Editing SSC/Stability/n1PolytropeLAWE/Pt3 (section)

===Second Attempt===

Up to this point we have been rather cavalier about the use of <math>~\xi</math> (and <math>~\xi_i</math>) to represent the envelope's dimensionless radius (and interface location).  Let's switch to <math>~\eta</math>,

<table border="0" align="center" cellpadding="8">

<tr>
  <td align="right">
<math>~r^*</math>
  </td>
  <td align="center"><math>~=</math></td>
  <td align="left">
<math>\biggl( \frac{\mu_e}{\mu_c} \biggr)^{-1} \theta^{-2}_i (2\pi)^{-1/2}\eta</math>
  </td>
</tr>
</table>

<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}{d\eta^2} + \biggl\{ 4 -\biggl(\frac{\rho^*}{P^*}\biggr)\frac{ M_r^*}{(r^*)}\biggr\}\frac{1}{\eta} \cdot \frac{dx}{d\eta} 
+ \frac{1}{2\pi \theta_i^4} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-2} \biggl(\frac{\rho^*}{ P^* } \biggr)\biggl\{ \frac{2\pi \sigma_c^2}{3\gamma_\mathrm{g}}  ~-~\frac{\alpha_\mathrm{g} M_r^*}{(r^*)^3}\biggr\}  x \, .
</math>
  </td>
</tr>
</table>

and, throughout the envelope we have,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{\rho^*}{P^*}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl( \frac{\mu_e}{\mu_c} \biggr) \theta^{-1}_i \phi(\eta)^{-1}
\, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{M_r^*}{r^*}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl( \frac{\mu_e}{\mu_c} \biggr)^{-2} \theta^{-1}_i \biggl( \frac{2}{\pi} \biggr)^{1/2} \biggl(-\eta^2 \frac{d\phi}{d\eta} \biggr)
\biggl[ \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-1} \theta^{-2}_i (2\pi)^{-1/2}\eta \biggr]^{-1}
=
2 \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-1} \theta_i  \eta \biggl(- \frac{d\phi}{d\eta} \biggr)
\, .
</math>
  </td>
</tr>
</table>

Hence, the LAWE relevant to the envelope 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}{d\eta^2} + \biggl\{ 4 -\biggl[ \frac{\rho^*}{P^*}\biggr] \biggl[ \frac{ M_r^*}{(r^*)} \biggr] \biggr\}\frac{1}{\eta} \cdot \frac{dx}{d\eta} 
+ \frac{1}{2\pi \theta_i^4} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-2} \biggl[ \frac{\rho^*}{ P^* } \biggr] \biggl\{ \frac{2\pi \sigma_c^2}{3\gamma_\mathrm{g}}  ~-~\frac{\alpha_e }{(r^*)^2} \biggl[ \frac{M_r^*}{r^*} \biggr] \biggr\}  x 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{d^2x}{d\eta^2} + \biggl\{ 4 -\biggl[ \biggl( \frac{\mu_e}{\mu_c} \biggr) \theta^{-1}_i \phi(\eta)^{-1} \biggr] \biggl[ 2 \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-1} \theta_i  \eta \biggl(- \frac{d\phi}{d\eta} \biggr) \biggr] \biggr\}\frac{1}{\eta} \cdot \frac{dx}{d\eta} 
+ \frac{1}{2\pi \theta_i^4} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-2} \biggl[ \biggl( \frac{\mu_e}{\mu_c} \biggr) \theta^{-1}_i \phi(\eta)^{-1} \biggr] \biggl\{ \frac{2\pi \sigma_c^2}{3\gamma_\mathrm{g}}  
~-~\alpha_e \biggl[ \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-1} \theta^{-2}_i (2\pi)^{-1/2}\eta \biggr]^{-2} \biggl[2 \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-1} \theta_i  \eta \biggl(- \frac{d\phi}{d\eta} \biggr) \biggr] \biggr\}  x 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{d^2x}{d\eta^2} + \biggl\{ 4 - \biggl[ \frac{2  \eta}{\phi} \biggl(- \frac{d\phi}{d\eta} \biggr) \biggr] \biggr\}\frac{1}{\eta} \cdot \frac{dx}{d\eta} 
+ \frac{1}{2\pi \theta_i^5 \phi} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-1}  \biggl\{ \frac{2\pi \sigma_c^2}{3\gamma_\mathrm{g}}  
~-~\alpha_e \biggl[ \biggl( \frac{\mu_e}{\mu_c} \biggr) \theta^{5}_i (4\pi) \eta^{-1} \biggr] \biggl(- \frac{d\phi}{d\eta} \biggr)  \biggr\}  x 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{d^2x}{d\eta^2} + \biggl\{ 4 - \biggl[ \frac{2  \eta}{\phi} \biggl(- \frac{d\phi}{d\eta} \biggr) \biggr] \biggr\}\frac{1}{\eta} \cdot \frac{dx}{d\eta} 
+ \frac{1}{2\pi \theta_i^5 \phi} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-1}  \biggl\{ \frac{2\pi \sigma_c^2}{3\gamma_\mathrm{g}}   \biggr\}  x 
~-~ \alpha_e  \biggl[ \frac{2\eta}{\phi} \biggl(- \frac{d\phi}{d\eta} \biggr) \biggr]   \frac{x}{\eta^2} \, .
</math>
  </td>
</tr>
</table>

