Appendix/Ramblings/PPToriPt1A

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.

---

Material that appears after this point in our presentation is under development and therefore
may contain incorrect mathematical equations and/or physical misinterpretations.
|   Go Home   |

---

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, $m$, we are referring both to the radial dependence of the distortion amplitude, $f_m(\varpi )$, and the radial dependence of the phase function, ${\varphi }_{\mathrm{m}\mathrm{a}\mathrm{x}}(\varpi )$ — 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:

• Papaloizou & Pringle (1987), MNRAS, 225, 267 (aka PPIII) — The dynamical stability of differentially rotating discs.   III.
• Blaes (1985), MNRAS, 216, 553 (aka Blaes85) — Oscillations of slender tori.
• Goldreich & Narayan (1985), MNRAS, 213, 7 (aka GGN86) — Non-axisymmetric instability in thin discs.

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 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,

$H=\frac{G{M}_{\mathrm{p}\mathrm{t}}}{{\varpi }_0}[({\chi }^2+{\zeta }^2{)}^{-1/2}-\frac{1}{2}{\chi }^{-2}-C_B^{\text{'}}].$

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

$H_0=\frac{G{M}_{\mathrm{p}\mathrm{t}}}{2{\varpi }_0}[1-2C_B^{\text{'}}]$

gives,

$[1-2C_B^{\text{'}}]\left(\frac{H}{H_0}\right)=2({\chi }^2+{\zeta }^2{)}^{-1/2}-{\chi }^{-2}-1+[1-2C_B^{\text{'}}].$

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,

${\mathfrak{𝔐}}_0^2\equiv \frac{(v_{\phi }{|}_0{)}^2}{(c_s{|}_0{)}^2}$	$=$	$\frac{2(n+1)}{\gamma }[\frac{1}{{\chi }_{-}}-1{]}^{-2}$
	$=$	$2n[1-2C_B^{\text{'}}{]}^{-1}$
$\Rightarrow [1-2C_B^{\text{'}}]$	$=$	$\frac{2n}{{\mathfrak{𝔐}}_0^2},$

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

${\beta }^2$

$\equiv$

$\frac{2n}{{\mathfrak{𝔐}}_0^2}.$

Implementing this parameter swap, the equilibrium expression becomes,

${\beta }^2\left(\frac{H}{H_0}\right)=2({\chi }^2+{\zeta }^2{)}^{-1/2}-{\chi }^{-2}-1+{\beta }^2,$

or,

$\frac{H}{H_0}$

$=$

$1-\frac{1}{{\beta }^2}[{\chi }^{-2}-2({\chi }^2+{\zeta }^2{)}^{-1/2}+1].$

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

$\chi =1-x\cos \theta$       and         $\zeta =x\sin \theta .$

Hence,

$\frac{H}{H_0}$	$=$	$1-\frac{1}{{\beta }^2}\{[1-x\cos \theta {]}^{-2}-2[(1-x\cos \theta {)}^2+x^2{\sin }^{}\theta {]}^{-1/2}+1\}$
	$=$	$1-\frac{1}{{\beta }^2}\{[1-x\cos \theta {]}^{-2}-2[(1-2x\cos \theta +x^2{\cos }^{}\theta )+x^2(1-{\cos }^{}\theta ){]}^{-1/2}+1\}$
	$=$	$1-\frac{1}{{\beta }^2}[(1-x\cos \theta {)}^{-2}-2(1-2x\cos \theta +x^2{)}^{-1/2}+1].$

This matches equation (2.2) of Blaes85, if we acknowledge that Blaes uses $f$ — instead of the parameter notation, ${\mathrm{\Theta }}_H$, found in our discussion of equilibrium polytropic configurations — to denote the normalized enthalpy; that is,

$f_{\mathrm{B}\mathrm{l}\mathrm{a}\mathrm{e}\mathrm{s}\mathrm{8}\mathrm{5}}={\mathrm{\Theta }}_H\equiv \frac{H}{H_0}.$

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

${\mathrm{\Theta }}_H$

$=$

$1-\frac{1}{{\beta }^2}[x^2+x^3(3\cos \theta -{\cos }^{}\theta )+{𝒪}(x^4)].$

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

${\eta }^2\equiv 1-{\mathrm{\Theta }}_H,$

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

$\eta$

$=$

$[1-{\mathrm{\Theta }}_H{]}^{1/2}\approx \frac{x}{\beta }.$

Cubic Equation Solution

For later use, let's invert the cubic relation to obtain a more broadly applicable $x(\eta )$ function. Because we are only interested in radial profiles in the equatorial plane — that is, only for the values of $\theta =0$ or $\theta =\pi$ — the relation to be inverted is,

$x^2\pm 2x^3$	$=$	$(\beta \eta {)}^2$
$\Rightarrow x^3\pm \frac{1}{2}{x}^2\mp \frac{1}{2}(\beta \eta {)}^2$	$=$	$0.$

Table 1:  Example Parameter Values determined by iterative solution for $\beta =\frac{1}{10}$
$\eta$	${\mathrm{\Gamma }}^2=54{\beta }^2{\eta }^2$	Inner solution $(\theta =0)$ [Superior sign in cubic eq.]		Outer solution $(\theta =\pi )$ [Inferior sign in cubic eq.]
		${†}^{}\left(\frac{x_{\mathrm{r}\mathrm{o}\mathrm{o}\mathrm{t}}}{\beta }\right)$	$6(S+T)$	${†}^{}\left(\frac{x_{\mathrm{r}\mathrm{o}\mathrm{o}\mathrm{t}}}{\beta }\right)$	$6(S+T)$
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, $x_{\mathrm{r}\mathrm{o}\mathrm{o}\mathrm{t}}$ 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,

$x^3+a_2{x}^2+a_{1}x+a_0=0,$

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:

$x_1$	$=$	$-\frac{1}{3}{a}_2+(S+T),$
$x_2$	$=$	$-\frac{1}{3}{a}_2-\frac{1}{2}(S+T)+\frac{1}{2}{i}\sqrt{{3}}{(}{S}{-}{T}{)},$
$x_3$	$=$	$-\frac{1}{3}{a}_2-\frac{1}{2}(S+T)-\frac{1}{2}{i}\sqrt{{3}}{(}{S}{-}{T}{)},$

where,

$S$	$\equiv$	$[R+\sqrt{D}{]}^{1/3},$
$T$	$\equiv$	$[R-\sqrt{D}{]}^{1/3},$
$D$	$\equiv$	$Q^3+R^2,$
$Q$	$\equiv$	$\frac{3a_1-a_2^2}{3^2},$
$R$	$\equiv$	$\frac{3^2{a}_2{a}_1-3^3{a}_0-2a_2^3}{2\cdot 3^3}.$

Outer [inferior sign] Solution

Focusing, first, on the inferior sign convention, which corresponds to the "outer" solution $(\theta =\pi )$, we see that the coefficients that lead to our specific cubic equation are:

$a_2$	$=$	$-\frac{1}{2},$
$a_1$	$=$	$0,$
$a_0$	$=$	$\frac{1}{2}(\beta \eta {)}^2.$

Applying Wolfram's definitions of the $Q$ and $R$ parameters to our particular problem gives,

$Q$	$\equiv$	$\frac{3a_1-a_2^2}{3^2}=-(\frac{a_2}{3}{)}^2=-\frac{1}{2^2\cdot 3^2};$
$R$	$\equiv$	$\frac{3^2{a}_2{a}_1-3^3{a}_0-2a_2^3}{2\cdot 3^3}=\frac{1}{2\cdot 3^3}[-\frac{3^3}{2}(\beta \eta {)}^2+\frac{1}{2^2}]$
	$\equiv$	$\frac{1}{2^3\cdot 3^3}[1-2\cdot 3^3(\beta \eta {)}^2].$

Defining the parameter,

${\mathrm{\Gamma }}^2$

$\equiv$

$2\cdot 3^3(\beta \eta {)}^2,$

we therefore have,

$(2\cdot 3{)}^{6}D$	$=$	$(1-{\mathrm{\Gamma }}^2{)}^2-1,$
$(2\cdot 3{)}^3{S}^3$	$\equiv$	$(2\cdot 3{)}^{3}R+\sqrt{(2\cdot 3{)}^{6}D}$
	$\equiv$	$(1-{\mathrm{\Gamma }}^2)+\sqrt{(1-{\mathrm{\Gamma }}^2{)}^2-1}$
	$\equiv$	$(1-{\mathrm{\Gamma }}^2)+i\sqrt{1-(1-{\mathrm{\Gamma }}^2{)}^2};$
$(2\cdot 3{)}^3{T}^3$	$\equiv$	$(1-{\mathrm{\Gamma }}^2)-i\sqrt{1-(1-{\mathrm{\Gamma }}^2{)}^2}.$

ASIDE:  The cube root of an imaginary number … ${\ell }^3$ $=$ $A\pm i\sqrt{1-A^2}$   $=$ $r_{\ell }{e}^{i{\theta }_{\ell }},$ where, $r_{\ell }$ $=$ $(A^2+1-A^2{)}^{1/2}=1,$ and, ${\theta }_{\ell }$ $=$ $\pm {\tan }^{-1}\left(\frac{\sqrt{1-A^2}}{A}\right)=\pm {\cos }^{-1}A.$ Now, according to this online resource, the three roots $(j=0,1,2)$ of ${\ell }^3$ are, ${\ell }_j=r_{\ell }^{1/3}{e}^{i({\theta }_{\ell }+2j\pi )/3)},$ which, for our specific problem gives, ${\ell }_j$ $=$ $e^{i{\theta }_{\pm }/3}\cdot e^{i(2j\pi /3)},$ where the subscript on $\theta$ refers to the $\pm$ in our original expression for $\ell$.

In our particular case, after associating $A\leftrightarrow (1-{\mathrm{\Gamma }}^2)$, we can write,

$2\cdot 3(S+T)$	$=$	$[(1-{\mathrm{\Gamma }}^2)+i\sqrt{1-(1-{\mathrm{\Gamma }}^2{)}^2}{]}^{1/3}+[(1-{\mathrm{\Gamma }}^2)-i\sqrt{1-(1-{\mathrm{\Gamma }}^2{)}^2}{]}^{1/3}$
	$=$	$e^{i{\theta }_{+}/3}\cdot e^{i(2j\pi /3)}+e^{i{\theta }_{-}/3}\cdot e^{i(2j\pi /3)}$
	$=$	$e^{i(2j\pi /3)}\{e^{i[{\cos }^{-1}(1-{\mathrm{\Gamma }}^2)]/3}+e^{-i[{\cos }^{-1}(1-{\mathrm{\Gamma }}^2)]/3}\}$
	$=$	$e^{i(2j\pi /3)}\{\cos [\frac{1}{3}{\cos }^{}(1-{\mathrm{\Gamma }}^2)]+i\sin [\frac{1}{3}{\cos }^{}(1-{\mathrm{\Gamma }}^2)]$
		$+\cos [\frac{1}{3}{\cos }^{}(1-{\mathrm{\Gamma }}^2)]-i\sin [\frac{1}{3}{\cos }^{}(1-{\mathrm{\Gamma }}^2)]\}$
	$=$	$2e^{i(2j\pi /3)}\cdot \cos [\frac{1}{3}{\cos }^{}(1-{\mathrm{\Gamma }}^2)].$

Similarly, we can write,

$2\cdot 3(S-T)$	$=$	$[(1-{\mathrm{\Gamma }}^2)+i\sqrt{1-(1-{\mathrm{\Gamma }}^2{)}^2}{]}^{1/3}-[(1-{\mathrm{\Gamma }}^2)-i\sqrt{1-(1-{\mathrm{\Gamma }}^2{)}^2}{]}^{1/3}$
	$=$	$e^{i{\theta }_{+}/3}\cdot e^{i(2j\pi /3)}-e^{i{\theta }_{-}/3}\cdot e^{i(2j\pi /3)}$
	$=$	$e^{i(2j\pi /3)}\{e^{i[{\cos }^{-1}(1-{\mathrm{\Gamma }}^2)]/3}-e^{-i[{\cos }^{-1}(1-{\mathrm{\Gamma }}^2)]/3}\}$
	$=$	$e^{i(2j\pi /3)}\{\cos [\frac{1}{3}{\cos }^{}(1-{\mathrm{\Gamma }}^2)]+i\sin [\frac{1}{3}{\cos }^{}(1-{\mathrm{\Gamma }}^2)]$
		$-\cos [\frac{1}{3}{\cos }^{}(1-{\mathrm{\Gamma }}^2)]+i\sin [\frac{1}{3}{\cos }^{}(1-{\mathrm{\Gamma }}^2)]\}$
	$=$	$2i{e}^{i(2j\pi /3)}\cdot \sin [\frac{1}{3}{\cos }^{}(1-{\mathrm{\Gamma }}^2)].$

