Editing SSC/StabilityConjecture/Bipolytrope51 (section)

====Expressions for q and &nu;====
Following through the numbered steps that we have used to [[SSC/Structure/BiPolytropes/Analytic51|construct a bipolytrope with]] <math>(n_c, n_e) = (5, 1)</math>, and adopting the substitute notation,
<div align="center">
<math>
\ell_i \equiv \frac{\xi_i}{\sqrt{3}}  \, ;
</math>
&nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp; 
<math>
m_3 \equiv 3 \biggl( \frac{\mu_e}{\mu_c} \biggr) \, ,
</math>
</div>
we seek expressions for <math>\nu(m_3,\ell_i)</math> and  <math>q(m_3,\ell_i)</math>.  ['''Example #1''' numerical evaluation is for <math>\mu_e/\mu_c = 0.25</math> and <math>\xi_i = 0.5</math>, which implies that <math>m_3 = 0.75</math> and <math>\ell_i = (12)^{-1 / 2}</math>.] 

Focusing, first, on the core, we find,
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right">
<math>\theta_i</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>(1 + \ell_i^2)^{-1 / 2} = 0.960768923</math>,
  </td>
</tr>

<tr>
  <td align="right">
<math>\biggl(\frac{d\theta}{d\xi}\biggr)_i</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>-\frac{\ell_i}{\sqrt{3}}(1 + \ell_i^2)^{-3 / 2} = -0.147810603</math>,
  </td>
</tr>
</table>

<!-- The radius and mass of the core -->
<table border="1" width="80%" align="center" cellpadding="8">
<tr><td align="center">
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right">
<math>\biggl[ \frac{G \rho_0^{4/5}}{K_c} \biggr]^{1 / 2} r_\mathrm{core}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\biggl(\frac{3^2}{2\pi} \biggr)^{1 / 2} \ell_i  = 0.345494149</math>,
  </td>
</tr>

<tr>
  <td align="right">
<math>\biggl[ \frac{G^3 \rho_0^{2/5}}{K_c^3} \biggr]^{1 / 2} M_\mathrm{core}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>3^2\biggl(\frac{2}{\pi} \biggr)^{1 / 2}\ell_i^3(1 + \ell_i^2)^{-3 / 2}  = 0.153203096</math>,
  </td>
</tr>
</table>
</td></tr>
</table>
Then moving across the interface, through the envelope, and ultimately to the surface of the configuration, we find,
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right">
<math>\eta_i</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>m_3\ell_i \theta_i^2 = \frac{m_3 \ell_i}{(1+\ell_i^2)} = 0.199852016</math>,
  </td>
</tr>

<tr>
  <td align="right">
<math>\biggl(\frac{d\phi}{d\eta}\biggr)_i</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>3^{1 / 2} \theta_i^{-3} \biggl( \frac{d\theta}{d\xi}\biggr)_i = -0.288675135</math>,
  </td>
</tr>

<tr>
  <td align="right">
<math>\Lambda_i</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{1}{\eta_i} + \biggl(\frac{d\phi}{d\eta}\biggr)_i
=
\frac{1}{m_3\theta_i^2 \ell_i} - \ell_i
=
\frac{1}{m_3\ell_i}\biggl[(1+\ell_i^2) - m_3\ell_i^2 \biggr]
=
\frac{1}{m_3\ell_i}\biggl[1 + (1 - m_3)\ell_i^2 \biggr] = 4.715027199\, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\eta_s</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl(\frac{\pi}{2} +  \tan^{-1}\Lambda_i\biggr) + \eta_i 
=
\biggl(\frac{\pi}{2} +  \tan^{-1}\Lambda_i\biggr) +  \frac{m_3\ell_i }{(1 + \ell_i^2)} 
= 3.132453649\, , 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>B</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\eta_s - \pi
= -0.009139005\, , 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>A</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\eta_i(1+\Lambda_i^2)^{1 / 2}
=
\frac{m_3\ell_i}{(1+\ell_i^2)}\biggl\{
1 + \frac{1}{m_3^2 \ell_i^2}\biggl[1 + (1 - m_3)\ell_i^2 \biggr]^2
\biggr\}^{1 / 2}
=
0.963267676\, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\biggl( \frac{d\phi}{d\eta}\biggr)_s</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{A}{\eta_s^2} \biggl[\eta_s\cos(\pi) - \sin(\pi) \biggr]
=
-\frac{A}{\eta_s} 
=
-0.307512188\, ,
</math>
  </td>
</tr>
</table>

