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=Stability Analyses of PP Tori=
=Stability Analyses of PP Tori=
<font color="red"><b>[Comment by J. E. Tohline on 24 May 2016]</b></font> &nbsp; This chapter contains a set of technical notes and accompanying discussion that I put together several months ago as I was trying to gain a foundational understanding of the results of a large study of instabilities in self-gravitating tori published by the Imamura &amp; Hadley collaboration.  I have come to appreciate that some of the logic and interpretation of published results that are presented, below, has serious flaws.  Therefore, anyone reading this should be quite cautious in deciding what subsections provide useful insight.  I have written a separate chapter titled, "[[User:Tohline/Apps/ImamuraHadleyCollaboration#Characteristics_of_Unstable_Eigenvectors_in_Self-Gravitating_Tori|Characteristics of Unstable Eigenvectors in Self-Gravitating Tori]]," that contains a much more trustworthy analysis of this very interesting problem.
{{LSU_WorkInProgress}}
{{LSU_HBook_header}}
{{LSU_HBook_header}}


As has been summarized in [[User:Tohline/Appendix/Ramblings/To_Hadley_and_Imamura#Summary_for_Hadley_.26_Imamura|an accompanying chapter]] &#8212; also see our related [[User:Tohline/Appendix/Ramblings/Azimuthal_Distortions#Analyzing_Azimuthal_Distortions|detailed notes]] &#8212; we have been trying to understand why unstable nonaxisymmetric eigenvectors have the shapes that they do in rotating toroidal configurations.  For any azimuthal mode, <math>~m</math>, we are referring both to the radial dependence of the distortion amplitude, <math>~f_m(\varpi)</math>, and the radial dependence of the phase function, <math>~\phi_m(\varpi)</math> &#8212; the latter is what the [[#See_Also|Imamura and Hadley collaboration]] refer to as a "constant phase locus."  Some old videos showing the development over time of various self-gravitating "constant phase loci" can be found [[User:Tohline/Apps/WoodwardTohlineHachisu94#Online_Movies|here]]; these videos supplement the published work of [http://adsabs.harvard.edu/abs/1994ApJ...420..247W Woodward, Tohline &amp; Hachisu (1994)].
As has been summarized in [[User:Tohline/Appendix/Ramblings/To_Hadley_and_Imamura#Summary_for_Hadley_.26_Imamura|an accompanying chapter]] &#8212; also see our related [[User:Tohline/Appendix/Ramblings/Azimuthal_Distortions#Analyzing_Azimuthal_Distortions|detailed notes]] &#8212; we have been trying to understand why unstable nonaxisymmetric eigenvectors have the shapes that they do in rotating toroidal configurations.  For any azimuthal mode, <math>~m</math>, we are referring both to the radial dependence of the distortion amplitude, <math>~f_m(\varpi)</math>, and the radial dependence of the phase function, <math>~\phi_\mathrm{max}(\varpi)</math> &#8212; the latter is what the [[#See_Also|Imamura and Hadley collaboration]] refer to as a "constant phase locus."  Some old videos showing the development over time of various self-gravitating "constant phase loci" can be found [[User:Tohline/Apps/WoodwardTohlineHachisu94#Online_Movies|here]]; these videos supplement the published work of [http://adsabs.harvard.edu/abs/1994ApJ...420..247W Woodward, Tohline &amp; Hachisu (1994)].
 


Here, we focus specifically on instabilities that arise in so-called (non-self-gravitating) [[User:Tohline/Apps/PapaloizouPringleTori#Massless_Polytropic_Tori|Papaloizou-Pringle tori]] and will draw heavily from two publications:  (1) [http://adsabs.harvard.edu/abs/1987MNRAS.225..267P Papaloizou &amp; Pringle (1987), MNRAS, 225, 267] &#8212; ''The dynamical stability of differentially rotating discs. &nbsp; III.'' &#8212; hereafter, PPIII &#8212; and (2) [http://adsabs.harvard.edu/abs/1985MNRAS.216..553B Blaes (1985), MNRAS, 216, 553] &#8212; ''Oscillations of slender tori''.
Here, we focus specifically on instabilities that arise in so-called (non-self-gravitating) [[User:Tohline/Apps/PapaloizouPringleTori#Massless_Polytropic_Tori|Papaloizou-Pringle tori]] and will draw heavily from three publications:   
[http://adsabs.harvard.edu/abs/1987MNRAS.225..267P Papaloizou &amp; Pringle (1987), MNRAS, 225, 267] (''aka'' PPIII) &#8212; ''The dynamical stability of differentially rotating discs. &nbsp; III.''
* [http://adsabs.harvard.edu/abs/1985MNRAS.216..553B Blaes (1985), MNRAS, 216, 553] (''aka'' Blaes85) &#8212; ''Oscillations of slender tori.''
*  [http://adsabs.harvard.edu/abs/1985MNRAS.213P...7G Goldreich &amp; Narayan (1985), MNRAS, 213, 7] (''aka'' GGN86) &#8212; ''Non-axisymmetric instability in thin discs.''


==PP III==
==PP III==


<div align="center">
<div align="center">
<table border="1" cellpadding="5" width="75%">
<table border="1" cellpadding="5" width="50%">
<tr><td align="center">
<tr><td align="center">
Figure 2 extracted without modification from p. 274 of [http://adsabs.harvard.edu/abs/1987MNRAS.225..267P  J. C. B. Papaloizou &amp; J. E. Pringle (1987)]<p></p>
Figure 2 extracted without modification from p. 274 of [http://adsabs.harvard.edu/abs/1987MNRAS.225..267P  J. C. B. Papaloizou &amp; J. E. Pringle (1987)]<p></p>
Line 19: Line 29:
<tr>
<tr>
   <td align="center">
   <td align="center">
<!-- [[File:NormanWilson78D.png|600px|center|Norman &amp; Wilson (1978)]] -->
<!-- [[File:NormanWilson78D.png|450px|center|Norman &amp; Wilson (1978)]] -->
[[File:PPIIIFig2.png|Figure 2 from PP III]]
[[File:PPIIIFig2.png|350px|center|Figure 2 from PP III]]
   </td>
   </td>
</tr>
</tr>
Line 26: Line 36:
</div>
</div>


==Blaes (1985)==
===Equilibrium Configuration===
In our [[User:Tohline/Apps/PapaloizouPringleTori#Solution|separate discussion of PP84]], we showed that the equilibrium structure of a PP-torus is defined by the enthalpy distribution,
<div align="center">
<math>
H = \frac{GM_\mathrm{pt}}{\varpi_0} \biggl[ (\chi^2 + \zeta^2)^{-1/2} - \frac{1}{2}\chi^{-2}  - C_B^' \biggr] .
</math>
</div>
Normalizing this expression by the enthalpy at the "center" &#8212; ''i.e.,'' at the pressure maximum &#8212; of the torus which, as we have [[User:Tohline/Apps/PapaloizouPringleTori#Pressure_Maximum|already shown]], is
<div align="center">
<math>
H_0 = \frac{GM_\mathrm{pt}}{2\varpi_0} [1-2C_B^' ] \,
</math>
</div>
gives,
<div align="center">
<math>
[1-2C_B^' ]\biggl(\frac{H}{H_0}\biggr) = 2(\chi^2 + \zeta^2)^{-1/2} - \chi^{-2} -1  + [1 - 2C_B^' ].
</math>
</div>


==Blaes85==
Now, in our review of [[User:Tohline/Apps/PapaloizouPringleTori#Model_as_Described_by_Kojima|Kojima's (1986)]] work, we showed that the square of the Mach number at the "center" of the torus is,


===His Notation===
[http://adsabs.harvard.edu/abs/1985MNRAS.216..553B Blaes (1985)] adopts a [[User:Tohline/SR#Barotropic_Structure|polytropic equation of state]],
<div align="center">
<math>~\frac{\rho}{\rho_c} = \Theta_H^n \, ,</math>
</div>
which gives rise to (slim tori) equilibrium structures for which (see his equation 1.3),
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 40: Line 65:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\Theta_H</math>
<math>~\mathfrak{M}_0^2 \equiv \frac{(v_\varphi|_0)^2}{(c_s|_0)^2}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 46: Line 71:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~1 - \frac{1}{\beta^2}\biggl[x^2 + x^3(3\cos\theta - \cos^3\theta) + \mathcal{O}(x^4)  \biggr] \, ,</math>
<math>~\frac{2(n+1)}{\gamma}\biggl[ \frac{1}{\chi_-} - 1 \biggr]^{-2}</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
where, the (constant) model parameter,
<div align="center">
<math>\beta \equiv \frac{(2n)^{1/2}}{\mathcal{M}_0} \, ,</math>
</div>
and <math>~\mathcal{M}_0</math> is the Mach number of the rotational velocity at the torus center.  Blaes then adopts a related parameter that is constant on isobaric surfaces, namely,
<div align="center">
<math>\eta^2 \equiv 1 - \Theta_H \, ,</math>
</div>
which is unity at the surface of the torus and which goes to zero at the cross-sectional center of the torus.  Notice that <math>~\eta</math> tracks the "radial" coordinate that measures the distance from the center of the torus; in particular, keeping only the leading-order term in <math>~x</math>, there is a simple linear relationship between  <math>~\eta</math> and <math>~x</math>, namely,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\eta</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
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   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~[1 - \Theta_H]^{1/2} \approx \frac{x}{\beta} \, .</math>
<math>~2n [ 1- 2C_B^' ]^{-1} </math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
===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>
<tr>
   <td align="right">
   <td align="right">
<math>~x^2 \pm 2 x^3</math>
<math>~\Rightarrow ~~~~ [1 - 2C_B^'] </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
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   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~(\beta\eta)^2</math>
<math>~\frac{2n}{\mathfrak{M}_0^2} \, , </math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
where, in obtaining this last expression we have related the adiabatic exponent to the polytropic index via the relation, <math>~\gamma = (n+1)/n</math>.  Instead of specifying the system's Mach number, [http://adsabs.harvard.edu/abs/1985MNRAS.216..553B Blaes (1985)] defines the dimensionless parameter,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\Rightarrow ~~~~ x^3 \pm \tfrac{1}{2}x^2 \mp \tfrac{1}{2}(\beta\eta)^2</math>
<math>~\beta^2 </math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\frac{2n}{\mathfrak{M}_0^2} \, .</math>
  </td>
</tr>
 
</table>
</div>
 
Implementing this parameter swap, the equilibrium expression becomes,
<div align="center">
<math>
\beta^2 \biggl(\frac{H}{H_0}\biggr) = 2(\chi^2 + \zeta^2)^{-1/2} - \chi^{-2} -1  + \beta^2 \, ,
</math>
</div>
or,
 
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\frac{H}{H_0} </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 103: Line 139:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~0 \, .</math>
<math>~1 - \frac{1}{\beta^2}\biggl[\chi^{-2} - 2(\chi^2 + \zeta^2)^{-1/2} + 1 \biggr] \, .</math>
   </td>
   </td>
</tr>
</tr>
Line 109: Line 145:
</div>
</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,
Looking at Figure 1 of Blaes85 &#8212; see also the coordinate definitions given in his equation (2.1) &#8212; I conclude that,


<div align="center">
<div align="center">
<math>~x^3 + a_2x^2 + a_1x + a_0 = 0 \, .</math>
<math>~\chi = 1 - x\cos\theta</math>
&nbsp; &nbsp;&nbsp;&nbsp; and &nbsp; &nbsp; &nbsp; &nbsp;
<math>\zeta = x\sin\theta \, .</math>
</div>
</div>


Hence,


Doing this &#8212; and focusing, first, on the ''superior'' sign convention, which corresponds to the "inner" solution <math>~(\theta = 0)</math> &#8212; we see that the coefficients that lead to our specific cubic equation are:
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 122: Line 160:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~a_2</math>
<math>~\frac{H}{H_0} </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 128: Line 166:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\tfrac{1}{2} \, ,</math>
<math>~1 - \frac{1}{\beta^2}\biggl\{ [1 - x\cos\theta]^{-2} - 2[(1 - x\cos\theta)^2 + x^2\sin^2\theta]^{-1/2} + 1 \biggr\} </math>
   </td>
   </td>
</tr>
</tr>
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<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~a_1</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
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   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~0 \, ,</math>
<math>~1 - \frac{1}{\beta^2}\biggl\{ [1 - x\cos\theta]^{-2} - 2[(1 - 2x\cos\theta + x^2\cos^2\theta) + x^2(1-\cos^2\theta)]^{-1/2} + 1 \biggr\} </math>
   </td>
   </td>
</tr>
</tr>
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<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~a_0</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
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   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~- \tfrac{1}{2}(\beta\eta)^2 \, .</math>
<math>~1 - \frac{1}{\beta^2}\biggl[ (1 - x\cos\theta)^{-2} - 2(1 - 2x\cos\theta + x^2)^{-1/2} + 1 \biggr] \, .</math>
   </td>
   </td>
</tr>
</tr>
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</div>
</div>


<table border="1" align="center" cellpadding="5" width="40%">
This matches equation (2.2) of Blaes85, if we acknowledge that Blaes uses <math>~f</math> &#8212; instead of the parameter notation, <math>~\Theta_H</math>, found in [[User:Tohline/SSC/Structure/Polytropes#Governing_Relations|our discussion of equilibrium polytropic configurations]] &#8212; to denote the normalized enthalpy; that is,
<tr><td align="left">
 
Note:  If <math>~\beta = 0.1</math>, this "inner" solution should give <math>~x = 0.0244112</math> for <math>~\eta = 0.25</math>, and it should give <math>~x = 0.091909</math> for <math>~\eta = 1</math>.]
</td></tr>
</table>
 
<div align="center">
<div align="center">
<table border="1" align="center" cellpadding="8">
<math>~f_\mathrm{Blaes85} = \Theta_H \equiv \frac{H}{H_0} \, .</math>
<tr>
  <td align="center" colspan="3">For <math>~\beta = \tfrac{1}{10}</math></td>
</tr>
<tr>
  <td align="center"><math>~\eta</math></td>
  <td align="center"><math>~\Gamma^2 = 54\beta^2\eta^2</math></td>
  <td align="center"><math>~x_1</math></td>
</tr>
<tr>
  <td align="right">0.25</td>
  <td align="center">0.03375</td>
  <td align="right">0.0244112</td>
</tr>
<tr>
  <td align="right">1.0</td>
  <td align="center">0.54</td>
  <td align="right">0.091909</td>
</tr>
</table>
</div>
</div>


 
This expression for the enthalpy throughout a Papaloizou-Pringle torus is valid for tori of arbitrary thickness <math>~(0 < \beta < 1)</math>.  When considering only slim tori, [http://adsabs.harvard.edu/abs/1985MNRAS.216..553B Blaes (1985)] points out that this expression can be written in terms of the following power series in <math>~x</math> (see his equation 1.3):
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">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
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<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~x_1</math>
<math>~\Theta_H</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 200: Line 214:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~1 - \frac{1}{\beta^2}\biggl[x^2 + x^3(3\cos\theta - \cos^3\theta) + \mathcal{O}(x^4) \biggr] \, .</math>
-\frac{1}{3}a_2 + (S + T) \, ,
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
Blaes then adopts a related parameter that is constant on isobaric surfaces, namely,
<div align="center">
<math>\eta^2 \equiv 1 - \Theta_H \, ,</math>
</div>
which is unity at the surface of the torus and which goes to zero at the cross-sectional center of the torus.  Notice that <math>~\eta</math> tracks the "radial" coordinate that measures the distance from the center of the torus; in particular, keeping only the leading-order term in <math>~x</math>, there is a simple linear relationship between  <math>~\eta</math> and <math>~x</math>, namely,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~x_2</math>
<math>~\eta</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 214: Line 236:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~[1 - \Theta_H]^{1/2} \approx \frac{x}{\beta} \, .</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>
   </td>
</tr>
</tr>
</table>
</table>
</div>
</div>
where,
 
===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">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 241: Line 250:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~S</math>
<math>~x^2 \pm 2 x^3</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\equiv</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~[R + \sqrt{D}]^{1/3} \, ,</math>
<math>~(\beta\eta)^2</math>
   </td>
   </td>
</tr>
</tr>
Line 253: Line 262:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~T</math>
<math>~\Rightarrow ~~~~ x^3 \pm \tfrac{1}{2}x^2 \mp \tfrac{1}{2}(\beta\eta)^2</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\equiv</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~[R - \sqrt{D}]^{1/3} \, ,</math>
<math>~0 \, .</math>
   </td>
   </td>
</tr>
</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>
<tr>
   <td align="right">
   <td align="center" rowspan="2"><math>~\eta</math></td>
<math>~D</math>
  <td align="center" rowspan="2"><math>~\Gamma^2 = 54\beta^2\eta^2</math></td>
   </td>
   <td align="center" colspan="2">Inner solution <math>~(\theta = 0)</math>
   <td align="center">
    <p></p>[''Superior'' sign in cubic eq.]</td>
<math>~\equiv</math>
  <td align="center" colspan="2">Outer solution <math>~(\theta = \pi)</math>
   </td>
    <p></p>[''Inferior'' sign in cubic eq.]</td>
   <td align="left">
</tr>
<math>~Q^3 + R^2 \, .</math>
<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>
   </td>
</tr>
</tr>
Line 277: Line 319:
</div>
</div>


Applying Wolfram's definitions of the <math>~Q</math> and <math>~R</math> parameters to our particular problem gives,
 
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">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 283: Line 332:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~Q</math>
<math>~x_1</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\equiv</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <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>
<math>~
-\frac{1}{3}a_2 + (S + T) \, ,
</math>
   </td>
   </td>
</tr>
</tr>
Line 295: Line 346:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~R</math>
<math>~x_2</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\equiv</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{3^2a_2 a_1 - 3^3a_0 - 2a_2^3}{2\cdot 3^3}
<math>~
= - \frac{1}{2\cdot 3^3} \biggl[ -\frac{3^3}{2}(\beta\eta)^2  + \frac{1}{2^2} \biggr] </math>
-\frac{1}{3}a_2 - \frac{1}{2} (S + T) + \frac{1}{2} \it{i} \sqrt{3} (S-T)\, ,
</math>
   </td>
   </td>
</tr>
</tr>
Line 308: Line 360:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~x_3</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 315: Line 367:
   <td align="left">
   <td align="left">
<math>~
<math>~
- \frac{1}{2^3\cdot 3^3} \biggl[ 1-2\cdot 3^3(\beta\eta)^2 \biggr]\, . </math>
-\frac{1}{3}a_2 - \frac{1}{2} (S + T) - \frac{1}{2} \it{i} \sqrt{3} (S-T)\, ,
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</div>
</div>
Defining the parameter,
where,
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 326: Line 379:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\Gamma^2</math>
<math>~S</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 332: Line 385:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ 2\cdot 3^3(\beta\eta)^2 \, ,</math>
<math>~[R + \sqrt{D}]^{1/3} \, ,</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
we therefore have,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~2^6\cdot 3^6D</math>
<math>~T</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~\equiv</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~[ 1-2\cdot 3^3(\beta\eta)^2 ]^2-1 </math>
<math>~[R - \sqrt{D}]^{1/3} \, ,</math>
   </td>
   </td>
</tr>
</tr>
Line 356: Line 403:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~D</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~\equiv</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~[ 1-\Gamma^2 ]^2-1 \, ;</math>
<math>~Q^3 + R^2 \, ,</math>
   </td>
   </td>
</tr>
</tr>
Line 368: Line 415:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~2\cdot 3S</math>
<math>~Q</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~\equiv</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~[2^3\cdot 3^3R + \sqrt{2^6\cdot 3^6D}]^{1/3} </math>
<math>~\frac{3a_1 - a_2^2}{3^2} \, ,</math>
   </td>
   </td>
</tr>
</tr>
Line 380: Line 427:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~R</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~\equiv</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl\{
<math>~\frac{3^2a_2 a_1 - 3^3a_0 - 2a_2^3}{2\cdot 3^3} \, .
[ \Gamma^2 -1 ] + \sqrt{[ 1-\Gamma^2 ]^2-1}
\biggr\}^{1/3}  
</math>
</math>
   </td>
   </td>
</tr>
</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>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~a_2</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 401: Line 453:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl\{
<math>~- \tfrac{1}{2} \, ,</math>
[ \Gamma^2 -1 ] + i \sqrt{1 - [ 1-\Gamma^2 ]^2}  
\biggr\}^{1/3} \, ;
</math>
   </td>
   </td>
</tr>
</tr>
Line 410: Line 459:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~2\cdot 3T</math>
<math>~a_1</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 416: Line 465:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl\{
<math>~0 \, ,</math>
[ \Gamma^2 -1 ] - i \sqrt{ 1 - [ 1-\Gamma^2 ]^2 }
\biggr\}^{1/3}  \, .
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
Let's rewrite this last expression as,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~2\cdot 3T</math>
<math>~a_0</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 437: Line 477:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl\{r_T e^{i\theta_T}  
<math>~\tfrac{1}{2}(\beta\eta)^2 \, .</math>
\biggr\}^{1/3}  \, ,
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</div>
</div>
in which case,
 
Applying Wolfram's definitions of the <math>~Q</math> and <math>~R</math> parameters to our particular problem gives,
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 450: Line 489:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~r_T^2</math>
<math>~Q</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~\equiv</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~\frac{3a_1 - a_2^2}{3^2} = -\biggl(\frac{a_2}{3}\biggr)^2 = - \frac{1}{2^2\cdot 3^2} \, ;</math>
[ \Gamma^2 -1 ]^2+ 1 - [ 1-\Gamma^2 ]^2 = 1 \, ,
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
and,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\theta_T</math>
<math>~R</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~\equiv</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\tan^{-1}\biggl\{ -\biggl[ \frac{1 - [ 1-\Gamma^2 ]^2}{[ \Gamma^2 -1 ]^2} \biggr]^{1/2}\biggr\}
<math>~\frac{3^2a_2 a_1 - 3^3a_0 - 2a_2^3}{2\cdot 3^3}
= - \cos^{-1}[1-\Gamma^2] \, .</math>
= \frac{1}{2\cdot 3^3} \biggl[   -\frac{ 3^3}{2}(\beta\eta)^2 + \frac{1}{2^2}\biggr]
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
Next, according to [http://math.stackexchange.com/questions/8760/what-are-the-three-cube-roots-of-1 this online resource], the three roots are,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~r_T^{1/3}e^{i\theta_T/3}</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp;
<math>~\equiv</math>
   </td>
   </td>
   <td align="center">
   <td align="left">
<math>~r_T^{1/3}e^{i(\theta_T/3 + 2\pi/3)}</math>
<math>~\frac{1}{2^3\cdot 3^3} \biggl[ 1  - 2\cdot 3^3(\beta\eta)^2\biggr]  \, .
  </td>
</math>
  <td align="center">
&nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp;
  </td>
  <td align="left">
<math>~r_T^{1/3}e^{i(\theta_T/3 + 4\pi/3)} \, .</math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</div>
</div>


Similarly, we can write,
Defining the parameter,
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 512: Line 535:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~2\cdot 3S</math>
<math>~\Gamma^2</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~\equiv</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl\{r_S e^{i\theta_S}
<math>~ 2\cdot 3^3(\beta\eta)^2 \, ,</math>
\biggr\}^{1/3}  \, ,
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</div>
</div>
in which case,
 
we therefore have,
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 531: Line 553:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~r_S^2</math>
<math>~(2\cdot 3)^6 D</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 537: Line 559:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~( 1 - \Gamma^2 )^2-1 \, ,</math>
[ \Gamma^2 -1 ]^2+ 1 - [ 1-\Gamma^2 ]^2 = 1 \, ,
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
and,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\theta_S</math>
<math>~(2\cdot 3)^3S^3</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~\equiv</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\tan^{-1}\biggl\{ +\biggl[\frac{1 - [ 1-\Gamma^2 ]^2}{[ \Gamma^2 -1 ]^2} \biggr]^{1/2}\biggr\}
<math>~(2\cdot 3)^3 R + \sqrt{(2\cdot 3)^6D} </math>
= \cos^{-1}[1-\Gamma^2] \, .</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
And, just as before, the three roots are,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~r_S^{1/3}e^{i\theta_S/3}</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp;
<math>~\equiv</math>
  </td>
  <td align="center">
<math>~r_S^{1/3}e^{i(\theta_S/3 + 2\pi/3)}</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp;
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~r_S^{1/3}e^{i(\theta_S/3 + 4\pi/3)} \, .</math>
<math>~(1-\Gamma^2) + \sqrt{( 1  - \Gamma^2 )^2-1}</math>
   </td>
   </td>
</tr>
</tr>
</table>
 
</div>
<tr>
 
  <td align="right">
The first (and only real) root of our cubic equation is, therefore,
&nbsp;
<div align="center">
  </td>
<table border="0" cellpadding="5" align="center">
  <td align="center">
 
<math>~\equiv</math>
<tr>
  </td>
   <td align="right">
   <td align="left">
<math>~2\cdot 3 x_1</math>
<math>~(1-\Gamma^2) + i\sqrt{1-( 1  - \Gamma^2 )^2} \, ;</math>
   </td>
  </td>
   <td align="center">
</tr>
<math>~=</math>
 
   </td>
<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 [[User:Tohline/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>
 
===Analytically Prescribed Eigenvector===
====Our Notation====
As is explicitly defined in [[User:Tohline/Appendix/Ramblings/Azimuthal_Distortions#Figure1|Figure 1 of our accompanying detailed notes]], we have chosen to represent the spatial structure of an eigenfunction in the equatorial-plane of toroidal-like configurations via the expression,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\biggl[ \frac{\delta\rho}{\rho_0}\biggr]_\mathrm{spatial} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl\{ f_m(\varpi)e^{-im\phi_m} \biggr\}  \, .</math>
  </td>
</tr>
</table>
</div>
In general, we should assume that the function that delineates the radial dependence of the eigenfunction has both a real and an imaginary component, that is, we should assume that,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~f_m(\varpi)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\mathcal{A}(\varpi) + i\mathcal{B}(\varpi) \, ,</math>
  </td>
</tr>
</table>
</div>
in which case the square of the modulus of the function is,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~|f_m|^2 \equiv f_m \cdot f^*_m </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\mathcal{A}^2 + \mathcal{B}^2 \, .</math>
  </td>
</tr>
</table>
</div>
We can rewrite this complex function in the form,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~f_m(\varpi)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~|f_m|e^{-i[\alpha(\varpi) + \pi/2]} \, ,</math>
  </td>
</tr>
</table>
</div>
if the angle, <math>~\alpha(\varpi)</math> is defined such that,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\sin\alpha = \frac{\mathcal{A}}{\sqrt{\mathcal{A}^2 + \mathcal{B}^2}}</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp; &nbsp;
  </td>
  <td align="left">
<math>~\cos\alpha = \frac{\mathcal{B}}{\sqrt{\mathcal{A}^2 + \mathcal{B}^2}}</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\Rightarrow ~~~~ \alpha</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\tan^{-1}\biggl(\frac{\mathcal{A}}{\mathcal{B}}\biggr)
= \tan^{-1}\biggl[ \frac{\mathrm{Re}(f_m)}{\mathrm{Im}(f_m)} \biggr]
\, .</math>
  </td>
</tr>
</table>
</div>
 
Hence, the spatial structure of the eigenfunction can be rewritten as,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\biggl[ \frac{\delta\rho}{\rho_0}\biggr]_\mathrm{spatial} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~|f_m(\varpi)|e^{-i[\alpha(\varpi) + \pi/2+ m\phi_m]} \, . </math>
  </td>
</tr>
</table>
</div>
 
From this representation we can see that, at each radial location, <math>~\varpi</math>, the phase angle(s) at which the fractional perturbation exhibits its maximum amplitude, <math>~|f_m|</math>, is identified by setting the exponent of the exponential to zero.  That is,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\phi_m = \phi_\mathrm{max}(\varpi)</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~-\frac{1}{m}\biggl[\alpha(\varpi) +\frac{\pi}{2}\biggr]
= -\frac{1}{m}\biggl\{ \tan^{-1}\biggl[ \frac{\mathrm{Re}(f_m)}{\mathrm{Im}(f_m)} \biggr] +\frac{\pi}{2} \biggr\}
\, .</math>
  </td>
</tr>
</table>
</div>
 
An equatorial-plane plot of <math>~\phi_\mathrm{max}(\varpi)</math> should produce the "constant phase locus" referenced, for example, in recent papers from the [[User:Tohline/Appendix/Ramblings/To_Hadley_and_Imamura#Summary_for_Hadley_.26_Imamura|Imamura &amp; Hadley collaboration]].
 
<!-- SECOND ATTEMPT
This is the form that has been adopted broadly by the numerical simulation community, as graphical displays of <math>~f_m(\varpi)</math> and <math>~\phi_m(\varpi)</math> have been used to study the structure of unstable eigenmodes &#8212; see, for example, our discussion of [[User:Tohline/Appendix/Ramblings/To_Hadley_and_Imamura#Summary_for_Hadley_.26_Imamura|simulation results published by the Imamura &amp; Hadley collaboration]].  Multiplying this expression through by its complex conjugate gives the square of the modulus of the function.
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\biggl| \frac{\delta\rho}{\rho_0}\biggr|^2_\mathrm{spatial} \equiv
\biggl[ \frac{\delta\rho}{\rho_0}\biggr] \cdot \biggl[ \frac{\delta\rho}{\rho_0}\biggr]^*
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~f^2_m(\varpi)e^{-im[\phi_m(\varpi)]} \cdot e^{+im[\phi_m(\varpi)]} = f^2_m(\varpi) \, .</math>
  </td>
</tr>
</table>
</div>
We see, therefore, that written in this manner, <math>~f_m</math> is the modulus of the eigenfunction.
 
Alternatively, we could choose to omit explicit reference to an angular phase function and write the perturbation amplitude as a function with an imaginary as well as a real part, say,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\biggl[ \frac{\delta\rho}{\rho_0}\biggr]_\mathrm{spatial} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\mathcal{A}(\varpi) - i\mathcal{B}(\varpi)  \, .</math>
  </td>
</tr>
</table>
</div>
This is the form often used in research papers that seek to find analytic expressions for the structure of unstable eigenmodes, such as the works of [http://adsabs.harvard.edu/abs/1985MNRAS.216..553B Blaes85] and [http://adsabs.harvard.edu/abs/1986MNRAS.221..339G GGN86].  A mapping from one representation to the other is accomplished by, first, constructing the modulus of the complex perturbation amplitude and equating it to <math>~f_m</math>, that is,
 
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~f_m(\varpi) = \sqrt{\biggl[ \frac{\delta\rho}{\rho_0}\biggr] \cdot \biggl[ \frac{\delta\rho}{\rho_0}\biggr]^* }</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\sqrt{\mathcal{A}^2(\varpi) + \mathcal{B}^2(\varpi)} \, .</math>
  </td>
</tr>
</table>
</div>
Second, the phase function is obtained via the relation,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~m\phi_m(\varpi)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\tan^{-1}\biggl[ \frac{\mathcal{B}(\varpi)}{\mathcal{A}(\varpi)} \biggr] \, ,</math>
  </td>
</tr>
</table>
</div>
where, more thoroughly it must be understood that,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\cos(m\phi_m) = \biggl[ \frac{\mathcal{A}}{\sqrt{\mathcal{A}^2 + \mathcal{B}^2}} \biggr]</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp; &nbsp;
  </td>
  <td align="left">
<math>~\sin(m\phi_m) = \biggl[ \frac{\mathcal{B}}{\sqrt{\mathcal{A}^2 + \mathcal{B}^2}} \biggr] \, .</math>
  </td>
</tr>
</table>
</div>
 
 
To demonstrate that this is the proper mapping for the phase function, let's plug the expression for <math>~f_m</math> along with the expressions for <math>~\cos(m\phi_m)</math> and <math>~\sin(m\phi_m)</math> into our first relation, that is,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\biggl[ \frac{\delta\rho}{\rho_0}\biggr]_\mathrm{spatial} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~f_m(\varpi) \{ \cos[m\phi_m(\varpi)] - i\sin[m\phi_m(\varpi)] \}</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\sqrt{\mathcal{A}^2 + \mathcal{B}^2}\biggl\{ \biggl[ \frac{\mathcal{A}}{\sqrt{\mathcal{A}^2 + \mathcal{B}^2}} \biggr]
- i\biggl[ \frac{\mathcal{B}}{\sqrt{\mathcal{A}^2 + \mathcal{B}^2}} \biggr] \biggr\}</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \mathcal{A} - i\mathcal{B} \, .</math>
  </td>
</tr>
</table>
</div>
 
Q.E.D.
-->
<!-- USEFUL (BUT NOT FULLY CORRECT) MANIPULATION OF COMPLEX EIGENFUNCTION EXPRESSIONS ...
 
