SSC/Structure/BiPolytropes/Analytic51Renormalize

BiPolytrope with (n_c, n_e) = (5, 1) Renormalized

Part I:   (5, 1) Analytic Renormalize

Part II:  Envelope

III:  Interface Pressure Gradient

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

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

---

Specify: $K_c$ and $M_{\mathrm{t}\mathrm{o}\mathrm{t}}\Rightarrow$

$\rho$	$=$	$\left\{{{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}\right(\frac{K_c}{G}{)}^{3/2}\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{-1}{\}}^5\right(1+\frac{1}{3}{\xi }^2{)}^{-5/2}$
	$=$	$\left(\frac{{{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{)}^5\right(\frac{K_c}{G}{)}^{15/2}\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-10}\right(1+\frac{1}{3}{\xi }^2{)}^{-5/2};$
$P$	$=$	$K_c\left\{{{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}\right(\frac{K_c}{G}{)}^{3/2}\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{-1}{\}}^6\right(1+\frac{1}{3}{\xi }^2{)}^{-3}$
	$=$	$\left(\frac{{{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{)}^6{K}_c^{10}{G}^{-9}\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-12}(1+\frac{1}{3}{\xi }^2{)}^{-3};$
$r$	$=$	$\left\{{{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}\right(\frac{K_c}{G}{)}^{3/2}\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{-1}{\}}^{-2}\right[\frac{K_c}{G}{]}^{1/2}(\frac{3}{2\pi }{)}^{1/2}\xi$
	$=$	$\left(\frac{{{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{)}^{-2}\right(\frac{K_c}{G}{)}^{-5/2}\left(\frac{{\mu }_e}{{\mu }_c}{)}^4\right(\frac{3}{2\pi }{)}^{1/2}\xi ;$
$M_r$	$=$	$\left\{{{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}\right(\frac{K_c}{G}{)}^{3/2}\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{-1}{\}}^{-1}\right[\frac{K_c^3}{G^3}{]}^{1/2}\left(\frac{2\cdot 3}{\pi }{)}^{1/2}\right[{\xi }^3(1+\frac{1}{3}{\xi }^2{)}^{-3/2}]$
	$=$	$\left(\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{{{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}}\right)\left(\frac{{\mu }_e}{{\mu }_c}{)}^2\right(\frac{2\cdot 3}{\pi }{)}^{1/2}\left[{\xi }^3\right(1+\frac{1}{3}{\xi }^2{)}^{-3/2}].$

 

New Normalization $\tilde{\rho }$ $\equiv$ $\rho \left[\right(\frac{K_c}{G}{)}^{3/2}\frac{1}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{]}^{-5};$ $\tilde{P}$ $\equiv$ $P\left[K_c^{-10}{G}^9{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^6\right];$ $\tilde{r}$ $\equiv$ $r\left[\right(\frac{K_c}{G}{)}^{5/2}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{-2}],$ ${\tilde{M}}_r$ $\equiv$ $\frac{M_r}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}};$ $\tilde{H}$ $\equiv$ $H\left[K_c^{-5/2}{G}^{3/2}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}\right].$

After applying this new normalization, we have throughout the core,

$\tilde{\rho }$	$=$	${{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^5\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-10}\right(1+\frac{1}{3}{\xi }^2{)}^{-5/2};$
$\tilde{P}$	$=$	${{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^6\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-12}\right(1+\frac{1}{3}{\xi }^2{)}^{-3};$
$\tilde{r}$	$=$	${{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^{-2}\left(\frac{{\mu }_e}{{\mu }_c}{)}^4\right(\frac{3}{2\pi }{)}^{1/2}\xi ;$
${\tilde{M}}_r$	$=$	${{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^{-1}\left(\frac{{\mu }_e}{{\mu }_c}{)}^2\right(\frac{2\cdot 3}{\pi }{)}^{1/2}\left[{\xi }^3\right(1+\frac{1}{3}{\xi }^2{)}^{-3/2}].$