Focusing specifically on the "j=0" root, and setting $a_2=-\frac{1}{2}$, we therefore have,

$6x_1-1$	$=$	$6(S+T)$
	$=$	$2\cos [\frac{1}{3}{\cos }^{}(1-{\mathrm{\Gamma }}^2)];$
$6x_2-1$	$=$	$-\frac{1}{2}\{2\cos [\frac{1}{3}{\cos }^{}(1-{\mathrm{\Gamma }}^2)]-i\sqrt{3}\cdot 2i\sin [\frac{1}{3}{\cos }^{}(1-{\mathrm{\Gamma }}^2)\left]\right\}$
	$=$	$-2\{\frac{1}{2}\cdot \cos [\frac{1}{3}{\cos }^{}(1-{\mathrm{\Gamma }}^2)]+\frac{\sqrt{3}}{2}\cdot \sin [\frac{1}{3}{\cos }^{}(1-{\mathrm{\Gamma }}^2)\left]\right\}$
	$=$	$2\{\cos [\frac{1}{3}{\cos }^{}(1-{\mathrm{\Gamma }}^2)+\frac{2\pi }{3}\left]\right\};$
$6x_3-1$	$=$	$-\frac{1}{2}\{2\cos [\frac{1}{3}{\cos }^{}(1-{\mathrm{\Gamma }}^2)]+i\sqrt{3}\cdot 2i\sin [\frac{1}{3}{\cos }^{}(1-{\mathrm{\Gamma }}^2)\left]\right\}$
	$=$	$-2\{\frac{1}{2}\cdot \cos [\frac{1}{3}{\cos }^{}(1-{\mathrm{\Gamma }}^2)]-\frac{\sqrt{3}}{2}\cdot \sin [\frac{1}{3}{\cos }^{}(1-{\mathrm{\Gamma }}^2)\left]\right\}$
	$=$	$2\{\cos [\frac{1}{3}{\cos }^{}(1-{\mathrm{\Gamma }}^2)-\frac{2\pi }{3}\left]\right\}.$

Table 1:  Analytically Evaluated Roots determined for $\beta =\frac{1}{10}$
$\eta$	${\mathrm{\Gamma }}^2=54{\beta }^2{\eta }^2$	Inner solution $(\theta =0)$ [Superior sign in cubic eq.]			Outer solution $(\theta =\pi )$ [Inferior sign in cubic eq.]
		$x_1/\beta$	$x_2/\beta$	$x_3/\beta$	$x_1/\beta$	$x_2/\beta$	$x_3/\beta$
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, $x^2+2x^3=(\beta \eta {)}^2$			CONFIRMATION: In all cases, $x^2-2x^3=(\beta \eta {)}^2$

Inner [superior sign] Solution

Next, examing the superior sign convention, which corresponds to the "inner" solution $(\theta =0)$, we see that the coefficients that lead to our specific cubic equation are:

$a_2$	$=$	$\frac{1}{2},$
$a_1$	$=$	$0,$
$a_0$	$=$	$-\frac{1}{2}(\beta \eta {)}^2.$

Following the same set of steps that were followed in determining the "outer" solution, here we find: $Q$ remains the same; $R$ has the same magnitude, but changes sign; and, hence, $D$ remains the same. We therefore have,

$(2\cdot 3{)}^3{S}^3$	$=$	$-(1-{\mathrm{\Gamma }}^2)+i\sqrt{1-(1-{\mathrm{\Gamma }}^2{)}^2};$
$(2\cdot 3{)}^3{T}^3$	$=$	$-(1-{\mathrm{\Gamma }}^2)-i\sqrt{1-(1-{\mathrm{\Gamma }}^2{)}^2},$

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

$6x_1+1$	$=$	$-2\cos [\frac{1}{3}{\cos }^{}(1-{\mathrm{\Gamma }}^2)];$
$6x_2+1$	$=$	$-2\{\cos [\frac{1}{3}{\cos }^{}(1-{\mathrm{\Gamma }}^2)+\frac{2\pi }{3}\left]\right\};$
$6x_3+1$	$=$	$-2\{\cos [\frac{1}{3}{\cos }^{}(1-{\mathrm{\Gamma }}^2)-\frac{2\pi }{3}\left]\right\}.$

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,

$[\frac{\delta \rho }{{\rho }_0}{]}_{\mathrm{s}\mathrm{p}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{a}\mathrm{l}}$

$=$

$\{f_m(\varpi )e^{-im{\varphi }_m}\}.$

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,

$f_m(\varpi )$

$=$

${𝒜}(\varpi )+i{ℬ}(\varpi ),$

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

$|f_m{|}^2\equiv f_m\cdot f_m^{*}$

$=$

${{𝒜}}^2+{{ℬ}}^2.$

We can rewrite this complex function in the form,

$f_m(\varpi )$

$=$

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

if the angle, $\alpha (\varpi )$ is defined such that,

$\sin \alpha =\frac{{𝒜}}{\sqrt{{{𝒜}}^2+{{ℬ}}^2}}$	and	$\cos \alpha =\frac{{ℬ}}{\sqrt{{{𝒜}}^2+{{ℬ}}^2}}$
$\Rightarrow \alpha$	$\equiv$	${\tan }^{-1}\left(\frac{{𝒜}}{{ℬ}}\right)={\tan }^{-1}\left[\frac{\mathrm{R}\mathrm{e}(f_m)}{\mathrm{I}\mathrm{m}(f_m)}\right].$

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

$[\frac{\delta \rho }{{\rho }_0}{]}_{\mathrm{s}\mathrm{p}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{a}\mathrm{l}}$

$=$

$|f_m(\varpi )|e^{-i[\alpha (\varpi )+\pi /2+m{\varphi }_m]}.$

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

${\varphi }_m={\varphi }_{\mathrm{m}\mathrm{a}\mathrm{x}}(\varpi )$

$\equiv$

$-\frac{1}{m}[\alpha (\varpi )+\frac{\pi }{2}]=-\frac{1}{m}\left\{{\tan }^{-1}\right[\frac{\mathrm{R}\mathrm{e}(f_m)}{\mathrm{I}\mathrm{m}(f_m)}]+\frac{\pi }{2}\}.$

An equatorial-plane plot of ${\varphi }_{\mathrm{m}\mathrm{a}\mathrm{x}}(\varpi )$ 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,

$\frac{\delta W}{W_0}\equiv [\frac{W(\eta ,\theta )}{C}-1]e^{im{\mathrm{\Omega }}_{p}t}{e}^{-y_2({\mathrm{\Omega }}_{0}t)}$

$=$

$\{f_m(\eta ,\theta )e^{-i[m{\varphi }_m+k\theta ]}\},$

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 $m$," and its complex amplitude takes the following form (see his equation 1.10):

$f_m(\eta ,\theta )$

$=$

${\beta }^2{m}^2[2{\eta }^2{\cos }^2\theta -\frac{3{\eta }^2}{4(n+1)}-\frac{(4n+1)}{4(n+1{)}^2}\pm 4i(\frac{3}{2n+2}{)}^{1/2}\eta \cos \theta ]+{𝒪}({\beta }^3).$

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

$\left|\frac{\delta W}{W_0}\right|$

$=$

$\sqrt{[\mathrm{R}\mathrm{e}(f_m){]}^2+[\mathrm{I}\mathrm{m}(f_m){]}^2};$

and the associated phase function is,

$m{\varphi }_{\mathrm{m}\mathrm{a}\mathrm{x}}(\varpi )$

$=$

$-\left\{{\tan }^{-1}\right[\frac{\mathrm{R}\mathrm{e}(f_m)}{\mathrm{I}\mathrm{m}(f_m)}]+\frac{\pi }{2}+k\theta \}.$

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,

$\cos \theta \to \pm 1.$

Hence, we have,

$\frac{1}{{\beta }^4{m}^4}|\frac{\delta W}{W_0}{|}^2$	$=$	$[2{\eta }^2-\frac{3{\eta }^2}{4(n+1)}-\frac{(4n+1)}{4(n+1{)}^2}{]}^2+16[(\frac{3}{2n+2}{)}^{1/2}\eta {]}^2$
	$=$	$\frac{1}{[2^2(n+1{)}^2{]}^2}[2^3(n+1{)}^2{\eta }^2-3(n+1){\eta }^2-(4n+1){]}^2+\frac{2^3\cdot 3{\eta }^2}{(n+1)}$
	$=$	$\frac{1}{[2^2(n+1{)}^2{]}^2}[(4n+1)-(n+1)[2^3(n+1)-3]{\eta }^2{]}^2+\frac{2^3\cdot 3{\eta }^2}{(n+1)}$
$\Rightarrow \left[\frac{2(n+1)}{\beta m}{]}^4\right|\frac{\delta W}{W_0}{|}^2$	$=$	$[(4n+1)-(n+1)[2^3(n+1)-3]{\eta }^2{]}^2+2^7\cdot 3(n+1{)}^3{\eta }^2.$

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

	$m{\varphi }_{\mathrm{m}\mathrm{a}\mathrm{x}}$	$=$	$-\left\{{\tan }^{-1}\right[\frac{(4n+1)-(n+1)[2^3(n+1)-3]{\eta }^2}{2^{7/2}\cdot 3^{1/2}(n+1{)}^{3/2}\eta }]+\frac{\pi }{2}\}$	over	inner $(\theta =0)$ region of the torus;
while
	$m{\varphi }_{\mathrm{m}\mathrm{a}\mathrm{x}}$	$=$	$-\left\{{\tan }^{-1}\right[-\frac{(4n+1)-(n+1)[2^3(n+1)-3]{\eta }^2}{2^{7/2}\cdot 3^{1/2}(n+1{)}^{3/2}\eta }\}+\frac{3\pi }{2}\}$	over	outer $(\theta =\pi )$ region of torus.

Incompressible Slim Tori

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

$f_m(\eta ,\theta )$

$=$

${\beta }^2{m}^2[2{\eta }^2{\cos }^2\theta -\frac{3{\eta }^2}{4}-\frac{1}{4}\pm 4i(\frac{3}{2}{)}^{1/2}\eta \cos \theta ]+{\xcancel{{𝒪}({\beta }^3)}}^0.$

Check Validity of Blaes85 Eigenvector

Step 1

Equation (2.6) of Blaes85 states that,

${\beta }^2{\eta }^2=[x^2+x^3(3\cos \theta -{\cos }^{}\theta )]$

$\Rightarrow$

${\beta }^2(1-{\eta }^2)=[{\beta }^2-x^2-x^3(3\cos \theta -{\cos }^{}\theta )].$

This means that,

$\frac{\partial {\eta }^2}{\partial x}$

$=$

$\frac{1}{{\beta }^2}[2x+3x^2(3\cos \theta -{\cos }^{}\theta )];$

and,

$\frac{\partial {\eta }^2}{\partial \theta }$	$=$	$\frac{x^3}{{\beta }^2}[-3\sin \theta +3\sin \theta {\cos }^{}\theta ]$
	$=$	$\frac{3x^3\sin \theta }{{\beta }^2}[{\cos }^2\theta -1].$

Furthermore,

$\frac{\partial \eta }{\partial x}$

$=$

$\frac{1}{2\beta }[x^2+x^3(3\cos \theta -{\cos }^{}\theta ){]}^{-1/2}[2x+3x^2(3\cos \theta -{\cos }^{}\theta )];$

and,

$\frac{\partial \eta }{\partial \theta }$

$=$

$\frac{1}{2\beta }[x^2+x^3(3\cos \theta -{\cos }^{}\theta ){]}^{-1/2}[-3\sin \theta +3\sin \theta {\cos }^{}\theta ].$