Recognizing that the leading factor, <math>~f_m</math>, is, in general, composed of both a real part and an imaginary part, we can rewrite this expression as,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\biggl[ \frac{\delta\rho}{\rho_0}\biggr]_\mathrm{spatial} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[\mathrm{Re}(f_m) + i\mathrm{Im}(f_m)\biggr] \biggl[\cos(m\phi_m) - i\sin(m\phi_m) \biggr]</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[\mathrm{Re}(f_m) \cos(m\phi_m) + \mathrm{Im}(f_m)\sin(m\phi_m)\biggr]
+ i \biggl[- \mathrm{Re}(f_m)\sin(m\phi_m) + \mathrm{Im}(f_m)\cos(m\phi_m)\biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\sqrt{[\mathrm{Re}(f_m)]^2 + [\mathrm{Im}(f_m)]^2} \biggl\{
\biggl[\frac{\mathrm{Re}(f_m) \cos(m\phi_m) + \mathrm{Im}(f_m)\sin(m\phi_m)}{\sqrt{[\mathrm{Re}(f_m)]^2 + [\mathrm{Im}(f_m)]^2} }\biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ i \biggl[\frac{- \mathrm{Re}(f_m)\sin(m\phi_m) + \mathrm{Im}(f_m)\cos(m\phi_m)}{\sqrt{[\mathrm{Re}(f_m)]^2 + [\mathrm{Im}(f_m)]^2} }\biggr]
\biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl| \frac{\delta\rho}{\rho_0}\biggr|_\mathrm{spatial} \biggl\{\sin(\beta_f + m\phi_m) + i \cos(\beta_f + m\phi_m)\biggr\} \, ,
</math>
  </td>
</tr>
</table>
</div>
 
where, the modulus of this function is,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\biggl| \frac{\delta\rho}{\rho_0}\biggr|_\mathrm{spatial} \equiv
\sqrt{\biggl[ \frac{\delta\rho}{\rho_0}\biggr]\cdot\biggl[ \frac{\delta\rho}{\rho_0}\biggr]^*}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \biggl\{
\biggl[\mathrm{Re}(f_m) \cos(m\phi_m) + \mathrm{Im}(f_m)\sin(m\phi_m)\biggr]^2
+ \biggl[- \mathrm{Re}(f_m)\sin(m\phi_m) + \mathrm{Im}(f_m)\cos(m\phi_m)\biggr]^2
\biggr\}^{1/2}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \biggl\{
[\mathrm{Re}(f_m) \cos(m\phi_m)]^2 + 2\mathrm{Re}(f_m)\mathrm{Im}(f_m)\sin(m\phi_m)\cos(m\phi_m)+ [\mathrm{Im}(f_m)\sin(m\phi_m)]^2
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ [\mathrm{Re}(f_m)\sin(m\phi_m)]^2 - 2\mathrm{Re}(f_m)\mathrm{Im}(f_m)\sin(m\phi_m)\cos(m\phi_m) + [\mathrm{Im}(f_m)\cos(m\phi_m)]^2 \biggr\}^{1/2}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \sqrt{
[\mathrm{Re}(f_m)]^2 + [\mathrm{Im}(f_m)]^2} \, ,
</math>
  </td>
</tr>
</table>
</div>
 
and where we have introduced a new angle, <math>~\beta_f</math>, such that,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\sin\beta_f = \frac{\mathrm{Re}(f_m)}{\sqrt{[\mathrm{Re}(f_m)]^2 + [\mathrm{Im}(f_m)]^2}}</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp;
  </td>
  <td align="left">
<math>~\cos\beta_f = \frac{\mathrm{Im}(f_m)}{\sqrt{[\mathrm{Re}(f_m)]^2 + [\mathrm{Im}(f_m)]^2}} </math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\Rightarrow ~~~~ \beta_f </math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\tan^{-1}\biggl[\frac{\mathrm{Re}(f_m)}{\mathrm{Im}(f_m)} \biggr] \, .</math>
  </td>
</tr>
</table>
</div>
 
Alternatively, we could choose to write,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\biggl[ \frac{\delta\rho}{\rho_0}\biggr]_\mathrm{spatial} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl| \frac{\delta\rho}{\rho_0}\biggr|_\mathrm{spatial} \biggl\{ \cos(\alpha_f - m\phi_m) + i \sin(\alpha_f - m\phi_m)\biggr\}
= \biggl| \frac{\delta\rho}{\rho_0}\biggr|_\mathrm{spatial} e^{i(\alpha_f -m\phi_m)} \, ,
</math>
  </td>
</tr>
</table>
</div>
where we have introduced a new angle, <math>~\alpha_f</math>, such that,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\cos\alpha_f = \frac{\mathrm{Re}(f_m)}{\sqrt{[\mathrm{Re}(f_m)]^2 + [\mathrm{Im}(f_m)]^2}}</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp;
  </td>
  <td align="left">
<math>~\sin\alpha_f = \frac{\mathrm{Im}(f_m)}{\sqrt{[\mathrm{Re}(f_m)]^2 + [\mathrm{Im}(f_m)]^2}}  </math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\Rightarrow ~~~~ \alpha_f </math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\tan^{-1}\biggl[\frac{\mathrm{Im}(f_m)}{\mathrm{Re}(f_m)} \biggr] \, .</math>
  </td>
</tr>
</table>
</div>
 
Notice that, when written in this form, it is clear from taking the complex conjugate of the function that,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\alpha_f - m\phi_m</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~0 </math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\Rightarrow ~~~~ m\phi_m</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\tan^{-1}\biggl[\frac{\mathrm{Im}(f_m)}{\mathrm{Re}(f_m)} \biggr] \, .</math>
  </td>
</tr>
</table>
</div>
 
FINISHED EXTRACTION OF COMPLEX FUNCTION MANIPULATION  -->
 
====General Formulation====
From my initial focused reading of the analysis presented by [http://adsabs.harvard.edu/abs/1985MNRAS.216..553B Blaes (1985)], I conclude that, in the infinitely slender torus case, unstable modes in PP tori exhibit eigenvectors of the form,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\frac{\delta W}{W_0} \equiv \biggl[ \frac{W(\eta,\theta)}{C} - 1 \biggr]e^{im\Omega_p t}e^{-y_2 (\Omega_0 t)} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl\{ f_m(\eta,\theta)e^{-i[m\phi_m + k\theta]} \biggr\}  \, ,</math>
  </td>
</tr>
</table>
</div>
 
<!-- [<font color="red"><b>NOTE:</b></font>  Initially, I wrote "+ k" rather than "- k" in the exponent of the exponential term on the RHS of this expression; but experience shows that "- k" is required to achieve proper behavior of the "constant phase locus" plot, as displayed below.]  -->
where we have written the perturbation amplitude in a manner that conforms with the notation that we have used in  [[User:Tohline/Appendix/Ramblings/Azimuthal_Distortions#Figure1|Figure 1 of a related, but more general discussion]].  As is summarized in &sect;1.3 of Blaes (1985), for "thick" (but, actually, still quite thin) tori, "exactly one exponentially growing mode exists for each value of the azimuthal wavenumber <math>~m</math>," and its complex amplitude takes the following form (see his equation 1.10):
 
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~f_m(\eta,\theta)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
~\beta^2 m^2 \biggl[ 2\eta^2 \cos^2\theta - \frac{3\eta^2}{4(n+1)} - \frac{(4n+1)}{4(n+1)^2}
\pm 4i\biggl(\frac{3}{2n+2}\biggr)^{1/2} \eta\cos\theta\biggr]
+ \mathcal{O}(\beta^3) \, .
</math>
  </td>
</tr>
</table>
</div>
 
Aside from an arbitrary leading scale factor, we should therefore find that the amplitude (modulus) of the enthalpy perturbation is,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\biggl|\frac{\delta W}{W_0} \biggr|</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\sqrt{[\mathrm{Re}(f_m)]^2+ [\mathrm{Im}(f_m)]^2} \, ;</math>
  </td>
</tr>
</table>
</div>
and the associated phase function is,
<!--
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~m\phi_m - k\theta</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\tan^{-1} \biggl\{ \frac{\mathrm{Re}(f_m)}{\mathrm{Im}(f_m)} \biggr\} \, .</math>
  </td>
</tr>
</table>
</div>
[<font color="red"><b>NOTE:</b></font>  Initially, I expected the argument inside the arctan function to be the ratio, <math>~\mathrm{Im}(f_m)/\mathrm{Re}(f_m)</math>; but experience shows that the reciprocal of this ratio is required to achieve proper behavior of the "constant phase locus" plot, as displayed below.]
-->
 
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~m\phi_\mathrm{max}(\varpi)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-\biggl\{ \tan^{-1}\biggl[ \frac{\mathrm{Re}(f_m)}{\mathrm{Im}(f_m)} \biggr] +\frac{\pi}{2} +k\theta \biggr\}
\, .</math>
  </td>
</tr>
</table>
</div>
 
Now, keeping in mind that, for the time being, we are only interested in examining the shape of the unstable eigenvector in the ''equatorial plane'' of the torus, we can set,
<div align="center">
<math>~\cos\theta ~~ \rightarrow ~~ \pm 1 \, .</math>
</div>
Hence, we have,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\frac{1}{\beta^4 m^4}\biggl|\frac{\delta W}{W_0} \biggr|^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[2\eta^2  - \frac{3\eta^2}{4(n+1)} - \frac{(4n+1)}{4(n+1)^2}\biggr]^2
+ 16\biggl[\biggl(\frac{3}{2n+2}\biggr)^{1/2} \eta \biggr]^2 </math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{[2^2(n+1)^2]^2}\biggl[2^3(n+1)^2\eta^2  - 3(n+1)\eta^2 - (4n+1) \biggr]^2
+ \frac{2^3 \cdot 3\eta^2}{(n+1)}  </math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{[2^2(n+1)^2]^2}\biggl[(4n+1) - (n+1)[2^3(n+1)-3]\eta^2  \biggr]^2
+ \frac{2^3 \cdot 3\eta^2}{(n+1)}  </math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow ~~~~ \biggl[\frac{2(n+1)}{\beta m} \biggr]^4 \biggl|\frac{\delta W}{W_0} \biggr|^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[(4n+1) - (n+1)[2^3(n+1)-3]\eta^2  \biggr]^2
+ 2^7 \cdot 3(n+1)^3\eta^2 \, .  </math>
  </td>
</tr>
</table>
</div>
Also, keeping in mind that, because of the <math>~\cos\theta</math> factor, the sign on the imaginary term flips its sign when switching from the "inner" region to the "outer" region of the torus,
 
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="center" bgcolor="blue">&nbsp;</td>
  <td align="right">
<math>~m\phi_\mathrm{max}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- \biggl\{ \tan^{-1}\biggl[
\frac{(4n+1) - (n+1)[2^3(n+1)-3]\eta^2  }{2^{7/2}\cdot 3^{1/2}(n+1)^{3/2}\eta}
\biggr]+\frac{\pi}{2}\biggr\}</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; &nbsp; over &nbsp; &nbsp; &nbsp; &nbsp;
  </td>
  <td align="left">
inner <math>~(\theta=0)</math> region of the torus;
  </td>
</tr>
<tr><td colspan="6" align="center">while</td></tr>
<tr>
  <td align="center" bgcolor="green">&nbsp;</td>
  <td align="right">
<math>~m\phi_\mathrm{max}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- \biggl\{ \tan^{-1}\biggl[- 
\frac{(4n+1) - (n+1)[2^3(n+1)-3]\eta^2  }{2^{7/2}\cdot 3^{1/2}(n+1)^{3/2}\eta}\biggr\}  +\frac{3\pi}{2}
\biggr\}</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; &nbsp; over &nbsp; &nbsp; &nbsp; &nbsp;
  </td>
  <td align="left">
outer <math>~(\theta=\pi)</math> region of torus.
  </td>
</tr>
</table>
</div>
 
====Incompressible Slim Tori====
 
If we specifically consider geometrically slim, incompressible tori &#8212; that is, if we set the polytropic index, <math>~n=0</math> &#8212; to lowest order the eigenfunction derived by [http://adsabs.harvard.edu/abs/1985MNRAS.216..553B Blaes (1985)] takes the form,
 
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~f_m(\eta,\theta)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
~\beta^2 m^2 \biggl[ 2\eta^2 \cos^2\theta - \frac{3\eta^2}{4} - \frac{1}{4}
\pm 4i\biggl(\frac{3}{2}\biggr)^{1/2} \eta\cos\theta\biggr]
+ \cancelto{0}{\mathcal{O}(\beta^3)} \, .
</math>
  </td>
</tr>
</table>
</div>
 
====Check Validity of Blaes85 Eigenvector====
 
=====Step 1=====
Equation (2.6) of Blaes85 states that,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\beta^2 \eta^2 = [x^2 + x^3(3\cos\theta - \cos^3\theta) ]</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; &nbsp;
<math>~\Rightarrow</math>
&nbsp; &nbsp; &nbsp; &nbsp;
  </td>
  <td align="left">
<math>~
\beta^2(1 - \eta^2) = [ \beta^2 - x^2 - x^3(3\cos\theta - \cos^3\theta) ]
\, .</math>
  </td>
</tr>
</table>
</div>
 
This means that,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\frac{\partial \eta^2}{\partial x}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{\beta^2}\biggl[ 
2x +3x^2(3\cos\theta - \cos^3\theta)
\biggr] \, ;
</math>
  </td>
</tr>
</table>
</div>
and,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\frac{\partial \eta^2}{\partial\theta}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{x^3}{\beta^2}\biggl[
-3\sin\theta + 3\sin\theta \cos^2\theta
\biggr] </math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{3x^3 \sin\theta}{\beta^2}\biggl[\cos^2\theta -1\biggr] \, .</math>
  </td>
</tr>
</table>
</div>
 
Furthermore,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\frac{\partial \eta}{\partial x}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{2\beta}\biggl[x^2 + x^3(3\cos\theta - \cos^3\theta)\biggr]^{-1/2}\biggl[ 
2x +3x^2(3\cos\theta - \cos^3\theta)
\biggr] \, ;
</math>
  </td>
</tr>
</table>
</div>
and,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\frac{\partial \eta}{\partial\theta}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{2\beta}\biggl[x^2 + x^3(3\cos\theta - \cos^3\theta)\biggr]^{-1/2}\biggl[
-3\sin\theta + 3\sin\theta \cos^2\theta
\biggr] \, .</math>
  </td>
</tr>
</table>
</div>
 
=====Step 2=====
Equation (4.14) of Blaes85 states that,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\nu + m</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\pm im\biggl[\frac{3}{2(n+1)}\biggr]^{1/2} \beta \, ;
</math>
  </td>
</tr>
</table>
</div>
 
and equation (4.13) of Blaes85 states that,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\frac{\delta W}{A_{00}}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~1 + \beta^2 m^2\biggl[
2\eta^2 \cos^2\theta - \frac{3\eta^2}{4(n+1)} - \frac{(4n+1)}{4(n+1)^2} \pm 4i\biggl( \frac{3}{2n+2} \biggr)^{1/2} \eta\cos\theta
\biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\Rightarrow~~~~\frac{1}{m^2}\biggl[\frac{\delta W}{A_{00}}-1\biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\beta^2 \biggl\{- \frac{(4n+1)}{4(n+1)^2} +
\eta^2\biggl[ 2\cos^2\theta - \frac{3}{4(n+1)}\biggr] \pm i\biggl( \frac{2^3\cdot 3}{n+1} \biggr)^{1/2} \eta\cos\theta
\biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\beta^2}{2^2(n+1)^2} \biggl\{- (4n+1) +
\eta^2 [ 2^3(n+1)^2\cos^2\theta - 3(n+1)] \pm i ~[ 2^7\cdot 3(n+1)^3 ]^{1/2} \eta\cos\theta
\biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\Rightarrow~~~~\Lambda \equiv \frac{2^2(n+1)^2}{m^2}\biggl[\frac{\delta W}{A_{00}}-1\biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- (4n+1)\beta^2 +
[ 2^3(n+1)^2\cos^2\theta - 3(n+1)] \cdot [x^2 + x^3(3\cos\theta - \cos^3\theta)]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~\pm ~i~\beta [ 2^7\cdot 3(n+1)^3 ]^{1/2} [x^2\cos^2\theta + x^3(3\cos^3\theta - \cos^5\theta) ]^{1/2}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- (4n+1)\beta^2 +
(n+1)x^2[ 2^3(n+1)\cos^2\theta - 3] \cdot [1 + xb] ~\pm ~i~\beta [ 2^7\cdot 3(n+1)^3 ]^{1/2} x\cos\theta [1 + x b ]^{1/2}
</math>
  </td>
</tr>
</table>
</div>
 
<!-- SEPARATE EVALUATION OF LAMBDA -->
<div align="center">
<table border="1" cellpadding="8" align="center">
<tr><td align="center" bgcolor="purple"><font color="white">'''Through a separate white-board derivation I have obtained &hellip;'''</font></td></tr>
<tr><td align="center">
 
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\Lambda </math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
\frac{2^2(n+1)^2}{m^2}\biggl[\frac{\delta W}{A_{00}}-1\biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-(4n+1)\beta^2 + (\beta\eta)^2(n+1)[2^3(n+1)\cos^2\theta - 3] ~~~\pm ~~i~\beta\cos\theta [2^7\cdot 3 (n+1)^3]^{1/2} (\beta\eta) \, ,
</math>
  </td>
</tr>
 
<tr><td colspan="3" align="left">where,</td></tr>
 
<tr>
  <td align="right">
<math>~(\beta\eta)^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~x^2(1+xb) \, ,</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~b</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~3\cos\theta - \cos^3\theta \, .</math>
  </td>
</tr>
 
</table>
 
</td></tr></table>
</div>
 
Note that, differentiating the left-hand-side with respect to either coordinate <math>~(x</math> or <math>~\theta)</math> gives,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\frac{\partial \Lambda}{\partial x_i} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2^2(n+1)^2}{m^2}\cdot \frac{\partial }{\partial x_i} \biggl(\frac{\delta W}{A_{00}} \biggr) </math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\Rightarrow~~~~\frac{\partial }{\partial x_i} \biggl(\frac{\delta W}{A_{00}} \biggr)  </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[\frac{m}{2(n+1)}\biggr]^2 \cdot \frac{\partial \Lambda}{\partial x_i} \, .</math>
  </td>
</tr>
</table>
</div>
 
 
Given that <math>~\nu</math> has both real and imaginary parts, presumably,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\nu^2 \equiv \nu \cdot \nu^*</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl\{ -m \pm im\biggl[\frac{3}{2(n+1)}\biggr]^{1/2} \beta\biggr\} \biggl\{ -m \mp im\biggl[\frac{3}{2(n+1)}\biggr]^{1/2} \beta\biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~m^2 + m^2\biggl[\frac{3}{2(n+1)}\biggr]\beta^2 \, .
</math>
  </td>
</tr>
</table>
</div>
 
For later reference, let's take the relevant partial derivatives of the function, <math>~\Lambda(x,\theta)</math>.  Adopting the shorthand notation,
<div align="center">
<math>~b \equiv (3\cos\theta-\cos^3\theta) \, ,</math>
</div>
 
we have,
 
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\frac{\partial \Lambda}{\partial x}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~[ 2^3(n+1)^2\cos^2\theta - 3(n+1)] \cdot \frac{\partial }{\partial x} \biggl\{
[x^2 + x^3b]
\biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~\pm ~i~\beta [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cos\theta \cdot \frac{\partial }{\partial x} \biggl\{ [x^2 + x^3b ]^{1/2} 
\biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~[ 2^3(n+1)^2\cos^2\theta - 3(n+1)] \cdot [2x + 3x^2b]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~\pm ~i~\beta [ 2^5\cdot 3(n+1)^3 ]^{1/2} \cos\theta \cdot [1 + xb ]^{-1/2} \cdot [2 + 3xb ] 
</math>
  </td>
</tr>
</table>
</div>
 
<!-- SEPARATE EVALUATION OF \partial\Lambda/\partial x -->
<div align="center">
<table border="1" cellpadding="8" align="center">
<tr><td align="center" bgcolor="purple"><font color="white">'''Through a separate white-board derivation I have obtained &hellip;'''</font></td></tr>
<tr><td align="center">
 
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\frac{\partial\Lambda}{\partial x}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(n+1)[2^3(n+1)\cos^2\theta -3]x(2+3xb) ~~~\pm ~~i~\beta\cos\theta [2^5\cdot 3 (n+1)^3]^{1/2} \cdot \frac{x(2+3xb)}{(\beta\eta)} \, ,
</math>
  </td>
</tr>
 
<tr><td colspan="3" align="left">which is the same.</td></tr>
 
</table>
 
</td></tr></table>
</div>
 
 
 
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\frac{\partial^2 \Lambda}{\partial x^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~[ 2^3(n+1)^2\cos^2\theta - 3(n+1)] \cdot \frac{\partial }{\partial x}[2x + 3x^2b]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~\pm ~i~\beta [ 2^5\cdot 3(n+1)^3 ]^{1/2} \cos\theta \cdot \frac{\partial }{\partial x}\biggl\{[1 + xb ]^{-1/2} \cdot [2 + 3xb ]  \biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~[ 2^3(n+1)^2\cos^2\theta - 3(n+1)] \cdot [2 + 6xb]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~\pm ~i~\beta [ 2^5\cdot 3(n+1)^3 ]^{1/2} \cos\theta \cdot \biggl\{
-\tfrac{b}{2}[1 + xb ]^{-3/2} \cdot [2 + 3xb ]  + [1 + xb ]^{-1/2} \cdot [3b ]
\biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2[ 2^3(n+1)^2\cos^2\theta - 3(n+1)] \cdot [1 + 3xb]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~\pm ~i~\beta [ 2^5\cdot 3(n+1)^3 ]^{1/2} \cos\theta \cdot \biggl\{
- [2b + 3xb^2 ]  + [6b + 6xb^2 ]
\biggr\} \frac{1}{2(1+xb)^{3/2}}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2[ 2^3(n+1)^2\cos^2\theta - 3(n+1)] \cdot [1 + 3xb]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~\pm ~i~\beta [ 2^3\cdot 3(n+1)^3 ]^{1/2} \cos\theta \cdot b[1 + xb ]^{-3/2} [4 + 3xb ] 
</math>
  </td>
</tr>
</table>
</div>
 
 
<!-- SEPARATE EVALUATION OF 2nd DERIVATIVE wrt "x" -->
<div align="center">
<table border="1" cellpadding="8" align="center">
<tr><td align="center" bgcolor="purple"><font color="white">'''Through a separate white-board derivation I have obtained &hellip;'''</font></td></tr>
<tr><td align="center">
 
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\frac{\partial^2\Lambda}{\partial x^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2(n+1)[2^3(n+1)\cos^2\theta -3](1+3xb) ~\pm~~i~\beta\cos\theta [2^3\cdot 3(n+1)^3]^{1/2} \biggl[ \frac{b(4+3xb)}{(1+xb)^{3/2}} \biggr] \, ,
</math>
  </td>
</tr>
 
<tr><td colspan="3" align="left">which is the same.</td></tr>
 
</table>
 
</td></tr></table>
</div>
 
 
 
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\frac{\partial \Lambda}{\partial \theta}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\partial }{\partial \theta} \biggl\{
[ 2^3(n+1)^2\cos^2\theta - 3(n+1)] \cdot [x^2 + x^3(3\cos\theta - \cos^3\theta)]
\biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~\pm ~i~\frac{\partial }{\partial \theta} \biggl\{ \beta [ 2^7\cdot 3(n+1)^3 ]^{1/2} [x^2\cos^2\theta + x^3(3\cos^3\theta - \cos^5\theta) ]^{1/2} 
\biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
[ 2^3(n+1)^2\cos^2\theta - 3(n+1)] \cdot x^3 \cdot  [\sin\theta (-3 + 3\cos^2\theta)]
+ [x^2 + x^3(3\cos\theta - \cos^3\theta)] \cdot [ -2^4(n+1)^2 \sin\theta \cos\theta ]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
~\pm ~i~\beta [ 2^5\cdot 3(n+1)^3 ]^{1/2} \cdot [x^2\cos^2\theta + x^3(3\cos^3\theta - \cos^5\theta) ]^{-1/2}  \cdot
(-\sin\theta) [2x^2\cos\theta + x^3(9\cos^2\theta - 5\cos^4\theta) ]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-3 x^3 \sin^3\theta [ 2^3(n+1)^2\cos^2\theta - 3(n+1)] 
-2^4(n+1)^2 x^2\sin\theta \cos\theta [1 + x(3\cos\theta-\cos^3\theta) ]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
~\pm ~i~\beta (-1) [ 2^5\cdot 3(n+1)^3 ]^{1/2} \cdot [1 + x(3\cos\theta - \cos^3\theta) ]^{-1/2}  \cdot
x~\sin\theta [2 + x(9\cos\theta - 5\cos^3\theta) ]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-3 x^3 \sin^3\theta [ 2^3(n+1)^2\cos^2\theta - 3(n+1)] 
-2^4(n+1)^2 x^2\sin\theta \cos\theta [1 + xb ]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
~\pm ~i~\beta (-1) [ 2^5\cdot 3(n+1)^3 ]^{1/2} \cdot [1 + xb ]^{-1/2}  \cdot
x~\sin\theta [2 + x(9\cos\theta - 5\cos^3\theta) ]
</math>
  </td>
</tr>
</table>
</div>
 
 
<!-- SEPARATE EVALUATION OF \partial\Lambda/\partial \theta -->
<div align="center">
<table border="1" cellpadding="8" align="center">
<tr><td align="center" bgcolor="purple"><font color="white">'''Through a separate white-board derivation I have obtained &hellip;'''</font></td></tr>
<tr><td align="center">
 
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\frac{\partial\Lambda}{\partial \theta}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
9(n+1)x^3\sin^3\theta - 2^3\cdot 3(n+1)^2x^3\sin^3\theta \cos^2\theta -2^4 (n+1)^2 (\beta\eta)^2 \sin\theta\cos\theta
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
\pm~~i~(-1)\beta [2^7\cdot 3 (n+1)^3 ]^{1/2}\biggl\{ (\beta\eta)\sin\theta +\frac{3x^3}{2}\cdot\biggl[ \frac{\sin^3\theta \cos\theta}{(\beta\eta)} \biggr]\biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(n+1)\sin\theta \biggl\{
-2^4 (n+1) (\beta\eta)^2 \cos\theta + 3x^3 \sin^2\theta \biggl[3 - 2^3(n+1)\cos^2\theta \biggr]
\biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
\pm~~i~(-1)\beta \sin\theta [2^7\cdot 3 (n+1)^3 (\beta\eta)^2]^{1/2}\biggl\{ 1 +\frac{3x^3}{2}\cdot\biggl[ \frac{\sin^2\theta \cos\theta}{(\beta\eta)^2} \biggr]\biggr\} \, ,
</math>
  </td>
</tr>
 
<tr><td colspan="3" align="left">which is the same.</td></tr>
 
</table>
 
</td></tr></table>
</div>
 
 
 
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\frac{\partial^2 \Lambda}{\partial \theta^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-3 x^3 \frac{\partial }{\partial \theta} \cdot \biggl\{\sin^3\theta [ 2^3(n+1)^2\cos^2\theta - 3(n+1)]  \biggr\}
-2^4(n+1)^2 x^2 \cdot \frac{\partial }{\partial \theta} \biggl\{\sin\theta \cos\theta [1 + x(3\cos\theta-\cos^3\theta) ] \biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
~\pm ~i~\beta (-1)x [ 2^5\cdot 3(n+1)^3 ]^{1/2} \cdot \frac{\partial }{\partial \theta} \biggl\{
\sin\theta [2 + x(9\cos\theta - 5\cos^3\theta) ] \cdot [1 + x(3\cos\theta - \cos^3\theta) ]^{-1/2} 
\biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-3 x^3 \cdot \biggl\{3\sin^2\theta \cos\theta [ 2^3(n+1)^2\cos^2\theta - 3(n+1)] - \sin^3\theta [ 2^3(n+1)^2 \cdot 2\sin\theta\cos\theta ]  \biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
-2^4(n+1)^2 x^2 \cdot
\biggl\{  (\cos^2\theta - \sin^2\theta ) [1 + x(3\cos\theta-\cos^3\theta) ] - 3x\sin^4\theta \cos\theta
\biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
~\pm ~i~\beta (-1)x [ 2^5\cdot 3(n+1)^3 ]^{1/2} \cdot \biggl\{
\cos\theta \cdot [2 + x(9\cos\theta - 5\cos^3\theta) ] \cdot [1 + x(3\cos\theta - \cos^3\theta) ]^{-1/2} 
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
-  3x\sin^2\theta [1 + x(3\cos\theta - \cos^3\theta) ]^{-1/2} \cdot (3 - 5\cos^2\theta)
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
+ \frac{3}{2} \cdot x\sin^4\theta [2 + x(9\cos\theta - 5\cos^3\theta) ] \cdot  [1 + x(3\cos\theta - \cos^3\theta) ]^{-3/2}
\biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-3 x^3 \cdot \biggl\{3\sin^2\theta \cos\theta [ 2^3(n+1)^2\cos^2\theta - 3(n+1)] - \sin^3\theta [ 2^3(n+1)^2 \cdot 2\sin\theta\cos\theta ]  \biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
-2^4(n+1)^2 x^2 \cdot
\biggl\{  (1 - 2\sin^2\theta )  + x (3\cos\theta -\cos^3\theta - 6\sin^2\theta\cos\theta + 2\sin^2\theta\cos^3\theta  - 3\sin^4\theta \cos\theta )
\biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
~\pm ~i~\beta (-1)x [ 2^5\cdot 3(n+1)^3 ]^{1/2} \cdot [1 + xb ]^{-3/2} \cdot \biggl\{
\cos\theta \cdot [2 + x(9\cos\theta - 5\cos^3\theta) ] \cdot [1 + xb ] 
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
-  3x\sin^2\theta [1 + xb ] \cdot (3 - 5\cos^2\theta)
+ \frac{3}{2} \cdot x\sin^4\theta [2 + x(9\cos\theta - 5\cos^3\theta) ] 
\biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-2^4(n+1)^2 x^2 (1 - 2\sin^2\theta ) 
- x^3 \cdot \biggl\{ [ 2^3\cdot 3^2(n+1)^2\sin^2\theta \cos^3\theta - 3^3(n+1)\sin^2\theta \cos\theta] - 2^4\cdot 3(n+1)^2 \sin^4\theta\cos\theta 
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ 2^4(n+1)^2 b + 2^4(n+1)^2 \cdot( - 6\sin^2\theta\cos\theta + 2\sin^2\theta\cos^3\theta  - 3\sin^4\theta \cos\theta )\biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
~\pm ~i~\beta (-1)x [ 2^3\cdot 3(n+1)^3 ]^{1/2} \cdot [1 + xb ]^{-3/2} \cdot \biggl\{
2\cos\theta \cdot [2 + x(15\cos\theta - 7\cos^3\theta) + x^2 b(9\cos\theta - 5\cos^3\theta)]   
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
-  6x\sin^2\theta (1 + xb ) \cdot (3 - 5\cos^2\theta)
+ 3 \cdot x\sin^4\theta [2 + x(9\cos\theta - 5\cos^3\theta) ] 
\biggr\}
</math>
  </td>
</tr>
</table>
</div>
 
 
<!-- SEPARATE EVALUATION OF 2nd DERIVATIVE wrt "theta" -->
<div align="center">
<table border="1" cellpadding="8" align="center">
<tr><td align="center" bgcolor="purple"><font color="white">'''Through a separate white-board derivation I have obtained &hellip;'''</font></td></tr>
<tr><td align="center">
 
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\frac{\partial^2\Lambda}{\partial \theta^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
x^2 \biggl\{2^4(n+1)^2(\sin^2\theta - \cos^2\theta)
\biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ x^3\biggl\{-2^4\cdot 3 (n+1)^2\cos^3\theta + 2^4(n+1)^2\cos^5\theta + 3^2(n+1)(16n +19)\sin^2\theta \cos\theta -2^3\cdot 23 (n+1)^2\sin^2\theta \cos^3\theta
\biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
\pm~~i~(-1)\beta [2^7\cdot 3 (n+1)^3 ]^{1/2} \biggl\{
(\beta\eta)\cos\theta + \frac{3x^3\sin^2\theta}{2(\beta\eta)}(5\cos^2\theta -2) + \frac{3^2x^6\sin^6\theta\cos\theta}{2^2(\beta\eta)^3}
\biggr\} \, .
</math>
  </td>
</tr>
 
</table>
 
</td></tr></table>
</div>
 
=====Step 3=====
From our [[User:Tohline/Apps/PapaloizouPringle84#isolatingBlaes85|accompanying discussion of the Blaes85 derivation]], we expect the following equality to hold (see his equations 4.1 and 4.2):
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\hat{L} (\delta W)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-2n(1-\Theta_H)(M\nu^2 + N\nu)(\delta W) \, ,</math>
  </td>
</tr>
</table>
</div>
where,
 
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\hat{L} (\delta W)</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
\Theta_H x^2\cdot \frac{\partial^2(\delta W)}{\partial x^2}
+\Theta_H \cdot \frac{\partial^2(\delta W)}{\partial\theta^2}
+ \biggl\{\Theta_H x \biggl[\frac{1-2x \cos\theta}{ 1-x\cos\theta}\biggr] + nx^2 \cdot \frac{\partial\Theta_H}{\partial x}
\biggr\} \cdot \frac{\partial (\delta W)}{\partial x}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \biggl[\frac{\Theta_Hx\sin\theta}{ (1-x\cos\theta) } +
n\cdot \frac{\partial \Theta_H}{\partial\theta}  \biggr]
\cdot \frac{\partial (\delta W)}{\partial\theta}
+ \biggl[ \frac{2n x^2 m^2}{\beta^2(1-x\cos\theta)^4}  -  \frac{m^2 x^2\Theta_H}{(1-x\cos\theta)^2} \biggr]\delta W \, ,
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~M</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\frac{x^2}{(1-\Theta_H)\beta^2} \, ,</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~N</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\frac{2mx^2}{(1-\Theta_H)\beta^2(1-x\cos\theta)^2} \, .</math>
  </td>
</tr>
 
</table>
</div>
 
Immediately evaluating the right-hand-side (RHS) of the equality, we have,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\biggl[\frac{2(n+1)}{m}\biggr]^2 \frac{RHS}{A_{00}}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-2n(1-\Theta_H)(M\nu^2 + N\nu)\cdot \biggl[\frac{2(n+1)}{m}\biggr]^2 \frac{\delta W}{A_{00}}</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-\frac{2nx^2}{\beta^2}\biggl[
\nu^2 +  \frac{2m\nu}{(1-x\cos\theta)^2}
\biggr] \cdot \biggl[\frac{2(n+1)}{m}\biggr]^2 \frac{\delta W}{A_{00}}</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-\frac{2nx^2}{\beta^2}\biggl[
\nu^2 +  \frac{2m\nu}{(1-x\cos\theta)^2}
\biggr] \cdot \biggl\{\Lambda + \biggl[ \frac{2(n+1)}{m} \biggr]^2  \biggr\} \, .</math>
  </td>
</tr>
</table>
</div>
 
And the similarly modified LHS is:
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\biggl[\frac{2(n+1)}{m}\biggr]^2 \frac{LHS}{A_{00}}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\Theta_H x^2\cdot \frac{\partial^2 \Lambda}{\partial x^2}
+\Theta_H \cdot \frac{\partial^2 \Lambda}{\partial\theta^2}
+ \biggl\{\Theta_H x \biggl[\frac{1-2x \cos\theta}{ 1-x\cos\theta}\biggr] + nx^2 \cdot \frac{\partial\Theta_H}{\partial x}
\biggr\} \cdot \frac{\partial  \Lambda }{\partial x}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \biggl[\frac{\Theta_Hx\sin\theta}{ (1-x\cos\theta) } +
n\cdot \frac{\partial \Theta_H}{\partial\theta}  \biggr]
\cdot \frac{\partial\Lambda}{\partial\theta}
+ \biggl[ \frac{2n x^2 m^2}{\beta^2(1-x\cos\theta)^4}  -  \frac{m^2 x^2\Theta_H}{(1-x\cos\theta)^2} \biggr]
\cdot \biggl[\frac{2(n+1)}{m}\biggr]^2 \frac{\delta W}{A_{00}}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(1-\eta^2) x^2\cdot \frac{\partial^2 \Lambda}{\partial x^2}
+(1-\eta^2) \cdot \frac{\partial^2 \Lambda}{\partial\theta^2}
+ \biggl\{(1-\eta^2) x \biggl[\frac{1-2x \cos\theta}{ 1-x\cos\theta}\biggr] + nx^2 \cdot \frac{\partial (1-\eta^2) }{\partial x}
\biggr\} \cdot \frac{\partial  \Lambda }{\partial x}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \biggl[\frac{(1-\eta^2) x\sin\theta}{ (1-x\cos\theta) } +
n\cdot \frac{\partial (1-\eta^2) }{\partial\theta}  \biggr]
\cdot \frac{\partial\Lambda}{\partial\theta}
+ \biggl[ \frac{2n x^2 m^2}{\beta^2(1-x\cos\theta)^4}  -  \frac{m^2 x^2(1-\eta^2) }{(1-x\cos\theta)^2} \biggr]
\cdot \biggl\{\Lambda + \biggl[ \frac{2(n+1)}{m} \biggr]^2  \biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(1-\eta^2) x^2\cdot \frac{\partial^2 \Lambda}{\partial x^2}
+(1-\eta^2) \cdot \frac{\partial^2 \Lambda}{\partial\theta^2}
+ \biggl\{(1-\eta^2) x \biggl[\frac{1-2x \cos\theta}{ 1-x\cos\theta}\biggr] - nx^2 \cdot \frac{\partial \eta^2 }{\partial x}
\biggr\} \cdot \frac{\partial  \Lambda }{\partial x}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \biggl[\frac{(1-\eta^2) x\sin\theta}{ (1-x\cos\theta) } -
n\cdot \frac{\partial \eta^2 }{\partial\theta}  \biggr]
\cdot \frac{\partial\Lambda}{\partial\theta}
+ \biggl[ \frac{2n x^2 m^2}{\beta^2(1-x\cos\theta)^4}  -  \frac{m^2 x^2(1-\eta^2) }{(1-x\cos\theta)^2} \biggr]
\cdot \biggl\{\Lambda + \biggl[ \frac{2(n+1)}{m} \biggr]^2  \biggr\}
</math>
  </td>
</tr>
</table>
</div>
 
Now multiply both sides by &hellip;
<div align="center">
<math>~\beta^2 m^2 (1-x\cos\theta)^4 \, .</math>
</div>
 
 
We have,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~m^2\mathcal{L}_{RHS} \equiv \beta^2 (1-x\cos\theta)^4 \biggl[2(n+1)\biggr]^2 \frac{RHS}{A_{00}}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- 2n m^2 x^2(1-x\cos\theta)^4  \biggl[
\nu^2 +  \frac{2m\nu}{(1-x\cos\theta)^2}
\biggr] \cdot \biggl\{\Lambda + \biggl[ \frac{2(n+1)}{m} \biggr]^2  \biggr\} </math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- 2n x^2 (1-x\cos\theta)^2  \biggl[ (1-x\cos\theta)^2 \nu^2 +  (2m\nu)
\biggr] \cdot \{m^2 \Lambda + [ 2(n+1) ]^2  \} </math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- 2n x^2 (1-x\cos\theta)^2 \{m^2 \Lambda + [ 2(n+1) ]^2  \}  \biggl\{
m^2 (1-x\cos\theta)^2 \biggl[ 1+\frac{3\beta^2}{2(n+1)} \biggr]
+  2m^2 \biggl[  -1 ~\pm~ i~\biggl[ \frac{3\beta^2}{2(n+1)} \biggr]^{1/2} \biggr]
\biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \frac{n m^2 x^2}{(n+1)} \cdot  (1-x\cos\theta)^2\{m^2 \Lambda + [ 2(n+1) ]^2  \} \biggl\{ (1-x\cos\theta)^2 [ 2(n+1) + 3\beta^2 ]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
-  4 (n+1)  \pm i~ [2^3\cdot 3 (n+1)\beta^2 ]^{1/2} \biggr\} \, .
</math>
  </td>
</tr>
</table>
</div>
 
And,
 
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~m^2\mathcal{L}_{LHS} \equiv \beta^2 (1-x\cos\theta)^4 \biggl[2(n+1)\biggr]^2 \frac{LHS}{A_{00}}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\beta^2 m^2 (1-x\cos\theta)^4 (1-\eta^2) x^2\cdot \frac{\partial^2 \Lambda}{\partial x^2}
+\beta^2 m^2 (1-x\cos\theta)^4 (1-\eta^2) \cdot \frac{\partial^2 \Lambda}{\partial\theta^2}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \beta^2 m^2 (1-x\cos\theta)^3 \biggl[ (1-\eta^2) x (1-2x \cos\theta ) - nx^2  (1-x\cos\theta)\cdot \frac{\partial \eta^2 }{\partial x}
\biggr] \cdot \frac{\partial  \Lambda }{\partial x}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \beta^2 m^2 (1-x\cos\theta)^3 \biggl[(1-\eta^2) x\sin\theta -
n (1-x\cos\theta)\cdot \frac{\partial \eta^2 }{\partial\theta}  \biggr]
\cdot \frac{\partial\Lambda}{\partial\theta}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ m^2 \biggl[ 2n x^2  -  (1-x\cos\theta)^2 \beta^2 x^2(1-\eta^2) \biggr]
\cdot \biggl\{m^2 \Lambda + \biggl[ 2(n+1) \biggr]^2  \biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
m^2 (1-x\cos\theta)^4 [ \beta^2 - x^2 - x^3(3\cos\theta - \cos^3\theta) ] x^2\cdot \frac{\partial^2 \Lambda}{\partial x^2}
+m^2 (1-x\cos\theta)^4 [ \beta^2 - x^2 - x^3(3\cos\theta - \cos^3\theta) ]\cdot \frac{\partial^2 \Lambda}{\partial\theta^2}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ m^2 (1-x\cos\theta)^3 \biggl\{ [ \beta^2 - x^2 - x^3(3\cos\theta - \cos^3\theta) ] x (1-2x \cos\theta ) - nx^2  (1-x\cos\theta)\cdot [ 
2x +3x^2(3\cos\theta - \cos^3\theta)] \biggr\} \cdot \frac{\partial  \Lambda }{\partial x}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ m^2 (1-x\cos\theta)^3 \biggl\{[ \beta^2 - x^2 - x^3(3\cos\theta - \cos^3\theta) ] x\sin\theta -
3n (1-x\cos\theta)\cdot ( \cos^2\theta -1 )x^3 \sin\theta  \biggr\}
\cdot \frac{\partial\Lambda}{\partial\theta}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ m^2 \biggl\{ 2n x^2  -  (1-x\cos\theta)^2 x^2 [ \beta^2 - x^2 - x^3(3\cos\theta - \cos^3\theta) ] \biggr\}
\cdot \biggl[m^2 \Lambda + 2^2(n+1)^2  \biggr] \, .
</math>
  </td>
</tr>
</table>
</div>
 