Step 8: Throughout the envelope

Given (from above) that,

${\rho }_0$

$=$

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

we have throughout the envelope,

$\rho$	$=$	${\rho }_0\left(\frac{{\mu }_e}{{\mu }_c}\right){\theta }_i^5\varphi$
	$=$	$\left\{{{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}\right(\frac{K_c}{G}{)}^{3/2}\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{-1}{\}}^5\right(\frac{{\mu }_e}{{\mu }_c}){\theta }_i^5\varphi$
	$=$	$\left\{\right(\frac{K_c}{G}{)}^{15/2}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{-5}\left\}{{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^5\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-9}{\theta }_i^5\varphi ;$
$P$	$=$	$K_c{\rho }_0^{6/5}{\theta }_i^6{\varphi }^2$
	$=$	$K_c\left\{{{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}\right(\frac{K_c}{G}{)}^{3/2}(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{-1}{\}}^6{\theta }_i^6{\varphi }^2$
	$=$	$\left\{K_c^{10}{G}^{-9}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{-6}\right\}{{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^6(\frac{{\mu }_e}{{\mu }_c}{)}^{-12}{\theta }_i^6{\varphi }^2;$
$r$	$=$	$\left[\frac{K_c}{G{\rho }_0^{4/5}}{]}^{1/2}\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-1}{\theta }_i^{-2}(2\pi {)}^{-1/2}\eta$
	$=$	$\left[\frac{K_c}{G}{]}^{1/2}\right\{{{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}\left(\frac{K_c}{G}{)}^{3/2}\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{-1}{\}}^{-2}(\frac{{\mu }_e}{{\mu }_c}{)}^{-1}{\theta }_i^{-2}(2\pi {)}^{-1/2}\eta$
	$=$	$\left\{\right(\frac{K_c}{G}{)}^{-5/2}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2\left\}{{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^{-2}\right(\frac{{\mu }_e}{{\mu }_c}{)}^3{\theta }_i^{-2}(2\pi {)}^{-1/2}\eta ;$
$M_r$	$=$	$\left[\frac{K_c^3}{G^3{\rho }_0^{2/5}}{]}^{1/2}\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}{\theta }_i^{-1}\left(\frac{2}{\pi }{)}^{1/2}\right(-{\eta }^2\frac{d\varphi }{d\eta })$
	$=$	$\left[\frac{K_c^3}{G^3}{]}^{1/2}\right\{{{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}\left(\frac{K_c}{G}{)}^{3/2}\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{-1}{\}}^{-1}\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}{\theta }_i^{-1}\right(\frac{2}{\pi }{)}^{1/2}(-{\eta }^2\frac{d\varphi }{d\eta })$
	$=$	$M_{\mathrm{t}\mathrm{o}\mathrm{t}}{{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^{-1}{\theta }_i^{-1}\left(\frac{2}{\pi }{)}^{1/2}\right(-{\eta }^2\frac{d\varphi }{d\eta }).$

Adopting the new normalization then gives,

$\tilde{\rho }$	$=$	${{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^5(\frac{{\mu }_e}{{\mu }_c}{)}^{-9}{\theta }_i^5\varphi ;$
$\tilde{P}$	$=$	${{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^6(\frac{{\mu }_e}{{\mu }_c}{)}^{-12}{\theta }_i^6{\varphi }^2;$
$\tilde{r}$	$=$	${{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^{-2}(\frac{{\mu }_e}{{\mu }_c}{)}^3{\theta }_i^{-2}(2\pi {)}^{-1/2}\eta ;$
${\tilde{M}}_r$	$=$	${{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^{-1}{\theta }_i^{-1}\left(\frac{2}{\pi }{)}^{1/2}\right(-{\eta }^2\frac{d\varphi }{d\eta }).$

Behavior of Central Density Along Equilibrium Sequence

Each equilibrium sequence will be defined as a sequence of models having the same jump in the mean-molecular weight, ${\mu }_e/{\mu }_c$. Along a given sequence, we vary the location of the core/envelope interface, ${\xi }_i$. Our desire is to analyze the behavior of the central density, while holding the total mass fixed, as the location of the interface is varied.

The central density is given by the expression,

${\tilde{\rho }}_c$

$=$

${{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^5\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-10}\right[(1+\frac{1}{3}{\xi }^2{)}^{-5/2}{]}_{\xi =0}={{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^5(\frac{{\mu }_e}{{\mu }_c}{)}^{-10},$

where,

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

$=$

$(\frac{2}{\pi }{)}^{1/2}\frac{A{\eta }_s}{{\theta }_i}.$

In order to evaluate ${{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}$ for a given specification of the interface location, ${\xi }_i$, we need to know that,

