c}{)}^{3/2}\left(\frac{{\rho }_e}{{\rho }_0}\right)\left(\frac{2}{\pi }{)}^{1/2}\right(-{\eta }^2\frac{d\varphi }{d\eta })$	$=$	$\left[\frac{K_c^3}{G^3{\rho }_0^{2/5}}{]}^{1/2}\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}{\theta }_i^{-1}\left(\frac{2}{\pi }{)}^{1/2}\right(-{\eta }^2\frac{d\varphi }{d\eta })$

n = 1 Polytropic Envelope Let's verify the expression for the pressure by integrating the hydrostatic-balance equation, $\frac{dP}{dr}=-\frac{G{M}_r\rho }{r^2}$ From our introductory discussion of the Lane-Emden Equation $\frac{1}{{\xi }^2}\frac{d}{d\xi }\left({\xi }^2\frac{d{\mathrm{\Theta }}_H}{d\xi }\right)=-{\mathrm{\Theta }}_H^n$ we appreciate that, for this $n=1$ envelope, $\rho ={\rho }_c{\mathrm{\Theta }}_H=\varphi =A\left[\frac{\sin (\eta -B)}{\eta }\right],$ and, $r=[\frac{K_1}{2\pi G}{]}^{1/2}\xi$,            in which case, $[\frac{K_1}{2\pi G}{]}^{1/2}=\frac{R}{\pi }.$ Combining these expressions with our above-derived expression for $M_r$, namely, $M_r(\xi )$ $=$ $\underset{0}{\overset{r}{\int }}4\pi r^2\rho dr$

An examination of their equations (A3) reveals that EFC98 continue to use $\theta$ to represent the polytropic function throughout the envelope — for clarity, we will attach the subscript ${\theta }_{\mathrm{e}\mathrm{f}\mathrm{c}}$ — whereas we use $\varphi$. Henceforth we will assume that these functions are interchangeable, that is, ${\theta }_{\mathrm{e}\mathrm{f}\mathrm{c}}\leftrightarrow \varphi$, and examine whether or not their various physical parameter expressions match ours. Comment by J. E. Tohline: As detailed in the text, there appears to be a type-setting error in both of these expressions; as published by EFC98, the exponent on the coefficient of theta_i should be 6 and 5, respectively, whereas it appears as 4. In their Eqs. (A3), EFC98 state that, throughout the envelope, $P$ $=$ $p_0{\theta }_c^4{\theta }_{\mathrm{e}\mathrm{f}\mathrm{c}}^2=K_c{\rho }_0^{6/5}{\theta }_i^4{\varphi }^2$         and, $\rho$ $=$ $\frac{{\rho }_0}{\alpha }\cdot {\theta }_c^4{\theta }_{\mathrm{e}\mathrm{f}\mathrm{c}}={\rho }_0\left(\frac{{\mu }_e}{{\mu }_c}\right){\theta }_i^4\varphi ,$ where, in both expressions, we have replaced their label for the value of the polytropic function at the core-envelope interface, ${\theta }_c$, with our label, ${\theta }_i$. Both of their expressions match ours EXCEPT … NOTE:   in both of their expressions, ${\theta }_i$ is raised to the 4th power whereas, according to our derivation, this interface value should be raised to the 6th power in the expression for pressure and it should be raised to the 5th power in the expression for the density. We are confident that our expressions are the correct ones and therefore presume that type-setting errors are present in both of the EFC98 expressions. We state that the envelope's polytropic function has the form, $\varphi$ $=$ $-\frac{A}{\eta }\cdot \sin (B-\eta ),$ where, $\eta$ $=$ $\left[\frac{G{\rho }_0^{4/5}}{K_c}{]}^{1/2}\right(\frac{{\mu }_e}{{\mu }_c}){\theta }_i^2(2\pi {)}^{1/2}r.$ EFC98 state that, ${\theta }_{\mathrm{e}\mathrm{f}\mathrm{c}}$ $=$ $\frac{B_{\mathrm{e}\mathrm{f}\mathrm{c}}}{r}\cdot \sin [\beta (r_s-r)],$ where, $\beta r$ $=$ $\frac{3{\theta }_c^2}{\alpha }[a_{\mathrm{e}\mathrm{f}\mathrm{c}}{]}^{-1}r$   $=$ $3{\theta }_c^2\left(\frac{{\mu }_e}{{\mu }_c}\right)\left\{\sqrt{3}\right[\frac{K_c}{G{\rho }_0^{4/5}}{]}^{1/2}(\frac{3}{2\pi }{)}^{1/2}{\}}^{-1}r$   $=$ $\left[\frac{G{\rho }_0^{4/5}}{K_c}{]}^{1/2}\right(\frac{{\mu }_e}{{\mu }_c}){\theta }_i^2(2\pi {)}^{1/2}r.$ We therefore conclude that $\beta r=\eta$, and $\beta r_s=B$. If, as we assume to be the case, ${\theta }_{\mathrm{e}\mathrm{f}\mathrm{c}}=\varphi$, it must also be the case that, $\frac{B_{\mathrm{e}\mathrm{f}\mathrm{c}}}{r}$ $=$ $-\frac{A}{\eta }$ $\Rightarrow B_{\mathrm{e}\mathrm{f}\mathrm{c}}$ $=$ $-\frac{A}{\beta }.$ Our expression for the integrated mass throughout the envelope is, $M_r$ $=$ $\left[\frac{K_c^3}{G^3{\rho }_0^{2/5}}{]}^{1/2}\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}{\theta }_i^{-1}\left(\frac{2}{\pi }{)}^{1/2}\right\{-A[\eta \cos (\eta -B)-\sin (\eta -B)]\}$   $=$ $-A\left[\frac{K_c^3}{G^3{\rho }_0^{2/5}}{]}^{1/2}\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}{\theta }_i^{-1}\left(\frac{2}{\pi }{)}^{1/2}\right[\sin (B-\eta )+\eta \cos (B-\eta )].$ According to EFC98, $M_r$ $=$ $\frac{4\pi {\rho }_0\alpha }{9}\left[B_{\mathrm{e}\mathrm{f}\mathrm{c}}{a}_{\mathrm{e}\mathrm{f}\mathrm{c}}^2\right]\{\sin [\beta (r_s-r)]+\beta r\cos [\beta (r_s-r)]\}$   $=$ $\frac{4\pi {\rho }_0}{9}\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-1}\right[-\frac{A{a}_{\mathrm{e}\mathrm{f}\mathrm{c}}^2}{\beta }\left]\right[\sin (B-\eta )+\eta \cos (B-\eta )]$   $=$ $-A\cdot \frac{4\pi {\rho }_0}{9}\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-1}\right[\sin (B-\eta )+\eta \cos (B-\eta )\left]\right\{\left[\frac{G{\rho }_0^{4/5}}{K_c}{]}^{1/2}\right(\frac{{\mu }_e}{{\mu }_c}){\theta }_i^2(2\pi {)}^{1/2}{\}}^{-1}\left\{3\right[\frac{K_c}{G{\rho }_0^{4/5}}\cdot \left(\frac{3}{2\pi }\right)\left]\right\}$   $=$ $-A\cdot \left(\frac{2}{\pi }{)}^{1/2}\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}{\theta }_i^{-2}[\sin (B-\eta )+\eta \cos (B-\eta )][\frac{K_c^3}{G^3{\rho }_0^{2/5}}{]}^{1/2}$

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