=====Step 4=====
 
Let's divide both sides by <math>~m^2</math> and swap a couple of terms between the sides in order to group, on the right, terms with no explicit mention of <math>~\Lambda</math>.
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\mathcal{L}_{RHS} </math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
\frac{\beta^2}{m^2} (1-x\cos\theta)^4 \biggl[2(n+1)\biggr]^2 \frac{RHS}{A_{00}} ~\pm~\mathrm{swaps}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \frac{n x^2}{(n+1)} \cdot  (1-x\cos\theta)^2\{[ 2(n+1) ]^2  \} \biggl\{ (1-x\cos\theta)^2 [ 2(n+1) + 3\beta^2 ]
-  4 (n+1)  \pm i~ [2^3\cdot 3 (n+1)\beta^2 ]^{1/2} \biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
- \biggl\{ 2n x^2  -  (1-x\cos\theta)^2 x^2 [ \beta^2 - x^2 - x^3(3\cos\theta - \cos^3\theta) ] \biggr\}
\cdot \biggl[2^2(n+1)^2  \biggr]
</math>
  </td>
</tr>
</table>
</div>
 
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\Rightarrow ~~~~ -~\frac{\mathcal{L}_{RHS} }{2^2(n+1)x^2} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
n \cdot  (1-x\cos\theta)^2 \biggl\{ (1-x\cos\theta)^2 [ 2(n+1) + 3\beta^2 ]
-  4 (n+1)  \pm i~ [2^3\cdot 3 (n+1)\beta^2 ]^{1/2} \biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ (n+1)\biggl\{ 2n -  (1-x\cos\theta)^2 [ \beta^2 - x^2 - x^3(3\cos\theta - \cos^3\theta) ] \biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2n(n+1) -  (1-x\cos\theta)^2 (n+1)[ \beta^2 - x^2 - x^3(3\cos\theta - \cos^3\theta) ]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ n(1-x\cos\theta)^4 [ 2(n+1) + 3\beta^2 ] -  4 n(n+1)  (1-x\cos\theta)^2
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
\pm~i~  (1-x\cos\theta)^2 [2^3\cdot 3 n^2(n+1)\beta^2 ]^{1/2}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2n(n+1) -  (n+1)(1-x\cos\theta)^2[ 4n + \beta^2 - x^2 - x^3(3\cos\theta - \cos^3\theta) ] + n(1-x\cos\theta)^4 [ 2(n+1) + 3\beta^2 ]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
\pm~i~  (1-x\cos\theta)^2 [2^3\cdot 3 n^2(n+1)\beta^2 ]^{1/2} \, .
</math>
  </td>
</tr>
</table>
</div>
 
 
<!-- SEPARATE DEFINITION OF RHS_3 -->
<div align="center">
<table border="1" cellpadding="8" align="center">
<tr><td align="center" bgcolor="purple"><font color="white">'''Through a separate white-board derivation I have obtained &hellip;'''</font></td></tr>
<tr><td align="center">
 
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\mathrm{RHS}_3</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
\beta^2 (1-x\cos\theta)^4 \cdot \biggl\{ \mathrm{RHS}_1 ~~\pm~~ \mathrm{swaps} \biggr\}
= \beta^2 (1-x\cos\theta)^4 \cdot \biggl\{ \frac{1}{A_{00}}\biggl[ \frac{2(n+1)}{m} \biggr]^2\mathrm{RHS}_0 ~~\pm~~ \mathrm{swaps} \biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-~
2^2(n+1)^2 x^2 \cdot \mathcal{A}  \, ,
</math>
  </td>
</tr>
 
<tr><td colspan="3" align="left">where,</td></tr>
 
<tr>
  <td align="right">
<math>~\mathcal{A}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
2n  + (1-x\cos\theta)^2\biggl[ 4n \biggl( \frac{\nu}{m}\biggr) -  \beta^2 + x^2(1+xb)\biggr]
+ 2n (1-x\cos\theta)^4 \biggl(\frac{\nu}{m}\biggr)^2  \, ,
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\frac{\nu}{m}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-~1~~
\pm~~ i~\biggl[\frac{3}{2(n+1)}\biggr]^{1/2} \beta \, .
</math>
  </td>
</tr>
 
</table>
</td></tr>
 
<tr><td align="center">
 
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr><td colspan="3" align="left">
<font color="red" size="+1"><b>
Case A:
</b></font>&nbsp; &nbsp; &nbsp;
<math>~\biggl(\frac{\nu}{m}\biggr)^2 = \frac{\nu \cdot \nu^*}{m^2} = 1+\frac{3\beta^2}{2(n+1)} ~~~\Rightarrow</math>
</td></tr>
 
<tr>
  <td align="right">
<math>~(n+1)\cdot \mathcal{A}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2n  (n+1)
+ (n+1)(1-x\cos\theta)^2 [  x^2(1+xb) -  \beta^2 - 4n ]
+ (1-x\cos\theta)^4 [2n(n+1) + 3n\beta^2  ]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
\pm~~i~(1-x\cos\theta)^2 [ 2^3\cdot 3 n^2\beta^2(n+1)  ]^{1/2} \, .
</math>
  </td>
</tr>
</table>
</div>
</td></tr>
 
 
<tr><td align="center">
 
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr><td colspan="3" align="left">
<font color="red" size="+1"><b>
Case B:
</b></font>&nbsp; &nbsp; &nbsp;
<math>~\biggl(\frac{\nu}{m}\biggr)^2 = \frac{\nu }{m} \cdot  \frac{\nu }{m}
= 1 - \frac{3\beta^2}{2(n+1)} ~~\pm~i~(-1)\biggl[\frac{2\cdot 3\beta^2}{(n+1)}\biggr]^{1/2} ~~~\Rightarrow</math>
</td></tr>
 
<tr>
  <td align="right">
<math>~(n+1)\cdot \mathcal{A}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2n  (n+1)
+ (n+1)(1-x\cos\theta)^2 [  x^2(1+xb) -  \beta^2 - 4n ]
+ (1-x\cos\theta)^4 [2n(n+1) - 3n\beta^2  ]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
\pm~~i~x\cos\theta (1-x\cos\theta)^2 [ 2^3\cdot 3 n^2\beta^2(n+1)  ]^{1/2} \, .
</math>
  </td>
</tr>
</table>
</div>
 
</td></tr></table>
</div>
 
 
And,
 
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\mathcal{L}_{LHS} </math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
\frac{\beta^2}{m^2} (1-x\cos\theta)^4 \biggl[2(n+1)\biggr]^2 \frac{LHS}{A_{00}} ~\mp~\mathrm{swaps}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(1-x\cos\theta)^4 [ \beta^2 - x^2 - x^3(3\cos\theta - \cos^3\theta) ] x^2\cdot \frac{\partial^2 \Lambda}{\partial x^2}
+m^2 (1-x\cos\theta)^4 [ \beta^2 - x^2 - x^3(3\cos\theta - \cos^3\theta) ]\cdot \frac{\partial^2 \Lambda}{\partial\theta^2}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ (1-x\cos\theta)^3 \biggl\{ [ \beta^2 - x^2 - x^3(3\cos\theta - \cos^3\theta) ] x (1-2x \cos\theta ) - nx^2  (1-x\cos\theta)\cdot [ 
2x +3x^2(3\cos\theta - \cos^3\theta)] \biggr\} \cdot \frac{\partial  \Lambda }{\partial x}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ (1-x\cos\theta)^3 \biggl\{[ \beta^2 - x^2 - x^3(3\cos\theta - \cos^3\theta) ] x\sin\theta -
3n (1-x\cos\theta)\cdot ( \cos^2\theta -1 )x^3 \sin\theta  \biggr\}
\cdot \frac{\partial\Lambda}{\partial\theta}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \biggl\{ 2n x^2  -  (1-x\cos\theta)^2 x^2 [ \beta^2 - x^2 - x^3(3\cos\theta - \cos^3\theta) ] \biggr\}
\cdot \biggl[m^2 \Lambda \biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \frac{n x^2}{(n+1)} \cdot  (1-x\cos\theta)^2\{m^2 \Lambda  \} \biggl\{ (1-x\cos\theta)^2 [ 2(n+1) + 3\beta^2 ]
-  4 (n+1)  \pm i~ [2^3\cdot 3 (n+1)\beta^2 ]^{1/2} \biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(1-x\cos\theta)^4 [ \beta^2 - x^2 - x^3b] \biggl\{ x^2\cdot \frac{\partial^2 \Lambda}{\partial x^2} +m^2 \cdot \frac{\partial^2 \Lambda}{\partial\theta^2} \biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ (1-x\cos\theta)^3 \biggl\{ x( \beta^2 - x^2 - x^3b ) \biggl[ (1-2x \cos\theta )\cdot \frac{\partial  \Lambda }{\partial x} + \sin\theta \cdot \frac{\partial\Lambda}{\partial\theta}\biggr]
- n (1-x\cos\theta)\cdot x^3 \biggl[ (  2 +3xb )\cdot \frac{\partial  \Lambda }{\partial x} 
- 3\sin^3\theta \cdot \frac{\partial\Lambda}{\partial\theta} \biggr] \biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \biggl\{ 2n  -  (1-x\cos\theta)^2 [ \beta^2 - x^2 - x^3b ] + (1-x\cos\theta)^4 \biggl[ 2 + \frac{3\beta^2}{(n+1)} \biggr] n -  4 n  (1-x\cos\theta)^2 \biggr\}
\cdot m^2 x^2\Lambda
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
\pm i~\biggl[\frac{2^3\cdot 3 n^2 m^4 \beta^2}{(n+1)} \biggr]^{1/2}  x^2 \cdot  (1-x\cos\theta)^2  \Lambda
</math>
  </td>
</tr>
</table>
</div>
 
 
<!-- SEPARATE DEFINITION OF LHS_3 -->
<div align="center">
<table border="1" cellpadding="8" align="center">
<tr><td align="center" bgcolor="purple"><font color="white">'''Through a separate white-board derivation I have obtained &hellip;'''</font></td></tr>
<tr><td align="center">
 
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\mathrm{LHS}_3</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
\beta^2 (1-x\cos\theta)^4 \cdot \biggl\{ \mathrm{LHS}_1 ~~\mp~~ \mathrm{swaps} \biggr\}
= \beta^2 (1-x\cos\theta)^4 \cdot \biggl\{ \frac{1}{A_{00}}\biggl[ \frac{2(n+1)}{m} \biggr]^2\mathrm{LHS}_0 ~~\mp~~ \mathrm{swaps} \biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\beta^2(1-\eta^2) (1-x\cos\theta)^3 \biggl\{ (1-x\cos\theta)\biggl[x^2 \frac{\partial^2\Lambda}{\partial x^2} + \frac{\partial^2\Lambda}{\partial \theta^2} \biggr]
+ x\biggl[(1-2x\cos\theta)\frac{\partial\Lambda}{\partial x} + \sin\theta \frac{\partial\Lambda}{\partial\theta}  \biggr]
\biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
-~n x^3 (1-x\cos\theta)^4\biggl[(2+3xb) \frac{\partial\Lambda}{\partial x} -3\sin^3\theta \frac{\partial\Lambda}{\partial\theta} \biggr]
+ m^2 x^2 \Lambda \cdot \mathcal{A} \, ,</math>
  </td>
</tr>
 
<tr><td colspan="3" align="left">where, as above in the definition of <math>~\mathrm{RHS_3} \, ,</math></td></tr>
 
<tr>
  <td align="right">
<math>~\mathcal{A}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
2n  + (1-x\cos\theta)^2\biggl[ 4n \biggl( \frac{\nu}{m}\biggr) -  \beta^2 + x^2(1+xb)\biggr]
+ 2n (1-x\cos\theta)^4 \biggl(\frac{\nu}{m}\biggr)^2  \, .
</math>
  </td>
</tr>
</table>
 
</td></tr></table>
</div>
 
 
The remaining question is, does <math>~\mathcal{L}_{LHS} = \mathcal{L}_{RHS}</math> &#8212; at least to lowest order(s) in <math>~x</math> &#8212; after the Blaes85 expression for the eigenfunction, <math>~\Lambda</math> (and its derivatives), is inserted into the LHS expression?
 
=====Step 5=====
Now let's evaluate the LHS terms, keeping only leading-orders in <math>~x</math> before plugging derivatives of <math>~\Lambda</math> into each term.  For example,
 
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~
x^2\cdot \frac{\partial^2 \Lambda}{\partial x^2} +m^2 \cdot \frac{\partial^2 \Lambda}{\partial\theta^2}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
x^2 \biggl\{2[ 2^3(n+1)^2\cos^2\theta - 3(n+1)] \cdot [1 + \cancelto{0}{3xb}] \pm ~i~\beta [ 2^3\cdot 3(n+1)^3 ]^{1/2} \cos\theta \cdot b[1 + \cancelto{0}{xb} ]^{-3/2} [4 + \cancelto{0}{9xb} ]
\biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~m^2\cdot \biggl\{
-2^4(n+1)^2 x^2 (1 - 2\sin^2\theta ) 
- \cancelto{0}{x^3 }\cdot \biggl[ [ 2^3\cdot 3^2(n+1)^2\sin^2\theta \cos^3\theta - 3^3(n+1)\sin^2\theta \cos\theta] - 2^4\cdot 3(n+1)^2 \sin^4\theta\cos\theta 
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ 2^4(n+1)^2 b + 2^4(n+1)^2 \cdot( - 6\sin^2\theta\cos\theta + 2\sin^2\theta\cos^3\theta  - 3\sin^4\theta \cos\theta )\biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
~\pm ~i~\beta (-1)x [ 2^3\cdot 3(n+1)^3 ]^{1/2} \cdot [1 + xb ]^{-3/2} \cdot \biggl[
2\cos\theta \cdot [2 + x(15\cos\theta - 7\cos^3\theta) + \cancelto{0}{x^2 b}(9\cos\theta - 5\cos^3\theta)]   
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
-  6x\sin^2\theta (1 + \cancelto{0}{xb }) \cdot (3 - 5\cos^2\theta)
+ 3 \cdot x\sin^4\theta [2 + \cancelto{0}{x}(9\cos\theta - 5\cos^3\theta) ] 
\biggr] \biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~
x^2 \biggl[ 2^4(n+1)^2\cos^2\theta - 6(n+1) -2^4 m^2 (n+1)^2 (1 - 2\sin^2\theta ) \biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
~\pm ~i~m^2\beta (-1)x [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cdot [1 + xb ]^{-3/2} \cos\theta   
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
~\pm ~i~\beta x^2 [ 2^3\cdot 3(n+1)^3 ]^{1/2} \cdot \biggl[(12\cos^2\theta - 4\cos^4\theta)~~ - ~~m^2\cdot (1 + xb )^{-3/2}
[2\cos^2\theta(15 - 7\cos^2\theta)    -  6\sin^2\theta (3 - 5\cos^2\theta) + 6 \sin^4\theta]  \biggr]
</math>
  </td>
</tr>
</table>
</div>
 
Next,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~ (1-2x \cos\theta )\cdot \frac{\partial  \Lambda }{\partial x} + \sin\theta \cdot \frac{\partial\Lambda}{\partial\theta}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(1-2x \cos\theta )\cdot \biggl\{
[ 2^3(n+1)^2\cos^2\theta - 3(n+1)] \cdot [2x + \cancelto{0}{3x^2b}] ~~\pm ~~i~\beta [ 2^5\cdot 3(n+1)^3 ]^{1/2} \cos\theta \cdot [1 + xb ]^{-1/2} \cdot [2 + \cancelto{0}{3xb }] 
\biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \sin\theta \cdot \biggl\{
-2^4(n+1)^2 \cancelto{0}{x^2}\sin\theta \cos\theta [1 + xb ]
-3 \cancelto{0}{x^3} \sin^3\theta [ 2^3(n+1)^2\cos^2\theta - 3(n+1)] 
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
~~\pm ~~i~\beta (-1) [ 2^5\cdot 3(n+1)^3 ]^{1/2} \cdot [1 + xb ]^{-1/2}  \cdot
x~\sin\theta [2 + \cancelto{0}{x}(9\cos\theta - 5\cos^3\theta) ]
\biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~
(1-2x \cos\theta )\cdot \biggl[ 2x(n+1)
[ 2^3 (n+1)\cos^2\theta - 3] ~~\pm ~~i~\beta [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cos\theta \cdot (1 + xb )^{-1/2} 
\biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
~~\pm ~~i~\beta x (-1) [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cdot [1 + xb ]^{-1/2}  \cdot \sin^2\theta 
</math>
  </td>
</tr>
</table>
</div>
 
 
<!-- EVALUATE 1st CROSS-TERM -->
<div align="center">
<table border="1" cellpadding="8" align="center">
<tr><td align="center" bgcolor="purple"><font color="white">'''Through a separate white-board derivation I have obtained &hellip;'''</font></td></tr>
<tr><td align="center">
 
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~ (1-2x \cos\theta )\cdot \frac{\partial  \Lambda }{\partial x} + \sin\theta \cdot \frac{\partial\Lambda}{\partial\theta}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
x(n+1)[-6 + 2^4(n+1)\cos^2\theta]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
- x^2(n+1)\cos\theta \{ [ 15 + 2^4(n+1) ] -\cos^2\theta[9 + 2^3\cdot 7 (n+1)] +2^3\cdot 3(n+1)\cos^4\theta \}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+x^3(n+1) \{9 - 2^2\cdot 3^2(1+2n)\cos^2\theta - [9 + 32(n+1)]\cos^4\theta +2^3(n+1)\cos^6\theta
\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~\pm~~i~\beta~\biggl[ \frac{2^5\cdot 3 (n+1)^3}{1+x(3\cos\theta-\cos^3\theta)} \biggr]^{1/2} \biggl\{ 2\cos\theta
- x[2 - 7\cos^2\theta + 3\cos^4\theta ] </math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~- x^2 \cos\theta [ 9 +4\cos^2\theta -\cos^4\theta ]
\biggr\}</math>
  </td>
</tr>
 
</table>
 
</td></tr></table>
</div>
 
 
 
Also, from above,
 
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\Lambda </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- (4n+1)\beta^2 +
(n+1)\cancelto{0}{x^2}[ 2^3(n+1)\cos^2\theta - 3] \cdot [1 + xb] ~\pm ~i~\beta [ 2^7\cdot 3(n+1)^3 ]^{1/2} x\cos\theta [1 + x b ]^{1/2}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~- (4n+1)\beta^2  ~\pm ~i~\beta [ 2^7\cdot 3(n+1)^3 ]^{1/2} x\cos\theta [1 + x b ]^{1/2}
</math>
  </td>
</tr>
</table>
</div>
 
 
<!-- EVALUATE 2nd CROSS-TERM -->
<div align="center">
<table border="1" cellpadding="8" align="center">
<tr><td align="center" bgcolor="purple"><font color="white">'''Through a separate white-board derivation I have obtained &hellip;'''</font></td></tr>
<tr><td align="center">
 
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~ (2+3xb )\cdot \frac{\partial  \Lambda }{\partial x} - 3\sin^3\theta \cdot \frac{\partial\Lambda}{\partial\theta}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
x\cdot 2^2(n+1)[2^3(n+1)\cos^2\theta -3]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ x^2\cdot 2^2\cdot 3(n+1)\cos\theta \{ [2^2n -5] +\cos^2\theta[2^4n + 19] - 2^2(n+1)\cos^4\theta \}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+x^3 \cdot 3^2(n+1)\cos^2\theta \{ -3^3 + 2\cdot 3^2\cos^2\theta[2^2n+5] - 99\cdot \cos^4\theta + 2^3(n+1)\cos^6\theta \}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+x^3 (n+1)\sin^4\theta \{  -3^3 + 3^2\cos^2\theta [ 2^3\cdot 3 n + 27] -2^3\cdot 3\cdot 5(n+1)\cos^4\theta \}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~\pm~~i~\beta~ \biggl[ \frac{2^5\cdot 3(n+1)^3}{1+xb} \biggr]^{1/2} \biggl\{
4\cos\theta + 6x(2b\cos\theta + \sin^4\theta) + 3x^2(3b^2\cos\theta +2b\sin^4\theta + 3\sin^6\theta\cos\theta)
\biggr\} </math>
  </td>
</tr>
 
</table>
 
</td></tr></table>
</div>
 
 
 
Taken together, then, we have,
 
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\mathcal{L}_{LHS} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(1-x\cos\theta)^4 [ \beta^2 - x^2 - \cancelto{0}{x^3b}] \biggl\{ x^2\cdot \frac{\partial^2 \Lambda}{\partial x^2} +m^2 \cdot \frac{\partial^2 \Lambda}{\partial\theta^2} \biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ (1-x\cos\theta)^3 \biggl\{ x( \beta^2 - \cancelto{0}{x^2} - \cancelto{0}{x^3b} ) \biggl[ (1-2x \cos\theta )\cdot \frac{\partial  \Lambda }{\partial x} + \sin\theta \cdot \frac{\partial\Lambda}{\partial\theta}\biggr]
- n (1-x\cos\theta)\cdot \cancelto{0}{x^3} \biggl[ (  2 +3xb )\cdot \frac{\partial  \Lambda }{\partial x} 
- 3\sin^3\theta \cdot \frac{\partial\Lambda}{\partial\theta} \biggr] \biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \biggl\{ 2n  -  (1-x\cos\theta)^2 [ \beta^2 - x^2 - \cancelto{0}{x^3}b ] + (1-x\cos\theta)^4 \biggl[ 2 + \frac{3\beta^2}{(n+1)} \biggr] n -  4 n  (1-x\cos\theta)^2 \biggr\}
\cdot m^2 x^2\Lambda
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
\pm i~\biggl[\frac{2^3\cdot 3 n^2 m^4 \beta^2}{(n+1)} \biggr]^{1/2}  x^2 \cdot  (1-x\cos\theta)^2  \Lambda
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~
(1-x\cos\theta)^4 [ \beta^2 - x^2 ] \biggl\{ x^2\cdot \frac{\partial^2 \Lambda}{\partial x^2} +m^2 \cdot \frac{\partial^2 \Lambda}{\partial\theta^2} \biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ (1-x\cos\theta)^3 \biggl\{ x( \beta^2) \biggl[ (1-2x \cos\theta )\cdot \frac{\partial  \Lambda }{\partial x} + \sin\theta \cdot \frac{\partial\Lambda}{\partial\theta}\biggr]
\biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \biggl\{ 2n  -  (1-x\cos\theta)^2 [ \beta^2 - x^2  ] + (1-x\cos\theta)^4 \biggl[ 2 + \frac{3\beta^2}{(n+1)} \biggr] n -  4 n  (1-x\cos\theta)^2 \biggr\}
\cdot m^2 x^2\Lambda
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
\pm i~\biggl[\frac{2^3\cdot 3 n^2 m^4 \beta^2}{(n+1)} \biggr]^{1/2}  x^2 \cdot  (1-x\cos\theta)^2  \Lambda
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~
(1-x\cos\theta)^4 [ \beta^2 - x^2 ] \biggl\{ x^2 \biggl[ 2^4(n+1)^2\cos^2\theta - 6(n+1) -2^4 m^2 (n+1)^2 (1 - 2\sin^2\theta ) \biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
~\pm ~i~m^2\beta (-1)x [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cdot [1 + xb ]^{-3/2} \cos\theta   
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
~\pm ~i~\beta x^2 [ 2^3\cdot 3(n+1)^3 ]^{1/2} \cdot \biggl[(12\cos^2\theta - 4\cos^4\theta)~~ - ~~m^2\cdot (1 + xb )^{-3/2}
[2\cos^2\theta(15 - 7\cos^2\theta)    -  6\sin^2\theta (3 - 5\cos^2\theta) + 6 \sin^4\theta]  \biggr]
\biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ x\beta^2(1-x\cos\theta)^3 \biggl\{
(1-2x \cos\theta )\cdot \biggl[ 2x(n+1)[ 2^3 (n+1)\cos^2\theta - 3] ~~\pm ~~i~\beta [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cos\theta \cdot (1 + xb )^{-1/2}  \biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
~~\pm ~~i~\beta x (-1) [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cdot [1 + xb ]^{-1/2}  \cdot \sin^2\theta  \biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ m^2 x^2\biggl\{ 2n  -  (1-x\cos\theta)^2 [ \beta^2 - x^2  ] + (1-x\cos\theta)^4 \biggl[ 2 + \frac{3\beta^2}{(n+1)} \biggr] n -  4 n  (1-x\cos\theta)^2 \biggr\}
\cdot \biggl\{- (4n+1)\beta^2  ~\pm ~i~\beta [ 2^7\cdot 3(n+1)^3 ]^{1/2} x\cos\theta [1 + x b ]^{1/2}\biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
\pm i~\biggl[\frac{2^3\cdot 3 n^2 m^4 \beta^2}{(n+1)} \biggr]^{1/2}  x^2 \cdot  (1-x\cos\theta)^2  \cdot \biggl\{
- (4n+1)\beta^2  ~\pm ~i~\beta [ 2^7\cdot 3(n+1)^3 ]^{1/2} x\cos\theta [1 + x b ]^{1/2}
\biggr\}
</math>
  </td>
</tr>
</table>
</div>
 
Let's further simplify:
 
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\mathcal{L}_{LHS} </math>
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~
x^2\biggl\{ (1-x\cos\theta)^4 [ \beta^2 - x^2 ] \biggl[ 2^4(n+1)^2\cos^2\theta - 6(n+1) -2^4 m^2 (n+1)^2 (1 - 2\sin^2\theta ) \biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
-~\biggl[\frac{2^3\cdot 3 n^2 m^4 \beta^2}{(n+1)} \biggr]^{1/2} \cdot  (1-x\cos\theta)^2  \cdot
\beta [ 2^7\cdot 3(n+1)^3 ]^{1/2} x\cos\theta [1 + x b ]^{1/2}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \beta^2(1-x\cos\theta)^3
(1-2x \cos\theta )\cdot \biggl[ 2(n+1)[ 2^3 (n+1)\cos^2\theta - 3]  \biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
- 2n(4n+1)\beta^2  m^2
+ (4n+1)\beta^2  m^2  [4n + \beta^2 - x^2 ] (1-x\cos\theta)^2
- (4n+1)\beta^2  m^2 n (1-x\cos\theta)^4 \biggl[ 2 + \frac{3\beta^2}{(n+1)} \biggr]
\biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
~~\pm ~~i x\beta^3 (1-x\cos\theta)^3
(1-2x \cos\theta )\cdot [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cos\theta \cdot (1 + xb )^{-1/2} 
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
~\pm ~i~m^2\beta (-1)x [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cdot [1 + xb ]^{-3/2} \cos\theta  (1-x\cos\theta)^4 [ \beta^2 - x^2 ] 
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
\pm ~i~\beta m^2 x^2\biggl\{ 2n  -  (1-x\cos\theta)^2 [ \beta^2 - x^2  ] + (1-x\cos\theta)^4 \biggl[ 2 + \frac{3\beta^2}{(n+1)} \biggr] n -  4 n  (1-x\cos\theta)^2 \biggr\}
\cdot [ 2^7\cdot 3(n+1)^3 ]^{1/2} x\cos\theta [1 + x b ]^{1/2}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
~\pm ~i~\beta x^2 [ 2^3\cdot 3(n+1)^3 ]^{1/2} (1-x\cos\theta)^4 [ \beta^2 - x^2 ] \cdot \biggl[(12\cos^2\theta - 4\cos^4\theta)~~ - ~~m^2\cdot (1 + xb )^{-3/2}
[2\cos^2\theta(15 - 7\cos^2\theta)    -  6\sin^2\theta (3 - 5\cos^2\theta) + 6 \sin^4\theta]  \biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
~~\pm ~~i~\beta x^2 (-1) [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cdot \beta^2(1-x\cos\theta)^3  [1 + xb ]^{-1/2}  \cdot \sin^2\theta 
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
\pm i~x^2(-1) (4n+1)\beta^2 \biggl[\frac{2^3\cdot 3 n^2 m^4 \beta^2}{(n+1)} \biggr]^{1/2}  \cdot  (1-x\cos\theta)^2 
</math>
  </td>
</tr>
</table>
</div>
 
=====Step 6=====
Hence, to lowest order we want to compare the following two expressions:
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~~ -~\frac{\mathcal{L}_{RHS} }{2^2(n+1)x^2} </math>
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~
2n(n+1) -  (n+1)(1-\cancelto{0}{x}\cos\theta)^2[ 4n + \beta^2 - \cancelto{0}{x^2} - \cancelto{0}{x^3}(3\cos\theta - \cos^3\theta) ] + n(1-\cancelto{0}{x}\cos\theta)^4 [ 2(n+1) + 3\beta^2 ]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
\pm~i~  (1-\cancelto{0}{x}\cos\theta)^2 [2^3\cdot 3 n^2(n+1)\beta^2 ]^{1/2}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~
2n(n+1) -  (n+1)[ 4n + \beta^2  ] + n [ 2(n+1) + 3\beta^2 ] \pm~i~  [2^3\cdot 3 n^2(n+1)\beta^2 ]^{1/2}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~3n\beta^2 + (n+1)\biggl[
2n -  4n - \beta^2  + 2n \biggr] \pm~i~  [2^3\cdot 3 n^2(n+1)\beta^2 ]^{1/2}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~\beta^2 (2n-1) ~\pm~i~  [2^3\cdot 3 n^2(n+1)\beta^2 ]^{1/2}
</math>
  </td>
</tr>
</table>
</div>
 
 
 
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\mathrm{Re}\biggl[\frac{\mathcal{L}_{LHS}}{x^2}\biggr] </math>
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~
(1-\cancelto{0}{x}\cos\theta)^4 [ \beta^2 - \cancelto{0}{x^2} ] \biggl[ 2^4(n+1)^2\cos^2\theta - 6(n+1) -2^4 m^2 (n+1)^2 (1 - 2\sin^2\theta ) \biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
-~\biggl[\frac{2^3\cdot 3 n^2 m^4 \beta^2}{(n+1)} \biggr]^{1/2} \cdot  (1-x\cos\theta)^2  \cdot
\beta [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cancelto{0}{x}\cos\theta [1 + x b ]^{1/2}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \beta^2(1-\cancelto{0}{x}\cos\theta)^3
(1-\cancelto{0}{2x} \cos\theta )\cdot \biggl[ 2(n+1)[ 2^3 (n+1)\cos^2\theta - 3]  \biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
- 2n(4n+1)\beta^2  m^2
+ (4n+1)\beta^2  m^2  [4n + \beta^2 - \cancelto{0}{x^2} ] (1-\cancelto{0}{x}\cos\theta)^2
- (4n+1)\beta^2  m^2 n (1-\cancelto{0}{x}\cos\theta)^4 \biggl[ 2 + \frac{3\beta^2}{(n+1)} \biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~
\beta^2\biggl[ 2^4(n+1)^2\cos^2\theta - 6(n+1) -2^4 m^2 (n+1)^2 (1 - 2\sin^2\theta ) \biggr]
+ \beta^2\biggl[ 2^4 (n+1)^2\cos^2\theta  -6(n+1)\biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~+ \beta^2\biggl\{- 2n(4n+1)  m^2+ (4n+1)m^2  [4n  ] - 2(4n+1)  m^2 n
\biggr\} + \beta^4\biggl\{(4n+1) m^2 - (4n+1) m^2 n \biggl[ \frac{3}{(n+1)} \biggr]
\biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~
\beta^2\biggl[ 2^4(n+1)^2\cos^2\theta - 6(n+1)  \biggr]
+ \beta^2\biggl[ 2^4 (n+1)^2\cos^2\theta  -6(n+1)\biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ m^2\beta^2\biggl[  -2^4(n+1)^2 (1 - 2\sin^2\theta ) \biggr]
+ m^2\beta^2\biggl\{- 2n(4n+1)  + 4n(4n+1)  - 2n(4n+1) \biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ (4n+1) m^2\beta^4\biggl[1 -  \frac{3n}{(n+1)} \biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~
2\beta^2\biggl[ 2^4(n+1)^2\cos^2\theta - 6(n+1)  \biggr]
+ m^2\beta^2\biggl[  2^4(n+1)^2 (1 - 2\cos^2\theta ) \biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ (4n+1) m^2\beta^4\biggl[1 -  \frac{3n}{(n+1)} \biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~
(1-m^2)2^5\beta^2(n+1)^2\cos^2\theta
+ 2^2(n+1)\beta^2\biggl[ 4m^2(n+1)  -3\biggr]
+ (4n+1) m^2\beta^4\biggl[1 -  \frac{3n}{(n+1)} \biggr] \, .
</math>
  </td>
</tr>
</table>
</div>
 
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\pm~\mathrm{Im}\biggl[\mathcal{L}_{LHS} \biggr]</math>
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~
~x\biggl\{\beta^3 (1-\cancelto{0}{x}\cos\theta)^3
(1-\cancelto{0}{2x} \cos\theta )\cdot [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cos\theta \cdot (1 + \cancelto{0}{xb} )^{-1/2} 
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
-~m^2\beta  [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cdot [1 + \cancelto{0}{xb} ]^{-3/2} \cos\theta  (1-\cancelto{0}{x}\cos\theta)^4 [ \beta^2 - \cancelto{0}{x}^2 ]  \biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+~x^2 \biggl\{ \beta m^2 \biggl\{ 2n  -  (1-x\cos\theta)^2 [ \beta^2 - x^2  ] + (1-x\cos\theta)^4 \biggl[ 2 + \frac{3\beta^2}{(n+1)} \biggr] n -  4 n  (1-x\cos\theta)^2 \biggr\}
\cdot [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cancelto{0}{x}\cos\theta [1 + x b ]^{1/2}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
+~\beta [ 2^3\cdot 3(n+1)^3 ]^{1/2} (1-\cancelto{0}{x}\cos\theta)^4 [ \beta^2 - \cancelto{0}{x^2} ] \cdot \biggl[(12\cos^2\theta - 4\cos^4\theta)~~ - ~~m^2\cdot (1 + \cancelto{0}{xb} )^{-3/2}
[2\cos^2\theta(15 - 7\cos^2\theta)    -  6\sin^2\theta (3 - 5\cos^2\theta) + 6 \sin^4\theta]  \biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
-~\beta  [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cdot \beta^2(1-\cancelto{0}{x}\cos\theta)^3  [1 + \cancelto{0}{xb} ]^{-1/2}  \cdot \sin^2\theta 
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
-~ (4n+1)\beta^2 \biggl[\frac{2^3\cdot 3 n^2 m^4 \beta^2}{(n+1)} \biggr]^{1/2}  \cdot  (1-\cancelto{0}{x}\cos\theta)^2  \biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~
~x(1-m^2)\cdot \beta^3  [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cos\theta
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
+~~x^2 \beta^3 [ 2^3\cdot 3(n+1)^3 ]^{1/2}\biggl\{(12\cos^2\theta - 4\cos^4\theta) - 4\sin^2\theta
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~- m^2 \biggl[2\cos^2\theta(15 - 7\cos^2\theta)    -  6\sin^2\theta (3 - 5\cos^2\theta) + 6 \sin^4\theta
-~ \frac{n (4n+1)}{(n+1)^2} \biggr]\biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~
~x(1-m^2)\cdot \beta^3  [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cos\theta
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
+~~x^2 \beta^3 [ 2^7\cdot 3(n+1)^3 ]^{1/2}\biggl\{4\cos^2\theta - \cos^4\theta -1
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~- \frac{m^2}{2} \biggl[\cos^2\theta(15 - 7\cos^2\theta)  -12\sin^4\theta +6 \sin^2\theta
-~ \frac{n (4n+1)}{2(n+1)^2} \biggr]\biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~
~x(1-m^2)\cdot \beta^3  [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cos\theta
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
+~~x^2 \beta^3 [ 2^7\cdot 3(n+1)^3 ]^{1/2}\biggl\{3-(1+\sin^2\theta)^2
- \frac{m^2}{2} \biggl[33\cos^2\theta-19 \cos^4\theta -6
-~ \frac{n (4n+1)}{2(n+1)^2} \biggr]\biggr\}
</math>
  </td>
</tr>
</table>
</div>
 
 
=====Examples=====
 
Evaluate various expressions using the parameter set:&nbsp;&nbsp;
<math>~(n, \theta, x) = (1, \tfrac{\pi}{3}, \tfrac{1}{4})</math>
 
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~b</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{3}{2} - \frac{1}{8} = \frac{11}{8} </math>
  </td>
  <td align="right">
1.375000000
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~(\beta\eta)^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl(\frac{1}{2^2}\biggr)^2\biggl[ 1 + \frac{11}{2^5} \biggr] = \frac{2^5 + 11}{2^9} = \frac{43}{2^9} </math>
  </td>
  <td align="right">
0.083984375
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\mathrm{Re}(\Lambda)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-5\beta^2 + \frac{43}{2^9} \biggl[ 2^3 - 6 \biggr]
= -5\beta^2 + \frac{43}{2^8}
</math>
  </td>
  <td align="right">
<math>~- 5\beta^2</math> + 0.167968750
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\mathrm{Im}(\Lambda)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{\beta}{2}\biggl[ 2^{10}\cdot 3 \cdot \frac{43}{2^9} \biggr]^{1/2}
= \beta \biggl[ \frac{3\cdot 43}{2} \biggr]^{1/2}
</math>
  </td>
  <td align="right">
8.031189202 <math>~\beta</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\mathrm{Re}\biggl( \frac{\partial\Lambda}{\partial x} \biggr)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ 2^3 - 6 \biggr]\frac{1}{2^2}\biggl( 2 + \frac{3\cdot 11}{2^2\cdot 2^3} \biggr)
= \biggl( 1 + \frac{33}{2^6} \biggr)
</math>
  </td>
  <td align="right">
1.515625000
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\mathrm{Im}\biggl( \frac{\partial\Lambda}{\partial x} \biggr)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{\beta}{2} \biggl[ \frac{2^9\cdot 2^8\cdot 3}{43} \biggr]^{1/2} \biggl[ \frac{1}{2^2}\cdot \biggl(2 + \frac{3\cdot 11}{2^2\cdot 2^3}\biggr) \biggr]
= \beta \biggl[ \frac{2^{13}\cdot 3}{43} \biggr]^{1/2} \biggl(1 + \frac{3\cdot 11}{2^{6}}\biggr)
</math>
  </td>
  <td align="right">
36.23373732 <math>~\beta</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\mathrm{Re}\biggl( \frac{\partial\Lambda}{\partial \theta} \biggr)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2\biggl(\frac{3}{4}\biggr)^{1/2} \biggl\{-2^4 \cdot \frac{43}{2^9} +\frac{3}{2^6}\biggl(\frac{3}{4}\biggr)\biggl[3-4\biggr]
\biggr\}
= ~-~\frac{\sqrt{3}}{2^8} \cdot (2^3\cdot 43 + 9)
</math>
  </td>
  <td align="right">
-2.388335684
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\mathrm{Im}\biggl( \frac{\partial\Lambda}{\partial \theta} \biggr)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(-1) \beta~\biggl(\frac{3}{4}\biggr)^{1/2} \cdot \biggl[ \frac{2^{10} \cdot 3\cdot 43}{2^9} \biggr]^{1/2}
\biggl\{1 + \frac{3\cdot 2^9}{2^7 \cdot 43} \biggl(\frac{3}{2^3}\biggr)\biggr\} =
(-1) \beta~\biggl(\frac{3}{4}\biggr)^{1/2} \cdot [ 2\cdot 3\cdot 43  ]^{1/2}
\biggl\{1 + \frac{3^2}{2\cdot 43} \biggr\}
</math>
  </td>
  <td align="right">
(-1) &times; 15.36617018 <math>~\beta</math>
  </td>
</tr>
</table>
</div>
 
