Editing Appendix/Ramblings/PPToriPt1A (section)

===Cubic Equation Solution===
For later use, let's invert the cubic relation to obtain a more broadly applicable <math>~x(\eta)</math> function.  Because we are only interested in radial profiles in the equatorial plane &#8212; that is, only for the values of <math>~\theta = 0</math> or <math>~\theta=\pi</math> &#8212; the relation to be inverted is,

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

<tr>
  <td align="right">
<math>~x^2 \pm 2 x^3</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(\beta\eta)^2</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~~ x^3 \pm \tfrac{1}{2}x^2 \mp \tfrac{1}{2}(\beta\eta)^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~0 \, .</math>
  </td>
</tr>
</table>
</div>


<div align="center">
<table border="1" align="center" cellpadding="8">
<tr>
  <td align="center" colspan="6"><font size="+1"><b>Table 1: &nbsp;Example Parameter Values</b></font><p></p>
determined by iterative solution for <math>~\beta = \tfrac{1}{10}</math></td>
</tr>
<tr>
  <td align="center" rowspan="2"><math>~\eta</math></td>
  <td align="center" rowspan="2"><math>~\Gamma^2 = 54\beta^2\eta^2</math></td>
  <td align="center" colspan="2">Inner solution <math>~(\theta = 0)</math>
    <p></p>[''Superior'' sign in cubic eq.]</td>
  <td align="center" colspan="2">Outer solution <math>~(\theta = \pi)</math>
    <p></p>[''Inferior'' sign in cubic eq.]</td>
</tr>
<tr>
  <td align="center"><math>^\dagger\biggl(\frac{x_\mathrm{root}}{\beta}\biggr)</math></td>
  <td align="center"><math>~6(S + T)</math></td>
  <td align="center"><math>^\dagger\biggl(\frac{x_\mathrm{root}}{\beta}\biggr)</math></td>
  <td align="center"><math>~6(S + T)</math></td>
</tr>
<tr>
  <td align="right">0.25</td>
  <td align="center">0.03375</td>
  <td align="right">0.244112</td>
  <td align="right">1.14647</td>
  <td align="right">0.256675</td>
  <td align="right">-0.84600</td>
</tr>
<tr>
  <td align="right">1.0</td>
  <td align="center">0.54</td>
  <td align="right">0.91909</td>
  <td align="right">1.55145</td>
  <td align="right">1.1378</td>
  <td align="right">-0.31732</td>
</tr>
<tr>
  <td align="left" colspan="6">
<sup>&dagger;</sup>Here, <math>~x_\mathrm{root}</math> has been determined via a brute-force, iterative technique.
  </td>
</tr>
</table>
</div>


Following [http://mathworld.wolfram.com/CubicFormula.html Wolfram's discussion of the cubic formula], we should view this expression as a specific case of the general formula,

<div align="center">
<math>~x^3 + a_2x^2 + a_1x + a_0 = 0 \, ,</math>
</div>

in which case, as is detailed in equations (54) - (56) of [http://mathworld.wolfram.com/CubicFormula.html Wolfram's discussion of the cubic formula], the three roots of any cubic equation are:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~x_1</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-\frac{1}{3}a_2 + (S + T) \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~x_2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-\frac{1}{3}a_2 - \frac{1}{2} (S + T) + \frac{1}{2} \it{i} \sqrt{3} (S-T)\, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~x_3</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-\frac{1}{3}a_2 - \frac{1}{2} (S + T) - \frac{1}{2} \it{i} \sqrt{3} (S-T)\, ,
</math>
  </td>
</tr>
</table>
</div>
where,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~S</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~[R + \sqrt{D}]^{1/3} \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~T</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~[R - \sqrt{D}]^{1/3} \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~D</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~Q^3 + R^2 \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~Q</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\frac{3a_1 - a_2^2}{3^2} \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~R</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\frac{3^2a_2 a_1 - 3^3a_0 - 2a_2^3}{2\cdot 3^3} \, .
</math>
  </td>
</tr>
</table>
</div>

====Outer [inferior sign] Solution====
Focusing, first, on the ''inferior'' sign convention, which corresponds to the "outer" solution <math>~(\theta = \pi)</math>, we see that the coefficients that lead to our specific cubic equation are:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~a_2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- \tfrac{1}{2} \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~a_1</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~0 \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~a_0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\tfrac{1}{2}(\beta\eta)^2 \, .</math>
  </td>
</tr>
</table>
</div>

