SSC/Structure/BiPolytropes/Analytic51Renormalize

BiPolytrope with $n_c=5$ and $n_e=1$

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 ($0\le \xi \le {\xi }_i$)

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

---

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 (${\eta }_i\le \eta \le {\eta }_s$)

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

Model Pairings

Here 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 = K-BK74 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

Core
$m_r$	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

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,

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

$=$

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

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,

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

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

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