${\theta }_i$	$=$	$(1+\frac{1}{3}{\xi }_i^2{)}^{-1/2},$
${\eta }_i$	$=$	$\left(\frac{{\mu }_e}{{\mu }_c}\right)\sqrt{3}{\theta }_i^2{\xi }_i,$
${\mathrm{\Lambda }}_i$	$=$	$\frac{{\xi }_i}{\sqrt{3}}\left[\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-1}\frac{1}{{\theta }_i^2{\xi }_i^2}-1],$
$A$	$=$	${\eta }_i(1+{\mathrm{\Lambda }}_i^2{)}^{1/2},$
${\eta }_s$	$=$	$\frac{\pi }{2}+{\eta }_i+{\tan }^{-1}({\mathrm{\Lambda }}_i).$

Keep in mind, as well, that,

$\nu \equiv \frac{M_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}$	$=$	$\left(\frac{{\mu }_e}{{\mu }_c}{)}^2\sqrt{3}\right[\frac{{\xi }_i^3{\theta }_i^4}{A{\eta }_s}],$
$q\equiv \frac{r_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}}{R}$	$=$	$\left(\frac{{\mu }_e}{{\mu }_c}\right)\sqrt{3}\left[\frac{{\xi }_i{\theta }_i^2}{{\eta }_s}\right].$

 

Bipolytropic (5, 1) Equilibrium Sequences

Central Density versus xi_i (mu_ratio = 0.3100)

Model Pairings

Here we work in the context of the B-KB74 conjecture. We will stick with the sequence corresponding to ${\mu }_e/{\mu }_c=0.31$, and continue to examine the model pairings (B1 and B2) associated with the degenerate model (A) at ${\nu }_{\mathrm{m}\mathrm{a}\mathrm{x}}$. Specifically …

file = Dropbox/WorkFolder/Wiki edits/Bipolytrope/Stability/qAndNuMax.xlsx --- worksheet = B-KB74 thru MinuPreparation Bipolytrope with $(n_c,n_e)=(5,1)$ Selected Pairings along the ${\mu }_e/{\mu }_c=0.31$ Sequence
Pairing	${\xi }_i$	${\mathrm{\Lambda }}_i$	$\nu$	$q$
A	$9.014959766$	$0.59835053$	$0.3372170064$	$0.0755022550$
B1	$9.12744$	$0.60069262$	$0.3372001445$	$0.0746451491$
B2	$8.90394$	$0.59610192$	$0.33720014467$	$0.0763642133$

Bipolytropic (5, 1) Equilibrium Sequences Bipolytropic (5, 1) Equilibrium Sequences

Bipolytropic (5, 1) Equilibrium Sequences

Core						file = Dropbox/WorkFolder/Wiki edits/BiPolytrope/qAndNuMaxAug21.xlsx --- worksheet = B1renorm B-KB74 Eigenfunction
$m_r$	B1		B2		$\frac{\delta r}{r}=\frac{{\tilde{r}}_{\mathrm{B}\mathrm{1}}-{\tilde{r}}_{\mathrm{B}\mathrm{2}}}{2({\tilde{r}}_{\mathrm{B}\mathrm{1}}+{\tilde{r}}_{\mathrm{B}\mathrm{2}})}$
	$\xi$	$\tilde{r}(m_r)$	$\xi$	$\tilde{r}(m_r)$
0.0	0.0	0.0	0.0	0.0	--
0.005	0.430797	0.0007299	0.430395	0.0007331	-0.00109
0.05	1.054468	0.0017865	1.053194	0.0017938	-0.00102
0.1	1.502081	0.0025449	1.499761	0.0025544	-0.00093
0.15	1.963871	0.0033273	1.959917	0.0033382	-0.00082
0.20	2.5329785	0.0042915	2.525985	0.0043023	-0.00063
0.25	3.366385	0.0057034	3.352268	0.0057097	-0.00028
0.30	5.000525	0.0084721	4.959790	0.0084477	+0.00072
0.33715	9.11445	0.015442	8.89185	0.0151449	+0.00486
0.3372001	9.12744	0.015464	8.90394	0.0151654	+0.00487

