Editing SSC/StabilityConjecture/Bipolytrope51 (section)

====Differentiate &nu; with Respect to &ell;<sub>I</sub>====
In order to determine the maximum value of the fractional core mass, we next need to determine the derivative of <math>\nu</math> with respect to <math>\ell_i</math>. ['''Example #2:''' Borrowing from [[#Table1|Table 1, above]], in this case our numerical evaluation is for <math>\mu_e/\mu_c = 0.25</math> and <math>\xi_i = 4.93827</math>, for which the expected maximum mass-fraction is, <math>\nu_\mathrm{max} = 0.1394</math>.  This implies that <math>m_3 = 0.75</math> and <math>\ell_i = 2.85111</math>.]

Let's rewrite the function as,

<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right">
<math>\nu</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{m_3^2 \ell_i^3}{F^{1 / 2} \cdot H} 
= 0.139370157 (8)\, ,
</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] = 1.418024375 (7) \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>F</math>
  </td>
  <td align="center">
<math>\equiv</math>
  </td>
  <td align="left">
<math>
m_3^2 \ell_i^2 + [1 + (1 - m_3)\ell_i^2 ]^2
= 13.76676346 (4)
\, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>H</math>
  </td>
  <td align="center">
<math>\equiv</math>
  </td>
  <td align="left">
<math>
\biggl[\frac{\pi}{2} +  \tan^{-1}\Lambda_i\biggr](1 + \ell_i^2) +  m_3\ell_i 
= 25.21038191 (5)
\, .
</math>
  </td>
</tr>

<tr>
  <td align="right">
NOTE: &nbsp;<math>q</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{m_3 \ell_i}{H} 
= 0.084820
\, .
</math>
  </td>
</tr>
</table>
Then we have,

<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right">
<math>\frac{1}{m_3^2} \cdot \frac{d\nu}{d\ell_i}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{3\ell_i^2}{F^{1 / 2} \cdot H} 
-\frac{1}{2}\biggl[ \frac{\ell_i^3}{F^{3 / 2} \cdot H}\biggr] \frac{dF}{d\ell_i}
- \biggl[ \frac{\ell_i^3}{F^{1 / 2} \cdot H^2} \biggr] \frac{dH}{d\ell_i}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\frac{\ell_i^2}{F^{3 / 2} \cdot H^2} \biggl\{
3FH - \frac{\ell_i H}{2}\cdot \frac{dF}{d\ell_i}  - \ell_i F \cdot \frac{dH}{d\ell_i}
\biggr\} 
\, .
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{\ell_i^2}{F^{3 / 2} \cdot H^2} \biggl\{ 1041.196093 - 425.9706908 - 615.2543231 \biggr\} 
= \frac{\ell_i^2}{F^{3 / 2} \cdot H^2} \biggl\{-0.028921\biggr\}\, .
</math>
&nbsp; &nbsp; <font color="red">EXCELLENT!</font>  
  </td>
</tr>
</table>
Furthermore,

<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right">
<math>\frac{dF}{d\ell_i}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
2m_3^2 \ell_i + 4[1 + (1 - m_3)\ell_i^2 ](1-m_3)\ell_i
= 11.85266706 (6)\, ,
</math>
  </td>
</tr>
</table>
and,
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right">
<math>\frac{dH}{d\ell_i}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
(1 + \ell_i^2) \frac{d}{d\ell_i} \biggl(\tan^{-1}\Lambda_i \biggr) 
+ 2\ell_i\biggl[\frac{\pi}{2} + \tan^{-1}\Lambda_i\biggr] + m_3
= 15.67503863 (9)\, ,
</math>
  </td>
</tr>
</table>
and,
<table border="0" align="center" cellpadding="5">