If we assume that, <math>~\alpha_e = (3 - 4/2) = 1</math> and <math>~\sigma_c^2 = 0</math>, then 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}{d\eta^2} + \biggl\{ 4 -2Q \biggr\}\frac{1}{\eta} \cdot \frac{dx}{d\eta} 
~-~  \biggl[ 2 Q \biggr]   \frac{x}{\eta^2} \, ,
</math>
  </td>
</tr>
</table>
where,
<div align="center">
<math>~
Q \equiv - \frac{d \ln \phi}{ d\ln \eta} \, .
</math>
</div>

<span id="Consider">Now consider</span> the,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="center" colspan="3"><font color="maroon"><b>Precise Solution to the Polytropic LAWE</b></font></td>
</tr>

<tr>
  <td align="right">
<math>~x_P</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{b(n-1)}{2n}\biggl[1 + \biggl(\frac{n-3}{n-1}\biggr) \biggl( \frac{1}{\eta \phi^{n}}\biggr) \frac{d\phi}{d\eta}\biggr]</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-b\biggl[ \biggl( \frac{1}{\eta \phi}\biggr) \frac{d\phi}{d\eta}\biggr]</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{b}{\eta^2}\biggl[ -\frac{d\ln \phi}{d\ln \eta}\biggr] </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{bQ}{\eta^2} \, .</math>
  </td>
</tr>
</table>
</div>

From our [[SSC/Structure/BiPolytropes/Analytic51#Step_6:__Envelope_Solution|accompanying discussion]], we recall that the most general solution to the <math>n=1</math> Lane-Emden equation can be written in the form,
<div align="center">
<math>
\phi = A \biggl[ \frac{\sin(\eta - B)}{\eta} \biggr] \, ,
</math>
</div>
where <math>A</math> and <math>B</math> are constants whose values can be obtained from our [[SSC/Structure/BiPolytropes/Analytic51#Parameter_Values|accompanying parameter table]].  The first derivative of this function is,
<div align="center">
<math>
\frac{d\phi}{d\eta} = \frac{A}{\eta^2} \biggl[ \eta\cos(\eta-B) - \sin(\eta-B) \biggr] \, .
</math>
</div>
Hence,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~Q = -\frac{d\ln\phi}{d\ln\eta}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \frac{\eta}{\phi} \cdot \frac{A}{\eta^2} \biggl[ \eta\cos(\eta-B) - \sin(\eta-B) \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[1- \eta\cot(\eta-B) \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ x_P</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{b}{\eta^2} \biggl[1- \eta\cot(\eta-B) \biggr] \, .
</math>
  </td>
</tr>
</table>
What is this in terms of the dimensionless radius, <math>~r^*/R^*</math>?  Well,

<table border="0" align="center" cellpadding="8">

<tr>
  <td align="right">
<math>\frac{~r^*}{R^*}</math>
  </td>
  <td align="center"><math>~=</math></td>
  <td align="left">
<math>\biggl( \frac{\mu_e}{\mu_c} \biggr)^{-1} \theta^{-2}_i (2\pi)^{-1/2}\eta \biggl[\frac{\sqrt{2\pi}~\theta_i^2}{\eta_s} \biggl(\frac{\mu_e}{\mu_c}\biggr)\biggr]</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center"><math>~=</math></td>
  <td align="left">
<math>\frac{\eta}{\eta_s} = \frac{\eta}{(\pi + B)} </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \eta</math>
  </td>
  <td align="center"><math>~=</math></td>
  <td align="left">
<math>\frac{~r^*}{R^*}\biggl(\pi + B \biggr) \, .</math>
  </td>
</tr>
</table>
Also,