Envelope
$m_r$	B1		B2		$\frac{\delta r}{r}=\frac{{\tilde{r}}_{\mathrm{B}\mathrm{1}}-{\tilde{r}}_{\mathrm{B}\mathrm{2}}}{2({\tilde{r}}_{\mathrm{B}\mathrm{1}}+{\tilde{r}}_{\mathrm{B}\mathrm{2}})}$
	$\eta$	$\tilde{r}(m_r)$	$\eta$	$\tilde{r}(m_r)$
0.3372001	0.1703455	0.015464	0.1743134	0.0151654	+0.00487
0.35	0.3073375	0.0279002	0.309463	0.0269236	+0.00891
0.40	0.5753765	0.0522328	0.576515	0.0501574	+0.01013
0.45	0.748189	0.0679208	0.749101	0.0651726	+0.01032
0.50	0.8885645	0.0806641	0.8893695	0.0773761	+0.01040
0.55	1.0122575	0.091893	1.012999	0.088132	+0.01045
0.60	1.126297	0.1022455	1.1269968	0.0980499	+0.01047
0.65	1.2347644	0.1120922	1.2354345	0.1074841	+0.01049
0.70	1.3405518	0.1216956	1.3411998	0.1166858	+0.01051
0.75	1.4461523	0.131282	1.4467833	0.1258716	+0.01052
0.80	1.5542198	0.1410924	1.5548378	0.1352725	+0.01053
0.85	1.6683004	0.1514487	1.668908	0.1451967	+0.01054
0.90	1.794487	0.1629039	1.7950862	0.1561743	+0.01055
0.95	1.94764	0.1768072	1.9482325	0.1694982	+0.01055
1.00	2.2820704	0.2071669	2.282658	0.1985936	+0.01056

B-KB74 Eigenfunction

Attempt at Constructing Analytic Eigenfunction Expression

Background

In our accompanying discussion of eigenvectors associated with the radial oscillation of pressure-truncated polytropes, we derived the following,

Exact Solution to the Polytropic LAWE
${\sigma }_c^2=0$	and	$x_P\equiv \frac{3(n-1)}{2n}[1+(\frac{n-3}{n-1}\left)\right(\frac{1}{\xi {\theta }^n}\left)\frac{d\theta }{d\xi }\right].$

Drawing on the definition of $\theta (\xi )$ for n = 5 polytropes, as given in an accompanying chapter, we deduce that,

$x_P{|}_{n=5}$	$=$	$\frac{6}{5}[1+\frac{1}{2}(\frac{1}{\xi {\theta }^5})\frac{d\theta }{d\xi }{]}_{n=5}$
	$=$	$\frac{6}{5}-\frac{3}{5\xi }(1+\frac{{\xi }^2}{3}{)}^{5/2}\frac{\xi }{3}(1+\frac{{\xi }^2}{3}{)}^{-3/2}$
	$=$	$\frac{6}{5}-\frac{1}{5}(1+\frac{{\xi }^2}{3})$
	$=$	$1-\frac{{\xi }^2}{15}.$

And, given that for n = 1 polytropes,

$\theta (\xi )=\frac{\sin \xi }{\xi },$

we also find,

$x_P{|}_{n=1}$	$=$	$-3\left[\right(\frac{1}{\xi \theta })\frac{d\theta }{d\xi }{]}_{n=1}$
	$=$	$\frac{3}{\xi }\left(\frac{\xi }{\sin \xi }\right)[\frac{\sin \xi }{{\xi }^2}-\frac{\cos \xi }{\xi }]$
	$=$	$\frac{3}{{\xi }^2}[1-\xi \cot \xi ].$

Core

Allowing for an overall leading scale factor, $\alpha$, a viable displacement function for the $(n=5)$ core of our bipolytropic configuration is,

$\frac{x_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}}{\alpha }$

$=$

$[1-\frac{{\xi }^2}{15}].$

Throughout the core, the corresponding Lagrangian radial coordinate, $\tilde{r}$, is given by the expression,

${\tilde{r}}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$

$=$

${{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^{-2}\left(\frac{{\mu }_e}{{\mu }_c}{)}^4\right(\frac{3}{2\pi }{)}^{1/2}\xi .$

For "model A" the range is,

$0\le \xi \le {\xi }_i=9.0149598.$

SLOPE:  What is the slope of the function, $x_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}(\tilde{r})$, at the interface? $\frac{d{x}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}}{d\xi }{|}_i$ $=$ $-\frac{2\alpha {\xi }_i}{15},$ $\Rightarrow \frac{d{x}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}}{d\tilde{r}}{|}_i$ $=$ $-\frac{2\alpha {\xi }_i}{15}\left[{{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^{-2}\right(\frac{{\mu }_e}{{\mu }_c}{)}^4(\frac{3}{2\pi }{)}^{1/2}{]}^{-1}=-707.53765\alpha ,$ where, for "model A," we have set $({\mu }_e/{\mu }_c)=0.31$ and ${{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}=1.938127063$. Note as well that, $\frac{d\ln (x_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}})}{d\ln \tilde{r}}{|}_i=\frac{d\ln (x_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}})}{d\ln \xi }{|}_i$ $=$ $-\frac{2{\xi }_i^2}{15}[1-\frac{{\xi }_i^2}{15}{]}^{-1}=+2.452697.$