 
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\mathrm{Re}\biggl( \frac{\partial^2\Lambda}{\partial x^2}\biggr)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2^2[2^2-3]\biggl[1 + \frac{3}{2^2}\cdot \frac{11}{2^3}  \biggr] = 2^2 + \frac{33}{2^3} = \frac{65}{8}
</math>
  </td>
  <td align="right">
&nbsp;  &nbsp;  &nbsp;  &nbsp;  &nbsp;  8.125000000
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\mathrm{Im}\biggl( \frac{\partial^2\Lambda}{\partial x^2}\biggr)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\beta~
2^2\cdot \sqrt{3} \biggl[\frac{11}{2^3} \biggl(2^2+\frac{3}{2^2}\cdot \frac{11}{2^3} \biggr)\biggr] \biggl[ \frac{2^4}{43}\biggr]^{3/2}
=\biggl[ \frac{2^3\cdot 3}{43^3}\biggr]^{1/2} \biggl[11\cdot (2^7+33)\biggr] \beta
</math>
  </td>
  <td align="right">
&nbsp;  &nbsp;  &nbsp;  &nbsp;  &nbsp;  30.76957507<math>~\beta</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\mathrm{Re}\biggl(\frac{\partial^2\Lambda}{\partial \theta^2}\biggr)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{2^4} \biggl\{2^6 \biggl(\frac{3}{2^2} - \frac{1}{2^2} \biggr)
\biggr\} + \frac{1}{2^6}\biggl\{-2^3\cdot 3  + 2 + \frac{3^3\cdot 5\cdot 7}{2^2}  -3\cdot 23
\biggr\} = 2 + \frac{1}{2^8}\biggl\{2^3 + 3^3\cdot 5\cdot 7  - 2^2\cdot 3\cdot 31
\biggr\}
</math>
  </td>
  <td align="right">
&nbsp;  &nbsp;  &nbsp;  &nbsp;  &nbsp;  4.269531250
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\mathrm{Im}\biggl( \frac{\partial^2\Lambda}{\partial \theta^2}\biggr)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(-1)\beta [2^{10}\cdot 3]^{1/2} \biggl( \frac{43}{2^9} \biggr)^{1/2} \biggl\{
\frac{1}{2} + \frac{3}{2}\cdot \frac{2^9}{43} \cdot \frac{1}{2^6}\cdot \frac{3}{2^2}  \biggl(\frac{5}{2^2} -2 \biggr)
+ \biggl( \frac{3}{2}\cdot \frac{1}{2^6} \cdot \frac{2^9}{43}\biggr)^2 \biggl( \frac{3}{2^2} \biggr)^3 \frac{1}{2}
\biggr\}
</math>
  </td>
  <td align="right">
&nbsp;  &nbsp;  &nbsp;  &nbsp;  &nbsp;
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(-1)\beta\biggl( \frac{3\cdot 43}{2} \biggr)^{1/2} \biggl\{
1 - \frac{3^3}{2\cdot 43} + \frac{3^5}{(2\cdot 43)^2}
\biggr\}
</math>
  </td>
  <td align="right">
(-1) &times; 5.773638858 <math>~\beta</math>
  </td>
</tr>
 
</table>
</div>
 
 
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~ \mathrm{Re}\biggl\{ (1-2x \cos\theta )\cdot \frac{\partial  \Lambda }{\partial x} + \sin\theta \cdot \frac{\partial\Lambda}{\partial\theta} \biggr\}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl( 1 - \frac{1}{2^2} \biggr)\biggl( 1 + \frac{33}{2^6} \biggr)
-~\frac{\sqrt{3}}{2}\cdot \frac{\sqrt{3}}{2^8} \cdot (2^3\cdot 43 + 9)
</math>
  </td>
  <td align="right">
&nbsp;
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
~\frac{(2\cdot 3\cdot 97)-3\cdot (2^3\cdot 43 + 9)}{2^9}
</math>
  </td>
  <td align="right">
-0.931640625
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~ \mathrm{Im}\biggl\{ (1-2x \cos\theta )\cdot \frac{\partial  \Lambda }{\partial x} + \sin\theta \cdot \frac{\partial\Lambda}{\partial\theta} \biggr\}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl( 1 - \frac{1}{2^2} \biggr)\beta \biggl[ \frac{2^{13}\cdot 3}{43} \biggr]^{1/2} \biggl(1 + \frac{3\cdot 11}{2^{6}}\biggr)
+~\frac{\sqrt{3}}{2}\cdot (-1) \beta~\frac{\sqrt{3}}{2}\cdot[ 2\cdot 3\cdot 43  ]^{1/2}
\biggl\{1 + \frac{3^2}{2\cdot 43} \biggr\}
</math>
  </td>
  <td align="right">
&nbsp;
  </td>
</tr>
<!--
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\beta \biggl\{ \biggl[ \frac{3^3}{2^{3}\cdot 43} \biggr]^{1/2} (2^{6} + 3\cdot 11 )
-~\biggl[ \frac{3^4}{2^5\cdot 43} \biggr]^{1/2}
( 2\cdot 43 + 3^2 ) \biggr\}
</math>
  </td>
  <td align="right">
NEW:&nbsp;13.86780926 <math>~\beta</math>
  </td>
</tr>
-->
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\beta \biggl[ \frac{3^3}{2^5\cdot 43} \biggr]^{1/2} [2 (2^6 + 33) - (2\cdot 43 + 3^2) ]
</math>
  </td>
  <td align="right">
13.86780926 <math>~\beta</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~ \mathrm{Re}\biggl\{ (2+3xb )\cdot \frac{\partial  \Lambda }{\partial x} - 3\sin^3\theta \cdot \frac{\partial\Lambda}{\partial\theta} \biggr\}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl( 2 + \frac{3}{2^2}\cdot \frac{11}{2^3} \biggr)\biggl( 1 + \frac{33}{2^6} \biggr)
+~3\cdot\biggl(\frac{3}{2^2}\biggr)^{3/2}\cdot \frac{\sqrt{3}}{2^8} \cdot (2^3\cdot 43 + 9)
</math>
  </td>
  <td align="right">
&nbsp;
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{97^2}{2^{11}} +~\frac{3^3}{2^{11}}\cdot (2^3\cdot 43 + 9)
</math>
  </td>
  <td align="right">
9.248046874
  </td>
</tr>
 
 
<tr>
  <td align="right">
<math>~ \mathrm{Im}\biggl\{ (2+3xb )\cdot \frac{\partial  \Lambda }{\partial x} - 3\sin^3\theta \cdot \frac{\partial\Lambda}{\partial\theta} \biggr\}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\beta \biggl\{ \biggl( 2 + \frac{3}{2^2}\cdot \frac{11}{2^3} \biggr) \biggl[ \frac{2^{13}\cdot 3}{43} \biggr]^{1/2} \biggl(1 + \frac{3\cdot 11}{2^{6}}\biggr)
+~3\cdot\biggl(\frac{3}{2^2}\biggr)^{3/2}\cdot \biggl(\frac{3}{4}\biggr)^{1/2} \cdot [ 2\cdot 3\cdot 43  ]^{1/2}
\biggl[1 + \frac{3^2}{2\cdot 43} \biggr]\biggr\}
</math>
  </td>
  <td align="right">
&nbsp;
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\beta \biggl\{ \biggl( \frac{3}{2^9\cdot 43} \biggr)^{1/2}  \biggl[ ( 2^6 + 33)^2
+~3^3(2\cdot 43 +3^2 ) \biggr]\biggr\}
</math>
  </td>
  <td align="right">
139.7753772
  </td>
</tr>
</table>
</div>
 
=====Step 7=====
Let's begin by slightly redefining the LHS and RHS collections of terms.
 
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\mathrm{RHS}_4</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\mathrm{RHS}_3 - m^2 x^2 \Lambda \cdot \mathcal{A} </math>
  </td>
</tr>
 
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-~x^2 ~ [ 2^2(n+1)^2  - m^2 \Lambda ] \cdot \mathcal{A} \, ,
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\mathrm{LHS}_4</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\mathrm{LHS}_3 - m^2 x^2 \Lambda \cdot \mathcal{A} </math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\beta^2(1-\eta^2) (1-x\cos\theta)^3 \biggl\{ (1-x\cos\theta)\biggl[x^2 \frac{\partial^2\Lambda}{\partial x^2} + \frac{\partial^2\Lambda}{\partial \theta^2} \biggr]
+ x\biggl[(1-2x\cos\theta)\frac{\partial\Lambda}{\partial x} + \sin\theta \frac{\partial\Lambda}{\partial\theta}  \biggr]
\biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
-~n x^3 (1-x\cos\theta)^4\biggl[(2+3xb) \frac{\partial\Lambda}{\partial x} -3\sin^3\theta \frac{\partial\Lambda}{\partial\theta} \biggr]
</math>
  </td>
</tr>
 
</table>
 
<!-- Early attempt at lowest order ****************
 
======To Lowest Order======
 
Now, let's gather together only the lowest order components of all the LHS terms.
 
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\mathrm{LHS}_4</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
[ \beta^2 - \cancelto{0}{x^2} - \cancelto{0}{x^3}(3\cos\theta - \cos^3\theta) ] (1-\cancelto{0}{x}\cos\theta)^3 \biggl\{ (1-\cancelto{0}{x}\cos\theta)
\biggl[x^2 \frac{\partial^2\Lambda}{\partial x^2} + \frac{\partial^2\Lambda}{\partial \theta^2} \biggr]
+ x\biggl[(1-2x\cos\theta)\frac{\partial\Lambda}{\partial x} + \sin\theta \frac{\partial\Lambda}{\partial\theta}  \biggr]
\biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
-~n \cancelto{0}{x^3} (1-x\cos\theta)^4\biggl[(2+3xb) \frac{\partial\Lambda}{\partial x} -3\sin^3\theta \frac{\partial\Lambda}{\partial\theta} \biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~
\beta^2~\biggl\{
\biggl[x^2 \frac{\partial^2\Lambda}{\partial x^2} + \frac{\partial^2\Lambda}{\partial \theta^2} \biggr]
+ x\biggl[(1-2x\cos\theta)\frac{\partial\Lambda}{\partial x} + \sin\theta \frac{\partial\Lambda}{\partial\theta}  \biggr]
\biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\Rightarrow~~~~\frac{1}{\beta^2} \cdot \mathrm{LHS}_4</math>
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~
x^2 \biggl\{ 2(n+1)[2^3(n+1)\cos^2\theta -3](1+3\cancelto{0}{x}b) ~~\pm~~i~\beta\cos\theta [2^3\cdot 3(n+1)^3]^{1/2} \biggl[ \frac{b(4+3\cancelto{0}{x}b)}{(1+\cancelto{0}{x}b)^{3/2}} \biggr] \biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~ + x^2 \biggl\{2^4(n+1)^2(\sin^2\theta - \cos^2\theta)\biggr\}
~~\pm~~i~(-1)\beta [2^7\cdot 3 (n+1)^3 ]^{1/2} \biggl\{
\cancelto{x}{(\beta\eta)}\cos\theta \biggr\} 
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ x\biggl\{
x(n+1)[-6 + 2^4(n+1)\cos^2\theta]
~~\pm~~i~\beta~\biggl[ \frac{2^5\cdot 3 (n+1)^3}{1+\cancelto{0}{x}(3\cos\theta-\cos^3\theta)} \biggr]^{1/2} \biggl[ 2\cos\theta
- \cancelto{0}{x}(2 - 7\cos^2\theta + 3\cos^4\theta )\biggr]
\biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~
x^2 \biggl\{ 2(n+1)[2^3(n+1)\cos^2\theta -3] + 2^4(n+1)^2(\sin^2\theta - \cos^2\theta) + (n+1)[-6 + 2^4(n+1)\cos^2\theta] \biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
\pm~~i~\biggl\{(-1)\beta [2^7\cdot 3 (n+1)^3 ]^{1/2} \cdot x\cos\theta
+\beta x [ 2^5\cdot 3 (n+1)^3 ]^{1/2} \cdot 2\cos\theta
\biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~
x^2 \cdot 2(n+1)\biggl\{ [2^3(n+1)\cos^2\theta -3] + 2^3(n+1)(\sin^2\theta - \cos^2\theta) + [-3 + 2^3(n+1)\cos^2\theta] \biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
\pm~~i~x\beta \cos\theta\biggl\{
[ 2^5\cdot 3 (n+1)^3 ]^{1/2} \cdot 2 - [2^7\cdot 3 (n+1)^3 ]^{1/2}
\biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~
x^2 \cdot 2(n+1)\biggl\{2^3(n+1)-6\biggr\}
\pm~~i~x\beta \cos\theta\biggl\{ ~0~ \biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~
x^2 \cdot 4(n+1)(4n+1) \, .
</math>
  </td>
</tr>
 
</table>
 
Contrast this with,
 
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\mathrm{RHS}_4</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~x^2 ~ [m^2 \Lambda - 2^2(n+1)^2 ] \cdot \mathcal{A}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~x^2 ~ \biggl\{ m^2 \biggl[ -(4n+1)\beta^2 ~~\pm~i~\cancelto{0}{x}\beta \cos\theta [2^7\cdot 3(n+1)^3]^{1/2}\biggr]
- 2^2(n+1)^2 \biggr\} \cdot \mathcal{A}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~x^2 ~ \biggl\{ m^2 \biggl[ -(4n+1)\beta^2 \biggr]
- 2^2(n+1)^2 \biggr\} \cdot \mathcal{A}
</math>
  </td>
</tr>
</table>
 
Now, to lowest order [<font color="red">Case B</font>],
 
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~(n+1)\cdot \mathcal{A}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2n  (n+1)
+ (n+1)(1-\cancelto{0}{x}\cos\theta)^2 [ \cancelto{0}{ x^2}(1+xb) -  \beta^2 - 4n ]
+ (1-\cancelto{0}{x}\cos\theta)^4 [2n(n+1) - 3n\beta^2  ]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
\pm~~i~\cancelto{0}{x}\cos\theta (1-x\cos\theta)^2 [ 2^3\cdot 3 n^2\beta^2(n+1)  ]^{1/2}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~
2n  (n+1) + (n+1)[ -  \beta^2 - 4n ]+ [2n(n+1) - 3n\beta^2  ]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~-  \beta^2(4n+1)
</math>
  </td>
</tr>
 
</table>
 
Hence,
 
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\frac{1}{\beta^2}\cdot \mathrm{RHS}_4</math>
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~x^2 ~ \biggl\{ m^2 \biggl[ -(4n+1)\cancelto{0}{\beta^2} \biggr]
- 2^2(n+1)^2 \biggr\} \cdot \biggl[ \frac{-(4n+1)}{(n+1)} \biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~x^2 ~ \biggl\{ 
2^2(n+1) \biggr\} \cdot (4n+1) \, ,
</math>
  </td>
</tr>
</table>
and we see that, to lowest order, the two sides do match.  Notice that the mode number, <math>~m</math>, drops out in this lowest order approximation.
 
***************** -->
 
======Next Lowest Order======
 
Let's begin with the RHS (<font color="red">Case B</font>).
 
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~(n+1)\cdot \mathcal{A}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2n  (n+1)
+ (n+1)(1-x\cos\theta)^2 [  x^2(1+xb) -  \beta^2 - 4n ]
+ (1-x\cos\theta)^4 [2n(n+1) - 3n\beta^2  ]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
\pm~~i~x\cos\theta (1-x\cos\theta)^2 [ 2^3\cdot 3 n^2\beta^2(n+1)  ]^{1/2}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2n  (n+1)
+ (n+1)[1-2x\cos\theta + x^2\cos^2\theta + \mathcal{O}(x^3) ] [  x^2(1+xb) -  \beta^2 - 4n ]
+ [1-4x\cos\theta + 6x^2\cos^2\theta + \mathcal{O}(x^3) ] [2n(n+1) - 3n\beta^2  ]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
\pm~~i~x\cos\theta [1-2x\cos\theta + x^2\cos^2\theta + \mathcal{O}(x^3) ] [ 2^3\cdot 3 n^2\beta^2(n+1)  ]^{1/2}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~x^0\biggl\{2n  (n+1) -4n(n+1) + 2n(n+1) \biggr\}
+ x^1\biggl\{8n(n+1)\cos\theta - 8n(n+1)\cos\theta\biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ x^2\biggl\{-4n(n+1)\cos^2\theta + (n+1)\biggl[  1 -  \biggl(\frac{\beta}{x}\biggr)^2 \biggr] + 12n(n+1)\cos^2\theta - 3n \biggl(\frac{\beta}{x}\biggr)^2 \biggr\}
+ \mathcal{O}(x^3)
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
\pm~~i~x^2\biggl(\frac{\beta}{x}\biggr) \frac{nb_0}{2^2(n+1)} \biggl[1-2x\cos\theta + x^2\cos^2\theta + \mathcal{O}(x^3) \biggr] 
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~x^0\biggl\{0 \biggr\}
+ x^1\biggl\{0\biggr\}
+ x^2\biggl\{(n+1)[8n\cos^2\theta + 1] - (4n+1)\biggl(\frac{\beta}{x}\biggr)^2  \biggr\}
+ \mathcal{O}(x^3)
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
\pm~~i~x^2\biggl(\frac{\beta}{x}\biggr) \frac{nb_0}{2^2(n+1)} \biggl[1-2x\cos\theta + x^2\cos^2\theta + \mathcal{O}(x^3) \biggr] 
</math>
  </td>
</tr>
</table>
</div>
 
where,
<table border="0" align="center" cellpadding="8">
<tr><td align="center">
<math>b_0 \equiv [ 2^7\cdot 3 (n+1)^3 \cos^2\theta ]^{1/2} \, .</math>
</td></tr>
</table>
 