Step 2

Equation (4.14) of Blaes85 states that,

$\nu +m$

$=$

$\pm im[\frac{3}{2(n+1)}{]}^{1/2}\beta ;$

and equation (4.13) of Blaes85 states that,

$\frac{\delta W}{A_{00}}$	$=$	$1+{\beta }^2{m}^2[2{\eta }^2{\cos }^2\theta -\frac{3{\eta }^2}{4(n+1)}-\frac{(4n+1)}{4(n+1{)}^2}\pm 4i(\frac{3}{2n+2}{)}^{1/2}\eta \cos \theta ]$
$\Rightarrow \frac{1}{m^2}[\frac{\delta W}{A_{00}}-1]$	$=$	${\beta }^2\{-\frac{(4n+1)}{4(n+1{)}^2}+{\eta }^2[2{\cos }^2\theta -\frac{3}{4(n+1)}]\pm i(\frac{2^3\cdot 3}{n+1}{)}^{1/2}\eta \cos \theta \}$
	$=$	$\frac{{\beta }^2}{2^2(n+1{)}^2}\{-(4n+1)+{\eta }^2[2^3(n+1{)}^2{\cos }^2\theta -3(n+1)]\pm i[2^7\cdot 3(n+1{)}^3{]}^{1/2}\eta \cos \theta \}$
$\Rightarrow \mathrm{\Lambda }\equiv \frac{2^2(n+1{)}^2}{m^2}[\frac{\delta W}{A_{00}}-1]$	$=$	$-(4n+1){\beta }^2+[2^3(n+1{)}^2{\cos }^2\theta -3(n+1)]\cdot [x^2+x^3(3\cos \theta -{\cos }^{}\theta )]$
		$\pm i\beta [2^7\cdot 3(n+1{)}^3{]}^{1/2}[x^2{\cos }^2\theta +x^3(3{\cos }^3\theta -{\cos }^5\theta ){]}^{1/2}$
	$=$	$-(4n+1){\beta }^2+(n+1)x^2[2^3(n+1){\cos }^2\theta -3]\cdot [1+xb]\pm i\beta [2^7\cdot 3(n+1{)}^3{]}^{1/2}x\cos \theta [1+xb{]}^{1/2}$

Through a separate white-board derivation I have obtained …

$\mathrm{\Lambda }$ $\equiv$ $\frac{2^2(n+1{)}^2}{m^2}[\frac{\delta W}{A_{00}}-1]$   $=$ $-(4n+1){\beta }^2+(\beta \eta {)}^2(n+1)[2^3(n+1){\cos }^2\theta -3]\pm i\beta \cos \theta [2^7\cdot 3(n+1{)}^3{]}^{1/2}(\beta \eta ),$ where, $(\beta \eta {)}^2$ $=$ $x^2(1+xb),$ $b$ $\equiv$ $3\cos \theta -{\cos }^{}\theta .$

Note that, differentiating the left-hand-side with respect to either coordinate $(x$ or $\theta )$ gives,

$\frac{\partial \mathrm{\Lambda }}{\partial x_i}$	$=$	$\frac{2^2(n+1{)}^2}{m^2}\cdot \frac{\partial }{\partial x_i}\left(\frac{\delta W}{A_{00}}\right)$
$\Rightarrow \frac{\partial }{\partial x_i}\left(\frac{\delta W}{A_{00}}\right)$	$=$	$[\frac{m}{2(n+1)}{]}^2\cdot \frac{\partial \mathrm{\Lambda }}{\partial x_i}.$

Given that $\nu$ has both real and imaginary parts, presumably,

${\nu }^2\equiv \nu \cdot {\nu }^{*}$	$=$	$\{-m\pm im[\frac{3}{2(n+1)}{]}^{1/2}\beta \left\}\right\{-m\mp im\left[\frac{3}{2(n+1)}{]}^{1/2}\beta \right\}$
	$=$	$m^2+m^2\left[\frac{3}{2(n+1)}\right]{\beta }^2.$

For later reference, let's take the relevant partial derivatives of the function, $\mathrm{\Lambda }(x,\theta )$. Adopting the shorthand notation,

$b\equiv (3\cos \theta -{\cos }^{}\theta ),$

we have,

$\frac{\partial \mathrm{\Lambda }}{\partial x}$	$=$	$[2^3(n+1{)}^2{\cos }^2\theta -3(n+1)]\cdot \frac{\partial }{\partial x}\{[x^2+x^{3}b]\}$
		$\pm i\beta [2^7\cdot 3(n+1{)}^3{]}^{1/2}\cos \theta \cdot \frac{\partial }{\partial x}\{[x^2+x^{3}b{]}^{1/2}\}$
	$=$	$[2^3(n+1{)}^2{\cos }^2\theta -3(n+1)]\cdot [2x+3x^{2}b]$
		$\pm i\beta [2^5\cdot 3(n+1{)}^3{]}^{1/2}\cos \theta \cdot [1+xb{]}^{-1/2}\cdot [2+3xb]$

Through a separate white-board derivation I have obtained …

$\frac{\partial \mathrm{\Lambda }}{\partial x}$ $=$ $(n+1)[2^3(n+1){\cos }^2\theta -3]x(2+3xb)\pm i\beta \cos \theta [2^5\cdot 3(n+1{)}^3{]}^{1/2}\cdot \frac{x(2+3xb)}{(\beta \eta )},$ which is the same.

$\frac{{\partial }^2\mathrm{\Lambda }}{\partial x^2}$	$=$	$[2^3(n+1{)}^2{\cos }^2\theta -3(n+1)]\cdot \frac{\partial }{\partial x}[2x+3x^{2}b]$
		$\pm i\beta [2^5\cdot 3(n+1{)}^3{]}^{1/2}\cos \theta \cdot \frac{\partial }{\partial x}\{[1+xb{]}^{-1/2}\cdot [2+3xb]\}$
	$=$	$[2^3(n+1{)}^2{\cos }^2\theta -3(n+1)]\cdot [2+6xb]$
		$\pm i\beta [2^5\cdot 3(n+1{)}^3{]}^{1/2}\cos \theta \cdot \{-\frac{b}{2}[1+xb{]}^{-3/2}\cdot [2+3xb]+[1+xb{]}^{-1/2}\cdot [3b]\}$
	$=$	$2[2^3(n+1{)}^2{\cos }^2\theta -3(n+1)]\cdot [1+3xb]$
		$\pm i\beta [2^5\cdot 3(n+1{)}^3{]}^{1/2}\cos \theta \cdot \{-[2b+3x{b}^2]+[6b+6x{b}^2]\}\frac{1}{2(1+xb{)}^{3/2}}$
	$=$	$2[2^3(n+1{)}^2{\cos }^2\theta -3(n+1)]\cdot [1+3xb]$
		$\pm i\beta [2^3\cdot 3(n+1{)}^3{]}^{1/2}\cos \theta \cdot b[1+xb{]}^{-3/2}[4+3xb]$

Through a separate white-board derivation I have obtained …

$\frac{{\partial }^2\mathrm{\Lambda }}{\partial x^2}$ $=$ $2(n+1)[2^3(n+1){\cos }^2\theta -3](1+3xb)\pm i\beta \cos \theta [2^3\cdot 3(n+1{)}^3{]}^{1/2}\left[\frac{b(4+3xb)}{(1+xb{)}^{3/2}}\right],$ which is the same.

$\frac{\partial \mathrm{\Lambda }}{\partial \theta }$	$=$	$\frac{\partial }{\partial \theta }\{[2^3(n+1{)}^2{\cos }^2\theta -3(n+1)]\cdot [x^2+x^3(3\cos \theta -{\cos }^{}\theta )]\}$
		$\pm i\frac{\partial }{\partial \theta }\{\beta [2^7\cdot 3(n+1{)}^3{]}^{1/2}[x^2{\cos }^2\theta +x^3(3{\cos }^3\theta -{\cos }^5\theta ){]}^{1/2}\}$
	$=$	$[2^3(n+1{)}^2{\cos }^2\theta -3(n+1)]\cdot x^3\cdot [\sin \theta (-3+3{\cos }^{}\theta )]+[x^2+x^3(3\cos \theta -{\cos }^{}\theta )]\cdot [-2^4(n+1{)}^2\sin \theta \cos \theta ]$
		$\pm i\beta [2^5\cdot 3(n+1{)}^3{]}^{1/2}\cdot [x^2{\cos }^2\theta +x^3(3{\cos }^3\theta -{\cos }^5\theta ){]}^{-1/2}\cdot (-\sin \theta )[2x^2\cos \theta +x^3(9{\cos }^{}\theta -5{\cos }^{}\theta )]$
	$=$	$-3x^3{\sin }^3\theta [2^3(n+1{)}^2{\cos }^2\theta -3(n+1)]-2^4(n+1{)}^2{x}^2\sin \theta \cos \theta [1+x(3\cos \theta -{\cos }^{}\theta )]$
		$\pm i\beta (-1)[2^5\cdot 3(n+1{)}^3{]}^{1/2}\cdot [1+x(3\cos \theta -{\cos }^{}\theta ){]}^{-1/2}\cdot x\sin \theta [2+x(9\cos \theta -5{\cos }^{}\theta )]$
	$=$	$-3x^3{\sin }^3\theta [2^3(n+1{)}^2{\cos }^2\theta -3(n+1)]-2^4(n+1{)}^2{x}^2\sin \theta \cos \theta [1+xb]$
		$\pm i\beta (-1)[2^5\cdot 3(n+1{)}^3{]}^{1/2}\cdot [1+xb{]}^{-1/2}\cdot x\sin \theta [2+x(9\cos \theta -5{\cos }^{}\theta )]$

Through a separate white-board derivation I have obtained …

$\frac{\partial \mathrm{\Lambda }}{\partial \theta }$ $=$ $9(n+1)x^3{\sin }^3\theta -2^3\cdot 3(n+1{)}^2{x}^3{\sin }^3\theta {\cos }^2\theta -2^4(n+1{)}^2(\beta \eta {)}^2\sin \theta \cos \theta$     $\pm i(-1)\beta [2^7\cdot 3(n+1{)}^3{]}^{1/2}\{(\beta \eta )\sin \theta +\frac{3x^3}{2}\cdot [\frac{{\sin }^3\theta \cos \theta }{(\beta \eta )}\left]\right\}$   $=$ $(n+1)\sin \theta \{-2^4(n+1)(\beta \eta {)}^2\cos \theta +3x^3{\sin }^{}\theta [3-2^3(n+1){\cos }^{}\theta ]\}$     $\pm i(-1)\beta \sin \theta [2^7\cdot 3(n+1{)}^3(\beta \eta {)}^2{]}^{1/2}\{1+\frac{3x^3}{2}\cdot \left[\frac{{\sin }^2\theta \cos \theta }{(\beta \eta {)}^2}\right]\},$ which is the same.

