Editing SSC/Stability/GammaVariation (section)

====First Try====
What about, in terms of the entropy?  Well, [[#Entropy_Distribution|from above]], once the value of <math>\gamma_\mathrm{g}</math> has been specified, to within an additive constant, the dimensionless entropy, <math>\Sigma</math>, is given by the relation,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\Sigma \equiv \frac{s}{\Re/\bar{\mu}}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{1}{(\gamma_g-1)}\ln \biggl[ \biggl(\frac{P}{P_c}\biggr) \biggl(\frac{\rho}{\rho_c}\biggr)^{-\gamma_\mathrm{g}} \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{1}{(\gamma_g-1)}\ln \biggl[ \Theta_H^{n+1} \biggl(\Theta_H^n\biggr)^{-\gamma_\mathrm{g}} \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\ln \biggl[ \Theta_H^{n+1 - n\gamma_\mathrm{g}} \biggr]^{1/(\gamma_\mathrm{g}-1)} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ e^\Sigma</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[ \Theta_H^{(n+1 - n\gamma_\mathrm{g})/(\gamma_\mathrm{g}-1)} \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ \Theta_\mathrm{H}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math> 
e^{\Sigma(\gamma_\mathrm{g}-1)/(n+1 - n\gamma_\mathrm{g})} \, .
</math>
  </td>
</tr>
</table>

<table border="1" align="center" width="80%" cellpadding="8"><tr><td align="left">
<font color="red"><b>NOTE:</b></font> &nbsp; If we set <math>\gamma_\mathrm{g} = (1 + 1/n)</math>, then the exponent,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>\biggl[ \frac{n + 1 - n\gamma_\mathrm{g}}{\gamma_\mathrm{g}-1} \biggr]_{\gamma_\mathrm{g} = (1 + 1/n)}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math> 
n\biggl[n+1 - (n+1) \biggr] = 0 \, ,
</math>
  </td>
</tr>
</table>
which means that, independent of the functional behavior of the dimensionless enthalpy,

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

<tr>
  <td align="right">
<math>e^\Sigma</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\Theta_H^{0} = \mathrm{constant} \, ,
</math>
  </td>
</tr>
</table>
that is, the entropy is uniform throughout the equilibrium configuration.
</td></tr></table>

Generally, then, in terms of the dimensionless entropy, the Lane-Emden equation may be rewritten as,

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

<tr>
  <td align="right">
<math>\frac{1}{\xi^2} \frac{d}{d\xi}\biggl[\xi^2 \frac{d}{d\xi}\biggl( e^{\Sigma(\gamma_\mathrm{g}-1)/(n+1 - n\gamma_\mathrm{g})} \biggr)\biggr]</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
-\biggl[e^{\Sigma(\gamma_\mathrm{g}-1)/(n+1 - n\gamma_\mathrm{g})}\biggr]^n \, .
</math>
  </td>
</tr>
</table>
That is,

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

<tr>
  <td align="right">
<math>
-e^{n\Sigma(\gamma_\mathrm{g}-1)/(n+1 - n\gamma_\mathrm{g})} 
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[ \frac{\gamma_\mathrm{g} - 1}{n+1-n\gamma_\mathrm{g}} \biggr]
\frac{1}{\xi^2} \frac{d}{d\xi}\biggl[\xi^2 \biggl( e^{\Sigma(\gamma_\mathrm{g}-1)/(n+1 - n\gamma_\mathrm{g})} \biggr)\frac{d\Sigma}{d\xi}\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[ \frac{\gamma_\mathrm{g} - 1}{n+1-n\gamma_\mathrm{g}} \biggr]
\frac{1}{\xi^2} \biggl\{
2\xi \biggl( e^{\Sigma(\gamma_\mathrm{g}-1)/(n+1 - n\gamma_\mathrm{g})} \biggr)\frac{d\Sigma}{d\xi}
+
\xi^2 \biggl( e^{\Sigma(\gamma_\mathrm{g}-1)/(n+1 - n\gamma_\mathrm{g})} \biggr) \frac{d^2\Sigma}{d\xi^2}
+
\xi^2 \biggl[ \frac{\gamma_\mathrm{g} - 1}{n+1-n\gamma_\mathrm{g}} \biggr]
\biggl( e^{\Sigma(\gamma_\mathrm{g}-1)/(n+1 - n\gamma_\mathrm{g})} \biggr) \biggl(\frac{d\Sigma}{d\xi}\biggr)^2
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[ \frac{\gamma_\mathrm{g} - 1}{n+1-n\gamma_\mathrm{g}} \biggr]
\biggl\{
\frac{2}{\xi} \cdot \frac{d\Sigma}{d\xi}
+
\frac{d^2\Sigma}{d\xi^2}
+
\biggl[ \frac{\gamma_\mathrm{g} - 1}{n+1-n\gamma_\mathrm{g}} \biggr]
\biggl(\frac{d\Sigma}{d\xi}\biggr)^2
\biggr\}\biggl( e^{\Sigma(\gamma_\mathrm{g}-1)/(n+1 - n\gamma_\mathrm{g})} \biggr)
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow~~~
-e^{n} 
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[ \frac{\gamma_\mathrm{g} - 1}{n+1-n\gamma_\mathrm{g}} \biggr]
\biggl\{
\frac{2}{\xi} \cdot \frac{d\Sigma}{d\xi}
+
\frac{d^2\Sigma}{d\xi^2}
+
\biggl[ \frac{\gamma_\mathrm{g} - 1}{n+1-n\gamma_\mathrm{g}} \biggr]
\biggl(\frac{d\Sigma}{d\xi}\biggr)^2
\biggr\} \, .
</math>
  </td>
</tr>
</table>
Setting,
<div align="center">
<math>\Upsilon \equiv \biggl[ \frac{\gamma_\mathrm{g} - 1}{n+1-n\gamma_\mathrm{g}} \biggr]\Sigma \, ,</math>
</div>
the statement of hydrostatic balance becomes,