Envelope

As we have demonstrated in a separate structure discussion, the radial profile of the $(n=1)$ envelope of our bipolytropic configuration is governed by the modified sinc-function,

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

where, for "model A," $A=0.200812422$ and $B=-0.859270052$.

Again allowing for an overall leading scale factor, $\beta$, a viable displacement function for the $(n=1)$ envelope of our bipolytropic configuration is,

$\frac{x_{\mathrm{e}\mathrm{n}\mathrm{v}}}{3\beta }$	$=$	$\frac{1}{\eta \varphi }(-\frac{d\varphi }{d\eta })$
	$=$	$\frac{A}{{\eta }^3}[\sin (\eta -B)-\eta \cos (\eta -B)]\cdot \left[\frac{\eta }{A\sin (\eta -B)}\right]$
	$=$	$\frac{1}{{\eta }^2}[1-\eta \cot (\eta -B)].$

Throughout the envelope, the corresponding Lagrangian radial coordinate is,

${\tilde{r}}_{\mathrm{e}\mathrm{n}\mathrm{v}}$

$=$

${{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^{-2}(\frac{{\mu }_e}{{\mu }_c}{)}^3{\theta }_i^{-2}(2\pi {)}^{-1/2}\eta .$

For "model A" the range is,

$({\eta }_i=0.1723205)\le \eta \le ({\eta }_s=2.282322601).$

SLOPE:  As we have detailed elsewhere, the slope of the function, $x_{\mathrm{e}\mathrm{n}\mathrm{v}}(\tilde{r})$, is related to the slope of $x_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}(\tilde{r})$ at the interface via the expression, $\{\frac{d\ln x}{d\ln r}{|}_i{\}}_{\mathrm{e}\mathrm{n}\mathrm{v}}=\{\frac{d\ln x}{d\ln \eta }{|}_i{\}}_{\mathrm{e}\mathrm{n}\mathrm{v}}$ $=$ $3(\frac{{\gamma }_c}{{\gamma }_e}-1)+\frac{{\gamma }_c}{{\gamma }_e}\{\frac{d\ln x}{d\ln \xi }{|}_i{\}}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}.$ In our case, ${\gamma }_c=6/5$ and ${\gamma }_e=2\Rightarrow {\gamma }_c/{\gamma }_e=3/5$. Hence, from the point of view of the envelope displacement function, at the interface, $[\frac{\tilde{r}}{x_{\mathrm{e}\mathrm{n}\mathrm{v}}}\cdot \frac{d{x}_{\mathrm{e}\mathrm{n}\mathrm{v}}}{d\tilde{r}}{]}_i$ $=$ $\frac{3}{5}\left\{\right[\frac{d\ln (x_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}})}{d\ln \xi }{]}_i-2\}$   $=$ $\frac{3}{5}\left\{\right[2.452697-2\}=0.271618.$ Now, at the interface of any bipolytrope, the ratio $\tilde{r}/x$ should have the same numerical value whether it is viewed from the point of view of the core or the envelope. Given that, for our particular "model A", $[\frac{\tilde{r}}{x}{]}_i=\frac{0.015315}{0.00485976}=3.15139,$ we should expect the slope of the envelope's displacement function at the interface to be, $\frac{d{x}_{\mathrm{e}\mathrm{n}\mathrm{v}}}{d\tilde{r}}{|}_i$ $=$ $0.08619.$

Trial Displacement Function

The blue curve in the following figure results from plotting $x_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$ versus ${\tilde{r}}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$ after setting the leading coefficient, $\alpha =-0.0011$. The red-dotted curve results from plotting $(x_{\mathrm{e}\mathrm{n}\mathrm{v}}+x_{\mathrm{s}\mathrm{h}\mathrm{i}\mathrm{f}\mathrm{t}})$ versus ${\tilde{r}}_{\mathrm{e}\mathrm{n}\mathrm{v}}$ after setting the leading coefficient, $\beta =-0.000062$, and $x_{\mathrm{s}\mathrm{h}\mathrm{i}\mathrm{f}\mathrm{t}}=+0.0105$.