$\frac{{\partial }^2\mathrm{\Lambda }}{\partial {\theta }^2}$	$=$	$-3x^3\frac{\partial }{\partial \theta }\cdot \{{\sin }^3\theta [2^3(n+1{)}^2{\cos }^2\theta -3(n+1)]\}-2^4(n+1{)}^2{x}^2\cdot \frac{\partial }{\partial \theta }\{\sin \theta \cos \theta [1+x(3\cos \theta -{\cos }^{}\theta )]\}$
		$\pm i\beta (-1)x[2^5\cdot 3(n+1{)}^3{]}^{1/2}\cdot \frac{\partial }{\partial \theta }\{\sin \theta [2+x(9\cos \theta -5{\cos }^{}\theta )]\cdot [1+x(3\cos \theta -{\cos }^{}\theta ){]}^{-1/2}\}$
	$=$	$-3x^3\cdot \{3{\sin }^2\theta \cos \theta [2^3(n+1{)}^2{\cos }^{}\theta -3(n+1)]-{\sin }^{}\theta [2^3(n+1{)}^2\cdot 2\sin \theta \cos \theta ]\}$
		$-2^4(n+1{)}^2{x}^2\cdot \{({\cos }^2\theta -{\sin }^2\theta )[1+x(3\cos \theta -{\cos }^{}\theta )]-3x{\sin }^{}\theta \cos \theta \}$
		$\pm i\beta (-1)x[2^5\cdot 3(n+1{)}^3{]}^{1/2}\cdot \{\cos \theta \cdot [2+x(9\cos \theta -5{\cos }^{}\theta )]\cdot [1+x(3\cos \theta -{\cos }^{}\theta ){]}^{-1/2}$
		$-3x{\sin }^2\theta [1+x(3\cos \theta -{\cos }^{}\theta ){]}^{-1/2}\cdot (3-5{\cos }^{}\theta )$
		$+\frac{3}{2}\cdot x{\sin }^4\theta [2+x(9\cos \theta -5{\cos }^{}\theta )]\cdot [1+x(3\cos \theta -{\cos }^{}\theta ){]}^{-3/2}\}$
	$=$	$-3x^3\cdot \{3{\sin }^2\theta \cos \theta [2^3(n+1{)}^2{\cos }^{}\theta -3(n+1)]-{\sin }^{}\theta [2^3(n+1{)}^2\cdot 2\sin \theta \cos \theta ]\}$
		$-2^4(n+1{)}^2{x}^2\cdot \{(1-2{\sin }^2\theta )+x(3\cos \theta -{\cos }^{}\theta -6{\sin }^{}\theta \cos \theta +2{\sin }^{}\theta {\cos }^{}\theta -3{\sin }^{}\theta \cos \theta )\}$
		$\pm i\beta (-1)x[2^5\cdot 3(n+1{)}^3{]}^{1/2}\cdot [1+xb{]}^{-3/2}\cdot \{\cos \theta \cdot [2+x(9\cos \theta -5{\cos }^{}\theta )]\cdot [1+xb]$
		$-3x{\sin }^2\theta [1+xb]\cdot (3-5{\cos }^2\theta )+\frac{3}{2}\cdot x{\sin }^4\theta [2+x(9\cos \theta -5{\cos }^{}\theta )]\}$
	$=$	$-2^4(n+1{)}^2{x}^2(1-2{\sin }^2\theta )-x^3\cdot \{[2^3\cdot 3^2(n+1{)}^2{\sin }^2\theta {\cos }^3\theta -3^3(n+1){\sin }^2\theta \cos \theta ]-2^4\cdot 3(n+1{)}^2{\sin }^{}\theta \cos \theta$
		$+2^4(n+1{)}^{2}b+2^4(n+1{)}^2\cdot (-6{\sin }^2\theta \cos \theta +2{\sin }^{}\theta {\cos }^{}\theta -3{\sin }^{}\theta \cos \theta )\}$
		$\pm i\beta (-1)x[2^3\cdot 3(n+1{)}^3{]}^{1/2}\cdot [1+xb{]}^{-3/2}\cdot \{2\cos \theta \cdot [2+x(15\cos \theta -7{\cos }^{}\theta )+x^{2}b(9\cos \theta -5{\cos }^{}\theta )]$
		$-6x{\sin }^2\theta (1+xb)\cdot (3-5{\cos }^2\theta )+3\cdot x{\sin }^4\theta [2+x(9\cos \theta -5{\cos }^{}\theta )]\}$

Through a separate white-board derivation I have obtained …

$\frac{{\partial }^2\mathrm{\Lambda }}{\partial {\theta }^2}$ $=$ $x^2\{2^4(n+1{)}^2({\sin }^2\theta -{\cos }^2\theta )\}$     $+x^3\{-2^4\cdot 3(n+1{)}^2{\cos }^3\theta +2^4(n+1{)}^2{\cos }^5\theta +3^2(n+1)(16n+19){\sin }^2\theta \cos \theta -2^3\cdot 23(n+1{)}^2{\sin }^{}\theta {\cos }^{}\theta \}$     $\pm i(-1)\beta [2^7\cdot 3(n+1{)}^3{]}^{1/2}\{(\beta \eta )\cos \theta +\frac{3x^3{\sin }^2\theta }{2(\beta \eta )}(5{\cos }^{}\theta -2)+\frac{3^2{x}^6{\sin }^6\theta \cos \theta }{2^2(\beta \eta {)}^3}\}.$

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):

$\hat{L}(\delta W)$

$=$

$-2n(1-{\mathrm{\Theta }}_H)(M{\nu }^2+N\nu )(\delta W),$

where,

$\hat{L}(\delta W)$	$\equiv$	${\mathrm{\Theta }}_H{x}^2\cdot \frac{{\partial }^2(\delta W)}{\partial x^2}+{\mathrm{\Theta }}_H\cdot \frac{{\partial }^2(\delta W)}{\partial {\theta }^2}+\left\{{\mathrm{\Theta }}_{H}x\right[\frac{1-2x\cos \theta }{1-x\cos \theta }]+n{x}^2\cdot \frac{\partial {\mathrm{\Theta }}_H}{\partial x}\}\cdot \frac{\partial (\delta W)}{\partial x}$
		$+[\frac{{\mathrm{\Theta }}_{H}x\sin \theta }{(1-x\cos \theta )}+n\cdot \frac{\partial {\mathrm{\Theta }}_H}{\partial \theta }]\cdot \frac{\partial (\delta W)}{\partial \theta }+[\frac{2n{x}^2{m}^2}{{\beta }^2(1-x\cos \theta {)}^4}-\frac{m^2{x}^2{\mathrm{\Theta }}_H}{(1-x\cos \theta {)}^2}]\delta W,$
$M$	$\equiv$	$\frac{x^2}{(1-{\mathrm{\Theta }}_H){\beta }^2},$
$N$	$\equiv$	$\frac{2m{x}^2}{(1-{\mathrm{\Theta }}_H){\beta }^2(1-x\cos \theta {)}^2}.$

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

$[\frac{2(n+1)}{m}{]}^2\frac{RHS}{A_{00}}$	$=$	$-2n(1-{\mathrm{\Theta }}_H)(M{\nu }^2+N\nu )\cdot [\frac{2(n+1)}{m}{]}^2\frac{\delta W}{A_{00}}$
	$=$	$-\frac{2n{x}^2}{{\beta }^2}[{\nu }^2+\frac{2m\nu }{(1-x\cos \theta {)}^2}]\cdot [\frac{2(n+1)}{m}{]}^2\frac{\delta W}{A_{00}}$
	$=$	$-\frac{2n{x}^2}{{\beta }^2}[{\nu }^2+\frac{2m\nu }{(1-x\cos \theta {)}^2}]\cdot \{\mathrm{\Lambda }+[\frac{2(n+1)}{m}{]}^2\}.$

And the similarly modified LHS is:

$[\frac{2(n+1)}{m}{]}^2\frac{LHS}{A_{00}}$	$=$	${\mathrm{\Theta }}_H{x}^2\cdot \frac{{\partial }^2\mathrm{\Lambda }}{\partial x^2}+{\mathrm{\Theta }}_H\cdot \frac{{\partial }^2\mathrm{\Lambda }}{\partial {\theta }^2}+\left\{{\mathrm{\Theta }}_{H}x\right[\frac{1-2x\cos \theta }{1-x\cos \theta }]+n{x}^2\cdot \frac{\partial {\mathrm{\Theta }}_H}{\partial x}\}\cdot \frac{\partial \mathrm{\Lambda }}{\partial x}$
		$+[\frac{{\mathrm{\Theta }}_{H}x\sin \theta }{(1-x\cos \theta )}+n\cdot \frac{\partial {\mathrm{\Theta }}_H}{\partial \theta }]\cdot \frac{\partial \mathrm{\Lambda }}{\partial \theta }+[\frac{2n{x}^2{m}^2}{{\beta }^2(1-x\cos \theta {)}^4}-\frac{m^2{x}^2{\mathrm{\Theta }}_H}{(1-x\cos \theta {)}^2}]\cdot [\frac{2(n+1)}{m}{]}^2\frac{\delta W}{A_{00}}$
	$=$	$(1-{\eta }^2)x^2\cdot \frac{{\partial }^2\mathrm{\Lambda }}{\partial x^2}+(1-{\eta }^2)\cdot \frac{{\partial }^2\mathrm{\Lambda }}{\partial {\theta }^2}+\{(1-{\eta }^2)x[\frac{1-2x\cos \theta }{1-x\cos \theta }]+n{x}^2\cdot \frac{\partial (1-{\eta }^2)}{\partial x}\}\cdot \frac{\partial \mathrm{\Lambda }}{\partial x}$
		$+[\frac{(1-{\eta }^2)x\sin \theta }{(1-x\cos \theta )}+n\cdot \frac{\partial (1-{\eta }^2)}{\partial \theta }]\cdot \frac{\partial \mathrm{\Lambda }}{\partial \theta }+[\frac{2n{x}^2{m}^2}{{\beta }^2(1-x\cos \theta {)}^4}-\frac{m^2{x}^2(1-{\eta }^2)}{(1-x\cos \theta {)}^2}]\cdot \{\mathrm{\Lambda }+[\frac{2(n+1)}{m}{]}^2\}$
	$=$	$(1-{\eta }^2)x^2\cdot \frac{{\partial }^2\mathrm{\Lambda }}{\partial x^2}+(1-{\eta }^2)\cdot \frac{{\partial }^2\mathrm{\Lambda }}{\partial {\theta }^2}+\{(1-{\eta }^2)x[\frac{1-2x\cos \theta }{1-x\cos \theta }]-n{x}^2\cdot \frac{\partial {\eta }^2}{\partial x}\}\cdot \frac{\partial \mathrm{\Lambda }}{\partial x}$
		$+[\frac{(1-{\eta }^2)x\sin \theta }{(1-x\cos \theta )}-n\cdot \frac{\partial {\eta }^2}{\partial \theta }]\cdot \frac{\partial \mathrm{\Lambda }}{\partial \theta }+[\frac{2n{x}^2{m}^2}{{\beta }^2(1-x\cos \theta {)}^4}-\frac{m^2{x}^2(1-{\eta }^2)}{(1-x\cos \theta {)}^2}]\cdot \{\mathrm{\Lambda }+[\frac{2(n+1)}{m}{]}^2\}$

Now multiply both sides by …

${\beta }^2{m}^2(1-x\cos \theta {)}^4.$

We have,

$m^2{{ℒ}}_{RHS}\equiv {\beta }^2(1-x\cos \theta {)}^4[2(n+1){]}^2\frac{RHS}{A_{00}}$	$=$	$-2n{m}^2{x}^2(1-x\cos \theta {)}^4[{\nu }^2+\frac{2m\nu }{(1-x\cos \theta {)}^2}]\cdot \{\mathrm{\Lambda }+\left[\frac{2(n+1)}{m}{]}^2\right\}$
	$=$	$-2n{x}^2(1-x\cos \theta {)}^2[(1-x\cos \theta {)}^2{\nu }^2+(2m\nu )]\cdot \{m^2\mathrm{\Lambda }+[2(n+1){]}^2\}$
	$=$	$-2n{x}^2(1-x\cos \theta {)}^2\{m^2\mathrm{\Lambda }+[2(n+1){]}^2\}\{m^2(1-x\cos \theta {)}^2[1+\frac{3{\beta }^2}{2(n+1)}]+2m^2[-1\pm i[\frac{3{\beta }^2}{2(n+1)}{]}^{1/2}\left]\right\}$
	$=$	$-\frac{n{m}^2{x}^2}{(n+1)}\cdot (1-x\cos \theta {)}^2\{m^2\mathrm{\Lambda }+[2(n+1){]}^2\}\{(1-x\cos \theta {)}^2[2(n+1)+3{\beta }^2]$
		$-4(n+1)\pm i[2^3\cdot 3(n+1){\beta }^2{]}^{1/2}\}.$