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl\{ \frac{m_3^2 \ell_i^2 }{[1 + (1 - m_3)\ell_i^2 ]^2 + m_3^2 \ell_i^2} \biggr\} \biggl\{
-\frac{1}{m_3\ell_i^2}\biggl[1 + (1 - m_3)\ell_i^2 \biggr]
+
\frac{2(1-m_3)\ell_i}{m_3\ell_i}
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl\{ [1 + (1 - m_3)\ell_i^2 ]^2 + m_3^2 \ell_i^2 \biggr\}^{-1} \biggl\{
2m_3(1-m_3)\ell_i^2
- m_3 [1 + (1 - m_3)\ell_i^2 ]
\biggr\} 
= 0.056233763 (8)\, .
</math>
  </td>
</tr>
</table>
Now, along an equilibrium sequence of fixed <math>m_3</math>, the point of maximum core mass is located at the point where,

<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right">
<math>\frac{d\nu}{d\ell_i}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>0 </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow~~~0</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\frac{\ell_i^2}{F^{3 / 2} \cdot H^2} \biggl\{
3FH - \frac{\ell_i H}{2}\cdot \frac{dF}{d\ell_i}  - \ell_i F \cdot \frac{dH}{d\ell_i}
\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow~~~ 0</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
H \biggl\{ 3F - \frac{\ell_i}{2} \cdot \frac{dF}{d\ell_i}\biggr\}
-\ell_i F \cdot \frac{dH}{d\ell_i}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>1041.196093 - 425.9706908 - 615.2543231 = -0.0289209
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl\{ \biggl[\frac{\pi}{2} +  \tan^{-1}\Lambda_i\biggr](1 + \ell_i^2) +  m_3\ell_i \biggr\}
\biggl\{ 3F - \ell_i \biggl[ m_3^2 \ell_i + 2[1 + (1 - m_3)\ell_i^2 ](1-m_3)\ell_i\biggr] \biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
- \ell_i F \biggl\{ 
(1 + \ell_i^2) \frac{d}{d\ell_i} \biggl(\tan^{-1}\Lambda_i \biggr) 
+ 2\ell_i\biggl[\frac{\pi}{2} + \tan^{-1}\Lambda_i\biggr] + m_3
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl\{ \biggl[\frac{\pi}{2} +  \tan^{-1}\Lambda_i\biggr](1 + \ell_i^2) \biggr\}
\biggl\{ 3F - \ell_i \biggl[ m_3^2 \ell_i + 2[1 + (1 - m_3)\ell_i^2 ](1-m_3)\ell_i\biggr] \biggr\}

- 2\ell_i^2 F \biggl[\frac{\pi}{2} + \tan^{-1}\Lambda_i\biggr]  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
- m_3\ell_i F
+ m_3\ell_i 
\biggl\{ 3F - \ell_i \biggl[ m_3^2 \ell_i + 2[1 + (1 - m_3)\ell_i^2 ](1-m_3)\ell_i\biggr] \biggr\} 
- \ell_i F (1 + \ell_i^2) \frac{d}{d\ell_i} \biggl(\tan^{-1}\Lambda_i \biggr) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[\frac{\pi}{2} +  \tan^{-1}\Lambda_i\biggr] 
\biggl\{ 
3F(1 + \ell_i^2) 
- 2\ell_i^2 F 
- m_3^2 \ell_i \cdot \ell_i(1 + \ell_i^2)  
- 2[1 + (1 - m_3)\ell_i^2 ](1-m_3)\ell_i^2 (1 + \ell_i^2)  
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
+  
3Fm_3\ell_i 
- m_3\ell_i F
- \ell_i F (1 + \ell_i^2) \frac{d}{d\ell_i} \biggl(\tan^{-1}\Lambda_i \biggr) 
- m_3\ell_i^2 \biggl[ m_3^2 \ell_i + 2[1 + (1 - m_3)\ell_i^2 ](1-m_3)\ell_i\biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[\frac{\pi}{2} +  \tan^{-1}\Lambda_i\biggr] 
\biggl\{ 
F(3 + \ell_i^2) 
- m_3^2 \ell_i^2(1 + \ell_i^2)  
- 2\biggl[1 + (1 - m_3)\ell_i^2 \biggr](1-m_3)\ell_i^2 (1 + \ell_i^2)  
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
+  
2Fm_3\ell_i 
- \ell_i F (1 + \ell_i^2) \frac{d}{d\ell_i} \biggl(\tan^{-1}\Lambda_i \biggr) 
- m_3\ell_i^2 \biggl[ m_3^2 \ell_i + 2[1 + (1 - m_3)\ell_i^2 ](1-m_3)\ell_i\biggr] 
\, .
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
2.527380938 
\biggl\{ 111.4667252 -112.5053101 \biggr\} + 38.72662783  -36.13064883 
=
-2.624899679 + 2.595979 = -0.028920679 \, .
</math>
&nbsp; &nbsp; <font color="red">EXCELLENT!</font>
  </td>
</tr>
</table>