<table border="0" align="center" cellpadding="8">

<tr>
  <td align="right">
<math>~\eta-B</math>
  </td>
  <td align="center"><math>~=</math></td>
  <td align="left">
<math>\frac{~r^*}{R^*}\biggl(\pi + B \biggr) -B = \pi \biggl( \frac{r^*}{R^*}\biggr) - B\biggl[1-\biggl( \frac{r^*}{R^*}\biggr)\biggr]</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center"><math>~=</math></td>
  <td align="left">
<math>\pi + \pi \biggl[ \biggl( \frac{r^*}{R^*}\biggr)-1\biggr] - B\biggl[1-\biggl( \frac{r^*}{R^*}\biggr)\biggr]</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center"><math>~=</math></td>
  <td align="left">
<math>\pi  - (\pi + B)\biggl[1-\biggl( \frac{r^*}{R^*}\biggr)\biggr] \, .</math>
  </td>
</tr>
</table>


<font color="red">'''[12 January 2019]:'''</font>  Here's what appears to work pretty well, empirically:

<table border="1" width="60%" align="center" cellpadding="8"><tr><td align="left">
<table border="0" align="center" cellpadding="8">

<tr>
  <td align="right">
<math>~x_P</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{\eta^2} \biggl\{1- \eta\cot[\eta-(\pi - 0.8)] \biggr\} \, .
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\eta</math>
  </td>
  <td align="center"><math>~=</math></td>
  <td align="left">
<math>\frac{~r^*}{R^*}\biggl(\pi - 0.6\pi \biggr) \, .</math>
  </td>
</tr>
</table>
</td></tr></table>


<span id="tagJanuary2019">
Let's work through the analytic derivatives again.  Keeping in mind that,</span>
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{d}{d\eta}\biggl[\cot(\eta - B) \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  <td align="left">
<math>~
- \biggl[ 1 + \cot^2(\eta - B)\biggr] \, ,
</math>
  </td>
</tr>
</table>
and starting with the ''guess'',

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

<tr>
  <td align="right">
<math>~x_P</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{b}{\eta^2} \biggl[1- \eta\cot(\eta-B) \biggr] \, ,
</math>
  </td>
</tr>
</table>
we have,

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

<tr>
  <td align="right">
<math>~ \frac{dx_P}{d\eta}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-\frac{2b}{\eta^3} \biggl[1- \eta\cot(\eta-B) \biggr] 
- \frac{b}{\eta^2} \biggl\{ 
\cot(\eta-B) - \eta \biggl[ 1 + \cot^2(\eta - B)\biggr]
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~ \Rightarrow ~~~ \biggl( \frac{\eta^3}{b} \biggr) \frac{dx_P}{d\eta}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-  \biggl[2- 2\eta\cot(\eta-B) \biggr] 
- \biggl\{ 
\eta \cot(\eta-B) - \eta^2 \biggl[ 1 + \cot^2(\eta - B)\biggr]
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\eta^2 -2 + \eta\cot(\eta-B)  +  \eta^2\cot^2(\eta - B) \, .
</math>
  </td>
</tr>
</table>
The second derivative then gives,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~ \frac{d^2x_P}{d\eta^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{d}{d\eta}\biggl\{
\frac{b}{\eta^3} \biggl[
\eta^2 -2 + \eta\cot(\eta-B)  +  \eta^2\cot^2(\eta - B) 
\biggr]
\biggr\}
</math>
  </td>
</tr>