And,

$m^2{{ℒ}}_{LHS}\equiv {\beta }^2(1-x\cos \theta {)}^4[2(n+1){]}^2\frac{LHS}{A_{00}}$	$=$	${\beta }^2{m}^2(1-x\cos \theta {)}^4(1-{\eta }^2)x^2\cdot \frac{{\partial }^2\mathrm{\Lambda }}{\partial x^2}+{\beta }^2{m}^2(1-x\cos \theta {)}^4(1-{\eta }^2)\cdot \frac{{\partial }^2\mathrm{\Lambda }}{\partial {\theta }^2}$
		$+{\beta }^2{m}^2(1-x\cos \theta {)}^3[(1-{\eta }^2)x(1-2x\cos \theta )-n{x}^2(1-x\cos \theta )\cdot \frac{\partial {\eta }^2}{\partial x}]\cdot \frac{\partial \mathrm{\Lambda }}{\partial x}$
		$+{\beta }^2{m}^2(1-x\cos \theta {)}^3[(1-{\eta }^2)x\sin \theta -n(1-x\cos \theta )\cdot \frac{\partial {\eta }^2}{\partial \theta }]\cdot \frac{\partial \mathrm{\Lambda }}{\partial \theta }$
		$+m^2[2n{x}^2-(1-x\cos \theta {)}^2{\beta }^2{x}^2(1-{\eta }^2)]\cdot \{m^2\mathrm{\Lambda }+[2(n+1){]}^2\}$
	$=$	$m^2(1-x\cos \theta {)}^4[{\beta }^2-x^2-x^3(3\cos \theta -{\cos }^{}\theta )]x^2\cdot \frac{{\partial }^2\mathrm{\Lambda }}{\partial x^2}+m^2(1-x\cos \theta {)}^4[{\beta }^2-x^2-x^3(3\cos \theta -{\cos }^{}\theta )]\cdot \frac{{\partial }^2\mathrm{\Lambda }}{\partial {\theta }^2}$
		$+m^2(1-x\cos \theta {)}^3\{[{\beta }^2-x^2-x^3(3\cos \theta -{\cos }^{}\theta )]x(1-2x\cos \theta )-n{x}^2(1-x\cos \theta )\cdot [2x+3x^2(3\cos \theta -{\cos }^{}\theta )]\}\cdot \frac{\partial \mathrm{\Lambda }}{\partial x}$
		$+m^2(1-x\cos \theta {)}^3\{[{\beta }^2-x^2-x^3(3\cos \theta -{\cos }^{}\theta )]x\sin \theta -3n(1-x\cos \theta )\cdot ({\cos }^{}\theta -1)x^3\sin \theta \}\cdot \frac{\partial \mathrm{\Lambda }}{\partial \theta }$
		$+m^2\{2n{x}^2-(1-x\cos \theta {)}^2{x}^2[{\beta }^2-x^2-x^3(3\cos \theta -{\cos }^{}\theta )]\}\cdot [m^2\mathrm{\Lambda }+2^2(n+1{)}^2].$

Step 4

Let's divide both sides by $m^2$ and swap a couple of terms between the sides in order to group, on the right, terms with no explicit mention of $\mathrm{\Lambda }$.

${{ℒ}}_{RHS}$	$\equiv$	$\frac{{\beta }^2}{m^2}(1-x\cos \theta {)}^4[2(n+1){]}^2\frac{RHS}{A_{00}}\pm \mathrm{s}\mathrm{w}\mathrm{a}\mathrm{p}\mathrm{s}$
	$=$	$-\frac{n{x}^2}{(n+1)}\cdot (1-x\cos \theta {)}^2\{[2(n+1){]}^2\}\{(1-x\cos \theta {)}^2[2(n+1)+3{\beta }^2]-4(n+1)\pm i[2^3\cdot 3(n+1){\beta }^2{]}^{1/2}\}$
		$-\{2n{x}^2-(1-x\cos \theta {)}^2{x}^2[{\beta }^2-x^2-x^3(3\cos \theta -{\cos }^{}\theta )]\}\cdot [2^2(n+1{)}^2]$

$\Rightarrow -\frac{{{ℒ}}_{RHS}}{2^2(n+1)x^2}$	$=$	$n\cdot (1-x\cos \theta {)}^2\{(1-x\cos \theta {)}^2[2(n+1)+3{\beta }^2]-4(n+1)\pm i[2^3\cdot 3(n+1){\beta }^2{]}^{1/2}\}$
		$+(n+1)\{2n-(1-x\cos \theta {)}^2[{\beta }^2-x^2-x^3(3\cos \theta -{\cos }^{}\theta )]\}$
	$=$	$2n(n+1)-(1-x\cos \theta {)}^2(n+1)[{\beta }^2-x^2-x^3(3\cos \theta -{\cos }^{}\theta )]$
		$+n(1-x\cos \theta {)}^4[2(n+1)+3{\beta }^2]-4n(n+1)(1-x\cos \theta {)}^2$
		$\pm i(1-x\cos \theta {)}^2[2^3\cdot 3n^2(n+1){\beta }^2{]}^{1/2}$
	$=$	$2n(n+1)-(n+1)(1-x\cos \theta {)}^2[4n+{\beta }^2-x^2-x^3(3\cos \theta -{\cos }^{}\theta )]+n(1-x\cos \theta {)}^4[2(n+1)+3{\beta }^2]$
		$\pm i(1-x\cos \theta {)}^2[2^3\cdot 3n^2(n+1){\beta }^2{]}^{1/2}.$

Through a separate white-board derivation I have obtained …

${\mathrm{R}\mathrm{H}\mathrm{S}}_3$ $\equiv$ ${\beta }^2(1-x\cos \theta {)}^4\cdot \{{\mathrm{R}\mathrm{H}\mathrm{S}}_1\pm \mathrm{s}\mathrm{w}\mathrm{a}\mathrm{p}\mathrm{s}\}={\beta }^2(1-x\cos \theta {)}^4\cdot \left\{\frac{1}{A_{00}}\right[\frac{2(n+1)}{m}{]}^2{\mathrm{R}\mathrm{H}\mathrm{S}}_0\pm \mathrm{s}\mathrm{w}\mathrm{a}\mathrm{p}\mathrm{s}\}$   $=$ $-2^2(n+1{)}^2{x}^2\cdot {𝒜},$ where, ${𝒜}$ $\equiv$ $2n+(1-x\cos \theta {)}^2[4n\left(\frac{\nu }{m}\right)-{\beta }^2+x^2(1+xb)]+2n(1-x\cos \theta {)}^4(\frac{\nu }{m}{)}^2,$ $\frac{\nu }{m}$ $=$ $-1\pm i[\frac{3}{2(n+1)}{]}^{1/2}\beta .$

Case A:       $(\frac{\nu }{m}{)}^2=\frac{\nu \cdot {\nu }^{*}}{m^2}=1+\frac{3{\beta }^2}{2(n+1)}\Rightarrow$ $(n+1)\cdot {𝒜}$ $=$ $2n(n+1)+(n+1)(1-x\cos \theta {)}^2[x^2(1+xb)-{\beta }^2-4n]+(1-x\cos \theta {)}^4[2n(n+1)+3n{\beta }^2]$     $\pm i(1-x\cos \theta {)}^2[2^3\cdot 3n^2{\beta }^2(n+1){]}^{1/2}.$

Case B:       $(\frac{\nu }{m}{)}^2=\frac{\nu }{m}\cdot \frac{\nu }{m}=1-\frac{3{\beta }^2}{2(n+1)}\pm i(-1)[\frac{2\cdot 3{\beta }^2}{(n+1)}{]}^{1/2}\Rightarrow$ $(n+1)\cdot {𝒜}$ $=$ $2n(n+1)+(n+1)(1-x\cos \theta {)}^2[x^2(1+xb)-{\beta }^2-4n]+(1-x\cos \theta {)}^4[2n(n+1)-3n{\beta }^2]$     $\pm ix\cos \theta (1-x\cos \theta {)}^2[2^3\cdot 3n^2{\beta }^2(n+1){]}^{1/2}.$

And,

${{ℒ}}_{LHS}$	$\equiv$	$\frac{{\beta }^2}{m^2}(1-x\cos \theta {)}^4[2(n+1){]}^2\frac{LHS}{A_{00}}\mp \mathrm{s}\mathrm{w}\mathrm{a}\mathrm{p}\mathrm{s}$
	$=$	$(1-x\cos \theta {)}^4[{\beta }^2-x^2-x^3(3\cos \theta -{\cos }^{}\theta )]x^2\cdot \frac{{\partial }^2\mathrm{\Lambda }}{\partial x^2}+m^2(1-x\cos \theta {)}^4[{\beta }^2-x^2-x^3(3\cos \theta -{\cos }^{}\theta )]\cdot \frac{{\partial }^2\mathrm{\Lambda }}{\partial {\theta }^2}$
		$+(1-x\cos \theta {)}^3\{[{\beta }^2-x^2-x^3(3\cos \theta -{\cos }^{}\theta )]x(1-2x\cos \theta )-n{x}^2(1-x\cos \theta )\cdot [2x+3x^2(3\cos \theta -{\cos }^{}\theta )]\}\cdot \frac{\partial \mathrm{\Lambda }}{\partial x}$
		$+(1-x\cos \theta {)}^3\{[{\beta }^2-x^2-x^3(3\cos \theta -{\cos }^{}\theta )]x\sin \theta -3n(1-x\cos \theta )\cdot ({\cos }^{}\theta -1)x^3\sin \theta \}\cdot \frac{\partial \mathrm{\Lambda }}{\partial \theta }$
		$+\{2n{x}^2-(1-x\cos \theta {)}^2{x}^2[{\beta }^2-x^2-x^3(3\cos \theta -{\cos }^{}\theta )]\}\cdot [m^2\mathrm{\Lambda }]$
		$+\frac{n{x}^2}{(n+1)}\cdot (1-x\cos \theta {)}^2\{m^2\mathrm{\Lambda }\}\{(1-x\cos \theta {)}^2[2(n+1)+3{\beta }^2]-4(n+1)\pm i[2^3\cdot 3(n+1){\beta }^2{]}^{1/2}\}$
	$=$	$(1-x\cos \theta {)}^4[{\beta }^2-x^2-x^{3}b]\{x^2\cdot \frac{{\partial }^2\mathrm{\Lambda }}{\partial x^2}+m^2\cdot \frac{{\partial }^2\mathrm{\Lambda }}{\partial {\theta }^2}\}$
		$+(1-x\cos \theta {)}^3\{x({\beta }^2-x^2-x^{3}b)[(1-2x\cos \theta )\cdot \frac{\partial \mathrm{\Lambda }}{\partial x}+\sin \theta \cdot \frac{\partial \mathrm{\Lambda }}{\partial \theta }]-n(1-x\cos \theta )\cdot x^3[(2+3xb)\cdot \frac{\partial \mathrm{\Lambda }}{\partial x}-3{\sin }^{}\theta \cdot \frac{\partial \mathrm{\Lambda }}{\partial \theta }]\}$
		$+\{2n-(1-x\cos \theta {)}^2[{\beta }^2-x^2-x^{3}b]+(1-x\cos \theta {)}^4[2+\frac{3{\beta }^2}{(n+1)}]n-4n(1-x\cos \theta {)}^2\}\cdot m^2{x}^2\mathrm{\Lambda }$
		$\pm i[\frac{2^3\cdot 3n^2{m}^4{\beta }^2}{(n+1)}{]}^{1/2}{x}^2\cdot (1-x\cos \theta {)}^2\mathrm{\Lambda }$

Through a separate white-board derivation I have obtained …

${\mathrm{L}\mathrm{H}\mathrm{S}}_3$ $\equiv$ ${\beta }^2(1-x\cos \theta {)}^4\cdot \{{\mathrm{L}\mathrm{H}\mathrm{S}}_1\mp \mathrm{s}\mathrm{w}\mathrm{a}\mathrm{p}\mathrm{s}\}={\beta }^2(1-x\cos \theta {)}^4\cdot \left\{\frac{1}{A_{00}}\right[\frac{2(n+1)}{m}{]}^2{\mathrm{L}\mathrm{H}\mathrm{S}}_0\mp \mathrm{s}\mathrm{w}\mathrm{a}\mathrm{p}\mathrm{s}\}$   $=$ ${\beta }^2(1-{\eta }^2)(1-x\cos \theta {)}^3\{(1-x\cos \theta )[x^2\frac{{\partial }^2\mathrm{\Lambda }}{\partial x^2}+\frac{{\partial }^2\mathrm{\Lambda }}{\partial {\theta }^2}]+x[(1-2x\cos \theta )\frac{\partial \mathrm{\Lambda }}{\partial x}+\sin \theta \frac{\partial \mathrm{\Lambda }}{\partial \theta }]\}$     $-n{x}^3(1-x\cos \theta {)}^4[(2+3xb)\frac{\partial \mathrm{\Lambda }}{\partial x}-3{\sin }^{}\theta \frac{\partial \mathrm{\Lambda }}{\partial \theta }]+m^2{x}^2\mathrm{\Lambda }\cdot {𝒜},$ where, as above in the definition of $\mathrm{R}\mathrm{H}{\mathrm{S}}_{\mathrm{3}},$ ${𝒜}$ $\equiv$ $2n+(1-x\cos \theta {)}^2[4n\left(\frac{\nu }{m}\right)-{\beta }^2+x^2(1+xb)]+2n(1-x\cos \theta {)}^4(\frac{\nu }{m}{)}^2.$