Hence,
 
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\frac{\mathrm{RHS}_4}{x^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-~ 2^2(n+1)^2\cdot \mathcal{A}  + m^2 \cdot \mathcal{A} \biggl\{
-(4n+1)\beta^2 + (\beta\eta)^2(n+1)[2^3(n+1)\cos^2\theta - 3] ~~~\pm ~~i~\beta\cos\theta [2^7\cdot 3 (n+1)^3]^{1/2} (\beta\eta)
\biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-~ 2^2(n+1)^2\cdot \mathcal{A}  + m^2 \cdot \mathcal{A} (n+1)\biggl\{
-x^2\biggl(\frac{4n+1}{n+1}\biggr)\cdot \biggl(\frac{\beta}{x}\biggr)^2 + x^2(1+\cancelto{0}{x}b)[2^3(n+1)\cos^2\theta - 3] ~~~\pm ~~i~x^2 \biggl(\frac{\beta}{x}\biggr) \frac{b_0}{(n+1)} (1+\cancelto{0}{x}b)^{1/2}
\biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~\mathcal{A} (n+1)\biggl\{-~ 2^2(n+1)  + m^2 \cancelto{0}{x^2} \biggl[
[2^3(n+1)\cos^2\theta - 3] -\biggl(\frac{4n+1}{n+1}\biggr)\cdot \biggl(\frac{\beta}{x}\biggr)^2 ~~~\pm ~~i~ \biggl(\frac{\beta}{x}\biggr) \frac{b_0}{(n+1)} 
\biggr] \biggl\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~-2^2(n+1)x^2\biggl\{(n+1)[8n\cos^2\theta + 1] - (4n+1)\biggl(\frac{\beta}{x}\biggr)^2 
~~~\pm~i~\biggl(\frac{\beta}{x}\biggr) \frac{nb_0}{2^2(n+1)} \biggl[1-2\cancelto{0}{x}\cos\theta + \cancelto{0}{x^2}\cos^2\theta + \cancelto{0}{\mathcal{O}}(x^3) \biggr] \biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\Rightarrow ~~~~\frac{\mathrm{RHS}_4}{x^4}</math>
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>2^2(n+1)(4n+1)\biggl(\frac{\beta}{x}\biggr)^2  ~-~2^2(n+1)^2[8n\cos^2\theta + 1]
~~\pm~i~(-1)\biggl(\frac{\beta}{x}\biggr) nb_0  \, .
</math>
  </td>
</tr>
</table>
 
This should be compared to,
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\frac{\mathrm{LHS}_4}{x^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \biggl(\frac{\beta}{x}\biggr)^2 - 1 - xb \biggr] (1-x\cos\theta)^3 \biggl\{ (1-x\cos\theta)
\biggl[x^2 \frac{\partial^2\Lambda}{\partial x^2} + \frac{\partial^2\Lambda}{\partial \theta^2} \biggr]
+ x\biggl[(1-2x\cos\theta)\frac{\partial\Lambda}{\partial x} + \sin\theta \frac{\partial\Lambda}{\partial\theta}  \biggr]
\biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
-~n x (1-x\cos\theta)^4\biggl[(2+3xb) \frac{\partial\Lambda}{\partial x} -3\sin^3\theta \frac{\partial\Lambda}{\partial\theta} \biggr] \, .
</math>
  </td>
</tr>
 
</table>
 
Now, from above, we can write,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\biggl[x^2 \frac{\partial^2\Lambda}{\partial x^2} + \frac{\partial^2\Lambda}{\partial \theta^2} \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~x^2\biggl\{
2(n+1)[2^3(n+1)\cos^2\theta -3](1+3\cancelto{0}{x}b) ~\pm~~i~\beta \biggl(\frac{b_0}{2^2}\biggr)\biggl[ \frac{b(4+3xb)}{(1+xb)^{3/2}} \biggr]
\biggr\} + x^2 \biggl\{2^4(n+1)^2(\sin^2\theta - \cos^2\theta)\biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \cancelto{0}{x^3}\biggl\{-2^4\cdot 3 (n+1)^2\cos^3\theta + 2^4(n+1)^2\cos^5\theta + 3^2(n+1)(16n +19)\sin^2\theta \cos\theta -2^3\cdot 23 (n+1)^2\sin^2\theta \cos^3\theta
\biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
\pm~~i~(-1)\beta b_0 ~x(1+\cancelto{0}{x}b)^{1/2}\biggl\{
1 + \frac{3\cancelto{0}{x}\sin^2\theta (5\cos^2\theta -2)}{2(1+xb)\cos\theta } + \frac{3^2\cancelto{0}{x^2}\sin^6\theta}{2^2(1+xb)^2}
\biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~x^2\biggl\{
2(n+1)[2^3(n+1)\cos^2\theta -3]
+ 2^4(n+1)^2(\sin^2\theta - \cos^2\theta)\biggr\}
~\pm~~i~x \biggl\{ \cancelto{0}{x\beta} \biggl(\frac{b_0}{2^2}\biggr)\biggl[ \frac{b(4+3xb)}{(1+xb)^{3/2}} \biggr]- \beta b_0\biggr\}</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~2(n+1)x^2\biggl\{ 2^3(n+1)\sin^2\theta -3 \biggr\}
~\pm~~i~x^2\biggl\{ - \biggl(\frac{\beta}{x}\biggr) b_0\biggr\} \, .</math>
  </td>
</tr>
</table>
</div>
 
Also,
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~ x\biggl[ (1-2x \cos\theta )\cdot \frac{\partial  \Lambda }{\partial x} + \sin\theta \cdot \frac{\partial\Lambda}{\partial\theta}\biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
x^2(n+1)[-6 + 2^4(n+1)\cos^2\theta]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
- \cancelto{0}{x^3}(n+1)\cos\theta \{ [ 15 + 2^4(n+1) ] -\cos^2\theta[9 + 2^3\cdot 7 (n+1)] +2^3\cdot 3(n+1)\cos^4\theta \}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+\cancelto{0}{x^4}(n+1) \{9 - 2^2\cdot 3^2(1+2n)\cos^2\theta - [9 + 32(n+1)]\cos^4\theta +2^3(n+1)\cos^6\theta
\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~\pm~~i~x^2 \biggl(\frac{\beta}{x}\biggr)~\biggl[ \frac{2^5\cdot 3 (n+1)^3}{1+\cancelto{0}{x}(3\cos\theta-\cos^3\theta)} \biggr]^{1/2} \biggl\{ 2\cos\theta
- \cancelto{0}{x}[2 - 7\cos^2\theta + 3\cos^4\theta ] </math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~- \cancelto{0}{x^2} \cos\theta [ 9 +4\cos^2\theta -\cos^4\theta ]
\biggr\}</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~
2(n+1)x^2[2^3(n+1)\cos^2\theta -3] ~\pm~~i~x^2 \biggl(\frac{\beta}{x}\biggr)b_0 \, .
</math>
  </td>
</tr>
 
</table>
 
Finally,
 
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~ nx\biggl[ (2+3xb )\cdot \frac{\partial  \Lambda }{\partial x} - 3\sin^3\theta \cdot \frac{\partial\Lambda}{\partial\theta} \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
x^2\cdot 2^2n(n+1)[2^3(n+1)\cos^2\theta -3]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ n \cancelto{0}{x^3}\cdot 2^2\cdot 3(n+1)\cos\theta \{ [2^2n -5] +\cos^2\theta[2^4n + 19] - 2^2(n+1)\cos^4\theta \}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+n \cancelto{0}{x^4} \cdot 3^2(n+1)\cos^2\theta \{ -3^3 + 2\cdot 3^2\cos^2\theta[2^2n+5] - 99\cdot \cos^4\theta + 2^3(n+1)\cos^6\theta \}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ n\cancelto{0}{x^4} (n+1)\sin^4\theta \{  -3^3 + 3^2\cos^2\theta [ 2^3\cdot 3 n + 27] -2^3\cdot 3\cdot 5(n+1)\cos^4\theta \}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~\pm~~i~nx^2\biggl(\frac{\beta}{x}\biggr)~ \biggl[ \frac{2^5\cdot 3(n+1)^3}{1+\cancelto{0}{x}b} \biggr]^{1/2} \biggl\{
4\cos\theta + 6\cancelto{0}{x}(2b\cos\theta + \sin^4\theta) + 3\cancelto{0}{x^2}(3b^2\cos\theta +2b\sin^4\theta + 3\sin^6\theta\cos\theta)
\biggr\} </math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~
x^2\cdot 2^2n(n+1)[2^3(n+1)\cos^2\theta -3] ~~\pm~~i~2nx^2\biggl(\frac{\beta}{x}\biggr)b_0 \, .
</math>
  </td>
</tr>
</table>
 
Inserting these three approximate expressions into the LHS_4 ensemble gives,
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\frac{\mathrm{LHS}_4}{x^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \biggl(\frac{\beta}{x}\biggr)^2 - 1 - \cancelto{0}{xb} \biggr] (1-\cancelto{0}{x}\cos\theta)^3 \biggl\{ (1-\cancelto{0}{x}\cos\theta)
\biggl[x^2 \frac{\partial^2\Lambda}{\partial x^2} + \frac{\partial^2\Lambda}{\partial \theta^2} \biggr]
+ x\biggl[(1-2x\cos\theta)\frac{\partial\Lambda}{\partial x} + \sin\theta \frac{\partial\Lambda}{\partial\theta}  \biggr]
\biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
-~n x (1-\cancelto{0}{x}\cos\theta)^4\biggl[(2+3xb) \frac{\partial\Lambda}{\partial x} -3\sin^3\theta \frac{\partial\Lambda}{\partial\theta} \biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~</math>
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~
\biggl[ \biggl(\frac{\beta}{x}\biggr)^2 - 1 \biggr]  \biggl\{
~2(n+1)x^2\biggl[ 2^3(n+1)\sin^2\theta -3 \biggr]
~\pm~~i~x^2\biggl[ - \biggl(\frac{\beta}{x}\biggr) b_0\biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~</math>
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ 2(n+1)x^2[2^3(n+1)\cos^2\theta -3] ~\pm~~i~x^2 \biggl(\frac{\beta}{x}\biggr)b_0
\biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
-~x^2\cdot 2^2n(n+1)[2^3(n+1)\cos^2\theta -3] ~~\pm~~i~2nx^2\biggl(\frac{\beta}{x}\biggr)b_0
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\Rightarrow~~~\frac{\mathrm{LHS}_4}{x^4}</math>
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~
\biggl[ \biggl(\frac{\beta}{x}\biggr)^2 - 1 \biggr]  \biggl\{
~2(n+1)\biggl[ 2^3(n+1)\sin^2\theta -3 \biggr] + 2(n+1)[2^3(n+1)\cos^2\theta -3] \biggr\} -~ 2^2n(n+1)[2^3(n+1)\cos^2\theta -3]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~</math>
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
~\pm~~i~2n\biggl(\frac{\beta}{x}\biggr)b_0
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~
2^2(n+1) \biggl(\frac{\beta}{x}\biggr)^2  [2^2(n+1)  - 3 ]
-(n+1) \biggl\{ [2^4(n+1)  -12]
+~ 2^2n[2^3(n+1)\cos^2\theta -3] \biggr\}
~\pm~~i~2n\biggl(\frac{\beta}{x}\biggr)b_0
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~
2^2(n+1) \biggl(\frac{\beta}{x}\biggr)^2  [4n+1 ]
- 2^2(n+1)^2 [ 1+~ 2^3n\cos^2\theta ] ~\pm~~i~2n\biggl(\frac{\beta}{x}\biggr)b_0
\, .
</math>
  </td>
</tr>
 
</table>
 
======Assessment======
 
The good news is that the real part of the <math>~\mathrm{LHS}_4</math> expression exactly matches the real part of the <math>~\mathrm{RHS}_4</math> expression.  But the imaginary differ by a factor of 2.  So, let's repeat the steps leading to the imaginary parts.
 
'''<font color="red" size="+1">Case B:</font>'''
 
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\mathrm{Im}\biggl[ \frac{\mathrm{RHS_4}}{x^2} \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-~2^2(n+1)\cdot \mathrm{Im}[(n+1)\mathcal{A}]
+\frac{m^2}{(n+1)}\biggl\{ \mathrm{Im}[(n+1)\mathcal{A}]\cdot \mathrm{Re}[\Lambda] + \mathrm{Re}[(n+1)\mathcal{A}]\cdot \mathrm{Im}[\Lambda]  \biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-~2^2(n+1)\cdot x\cos\theta (1-x\cos\theta)^2 [ 2^3\cdot 3 n^2\beta^2(n+1)  ]^{1/2}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+\frac{m^2}{(n+1)}\biggl\{ x\cos\theta (1-x\cos\theta)^2 [ 2^3\cdot 3 n^2\beta^2(n+1)  ]^{1/2} \biggr\} \cdot \biggl\{ -(4n+1)\beta^2 + (\beta\eta)^2(n+1)[2^3(n+1)\cos^2\theta - 3] \biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+\frac{m^2}{(n+1)}\biggl\{ 2n  (n+1)
+ (n+1)(1-x\cos\theta)^2 [  x^2(1+xb) -  \beta^2 - 4n ]
+ (1-x\cos\theta)^4 [2n(n+1) - 3n\beta^2  ]
\biggr\} \cdot \biggl\{\beta\cos\theta [2^7\cdot 3 (n+1)^3]^{1/2} (\beta\eta) \biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-~x^2 \biggl(\frac{\beta}{x}\biggr) \biggl\{ (1-x\cos\theta)^2 nb_0 \biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+m^2 x^4\biggl(\frac{\beta}{x}\biggr)\biggl\{ (1-x\cos\theta)^2 \biggl[\frac{n b_0}{2^2(n+1)^2} \biggr] \biggr\} \cdot
\biggl\{ -\biggl[ \frac{4n+1}{n+1}\biggr] \biggl(\frac{\beta}{x}\biggr)^2 + (1+xb)[2^3(n+1)\cos^2\theta - 3] \biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ m^2 x^2\biggl(\frac{\beta}{x}\biggr) \biggl\{ 2n
+ (1-x\cos\theta)^2 [  x^2(1+xb) -  \beta^2 - 4n ]
+ (1-x\cos\theta)^4\cdot \biggl[2n - \frac{3n\beta^2}{(n+1)}  \biggr]
\biggr\} \cdot \biggl\{(1+xb)^{1/2} \biggr\}
</math>
  </td>
</tr>
</table>
</div>
 
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\Rightarrow~~~\mathrm{Im}\biggl[ \frac{\mathrm{RHS_4}}{x^4} \biggr]\biggl(\frac{x}{\beta}\biggr) </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-~(1-x\cos\theta)^2 nb_0
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ m^2  \biggl\{ 2n
+ (1-x\cos\theta)^2 [  x^2(1+xb) -  \beta^2 - 4n ]
+ (1-x\cos\theta)^4\cdot \biggl[2n - \frac{3n\beta^2}{(n+1)}  \biggr]
\biggr\} \cdot (1+xb)^{1/2}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+m^2
\biggl\{ -\biggl[ \frac{4n+1}{n+1}\biggr] \beta^2 + x^2(1+xb)[2^3(n+1)\cos^2\theta - 3] \biggr\}
\cdot (1-x\cos\theta)^2 \biggl[\frac{n b_0}{2^2(n+1)^2} \biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-~(1-x\cos\theta)^2 nb_0
+ m^2  \biggl\{ 2n - 4n (1-x\cos\theta)^2 + 2n(1-x\cos\theta)^4 \biggr\} \cdot (1+xb)^{1/2}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ m^2 (1-x\cos\theta)^2 \biggl\{ x^2(1+xb)^{3/2} -  \beta^2\cdot (1+xb)^{1/2} \biggl[ 1
+ \frac{3n(1-x\cos\theta)^2}{(n+1)}  \biggr]
\biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+m^2  (1-x\cos\theta)^2
\biggl\{ x^2(1+xb)[2^3(n+1)\cos^2\theta - 3]\cdot\biggl[\frac{n b_0}{2^2(n+1)^2} \biggr] -\biggl[ \frac{nb_0(4n+1)}{2^2(n+1)^3}\biggr] \beta^2 \biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-~(1-x\cos\theta)^2 nb_0
+ m^2  \biggl\{ 2n - 4n [1-2x\cos\theta + x^2\cos^2\theta + \mathcal{O}(x^3)]
+ 2n[ 1-4x\cos\theta + 6x^2\cos^2\theta + \mathcal{O}(x^3)] \biggr\} \cdot (1+xb)^{1/2}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ m^2 x^2 (1-x\cos\theta)^2 \biggl\{ 2^2(n+1)^2(1+xb)^{1/2} + [2^3(n+1)\cos^2\theta - 3]\cdot n b_0
\biggr\} \cdot \frac{(1+xb)}{2^2(n+1)^2}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
- m^2 \beta^2 (1-x\cos\theta)^2
\biggl\{ nb_0(4n+1)  +  2^2(n+1)^2(1+xb)^{1/2} [ (n+1)
+ 3n(1-x\cos\theta)^2  ] \biggr\} \cdot \frac{1}{2^2(n+1)^3}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-~ nb_0 [1 -2x\cos\theta + x^2\cos^2\theta + \mathcal{O}(x^3)]
+ m^2  \biggl\{
8n x^2\cos^2\theta + \mathcal{O}(x^3)\biggr\} \cdot (1+\cancelto{0}{x}b)^{1/2}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ m^2 x^2 (1-\cancelto{0}{x}\cos\theta)^2 \biggl\{ 2^2(n+1)^2(1+\cancelto{0}{x}b)^{1/2} + [2^3(n+1)\cos^2\theta - 3]\cdot n b_0
\biggr\} \cdot \frac{(1+\cancelto{0}{x}b)}{2^2(n+1)^2}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
- m^2 \beta^2 (1-\cancelto{0}{x}\cos\theta)^2
\biggl\{ nb_0(4n+1)  +  2^2(n+1)^2(1+\cancelto{0}{x}b)^{1/2} [ (n+1)
+ 3n(1-\cancelto{0}{x}\cos\theta)^2  ] \biggr\} \cdot \frac{1}{2^2(n+1)^3}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~-~ nb_0 [1 -2x\cos\theta  ]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~-nb_0x^2\cos^2\theta
+ m^2 x^2 \biggl\{2^5n(n+1)^2 \cos^2\theta + 2^2(n+1)^2 + [2^3(n+1)\cos^2\theta - 3]\cdot n b_0
\biggr\} \cdot \frac{1}{2^2(n+1)^2}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
- m^2 \beta^2 
\biggl\{ nb_0(4n+1)  +  2^2(n+1)^2 [ (n+1)
+ 3n ] \biggr\} \cdot \frac{1}{2^2(n+1)^3}
</math>
  </td>
</tr>
</table>
</div>
 
==Goldreich, Goodman and Narayan (1986)==
 
===Unperturbed Slim Torus Structure===
[http://adsabs.harvard.edu/abs/1986MNRAS.221..339G Goldreich, Goodman &amp; Narayan (1986, MNRAS, 221, 339)] &#8212; hereafter, GGN86 &#8212; also used analytic techniques to analyze the properties of unstable, nonaxisymmetric eigenmodes in Papaloizou-Pringle tori.  They restricted their discussion to only the slimmest tori, so overlap between the GGN86 and Blaes85 work is easiest to recognize if we begin with the enthalpy distribution prescribed for a "slim torus" by Blaes (1985), as [[#Blaes85|discussed above]], namely,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~H = H_0\Theta_H</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~H_0 - \frac{H_0}{\beta^2}\biggl[r^2 + r^3(3\cos\theta - \cos^3\theta) + \mathcal{O}(r^4)  \biggr] \, .</math>
  </td>
</tr>
</table>
</div>
[<font color="red"><b>Note:</b></font> &nbsp; Here we have replaced the variable name, <math>~x</math>, as used in Blaes85, with the variable name, <math>~r</math>, in order (1) to emphasize that the variable represents a dimensionless ''radial'' coordinate, and (2) to avoid conflict with the GGN86 variable, <math>~x</math>, which is a Cartesian coordinate with the standard dimension of length.]
 
Now, from our [[#Equilibrium_Configuration|above discussion of equilibrium PP tori]] and recognizing that the Keplerian angular frequency at the location of the enthalpy maximum is,
 
<div align="center">
<math>\Omega_K \equiv \frac{GM_\mathrm{pt}}{\varpi_0^3} \, ,</math>
</div>
 
we can set,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~H_0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{GM_\mathrm{pt}\beta^2}{2\varpi_0} = \tfrac{1}{2} \Omega_K^2 \varpi_0^2 \beta^2 \, .</math>
  </td>
</tr>
</table>
</div>
Hence, for the slimmest tori &#8212; that is, keeping only the lowest order term in <math>~r</math> &#8212; the enthalpy distribution becomes,
 
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~H </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\tfrac{1}{2} \Omega_K^2 \varpi_0^2 \beta^2
- \tfrac{1}{2} \Omega_K^2 \varpi_0^2\biggl[r^2 + \cancelto{0}{r^3}(3\cos\theta - \cos^3\theta) + \cancelto{0}{\mathcal{O}(r^4)} ~~\biggr] </math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~\frac{\Omega_K^2}{2} [\varpi_0^2 \beta^2  -  \varpi_0^2 r^2] \, .</math>
  </td>
</tr>
</table>
</div>
 
Following GGN86, the surface of the torus &#8212; where the enthalpy drops to zero &#8212; occurs at <math>~r = a/\varpi_0</math>.  Hence, we recognize that,
 
<div align="center">
<math>\beta = \frac{a}{\varpi_0} \, ,</math>
</div>
 
and we can rewrite the expression for the unperturbed enthalpy distribution as,
 
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~H </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\Omega_K^2}{2} [a^2  - \varpi_0^2 r^2 ] \, .</math>
  </td>
</tr>
</table>
</div>
This expression exactly matches equation (2.13) of GGN86 &#8212; which, is,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~Q_0(x,z)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\Omega^2}{2} \biggl[ (2q-3)(a^2 - x^2) - z^2 \biggr] \, ,</math>
  </td>
</tr>
</table>
</div>
 
once it is appreciated that, in moving from the Blaes85 discussion to the GGN86 discussion, <math>~\varpi_0^2 r^2 \rightarrow (x^2 + z^2)</math>, and it is recognized that Blaes85 restricted his investigation to tori that have uniform specific angular momentum <math>~(q = 2)</math>.
 
===Additional Notation===
 
<div align="center">
<math>~(ky)_\mathrm{GGN} = \biggl( \frac{my}{\varpi_0} \biggr)_\mathrm{GGN} ~~\leftrightarrow ~~ (m\phi)_\mathrm{Blaes}</math>
</div>
 
<div align="center">
<math>~\beta_\mathrm{GGN} \equiv \biggl( \frac{ma}{\varpi_0} \biggr)_\mathrm{GGN} ~~\leftrightarrow ~~ m\beta_\mathrm{Blaes}</math>
</div>
 
From equation (5.16) of GGN86 we obtain "the lowest order [complex] expression for the [perturbed] velocity potential," namely,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\psi </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~1+\tfrac{1}{4} k^2(5x^2 - 3z^2) \mp 4i\biggl(\frac{3}{2}\biggr)^{1/2} k x \beta_\mathrm{GGN} \, .</math>
  </td>
</tr>
</table>
</div>
 
Working on the imaginary part of this expression to put it in the terminology of Blaes85, we find,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\mathrm{Im}(\psi)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\mp 4\biggl(\frac{3}{2}\biggr)^{1/2} k x \beta_\mathrm{GGN} </math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
   </td>
   <td align="center">
<math>~=</math>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~\mp 4\biggl(\frac{3}{2}\biggr)^{1/2} \biggl(\frac{m}{\varpi_0}\biggr) [\varpi_0 (\eta\beta_\mathrm{Blaes})\cos\theta ](m\beta_\mathrm{Blaes}) </math>
- 2 a_2 + 2\cdot 3(S + T)  
</math>
   </td>
   </td>
</tr>
</tr>
Line 613: Line 7,364:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~\mp 4\biggl(\frac{3}{2}\biggr)^{1/2} m^2\beta^2_\mathrm{Blaes} \eta\cos\theta \, ,</math>
- 1 + \biggl\{r_S e^{i\theta_S} \biggr\}^{1/3} + \biggl\{r_T e^{i\theta_T} \biggr\}^{1/3}
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>


<tr>
which exactly matches <math>~\mathrm{Im}(f_m)</math> as derived by Blaes85 and [[#Incompressible_Slim_Tori|summarized above]].  Similarly,
  <td align="right">
<div align="center">
&nbsp;
<table border="0" cellpadding="5" align="center">
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- 1 + e^{i\theta_S/3}  + e^{i\theta_T/3}
</math>
  </td>
</tr>


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\mathrm{Re}(\psi)</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 641: Line 7,382:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~1+\tfrac{1}{4} k^2(5x^2 - 3z^2) </math>
- 1 + e^{i\theta_S/3} + e^{-i\theta_S/3}
</math>
   </td>
   </td>
</tr>
</tr>
Line 655: Line 7,394:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~1+\frac{1}{4} \biggl(\frac{m}{\varpi_0}\biggr)^2[\varpi_0^2 r^2(5\cos^2\theta - 3\sin^2\theta)] </math>
-1 + \biggl[ \cos\biggl(\frac{\theta_S}{3}\biggr) + i\sin\biggl(\frac{\theta_S}{3}\biggr) \biggr]
+ \biggl[ \cos\biggl(\frac{\theta_S}{3}\biggr) - i\sin\biggl(\frac{\theta_S}{3}\biggr) \biggr]
</math>
   </td>
   </td>
</tr>
</tr>
Line 670: Line 7,406:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~1+\frac{1}{4} \eta^2 m^2 \beta^2_\mathrm{Blaes}[8\cos^2\theta - 3] </math>
-1 + 2\biggl[ \cos\biggl(\frac{\theta_S}{3}\biggr) \biggr]  
</math>
   </td>
   </td>
</tr>
</tr>
Line 684: Line 7,418:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~1+m^2 \beta^2_\mathrm{Blaes}\biggl[2\eta^2\cos^2\theta - \frac{3\eta^2}{4}\biggr] \, .</math>
2\cos\biggl[\frac{1}{3}\cos^{-1}(1-\Gamma^2)\biggr] -1 \, .
</math>
   </td>
   </td>
</tr>
</tr>
Line 692: Line 7,424:
</div>
</div>


===Analytically Prescribed Eigenvector===
This exactly matches <math>~\mathrm{Re}(f_m)</math> as derived by Blaes85 and [[#Incompressible_Slim_Tori|summarized above]].  This is in line with the following statement that appears in the acknowledgement section of GGN86: "We note that Omar Blaes &hellip; [has] independently derived many of the results reported in this paper."
From my initial focused reading of the analysis presented by [http://adsabs.harvard.edu/abs/1985MNRAS.216..553B Blaes (1985)], I conclude that, in the infinitely slender torus case, unstable modes in PP tori exhibit eigenvectors of the form,
 
==Summary Comparison==
 
For slim, incompressible tori with uniform specific angular momentum, [[#Incompressible_Slim_Tori|Blaes85 gives, to lowest order]]:
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 699: Line 7,434:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\frac{\delta W}{W_0} \equiv \biggl[ \frac{W(\eta,\theta)}{C} - 1 \biggr]e^{im\Omega_p t}e^{-y_2 (\Omega_0 t)} </math>
<math>~f_m(\eta,\theta) + \tfrac{1}{4} \beta^2 m^2</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 705: Line 7,440:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl\{ f_m(\eta,\theta)e^{-i[m\phi_m(\varpi) + k\theta]} \biggr\}  \, ,</math>
<math>
~\beta^2 m^2 \biggl[ 2\eta^2 \cos^2\theta - \frac{3\eta^2}{4} 
\pm 4i\biggl(\frac{3}{2}\biggr)^{1/2} \eta\cos\theta\biggr] \, .
</math>
   </td>
   </td>
</tr>
</tr>
Line 711: Line 7,449:
</div>
</div>


where we have written the perturbation amplitude in a manner that conforms with the notation that we have used in  [[User:Tohline/Appendix/Ramblings/Azimuthal_Distortions#Figure1|Figure 1 of a related, but more general discussion]]. As is summarized in &sect;1.3 of Blaes (1985), for "thick" (but, actually, still quite thin) tori, "exactly one exponentially growing mode exists for each value of the azimuthal wavenumber <math>~m</math>," and its complex amplitude takes the following form (see his equation 1.10):
By comparison, from [http://adsabs.harvard.edu/abs/1986MNRAS.221..339G GGN86] we obtain:
 
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 718: Line 7,455:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~f_m(\eta,\theta)</math>
<math>~\Psi(\eta,\theta) - 1</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 724: Line 7,461:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>
<math>~
~\beta^2 m^2 \biggl[ 2\eta^2 \cos^2\theta - \frac{3\eta^2}{4(n+1)} - \frac{(4n+1)}{4(n+1)^2}  
m^2 \beta^2_\mathrm{Blaes}\biggl[2\eta^2\cos^2\theta - \frac{3\eta^2}{4}  
\pm 4i\biggl(\frac{3}{2n+2}\biggr)^{1/2} \eta\cos\theta\biggr]  
\mp 4i\biggl(\frac{3}{2}\biggr)^{1/2} \eta\cos\theta \biggr] \, .
+ \mathcal{O}(\beta^3) \, .
</math>
</math>
   </td>
   </td>
Line 734: Line 7,470:
</div>
</div>


We should therefore find that the amplitude (modulus) of the enthalpy perturbation is,
To within an additive constant, these two functions are identical.  The resulting amplitude function is (to within an overall scale factor and to within an arbitrary additive constant),
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\biggl|\frac{\delta W}{W_0} \biggr|</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\sqrt{[\mathrm{Re}(f_m)]^2+ [\mathrm{Im}(f_m)]^2} \, ;</math>
  </td>
</tr>
</table>
</div>
and the associated phase function should be,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~m\phi_m + k\theta</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\tan^{-1} \biggl\{ \frac{-\mathrm{Re}(f_m)}{\mathrm{Im}(f_m)} \biggr\}</math>
  </td>
</tr>
</table>
</div>


Now, keeping in mind that, for the time being, we are only interested in examining the shape of the unstable eigenvector in the ''equatorial plane'' of the torus, we can set,
<div align="center">
<math>~\cos\theta ~~ \rightarrow ~~ \pm 1 \, .</math>
</div>
Hence, we have,
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 779: Line 7,477:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\frac{1}{\beta^4 m^4}\biggl|\frac{\delta W}{W_0} \biggr|^2</math>
<math>~F(\eta,\theta)</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 785: Line 7,483:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl[2\eta^2 - \frac{3\eta^2}{4(n+1)} - \frac{(4n+1)}{4(n+1)^2}\biggr]^2
<math>~\biggl\{ 
+ 16\biggl[\biggl(\frac{3}{2n+2}\biggr)^{1/2} \eta \biggr]^2 </math>
\biggl[ 2\eta^2\cos^2\theta - \frac{3\eta^2}{4} \biggr]^2 + \biggl[ 4\biggl(\frac{3}{2}\biggr)^{1/2} \eta\cos\theta \biggr]^2
\biggr\}^{1/2}</math>
   </td>
   </td>
</tr>
</tr>
Line 798: Line 7,497:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{1}{[2^2(n+1)^2]^2}\biggl[2^3(n+1)^2\eta^2  - 3(n+1)\eta^2 - (4n+1) \biggr]^2
<math>~\biggl\{  \eta^2 \biggl[ 4^2\biggl(\frac{3}{2}\biggr) \cos^2\theta \biggr] +
+ \frac{2^3 \cdot 3\eta^2}{(n+1)}   </math>
\frac{1}{4^2}\eta^4\biggl[ 8\cos^2\theta - 3\biggr]^2  
\biggr\}^{1/2}</math>
   </td>
   </td>
</tr>
</tr>
Line 811: Line 7,511:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{1}{[2^2(n+1)^2]^2}\biggl[(4n+1) - (n+1)[2^3(n+1)-3]\eta^2 \biggr]^2
<math>~\biggl\{  \biggl(\frac{\eta}{2}\biggr)^2 \biggl[ 2^5\cdot 3 \cos^2\theta \biggr] +
+ \frac{2^3 \cdot 3\eta^2}{(n+1)}  </math>
\biggl( \frac{\eta}{2}\biggr)^4\biggl[ 8\cos^2\theta - 3\biggr]^2  
  </td>
\biggr\}^{1/2} \, ;</math>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow ~~~~ \biggl[\frac{2(n+1)}{\beta m} \biggr]^4 \biggl|\frac{\delta W}{W_0} \biggr|^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[(4n+1) - (n+1)[2^3(n+1)-3]\eta^2  \biggr]^2
+ 2^7 \cdot 3(n+1)^3\eta^2 \, </math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</div>
</div>
Also,


and the associate phase function is,
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
  <td align="center" bgcolor="blue">&nbsp;</td>
   <td align="right">
   <td align="right">
<math>~m\phi_m</math>
<math>~m\phi_m </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 844: Line 7,531:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\tan^{-1}\biggl\{  
<math>~\tan^{-1} \biggl[  
\frac{(4n+1) - (n+1)[2^3(n+1)-3]\eta^2  }{2^{7/2}\cdot 3^{1/2}(n+1)^{3/2}\eta}
\frac{ 2^7 \cdot 3 \cos^2\theta }{\eta^2 ( 8\cos^2\theta - 3)^2}
\biggr\}</math>
\biggr] + k\theta \, .</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; &nbsp; over &nbsp; &nbsp; &nbsp; &nbsp;
  </td>
  <td align="left">
inner region of the torus;
  </td>
</tr>
<tr><td colspan="6" align="center">while</td></tr>
<tr>
  <td align="center" bgcolor="green">&nbsp;</td>
  <td align="right">
<math>~m\phi_m</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\tan^{-1}\biggl\{ 
\frac{(4n+1) - (n+1)[2^3(n+1)-3]\eta^2  }{2^{7/2}\cdot 3^{1/2}(n+1)^{3/2}\eta}
\biggr\} - k\pi</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; &nbsp; over &nbsp; &nbsp; &nbsp; &nbsp;
  </td>
  <td align="left">
outer region of torus.
   </td>
   </td>
</tr>
</tr>

Latest revision as of 23:18, 20 December 2016

Stability Analyses of PP Tori

[Comment by J. E. Tohline on 24 May 2016]   This chapter contains a set of technical notes and accompanying discussion that I put together several months ago as I was trying to gain a foundational understanding of the results of a large study of instabilities in self-gravitating tori published by the Imamura & Hadley collaboration. I have come to appreciate that some of the logic and interpretation of published results that are presented, below, has serious flaws. Therefore, anyone reading this should be quite cautious in deciding what subsections provide useful insight. I have written a separate chapter titled, "Characteristics of Unstable Eigenvectors in Self-Gravitating Tori," that contains a much more trustworthy analysis of this very interesting problem.


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Material that appears after this point in our presentation is under development and therefore
may contain incorrect mathematical equations and/or physical misinterpretations.
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As has been summarized in an accompanying chapter — also see our related detailed notes — we have been trying to understand why unstable nonaxisymmetric eigenvectors have the shapes that they do in rotating toroidal configurations. For any azimuthal mode, <math>~m</math>, we are referring both to the radial dependence of the distortion amplitude, <math>~f_m(\varpi)</math>, and the radial dependence of the phase function, <math>~\phi_\mathrm{max}(\varpi)</math> — the latter is what the Imamura and Hadley collaboration refer to as a "constant phase locus." Some old videos showing the development over time of various self-gravitating "constant phase loci" can be found here; these videos supplement the published work of Woodward, Tohline & Hachisu (1994).