Trial Analytic Eigenfunction

ASSESSMENT:

• Our analytically specified displacement function, $x_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$, appears to be an excellent match to the displacement function obtained throughout the core by implementing the B-KB74 conjecture.
• At first glance, the plot of $(x_{\mathrm{e}\mathrm{n}\mathrm{v}}+x_{\mathrm{s}\mathrm{h}\mathrm{i}\mathrm{f}\mathrm{t}})$ appears to provide a reasonably good fit to the approximate displacement function that we have obtained throughout the envelope by implementing the B-KB74 conjecture. But, in reality, there are two fatal flaws:
  1. We have presented the behavior of our analytically specified envelope displacement function only up to the radial coordinate, $\eta =2.19707({\tilde{r}}_{\mathrm{e}\mathrm{n}\mathrm{v}}=0.19526)$. Between this point and the surface, ${\eta }_s=2.2823226({\tilde{r}}_{\mathrm{e}\mathrm{n}\mathrm{v}}=0.2028415)$ — where the argument of the cotangent, $({\eta }_s-B)\to \pi$ — the analytic function dives steeply to negative infinity. This violently departs from the behavior derived via the B-KB74 conjecture.
  2. While our analytically specified displacement function, $x_{\mathrm{e}\mathrm{n}\mathrm{v}}$, satisfies the "n = 1" polytropic LAWE, this satisfaction is destroyed by adding $x_{\mathrm{s}\mathrm{h}\mathrm{i}\mathrm{f}\mathrm{t}}$ to the displacement function.

Let's examine the slope of the displacement function at the interface. From the perspective of the core, our analytic prescription for the displacement function matches the K-BK74-derived displacement function very well. An analytic evaluation of the slope at the inferface — as derived above — gives,

$\frac{d{x}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}}{d\tilde{r}}{|}_i=-707.53765\alpha =+0.77829$.

The black-dashed line segment that appears in the following figure has this slope and goes through the point of intersection; it appears to be tangent to the analytic displacement function, as expected. Alternatively, the orange-dashed line segment that appears in this same figure, also goes through the point of intersection, but it has a slope that matches our expectation for the envelope's displacement function; that is, it has a slope as derived of,

$\frac{d{x}_{\mathrm{e}\mathrm{n}\mathrm{v}}}{d\tilde{r}}{|}_i=+0.08619$.

This orange-dashed line segment does not appear to lie tangent to the K-BK74-derived displacement function for the envelope.

Trial Analytic Eigenfunction with Intersection Slopes

2^nd Trial

The relevant LAWE for the envelope is,

$\left[\frac{A\sin (\eta -B)}{\eta }\right]\frac{d^{2}x}{d{\eta }^2}$	$=$	$-\frac{2A}{\eta }\{\sin (\eta -B)+\eta \cos (\eta -B)\}\frac{1}{\eta }\cdot \frac{dx}{d\eta }+\frac{2A}{\eta }\{\sin (\eta -B)-\eta \cos (\eta -B)\}\frac{x}{{\eta }^2}-2\left(\frac{{\sigma }_c^2}{12}\right)x$
$\Rightarrow \frac{1}{3\beta }\cdot \frac{d^{2}x}{d{\eta }^2}$	$=$	$-\frac{2}{\eta }\{1+\eta \cot (\eta -B)\}[\frac{1}{3\beta }\cdot \frac{dx}{d\eta }]+2\{1-\eta \cot (\eta -B)\}\frac{x}{3\beta {\eta }^2}-2\left(\frac{{\sigma }_c^2}{12}\right)\left[\frac{\eta }{3\beta A\sin (\eta -B)}\right]x.$

Here, we will guess a displacement function, $x_{\mathrm{e}\mathrm{n}\mathrm{v}}$, of the form,

$\frac{x_{\mathrm{e}\mathrm{n}\mathrm{v}}}{3\beta }$

$=$

$\frac{1}{{\eta }^2}[1-\eta \cot (\eta -B_x)]=\frac{1}{{\eta }^2}-\frac{\cos (\eta -B_x)}{\eta \sin (\eta -B_x)},$

where we will assume, quite generally, that $B_x\ne B$. The first and second derivatives of $x_{\mathrm{e}\mathrm{n}\mathrm{v}}$ are,