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

Step 5

Now let's evaluate the LHS terms, keeping only leading-orders in $x$ before plugging derivatives of $\mathrm{\Lambda }$ into each term. For example,

$x^2\cdot \frac{{\partial }^2\mathrm{\Lambda }}{\partial x^2}+m^2\cdot \frac{{\partial }^2\mathrm{\Lambda }}{\partial {\theta }^2}$	$=$	$x^2\{2[2^3(n+1{)}^2{\cos }^2\theta -3(n+1)]\cdot [1+{\xcancel{3xb}}^0]\pm i\beta [2^3\cdot 3(n+1{)}^3{]}^{1/2}\cos \theta \cdot b[1+{\xcancel{xb}}^0{]}^{-3/2}[4+{\xcancel{9xb}}^0]\}$
		$m^2\cdot \{-2^4(n+1{)}^2{x}^2(1-2{\sin }^2\theta )-{\xcancel{x^3}}^0\cdot [[2^3\cdot 3^2(n+1{)}^2{\sin }^2\theta {\cos }^3\theta -3^3(n+1){\sin }^2\theta \cos \theta ]-2^4\cdot 3(n+1{)}^2{\sin }^{}\theta \cos \theta$
		$+2^4(n+1{)}^{2}b+2^4(n+1{)}^2\cdot (-6{\sin }^2\theta \cos \theta +2{\sin }^{}\theta {\cos }^{}\theta -3{\sin }^{}\theta \cos \theta )]$
		$\pm i\beta (-1)x[2^3\cdot 3(n+1{)}^3{]}^{1/2}\cdot [1+xb{]}^{-3/2}\cdot [2\cos \theta \cdot [2+x(15\cos \theta -7{\cos }^{}\theta )+{\xcancel{x^{2}b}}^0(9\cos \theta -5{\cos }^{}\theta )]$
		$-6x{\sin }^2\theta (1+{\xcancel{xb}}^0)\cdot (3-5{\cos }^2\theta )+3\cdot x{\sin }^4\theta [2+{\xcancel{x}}^0(9\cos \theta -5{\cos }^{}\theta )]\left]\right\}$
	$\approx$	$x^2[2^4(n+1{)}^2{\cos }^2\theta -6(n+1)-2^4{m}^2(n+1{)}^2(1-2{\sin }^2\theta )]$
		$\pm i{m}^2\beta (-1)x[2^7\cdot 3(n+1{)}^3{]}^{1/2}\cdot [1+xb{]}^{-3/2}\cos \theta$
		$\pm i\beta x^2[2^3\cdot 3(n+1{)}^3{]}^{1/2}\cdot [(12{\cos }^2\theta -4{\cos }^4\theta )-m^2\cdot (1+xb{)}^{-3/2}[2{\cos }^2\theta (15-7{\cos }^2\theta )-6{\sin }^2\theta (3-5{\cos }^2\theta )+6{\sin }^4\theta ]]$

Next,

$(1-2x\cos \theta )\cdot \frac{\partial \mathrm{\Lambda }}{\partial x}+\sin \theta \cdot \frac{\partial \mathrm{\Lambda }}{\partial \theta }$	$=$	$(1-2x\cos \theta )\cdot \{[2^3(n+1{)}^2{\cos }^{}\theta -3(n+1)]\cdot [2x+{\xcancel{3x^{2}b}}^0]\pm i\beta [2^5\cdot 3(n+1{)}^3{]}^{1/2}\cos \theta \cdot [1+xb{]}^{-1/2}\cdot [2+{\xcancel{3xb}}^0]\}$
		$+\sin \theta \cdot \{-2^4(n+1{)}^2{\xcancel{x^2}}^0\sin \theta \cos \theta [1+xb]-3{\xcancel{x^3}}^0{\sin }^{}\theta [2^3(n+1{)}^2{\cos }^{}\theta -3(n+1)]$
		$\pm i\beta (-1)[2^5\cdot 3(n+1{)}^3{]}^{1/2}\cdot [1+xb{]}^{-1/2}\cdot x\sin \theta [2+{\xcancel{x}}^0(9\cos \theta -5{\cos }^{}\theta )]\}$
	$\approx$	$(1-2x\cos \theta )\cdot [2x(n+1)[2^3(n+1){\cos }^{}\theta -3]\pm i\beta [2^7\cdot 3(n+1{)}^3{]}^{1/2}\cos \theta \cdot (1+xb{)}^{-1/2}]$
		$\pm i\beta x(-1)[2^7\cdot 3(n+1{)}^3{]}^{1/2}\cdot [1+xb{]}^{-1/2}\cdot {\sin }^2\theta$

Through a separate white-board derivation I have obtained …

$(1-2x\cos \theta )\cdot \frac{\partial \mathrm{\Lambda }}{\partial x}+\sin \theta \cdot \frac{\partial \mathrm{\Lambda }}{\partial \theta }$ $=$ $x(n+1)[-6+2^4(n+1){\cos }^2\theta ]$     $-x^2(n+1)\cos \theta \{[15+2^4(n+1)]-{\cos }^{}\theta [9+2^3\cdot 7(n+1)]+2^3\cdot 3(n+1){\cos }^{}\theta \}$     $+x^3(n+1)\{9-2^2\cdot 3^2(1+2n){\cos }^2\theta -[9+32(n+1)]{\cos }^4\theta +2^3(n+1){\cos }^6\theta \}$     $\pm i\beta \left[\frac{2^5\cdot 3(n+1{)}^3}{1+x(3\cos \theta -{\cos }^{}\theta )}{]}^{1/2}\right\{2\cos \theta -x[2-7{\cos }^{}\theta +3{\cos }^{}\theta ]$     $-x^2\cos \theta [9+4{\cos }^{}\theta -{\cos }^{}\theta ]\}$

Also, from above,

$\mathrm{\Lambda }$	$=$	$-(4n+1){\beta }^2+(n+1){\xcancel{x^2}}^0[2^3(n+1){\cos }^2\theta -3]\cdot [1+xb]\pm i\beta [2^7\cdot 3(n+1{)}^3{]}^{1/2}x\cos \theta [1+xb{]}^{1/2}$
	$\approx$	$-(4n+1){\beta }^2\pm i\beta [2^7\cdot 3(n+1{)}^3{]}^{1/2}x\cos \theta [1+xb{]}^{1/2}$

Through a separate white-board derivation I have obtained …

$(2+3xb)\cdot \frac{\partial \mathrm{\Lambda }}{\partial x}-3{\sin }^3\theta \cdot \frac{\partial \mathrm{\Lambda }}{\partial \theta }$ $=$ $x\cdot 2^2(n+1)[2^3(n+1){\cos }^2\theta -3]$     $+x^2\cdot 2^2\cdot 3(n+1)\cos \theta \{[2^{2}n-5]+{\cos }^{}\theta [2^{4}n+19]-2^2(n+1){\cos }^{}\theta \}$     $+x^3\cdot 3^2(n+1){\cos }^2\theta \{-3^3+2\cdot 3^2{\cos }^2\theta [2^{2}n+5]-99\cdot {\cos }^4\theta +2^3(n+1){\cos }^6\theta \}$     $+x^3(n+1){\sin }^4\theta \{-3^3+3^2{\cos }^2\theta [2^3\cdot 3n+27]-2^3\cdot 3\cdot 5(n+1){\cos }^4\theta \}$     $\pm i\beta \left[\frac{2^5\cdot 3(n+1{)}^3}{1+xb}{]}^{1/2}\right\{4\cos \theta +6x(2b\cos \theta +{\sin }^{}\theta )+3x^2(3b^2\cos \theta +2b{\sin }^{}\theta +3{\sin }^{}\theta \cos \theta )\}$

Taken together, then, we have,

${{ℒ}}_{LHS}$	$=$	$(1-x\cos \theta {)}^4[{\beta }^2-x^2-{\xcancel{x^{3}b}}^0]\{x^2\cdot \frac{{\partial }^2\mathrm{\Lambda }}{\partial x^2}+m^2\cdot \frac{{\partial }^2\mathrm{\Lambda }}{\partial {\theta }^2}\}$
		$+(1-x\cos \theta {)}^3\{x({\beta }^2-{\xcancel{x^2}}^0-{\xcancel{x^{3}b}}^0)[(1-2x\cos \theta )\cdot \frac{\partial \mathrm{\Lambda }}{\partial x}+\sin \theta \cdot \frac{\partial \mathrm{\Lambda }}{\partial \theta }]-n(1-x\cos \theta )\cdot {\xcancel{x^3}}^0[(2+3xb)\cdot \frac{\partial \mathrm{\Lambda }}{\partial x}-3{\sin }^{}\theta \cdot \frac{\partial \mathrm{\Lambda }}{\partial \theta }]\}$
		$+\{2n-(1-x\cos \theta {)}^2[{\beta }^2-x^2-{\xcancel{x^3}}^{0}b]+(1-x\cos \theta {)}^4[2+\frac{3{\beta }^2}{(n+1)}]n-4n(1-x\cos \theta {)}^2\}\cdot m^2{x}^2\mathrm{\Lambda }$
		$\pm i[\frac{2^3\cdot 3n^2{m}^4{\beta }^2}{(n+1)}{]}^{1/2}{x}^2\cdot (1-x\cos \theta {)}^2\mathrm{\Lambda }$
	$\approx$	$(1-x\cos \theta {)}^4[{\beta }^2-x^2]\{x^2\cdot \frac{{\partial }^2\mathrm{\Lambda }}{\partial x^2}+m^2\cdot \frac{{\partial }^2\mathrm{\Lambda }}{\partial {\theta }^2}\}$
		$+(1-x\cos \theta {)}^3\{x({\beta }^2)[(1-2x\cos \theta )\cdot \frac{\partial \mathrm{\Lambda }}{\partial x}+\sin \theta \cdot \frac{\partial \mathrm{\Lambda }}{\partial \theta }]\}$
		$+\{2n-(1-x\cos \theta {)}^2[{\beta }^2-x^2]+(1-x\cos \theta {)}^4[2+\frac{3{\beta }^2}{(n+1)}]n-4n(1-x\cos \theta {)}^2\}\cdot m^2{x}^2\mathrm{\Lambda }$
		$\pm i[\frac{2^3\cdot 3n^2{m}^4{\beta }^2}{(n+1)}{]}^{1/2}{x}^2\cdot (1-x\cos \theta {)}^2\mathrm{\Lambda }$
	$\approx$	$(1-x\cos \theta {)}^4[{\beta }^2-x^2]\{x^2[2^4(n+1{)}^2{\cos }^{}\theta -6(n+1)-2^4{m}^2(n+1{)}^2(1-2{\sin }^{}\theta )]$
		$\pm i{m}^2\beta (-1)x[2^7\cdot 3(n+1{)}^3{]}^{1/2}\cdot [1+xb{]}^{-3/2}\cos \theta$
		$\pm i\beta x^2[2^3\cdot 3(n+1{)}^3{]}^{1/2}\cdot [(12{\cos }^2\theta -4{\cos }^4\theta )-m^2\cdot (1+xb{)}^{-3/2}[2{\cos }^2\theta (15-7{\cos }^2\theta )-6{\sin }^2\theta (3-5{\cos }^2\theta )+6{\sin }^4\theta ]\left]\right\}$
		$+x{\beta }^2(1-x\cos \theta {)}^3\{(1-2x\cos \theta )\cdot [2x(n+1)[2^3(n+1){\cos }^{}\theta -3]\pm i\beta [2^7\cdot 3(n+1{)}^3{]}^{1/2}\cos \theta \cdot (1+xb{)}^{-1/2}]$
		$\pm i\beta x(-1)[2^7\cdot 3(n+1{)}^3{]}^{1/2}\cdot [1+xb{]}^{-1/2}\cdot {\sin }^2\theta \}$
		$+m^2{x}^2\{2n-(1-x\cos \theta {)}^2[{\beta }^2-x^2]+(1-x\cos \theta {)}^4[2+\frac{3{\beta }^2}{(n+1)}]n-4n(1-x\cos \theta {)}^2\}\cdot \{-(4n+1){\beta }^2\pm i\beta [2^7\cdot 3(n+1{)}^3{]}^{1/2}x\cos \theta [1+xb{]}^{1/2}\}$
		$\pm i[\frac{2^3\cdot 3n^2{m}^4{\beta }^2}{(n+1)}{]}^{1/2}{x}^2\cdot (1-x\cos \theta {)}^2\cdot \{-(4n+1){\beta }^2\pm i\beta [2^7\cdot 3(n+1{)}^3{]}^{1/2}x\cos \theta [1+xb{]}^{1/2}\}$