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

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

<tr>
  <td align="right">
<math>~\Rightarrow~~~ \frac{d^2x_P}{d\eta^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{b}{\eta^4} \biggl[
6 - 3\eta^2 - 3\eta\cot(\eta-B)  -  3\eta^2\cot^2(\eta - B) 
\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+\frac{b}{\eta^4} \biggl\{
2\eta^2 + \eta \cot(\eta-B)  +  2\eta^2 \cot^2(\eta - B) 
- \eta^2\biggl[ 1 + \cot^2(\eta - B)\biggr]  -  2\eta^3\cot(\eta-B)  \biggl[ 1 + \cot^2(\eta - B)\biggr]
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~\frac{\eta^4}{b}\cdot \frac{d^2x_P}{d\eta^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
6 - 3\eta^2 - 3\eta\cot(\eta-B)  -  3\eta^2\cot^2(\eta - B) 
</math>
  </td>
</tr>

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

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

<span id="Recalling">Recalling</span> that,
<div align="center">
<math>~Q = \biggl[1- \eta\cot(\eta-B) \biggr] \, ,</math>
</div>
plugging these expressions into the relevant envelope LAWE gives,

<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}{d\eta^2} + \biggl\{ 4 -2Q \biggr\}\frac{1}{\eta} \cdot \frac{dx}{d\eta} 
~-~  2 Q \cdot \frac{x}{\eta^2} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{d^2x}{d\eta^2} + \biggl\{ 4 -2 \biggl[1- \eta\cot(\eta-B) \biggr]\biggr\}\frac{1}{\eta} \cdot \frac{dx}{d\eta} 
~-~  \biggl[1- \eta\cot(\eta-B) \biggr]   \frac{2x}{\eta^2} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{b}{\eta^4} \biggl\{
\frac{\eta^4}{b} \cdot \frac{d^2x}{d\eta^2} + \biggl[ 1 + \eta\cot(\eta-B) \biggr] \frac{2\eta^3}{b} \cdot \frac{dx}{d\eta} 
~-~  \biggl[1- \eta\cot(\eta-B) \biggr]  \frac{2\eta^2 x}{b} 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2b}{\eta^4} \biggl\{
3 - \eta^2 
- (\eta + \eta^3)\cot(\eta-B)  
-  \eta^2\cot^2(\eta - B) 
-  \eta^3\cot^3(\eta-B) 
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2b}{\eta^4} \biggl\{
3 - \eta^2 
- (\eta + \eta^3)\cot(\eta-B)  
-  \eta^2\cot^2(\eta - B) 
-  \eta^3\cot^3(\eta-B) 
+ \biggl[\eta^2 -2 + \eta\cot(\eta-B)  +  \eta^2\cot^2(\eta - B)\biggr]
~-~  \biggl[1- \eta\cot(\eta-B) \biggr] 
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2b}{\eta^4} \biggl\{
- (\eta + \eta^3)\cot(\eta-B)  
-  \eta^3\cot^3(\eta-B) 
~+~2\eta\cot(\eta-B) 
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2b}{\eta^4} \biggl\{
[- \eta \cot(\eta-B)  
- \eta\cot(\eta-B) 
~+~2\eta\cot(\eta-B) ]
+ [\eta^3\cot(\eta-B) 
- \eta^3 \cot(\eta-B)  ]
</math>
  </td>
</tr>

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

Okay.  Now let's determine at what value of <math>~\eta</math> the logarithmic derivative of <math>~x_P</math> goes to negative one.
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{d\ln x_P}{d\ln \eta} = \frac{\eta}{x_P} \cdot \frac{dx_P}{d\eta}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\eta^3}{b }\biggl[1- \eta\cot(\eta-B) \biggr]^{-1}
\cdot \frac{dx_P}{d\eta}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[1- \eta\cot(\eta-B) \biggr]^{-1}
\biggl[ \eta^2 -2 + \eta\cot(\eta-B)  +  \eta^2\cot^2(\eta - B)  \biggr] \, .
</math>
  </td>
</tr>
</table>
Setting this to negative one, we have,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~
-\biggl[1- \eta\cot(\eta-B) \biggr]
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \eta^2 -2 + \eta\cot(\eta-B)  +  \eta^2\cot^2(\eta - B)  \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~
\Rightarrow~~~1
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\eta^2\biggl[ 1 +  \cot^2(\eta - B) \biggr] 
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
<math>~
\Rightarrow~~~1
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{\eta^2}{\sin^2(\eta - B)} \, .
</math>
  </td>
</tr>
</table>
And this occurs when,
<div align="center">
<math>~\biggl(\frac{A}{\phi } \biggr)^2 = 1 \, .</math>
</div>