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

<tr>
  <td align="right">
<math>
\frac{d^2\Upsilon}{d\xi^2}
+
\biggl(\frac{d\Upsilon}{d\xi}\biggr)^2
+
\frac{2}{\xi} \cdot \frac{d\Upsilon}{d\xi}
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
-e^{n} 
\, .
</math>
  </td>
</tr>
</table>

<font color="red"><b>What do I do with this???</b></font>


In our [[#Entropy_Distribution|above discussion of uniform-density configurations]], we found that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\Sigma = \frac{s}{\Re/\bar{\mu}}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
(\gamma_\mathrm{g} - 1)^{-1}\ln \biggl[(1 - \xi^2/6) \biggr] \, ,
</math>
  </td>
</tr>
</table>
where we have made the substitution, <math>\chi_0^2 = \xi^2/6</math>.  For this situation, we can write,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\Upsilon</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
m^{-1}\ln \biggl[1 - \frac{\xi^2}{6} \biggr] \, ,
</math>
  </td>
</tr>
</table>
where,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>m</math>
  </td>
  <td align="center">
<math>\equiv</math>
  </td>
  <td align="left">
<math>
n+1-n\gamma_\mathrm{g} \, .
</math>
  </td>
</tr>
</table>
In this case,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\frac{d\Upsilon}{d\xi}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
m^{-1} \biggl[1 - \frac{\xi^2}{6} \biggr]^{-1} \cdot \biggl(-\frac{\xi}{3}\biggr) \, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\biggl(\frac{d\Upsilon}{d\xi}\biggr)^2</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{ \xi^2 }{3^2m^2  } \biggl[1 - \frac{\xi^2}{6} \biggr]^{-2}  \, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\frac{d^2\Upsilon}{d\xi^2}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{\xi}{3m} \biggl[1 - \frac{\xi^2}{6} \biggr]^{-2} \cdot \biggl(-\frac{\xi}{3}\biggr) 
-
\frac{1}{3m} \biggl[1 - \frac{\xi^2}{6} \biggr]^{-1}  \, ;
</math>
  </td>
</tr>
</table>
that is to say,

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

<tr>
  <td align="right">
<math>
-e^{n} 
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{d^2\Upsilon}{d\xi^2}
+
\biggl(\frac{d\Upsilon}{d\xi}\biggr)^2
+
\frac{2}{\xi} \cdot \frac{d\Upsilon}{d\xi}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl\{
\frac{\xi}{3m} \biggl[1 - \frac{\xi^2}{6} \biggr]^{-2} \cdot \biggl(-\frac{\xi}{3}\biggr) 
-
\frac{1}{3m} \biggl[1 - \frac{\xi^2}{6} \biggr]^{-1} \biggr\}
+
\biggl\{  \frac{ \xi^2 }{3^2m^2  } \biggl[1 - \frac{\xi^2}{6} \biggr]^{-2}  \biggr\}
+
\frac{2}{\xi} \cdot \biggl\{  m^{-1} \biggl[1 - \frac{\xi^2}{6} \biggr]^{-1} \cdot \biggl(-\frac{\xi}{3}\biggr)\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[1 - \frac{\xi^2}{6} \biggr]^{-2} \biggl\{
\frac{\xi}{3m} \biggl(-\frac{\xi}{3}\biggr) 
-
\frac{1}{3m} \biggl[1 - \frac{\xi^2}{6} \biggr] 
+
\frac{ \xi^2 }{3^2m^2  }  
+
\frac{2}{\xi} \cdot  m^{-1} \biggl[1 - \frac{\xi^2}{6} \biggr] \cdot \biggl(-\frac{\xi}{3}\biggr)\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[1 - \frac{\xi^2}{6} \biggr]^{-2} \biggl\{
-\frac{\xi^2}{3^2m}  
- \frac{1}{3m} + \frac{\xi^2}{2\cdot 3^2m}  
+
\frac{ \xi^2 }{3^2m^2  }  
-\frac{2}{3m}
+
\frac{\xi^2}{3^2m} 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[1 - \frac{\xi^2}{6} \biggr]^{-2} \biggl\{
- \frac{1}{m} + \frac{\xi^2}{2\cdot 3^2m}  
+
\frac{ \xi^2 }{3^2m^2  }  
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[1 - \frac{\xi^2}{6} \biggr]^{-2} \biggl\{
- m 
+ \frac{\xi^2}{2\cdot 3^2}\biggl[ m  + 2 \biggr] 
\biggr\}\frac{1}{m^2} \, .
</math>
  </td>
</tr>
</table>
That is,

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

<tr>
  <td align="right">
<math>
- m 
+ \frac{\xi^2}{2\cdot 3^2}\biggl[ m  + 2 \biggr]  
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
-m^2 e^{n} \biggl[1 - \frac{\xi^2}{6} \biggr]^{2} \, .
</math>
  </td>
</tr>
</table>

<font color="red"><b>What do I do with this???</b></font>