SSC/Stability/BiPolytropes/51Models

BiPolytrope with n_c = 5 and n_e = 1

Here we construct and analyze the relative stability of a bipolytrope in which the core has an $n_c=5$ polytropic index and the envelope has an $n_e=1$ polytropic index.

Structure

1. Individual model profiles, taken from:
   • SSC/Structure/BiPolytropes/Analytic51#Examples
2. $(q,\nu )$ sequences of fixed ${\mu }_e/{\mu }_c$, taken from:
   • SSC/Structure/BiPolytropes/Analytic51#Model_Sequences
3. ${\nu }_{\mathrm{m}\mathrm{a}\mathrm{x}}$ model, taken from:
   • SSC/Structure/BiPolytropes/Analytic51#Limiting_Mass
      
     

     Maximum Fractional Core Mass, $\nu =M_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}/M_{\mathrm{t}\mathrm{o}\mathrm{t}}$ (solid green circular markers) for Equilibrium Sequences having Various Values of ${\mu }_e/{\mu }_c$
     $\frac{{\mu }_e}{{\mu }_c}$	${\xi }_i$	${\theta }_i$	${\eta }_i$	${\mathrm{\Lambda }}_i$	$A$	${\eta }_s$	LHS	RHS	$q\equiv \frac{r_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}}{R}$	$\nu \equiv \frac{M_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}$
     $\frac{1}{3}$	$\mathrm{\infty }$	---	---	---	---	---	---	---	0.0	$\frac{2}{\pi }$
     0.33	24.00496	0.0719668	0.0710624	0.2128753	0.0726547	1.8516032	-223.8157	-223.8159	0.038378833	0.52024552
     0.316943	10.744571	0.1591479	0.1493938	0.4903393	0.1663869	2.1760793	-31.55254	-31.55254	0.068652714	0.382383875
     0.31	9.014959766	---	---	0.59835053	---	---	---	---	0.0755022550	0.3372170064
     0.3090	8.8301772	0.1924833	0.1750954	0.6130669	0.2053811	2.2958639	-18.47809	-18.47808	0.076265588	0.331475715
     $\frac{1}{4}$	4.9379256	0.3309933	0.2342522	1.4179907	0.4064595	2.761622	-2.601255	-2.601257	0.084824137	0.139370157
     Recall that, ${\ell }_i\equiv \frac{{\xi }_i}{\sqrt{3}};$       and       $m_3\equiv 3\left(\frac{{\mu }_e}{{\mu }_c}\right).$

   
    
   
   • SSC/Structure/BiPolytropes/Analytic51Renormalize#Model_Pairings
      
     

     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

Stability

Here we solve the LAWE numerically (on a uniformly zoned mesh — different $\mathrm{\Delta }\tilde{r}$ for the separate core/envelope regions) using a 2^nd-order accurate, implicit integration scheme in which the LAWE is broken into a pair of 1^st-order ODEs. These results should be compared against a separate succinct discussion of our analysis obtained from integrating the LAWE in its standard 2^nd-order ODE form.

Model Sequence:  μ_e/μ_c = 1.00

Marginally Unstable Model

Numbers presented in the following table should be compared against our earlier determinations. Various things to note:

1. As discussed elsewhere — for example, here — when ${\sigma }_c^2=0$, the radial displacement function for the core — that is, for all $\xi \le {\xi }_i$ — should be given precisely by the expression,

   $x_P{|}_{n=5}$

   $=$

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

   Hence, given that ξ_i = 1.6639103365 as viewed from the perspective of the core, the magnitude of, and the logarithmic derivative of the radial displacement function should have the values, respectively,

   $x_i$

   $=$

   $0.8154268;$

   and

   $\{\frac{d\ln x}{d\ln \xi }{|}_i{\}}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$

   $=$

   $-\frac{2{\xi }^2}{15-{\xi }^2}=-0.45270322.$

2. As discussed elsewhere — for example, here — we expect,

   $\{\frac{d\ln x}{d\ln \tilde{r}}{|}_i{\}}_{\mathrm{e}\mathrm{n}\mathrm{v}}$	$=$	$3(\frac{{\gamma }_c}{{\gamma }_e}-1)+\frac{{\gamma }_c}{{\gamma }_e}\{\frac{d\ln x}{d\ln \xi }{|}_i{\}}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$
   	$=$	$3(\frac{3}{5}-1)+\frac{3}{5}\{\frac{d\ln x}{d\ln \xi }{|}_i{\}}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}=-1.471622.$

file = Dropbox/WorkFolder/Wiki edits/BiPolytrope/TwoFirstOrderODEs/Bipolytrope51.xlsx --- worksheet = MuRatio100Fund Our September 2023 Determinations for Marginally Unstable Model Having ${\mu }_e/{\mu }_c=1$   ${\xi }_i=1.6686460157$     NEW: ${\xi }_i=1.6639103365$
Mode	${\sigma }_c^2$	${\mathrm{\Omega }}^2\equiv \frac{{\sigma }_c^2}{2}\left(\frac{{\rho }_c}{\overline{\rho }}\right)$	$x_i$	$\frac{d\ln x}{d\ln r^{*}}{|}_i$		$x_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}$	$\frac{d\ln x}{d\ln r^{*}}{|}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}$		$\frac{r}{R}{|}_1$	$1-\frac{M_r}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{|}_1$	$\frac{r}{R}{|}_2$	$1-\frac{M_r}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{|}_2$	$\frac{r}{R}{|}_3$	$1-\frac{M_r}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{|}_3$
				core	env		expected	measured
1 (Fundamental)	0.00	0.00	+0.81437470 0.8154268	-0.455872 -0.452703	-1.473523 -1.471622	+0.3820 0.3849493	-1	-0.999999992 -1.00618	n/a	n/a	n/a	n/a	n/a	n/a
2	2.51513333 2.528013	10.7107538 10.720258	0.20482050 0.2069746	-7.09124	-5.4547441	- 0.9962 -1.018215	4.355376917 4.360129	4.35537692 4.3999485	0.64133	0.3502	n/a	n/a	n/a	n/a
3	5.72371888	24.3745901	-0.14269277	+8.046019	+3.627611	+0.9308	11.18729505	11.18729506	0.4837	0.5864	0.842	0.0854	n/a	n/a
4	10.3458476	44.0622916	-0.20845197	-0.6949966	-1.61699793	-1.1443	21.03114578	21.03114577	0.3939	0.7154	0.6902	0.2777	0.9115	0.0284

See Also