Applying Wolfram's definitions of the <math>~Q</math> and <math>~R</math> parameters to our particular problem gives,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~Q</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\frac{3a_1 - a_2^2}{3^2} = -\biggl(\frac{a_2}{3}\biggr)^2 = - \frac{1}{2^2\cdot 3^2} \, ;</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~R</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\frac{3^2a_2 a_1 - 3^3a_0 - 2a_2^3}{2\cdot 3^3}
= \frac{1}{2\cdot 3^3} \biggl[   -\frac{ 3^3}{2}(\beta\eta)^2 + \frac{1}{2^2}\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\frac{1}{2^3\cdot 3^3} \biggl[ 1  - 2\cdot 3^3(\beta\eta)^2\biggr]  \, .
</math>
  </td>
</tr>

</table>
</div>

Defining the parameter,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\Gamma^2</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~ 2\cdot 3^3(\beta\eta)^2 \, ,</math>
  </td>
</tr>
</table>
</div>

we therefore have,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~(2\cdot 3)^6 D</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~( 1  - \Gamma^2 )^2-1 \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~(2\cdot 3)^3S^3</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~(2\cdot 3)^3 R + \sqrt{(2\cdot 3)^6D} </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~(1-\Gamma^2) + \sqrt{( 1  - \Gamma^2 )^2-1}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~(1-\Gamma^2) + i\sqrt{1-( 1  - \Gamma^2 )^2} \, ;</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~(2\cdot 3)^3T^3</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~(1-\Gamma^2) - i\sqrt{1-( 1  - \Gamma^2 )^2} \, .</math>
  </td>
</tr>

</table>
</div>

<div align="center" id="CubeRootImaginary">
<table border="1" width="60%" cellpadding="8">
<tr><td align="left">
<font color="purple" size="+1"><b>ASIDE:</b></font>&nbsp;  The cube root of an imaginary number &hellip; 

<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\ell^3</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~A \pm i \sqrt{1-A^2}</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~r_\ell e^{i\theta_\ell} \, ,</math>
  </td>
</tr>
</table>
where,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~r_\ell</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~( A^2 + 1-A^2 )^{1/2} = 1 \, ,</math>
  </td>
</tr>
</table>
and,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\theta_\ell</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\pm \tan^{-1}\biggl( \frac{\sqrt{1-A^2}}{A} \biggr) = \pm \cos^{-1}A \, .</math>
  </td>
</tr>
</table>

Now, according to [http://math.stackexchange.com/questions/8760/what-are-the-three-cube-roots-of-1 this online resource], the three roots <math>~(j=0,1,2)</math> of <math>~\ell^3</math> are,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="center"><math>~\ell_j = r_\ell^{1/3}e^{i(\theta_\ell + 2j\pi)/3)} \, ,</math></td>
</tr>
</table>
</div>

which, for our specific problem gives,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\ell_j</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~e^{i\theta_\pm/3} \cdot e^{i(2j\pi/3)} \, ,</math>
  </td>
</tr>
</table>
</div>
where the subscript on <math>~\theta</math> refers to the <math>~\pm</math> in our original expression for <math>~\ell</math>.

<!--
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\ell_j</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\cos[(2j\pi +\theta_\ell)/3] + i \sin[(2j\pi + \theta_\ell)/3]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\cos\biggl[ {\frac{1}{3}\biggl(2j\pi \pm \cos^{-1}A\biggr)} \biggr] + i \sin\biggl[ {\frac{1}{3}\biggl(2j\pi \pm \cos^{-1}A\biggr)} \biggr]
</math>
  </td>
</tr>
</table>
</div>
-->