<!--  Total mass and radius -->
<table border="1" width="80%" align="center" cellpadding="8">
<tr><td align="center">
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right">
<math>\biggl[ \frac{G \rho_0^{4/5}}{K_c} \biggr]^{1 / 2} R</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\biggl( \frac{\mu_e}{\mu_c}\biggr)^{-1} \theta_i^{-2} (2\pi)^{-1 / 2} \eta_s = 5.415228878</math>,
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow~~~q \equiv \frac{r_\mathrm{core}}{R}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>0.063800470 </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\biggl[ \frac{G^3 \rho_0^{2/5}}{K_c^3} \biggr]^{1 / 2} M_\mathrm{tot}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl(\frac{\mu_e}{\mu_c}\biggr)^{-2} \biggl(\frac{2}{\pi}\biggr)^{1 / 2} \biggl[- \frac{\eta_s^2}{\theta_i} \cdot \biggl(\frac{d\phi}{d\eta}\biggr)_s \biggr]
= 40.0934</math>,
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow~~~\nu \equiv \frac{M_\mathrm{core}}{M_\mathrm{tot}}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>0.003821153 </math>
  </td>
</tr>
</table>
</td></tr>
</table>

Now, putting all these steps together, we can generate the pair of desired model-parameter expressions:
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right">
<math>q(m_3, \ell_i)</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{m_3 \ell_i \theta_i^2}{\eta_s}
=
\frac{m_3 \ell_i }{(1 + \ell_i^2)} \biggl\{
\biggl(\frac{\pi}{2} +  \tan^{-1}\Lambda_i\biggr) +  \frac{m_3\ell_i }{(1 + \ell_i^2)}
\biggr\}^{-1} 
=
\biggl\{
\biggl[\frac{\pi}{2} +  \tan^{-1}\Lambda_i\biggr]\frac{(1 + \ell_i^2)}{m_3\ell_i} +  1 
\biggr\}^{-1} 
= 0.063800470\, .
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\nu(m_3, \ell_i)</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\sqrt{3}\biggl[ \frac{\xi_i^3 \theta_i^4}{A\eta_s} \biggr]\frac{m_3^2}{3^2}
=
\frac{m_3 \ell_i \theta_i^2}{\eta_s}\biggl[ \frac{\xi_i^2 \theta_i^2}{A} \biggr]\frac{m_3}{3}
=
~q\ell_i \biggl[ \frac{m_3 \ell_i }{(1+\ell_i^2)A} \biggr]
=
~q\ell_i \biggl\{
1 + \frac{1}{m_3^2 \ell_i^2}\biggl[1 + (1 - m_3)\ell_i^2 \biggr]^2
\biggr\}^{-1 / 2} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
~q m_3 \ell_i^2 \biggl\{
m_3^2 \ell_i^2 + \biggl[1 + (1 - m_3)\ell_i^2 \biggr]^2
\biggr\}^{-1 / 2} 
=
(0.059892291)q = 0.00382116\, .
</math>
  </td>
</tr>
</table>
<!-- UNNECESSARY CALCULATION 
'''Example #1:'''  Trying, <math>\xi_i = 0.5 ~~\Rightarrow~~ \ell_i = (12)^{-1 / 2}</math>, and, <math>\mu_e/\mu_c = 0.25  ~~\Rightarrow~~ m_3 = 3/4</math>, we expect from [[SSC/Structure/BiPolytropes/Analytic51#Parameter_Values|Table 1 of our accompanying discussion]] that <math>(q, \nu) = (0.063720, 0.0033138)</math>.  Using our just-derived expressions, we obtain, <math>(\Lambda_i, q, \nu) = (4.71503, 0.063800, 0.0038211)</math>.
-->

Let's fully spell out the final <math>\nu(m_3, \ell_i)</math> function by incorporating the "q" function:
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right">
<math>\nu(m_3, \ell_i)</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
m_3^2 \ell_i^3 \biggl\{
m_3^2 \ell_i^2 + \biggl[1 + (1 - m_3)\ell_i^2 \biggr]^2
\biggr\}^{-1 / 2} 
\biggl\{
\biggl[\frac{\pi}{2} +  \tan^{-1}\Lambda_i\biggr](1 + \ell_i^2) +  m_3\ell_i 
\biggr\}^{-1} = 0.003821156\, ,
</math>
  </td>
</tr>
</table>
where,
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right">
<math>\Lambda_i</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{1}{m_3\ell_i}\biggl[1 + (1 - m_3)\ell_i^2 \biggr] = 4.715027198 \, .
</math>
  </td>
</tr>
</table>
For later use, note that,
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right">
<math>m_3^2 \ell_i^2 (1 + \Lambda_i^2)</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
[1 + (1 - m_3)\ell_i^2 ]^2 + m_3^2 \ell_i^2
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ \frac{1}{(1 + \Lambda_i^2)}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
m_3^2 \ell_i^2 \{[1 + (1 - m_3)\ell_i^2 ]^2 + m_3^2 \ell_i^2\}^{-1}
\, .
</math>
  </td>
</tr>
</table>