Hence,

<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right">
<math>
20.14923887 
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
-2.624899679 + 22.74521787
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>
\ell_i F (1 + \ell_i^2) \frac{d}{d\ell_i} \biggl(\tan^{-1}\Lambda_i \biggr) 
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[\frac{\pi}{2} +  \tan^{-1}\Lambda_i\biggr] 
\biggl\{ 
F(3 + \ell_i^2) 
- \ell_i^2(1 + \ell_i^2) \biggl[ m_3^2   
+ 2(1-m_3) + 2(1 - m_3)^2\ell_i^2 \biggr]  
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
+  
2Fm_3\ell_i 
- m_3\ell_i^3 \biggl[ m_3^2 + 2(1-m_3) + 2(1 - m_3)^2\ell_i^2 \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow~~~
\ell_i F (1 + \ell_i^2) 
\biggl\{ [1 + (1 - m_3)\ell_i^2 ]^2 + m_3^2 \ell_i^2 \biggr\}^{-1} \biggl\{
2m_3(1-m_3)\ell_i^2
- m_3 [1 + (1 - m_3)\ell_i^2 ]
\biggr\}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
F\biggl\{ \biggl[\frac{\pi}{2} +  \tan^{-1}\Lambda_i\biggr] 
(3 + \ell_i^2) 
+  
2m_3\ell_i \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
- \ell_i^2\biggl[ m_3^2   + 2(1-m_3) + 2(1 - m_3)^2\ell_i^2 \biggr]  
\biggl\{
\biggl[\frac{\pi}{2} +  \tan^{-1}\Lambda_i\biggr] 
(1 + \ell_i^2) +m_3\ell_i \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow~~~
\ell_i F (1 + \ell_i^2)^2 
\biggl\{
2m_3(1-m_3)\ell_i^2
- m_3 [1 + (1 - m_3)\ell_i^2 ]
\biggr\}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
F \cdot J \biggl\{ (H - m_3\ell_i) 
(3 + \ell_i^2) 
+  
2m_3\ell_i (1+\ell_i^2)\biggr\}
~- ~\ell_i^2(1 + \ell_i^2) H\cdot J\biggl[ m_3^2   + 2(1-m_3) + 2(1 - m_3)^2\ell_i^2 \biggr] \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>
2532.246281
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
56062.28134 -53533.66963 = 2528.61171
</math>
  </td>
</tr>
</table>
The difference between the LHS and RHS &#8212; <math>(2528.61171 - 2532.246281) = -3.634571 </math> &#8212; is larger than our previously obtained "difference" <math>(-0.028920679)</math> by the factor, <math>(1 + \ell_i^2)J = 125.6745377</math>.  We are therefore satisfied that, for a given value of <math>m_3</math>, the value of <math>\ell_i</math> associated with the model that has the maximum core mass-fraction is identified when the LHS and RHS of this final expression match.

Note that, in reaching this final expression, we have recognized that,
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right">
<math>
\biggl[\frac{\pi}{2} +  \tan^{-1}\Lambda_i\biggr] 
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[ \frac{H - m_3\ell_i}{1+\ell_i^2}\biggr]
\, ,
</math>
  </td>
</tr>
</table>
and have introduced the short-hand notation,
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right">
<math>
J 
</math>
  </td>
  <td align="center">
<math>\equiv</math>
  </td>
  <td align="left">
<math>
\biggl[1 + (1 - m_3)\ell_i^2 \biggr]^2 + m_3^2 \ell_i^2
= 13.76676346 (3)\, .
</math>
  </td>
</tr>
</table>