Let's further simplify:

${{ℒ}}_{LHS}$	$\approx$	$x^2\{(1-x\cos \theta {)}^4[{\beta }^2-x^2][2^4(n+1{)}^2{\cos }^{}\theta -6(n+1)-2^4{m}^2(n+1{)}^2(1-2{\sin }^{}\theta )]$
		$-[\frac{2^3\cdot 3n^2{m}^4{\beta }^2}{(n+1)}{]}^{1/2}\cdot (1-x\cos \theta {)}^2\cdot \beta [2^7\cdot 3(n+1{)}^3{]}^{1/2}x\cos \theta [1+xb{]}^{1/2}$
		$+{\beta }^2(1-x\cos \theta {)}^3(1-2x\cos \theta )\cdot [2(n+1)[2^3(n+1){\cos }^{}\theta -3]]$
		$-2n(4n+1){\beta }^2{m}^2+(4n+1){\beta }^2{m}^2[4n+{\beta }^2-x^2](1-x\cos \theta {)}^2-(4n+1){\beta }^2{m}^{2}n(1-x\cos \theta {)}^4[2+\frac{3{\beta }^2}{(n+1)}]\}$
		$\pm ix{\beta }^3(1-x\cos \theta {)}^3(1-2x\cos \theta )\cdot [2^7\cdot 3(n+1{)}^3{]}^{1/2}\cos \theta \cdot (1+xb{)}^{-1/2}$
		$\pm i{m}^2\beta (-1)x[2^7\cdot 3(n+1{)}^3{]}^{1/2}\cdot [1+xb{]}^{-3/2}\cos \theta (1-x\cos \theta {)}^4[{\beta }^2-x^2]$
		$\pm i\beta m^2{x}^2\{2n-(1-x\cos \theta {)}^2[{\beta }^2-x^2]+(1-x\cos \theta {)}^4[2+\frac{3{\beta }^2}{(n+1)}]n-4n(1-x\cos \theta {)}^2\}\cdot [2^7\cdot 3(n+1{)}^3{]}^{1/2}x\cos \theta [1+xb{]}^{1/2}$
		$\pm i\beta x^2[2^3\cdot 3(n+1{)}^3{]}^{1/2}(1-x\cos \theta {)}^4[{\beta }^2-x^2]\cdot [(12{\cos }^{}\theta -4{\cos }^{}\theta )-m^2\cdot (1+xb{)}^{-3/2}[2{\cos }^{}\theta (15-7{\cos }^{}\theta )-6{\sin }^{}\theta (3-5{\cos }^{}\theta )+6{\sin }^{}\theta ]]$
		$\pm i\beta x^2(-1)[2^7\cdot 3(n+1{)}^3{]}^{1/2}\cdot {\beta }^2(1-x\cos \theta {)}^3[1+xb{]}^{-1/2}\cdot {\sin }^{}\theta$
		$\pm i{x}^2(-1)(4n+1){\beta }^2[\frac{2^3\cdot 3n^2{m}^4{\beta }^2}{(n+1)}{]}^{1/2}\cdot (1-x\cos \theta {)}^2$

Step 6

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

$-\frac{{{ℒ}}_{RHS}}{2^2(n+1)x^2}$	$\approx$	$2n(n+1)-(n+1)(1-{\xcancel{x}}^0\cos \theta {)}^2[4n+{\beta }^2-{\xcancel{x^2}}^0-{\xcancel{x^3}}^0(3\cos \theta -{\cos }^{}\theta )]+n(1-{\xcancel{x}}^0\cos \theta {)}^4[2(n+1)+3{\beta }^2]$
		$\pm i(1-{\xcancel{x}}^0\cos \theta {)}^2[2^3\cdot 3n^2(n+1){\beta }^2{]}^{1/2}$
	$\approx$	$2n(n+1)-(n+1)[4n+{\beta }^2]+n[2(n+1)+3{\beta }^2]\pm i[2^3\cdot 3n^2(n+1){\beta }^2{]}^{1/2}$
	$\approx$	$3n{\beta }^2+(n+1)[2n-4n-{\beta }^2+2n]\pm i[2^3\cdot 3n^2(n+1){\beta }^2{]}^{1/2}$
	$\approx$	${\beta }^2(2n-1)\pm i[2^3\cdot 3n^2(n+1){\beta }^2{]}^{1/2}$

$\mathrm{R}\mathrm{e}\left[\frac{{{ℒ}}_{LHS}}{x^2}\right]$	$\approx$	$(1-{\xcancel{x}}^0\cos \theta {)}^4[{\beta }^2-{\xcancel{x^2}}^0][2^4(n+1{)}^2{\cos }^{}\theta -6(n+1)-2^4{m}^2(n+1{)}^2(1-2{\sin }^{}\theta )]$
		$-[\frac{2^3\cdot 3n^2{m}^4{\beta }^2}{(n+1)}{]}^{1/2}\cdot (1-x\cos \theta {)}^2\cdot \beta [2^7\cdot 3(n+1{)}^3{]}^{1/2}{\xcancel{x}}^0\cos \theta [1+xb{]}^{1/2}$
		$+{\beta }^2(1-{\xcancel{x}}^0\cos \theta {)}^3(1-{\xcancel{2x}}^0\cos \theta )\cdot [2(n+1)[2^3(n+1){\cos }^{}\theta -3]]$
		$-2n(4n+1){\beta }^2{m}^2+(4n+1){\beta }^2{m}^2[4n+{\beta }^2-{\xcancel{x^2}}^0](1-{\xcancel{x}}^0\cos \theta {)}^2-(4n+1){\beta }^2{m}^{2}n(1-{\xcancel{x}}^0\cos \theta {)}^4[2+\frac{3{\beta }^2}{(n+1)}]$
	$\approx$	${\beta }^2[2^4(n+1{)}^2{\cos }^2\theta -6(n+1)-2^4{m}^2(n+1{)}^2(1-2{\sin }^2\theta )]+{\beta }^2[2^4(n+1{)}^2{\cos }^2\theta -6(n+1)]$
		$+{\beta }^2\{-2n(4n+1)m^2+(4n+1)m^2[4n]-2(4n+1)m^{2}n\}+{\beta }^4\{(4n+1)m^2-(4n+1)m^{2}n[\frac{3}{(n+1)}\left]\right\}$
	$\approx$	${\beta }^2[2^4(n+1{)}^2{\cos }^2\theta -6(n+1)]+{\beta }^2[2^4(n+1{)}^2{\cos }^2\theta -6(n+1)]$
		$+m^2{\beta }^2[-2^4(n+1{)}^2(1-2{\sin }^2\theta )]+m^2{\beta }^2\{-2n(4n+1)+4n(4n+1)-2n(4n+1)\}$
		$+(4n+1)m^2{\beta }^4[1-\frac{3n}{(n+1)}]$
	$\approx$	$2{\beta }^2[2^4(n+1{)}^2{\cos }^2\theta -6(n+1)]+m^2{\beta }^2[2^4(n+1{)}^2(1-2{\cos }^2\theta )]$
		$+(4n+1)m^2{\beta }^4[1-\frac{3n}{(n+1)}]$
	$\approx$	$(1-m^2)2^5{\beta }^2(n+1{)}^2{\cos }^2\theta +2^2(n+1){\beta }^2[4m^2(n+1)-3]+(4n+1)m^2{\beta }^4[1-\frac{3n}{(n+1)}].$

$\pm \mathrm{I}\mathrm{m}\left[{{ℒ}}_{LHS}\right]$	$\approx$	$x\{{\beta }^3(1-{\xcancel{x}}^0\cos \theta {)}^3(1-{\xcancel{2x}}^0\cos \theta )\cdot [2^7\cdot 3(n+1{)}^3{]}^{1/2}\cos \theta \cdot (1+{\xcancel{xb}}^0{)}^{-1/2}$
		$-m^2\beta [2^7\cdot 3(n+1{)}^3{]}^{1/2}\cdot [1+{\xcancel{xb}}^0{]}^{-3/2}\cos \theta (1-{\xcancel{x}}^0\cos \theta {)}^4[{\beta }^2-{{\xcancel{x}}^0}^2]\}$
		$+x^2\left\{\beta m^2\right\{2n-(1-x\cos \theta {)}^2[{\beta }^2-x^2]+(1-x\cos \theta {)}^4[2+\frac{3{\beta }^2}{(n+1)}]n-4n(1-x\cos \theta {)}^2\}\cdot [2^7\cdot 3(n+1{)}^3{]}^{1/2}{\xcancel{x}}^0\cos \theta [1+xb{]}^{1/2}$
		$+\beta [2^3\cdot 3(n+1{)}^3{]}^{1/2}(1-{\xcancel{x}}^0\cos \theta {)}^4[{\beta }^2-{\xcancel{x^2}}^0]\cdot [(12{\cos }^{}\theta -4{\cos }^{}\theta )-m^2\cdot (1+{\xcancel{xb}}^0{)}^{-3/2}[2{\cos }^{}\theta (15-7{\cos }^{}\theta )-6{\sin }^{}\theta (3-5{\cos }^{}\theta )+6{\sin }^{}\theta ]]$
		$-\beta [2^7\cdot 3(n+1{)}^3{]}^{1/2}\cdot {\beta }^2(1-{\xcancel{x}}^0\cos \theta {)}^3[1+{\xcancel{xb}}^0{]}^{-1/2}\cdot {\sin }^{}\theta$
		$-(4n+1){\beta }^2[\frac{2^3\cdot 3n^2{m}^4{\beta }^2}{(n+1)}{]}^{1/2}\cdot (1-{\xcancel{x}}^0\cos \theta {)}^2\}$
	$\approx$	$x(1-m^2)\cdot {\beta }^3[2^7\cdot 3(n+1{)}^3{]}^{1/2}\cos \theta$
		$+x^2{\beta }^3[2^3\cdot 3(n+1{)}^3{]}^{1/2}\{(12{\cos }^2\theta -4{\cos }^4\theta )-4{\sin }^2\theta$
		$-m^2[2{\cos }^2\theta (15-7{\cos }^2\theta )-6{\sin }^2\theta (3-5{\cos }^2\theta )+6{\sin }^4\theta -\frac{n(4n+1)}{(n+1{)}^2}]\}$
	$\approx$	$x(1-m^2)\cdot {\beta }^3[2^7\cdot 3(n+1{)}^3{]}^{1/2}\cos \theta$
		$+x^2{\beta }^3[2^7\cdot 3(n+1{)}^3{]}^{1/2}\{4{\cos }^2\theta -{\cos }^4\theta -1$
		$-\frac{m^2}{2}[{\cos }^2\theta (15-7{\cos }^2\theta )-12{\sin }^4\theta +6{\sin }^2\theta -\frac{n(4n+1)}{2(n+1{)}^2}]\}$
	$\approx$	$x(1-m^2)\cdot {\beta }^3[2^7\cdot 3(n+1{)}^3{]}^{1/2}\cos \theta$
		$+x^2{\beta }^3[2^7\cdot 3(n+1{)}^3{]}^{1/2}\{3-(1+{\sin }^2\theta {)}^2-\frac{m^2}{2}[33{\cos }^2\theta -19{\cos }^4\theta -6-\frac{n(4n+1)}{2(n+1{)}^2}]\}$

Examples

Evaluate various expressions using the parameter set:   $(n,\theta ,x)=(1,\frac{\pi }{3},\frac{1}{4})$