$\frac{1}{3\beta }\left[\frac{d{x}_{\mathrm{e}\mathrm{n}\mathrm{v}}}{d\eta }\right]$	$=$	$-\frac{2}{{\eta }^3}+\frac{\cos (\eta -B_x)}{{\eta }^2\sin (\eta -B_x)}+\frac{\sin (\eta -B_x)}{\eta \sin (\eta -B_x)}+\frac{{\cos }^2(\eta -B_x)}{\eta {\sin }^2(\eta -B_x)}$
	$=$	$-\frac{2}{{\eta }^3}+\frac{1}{\eta }+\frac{\cos (\eta -B_x)}{{\eta }^2\sin (\eta -B_x)}+\frac{{\cos }^2(\eta -B_x)}{\eta {\sin }^2(\eta -B_x)};$
$\frac{1}{3\beta }\left[\frac{d^2{x}_{\mathrm{e}\mathrm{n}\mathrm{v}}}{d{\eta }^2}\right]$	$=$	$\frac{6}{{\eta }^4}-\frac{1}{{\eta }^2}-\frac{2\cos (\eta -B_x)}{{\eta }^3\sin (\eta -B_x)}+\frac{1}{{\eta }^2}[-\frac{\sin (\eta -B_x)}{\sin (\eta -B_x)}-\frac{{\cos }^2(\eta -B_x)}{{\sin }^2(\eta -B_x)}]-\frac{{\cos }^2(\eta -B_x)}{{\eta }^2{\sin }^2(\eta -B_x)}+\frac{1}{\eta }[-\frac{2\cos (\eta -B_x)}{\sin (\eta -B_x)}-\frac{2{\cos }^3(\eta -B_x)}{{\sin }^3(\eta -B_x)}]$
	$=$	$\frac{6}{{\eta }^4}-\frac{1}{{\eta }^2}-\frac{2\cos (\eta -B_x)}{{\eta }^3\sin (\eta -B_x)}-\frac{1}{{\eta }^2}[1+\frac{{\cos }^2(\eta -B_x)}{{\sin }^2(\eta -B_x)}]-\frac{{\cos }^2(\eta -B_x)}{{\eta }^2{\sin }^2(\eta -B_x)}-\frac{1}{\eta }[\frac{2\cos (\eta -B_x)}{\sin (\eta -B_x)}+\frac{2{\cos }^3(\eta -B_x)}{{\sin }^3(\eta -B_x)}]$
	$=$	$\frac{6}{{\eta }^4}-\frac{2}{{\eta }^2}-\frac{2\cos (\eta -B_x)}{{\eta }^3\sin (\eta -B_x)}-\frac{2}{{\eta }^2}\left[\frac{{\cos }^2(\eta -B_x)}{{\sin }^2(\eta -B_x)}\right]-\frac{2}{\eta }[\frac{\cos (\eta -B_x)}{\sin (\eta -B_x)}+\frac{{\cos }^3(\eta -B_x)}{{\sin }^3(\eta -B_x)}]$
	$=$	$\frac{6}{{\eta }^4}-\frac{2}{{\eta }^2}[1+{\cot }^2(\eta -B_x)]-\frac{2\cot (\eta -B_x)}{{\eta }^3}-\frac{2}{\eta }[\cot (\eta -B_x)+{\cot }^{}(\eta -B_x)].$

These match the expressions for $d{x}_P/d\eta$ and $d^2{x}_P/d{\eta }^2$ that we separately derived in a subsection labeled Attempt_4B of an accompanying discussion labeled. Plugging these three relations into the LAWE, then multiplying through by ${\eta }^4$, gives,