</td></tr>
</table>
</div>

In our particular case, after associating <math>~A \leftrightarrow (1-\Gamma^2)</math>,  we can write,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~
2\cdot 3(S + T) 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[(1-\Gamma^2) + i\sqrt{1-( 1  - \Gamma^2 )^2} \biggr]^{1/3} + 
\biggl[(1-\Gamma^2) - i\sqrt{1-( 1  - \Gamma^2 )^2} \biggr]^{1/3}  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
e^{i\theta_+/3} \cdot e^{i(2j\pi/3)}  + e^{i\theta_-/3} \cdot e^{i(2j\pi/3)}   
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~e^{i(2j\pi/3)} \biggl\{
e^{i[\cos^{-1}(1-\Gamma^2)]/3}   + e^{-i[\cos^{-1}(1-\Gamma^2)]/3} \biggr\}  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~e^{i(2j\pi/3)} \biggl\{\cos\biggl[ \tfrac{1}{3}\cos^{-1}(1-\Gamma^2)\biggr] 
+ i\sin\biggl[ \tfrac{1}{3} \cos^{-1}(1-\Gamma^2)\biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~+ \cos\biggl[ \tfrac{1}{3} \cos^{-1}(1-\Gamma^2)\biggr] 
- i\sin\biggl[ \tfrac{1}{3} \cos^{-1}(1-\Gamma^2)\biggr]
\biggr\}  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2 e^{i(2j\pi/3)} \cdot \cos\biggl[ \tfrac{1}{3}\cos^{-1}(1-\Gamma^2)\biggr] \, .</math>
  </td>
</tr>
</table>
</div>

Similarly, we can write,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~
2\cdot 3(S - T) 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[(1-\Gamma^2) + i\sqrt{1-( 1  - \Gamma^2 )^2} \biggr]^{1/3} - 
\biggl[(1-\Gamma^2) - i\sqrt{1-( 1  - \Gamma^2 )^2} \biggr]^{1/3}  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
e^{i\theta_+/3} \cdot e^{i(2j\pi/3)}  - e^{i\theta_-/3} \cdot e^{i(2j\pi/3)}   
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~e^{i(2j\pi/3)} \biggl\{
e^{i[\cos^{-1}(1-\Gamma^2)]/3}   - e^{-i[\cos^{-1}(1-\Gamma^2)]/3} \biggr\}  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~e^{i(2j\pi/3)} \biggl\{\cos\biggl[ \tfrac{1}{3}\cos^{-1}(1-\Gamma^2)\biggr] 
+ i\sin\biggl[ \tfrac{1}{3} \cos^{-1}(1-\Gamma^2)\biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~- \cos\biggl[ \tfrac{1}{3} \cos^{-1}(1-\Gamma^2)\biggr] 
+ i\sin\biggl[ \tfrac{1}{3} \cos^{-1}(1-\Gamma^2)\biggr]
\biggr\}  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2i e^{i(2j\pi/3)} \cdot \sin\biggl[ \tfrac{1}{3}\cos^{-1}(1-\Gamma^2)\biggr] \, .
</math>
  </td>
</tr>
</table>
</div>

Focusing specifically on the "j=0" root, and setting <math>~a_2 = -\tfrac{1}{2}</math>, we therefore have,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~6x_1-1</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
6(S + T) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2 \cos\biggl[ \tfrac{1}{3}\cos^{-1}(1-\Gamma^2)\biggr] \, ;</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~6x_2-1</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \frac{1}{2} \biggl\{ 2 \cos\biggl[ \tfrac{1}{3}\cos^{-1}(1-\Gamma^2)\biggr]  
- i\sqrt{3} \cdot 2i \sin\biggl[ \tfrac{1}{3}\cos^{-1}(1-\Gamma^2)\biggr]  \biggr\}
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2\biggl\{ \cos\biggl[ \tfrac{1}{3}\cos^{-1}(1-\Gamma^2) + \frac{2\pi}{3}\biggr]  \biggr\} \, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~6x_3-1</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \frac{1}{2} \biggl\{ 2 \cos\biggl[ \tfrac{1}{3}\cos^{-1}(1-\Gamma^2)\biggr]  
+ i\sqrt{3} \cdot 2i \sin\biggl[ \tfrac{1}{3}\cos^{-1}(1-\Gamma^2)\biggr]  \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-2\biggl\{ \frac{1}{2}\cdot \cos\biggl[ \tfrac{1}{3}\cos^{-1}(1-\Gamma^2)\biggr]  
-\frac{\sqrt{3}}{2} \cdot \sin\biggl[ \tfrac{1}{3}\cos^{-1}(1-\Gamma^2)\biggr]  \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2\biggl\{ \cos\biggl[ \tfrac{1}{3}\cos^{-1}(1-\Gamma^2) - \frac{2\pi}{3}\biggr]  \biggr\} \, .
</math>
  </td>
</tr>
</table>
</div>