Here, we focus specifically on instabilities that arise in so-called (non-self-gravitating) Papaloizou-Pringle tori and will draw heavily from three publications:

PP III

Figure 2 extracted without modification from p. 274 of J. C. B. Papaloizou & J. E. Pringle (1987)

"The Dynamical Stability of Differentially Rotating Discs.   III"

MNRAS, vol. 225, pp. 267-283 © The Royal Astronomical Society

Figure 2 from PP III

Blaes (1985)

Equilibrium Configuration

In our separate discussion of PP84, we showed that the equilibrium structure of a PP-torus is defined by the enthalpy distribution,

<math> H = \frac{GM_\mathrm{pt}}{\varpi_0} \biggl[ (\chi^2 + \zeta^2)^{-1/2} - \frac{1}{2}\chi^{-2} - C_B^' \biggr] . </math>

Normalizing this expression by the enthalpy at the "center" — i.e., at the pressure maximum — of the torus which, as we have already shown, is

<math> H_0 = \frac{GM_\mathrm{pt}}{2\varpi_0} [1-2C_B^' ] \, </math>

gives,

<math> [1-2C_B^' ]\biggl(\frac{H}{H_0}\biggr) = 2(\chi^2 + \zeta^2)^{-1/2} - \chi^{-2} -1 + [1 - 2C_B^' ]. </math>

Now, in our review of Kojima's (1986) work, we showed that the square of the Mach number at the "center" of the torus is,

<math>~\mathfrak{M}_0^2 \equiv \frac{(v_\varphi|_0)^2}{(c_s|_0)^2}</math>

<math>~=</math>

<math>~\frac{2(n+1)}{\gamma}\biggl[ \frac{1}{\chi_-} - 1 \biggr]^{-2}</math>

 

<math>~=</math>

<math>~2n [ 1- 2C_B^' ]^{-1} </math>

<math>~\Rightarrow ~~~~ [1 - 2C_B^'] </math>

<math>~=</math>

<math>~\frac{2n}{\mathfrak{M}_0^2} \, , </math>

where, in obtaining this last expression we have related the adiabatic exponent to the polytropic index via the relation, <math>~\gamma = (n+1)/n</math>. Instead of specifying the system's Mach number, Blaes (1985) defines the dimensionless parameter,

<math>~\beta^2 </math>

<math>~\equiv</math>

<math>~\frac{2n}{\mathfrak{M}_0^2} \, .</math>

Implementing this parameter swap, the equilibrium expression becomes,

<math> \beta^2 \biggl(\frac{H}{H_0}\biggr) = 2(\chi^2 + \zeta^2)^{-1/2} - \chi^{-2} -1 + \beta^2 \, , </math>

or,

<math>~\frac{H}{H_0} </math>

<math>~=</math>

<math>~1 - \frac{1}{\beta^2}\biggl[\chi^{-2} - 2(\chi^2 + \zeta^2)^{-1/2} + 1 \biggr] \, .</math>

Looking at Figure 1 of Blaes85 — see also the coordinate definitions given in his equation (2.1) — I conclude that,

<math>~\chi = 1 - x\cos\theta</math>       and         <math>\zeta = x\sin\theta \, .</math>

Hence,

<math>~\frac{H}{H_0} </math>

<math>~=</math>

<math>~1 - \frac{1}{\beta^2}\biggl\{ [1 - x\cos\theta]^{-2} - 2[(1 - x\cos\theta)^2 + x^2\sin^2\theta]^{-1/2} + 1 \biggr\} </math>

 

<math>~=</math>

<math>~1 - \frac{1}{\beta^2}\biggl\{ [1 - x\cos\theta]^{-2} - 2[(1 - 2x\cos\theta + x^2\cos^2\theta) + x^2(1-\cos^2\theta)]^{-1/2} + 1 \biggr\} </math>

 

<math>~=</math>

<math>~1 - \frac{1}{\beta^2}\biggl[ (1 - x\cos\theta)^{-2} - 2(1 - 2x\cos\theta + x^2)^{-1/2} + 1 \biggr] \, .</math>

This matches equation (2.2) of Blaes85, if we acknowledge that Blaes uses <math>~f</math> — instead of the parameter notation, <math>~\Theta_H</math>, found in our discussion of equilibrium polytropic configurations — to denote the normalized enthalpy; that is,

<math>~f_\mathrm{Blaes85} = \Theta_H \equiv \frac{H}{H_0} \, .</math>

This expression for the enthalpy throughout a Papaloizou-Pringle torus is valid for tori of arbitrary thickness <math>~(0 < \beta < 1)</math>. When considering only slim tori, Blaes (1985) points out that this expression can be written in terms of the following power series in <math>~x</math> (see his equation 1.3):

<math>~\Theta_H</math>

<math>~=</math>

<math>~1 - \frac{1}{\beta^2}\biggl[x^2 + x^3(3\cos\theta - \cos^3\theta) + \mathcal{O}(x^4) \biggr] \, .</math>

Blaes then adopts a related parameter that is constant on isobaric surfaces, namely,

<math>\eta^2 \equiv 1 - \Theta_H \, ,</math>

which is unity at the surface of the torus and which goes to zero at the cross-sectional center of the torus. Notice that <math>~\eta</math> tracks the "radial" coordinate that measures the distance from the center of the torus; in particular, keeping only the leading-order term in <math>~x</math>, there is a simple linear relationship between <math>~\eta</math> and <math>~x</math>, namely,

<math>~\eta</math>

<math>~=</math>

<math>~[1 - \Theta_H]^{1/2} \approx \frac{x}{\beta} \, .</math>

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 — that is, only for the values of <math>~\theta = 0</math> or <math>~\theta=\pi</math> — the relation to be inverted is,

<math>~x^2 \pm 2 x^3</math>

<math>~=</math>

<math>~(\beta\eta)^2</math>

<math>~\Rightarrow ~~~~ x^3 \pm \tfrac{1}{2}x^2 \mp \tfrac{1}{2}(\beta\eta)^2</math>

<math>~=</math>

<math>~0 \, .</math>


Table 1:  Example Parameter Values

determined by iterative solution for <math>~\beta = \tfrac{1}{10}</math>
<math>~\eta</math> <math>~\Gamma^2 = 54\beta^2\eta^2</math> Inner solution <math>~(\theta = 0)</math>

[Superior sign in cubic eq.]
Outer solution <math>~(\theta = \pi)</math>

[Inferior sign in cubic eq.]
<math>^\dagger\biggl(\frac{x_\mathrm{root}}{\beta}\biggr)</math> <math>~6(S + T)</math> <math>^\dagger\biggl(\frac{x_\mathrm{root}}{\beta}\biggr)</math> <math>~6(S + T)</math>
0.25 0.03375 0.244112 1.14647 0.256675 -0.84600
1.0 0.54 0.91909 1.55145 1.1378 -0.31732

Here, <math>~x_\mathrm{root}</math> has been determined via a brute-force, iterative technique.


Following Wolfram's discussion of the cubic formula, we should view this expression as a specific case of the general formula,

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

in which case, as is detailed in equations (54) - (56) of Wolfram's discussion of the cubic formula, the three roots of any cubic equation are:

<math>~x_1</math>

<math>~=</math>

<math>~ -\frac{1}{3}a_2 + (S + T) \, , </math>

<math>~x_2</math>

<math>~=</math>

<math>~ -\frac{1}{3}a_2 - \frac{1}{2} (S + T) + \frac{1}{2} \it{i} \sqrt{3} (S-T)\, , </math>

<math>~x_3</math>

<math>~=</math>

<math>~ -\frac{1}{3}a_2 - \frac{1}{2} (S + T) - \frac{1}{2} \it{i} \sqrt{3} (S-T)\, , </math>

where,

<math>~S</math>

<math>~\equiv</math>

<math>~[R + \sqrt{D}]^{1/3} \, ,</math>

<math>~T</math>

<math>~\equiv</math>

<math>~[R - \sqrt{D}]^{1/3} \, ,</math>

<math>~D</math>

<math>~\equiv</math>

<math>~Q^3 + R^2 \, ,</math>

<math>~Q</math>

<math>~\equiv</math>

<math>~\frac{3a_1 - a_2^2}{3^2} \, ,</math>

<math>~R</math>

<math>~\equiv</math>

<math>~\frac{3^2a_2 a_1 - 3^3a_0 - 2a_2^3}{2\cdot 3^3} \, . </math>

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:

<math>~a_2</math>

<math>~=</math>

<math>~- \tfrac{1}{2} \, ,</math>

<math>~a_1</math>

<math>~=</math>

<math>~0 \, ,</math>

<math>~a_0</math>

<math>~=</math>

<math>~\tfrac{1}{2}(\beta\eta)^2 \, .</math>

Applying Wolfram's definitions of the <math>~Q</math> and <math>~R</math> parameters to our particular problem gives,

<math>~Q</math>

<math>~\equiv</math>

<math>~\frac{3a_1 - a_2^2}{3^2} = -\biggl(\frac{a_2}{3}\biggr)^2 = - \frac{1}{2^2\cdot 3^2} \, ;</math>

<math>~R</math>

<math>~\equiv</math>

<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>

 

<math>~\equiv</math>

<math>~\frac{1}{2^3\cdot 3^3} \biggl[ 1 - 2\cdot 3^3(\beta\eta)^2\biggr] \, . </math>

Defining the parameter,

<math>~\Gamma^2</math>

<math>~\equiv</math>

<math>~ 2\cdot 3^3(\beta\eta)^2 \, ,</math>

we therefore have,

<math>~(2\cdot 3)^6 D</math>

<math>~=</math>

<math>~( 1 - \Gamma^2 )^2-1 \, ,</math>

<math>~(2\cdot 3)^3S^3</math>

<math>~\equiv</math>

<math>~(2\cdot 3)^3 R + \sqrt{(2\cdot 3)^6D} </math>

 

<math>~\equiv</math>

<math>~(1-\Gamma^2) + \sqrt{( 1 - \Gamma^2 )^2-1}</math>

 

<math>~\equiv</math>

<math>~(1-\Gamma^2) + i\sqrt{1-( 1 - \Gamma^2 )^2} \, ;</math>

<math>~(2\cdot 3)^3T^3</math>

<math>~\equiv</math>

<math>~(1-\Gamma^2) - i\sqrt{1-( 1 - \Gamma^2 )^2} \, .</math>

ASIDE:  The cube root of an imaginary number …

<math>~\ell^3</math>

<math>~=</math>

<math>~A \pm i \sqrt{1-A^2}</math>

 

<math>~=</math>

<math>~r_\ell e^{i\theta_\ell} \, ,</math>

where,

<math>~r_\ell</math>

<math>~=</math>

<math>~( A^2 + 1-A^2 )^{1/2} = 1 \, ,</math>

and,

<math>~\theta_\ell</math>

<math>~=</math>

<math>~\pm \tan^{-1}\biggl( \frac{\sqrt{1-A^2}}{A} \biggr) = \pm \cos^{-1}A \, .</math>

Now, according to this online resource, the three roots <math>~(j=0,1,2)</math> of <math>~\ell^3</math> are,

<math>~\ell_j = r_\ell^{1/3}e^{i(\theta_\ell + 2j\pi)/3)} \, ,</math>

which, for our specific problem gives,

<math>~\ell_j</math>

<math>~=</math>

<math>~e^{i\theta_\pm/3} \cdot e^{i(2j\pi/3)} \, ,</math>

where the subscript on <math>~\theta</math> refers to the <math>~\pm</math> in our original expression for <math>~\ell</math>.


In our particular case, after associating <math>~A \leftrightarrow (1-\Gamma^2)</math>, we can write,

<math>~ 2\cdot 3(S + T) </math>

<math>~=</math>

<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>

 

<math>~=</math>

<math>~ e^{i\theta_+/3} \cdot e^{i(2j\pi/3)} + e^{i\theta_-/3} \cdot e^{i(2j\pi/3)} </math>

 

<math>~=</math>

<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>

 

<math>~=</math>

<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>

 

 

<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>

 

<math>~=</math>

<math>~2 e^{i(2j\pi/3)} \cdot \cos\biggl[ \tfrac{1}{3}\cos^{-1}(1-\Gamma^2)\biggr] \, .</math>

Similarly, we can write,

<math>~ 2\cdot 3(S - T) </math>

<math>~=</math>

<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>

 

<math>~=</math>

<math>~ e^{i\theta_+/3} \cdot e^{i(2j\pi/3)} - e^{i\theta_-/3} \cdot e^{i(2j\pi/3)} </math>

 

<math>~=</math>

<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>

 

<math>~=</math>

<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>

 

 

<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>

 

<math>~=</math>

<math>~2i e^{i(2j\pi/3)} \cdot \sin\biggl[ \tfrac{1}{3}\cos^{-1}(1-\Gamma^2)\biggr] \, . </math>

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

<math>~6x_1-1</math>

<math>~=</math>

<math>~ 6(S + T) </math>

 

<math>~=</math>

<math>~2 \cos\biggl[ \tfrac{1}{3}\cos^{-1}(1-\Gamma^2)\biggr] \, ;</math>

<math>~6x_2-1</math>

<math>~=</math>

<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>

 

<math>~=</math>

<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>

 

<math>~=</math>

<math>~ 2\biggl\{ \cos\biggl[ \tfrac{1}{3}\cos^{-1}(1-\Gamma^2) + \frac{2\pi}{3}\biggr] \biggr\} \, ; </math>

<math>~6x_3-1</math>

<math>~=</math>

<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>

 

<math>~=</math>

<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>

 

<math>~=</math>

<math>~ 2\biggl\{ \cos\biggl[ \tfrac{1}{3}\cos^{-1}(1-\Gamma^2) - \frac{2\pi}{3}\biggr] \biggr\} \, . </math>


Table 1:  Analytically Evaluated Roots

determined for <math>~\beta = \tfrac{1}{10}</math>
<math>~\eta</math> <math>~\Gamma^2 = 54\beta^2\eta^2</math> Inner solution <math>~(\theta = 0)</math>

[Superior sign in cubic eq.]
Outer solution <math>~(\theta = \pi)</math>

[Inferior sign in cubic eq.]
<math>~x_1/\beta</math> <math>~x_2/\beta</math> <math>~x_3/\beta</math> <math>~x_1/\beta</math> <math>~x_2/\beta</math> <math>~x_3/\beta</math>
0.25 0.03375 -4.98744 0.24411 -0.25667 4.98744 -0.24411 0.25667
1.0 0.54 -4.78128 0.91909 -1.1378 4.78128 -0.91909 1.1378
  CONFIRMATION: In all cases,

<math>~x^2 + 2x^3 = (\beta\eta)^2</math>
CONFIRMATION: In all cases,

<math>~x^2 - 2x^3 = (\beta\eta)^2</math>


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:

<math>~a_2</math>

<math>~=</math>

<math>~\tfrac{1}{2} \, ,</math>

<math>~a_1</math>

<math>~=</math>

<math>~0 \, ,</math>

<math>~a_0</math>

<math>~=</math>

<math>~- \tfrac{1}{2}(\beta\eta)^2 \, .</math>

Following the same set of steps that were followed in 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,

<math>~(2\cdot 3)^3S^3</math>

<math>~=</math>

<math>~- (1-\Gamma^2) + i\sqrt{1-( 1 - \Gamma^2 )^2} \, ;</math>

<math>~(2\cdot 3)^3T^3</math>

<math>~=</math>

<math>~- (1-\Gamma^2) - i\sqrt{1-( 1 - \Gamma^2 )^2} \, ,</math>

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

<math>~6x_1+1</math>

<math>~=</math>

<math>~- 2 \cos\biggl[ \tfrac{1}{3}\cos^{-1}(1-\Gamma^2)\biggr] \, ;</math>

<math>~6x_2+1</math>

<math>~=</math>

<math>~ - 2\biggl\{ \cos\biggl[ \tfrac{1}{3}\cos^{-1}(1-\Gamma^2) + \frac{2\pi}{3}\biggr] \biggr\} \, ; </math>

<math>~6x_3+1</math>

<math>~=</math>

<math>~ - 2\biggl\{ \cos\biggl[ \tfrac{1}{3}\cos^{-1}(1-\Gamma^2) - \frac{2\pi}{3}\biggr] \biggr\} \, . </math>

Analytically Prescribed Eigenvector

Our Notation

As is explicitly defined in Figure 1 of our accompanying detailed notes, we have chosen to represent the spatial structure of an eigenfunction in the equatorial-plane of toroidal-like configurations via the expression,

<math>~\biggl[ \frac{\delta\rho}{\rho_0}\biggr]_\mathrm{spatial} </math>

<math>~=</math>

<math>~\biggl\{ f_m(\varpi)e^{-im\phi_m} \biggr\} \, .</math>

In general, we should assume that the function that delineates the radial dependence of the eigenfunction has both a real and an imaginary component, that is, we should assume that,

<math>~f_m(\varpi)</math>

<math>~=</math>

<math>~\mathcal{A}(\varpi) + i\mathcal{B}(\varpi) \, ,</math>

in which case the square of the modulus of the function is,

<math>~|f_m|^2 \equiv f_m \cdot f^*_m </math>

<math>~=</math>

<math>~\mathcal{A}^2 + \mathcal{B}^2 \, .</math>

We can rewrite this complex function in the form,

<math>~f_m(\varpi)</math>

<math>~=</math>

<math>~|f_m|e^{-i[\alpha(\varpi) + \pi/2]} \, ,</math>

if the angle, <math>~\alpha(\varpi)</math> is defined such that,

<math>~\sin\alpha = \frac{\mathcal{A}}{\sqrt{\mathcal{A}^2 + \mathcal{B}^2}}</math>

        and        

<math>~\cos\alpha = \frac{\mathcal{B}}{\sqrt{\mathcal{A}^2 + \mathcal{B}^2}}</math>

<math>~\Rightarrow ~~~~ \alpha</math>

<math>~\equiv</math>

<math>~\tan^{-1}\biggl(\frac{\mathcal{A}}{\mathcal{B}}\biggr) = \tan^{-1}\biggl[ \frac{\mathrm{Re}(f_m)}{\mathrm{Im}(f_m)} \biggr] \, .</math>

Hence, the spatial structure of the eigenfunction can be rewritten as,

<math>~\biggl[ \frac{\delta\rho}{\rho_0}\biggr]_\mathrm{spatial} </math>

<math>~=</math>

<math>~|f_m(\varpi)|e^{-i[\alpha(\varpi) + \pi/2+ m\phi_m]} \, . </math>

From this representation we can see that, at each radial location, <math>~\varpi</math>, the phase angle(s) at which the fractional perturbation exhibits its maximum amplitude, <math>~|f_m|</math>, is identified by setting the exponent of the exponential to zero. That is,

<math>~\phi_m = \phi_\mathrm{max}(\varpi)</math>

<math>~\equiv</math>

<math>~-\frac{1}{m}\biggl[\alpha(\varpi) +\frac{\pi}{2}\biggr] = -\frac{1}{m}\biggl\{ \tan^{-1}\biggl[ \frac{\mathrm{Re}(f_m)}{\mathrm{Im}(f_m)} \biggr] +\frac{\pi}{2} \biggr\} \, .</math>

An equatorial-plane plot of <math>~\phi_\mathrm{max}(\varpi)</math> should produce the "constant phase locus" referenced, for example, in recent papers from the Imamura & Hadley collaboration.


General Formulation

From my initial focused reading of the analysis presented by Blaes (1985), I conclude that, in the infinitely slender torus case, unstable modes in PP tori exhibit eigenvectors of the form,

<math>~\frac{\delta W}{W_0} \equiv \biggl[ \frac{W(\eta,\theta)}{C} - 1 \biggr]e^{im\Omega_p t}e^{-y_2 (\Omega_0 t)} </math>

<math>~=</math>

<math>~\biggl\{ f_m(\eta,\theta)e^{-i[m\phi_m + k\theta]} \biggr\} \, ,</math>

where we have written the perturbation amplitude in a manner that conforms with the notation that we have used in Figure 1 of a related, but more general discussion. As is summarized in §1.3 of Blaes (1985), for "thick" (but, actually, still quite thin) tori, "exactly one exponentially growing mode exists for each value of the azimuthal wavenumber <math>~m</math>," and its complex amplitude takes the following form (see his equation 1.10):

<math>~f_m(\eta,\theta)</math>

<math>~=</math>

<math> ~\beta^2 m^2 \biggl[ 2\eta^2 \cos^2\theta - \frac{3\eta^2}{4(n+1)} - \frac{(4n+1)}{4(n+1)^2} \pm 4i\biggl(\frac{3}{2n+2}\biggr)^{1/2} \eta\cos\theta\biggr] + \mathcal{O}(\beta^3) \, . </math>

Aside from an arbitrary leading scale factor, we should therefore find that the amplitude (modulus) of the enthalpy perturbation is,

<math>~\biggl|\frac{\delta W}{W_0} \biggr|</math>

<math>~=</math>

<math>~\sqrt{[\mathrm{Re}(f_m)]^2+ [\mathrm{Im}(f_m)]^2} \, ;</math>

and the associated phase function is,

<math>~m\phi_\mathrm{max}(\varpi)</math>

<math>~=</math>

<math>~ -\biggl\{ \tan^{-1}\biggl[ \frac{\mathrm{Re}(f_m)}{\mathrm{Im}(f_m)} \biggr] +\frac{\pi}{2} +k\theta \biggr\} \, .</math>

Now, keeping in mind that, for the time being, we are only interested in examining the shape of the unstable eigenvector in the equatorial plane of the torus, we can set,

<math>~\cos\theta ~~ \rightarrow ~~ \pm 1 \, .</math>

Hence, we have,

<math>~\frac{1}{\beta^4 m^4}\biggl|\frac{\delta W}{W_0} \biggr|^2</math>

<math>~=</math>

<math>~\biggl[2\eta^2 - \frac{3\eta^2}{4(n+1)} - \frac{(4n+1)}{4(n+1)^2}\biggr]^2 + 16\biggl[\biggl(\frac{3}{2n+2}\biggr)^{1/2} \eta \biggr]^2 </math>

 

<math>~=</math>

<math>~\frac{1}{[2^2(n+1)^2]^2}\biggl[2^3(n+1)^2\eta^2 - 3(n+1)\eta^2 - (4n+1) \biggr]^2 + \frac{2^3 \cdot 3\eta^2}{(n+1)} </math>

 

<math>~=</math>

<math>~\frac{1}{[2^2(n+1)^2]^2}\biggl[(4n+1) - (n+1)[2^3(n+1)-3]\eta^2 \biggr]^2 + \frac{2^3 \cdot 3\eta^2}{(n+1)} </math>

<math>~\Rightarrow ~~~~ \biggl[\frac{2(n+1)}{\beta m} \biggr]^4 \biggl|\frac{\delta W}{W_0} \biggr|^2</math>

<math>~=</math>

<math>~\biggl[(4n+1) - (n+1)[2^3(n+1)-3]\eta^2 \biggr]^2 + 2^7 \cdot 3(n+1)^3\eta^2 \, . </math>

Also, keeping in mind that, because of the <math>~\cos\theta</math> factor, the sign on the imaginary term flips its sign when switching from the "inner" region to the "outer" region of the torus,

 

<math>~m\phi_\mathrm{max}</math>

<math>~=</math>

<math>~- \biggl\{ \tan^{-1}\biggl[ \frac{(4n+1) - (n+1)[2^3(n+1)-3]\eta^2 }{2^{7/2}\cdot 3^{1/2}(n+1)^{3/2}\eta} \biggr]+\frac{\pi}{2}\biggr\}</math>

        over        

inner <math>~(\theta=0)</math> region of the torus;

while
 

<math>~m\phi_\mathrm{max}</math>

<math>~=</math>

<math>~- \biggl\{ \tan^{-1}\biggl[- \frac{(4n+1) - (n+1)[2^3(n+1)-3]\eta^2 }{2^{7/2}\cdot 3^{1/2}(n+1)^{3/2}\eta}\biggr\} +\frac{3\pi}{2} \biggr\}</math>

        over        

outer <math>~(\theta=\pi)</math> region of torus.

Incompressible Slim Tori

If we specifically consider geometrically slim, incompressible tori — that is, if we set the polytropic index, <math>~n=0</math> — to lowest order the eigenfunction derived by Blaes (1985) takes the form,

<math>~f_m(\eta,\theta)</math>

<math>~=</math>

<math> ~\beta^2 m^2 \biggl[ 2\eta^2 \cos^2\theta - \frac{3\eta^2}{4} - \frac{1}{4} \pm 4i\biggl(\frac{3}{2}\biggr)^{1/2} \eta\cos\theta\biggr] + \cancelto{0}{\mathcal{O}(\beta^3)} \, . </math>

Check Validity of Blaes85 Eigenvector

Step 1

Equation (2.6) of Blaes85 states that,

<math>~\beta^2 \eta^2 = [x^2 + x^3(3\cos\theta - \cos^3\theta) ]</math>

        <math>~\Rightarrow</math>        

<math>~ \beta^2(1 - \eta^2) = [ \beta^2 - x^2 - x^3(3\cos\theta - \cos^3\theta) ] \, .</math>

This means that,

<math>~\frac{\partial \eta^2}{\partial x}</math>

<math>~=</math>

<math>~ \frac{1}{\beta^2}\biggl[ 2x +3x^2(3\cos\theta - \cos^3\theta) \biggr] \, ; </math>

and,

<math>~\frac{\partial \eta^2}{\partial\theta}</math>

<math>~=</math>

<math>~\frac{x^3}{\beta^2}\biggl[ -3\sin\theta + 3\sin\theta \cos^2\theta \biggr] </math>

 

<math>~=</math>

<math>~\frac{3x^3 \sin\theta}{\beta^2}\biggl[\cos^2\theta -1\biggr] \, .</math>

Furthermore,

<math>~\frac{\partial \eta}{\partial x}</math>

<math>~=</math>

<math>~ \frac{1}{2\beta}\biggl[x^2 + x^3(3\cos\theta - \cos^3\theta)\biggr]^{-1/2}\biggl[ 2x +3x^2(3\cos\theta - \cos^3\theta) \biggr] \, ; </math>

and,

<math>~\frac{\partial \eta}{\partial\theta}</math>

<math>~=</math>

<math>~\frac{1}{2\beta}\biggl[x^2 + x^3(3\cos\theta - \cos^3\theta)\biggr]^{-1/2}\biggl[ -3\sin\theta + 3\sin\theta \cos^2\theta \biggr] \, .</math>

Step 2

Equation (4.14) of Blaes85 states that,

<math>~\nu + m</math>

<math>~=</math>

<math>~ \pm im\biggl[\frac{3}{2(n+1)}\biggr]^{1/2} \beta \, ; </math>

and equation (4.13) of Blaes85 states that,

<math>~\frac{\delta W}{A_{00}}</math>

<math>~=</math>

<math>~1 + \beta^2 m^2\biggl[ 2\eta^2 \cos^2\theta - \frac{3\eta^2}{4(n+1)} - \frac{(4n+1)}{4(n+1)^2} \pm 4i\biggl( \frac{3}{2n+2} \biggr)^{1/2} \eta\cos\theta \biggr] </math>

<math>~\Rightarrow~~~~\frac{1}{m^2}\biggl[\frac{\delta W}{A_{00}}-1\biggr]</math>

<math>~=</math>

<math>~\beta^2 \biggl\{- \frac{(4n+1)}{4(n+1)^2} + \eta^2\biggl[ 2\cos^2\theta - \frac{3}{4(n+1)}\biggr] \pm i\biggl( \frac{2^3\cdot 3}{n+1} \biggr)^{1/2} \eta\cos\theta \biggr\} </math>

 

<math>~=</math>

<math>~\frac{\beta^2}{2^2(n+1)^2} \biggl\{- (4n+1) + \eta^2 [ 2^3(n+1)^2\cos^2\theta - 3(n+1)] \pm i ~[ 2^7\cdot 3(n+1)^3 ]^{1/2} \eta\cos\theta \biggr\} </math>

<math>~\Rightarrow~~~~\Lambda \equiv \frac{2^2(n+1)^2}{m^2}\biggl[\frac{\delta W}{A_{00}}-1\biggr]</math>

<math>~=</math>

<math>~- (4n+1)\beta^2 + [ 2^3(n+1)^2\cos^2\theta - 3(n+1)] \cdot [x^2 + x^3(3\cos\theta - \cos^3\theta)] </math>

 

 

<math>~\pm ~i~\beta [ 2^7\cdot 3(n+1)^3 ]^{1/2} [x^2\cos^2\theta + x^3(3\cos^3\theta - \cos^5\theta) ]^{1/2} </math>

 

<math>~=</math>

<math>~- (4n+1)\beta^2 + (n+1)x^2[ 2^3(n+1)\cos^2\theta - 3] \cdot [1 + xb] ~\pm ~i~\beta [ 2^7\cdot 3(n+1)^3 ]^{1/2} x\cos\theta [1 + x b ]^{1/2} </math>

Through a separate white-board derivation I have obtained …

<math>~\Lambda </math>

<math>~\equiv</math>

<math>~ \frac{2^2(n+1)^2}{m^2}\biggl[\frac{\delta W}{A_{00}}-1\biggr] </math>

 

<math>~=</math>

<math>~ -(4n+1)\beta^2 + (\beta\eta)^2(n+1)[2^3(n+1)\cos^2\theta - 3] ~~~\pm ~~i~\beta\cos\theta [2^7\cdot 3 (n+1)^3]^{1/2} (\beta\eta) \, , </math>

where,

<math>~(\beta\eta)^2</math>

<math>~=</math>

<math>~x^2(1+xb) \, ,</math>

<math>~b</math>

<math>~\equiv</math>

<math>~3\cos\theta - \cos^3\theta \, .</math>

Note that, differentiating the left-hand-side with respect to either coordinate <math>~(x</math> or <math>~\theta)</math> gives,

<math>~\frac{\partial \Lambda}{\partial x_i} </math>

<math>~=</math>

<math>~\frac{2^2(n+1)^2}{m^2}\cdot \frac{\partial }{\partial x_i} \biggl(\frac{\delta W}{A_{00}} \biggr) </math>

<math>~\Rightarrow~~~~\frac{\partial }{\partial x_i} \biggl(\frac{\delta W}{A_{00}} \biggr) </math>

<math>~=</math>

<math>~\biggl[\frac{m}{2(n+1)}\biggr]^2 \cdot \frac{\partial \Lambda}{\partial x_i} \, .</math>


Given that <math>~\nu</math> has both real and imaginary parts, presumably,

<math>~\nu^2 \equiv \nu \cdot \nu^*</math>

<math>~=</math>

<math>~ \biggl\{ -m \pm im\biggl[\frac{3}{2(n+1)}\biggr]^{1/2} \beta\biggr\} \biggl\{ -m \mp im\biggl[\frac{3}{2(n+1)}\biggr]^{1/2} \beta\biggr\} </math>

 

<math>~=</math>

<math>~m^2 + m^2\biggl[\frac{3}{2(n+1)}\biggr]\beta^2 \, . </math>

For later reference, let's take the relevant partial derivatives of the function, <math>~\Lambda(x,\theta)</math>. Adopting the shorthand notation,

<math>~b \equiv (3\cos\theta-\cos^3\theta) \, ,</math>

we have,

<math>~\frac{\partial \Lambda}{\partial x}</math>

<math>~=</math>

<math>~[ 2^3(n+1)^2\cos^2\theta - 3(n+1)] \cdot \frac{\partial }{\partial x} \biggl\{ [x^2 + x^3b] \biggr\} </math>

 

 

<math>~\pm ~i~\beta [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cos\theta \cdot \frac{\partial }{\partial x} \biggl\{ [x^2 + x^3b ]^{1/2} \biggr\} </math>

 

<math>~=</math>

<math>~[ 2^3(n+1)^2\cos^2\theta - 3(n+1)] \cdot [2x + 3x^2b] </math>

 

 

<math>~\pm ~i~\beta [ 2^5\cdot 3(n+1)^3 ]^{1/2} \cos\theta \cdot [1 + xb ]^{-1/2} \cdot [2 + 3xb ] </math>

Through a separate white-board derivation I have obtained …

<math>~\frac{\partial\Lambda}{\partial x}</math>

<math>~=</math>

<math>~ (n+1)[2^3(n+1)\cos^2\theta -3]x(2+3xb) ~~~\pm ~~i~\beta\cos\theta [2^5\cdot 3 (n+1)^3]^{1/2} \cdot \frac{x(2+3xb)}{(\beta\eta)} \, , </math>

which is the same.


<math>~\frac{\partial^2 \Lambda}{\partial x^2}</math>

<math>~=</math>

<math>~[ 2^3(n+1)^2\cos^2\theta - 3(n+1)] \cdot \frac{\partial }{\partial x}[2x + 3x^2b] </math>

 

 

<math>~\pm ~i~\beta [ 2^5\cdot 3(n+1)^3 ]^{1/2} \cos\theta \cdot \frac{\partial }{\partial x}\biggl\{[1 + xb ]^{-1/2} \cdot [2 + 3xb ] \biggr\} </math>

 

<math>~=</math>

<math>~[ 2^3(n+1)^2\cos^2\theta - 3(n+1)] \cdot [2 + 6xb] </math>

 

 

<math>~\pm ~i~\beta [ 2^5\cdot 3(n+1)^3 ]^{1/2} \cos\theta \cdot \biggl\{ -\tfrac{b}{2}[1 + xb ]^{-3/2} \cdot [2 + 3xb ] + [1 + xb ]^{-1/2} \cdot [3b ] \biggr\} </math>

 

<math>~=</math>

<math>~2[ 2^3(n+1)^2\cos^2\theta - 3(n+1)] \cdot [1 + 3xb] </math>

 

 

<math>~\pm ~i~\beta [ 2^5\cdot 3(n+1)^3 ]^{1/2} \cos\theta \cdot \biggl\{ - [2b + 3xb^2 ] + [6b + 6xb^2 ] \biggr\} \frac{1}{2(1+xb)^{3/2}} </math>

 

<math>~=</math>

<math>~2[ 2^3(n+1)^2\cos^2\theta - 3(n+1)] \cdot [1 + 3xb] </math>

 

 

<math>~\pm ~i~\beta [ 2^3\cdot 3(n+1)^3 ]^{1/2} \cos\theta \cdot b[1 + xb ]^{-3/2} [4 + 3xb ] </math>


Through a separate white-board derivation I have obtained …

<math>~\frac{\partial^2\Lambda}{\partial x^2}</math>

<math>~=</math>

<math>~ 2(n+1)[2^3(n+1)\cos^2\theta -3](1+3xb) ~\pm~~i~\beta\cos\theta [2^3\cdot 3(n+1)^3]^{1/2} \biggl[ \frac{b(4+3xb)}{(1+xb)^{3/2}} \biggr] \, , </math>

which is the same.


<math>~\frac{\partial \Lambda}{\partial \theta}</math>

<math>~=</math>

<math>~\frac{\partial }{\partial \theta} \biggl\{ [ 2^3(n+1)^2\cos^2\theta - 3(n+1)] \cdot [x^2 + x^3(3\cos\theta - \cos^3\theta)] \biggr\} </math>

 

 

<math>~\pm ~i~\frac{\partial }{\partial \theta} \biggl\{ \beta [ 2^7\cdot 3(n+1)^3 ]^{1/2} [x^2\cos^2\theta + x^3(3\cos^3\theta - \cos^5\theta) ]^{1/2} \biggr\} </math>

 

<math>~=</math>

<math>~ [ 2^3(n+1)^2\cos^2\theta - 3(n+1)] \cdot x^3 \cdot [\sin\theta (-3 + 3\cos^2\theta)] + [x^2 + x^3(3\cos\theta - \cos^3\theta)] \cdot [ -2^4(n+1)^2 \sin\theta \cos\theta ] </math>

 

 

<math> ~\pm ~i~\beta [ 2^5\cdot 3(n+1)^3 ]^{1/2} \cdot [x^2\cos^2\theta + x^3(3\cos^3\theta - \cos^5\theta) ]^{-1/2} \cdot (-\sin\theta) [2x^2\cos\theta + x^3(9\cos^2\theta - 5\cos^4\theta) ] </math>

 

<math>~=</math>

<math>~ -3 x^3 \sin^3\theta [ 2^3(n+1)^2\cos^2\theta - 3(n+1)] -2^4(n+1)^2 x^2\sin\theta \cos\theta [1 + x(3\cos\theta-\cos^3\theta) ] </math>

 

 

<math> ~\pm ~i~\beta (-1) [ 2^5\cdot 3(n+1)^3 ]^{1/2} \cdot [1 + x(3\cos\theta - \cos^3\theta) ]^{-1/2} \cdot x~\sin\theta [2 + x(9\cos\theta - 5\cos^3\theta) ] </math>

 

<math>~=</math>

<math>~ -3 x^3 \sin^3\theta [ 2^3(n+1)^2\cos^2\theta - 3(n+1)] -2^4(n+1)^2 x^2\sin\theta \cos\theta [1 + xb ] </math>

 

 

<math> ~\pm ~i~\beta (-1) [ 2^5\cdot 3(n+1)^3 ]^{1/2} \cdot [1 + xb ]^{-1/2} \cdot x~\sin\theta [2 + x(9\cos\theta - 5\cos^3\theta) ] </math>


Through a separate white-board derivation I have obtained …

<math>~\frac{\partial\Lambda}{\partial \theta}</math>

<math>~=</math>

<math>~ 9(n+1)x^3\sin^3\theta - 2^3\cdot 3(n+1)^2x^3\sin^3\theta \cos^2\theta -2^4 (n+1)^2 (\beta\eta)^2 \sin\theta\cos\theta </math>

 

 

<math>~ \pm~~i~(-1)\beta [2^7\cdot 3 (n+1)^3 ]^{1/2}\biggl\{ (\beta\eta)\sin\theta +\frac{3x^3}{2}\cdot\biggl[ \frac{\sin^3\theta \cos\theta}{(\beta\eta)} \biggr]\biggr\} </math>

 

<math>~=</math>

<math>~(n+1)\sin\theta \biggl\{ -2^4 (n+1) (\beta\eta)^2 \cos\theta + 3x^3 \sin^2\theta \biggl[3 - 2^3(n+1)\cos^2\theta \biggr] \biggr\} </math>

 

 

<math>~ \pm~~i~(-1)\beta \sin\theta [2^7\cdot 3 (n+1)^3 (\beta\eta)^2]^{1/2}\biggl\{ 1 +\frac{3x^3}{2}\cdot\biggl[ \frac{\sin^2\theta \cos\theta}{(\beta\eta)^2} \biggr]\biggr\} \, , </math>

which is the same.


<math>~\frac{\partial^2 \Lambda}{\partial \theta^2}</math>

<math>~=</math>

<math>~ -3 x^3 \frac{\partial }{\partial \theta} \cdot \biggl\{\sin^3\theta [ 2^3(n+1)^2\cos^2\theta - 3(n+1)] \biggr\} -2^4(n+1)^2 x^2 \cdot \frac{\partial }{\partial \theta} \biggl\{\sin\theta \cos\theta [1 + x(3\cos\theta-\cos^3\theta) ] \biggr\} </math>

 

 

<math> ~\pm ~i~\beta (-1)x [ 2^5\cdot 3(n+1)^3 ]^{1/2} \cdot \frac{\partial }{\partial \theta} \biggl\{ \sin\theta [2 + x(9\cos\theta - 5\cos^3\theta) ] \cdot [1 + x(3\cos\theta - \cos^3\theta) ]^{-1/2} \biggr\} </math>

 

<math>~=</math>

<math>~ -3 x^3 \cdot \biggl\{3\sin^2\theta \cos\theta [ 2^3(n+1)^2\cos^2\theta - 3(n+1)] - \sin^3\theta [ 2^3(n+1)^2 \cdot 2\sin\theta\cos\theta ] \biggr\} </math>

 

 

<math>~ -2^4(n+1)^2 x^2 \cdot \biggl\{ (\cos^2\theta - \sin^2\theta ) [1 + x(3\cos\theta-\cos^3\theta) ] - 3x\sin^4\theta \cos\theta \biggr\} </math>

 

 

<math> ~\pm ~i~\beta (-1)x [ 2^5\cdot 3(n+1)^3 ]^{1/2} \cdot \biggl\{ \cos\theta \cdot [2 + x(9\cos\theta - 5\cos^3\theta) ] \cdot [1 + x(3\cos\theta - \cos^3\theta) ]^{-1/2} </math>

 

 

<math> - 3x\sin^2\theta [1 + x(3\cos\theta - \cos^3\theta) ]^{-1/2} \cdot (3 - 5\cos^2\theta) </math>

 

 