$\frac{6}{{\eta }^4}-\frac{2}{{\eta }^2}[1+{\cot }^2(\eta -B_x)]-\frac{2\cot (\eta -B_x)}{{\eta }^3}-\frac{2}{\eta }[\cot (\eta -B_x)+{\cot }^{}(\eta -B_x)]$	$=$	$-\frac{2}{\eta }\{1+\eta \cot (\eta -B)\}[-\frac{2}{{\eta }^3}+\frac{1}{\eta }+\frac{\cot (\eta -B_x)}{{\eta }^2}+\frac{{\cot }^2(\eta -B_x)}{\eta }]$
		$+\left\{2\right[1-\eta \cot (\eta -B)]\frac{1}{{\eta }^2}-2(\frac{{\sigma }_c^2}{12}\left)\right[\frac{\eta }{A\sin (\eta -B)}\left]\right\}[\frac{1}{{\eta }^2}-\frac{\cot (\eta -B_x)}{\eta }]$
$\Rightarrow 6-2{\eta }^2[1+{\cot }^2(\eta -B_x)]-2\eta \cot (\eta -B_x)-2{\eta }^3[\cot (\eta -B_x)+{\cot }^{}(\eta -B_x)]$	$=$	$-2\{1+\eta \cot (\eta -B)\}[-2+{\eta }^2+\eta \cot (\eta -B_x)+{\eta }^2{\cot }^{}(\eta -B_x)]$
		$+\left\{2\right[1-\eta \cot (\eta -B)]-2(\frac{{\sigma }_c^2}{12}\left)\right[\frac{{\eta }^3}{A\sin (\eta -B)}\left]\right\}[1-\eta \cot (\eta -B_x)]$

$\Rightarrow \left(\frac{{\sigma }_c^2}{6}\right)\left[\frac{{\eta }^3}{A\sin (\eta -B)}\right][1-\eta \cot (\eta -B_x)]$	$=$	$[1+\eta \cot (\eta -B)][4-2{\eta }^2-2\eta \cot (\eta -B_x)-2{\eta }^2{\cot }^{}(\eta -B_x)]$
		$+2[1-\eta \cot (\eta -B)][1-\eta \cot (\eta -B_x)]-6+2{\eta }^2[1+{\cot }^{}(\eta -B_x)]+2\eta \cot (\eta -B_x)+2{\eta }^3[\cot (\eta -B_x)+{\cot }^{}(\eta -B_x)]$
	$=$	$[4-2{\eta }^2-2\eta \cot (\eta -B_x)-2{\eta }^2{\cot }^{}(\eta -B_x)]+\eta \cot (\eta -B)[4-2{\eta }^2-2\eta \cot (\eta -B_x)-2{\eta }^2{\cot }^{}(\eta -B_x)]$
		$+[2-2\eta \cot (\eta -B)-2\eta \cot (\eta -B_x)+2{\eta }^2\cot (\eta -B)\cot (\eta -B_x)]-6+2{\eta }^2+2{\eta }^2{\cot }^{}(\eta -B_x)+2\eta \cot (\eta -B_x)+2{\eta }^3\cot (\eta -B_x)+2{\eta }^3{\cot }^{}(\eta -B_x)$
	$=$	$-2\eta \cot (\eta -B_x)-2{\eta }^2{\cot }^{}(\eta -B_x)+4\eta \cot (\eta -B)-2{\eta }^3\cot (\eta -B)-2{\eta }^2\cot (\eta -B)\cot (\eta -B_x)-2{\eta }^3\cot (\eta -B){\cot }^{}(\eta -B_x)$
		$-2\eta \cot (\eta -B)-2\eta \cot (\eta -B_x)+2{\eta }^2\cot (\eta -B)\cot (\eta -B_x)+2{\eta }^2{\cot }^{}(\eta -B_x)+2\eta \cot (\eta -B_x)+2{\eta }^3\cot (\eta -B_x)+2{\eta }^3{\cot }^{}(\eta -B_x)$
	$=$	$-2{\eta }^3[\cot (\eta -B)+\cot (\eta -B){\cot }^{}(\eta -B_x)-\cot (\eta -B_x)-{\cot }^{}(\eta -B_x)]+2\eta [\cot (\eta -B)-\cot (\eta -B_x)]$

Notice that if we set $B_x=B$, the RHS of this LAWE expression goes to zero. This [thankfully] is as expected.

See Also

• Prasad, C. (1953), Proc. Natn. Inst. Sci. India, Vol. 19, 739, Radial Oscillations of a Composite Model.
• Singh, Manmohan, (1969), Proc. Natn. Inst. Sci., India, Part A, Vol. 35, pp. 586 - 589, Radial Oscillations of Composite Polytropes — Part I
• Singh, Manmohan, (1969), Proc. Nat. Inst. Sci., India, Part A, Vol. 35, pp. 703 - 708, Radial Oscillations of Composite Polytropes — Part II
• Kumar, S., Saini, S., Singh, K. K., Bhatt, V., & Vashishta, L. (2021), Astronomical & Astrophysical Transactions, Vol. 32, Issue 4, pp. 371-382, Radial Pulsations of distorted Polytropes of Non-Uniform Density.