<div align="center">
<table border="1" align="center" cellpadding="8">
<tr>
  <td align="center" colspan="8"><font size="+1"><b>Table 1: &nbsp;Analytically Evaluated Roots</b></font><p></p>
determined for <math>~\beta = \tfrac{1}{10}</math></td>
</tr>
<tr>
  <td align="center" rowspan="2"><math>~\eta</math></td>
  <td align="center" rowspan="2"><math>~\Gamma^2 = 54\beta^2\eta^2</math></td>
  <td align="center" colspan="3">Inner solution <math>~(\theta = 0)</math>
    <p></p>[''Superior'' sign in cubic eq.]</td>
  <td align="center" colspan="3">Outer solution <math>~(\theta = \pi)</math>
    <p></p>[''Inferior'' sign in cubic eq.]</td>
</tr>
<tr>
  <td align="center"><math>~x_1/\beta</math></td>
  <td align="center"><math>~x_2/\beta</math></td>
  <td align="center"><math>~x_3/\beta</math></td>
  <td align="center"><math>~x_1/\beta</math></td>
  <td align="center"><math>~x_2/\beta</math></td>
  <td align="center"><math>~x_3/\beta</math></td>
</tr>
<tr>
  <td align="right">0.25</td>
  <td align="center">0.03375</td>
  <td align="right">-4.98744</td>
  <td align="right" bgcolor="yellow">0.24411</td>
  <td align="right">-0.25667</td>
  <td align="right">4.98744</td>
  <td align="right">-0.24411</td>
  <td align="right" bgcolor="yellow">0.25667</td>
</tr>
<tr>
  <td align="right">1.0</td>
  <td align="center">0.54</td>
  <td align="right">-4.78128</td>
  <td align="right" bgcolor="yellow">0.91909</td>
  <td align="right">-1.1378</td>
  <td align="right">4.78128</td>
  <td align="right">-0.91909</td>
  <td align="right" bgcolor="yellow">1.1378</td>
</tr>

<tr>
  <td align="center" colspan="2">&nbsp;</td>
  <td align="center" colspan="3">CONFIRMATION:  In all cases,
    <p></p><math>~x^2 + 2x^3 = (\beta\eta)^2</math></td>
  <td align="center" colspan="3">CONFIRMATION:  In all cases,
    <p></p><math>~x^2 - 2x^3 = (\beta\eta)^2</math></td>
</tr>

</table>
</div>

<!--  ************************* -->

====Inner [superior sign] Solution====
Next, examing the ''superior'' sign convention, which corresponds to the "inner" solution <math>~(\theta = 0)</math>, we see that the coefficients that lead to our specific cubic equation are:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~a_2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\tfrac{1}{2} \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~a_1</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~0 \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~a_0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- \tfrac{1}{2}(\beta\eta)^2 \, .</math>
  </td>
</tr>
</table>
</div>

Following the same set of steps that were followed in [[Appendix/Ramblings/PPTori#Outer_.5Binferior_sign.5D_Solution|determining the "outer" solution]], here we find:  <math>~Q</math> remains the same; <math>~R</math> has the same magnitude, but changes sign; and, hence, <math>~D</math> remains the same.  We therefore have,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~(2\cdot 3)^3S^3</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- (1-\Gamma^2) + i\sqrt{1-( 1  - \Gamma^2 )^2} \, ;</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~(2\cdot 3)^3T^3</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- (1-\Gamma^2) - i\sqrt{1-( 1  - \Gamma^2 )^2} \, ,</math>
  </td>
</tr>

</table>
</div>

which leads to the following expressions for the three "inner" roots:

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

<tr>
  <td align="right">
<math>~6x_1+1</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- 2 \cos\biggl[ \tfrac{1}{3}\cos^{-1}(1-\Gamma^2)\biggr] \, ;</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~6x_2+1</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- 2\biggl\{ \cos\biggl[ \tfrac{1}{3}\cos^{-1}(1-\Gamma^2) + \frac{2\pi}{3}\biggr]  \biggr\} \, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~6x_3+1</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- 2\biggl\{ \cos\biggl[ \tfrac{1}{3}\cos^{-1}(1-\Gamma^2) - \frac{2\pi}{3}\biggr]  \biggr\} \, .
</math>
  </td>
</tr>
</table>
</div>