<math> + \frac{3}{2} \cdot x\sin^4\theta [2 + x(9\cos\theta - 5\cos^3\theta) ] \cdot [1 + x(3\cos\theta - \cos^3\theta) ]^{-3/2} \biggr\} </math>

 

<math>~=</math>

<math>~ -3 x^3 \cdot \biggl\{3\sin^2\theta \cos\theta [ 2^3(n+1)^2\cos^2\theta - 3(n+1)] - \sin^3\theta [ 2^3(n+1)^2 \cdot 2\sin\theta\cos\theta ] \biggr\} </math>

 

 

<math>~ -2^4(n+1)^2 x^2 \cdot \biggl\{ (1 - 2\sin^2\theta ) + x (3\cos\theta -\cos^3\theta - 6\sin^2\theta\cos\theta + 2\sin^2\theta\cos^3\theta - 3\sin^4\theta \cos\theta ) \biggr\} </math>

 

 

<math> ~\pm ~i~\beta (-1)x [ 2^5\cdot 3(n+1)^3 ]^{1/2} \cdot [1 + xb ]^{-3/2} \cdot \biggl\{ \cos\theta \cdot [2 + x(9\cos\theta - 5\cos^3\theta) ] \cdot [1 + xb ] </math>

 

 

<math> - 3x\sin^2\theta [1 + xb ] \cdot (3 - 5\cos^2\theta) + \frac{3}{2} \cdot x\sin^4\theta [2 + x(9\cos\theta - 5\cos^3\theta) ] \biggr\} </math>

 

<math>~=</math>

<math>~-2^4(n+1)^2 x^2 (1 - 2\sin^2\theta ) - x^3 \cdot \biggl\{ [ 2^3\cdot 3^2(n+1)^2\sin^2\theta \cos^3\theta - 3^3(n+1)\sin^2\theta \cos\theta] - 2^4\cdot 3(n+1)^2 \sin^4\theta\cos\theta </math>

 

 

<math>~ + 2^4(n+1)^2 b + 2^4(n+1)^2 \cdot( - 6\sin^2\theta\cos\theta + 2\sin^2\theta\cos^3\theta - 3\sin^4\theta \cos\theta )\biggr\} </math>

 

 

<math> ~\pm ~i~\beta (-1)x [ 2^3\cdot 3(n+1)^3 ]^{1/2} \cdot [1 + xb ]^{-3/2} \cdot \biggl\{ 2\cos\theta \cdot [2 + x(15\cos\theta - 7\cos^3\theta) + x^2 b(9\cos\theta - 5\cos^3\theta)] </math>

 

 

<math> - 6x\sin^2\theta (1 + xb ) \cdot (3 - 5\cos^2\theta) + 3 \cdot x\sin^4\theta [2 + x(9\cos\theta - 5\cos^3\theta) ] \biggr\} </math>


Through a separate white-board derivation I have obtained …

<math>~\frac{\partial^2\Lambda}{\partial \theta^2}</math>

<math>~=</math>

<math>~ x^2 \biggl\{2^4(n+1)^2(\sin^2\theta - \cos^2\theta) \biggr\} </math>

 

 

<math>~ + x^3\biggl\{-2^4\cdot 3 (n+1)^2\cos^3\theta + 2^4(n+1)^2\cos^5\theta + 3^2(n+1)(16n +19)\sin^2\theta \cos\theta -2^3\cdot 23 (n+1)^2\sin^2\theta \cos^3\theta \biggr\} </math>

 

 

<math>~ \pm~~i~(-1)\beta [2^7\cdot 3 (n+1)^3 ]^{1/2} \biggl\{ (\beta\eta)\cos\theta + \frac{3x^3\sin^2\theta}{2(\beta\eta)}(5\cos^2\theta -2) + \frac{3^2x^6\sin^6\theta\cos\theta}{2^2(\beta\eta)^3} \biggr\} \, . </math>

Step 3

From our accompanying discussion of the Blaes85 derivation, we expect the following equality to hold (see his equations 4.1 and 4.2):

<math>~\hat{L} (\delta W)</math>

<math>~=</math>

<math>~-2n(1-\Theta_H)(M\nu^2 + N\nu)(\delta W) \, ,</math>

where,

<math>~\hat{L} (\delta W)</math>

<math>~\equiv</math>

<math>~ \Theta_H x^2\cdot \frac{\partial^2(\delta W)}{\partial x^2} +\Theta_H \cdot \frac{\partial^2(\delta W)}{\partial\theta^2} + \biggl\{\Theta_H x \biggl[\frac{1-2x \cos\theta}{ 1-x\cos\theta}\biggr] + nx^2 \cdot \frac{\partial\Theta_H}{\partial x} \biggr\} \cdot \frac{\partial (\delta W)}{\partial x} </math>

 

 

<math>~ + \biggl[\frac{\Theta_Hx\sin\theta}{ (1-x\cos\theta) } + n\cdot \frac{\partial \Theta_H}{\partial\theta} \biggr] \cdot \frac{\partial (\delta W)}{\partial\theta} + \biggl[ \frac{2n x^2 m^2}{\beta^2(1-x\cos\theta)^4} - \frac{m^2 x^2\Theta_H}{(1-x\cos\theta)^2} \biggr]\delta W \, , </math>

<math>~M</math>

<math>~\equiv</math>

<math>~\frac{x^2}{(1-\Theta_H)\beta^2} \, ,</math>

<math>~N</math>

<math>~\equiv</math>

<math>~\frac{2mx^2}{(1-\Theta_H)\beta^2(1-x\cos\theta)^2} \, .</math>

Immediately evaluating the right-hand-side (RHS) of the equality, we have,

<math>~\biggl[\frac{2(n+1)}{m}\biggr]^2 \frac{RHS}{A_{00}}</math>

<math>~=</math>

<math>~-2n(1-\Theta_H)(M\nu^2 + N\nu)\cdot \biggl[\frac{2(n+1)}{m}\biggr]^2 \frac{\delta W}{A_{00}}</math>

 

<math>~=</math>

<math>~-\frac{2nx^2}{\beta^2}\biggl[ \nu^2 + \frac{2m\nu}{(1-x\cos\theta)^2} \biggr] \cdot \biggl[\frac{2(n+1)}{m}\biggr]^2 \frac{\delta W}{A_{00}}</math>

 

<math>~=</math>

<math>~-\frac{2nx^2}{\beta^2}\biggl[ \nu^2 + \frac{2m\nu}{(1-x\cos\theta)^2} \biggr] \cdot \biggl\{\Lambda + \biggl[ \frac{2(n+1)}{m} \biggr]^2 \biggr\} \, .</math>

And the similarly modified LHS is:

<math>~\biggl[\frac{2(n+1)}{m}\biggr]^2 \frac{LHS}{A_{00}}</math>

<math>~=</math>

<math>~ \Theta_H x^2\cdot \frac{\partial^2 \Lambda}{\partial x^2} +\Theta_H \cdot \frac{\partial^2 \Lambda}{\partial\theta^2} + \biggl\{\Theta_H x \biggl[\frac{1-2x \cos\theta}{ 1-x\cos\theta}\biggr] + nx^2 \cdot \frac{\partial\Theta_H}{\partial x} \biggr\} \cdot \frac{\partial \Lambda }{\partial x} </math>

 

 

<math>~ + \biggl[\frac{\Theta_Hx\sin\theta}{ (1-x\cos\theta) } + n\cdot \frac{\partial \Theta_H}{\partial\theta} \biggr] \cdot \frac{\partial\Lambda}{\partial\theta} + \biggl[ \frac{2n x^2 m^2}{\beta^2(1-x\cos\theta)^4} - \frac{m^2 x^2\Theta_H}{(1-x\cos\theta)^2} \biggr] \cdot \biggl[\frac{2(n+1)}{m}\biggr]^2 \frac{\delta W}{A_{00}} </math>

 

<math>~=</math>

<math>~ (1-\eta^2) x^2\cdot \frac{\partial^2 \Lambda}{\partial x^2} +(1-\eta^2) \cdot \frac{\partial^2 \Lambda}{\partial\theta^2} + \biggl\{(1-\eta^2) x \biggl[\frac{1-2x \cos\theta}{ 1-x\cos\theta}\biggr] + nx^2 \cdot \frac{\partial (1-\eta^2) }{\partial x} \biggr\} \cdot \frac{\partial \Lambda }{\partial x} </math>

 

 

<math>~ + \biggl[\frac{(1-\eta^2) x\sin\theta}{ (1-x\cos\theta) } + n\cdot \frac{\partial (1-\eta^2) }{\partial\theta} \biggr] \cdot \frac{\partial\Lambda}{\partial\theta} + \biggl[ \frac{2n x^2 m^2}{\beta^2(1-x\cos\theta)^4} - \frac{m^2 x^2(1-\eta^2) }{(1-x\cos\theta)^2} \biggr] \cdot \biggl\{\Lambda + \biggl[ \frac{2(n+1)}{m} \biggr]^2 \biggr\} </math>

 

<math>~=</math>

<math>~ (1-\eta^2) x^2\cdot \frac{\partial^2 \Lambda}{\partial x^2} +(1-\eta^2) \cdot \frac{\partial^2 \Lambda}{\partial\theta^2} + \biggl\{(1-\eta^2) x \biggl[\frac{1-2x \cos\theta}{ 1-x\cos\theta}\biggr] - nx^2 \cdot \frac{\partial \eta^2 }{\partial x} \biggr\} \cdot \frac{\partial \Lambda }{\partial x} </math>

 

 

<math>~ + \biggl[\frac{(1-\eta^2) x\sin\theta}{ (1-x\cos\theta) } - n\cdot \frac{\partial \eta^2 }{\partial\theta} \biggr] \cdot \frac{\partial\Lambda}{\partial\theta} + \biggl[ \frac{2n x^2 m^2}{\beta^2(1-x\cos\theta)^4} - \frac{m^2 x^2(1-\eta^2) }{(1-x\cos\theta)^2} \biggr] \cdot \biggl\{\Lambda + \biggl[ \frac{2(n+1)}{m} \biggr]^2 \biggr\} </math>

Now multiply both sides by …

<math>~\beta^2 m^2 (1-x\cos\theta)^4 \, .</math>


We have,

<math>~m^2\mathcal{L}_{RHS} \equiv \beta^2 (1-x\cos\theta)^4 \biggl[2(n+1)\biggr]^2 \frac{RHS}{A_{00}}</math>

<math>~=</math>

<math>~- 2n m^2 x^2(1-x\cos\theta)^4 \biggl[ \nu^2 + \frac{2m\nu}{(1-x\cos\theta)^2} \biggr] \cdot \biggl\{\Lambda + \biggl[ \frac{2(n+1)}{m} \biggr]^2 \biggr\} </math>

 

<math>~=</math>

<math>~- 2n x^2 (1-x\cos\theta)^2 \biggl[ (1-x\cos\theta)^2 \nu^2 + (2m\nu) \biggr] \cdot \{m^2 \Lambda + [ 2(n+1) ]^2 \} </math>

 

<math>~=</math>

<math>~- 2n x^2 (1-x\cos\theta)^2 \{m^2 \Lambda + [ 2(n+1) ]^2 \} \biggl\{ m^2 (1-x\cos\theta)^2 \biggl[ 1+\frac{3\beta^2}{2(n+1)} \biggr] + 2m^2 \biggl[ -1 ~\pm~ i~\biggl[ \frac{3\beta^2}{2(n+1)} \biggr]^{1/2} \biggr] \biggr\} </math>

 

<math>~=</math>

<math>~ - \frac{n m^2 x^2}{(n+1)} \cdot (1-x\cos\theta)^2\{m^2 \Lambda + [ 2(n+1) ]^2 \} \biggl\{ (1-x\cos\theta)^2 [ 2(n+1) + 3\beta^2 ] </math>

 

 

<math>~ - 4 (n+1) \pm i~ [2^3\cdot 3 (n+1)\beta^2 ]^{1/2} \biggr\} \, . </math>

And,

<math>~m^2\mathcal{L}_{LHS} \equiv \beta^2 (1-x\cos\theta)^4 \biggl[2(n+1)\biggr]^2 \frac{LHS}{A_{00}}</math>

<math>~=</math>

<math>~ \beta^2 m^2 (1-x\cos\theta)^4 (1-\eta^2) x^2\cdot \frac{\partial^2 \Lambda}{\partial x^2} +\beta^2 m^2 (1-x\cos\theta)^4 (1-\eta^2) \cdot \frac{\partial^2 \Lambda}{\partial\theta^2} </math>

 

 

<math>~ + \beta^2 m^2 (1-x\cos\theta)^3 \biggl[ (1-\eta^2) x (1-2x \cos\theta ) - nx^2 (1-x\cos\theta)\cdot \frac{\partial \eta^2 }{\partial x} \biggr] \cdot \frac{\partial \Lambda }{\partial x} </math>

 

 

<math>~ + \beta^2 m^2 (1-x\cos\theta)^3 \biggl[(1-\eta^2) x\sin\theta - n (1-x\cos\theta)\cdot \frac{\partial \eta^2 }{\partial\theta} \biggr] \cdot \frac{\partial\Lambda}{\partial\theta} </math>

 

 

<math>~ + m^2 \biggl[ 2n x^2 - (1-x\cos\theta)^2 \beta^2 x^2(1-\eta^2) \biggr] \cdot \biggl\{m^2 \Lambda + \biggl[ 2(n+1) \biggr]^2 \biggr\} </math>

 

<math>~=</math>

<math>~ m^2 (1-x\cos\theta)^4 [ \beta^2 - x^2 - x^3(3\cos\theta - \cos^3\theta) ] x^2\cdot \frac{\partial^2 \Lambda}{\partial x^2} +m^2 (1-x\cos\theta)^4 [ \beta^2 - x^2 - x^3(3\cos\theta - \cos^3\theta) ]\cdot \frac{\partial^2 \Lambda}{\partial\theta^2} </math>

 

 

<math>~ + m^2 (1-x\cos\theta)^3 \biggl\{ [ \beta^2 - x^2 - x^3(3\cos\theta - \cos^3\theta) ] x (1-2x \cos\theta ) - nx^2 (1-x\cos\theta)\cdot [ 2x +3x^2(3\cos\theta - \cos^3\theta)] \biggr\} \cdot \frac{\partial \Lambda }{\partial x} </math>

 

 

<math>~ + m^2 (1-x\cos\theta)^3 \biggl\{[ \beta^2 - x^2 - x^3(3\cos\theta - \cos^3\theta) ] x\sin\theta - 3n (1-x\cos\theta)\cdot ( \cos^2\theta -1 )x^3 \sin\theta \biggr\} \cdot \frac{\partial\Lambda}{\partial\theta} </math>

 

 

<math>~ + m^2 \biggl\{ 2n x^2 - (1-x\cos\theta)^2 x^2 [ \beta^2 - x^2 - x^3(3\cos\theta - \cos^3\theta) ] \biggr\} \cdot \biggl[m^2 \Lambda + 2^2(n+1)^2 \biggr] \, . </math>

Step 4

Let's divide both sides by <math>~m^2</math> and swap a couple of terms between the sides in order to group, on the right, terms with no explicit mention of <math>~\Lambda</math>.

<math>~\mathcal{L}_{RHS} </math>

<math>~\equiv</math>

<math>~ \frac{\beta^2}{m^2} (1-x\cos\theta)^4 \biggl[2(n+1)\biggr]^2 \frac{RHS}{A_{00}} ~\pm~\mathrm{swaps} </math>

 

<math>~=</math>

<math>~ - \frac{n x^2}{(n+1)} \cdot (1-x\cos\theta)^2\{[ 2(n+1) ]^2 \} \biggl\{ (1-x\cos\theta)^2 [ 2(n+1) + 3\beta^2 ] - 4 (n+1) \pm i~ [2^3\cdot 3 (n+1)\beta^2 ]^{1/2} \biggr\} </math>

 

 

<math>~ - \biggl\{ 2n x^2 - (1-x\cos\theta)^2 x^2 [ \beta^2 - x^2 - x^3(3\cos\theta - \cos^3\theta) ] \biggr\} \cdot \biggl[2^2(n+1)^2 \biggr] </math>

<math>~\Rightarrow ~~~~ -~\frac{\mathcal{L}_{RHS} }{2^2(n+1)x^2} </math>

<math>~=</math>

<math>~ n \cdot (1-x\cos\theta)^2 \biggl\{ (1-x\cos\theta)^2 [ 2(n+1) + 3\beta^2 ] - 4 (n+1) \pm i~ [2^3\cdot 3 (n+1)\beta^2 ]^{1/2} \biggr\} </math>

 

 

<math>~ + (n+1)\biggl\{ 2n - (1-x\cos\theta)^2 [ \beta^2 - x^2 - x^3(3\cos\theta - \cos^3\theta) ] \biggr\} </math>

 

<math>~=</math>

<math>~ 2n(n+1) - (1-x\cos\theta)^2 (n+1)[ \beta^2 - x^2 - x^3(3\cos\theta - \cos^3\theta) ] </math>

 

 

<math>~ + n(1-x\cos\theta)^4 [ 2(n+1) + 3\beta^2 ] - 4 n(n+1) (1-x\cos\theta)^2 </math>

 

 

<math>~ \pm~i~ (1-x\cos\theta)^2 [2^3\cdot 3 n^2(n+1)\beta^2 ]^{1/2} </math>

 

<math>~=</math>

<math>~ 2n(n+1) - (n+1)(1-x\cos\theta)^2[ 4n + \beta^2 - x^2 - x^3(3\cos\theta - \cos^3\theta) ] + n(1-x\cos\theta)^4 [ 2(n+1) + 3\beta^2 ] </math>

 

 

<math>~ \pm~i~ (1-x\cos\theta)^2 [2^3\cdot 3 n^2(n+1)\beta^2 ]^{1/2} \, . </math>


Through a separate white-board derivation I have obtained …

<math>~\mathrm{RHS}_3</math>

<math>~\equiv</math>

<math>~ \beta^2 (1-x\cos\theta)^4 \cdot \biggl\{ \mathrm{RHS}_1 ~~\pm~~ \mathrm{swaps} \biggr\} = \beta^2 (1-x\cos\theta)^4 \cdot \biggl\{ \frac{1}{A_{00}}\biggl[ \frac{2(n+1)}{m} \biggr]^2\mathrm{RHS}_0 ~~\pm~~ \mathrm{swaps} \biggr\} </math>

 

<math>~=</math>

<math>~-~ 2^2(n+1)^2 x^2 \cdot \mathcal{A} \, , </math>

where,

<math>~\mathcal{A}</math>

<math>~\equiv</math>

<math>~ 2n + (1-x\cos\theta)^2\biggl[ 4n \biggl( \frac{\nu}{m}\biggr) - \beta^2 + x^2(1+xb)\biggr] + 2n (1-x\cos\theta)^4 \biggl(\frac{\nu}{m}\biggr)^2 \, , </math>

<math>~\frac{\nu}{m}</math>

<math>~=</math>

<math>~-~1~~ \pm~~ i~\biggl[\frac{3}{2(n+1)}\biggr]^{1/2} \beta \, . </math>

Case A:       <math>~\biggl(\frac{\nu}{m}\biggr)^2 = \frac{\nu \cdot \nu^*}{m^2} = 1+\frac{3\beta^2}{2(n+1)} ~~~\Rightarrow</math>

<math>~(n+1)\cdot \mathcal{A}</math>

<math>~=</math>

<math>~ 2n (n+1) + (n+1)(1-x\cos\theta)^2 [ x^2(1+xb) - \beta^2 - 4n ] + (1-x\cos\theta)^4 [2n(n+1) + 3n\beta^2 ] </math>

 

 

<math>~ \pm~~i~(1-x\cos\theta)^2 [ 2^3\cdot 3 n^2\beta^2(n+1) ]^{1/2} \, . </math>

Case B:       <math>~\biggl(\frac{\nu}{m}\biggr)^2 = \frac{\nu }{m} \cdot \frac{\nu }{m} = 1 - \frac{3\beta^2}{2(n+1)} ~~\pm~i~(-1)\biggl[\frac{2\cdot 3\beta^2}{(n+1)}\biggr]^{1/2} ~~~\Rightarrow</math>

<math>~(n+1)\cdot \mathcal{A}</math>

<math>~=</math>

<math>~ 2n (n+1) + (n+1)(1-x\cos\theta)^2 [ x^2(1+xb) - \beta^2 - 4n ] + (1-x\cos\theta)^4 [2n(n+1) - 3n\beta^2 ] </math>

 

 

<math>~ \pm~~i~x\cos\theta (1-x\cos\theta)^2 [ 2^3\cdot 3 n^2\beta^2(n+1) ]^{1/2} \, . </math>


And,

<math>~\mathcal{L}_{LHS} </math>

<math>~\equiv</math>

<math>~ \frac{\beta^2}{m^2} (1-x\cos\theta)^4 \biggl[2(n+1)\biggr]^2 \frac{LHS}{A_{00}} ~\mp~\mathrm{swaps} </math>

 

<math>~=</math>

<math>~ (1-x\cos\theta)^4 [ \beta^2 - x^2 - x^3(3\cos\theta - \cos^3\theta) ] x^2\cdot \frac{\partial^2 \Lambda}{\partial x^2} +m^2 (1-x\cos\theta)^4 [ \beta^2 - x^2 - x^3(3\cos\theta - \cos^3\theta) ]\cdot \frac{\partial^2 \Lambda}{\partial\theta^2} </math>

 

 

<math>~ + (1-x\cos\theta)^3 \biggl\{ [ \beta^2 - x^2 - x^3(3\cos\theta - \cos^3\theta) ] x (1-2x \cos\theta ) - nx^2 (1-x\cos\theta)\cdot [ 2x +3x^2(3\cos\theta - \cos^3\theta)] \biggr\} \cdot \frac{\partial \Lambda }{\partial x} </math>

 

 

<math>~ + (1-x\cos\theta)^3 \biggl\{[ \beta^2 - x^2 - x^3(3\cos\theta - \cos^3\theta) ] x\sin\theta - 3n (1-x\cos\theta)\cdot ( \cos^2\theta -1 )x^3 \sin\theta \biggr\} \cdot \frac{\partial\Lambda}{\partial\theta} </math>

 

 

<math>~ + \biggl\{ 2n x^2 - (1-x\cos\theta)^2 x^2 [ \beta^2 - x^2 - x^3(3\cos\theta - \cos^3\theta) ] \biggr\} \cdot \biggl[m^2 \Lambda \biggr] </math>

 

 

<math>~ + \frac{n x^2}{(n+1)} \cdot (1-x\cos\theta)^2\{m^2 \Lambda \} \biggl\{ (1-x\cos\theta)^2 [ 2(n+1) + 3\beta^2 ] - 4 (n+1) \pm i~ [2^3\cdot 3 (n+1)\beta^2 ]^{1/2} \biggr\} </math>

 

<math>~=</math>

<math>~ (1-x\cos\theta)^4 [ \beta^2 - x^2 - x^3b] \biggl\{ x^2\cdot \frac{\partial^2 \Lambda}{\partial x^2} +m^2 \cdot \frac{\partial^2 \Lambda}{\partial\theta^2} \biggr\} </math>

 

 

<math>~ + (1-x\cos\theta)^3 \biggl\{ x( \beta^2 - x^2 - x^3b ) \biggl[ (1-2x \cos\theta )\cdot \frac{\partial \Lambda }{\partial x} + \sin\theta \cdot \frac{\partial\Lambda}{\partial\theta}\biggr] - n (1-x\cos\theta)\cdot x^3 \biggl[ ( 2 +3xb )\cdot \frac{\partial \Lambda }{\partial x} - 3\sin^3\theta \cdot \frac{\partial\Lambda}{\partial\theta} \biggr] \biggr\} </math>

 

 

<math>~ + \biggl\{ 2n - (1-x\cos\theta)^2 [ \beta^2 - x^2 - x^3b ] + (1-x\cos\theta)^4 \biggl[ 2 + \frac{3\beta^2}{(n+1)} \biggr] n - 4 n (1-x\cos\theta)^2 \biggr\} \cdot m^2 x^2\Lambda </math>

 

 

<math>~ \pm i~\biggl[\frac{2^3\cdot 3 n^2 m^4 \beta^2}{(n+1)} \biggr]^{1/2} x^2 \cdot (1-x\cos\theta)^2 \Lambda </math>


Through a separate white-board derivation I have obtained …

<math>~\mathrm{LHS}_3</math>

<math>~\equiv</math>

<math>~ \beta^2 (1-x\cos\theta)^4 \cdot \biggl\{ \mathrm{LHS}_1 ~~\mp~~ \mathrm{swaps} \biggr\} = \beta^2 (1-x\cos\theta)^4 \cdot \biggl\{ \frac{1}{A_{00}}\biggl[ \frac{2(n+1)}{m} \biggr]^2\mathrm{LHS}_0 ~~\mp~~ \mathrm{swaps} \biggr\} </math>

 

<math>~=</math>

<math>~ \beta^2(1-\eta^2) (1-x\cos\theta)^3 \biggl\{ (1-x\cos\theta)\biggl[x^2 \frac{\partial^2\Lambda}{\partial x^2} + \frac{\partial^2\Lambda}{\partial \theta^2} \biggr] + x\biggl[(1-2x\cos\theta)\frac{\partial\Lambda}{\partial x} + \sin\theta \frac{\partial\Lambda}{\partial\theta} \biggr] \biggr\} </math>

 

 

<math>~ -~n x^3 (1-x\cos\theta)^4\biggl[(2+3xb) \frac{\partial\Lambda}{\partial x} -3\sin^3\theta \frac{\partial\Lambda}{\partial\theta} \biggr] + m^2 x^2 \Lambda \cdot \mathcal{A} \, ,</math>

where, as above in the definition of <math>~\mathrm{RHS_3} \, ,</math>

<math>~\mathcal{A}</math>

<math>~\equiv</math>

<math>~ 2n + (1-x\cos\theta)^2\biggl[ 4n \biggl( \frac{\nu}{m}\biggr) - \beta^2 + x^2(1+xb)\biggr] + 2n (1-x\cos\theta)^4 \biggl(\frac{\nu}{m}\biggr)^2 \, . </math>


The remaining question is, does <math>~\mathcal{L}_{LHS} = \mathcal{L}_{RHS}</math> — at least to lowest order(s) in <math>~x</math> — after the Blaes85 expression for the eigenfunction, <math>~\Lambda</math> (and its derivatives), is inserted into the LHS expression?

Step 5

Now let's evaluate the LHS terms, keeping only leading-orders in <math>~x</math> before plugging derivatives of <math>~\Lambda</math> into each term. For example,

<math>~ x^2\cdot \frac{\partial^2 \Lambda}{\partial x^2} +m^2 \cdot \frac{\partial^2 \Lambda}{\partial\theta^2} </math>

<math>~=</math>

<math>~ x^2 \biggl\{2[ 2^3(n+1)^2\cos^2\theta - 3(n+1)] \cdot [1 + \cancelto{0}{3xb}] \pm ~i~\beta [ 2^3\cdot 3(n+1)^3 ]^{1/2} \cos\theta \cdot b[1 + \cancelto{0}{xb} ]^{-3/2} [4 + \cancelto{0}{9xb} ] \biggr\} </math>

 

 

<math>~m^2\cdot \biggl\{ -2^4(n+1)^2 x^2 (1 - 2\sin^2\theta ) - \cancelto{0}{x^3 }\cdot \biggl[ [ 2^3\cdot 3^2(n+1)^2\sin^2\theta \cos^3\theta - 3^3(n+1)\sin^2\theta \cos\theta] - 2^4\cdot 3(n+1)^2 \sin^4\theta\cos\theta </math>

 

 

<math>~ + 2^4(n+1)^2 b + 2^4(n+1)^2 \cdot( - 6\sin^2\theta\cos\theta + 2\sin^2\theta\cos^3\theta - 3\sin^4\theta \cos\theta )\biggr] </math>

 

 

<math> ~\pm ~i~\beta (-1)x [ 2^3\cdot 3(n+1)^3 ]^{1/2} \cdot [1 + xb ]^{-3/2} \cdot \biggl[ 2\cos\theta \cdot [2 + x(15\cos\theta - 7\cos^3\theta) + \cancelto{0}{x^2 b}(9\cos\theta - 5\cos^3\theta)] </math>

 

 

<math> - 6x\sin^2\theta (1 + \cancelto{0}{xb }) \cdot (3 - 5\cos^2\theta) + 3 \cdot x\sin^4\theta [2 + \cancelto{0}{x}(9\cos\theta - 5\cos^3\theta) ] \biggr] \biggr\} </math>

 

<math>~\approx</math>

<math>~ x^2 \biggl[ 2^4(n+1)^2\cos^2\theta - 6(n+1) -2^4 m^2 (n+1)^2 (1 - 2\sin^2\theta ) \biggr] </math>

 

 

<math> ~\pm ~i~m^2\beta (-1)x [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cdot [1 + xb ]^{-3/2} \cos\theta </math>

 

 

<math> ~\pm ~i~\beta x^2 [ 2^3\cdot 3(n+1)^3 ]^{1/2} \cdot \biggl[(12\cos^2\theta - 4\cos^4\theta)~~ - ~~m^2\cdot (1 + xb )^{-3/2} [2\cos^2\theta(15 - 7\cos^2\theta) - 6\sin^2\theta (3 - 5\cos^2\theta) + 6 \sin^4\theta] \biggr] </math>

Next,

<math>~ (1-2x \cos\theta )\cdot \frac{\partial \Lambda }{\partial x} + \sin\theta \cdot \frac{\partial\Lambda}{\partial\theta}</math>

<math>~=</math>

<math>~ (1-2x \cos\theta )\cdot \biggl\{ [ 2^3(n+1)^2\cos^2\theta - 3(n+1)] \cdot [2x + \cancelto{0}{3x^2b}] ~~\pm ~~i~\beta [ 2^5\cdot 3(n+1)^3 ]^{1/2} \cos\theta \cdot [1 + xb ]^{-1/2} \cdot [2 + \cancelto{0}{3xb }] \biggr\} </math>

 

 

<math>~ + \sin\theta \cdot \biggl\{ -2^4(n+1)^2 \cancelto{0}{x^2}\sin\theta \cos\theta [1 + xb ] -3 \cancelto{0}{x^3} \sin^3\theta [ 2^3(n+1)^2\cos^2\theta - 3(n+1)] </math>

 

 

<math>~ ~~\pm ~~i~\beta (-1) [ 2^5\cdot 3(n+1)^3 ]^{1/2} \cdot [1 + xb ]^{-1/2} \cdot x~\sin\theta [2 + \cancelto{0}{x}(9\cos\theta - 5\cos^3\theta) ] \biggr\} </math>

 

<math>~\approx</math>

<math>~ (1-2x \cos\theta )\cdot \biggl[ 2x(n+1) [ 2^3 (n+1)\cos^2\theta - 3] ~~\pm ~~i~\beta [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cos\theta \cdot (1 + xb )^{-1/2} \biggr] </math>

 

 

<math>~ ~~\pm ~~i~\beta x (-1) [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cdot [1 + xb ]^{-1/2} \cdot \sin^2\theta </math>


Through a separate white-board derivation I have obtained …

<math>~ (1-2x \cos\theta )\cdot \frac{\partial \Lambda }{\partial x} + \sin\theta \cdot \frac{\partial\Lambda}{\partial\theta}</math>

<math>~=</math>

<math>~ x(n+1)[-6 + 2^4(n+1)\cos^2\theta] </math>

 

 

<math>~ - x^2(n+1)\cos\theta \{ [ 15 + 2^4(n+1) ] -\cos^2\theta[9 + 2^3\cdot 7 (n+1)] +2^3\cdot 3(n+1)\cos^4\theta \} </math>

 

 

<math>~ +x^3(n+1) \{9 - 2^2\cdot 3^2(1+2n)\cos^2\theta - [9 + 32(n+1)]\cos^4\theta +2^3(n+1)\cos^6\theta \} </math>

 

 

<math>~\pm~~i~\beta~\biggl[ \frac{2^5\cdot 3 (n+1)^3}{1+x(3\cos\theta-\cos^3\theta)} \biggr]^{1/2} \biggl\{ 2\cos\theta - x[2 - 7\cos^2\theta + 3\cos^4\theta ] </math>

 

 

<math>~- x^2 \cos\theta [ 9 +4\cos^2\theta -\cos^4\theta ] \biggr\}</math>


Also, from above,

<math>~\Lambda </math>

<math>~=</math>

<math>~- (4n+1)\beta^2 + (n+1)\cancelto{0}{x^2}[ 2^3(n+1)\cos^2\theta - 3] \cdot [1 + xb] ~\pm ~i~\beta [ 2^7\cdot 3(n+1)^3 ]^{1/2} x\cos\theta [1 + x b ]^{1/2} </math>

 

<math>~\approx</math>

<math>~- (4n+1)\beta^2 ~\pm ~i~\beta [ 2^7\cdot 3(n+1)^3 ]^{1/2} x\cos\theta [1 + x b ]^{1/2} </math>


Through a separate white-board derivation I have obtained …

<math>~ (2+3xb )\cdot \frac{\partial \Lambda }{\partial x} - 3\sin^3\theta \cdot \frac{\partial\Lambda}{\partial\theta}</math>

<math>~=</math>

<math>~ x\cdot 2^2(n+1)[2^3(n+1)\cos^2\theta -3] </math>

 

 

<math>~ + x^2\cdot 2^2\cdot 3(n+1)\cos\theta \{ [2^2n -5] +\cos^2\theta[2^4n + 19] - 2^2(n+1)\cos^4\theta \} </math>

 

 

<math>~ +x^3 \cdot 3^2(n+1)\cos^2\theta \{ -3^3 + 2\cdot 3^2\cos^2\theta[2^2n+5] - 99\cdot \cos^4\theta + 2^3(n+1)\cos^6\theta \} </math>

 

 

<math>~ +x^3 (n+1)\sin^4\theta \{ -3^3 + 3^2\cos^2\theta [ 2^3\cdot 3 n + 27] -2^3\cdot 3\cdot 5(n+1)\cos^4\theta \} </math>

 

 

<math>~\pm~~i~\beta~ \biggl[ \frac{2^5\cdot 3(n+1)^3}{1+xb} \biggr]^{1/2} \biggl\{ 4\cos\theta + 6x(2b\cos\theta + \sin^4\theta) + 3x^2(3b^2\cos\theta +2b\sin^4\theta + 3\sin^6\theta\cos\theta) \biggr\} </math>


Taken together, then, we have,

<math>~\mathcal{L}_{LHS} </math>

<math>~=</math>

<math>~ (1-x\cos\theta)^4 [ \beta^2 - x^2 - \cancelto{0}{x^3b}] \biggl\{ x^2\cdot \frac{\partial^2 \Lambda}{\partial x^2} +m^2 \cdot \frac{\partial^2 \Lambda}{\partial\theta^2} \biggr\} </math>

 

 

<math>~ + (1-x\cos\theta)^3 \biggl\{ x( \beta^2 - \cancelto{0}{x^2} - \cancelto{0}{x^3b} ) \biggl[ (1-2x \cos\theta )\cdot \frac{\partial \Lambda }{\partial x} + \sin\theta \cdot \frac{\partial\Lambda}{\partial\theta}\biggr] - n (1-x\cos\theta)\cdot \cancelto{0}{x^3} \biggl[ ( 2 +3xb )\cdot \frac{\partial \Lambda }{\partial x} - 3\sin^3\theta \cdot \frac{\partial\Lambda}{\partial\theta} \biggr] \biggr\} </math>

 

 

<math>~ + \biggl\{ 2n - (1-x\cos\theta)^2 [ \beta^2 - x^2 - \cancelto{0}{x^3}b ] + (1-x\cos\theta)^4 \biggl[ 2 + \frac{3\beta^2}{(n+1)} \biggr] n - 4 n (1-x\cos\theta)^2 \biggr\} \cdot m^2 x^2\Lambda </math>

 

 

<math>~ \pm i~\biggl[\frac{2^3\cdot 3 n^2 m^4 \beta^2}{(n+1)} \biggr]^{1/2} x^2 \cdot (1-x\cos\theta)^2 \Lambda </math>

 

<math>~\approx</math>

<math>~ (1-x\cos\theta)^4 [ \beta^2 - x^2 ] \biggl\{ x^2\cdot \frac{\partial^2 \Lambda}{\partial x^2} +m^2 \cdot \frac{\partial^2 \Lambda}{\partial\theta^2} \biggr\} </math>

 

 

<math>~ + (1-x\cos\theta)^3 \biggl\{ x( \beta^2) \biggl[ (1-2x \cos\theta )\cdot \frac{\partial \Lambda }{\partial x} + \sin\theta \cdot \frac{\partial\Lambda}{\partial\theta}\biggr] \biggr\} </math>

 

 

<math>~ + \biggl\{ 2n - (1-x\cos\theta)^2 [ \beta^2 - x^2 ] + (1-x\cos\theta)^4 \biggl[ 2 + \frac{3\beta^2}{(n+1)} \biggr] n - 4 n (1-x\cos\theta)^2 \biggr\} \cdot m^2 x^2\Lambda </math>

 

 

<math>~ \pm i~\biggl[\frac{2^3\cdot 3 n^2 m^4 \beta^2}{(n+1)} \biggr]^{1/2} x^2 \cdot (1-x\cos\theta)^2 \Lambda </math>

 

<math>~\approx</math>

<math>~ (1-x\cos\theta)^4 [ \beta^2 - x^2 ] \biggl\{ x^2 \biggl[ 2^4(n+1)^2\cos^2\theta - 6(n+1) -2^4 m^2 (n+1)^2 (1 - 2\sin^2\theta ) \biggr] </math>

 

 

<math> ~\pm ~i~m^2\beta (-1)x [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cdot [1 + xb ]^{-3/2} \cos\theta </math>

 

 

<math> ~\pm ~i~\beta x^2 [ 2^3\cdot 3(n+1)^3 ]^{1/2} \cdot \biggl[(12\cos^2\theta - 4\cos^4\theta)~~ - ~~m^2\cdot (1 + xb )^{-3/2} [2\cos^2\theta(15 - 7\cos^2\theta) - 6\sin^2\theta (3 - 5\cos^2\theta) + 6 \sin^4\theta] \biggr] \biggr\} </math>

 

 

<math>~ + x\beta^2(1-x\cos\theta)^3 \biggl\{ (1-2x \cos\theta )\cdot \biggl[ 2x(n+1)[ 2^3 (n+1)\cos^2\theta - 3] ~~\pm ~~i~\beta [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cos\theta \cdot (1 + xb )^{-1/2} \biggr] </math>

 

 

<math>~ ~~\pm ~~i~\beta x (-1) [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cdot [1 + xb ]^{-1/2} \cdot \sin^2\theta \biggr\} </math>

 

 

<math>~ + m^2 x^2\biggl\{ 2n - (1-x\cos\theta)^2 [ \beta^2 - x^2 ] + (1-x\cos\theta)^4 \biggl[ 2 + \frac{3\beta^2}{(n+1)} \biggr] n - 4 n (1-x\cos\theta)^2 \biggr\} \cdot \biggl\{- (4n+1)\beta^2 ~\pm ~i~\beta [ 2^7\cdot 3(n+1)^3 ]^{1/2} x\cos\theta [1 + x b ]^{1/2}\biggr\} </math>

 

 

<math>~ \pm i~\biggl[\frac{2^3\cdot 3 n^2 m^4 \beta^2}{(n+1)} \biggr]^{1/2} x^2 \cdot (1-x\cos\theta)^2 \cdot \biggl\{ - (4n+1)\beta^2 ~\pm ~i~\beta [ 2^7\cdot 3(n+1)^3 ]^{1/2} x\cos\theta [1 + x b ]^{1/2} \biggr\} </math>

Let's further simplify:

<math>~\mathcal{L}_{LHS} </math>

<math>~\approx</math>

<math>~ x^2\biggl\{ (1-x\cos\theta)^4 [ \beta^2 - x^2 ] \biggl[ 2^4(n+1)^2\cos^2\theta - 6(n+1) -2^4 m^2 (n+1)^2 (1 - 2\sin^2\theta ) \biggr] </math>

 

 

<math>~ -~\biggl[\frac{2^3\cdot 3 n^2 m^4 \beta^2}{(n+1)} \biggr]^{1/2} \cdot (1-x\cos\theta)^2 \cdot \beta [ 2^7\cdot 3(n+1)^3 ]^{1/2} x\cos\theta [1 + x b ]^{1/2} </math>

 

 

<math>~ + \beta^2(1-x\cos\theta)^3 (1-2x \cos\theta )\cdot \biggl[ 2(n+1)[ 2^3 (n+1)\cos^2\theta - 3] \biggr] </math>

 

 

<math>~ - 2n(4n+1)\beta^2 m^2 + (4n+1)\beta^2 m^2 [4n + \beta^2 - x^2 ] (1-x\cos\theta)^2 - (4n+1)\beta^2 m^2 n (1-x\cos\theta)^4 \biggl[ 2 + \frac{3\beta^2}{(n+1)} \biggr] \biggr\} </math>

 

 

<math>~ ~~\pm ~~i x\beta^3 (1-x\cos\theta)^3 (1-2x \cos\theta )\cdot [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cos\theta \cdot (1 + xb )^{-1/2} </math>

 

 

<math> ~\pm ~i~m^2\beta (-1)x [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cdot [1 + xb ]^{-3/2} \cos\theta (1-x\cos\theta)^4 [ \beta^2 - x^2 ] </math>

 

 

<math>~ \pm ~i~\beta m^2 x^2\biggl\{ 2n - (1-x\cos\theta)^2 [ \beta^2 - x^2 ] + (1-x\cos\theta)^4 \biggl[ 2 + \frac{3\beta^2}{(n+1)} \biggr] n - 4 n (1-x\cos\theta)^2 \biggr\} \cdot [ 2^7\cdot 3(n+1)^3 ]^{1/2} x\cos\theta [1 + x b ]^{1/2} </math>

 

 

<math> ~\pm ~i~\beta x^2 [ 2^3\cdot 3(n+1)^3 ]^{1/2} (1-x\cos\theta)^4 [ \beta^2 - x^2 ] \cdot \biggl[(12\cos^2\theta - 4\cos^4\theta)~~ - ~~m^2\cdot (1 + xb )^{-3/2} [2\cos^2\theta(15 - 7\cos^2\theta) - 6\sin^2\theta (3 - 5\cos^2\theta) + 6 \sin^4\theta] \biggr] </math>

 

 

<math>~ ~~\pm ~~i~\beta x^2 (-1) [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cdot \beta^2(1-x\cos\theta)^3 [1 + xb ]^{-1/2} \cdot \sin^2\theta </math>

 

 

<math>~ \pm i~x^2(-1) (4n+1)\beta^2 \biggl[\frac{2^3\cdot 3 n^2 m^4 \beta^2}{(n+1)} \biggr]^{1/2} \cdot (1-x\cos\theta)^2 </math>

Step 6

Hence, to lowest order we want to compare the following two expressions:

<math>~~ -~\frac{\mathcal{L}_{RHS} }{2^2(n+1)x^2} </math>

<math>~\approx</math>

<math>~ 2n(n+1) - (n+1)(1-\cancelto{0}{x}\cos\theta)^2[ 4n + \beta^2 - \cancelto{0}{x^2} - \cancelto{0}{x^3}(3\cos\theta - \cos^3\theta) ] + n(1-\cancelto{0}{x}\cos\theta)^4 [ 2(n+1) + 3\beta^2 ] </math>

 

 

<math>~ \pm~i~ (1-\cancelto{0}{x}\cos\theta)^2 [2^3\cdot 3 n^2(n+1)\beta^2 ]^{1/2} </math>

 

<math>~\approx</math>

<math>~ 2n(n+1) - (n+1)[ 4n + \beta^2 ] + n [ 2(n+1) + 3\beta^2 ] \pm~i~ [2^3\cdot 3 n^2(n+1)\beta^2 ]^{1/2} </math>

 

<math>~\approx</math>

<math>~3n\beta^2 + (n+1)\biggl[ 2n - 4n - \beta^2 + 2n \biggr] \pm~i~ [2^3\cdot 3 n^2(n+1)\beta^2 ]^{1/2} </math>

 

<math>~\approx</math>

<math>~\beta^2 (2n-1) ~\pm~i~ [2^3\cdot 3 n^2(n+1)\beta^2 ]^{1/2} </math>


<math>~\mathrm{Re}\biggl[\frac{\mathcal{L}_{LHS}}{x^2}\biggr] </math>

<math>~\approx</math>

<math>~ (1-\cancelto{0}{x}\cos\theta)^4 [ \beta^2 - \cancelto{0}{x^2} ] \biggl[ 2^4(n+1)^2\cos^2\theta - 6(n+1) -2^4 m^2 (n+1)^2 (1 - 2\sin^2\theta ) \biggr] </math>

 

 

<math>~ -~\biggl[\frac{2^3\cdot 3 n^2 m^4 \beta^2}{(n+1)} \biggr]^{1/2} \cdot (1-x\cos\theta)^2 \cdot \beta [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cancelto{0}{x}\cos\theta [1 + x b ]^{1/2} </math>

 

 

<math>~ + \beta^2(1-\cancelto{0}{x}\cos\theta)^3 (1-\cancelto{0}{2x} \cos\theta )\cdot \biggl[ 2(n+1)[ 2^3 (n+1)\cos^2\theta - 3] \biggr] </math>

 

 

<math>~ - 2n(4n+1)\beta^2 m^2 + (4n+1)\beta^2 m^2 [4n + \beta^2 - \cancelto{0}{x^2} ] (1-\cancelto{0}{x}\cos\theta)^2 - (4n+1)\beta^2 m^2 n (1-\cancelto{0}{x}\cos\theta)^4 \biggl[ 2 + \frac{3\beta^2}{(n+1)} \biggr] </math>

 

<math>~\approx</math>

<math>~ \beta^2\biggl[ 2^4(n+1)^2\cos^2\theta - 6(n+1) -2^4 m^2 (n+1)^2 (1 - 2\sin^2\theta ) \biggr] + \beta^2\biggl[ 2^4 (n+1)^2\cos^2\theta -6(n+1)\biggr] </math>

 

 

<math>~+ \beta^2\biggl\{- 2n(4n+1) m^2+ (4n+1)m^2 [4n ] - 2(4n+1) m^2 n \biggr\} + \beta^4\biggl\{(4n+1) m^2 - (4n+1) m^2 n \biggl[ \frac{3}{(n+1)} \biggr] \biggr\} </math>

 

<math>~\approx</math>

<math>~ \beta^2\biggl[ 2^4(n+1)^2\cos^2\theta - 6(n+1) \biggr] + \beta^2\biggl[ 2^4 (n+1)^2\cos^2\theta -6(n+1)\biggr] </math>

 

 

<math>~ + m^2\beta^2\biggl[ -2^4(n+1)^2 (1 - 2\sin^2\theta ) \biggr] + m^2\beta^2\biggl\{- 2n(4n+1) + 4n(4n+1) - 2n(4n+1) \biggr\} </math>

 

 

<math>~ + (4n+1) m^2\beta^4\biggl[1 - \frac{3n}{(n+1)} \biggr] </math>

 

<math>~\approx</math>

<math>~ 2\beta^2\biggl[ 2^4(n+1)^2\cos^2\theta - 6(n+1) \biggr] + m^2\beta^2\biggl[ 2^4(n+1)^2 (1 - 2\cos^2\theta ) \biggr] </math>

 

 

<math>~ + (4n+1) m^2\beta^4\biggl[1 - \frac{3n}{(n+1)} \biggr] </math>

 

<math>~\approx</math>

<math>~ (1-m^2)2^5\beta^2(n+1)^2\cos^2\theta + 2^2(n+1)\beta^2\biggl[ 4m^2(n+1) -3\biggr] + (4n+1) m^2\beta^4\biggl[1 - \frac{3n}{(n+1)} \biggr] \, . </math>

<math>~\pm~\mathrm{Im}\biggl[\mathcal{L}_{LHS} \biggr]</math>

<math>~\approx</math>

