Editing Appendix/Ramblings/BiPolytrope51AnalyticStability (section)

=====Is This Compatible With LAWE=====

In an effort to track the two <math>~Q(\eta)</math> functions separately, we will add a subscript zero to the one that applies to the ''structural'' properties of the underlying equilibrium configuration.  Again, we will be focused on finding a solution in the case where <math>~\sigma_c^2 = 0</math> and <math>~n = 1</math>, that is &#8212; see also our [[#Envelope:|above discussion]] &#8212; the relevant envelope LAWE is,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{d^2x_P}{d\eta^2} + \biggl[ 2 - Q_0 \biggr] \frac{2}{\eta}\cdot \frac{dx_P}{d\eta} -2\alpha_g Q_0 \cdot \frac{x_P}{\eta^2} \, ,
</math>
  </td>
</tr>
</table>
where, drawing from our [[SSC/Structure/BiPolytropes/Analytic51#Step_6:__Envelope_Solution|discussion of the n = 1 envelope's equilibrium structure]],
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~Q_0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ - \frac{d\ln\phi}{d\ln\eta} \biggr]_\mathrm{n=1} = - \frac{\eta}{\phi}\biggl[ \frac{d\phi}{d\eta} \biggr]_\mathrm{n=1}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- \biggl[ \frac{1}{\sin(\eta - B_0)} \biggr]\biggl[ \eta\cos(\eta-B_0) - \sin(\eta-B_0) \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[1 -  \eta\cot(\eta-B_0) \biggr] \, .</math>
  </td>
</tr>
</table>

Hence, after recognizing that for this specific case, <math>~\alpha_g = +1</math>, we have,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
LAWE
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{d^2x_P}{d\eta^2} + \biggl[ 2 - Q_0 \biggr] \frac{2}{\eta}\cdot \frac{dx_P}{d\eta} -2 Q_0 \cdot \frac{x_P}{\eta^2} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~</math>&nbsp; LAWE
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{3c_1}{\eta^4\cos^3(\eta-B)} \biggl[ 6\cos^3(\eta-B) 
+ 2\eta \sin(\eta-B)\cos^2(\eta-B)
- 2 \eta^2 \cos(\eta-B) + 2\eta^3 \sin(\eta-B)
\biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \biggl[ 2 - Q_0 \biggr] \frac{2}{\eta}\cdot \frac{3c_1}{\eta^3\cos^2(\eta-B)} \biggl[\eta^2 -2\cos^2(\eta-B) - \eta\sin(\eta-B)\cos(\eta-B)\biggr] 
- \frac{2 Q_0}{\eta^2} \cdot \frac{3c_1 }{\eta^2}\biggl[1 + \eta\tan(\eta-B)\biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~\biggl[ \frac{\eta^4\cos^3(\eta-B)}{6c_1} \biggr] \cdot</math>&nbsp; LAWE
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ 3\cos^3(\eta-B) 
+ \eta \sin(\eta-B)\cos^2(\eta-B)
-  \eta^2 \cos(\eta-B) + \eta^3 \sin(\eta-B)
\biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \biggl[ 2 - Q_0 \biggr]\cos(\eta - B)\biggl[\eta^2 -2\cos^2(\eta-B) - \eta\sin(\eta-B)\cos(\eta-B)\biggr] 
- Q_0 \cos^3(\eta - B) \biggl[1 + \eta\tan(\eta-B)\biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
3\cos^3(\eta-B) 
+ \eta \sin(\eta-B)\cos^2(\eta-B)
-  \eta^2 \cos(\eta-B) + \eta^3 \sin(\eta-B)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
-4\cos^3(\eta-B) - 2\eta\sin(\eta-B)\cos^2(\eta-B) + 2\eta^2\cos(\eta - B) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
- Q_0\biggl\{ \biggl[\eta^2\cos(\eta - B) -2\cos^3(\eta-B) - \eta\sin(\eta-B)\cos^2(\eta-B)\biggr] 
+ \biggl[\cos^3(\eta - B) + \eta\sin(\eta-B)\cos^2(\eta - B)\biggr]
\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-\cos^3(\eta-B) 
- \eta \sin(\eta-B)\cos^2(\eta-B)
+  \eta^2 \cos(\eta-B) + \eta^3 \sin(\eta-B)
+ Q_0\biggl[ \cos^3(\eta-B) - \eta^2\cos(\eta - B) \biggr] 
</math>
  </td>
</tr>
</table>

<table border="1" width="85%" cellpadding="10" align="center"><tr><td align="left">
<font color="red">'''Quick Check:'''</font> &nbsp; Now, if we set <math>~Q_0 = 1 + \eta\tan(\eta-B)</math>, these RHS terms should sum to zero.  Let's check.

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

<tr>
  <td align="right">
<math>~\biggl[ \frac{\eta^4\cos^3(\eta-B)}{6c_1} \biggr] \cdot</math>&nbsp; LAWE
  </td>
  <td align="center">
<math>~\rightarrow</math>
  </td>
  <td align="left">
<math>~
-\cos^3(\eta-B) 
- \eta \sin(\eta-B)\cos^2(\eta-B)
+  \eta^2 \cos(\eta-B) + \eta^3 \sin(\eta-B)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \biggl[\cos(\eta - B)+ \eta\sin(\eta-B)  \biggr] \cdot \biggl[ \cos^2(\eta-B) - \eta^2 \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
0 \, .
</math>
  </td>
</tr>
</table>

Excellent!
</td></tr></table>


Now, let's plug in the expression for the ''structural'' <math>~Q_0</math>.  Specifically, we want,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~Q_0</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\biggl[ 1 - \eta\cot(\eta-B_0) \biggr]</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\biggl[ 1 + \eta\tan(\eta-B_0 - \tfrac{\pi}{2} \pm m\pi) \biggr] \, ,</math>
  </td>
</tr>
</table>
in which case we have,

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

<tr>
  <td align="right">
<math>~\biggl[ \frac{\eta^4\cos^3(\eta-B)}{6c_1} \biggr] \cdot</math>&nbsp; LAWE
  </td>
  <td align="center">
<math>~\rightarrow</math>
  </td>
  <td align="left">
<math>~
-\cos^3(\eta-B) 
- \eta \sin(\eta-B)\cos^2(\eta-B)
+  \eta^2 \cos(\eta-B) + \eta^3 \sin(\eta-B)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \biggl[ 1 + \eta\tan(\eta-B_0 - \tfrac{\pi}{2} \pm m\pi) \biggr]  \cdot \biggl[ \cos^3(\eta-B) - \eta^2\cos(\eta-B) \biggr] \, .
</math>
  </td>
</tr>
</table>
As we have just shown, above, in the context of a "<font color="red">'''Quick Check'''</font>", the expression on the RHS will go to zero if we adopt the transformation, <math>~[B_0 + \tfrac{\pi}{2} \mp m\pi] \rightarrow B</math>.  Does this help shift the coordinate, <math>~\eta</math>?