<math>~ ~x\biggl\{\beta^3 (1-\cancelto{0}{x}\cos\theta)^3 (1-\cancelto{0}{2x} \cos\theta )\cdot [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cos\theta \cdot (1 + \cancelto{0}{xb} )^{-1/2} </math>

 

 

<math> -~m^2\beta [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cdot [1 + \cancelto{0}{xb} ]^{-3/2} \cos\theta (1-\cancelto{0}{x}\cos\theta)^4 [ \beta^2 - \cancelto{0}{x}^2 ] \biggr\} </math>

 

 

<math>~ +~x^2 \biggl\{ \beta m^2 \biggl\{ 2n - (1-x\cos\theta)^2 [ \beta^2 - x^2 ] + (1-x\cos\theta)^4 \biggl[ 2 + \frac{3\beta^2}{(n+1)} \biggr] n - 4 n (1-x\cos\theta)^2 \biggr\} \cdot [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cancelto{0}{x}\cos\theta [1 + x b ]^{1/2} </math>

 

 

<math> +~\beta [ 2^3\cdot 3(n+1)^3 ]^{1/2} (1-\cancelto{0}{x}\cos\theta)^4 [ \beta^2 - \cancelto{0}{x^2} ] \cdot \biggl[(12\cos^2\theta - 4\cos^4\theta)~~ - ~~m^2\cdot (1 + \cancelto{0}{xb} )^{-3/2} [2\cos^2\theta(15 - 7\cos^2\theta) - 6\sin^2\theta (3 - 5\cos^2\theta) + 6 \sin^4\theta] \biggr] </math>

 

 

<math>~ -~\beta [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cdot \beta^2(1-\cancelto{0}{x}\cos\theta)^3 [1 + \cancelto{0}{xb} ]^{-1/2} \cdot \sin^2\theta </math>

 

 

<math>~ -~ (4n+1)\beta^2 \biggl[\frac{2^3\cdot 3 n^2 m^4 \beta^2}{(n+1)} \biggr]^{1/2} \cdot (1-\cancelto{0}{x}\cos\theta)^2 \biggr\} </math>

 

<math>~\approx</math>

<math>~ ~x(1-m^2)\cdot \beta^3 [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cos\theta </math>

 

 

<math> +~~x^2 \beta^3 [ 2^3\cdot 3(n+1)^3 ]^{1/2}\biggl\{(12\cos^2\theta - 4\cos^4\theta) - 4\sin^2\theta </math>

 

 

<math>~- m^2 \biggl[2\cos^2\theta(15 - 7\cos^2\theta) - 6\sin^2\theta (3 - 5\cos^2\theta) + 6 \sin^4\theta -~ \frac{n (4n+1)}{(n+1)^2} \biggr]\biggr\} </math>

 

<math>~\approx</math>

<math>~ ~x(1-m^2)\cdot \beta^3 [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cos\theta </math>

 

 

<math> +~~x^2 \beta^3 [ 2^7\cdot 3(n+1)^3 ]^{1/2}\biggl\{4\cos^2\theta - \cos^4\theta -1 </math>

 

 

<math>~- \frac{m^2}{2} \biggl[\cos^2\theta(15 - 7\cos^2\theta) -12\sin^4\theta +6 \sin^2\theta -~ \frac{n (4n+1)}{2(n+1)^2} \biggr]\biggr\} </math>

 

<math>~\approx</math>

<math>~ ~x(1-m^2)\cdot \beta^3 [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cos\theta </math>

 

 

<math> +~~x^2 \beta^3 [ 2^7\cdot 3(n+1)^3 ]^{1/2}\biggl\{3-(1+\sin^2\theta)^2 - \frac{m^2}{2} \biggl[33\cos^2\theta-19 \cos^4\theta -6 -~ \frac{n (4n+1)}{2(n+1)^2} \biggr]\biggr\} </math>


Examples

Evaluate various expressions using the parameter set:   <math>~(n, \theta, x) = (1, \tfrac{\pi}{3}, \tfrac{1}{4})</math>

<math>~b</math>

<math>~=</math>

<math>~\frac{3}{2} - \frac{1}{8} = \frac{11}{8} </math>

1.375000000

<math>~(\beta\eta)^2</math>

<math>~=</math>

<math>~\biggl(\frac{1}{2^2}\biggr)^2\biggl[ 1 + \frac{11}{2^5} \biggr] = \frac{2^5 + 11}{2^9} = \frac{43}{2^9} </math>

0.083984375

<math>~\mathrm{Re}(\Lambda)</math>

<math>~=</math>

<math>~ -5\beta^2 + \frac{43}{2^9} \biggl[ 2^3 - 6 \biggr] = -5\beta^2 + \frac{43}{2^8} </math>

<math>~- 5\beta^2</math> + 0.167968750

<math>~\mathrm{Im}(\Lambda)</math>

<math>~=</math>

<math>~ \frac{\beta}{2}\biggl[ 2^{10}\cdot 3 \cdot \frac{43}{2^9} \biggr]^{1/2} = \beta \biggl[ \frac{3\cdot 43}{2} \biggr]^{1/2} </math>

8.031189202 <math>~\beta</math>

<math>~\mathrm{Re}\biggl( \frac{\partial\Lambda}{\partial x} \biggr)</math>

<math>~=</math>

<math>~ \biggl[ 2^3 - 6 \biggr]\frac{1}{2^2}\biggl( 2 + \frac{3\cdot 11}{2^2\cdot 2^3} \biggr) = \biggl( 1 + \frac{33}{2^6} \biggr) </math>

1.515625000

<math>~\mathrm{Im}\biggl( \frac{\partial\Lambda}{\partial x} \biggr)</math>

<math>~=</math>

<math>~ \frac{\beta}{2} \biggl[ \frac{2^9\cdot 2^8\cdot 3}{43} \biggr]^{1/2} \biggl[ \frac{1}{2^2}\cdot \biggl(2 + \frac{3\cdot 11}{2^2\cdot 2^3}\biggr) \biggr] = \beta \biggl[ \frac{2^{13}\cdot 3}{43} \biggr]^{1/2} \biggl(1 + \frac{3\cdot 11}{2^{6}}\biggr) </math>

36.23373732 <math>~\beta</math>

<math>~\mathrm{Re}\biggl( \frac{\partial\Lambda}{\partial \theta} \biggr)</math>

<math>~=</math>

<math>~ 2\biggl(\frac{3}{4}\biggr)^{1/2} \biggl\{-2^4 \cdot \frac{43}{2^9} +\frac{3}{2^6}\biggl(\frac{3}{4}\biggr)\biggl[3-4\biggr] \biggr\} = ~-~\frac{\sqrt{3}}{2^8} \cdot (2^3\cdot 43 + 9) </math>

-2.388335684

<math>~\mathrm{Im}\biggl( \frac{\partial\Lambda}{\partial \theta} \biggr)</math>

<math>~=</math>

<math>~ (-1) \beta~\biggl(\frac{3}{4}\biggr)^{1/2} \cdot \biggl[ \frac{2^{10} \cdot 3\cdot 43}{2^9} \biggr]^{1/2} \biggl\{1 + \frac{3\cdot 2^9}{2^7 \cdot 43} \biggl(\frac{3}{2^3}\biggr)\biggr\} = (-1) \beta~\biggl(\frac{3}{4}\biggr)^{1/2} \cdot [ 2\cdot 3\cdot 43 ]^{1/2} \biggl\{1 + \frac{3^2}{2\cdot 43} \biggr\} </math>

(-1) × 15.36617018 <math>~\beta</math>


<math>~\mathrm{Re}\biggl( \frac{\partial^2\Lambda}{\partial x^2}\biggr)</math>

<math>~=</math>

<math>~ 2^2[2^2-3]\biggl[1 + \frac{3}{2^2}\cdot \frac{11}{2^3} \biggr] = 2^2 + \frac{33}{2^3} = \frac{65}{8} </math>

          8.125000000

<math>~\mathrm{Im}\biggl( \frac{\partial^2\Lambda}{\partial x^2}\biggr)</math>

<math>~=</math>

<math>~\beta~ 2^2\cdot \sqrt{3} \biggl[\frac{11}{2^3} \biggl(2^2+\frac{3}{2^2}\cdot \frac{11}{2^3} \biggr)\biggr] \biggl[ \frac{2^4}{43}\biggr]^{3/2} =\biggl[ \frac{2^3\cdot 3}{43^3}\biggr]^{1/2} \biggl[11\cdot (2^7+33)\biggr] \beta </math>

          30.76957507<math>~\beta</math>

<math>~\mathrm{Re}\biggl(\frac{\partial^2\Lambda}{\partial \theta^2}\biggr)</math>

<math>~=</math>

<math>~ \frac{1}{2^4} \biggl\{2^6 \biggl(\frac{3}{2^2} - \frac{1}{2^2} \biggr) \biggr\} + \frac{1}{2^6}\biggl\{-2^3\cdot 3 + 2 + \frac{3^3\cdot 5\cdot 7}{2^2} -3\cdot 23 \biggr\} = 2 + \frac{1}{2^8}\biggl\{2^3 + 3^3\cdot 5\cdot 7 - 2^2\cdot 3\cdot 31 \biggr\} </math>

          4.269531250

<math>~\mathrm{Im}\biggl( \frac{\partial^2\Lambda}{\partial \theta^2}\biggr)</math>

<math>~=</math>

<math>~(-1)\beta [2^{10}\cdot 3]^{1/2} \biggl( \frac{43}{2^9} \biggr)^{1/2} \biggl\{ \frac{1}{2} + \frac{3}{2}\cdot \frac{2^9}{43} \cdot \frac{1}{2^6}\cdot \frac{3}{2^2} \biggl(\frac{5}{2^2} -2 \biggr) + \biggl( \frac{3}{2}\cdot \frac{1}{2^6} \cdot \frac{2^9}{43}\biggr)^2 \biggl( \frac{3}{2^2} \biggr)^3 \frac{1}{2} \biggr\} </math>

         

 

<math>~=</math>

<math>~(-1)\beta\biggl( \frac{3\cdot 43}{2} \biggr)^{1/2} \biggl\{ 1 - \frac{3^3}{2\cdot 43} + \frac{3^5}{(2\cdot 43)^2} \biggr\} </math>

(-1) × 5.773638858 <math>~\beta</math>


<math>~ \mathrm{Re}\biggl\{ (1-2x \cos\theta )\cdot \frac{\partial \Lambda }{\partial x} + \sin\theta \cdot \frac{\partial\Lambda}{\partial\theta} \biggr\}</math>

<math>~=</math>

<math>~\biggl( 1 - \frac{1}{2^2} \biggr)\biggl( 1 + \frac{33}{2^6} \biggr) -~\frac{\sqrt{3}}{2}\cdot \frac{\sqrt{3}}{2^8} \cdot (2^3\cdot 43 + 9) </math>

 

 

<math>~=</math>

<math>~ ~\frac{(2\cdot 3\cdot 97)-3\cdot (2^3\cdot 43 + 9)}{2^9} </math>

-0.931640625

<math>~ \mathrm{Im}\biggl\{ (1-2x \cos\theta )\cdot \frac{\partial \Lambda }{\partial x} + \sin\theta \cdot \frac{\partial\Lambda}{\partial\theta} \biggr\}</math>

<math>~=</math>

<math>~\biggl( 1 - \frac{1}{2^2} \biggr)\beta \biggl[ \frac{2^{13}\cdot 3}{43} \biggr]^{1/2} \biggl(1 + \frac{3\cdot 11}{2^{6}}\biggr) +~\frac{\sqrt{3}}{2}\cdot (-1) \beta~\frac{\sqrt{3}}{2}\cdot[ 2\cdot 3\cdot 43 ]^{1/2} \biggl\{1 + \frac{3^2}{2\cdot 43} \biggr\} </math>

 

 

<math>~=</math>

<math>~\beta \biggl[ \frac{3^3}{2^5\cdot 43} \biggr]^{1/2} [2 (2^6 + 33) - (2\cdot 43 + 3^2) ] </math>

13.86780926 <math>~\beta</math>

<math>~ \mathrm{Re}\biggl\{ (2+3xb )\cdot \frac{\partial \Lambda }{\partial x} - 3\sin^3\theta \cdot \frac{\partial\Lambda}{\partial\theta} \biggr\}</math>

<math>~=</math>

<math>~\biggl( 2 + \frac{3}{2^2}\cdot \frac{11}{2^3} \biggr)\biggl( 1 + \frac{33}{2^6} \biggr) +~3\cdot\biggl(\frac{3}{2^2}\biggr)^{3/2}\cdot \frac{\sqrt{3}}{2^8} \cdot (2^3\cdot 43 + 9) </math>

 

 

<math>~=</math>

<math>~\frac{97^2}{2^{11}} +~\frac{3^3}{2^{11}}\cdot (2^3\cdot 43 + 9) </math>

9.248046874

<math>~ \mathrm{Im}\biggl\{ (2+3xb )\cdot \frac{\partial \Lambda }{\partial x} - 3\sin^3\theta \cdot \frac{\partial\Lambda}{\partial\theta} \biggr\}</math>

<math>~=</math>

<math>~\beta \biggl\{ \biggl( 2 + \frac{3}{2^2}\cdot \frac{11}{2^3} \biggr) \biggl[ \frac{2^{13}\cdot 3}{43} \biggr]^{1/2} \biggl(1 + \frac{3\cdot 11}{2^{6}}\biggr) +~3\cdot\biggl(\frac{3}{2^2}\biggr)^{3/2}\cdot \biggl(\frac{3}{4}\biggr)^{1/2} \cdot [ 2\cdot 3\cdot 43 ]^{1/2} \biggl[1 + \frac{3^2}{2\cdot 43} \biggr]\biggr\} </math>

 

 

<math>~=</math>

<math>~\beta \biggl\{ \biggl( \frac{3}{2^9\cdot 43} \biggr)^{1/2} \biggl[ ( 2^6 + 33)^2 +~3^3(2\cdot 43 +3^2 ) \biggr]\biggr\} </math>

139.7753772

Step 7

Let's begin by slightly redefining the LHS and RHS collections of terms.

<math>~\mathrm{RHS}_4</math>

<math>~\equiv</math>

<math>~\mathrm{RHS}_3 - m^2 x^2 \Lambda \cdot \mathcal{A} </math>

 

<math>~=</math>

<math>~-~x^2 ~ [ 2^2(n+1)^2 - m^2 \Lambda ] \cdot \mathcal{A} \, , </math>

<math>~\mathrm{LHS}_4</math>

<math>~\equiv</math>

<math>~\mathrm{LHS}_3 - m^2 x^2 \Lambda \cdot \mathcal{A} </math>

 

<math>~=</math>

<math>~ \beta^2(1-\eta^2) (1-x\cos\theta)^3 \biggl\{ (1-x\cos\theta)\biggl[x^2 \frac{\partial^2\Lambda}{\partial x^2} + \frac{\partial^2\Lambda}{\partial \theta^2} \biggr] + x\biggl[(1-2x\cos\theta)\frac{\partial\Lambda}{\partial x} + \sin\theta \frac{\partial\Lambda}{\partial\theta} \biggr] \biggr\} </math>

 

 

<math>~ -~n x^3 (1-x\cos\theta)^4\biggl[(2+3xb) \frac{\partial\Lambda}{\partial x} -3\sin^3\theta \frac{\partial\Lambda}{\partial\theta} \biggr] </math>


Next Lowest Order

Let's begin with the RHS (Case B).

<math>~(n+1)\cdot \mathcal{A}</math>

<math>~=</math>

<math>~ 2n (n+1) + (n+1)(1-x\cos\theta)^2 [ x^2(1+xb) - \beta^2 - 4n ] + (1-x\cos\theta)^4 [2n(n+1) - 3n\beta^2 ] </math>

 

 

<math>~ \pm~~i~x\cos\theta (1-x\cos\theta)^2 [ 2^3\cdot 3 n^2\beta^2(n+1) ]^{1/2} </math>

 

<math>~=</math>

<math>~ 2n (n+1) + (n+1)[1-2x\cos\theta + x^2\cos^2\theta + \mathcal{O}(x^3) ] [ x^2(1+xb) - \beta^2 - 4n ] + [1-4x\cos\theta + 6x^2\cos^2\theta + \mathcal{O}(x^3) ] [2n(n+1) - 3n\beta^2 ] </math>

 

 

<math>~ \pm~~i~x\cos\theta [1-2x\cos\theta + x^2\cos^2\theta + \mathcal{O}(x^3) ] [ 2^3\cdot 3 n^2\beta^2(n+1) ]^{1/2} </math>

 

<math>~=</math>

<math>~x^0\biggl\{2n (n+1) -4n(n+1) + 2n(n+1) \biggr\} + x^1\biggl\{8n(n+1)\cos\theta - 8n(n+1)\cos\theta\biggr\} </math>

 

 

<math>~ + x^2\biggl\{-4n(n+1)\cos^2\theta + (n+1)\biggl[ 1 - \biggl(\frac{\beta}{x}\biggr)^2 \biggr] + 12n(n+1)\cos^2\theta - 3n \biggl(\frac{\beta}{x}\biggr)^2 \biggr\} + \mathcal{O}(x^3) </math>

 

 

<math>~ \pm~~i~x^2\biggl(\frac{\beta}{x}\biggr) \frac{nb_0}{2^2(n+1)} \biggl[1-2x\cos\theta + x^2\cos^2\theta + \mathcal{O}(x^3) \biggr] </math>

 

<math>~=</math>

<math>~x^0\biggl\{0 \biggr\} + x^1\biggl\{0\biggr\} + x^2\biggl\{(n+1)[8n\cos^2\theta + 1] - (4n+1)\biggl(\frac{\beta}{x}\biggr)^2 \biggr\} + \mathcal{O}(x^3) </math>

 

 

<math>~ \pm~~i~x^2\biggl(\frac{\beta}{x}\biggr) \frac{nb_0}{2^2(n+1)} \biggl[1-2x\cos\theta + x^2\cos^2\theta + \mathcal{O}(x^3) \biggr] </math>

where,

<math>b_0 \equiv [ 2^7\cdot 3 (n+1)^3 \cos^2\theta ]^{1/2} \, .</math>

Hence,

<math>~\frac{\mathrm{RHS}_4}{x^2}</math>

<math>~=</math>

<math>~-~ 2^2(n+1)^2\cdot \mathcal{A} + m^2 \cdot \mathcal{A} \biggl\{ -(4n+1)\beta^2 + (\beta\eta)^2(n+1)[2^3(n+1)\cos^2\theta - 3] ~~~\pm ~~i~\beta\cos\theta [2^7\cdot 3 (n+1)^3]^{1/2} (\beta\eta) \biggr\} </math>

 

<math>~=</math>

<math>~-~ 2^2(n+1)^2\cdot \mathcal{A} + m^2 \cdot \mathcal{A} (n+1)\biggl\{ -x^2\biggl(\frac{4n+1}{n+1}\biggr)\cdot \biggl(\frac{\beta}{x}\biggr)^2 + x^2(1+\cancelto{0}{x}b)[2^3(n+1)\cos^2\theta - 3] ~~~\pm ~~i~x^2 \biggl(\frac{\beta}{x}\biggr) \frac{b_0}{(n+1)} (1+\cancelto{0}{x}b)^{1/2} \biggr\} </math>

 

<math>~\approx</math>

<math>~\mathcal{A} (n+1)\biggl\{-~ 2^2(n+1) + m^2 \cancelto{0}{x^2} \biggl[ [2^3(n+1)\cos^2\theta - 3] -\biggl(\frac{4n+1}{n+1}\biggr)\cdot \biggl(\frac{\beta}{x}\biggr)^2 ~~~\pm ~~i~ \biggl(\frac{\beta}{x}\biggr) \frac{b_0}{(n+1)} \biggr] \biggl\} </math>

 

<math>~\approx</math>

<math>~-2^2(n+1)x^2\biggl\{(n+1)[8n\cos^2\theta + 1] - (4n+1)\biggl(\frac{\beta}{x}\biggr)^2 ~~~\pm~i~\biggl(\frac{\beta}{x}\biggr) \frac{nb_0}{2^2(n+1)} \biggl[1-2\cancelto{0}{x}\cos\theta + \cancelto{0}{x^2}\cos^2\theta + \cancelto{0}{\mathcal{O}}(x^3) \biggr] \biggr\} </math>

<math>~\Rightarrow ~~~~\frac{\mathrm{RHS}_4}{x^4}</math>

<math>~\approx</math>

<math>2^2(n+1)(4n+1)\biggl(\frac{\beta}{x}\biggr)^2 ~-~2^2(n+1)^2[8n\cos^2\theta + 1] ~~\pm~i~(-1)\biggl(\frac{\beta}{x}\biggr) nb_0 \, . </math>

This should be compared to,

<math>~\frac{\mathrm{LHS}_4}{x^2}</math>

<math>~=</math>

<math>~ \biggl[ \biggl(\frac{\beta}{x}\biggr)^2 - 1 - xb \biggr] (1-x\cos\theta)^3 \biggl\{ (1-x\cos\theta) \biggl[x^2 \frac{\partial^2\Lambda}{\partial x^2} + \frac{\partial^2\Lambda}{\partial \theta^2} \biggr] + x\biggl[(1-2x\cos\theta)\frac{\partial\Lambda}{\partial x} + \sin\theta \frac{\partial\Lambda}{\partial\theta} \biggr] \biggr\} </math>

 

 

<math>~ -~n x (1-x\cos\theta)^4\biggl[(2+3xb) \frac{\partial\Lambda}{\partial x} -3\sin^3\theta \frac{\partial\Lambda}{\partial\theta} \biggr] \, . </math>

Now, from above, we can write,

<math>~\biggl[x^2 \frac{\partial^2\Lambda}{\partial x^2} + \frac{\partial^2\Lambda}{\partial \theta^2} \biggr]</math>

<math>~=</math>

<math>~x^2\biggl\{ 2(n+1)[2^3(n+1)\cos^2\theta -3](1+3\cancelto{0}{x}b) ~\pm~~i~\beta \biggl(\frac{b_0}{2^2}\biggr)\biggl[ \frac{b(4+3xb)}{(1+xb)^{3/2}} \biggr] \biggr\} + x^2 \biggl\{2^4(n+1)^2(\sin^2\theta - \cos^2\theta)\biggr\} </math>

 

 

<math>~ + \cancelto{0}{x^3}\biggl\{-2^4\cdot 3 (n+1)^2\cos^3\theta + 2^4(n+1)^2\cos^5\theta + 3^2(n+1)(16n +19)\sin^2\theta \cos\theta -2^3\cdot 23 (n+1)^2\sin^2\theta \cos^3\theta \biggr\} </math>

 

 

<math>~ \pm~~i~(-1)\beta b_0 ~x(1+\cancelto{0}{x}b)^{1/2}\biggl\{ 1 + \frac{3\cancelto{0}{x}\sin^2\theta (5\cos^2\theta -2)}{2(1+xb)\cos\theta } + \frac{3^2\cancelto{0}{x^2}\sin^6\theta}{2^2(1+xb)^2} \biggr\} </math>

 

<math>~\approx</math>

<math>~x^2\biggl\{ 2(n+1)[2^3(n+1)\cos^2\theta -3] + 2^4(n+1)^2(\sin^2\theta - \cos^2\theta)\biggr\} ~\pm~~i~x \biggl\{ \cancelto{0}{x\beta} \biggl(\frac{b_0}{2^2}\biggr)\biggl[ \frac{b(4+3xb)}{(1+xb)^{3/2}} \biggr]- \beta b_0\biggr\}</math>

 

<math>~\approx</math>

<math>~2(n+1)x^2\biggl\{ 2^3(n+1)\sin^2\theta -3 \biggr\} ~\pm~~i~x^2\biggl\{ - \biggl(\frac{\beta}{x}\biggr) b_0\biggr\} \, .</math>

Also,

<math>~ x\biggl[ (1-2x \cos\theta )\cdot \frac{\partial \Lambda }{\partial x} + \sin\theta \cdot \frac{\partial\Lambda}{\partial\theta}\biggr]</math>

<math>~=</math>

<math>~ x^2(n+1)[-6 + 2^4(n+1)\cos^2\theta] </math>

 

 

<math>~ - \cancelto{0}{x^3}(n+1)\cos\theta \{ [ 15 + 2^4(n+1) ] -\cos^2\theta[9 + 2^3\cdot 7 (n+1)] +2^3\cdot 3(n+1)\cos^4\theta \} </math>

 

 

<math>~ +\cancelto{0}{x^4}(n+1) \{9 - 2^2\cdot 3^2(1+2n)\cos^2\theta - [9 + 32(n+1)]\cos^4\theta +2^3(n+1)\cos^6\theta \} </math>

 

 

<math>~\pm~~i~x^2 \biggl(\frac{\beta}{x}\biggr)~\biggl[ \frac{2^5\cdot 3 (n+1)^3}{1+\cancelto{0}{x}(3\cos\theta-\cos^3\theta)} \biggr]^{1/2} \biggl\{ 2\cos\theta - \cancelto{0}{x}[2 - 7\cos^2\theta + 3\cos^4\theta ] </math>

 

 

<math>~- \cancelto{0}{x^2} \cos\theta [ 9 +4\cos^2\theta -\cos^4\theta ] \biggr\}</math>

 

<math>~\approx</math>

<math>~ 2(n+1)x^2[2^3(n+1)\cos^2\theta -3] ~\pm~~i~x^2 \biggl(\frac{\beta}{x}\biggr)b_0 \, . </math>

Finally,

<math>~ nx\biggl[ (2+3xb )\cdot \frac{\partial \Lambda }{\partial x} - 3\sin^3\theta \cdot \frac{\partial\Lambda}{\partial\theta} \biggr]</math>

<math>~=</math>

<math>~ x^2\cdot 2^2n(n+1)[2^3(n+1)\cos^2\theta -3] </math>

 

 

<math>~ + n \cancelto{0}{x^3}\cdot 2^2\cdot 3(n+1)\cos\theta \{ [2^2n -5] +\cos^2\theta[2^4n + 19] - 2^2(n+1)\cos^4\theta \} </math>

 

 

<math>~ +n \cancelto{0}{x^4} \cdot 3^2(n+1)\cos^2\theta \{ -3^3 + 2\cdot 3^2\cos^2\theta[2^2n+5] - 99\cdot \cos^4\theta + 2^3(n+1)\cos^6\theta \} </math>

 

 

<math>~ + n\cancelto{0}{x^4} (n+1)\sin^4\theta \{ -3^3 + 3^2\cos^2\theta [ 2^3\cdot 3 n + 27] -2^3\cdot 3\cdot 5(n+1)\cos^4\theta \} </math>

 

 

<math>~\pm~~i~nx^2\biggl(\frac{\beta}{x}\biggr)~ \biggl[ \frac{2^5\cdot 3(n+1)^3}{1+\cancelto{0}{x}b} \biggr]^{1/2} \biggl\{ 4\cos\theta + 6\cancelto{0}{x}(2b\cos\theta + \sin^4\theta) + 3\cancelto{0}{x^2}(3b^2\cos\theta +2b\sin^4\theta + 3\sin^6\theta\cos\theta) \biggr\} </math>

 

<math>~\approx</math>

<math>~ x^2\cdot 2^2n(n+1)[2^3(n+1)\cos^2\theta -3] ~~\pm~~i~2nx^2\biggl(\frac{\beta}{x}\biggr)b_0 \, . </math>

Inserting these three approximate expressions into the LHS_4 ensemble gives,

<math>~\frac{\mathrm{LHS}_4}{x^2}</math>

<math>~=</math>

<math>~ \biggl[ \biggl(\frac{\beta}{x}\biggr)^2 - 1 - \cancelto{0}{xb} \biggr] (1-\cancelto{0}{x}\cos\theta)^3 \biggl\{ (1-\cancelto{0}{x}\cos\theta) \biggl[x^2 \frac{\partial^2\Lambda}{\partial x^2} + \frac{\partial^2\Lambda}{\partial \theta^2} \biggr] + x\biggl[(1-2x\cos\theta)\frac{\partial\Lambda}{\partial x} + \sin\theta \frac{\partial\Lambda}{\partial\theta} \biggr] \biggr\} </math>

 

 

<math>~ -~n x (1-\cancelto{0}{x}\cos\theta)^4\biggl[(2+3xb) \frac{\partial\Lambda}{\partial x} -3\sin^3\theta \frac{\partial\Lambda}{\partial\theta} \biggr] </math>

<math>~</math>

<math>~\approx</math>

<math>~ \biggl[ \biggl(\frac{\beta}{x}\biggr)^2 - 1 \biggr] \biggl\{ ~2(n+1)x^2\biggl[ 2^3(n+1)\sin^2\theta -3 \biggr] ~\pm~~i~x^2\biggl[ - \biggl(\frac{\beta}{x}\biggr) b_0\biggr] </math>

<math>~</math>

 

<math>~ + 2(n+1)x^2[2^3(n+1)\cos^2\theta -3] ~\pm~~i~x^2 \biggl(\frac{\beta}{x}\biggr)b_0 \biggr\} </math>

 

 

<math>~ -~x^2\cdot 2^2n(n+1)[2^3(n+1)\cos^2\theta -3] ~~\pm~~i~2nx^2\biggl(\frac{\beta}{x}\biggr)b_0 </math>

<math>~\Rightarrow~~~\frac{\mathrm{LHS}_4}{x^4}</math>

<math>~\approx</math>

<math>~ \biggl[ \biggl(\frac{\beta}{x}\biggr)^2 - 1 \biggr] \biggl\{ ~2(n+1)\biggl[ 2^3(n+1)\sin^2\theta -3 \biggr] + 2(n+1)[2^3(n+1)\cos^2\theta -3] \biggr\} -~ 2^2n(n+1)[2^3(n+1)\cos^2\theta -3] </math>

<math>~</math>

 

<math>~ ~\pm~~i~2n\biggl(\frac{\beta}{x}\biggr)b_0 </math>

 

<math>~\approx</math>

<math>~ 2^2(n+1) \biggl(\frac{\beta}{x}\biggr)^2 [2^2(n+1) - 3 ] -(n+1) \biggl\{ [2^4(n+1) -12] +~ 2^2n[2^3(n+1)\cos^2\theta -3] \biggr\} ~\pm~~i~2n\biggl(\frac{\beta}{x}\biggr)b_0 </math>

 

<math>~\approx</math>

<math>~ 2^2(n+1) \biggl(\frac{\beta}{x}\biggr)^2 [4n+1 ] - 2^2(n+1)^2 [ 1+~ 2^3n\cos^2\theta ] ~\pm~~i~2n\biggl(\frac{\beta}{x}\biggr)b_0 \, . </math>

Assessment

The good news is that the real part of the <math>~\mathrm{LHS}_4</math> expression exactly matches the real part of the <math>~\mathrm{RHS}_4</math> expression. But the imaginary differ by a factor of 2. So, let's repeat the steps leading to the imaginary parts.

Case B:

<math>~\mathrm{Im}\biggl[ \frac{\mathrm{RHS_4}}{x^2} \biggr]</math>

<math>~=</math>

<math>~-~2^2(n+1)\cdot \mathrm{Im}[(n+1)\mathcal{A}] +\frac{m^2}{(n+1)}\biggl\{ \mathrm{Im}[(n+1)\mathcal{A}]\cdot \mathrm{Re}[\Lambda] + \mathrm{Re}[(n+1)\mathcal{A}]\cdot \mathrm{Im}[\Lambda] \biggr\} </math>

 

<math>~=</math>

<math>~-~2^2(n+1)\cdot x\cos\theta (1-x\cos\theta)^2 [ 2^3\cdot 3 n^2\beta^2(n+1) ]^{1/2} </math>

 

 

<math>~ +\frac{m^2}{(n+1)}\biggl\{ x\cos\theta (1-x\cos\theta)^2 [ 2^3\cdot 3 n^2\beta^2(n+1) ]^{1/2} \biggr\} \cdot \biggl\{ -(4n+1)\beta^2 + (\beta\eta)^2(n+1)[2^3(n+1)\cos^2\theta - 3] \biggr\} </math>

 

 

<math>~ +\frac{m^2}{(n+1)}\biggl\{ 2n (n+1) + (n+1)(1-x\cos\theta)^2 [ x^2(1+xb) - \beta^2 - 4n ] + (1-x\cos\theta)^4 [2n(n+1) - 3n\beta^2 ] \biggr\} \cdot \biggl\{\beta\cos\theta [2^7\cdot 3 (n+1)^3]^{1/2} (\beta\eta) \biggr\} </math>

 

<math>~=</math>

<math>~-~x^2 \biggl(\frac{\beta}{x}\biggr) \biggl\{ (1-x\cos\theta)^2 nb_0 \biggr\} </math>

 

 

<math>~ +m^2 x^4\biggl(\frac{\beta}{x}\biggr)\biggl\{ (1-x\cos\theta)^2 \biggl[\frac{n b_0}{2^2(n+1)^2} \biggr] \biggr\} \cdot \biggl\{ -\biggl[ \frac{4n+1}{n+1}\biggr] \biggl(\frac{\beta}{x}\biggr)^2 + (1+xb)[2^3(n+1)\cos^2\theta - 3] \biggr\} </math>

 

 

<math>~ + m^2 x^2\biggl(\frac{\beta}{x}\biggr) \biggl\{ 2n + (1-x\cos\theta)^2 [ x^2(1+xb) - \beta^2 - 4n ] + (1-x\cos\theta)^4\cdot \biggl[2n - \frac{3n\beta^2}{(n+1)} \biggr] \biggr\} \cdot \biggl\{(1+xb)^{1/2} \biggr\} </math>

<math>~\Rightarrow~~~\mathrm{Im}\biggl[ \frac{\mathrm{RHS_4}}{x^4} \biggr]\biggl(\frac{x}{\beta}\biggr) </math>

<math>~=</math>

<math>~-~(1-x\cos\theta)^2 nb_0 </math>

 

 

<math>~ + m^2 \biggl\{ 2n + (1-x\cos\theta)^2 [ x^2(1+xb) - \beta^2 - 4n ] + (1-x\cos\theta)^4\cdot \biggl[2n - \frac{3n\beta^2}{(n+1)} \biggr] \biggr\} \cdot (1+xb)^{1/2} </math>

 

 

<math>~ +m^2 \biggl\{ -\biggl[ \frac{4n+1}{n+1}\biggr] \beta^2 + x^2(1+xb)[2^3(n+1)\cos^2\theta - 3] \biggr\} \cdot (1-x\cos\theta)^2 \biggl[\frac{n b_0}{2^2(n+1)^2} \biggr] </math>

 

<math>~=</math>

<math>~-~(1-x\cos\theta)^2 nb_0 + m^2 \biggl\{ 2n - 4n (1-x\cos\theta)^2 + 2n(1-x\cos\theta)^4 \biggr\} \cdot (1+xb)^{1/2} </math>

 

 

<math>~ + m^2 (1-x\cos\theta)^2 \biggl\{ x^2(1+xb)^{3/2} - \beta^2\cdot (1+xb)^{1/2} \biggl[ 1 + \frac{3n(1-x\cos\theta)^2}{(n+1)} \biggr] \biggr\} </math>

 

 

<math>~ +m^2 (1-x\cos\theta)^2 \biggl\{ x^2(1+xb)[2^3(n+1)\cos^2\theta - 3]\cdot\biggl[\frac{n b_0}{2^2(n+1)^2} \biggr] -\biggl[ \frac{nb_0(4n+1)}{2^2(n+1)^3}\biggr] \beta^2 \biggr\} </math>

 

<math>~=</math>

<math>~-~(1-x\cos\theta)^2 nb_0 + m^2 \biggl\{ 2n - 4n [1-2x\cos\theta + x^2\cos^2\theta + \mathcal{O}(x^3)] + 2n[ 1-4x\cos\theta + 6x^2\cos^2\theta + \mathcal{O}(x^3)] \biggr\} \cdot (1+xb)^{1/2} </math>

 

 

<math>~ + m^2 x^2 (1-x\cos\theta)^2 \biggl\{ 2^2(n+1)^2(1+xb)^{1/2} + [2^3(n+1)\cos^2\theta - 3]\cdot n b_0 \biggr\} \cdot \frac{(1+xb)}{2^2(n+1)^2} </math>

 

 

<math>~ - m^2 \beta^2 (1-x\cos\theta)^2 \biggl\{ nb_0(4n+1) + 2^2(n+1)^2(1+xb)^{1/2} [ (n+1) + 3n(1-x\cos\theta)^2 ] \biggr\} \cdot \frac{1}{2^2(n+1)^3} </math>

 

<math>~=</math>

<math>~-~ nb_0 [1 -2x\cos\theta + x^2\cos^2\theta + \mathcal{O}(x^3)] + m^2 \biggl\{ 8n x^2\cos^2\theta + \mathcal{O}(x^3)\biggr\} \cdot (1+\cancelto{0}{x}b)^{1/2} </math>

 

 

<math>~ + m^2 x^2 (1-\cancelto{0}{x}\cos\theta)^2 \biggl\{ 2^2(n+1)^2(1+\cancelto{0}{x}b)^{1/2} + [2^3(n+1)\cos^2\theta - 3]\cdot n b_0 \biggr\} \cdot \frac{(1+\cancelto{0}{x}b)}{2^2(n+1)^2} </math>

 

 

<math>~ - m^2 \beta^2 (1-\cancelto{0}{x}\cos\theta)^2 \biggl\{ nb_0(4n+1) + 2^2(n+1)^2(1+\cancelto{0}{x}b)^{1/2} [ (n+1) + 3n(1-\cancelto{0}{x}\cos\theta)^2 ] \biggr\} \cdot \frac{1}{2^2(n+1)^3} </math>

 

<math>~\approx</math>

<math>~-~ nb_0 [1 -2x\cos\theta ] </math>

 

 

<math>~-nb_0x^2\cos^2\theta + m^2 x^2 \biggl\{2^5n(n+1)^2 \cos^2\theta + 2^2(n+1)^2 + [2^3(n+1)\cos^2\theta - 3]\cdot n b_0 \biggr\} \cdot \frac{1}{2^2(n+1)^2} </math>

 

 

<math>~ - m^2 \beta^2 \biggl\{ nb_0(4n+1) + 2^2(n+1)^2 [ (n+1) + 3n ] \biggr\} \cdot \frac{1}{2^2(n+1)^3} </math>

Goldreich, Goodman and Narayan (1986)

Unperturbed Slim Torus Structure

Goldreich, Goodman & Narayan (1986, MNRAS, 221, 339) — hereafter, GGN86 — also used analytic techniques to analyze the properties of unstable, nonaxisymmetric eigenmodes in Papaloizou-Pringle tori. They restricted their discussion to only the slimmest tori, so overlap between the GGN86 and Blaes85 work is easiest to recognize if we begin with the enthalpy distribution prescribed for a "slim torus" by Blaes (1985), as discussed above, namely,

<math>~H = H_0\Theta_H</math>

<math>~=</math>

<math>~H_0 - \frac{H_0}{\beta^2}\biggl[r^2 + r^3(3\cos\theta - \cos^3\theta) + \mathcal{O}(r^4) \biggr] \, .</math>

[Note:   Here we have replaced the variable name, <math>~x</math>, as used in Blaes85, with the variable name, <math>~r</math>, in order (1) to emphasize that the variable represents a dimensionless radial coordinate, and (2) to avoid conflict with the GGN86 variable, <math>~x</math>, which is a Cartesian coordinate with the standard dimension of length.]

Now, from our above discussion of equilibrium PP tori and recognizing that the Keplerian angular frequency at the location of the enthalpy maximum is,

<math>\Omega_K \equiv \frac{GM_\mathrm{pt}}{\varpi_0^3} \, ,</math>

we can set,

<math>~H_0</math>

<math>~=</math>

<math>~\frac{GM_\mathrm{pt}\beta^2}{2\varpi_0} = \tfrac{1}{2} \Omega_K^2 \varpi_0^2 \beta^2 \, .</math>

Hence, for the slimmest tori — that is, keeping only the lowest order term in <math>~r</math> — the enthalpy distribution becomes,

<math>~H </math>

<math>~=</math>

<math>~\tfrac{1}{2} \Omega_K^2 \varpi_0^2 \beta^2 - \tfrac{1}{2} \Omega_K^2 \varpi_0^2\biggl[r^2 + \cancelto{0}{r^3}(3\cos\theta - \cos^3\theta) + \cancelto{0}{\mathcal{O}(r^4)} ~~\biggr] </math>

 

<math>~\approx</math>

<math>~\frac{\Omega_K^2}{2} [\varpi_0^2 \beta^2 - \varpi_0^2 r^2] \, .</math>

Following GGN86, the surface of the torus — where the enthalpy drops to zero — occurs at <math>~r = a/\varpi_0</math>. Hence, we recognize that,

<math>\beta = \frac{a}{\varpi_0} \, ,</math>

and we can rewrite the expression for the unperturbed enthalpy distribution as,

<math>~H </math>

<math>~=</math>

<math>~\frac{\Omega_K^2}{2} [a^2 - \varpi_0^2 r^2 ] \, .</math>

This expression exactly matches equation (2.13) of GGN86 — which, is,

<math>~Q_0(x,z)</math>

<math>~=</math>

<math>~\frac{\Omega^2}{2} \biggl[ (2q-3)(a^2 - x^2) - z^2 \biggr] \, ,</math>

once it is appreciated that, in moving from the Blaes85 discussion to the GGN86 discussion, <math>~\varpi_0^2 r^2 \rightarrow (x^2 + z^2)</math>, and it is recognized that Blaes85 restricted his investigation to tori that have uniform specific angular momentum <math>~(q = 2)</math>.

Additional Notation

<math>~(ky)_\mathrm{GGN} = \biggl( \frac{my}{\varpi_0} \biggr)_\mathrm{GGN} ~~\leftrightarrow ~~ (m\phi)_\mathrm{Blaes}</math>

<math>~\beta_\mathrm{GGN} \equiv \biggl( \frac{ma}{\varpi_0} \biggr)_\mathrm{GGN} ~~\leftrightarrow ~~ m\beta_\mathrm{Blaes}</math>

From equation (5.16) of GGN86 we obtain "the lowest order [complex] expression for the [perturbed] velocity potential," namely,

<math>~\psi </math>

<math>~=</math>

<math>~1+\tfrac{1}{4} k^2(5x^2 - 3z^2) \mp 4i\biggl(\frac{3}{2}\biggr)^{1/2} k x \beta_\mathrm{GGN} \, .</math>

Working on the imaginary part of this expression to put it in the terminology of Blaes85, we find,

<math>~\mathrm{Im}(\psi)</math>

<math>~=</math>

<math>~\mp 4\biggl(\frac{3}{2}\biggr)^{1/2} k x \beta_\mathrm{GGN} </math>

 

<math>~=</math>

<math>~\mp 4\biggl(\frac{3}{2}\biggr)^{1/2} \biggl(\frac{m}{\varpi_0}\biggr) [\varpi_0 (\eta\beta_\mathrm{Blaes})\cos\theta ](m\beta_\mathrm{Blaes}) </math>

 

<math>~=</math>

<math>~\mp 4\biggl(\frac{3}{2}\biggr)^{1/2} m^2\beta^2_\mathrm{Blaes} \eta\cos\theta \, ,</math>

which exactly matches <math>~\mathrm{Im}(f_m)</math> as derived by Blaes85 and summarized above. Similarly,

<math>~\mathrm{Re}(\psi)</math>

<math>~=</math>

<math>~1+\tfrac{1}{4} k^2(5x^2 - 3z^2) </math>

 

<math>~=</math>

<math>~1+\frac{1}{4} \biggl(\frac{m}{\varpi_0}\biggr)^2[\varpi_0^2 r^2(5\cos^2\theta - 3\sin^2\theta)] </math>

 

<math>~=</math>

<math>~1+\frac{1}{4} \eta^2 m^2 \beta^2_\mathrm{Blaes}[8\cos^2\theta - 3] </math>

 

<math>~=</math>

<math>~1+m^2 \beta^2_\mathrm{Blaes}\biggl[2\eta^2\cos^2\theta - \frac{3\eta^2}{4}\biggr] \, .</math>

This exactly matches <math>~\mathrm{Re}(f_m)</math> as derived by Blaes85 and summarized above. This is in line with the following statement that appears in the acknowledgement section of GGN86: "We note that Omar Blaes … [has] independently derived many of the results reported in this paper."

Summary Comparison

For slim, incompressible tori with uniform specific angular momentum, Blaes85 gives, to lowest order:

<math>~f_m(\eta,\theta) + \tfrac{1}{4} \beta^2 m^2</math>

<math>~=</math>

<math> ~\beta^2 m^2 \biggl[ 2\eta^2 \cos^2\theta - \frac{3\eta^2}{4} \pm 4i\biggl(\frac{3}{2}\biggr)^{1/2} \eta\cos\theta\biggr] \, . </math>

By comparison, from GGN86 we obtain:

<math>~\Psi(\eta,\theta) - 1</math>

<math>~=</math>

<math>~ m^2 \beta^2_\mathrm{Blaes}\biggl[2\eta^2\cos^2\theta - \frac{3\eta^2}{4} \mp 4i\biggl(\frac{3}{2}\biggr)^{1/2} \eta\cos\theta \biggr] \, . </math>

To within an additive constant, these two functions are identical. The resulting amplitude function is (to within an overall scale factor and to within an arbitrary additive constant),

<math>~F(\eta,\theta)</math>

<math>~=</math>

<math>~\biggl\{ \biggl[ 2\eta^2\cos^2\theta - \frac{3\eta^2}{4} \biggr]^2 + \biggl[ 4\biggl(\frac{3}{2}\biggr)^{1/2} \eta\cos\theta \biggr]^2 \biggr\}^{1/2}</math>

 

<math>~=</math>

<math>~\biggl\{ \eta^2 \biggl[ 4^2\biggl(\frac{3}{2}\biggr) \cos^2\theta \biggr] + \frac{1}{4^2}\eta^4\biggl[ 8\cos^2\theta - 3\biggr]^2 \biggr\}^{1/2}</math>

 

<math>~=</math>

<math>~\biggl\{ \biggl(\frac{\eta}{2}\biggr)^2 \biggl[ 2^5\cdot 3 \cos^2\theta \biggr] + \biggl( \frac{\eta}{2}\biggr)^4\biggl[ 8\cos^2\theta - 3\biggr]^2 \biggr\}^{1/2} \, ;</math>

and the associate phase function is,

<math>~m\phi_m </math>

<math>~=</math>

<math>~\tan^{-1} \biggl[ \frac{ 2^7 \cdot 3 \cos^2\theta }{\eta^2 ( 8\cos^2\theta - 3)^2} \biggr] + k\theta \, .</math>

See Also


Whitworth's (1981) Isothermal Free-Energy Surface

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