$b$	$=$	$\frac{3}{2}-\frac{1}{8}=\frac{11}{8}$	1.375000000
$(\beta \eta {)}^2$	$=$	$\left(\frac{1}{2^2}{)}^2\right[1+\frac{11}{2^5}]=\frac{2^5+11}{2^9}=\frac{43}{2^9}$	0.083984375
$\mathrm{R}\mathrm{e}(\mathrm{\Lambda })$	$=$	$-5{\beta }^2+\frac{43}{2^9}[2^3-6]=-5{\beta }^2+\frac{43}{2^8}$	$-5{\beta }^2$ + 0.167968750
$\mathrm{I}\mathrm{m}(\mathrm{\Lambda })$	$=$	$\frac{\beta }{2}[2^{10}\cdot 3\cdot \frac{43}{2^9}{]}^{1/2}=\beta [\frac{3\cdot 43}{2}{]}^{1/2}$	8.031189202 $\beta$
$\mathrm{R}\mathrm{e}\left(\frac{\partial \mathrm{\Lambda }}{\partial x}\right)$	$=$	$[2^3-6]\frac{1}{2^2}(2+\frac{3\cdot 11}{2^2\cdot 2^3})=(1+\frac{33}{2^6})$	1.515625000
$\mathrm{I}\mathrm{m}\left(\frac{\partial \mathrm{\Lambda }}{\partial x}\right)$	$=$	$\frac{\beta }{2}\left[\frac{2^9\cdot 2^8\cdot 3}{43}{]}^{1/2}\right[\frac{1}{2^2}\cdot (2+\frac{3\cdot 11}{2^2\cdot 2^3})]=\beta [\frac{2^{13}\cdot 3}{43}{]}^{1/2}(1+\frac{3\cdot 11}{2^6})$	36.23373732 $\beta$
$\mathrm{R}\mathrm{e}\left(\frac{\partial \mathrm{\Lambda }}{\partial \theta }\right)$	$=$	$2\left(\frac{3}{4}{)}^{1/2}\right\{-2^4\cdot \frac{43}{2^9}+\frac{3}{2^6}\left(\frac{3}{4}\right)[3-4]\}=-\frac{\sqrt{3}}{2^8}\cdot (2^3\cdot 43+9)$	-2.388335684
$\mathrm{I}\mathrm{m}\left(\frac{\partial \mathrm{\Lambda }}{\partial \theta }\right)$	$=$	$(-1)\beta (\frac{3}{4}{)}^{1/2}\cdot [\frac{2^{10}\cdot 3\cdot 43}{2^9}{]}^{1/2}\{1+\frac{3\cdot 2^9}{2^7\cdot 43}(\frac{3}{2^3}\left)\right\}=(-1)\beta (\frac{3}{4}{)}^{1/2}\cdot [2\cdot 3\cdot 43{]}^{1/2}\{1+\frac{3^2}{2\cdot 43}\}$	(-1) × 15.36617018 $\beta$

$\mathrm{R}\mathrm{e}\left(\frac{{\partial }^2\mathrm{\Lambda }}{\partial x^2}\right)$	$=$	$2^2[2^2-3][1+\frac{3}{2^2}\cdot \frac{11}{2^3}]=2^2+\frac{33}{2^3}=\frac{65}{8}$	8.125000000
$\mathrm{I}\mathrm{m}\left(\frac{{\partial }^2\mathrm{\Lambda }}{\partial x^2}\right)$	$=$	$\beta 2^2\cdot \sqrt{3}\left[\frac{11}{2^3}\right(2^2+\frac{3}{2^2}\cdot \frac{11}{2^3}\left)\right][\frac{2^4}{43}{]}^{3/2}=[\frac{2^3\cdot 3}{43^3}{]}^{1/2}[11\cdot (2^7+33)]\beta$	30.76957507$\beta$
$\mathrm{R}\mathrm{e}\left(\frac{{\partial }^2\mathrm{\Lambda }}{\partial {\theta }^2}\right)$	$=$	$\frac{1}{2^4}\left\{2^6\right(\frac{3}{2^2}-\frac{1}{2^2}\left)\right\}+\frac{1}{2^6}\{-2^3\cdot 3+2+\frac{3^3\cdot 5\cdot 7}{2^2}-3\cdot 23\}=2+\frac{1}{2^8}\{2^3+3^3\cdot 5\cdot 7-2^2\cdot 3\cdot 31\}$	4.269531250
$\mathrm{I}\mathrm{m}\left(\frac{{\partial }^2\mathrm{\Lambda }}{\partial {\theta }^2}\right)$	$=$	$(-1)\beta [2^{10}\cdot 3{]}^{1/2}(\frac{43}{2^9}{)}^{1/2}\{\frac{1}{2}+\frac{3}{2}\cdot \frac{2^9}{43}\cdot \frac{1}{2^6}\cdot \frac{3}{2^2}(\frac{5}{2^2}-2)+(\frac{3}{2}\cdot \frac{1}{2^6}\cdot \frac{2^9}{43}{)}^2\left(\frac{3}{2^2}{)}^3\frac{1}{2}\right\}$
	$=$	$(-1)\beta \left(\frac{3\cdot 43}{2}{)}^{1/2}\right\{1-\frac{3^3}{2\cdot 43}+\frac{3^5}{(2\cdot 43{)}^2}\}$	(-1) × 5.773638858 $\beta$

$\mathrm{R}\mathrm{e}\{(1-2x\cos \theta )\cdot \frac{\partial \mathrm{\Lambda }}{\partial x}+\sin \theta \cdot \frac{\partial \mathrm{\Lambda }}{\partial \theta }\}$	$=$	$(1-\frac{1}{2^2})(1+\frac{33}{2^6})-\frac{\sqrt{3}}{2}\cdot \frac{\sqrt{3}}{2^8}\cdot (2^3\cdot 43+9)$
	$=$	$\frac{(2\cdot 3\cdot 97)-3\cdot (2^3\cdot 43+9)}{2^9}$	-0.931640625
$\mathrm{I}\mathrm{m}\{(1-2x\cos \theta )\cdot \frac{\partial \mathrm{\Lambda }}{\partial x}+\sin \theta \cdot \frac{\partial \mathrm{\Lambda }}{\partial \theta }\}$	$=$	$(1-\frac{1}{2^2})\beta \left[\frac{2^{13}\cdot 3}{43}{]}^{1/2}\right(1+\frac{3\cdot 11}{2^6})+\frac{\sqrt{3}}{2}\cdot (-1)\beta \frac{\sqrt{3}}{2}\cdot [2\cdot 3\cdot 43{]}^{1/2}\{1+\frac{3^2}{2\cdot 43}\}$
	$=$	$\beta [\frac{3^3}{2^5\cdot 43}{]}^{1/2}[2(2^6+33)-(2\cdot 43+3^2)]$	13.86780926 $\beta$
$\mathrm{R}\mathrm{e}\{(2+3xb)\cdot \frac{\partial \mathrm{\Lambda }}{\partial x}-3{\sin }^3\theta \cdot \frac{\partial \mathrm{\Lambda }}{\partial \theta }\}$	$=$	$(2+\frac{3}{2^2}\cdot \frac{11}{2^3})(1+\frac{33}{2^6})+3\cdot (\frac{3}{2^2}{)}^{3/2}\cdot \frac{\sqrt{3}}{2^8}\cdot (2^3\cdot 43+9)$
	$=$	$\frac{97^2}{2^{11}}+\frac{3^3}{2^{11}}\cdot (2^3\cdot 43+9)$	9.248046874
$\mathrm{I}\mathrm{m}\{(2+3xb)\cdot \frac{\partial \mathrm{\Lambda }}{\partial x}-3{\sin }^3\theta \cdot \frac{\partial \mathrm{\Lambda }}{\partial \theta }\}$	$=$	$\beta \left\{\right(2+\frac{3}{2^2}\cdot \frac{11}{2^3}\left)\right[\frac{2^{13}\cdot 3}{43}{]}^{1/2}(1+\frac{3\cdot 11}{2^6})+3\cdot (\frac{3}{2^2}{)}^{3/2}\cdot (\frac{3}{4}{)}^{1/2}\cdot [2\cdot 3\cdot 43{]}^{1/2}[1+\frac{3^2}{2\cdot 43}\left]\right\}$
	$=$	$\beta \left\{\right(\frac{3}{2^9\cdot 43}{)}^{1/2}[(2^6+33{)}^2+3^3(2\cdot 43+3^2)\left]\right\}$	139.7753772

Step 7

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

${\mathrm{R}\mathrm{H}\mathrm{S}}_4$	$\equiv$	${\mathrm{R}\mathrm{H}\mathrm{S}}_3-m^2{x}^2\mathrm{\Lambda }\cdot {𝒜}$
	$=$	$-x^2[2^2(n+1{)}^2-m^2\mathrm{\Lambda }]\cdot {𝒜},$
${\mathrm{L}\mathrm{H}\mathrm{S}}_4$	$\equiv$	${\mathrm{L}\mathrm{H}\mathrm{S}}_3-m^2{x}^2\mathrm{\Lambda }\cdot {𝒜}$
	$=$	${\beta }^2(1-{\eta }^2)(1-x\cos \theta {)}^3\{(1-x\cos \theta )[x^2\frac{{\partial }^2\mathrm{\Lambda }}{\partial x^2}+\frac{{\partial }^2\mathrm{\Lambda }}{\partial {\theta }^2}]+x[(1-2x\cos \theta )\frac{\partial \mathrm{\Lambda }}{\partial x}+\sin \theta \frac{\partial \mathrm{\Lambda }}{\partial \theta }]\}$
		$-n{x}^3(1-x\cos \theta {)}^4[(2+3xb)\frac{\partial \mathrm{\Lambda }}{\partial x}-3{\sin }^{}\theta \frac{\partial \mathrm{\Lambda }}{\partial \theta }]$

Next Lowest Order

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

$(n+1)\cdot {𝒜}$	$=$	$2n(n+1)+(n+1)(1-x\cos \theta {)}^2[x^2(1+xb)-{\beta }^2-4n]+(1-x\cos \theta {)}^4[2n(n+1)-3n{\beta }^2]$
		$\pm ix\cos \theta (1-x\cos \theta {)}^2[2^3\cdot 3n^2{\beta }^2(n+1){]}^{1/2}$
	$=$	$2n(n+1)+(n+1)[1-2x\cos \theta +x^2{\cos }^{}\theta +{𝒪}(x^3)][x^2(1+xb)-{\beta }^2-4n]+[1-4x\cos \theta +6x^2{\cos }^{}\theta +{𝒪}(x^3)][2n(n+1)-3n{\beta }^2]$
		$\pm ix\cos \theta [1-2x\cos \theta +x^2{\cos }^{}\theta +{𝒪}(x^3)][2^3\cdot 3n^2{\beta }^2(n+1){]}^{1/2}$
	$=$	$x^0\{2n(n+1)-4n(n+1)+2n(n+1)\}+x^1\{8n(n+1)\cos \theta -8n(n+1)\cos \theta \}$
		$+x^2\{-4n(n+1){\cos }^2\theta +(n+1)[1-\left(\frac{\beta }{x}{)}^2\right]+12n(n+1){\cos }^2\theta -3n\left(\frac{\beta }{x}{)}^2\right\}+{𝒪}(x^3)$
		$\pm i{x}^2\left(\frac{\beta }{x}\right)\frac{n{b}_0}{2^2(n+1)}[1-2x\cos \theta +x^2{\cos }^{}\theta +{𝒪}(x^3)]$
	$=$	$x^0\left\{0\right\}+x^1\left\{0\right\}+x^2\{(n+1)[8n{\cos }^2\theta +1]-(4n+1)(\frac{\beta }{x}{)}^2\}+{𝒪}(x^3)$
		$\pm i{x}^2\left(\frac{\beta }{x}\right)\frac{n{b}_0}{2^2(n+1)}[1-2x\cos \theta +x^2{\cos }^{}\theta +{𝒪}(x^3)]$

where,

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

Hence,

$\frac{{\mathrm{R}\mathrm{H}\mathrm{S}}_4}{x^2}$	$=$	$-2^2(n+1{)}^2\cdot {𝒜}+m^2\cdot {𝒜}\{-(4n+1){\beta }^2+(\beta \eta {)}^2(n+1)[2^3(n+1){\cos }^2\theta -3]\pm i\beta \cos \theta [2^7\cdot 3(n+1{)}^3{]}^{1/2}(\beta \eta )\}$
	$=$	$-2^2(n+1{)}^2\cdot {𝒜}+m^2\cdot {𝒜}(n+1)\{-x^2\left(\frac{4n+1}{n+1}\right)\cdot (\frac{\beta }{x}{)}^2+x^2(1+{\xcancel{x}}^{0}b)[2^3(n+1){\cos }^2\theta -3]\pm i{x}^2(\frac{\beta }{x})\frac{b_0}{(n+1)}(1+{\xcancel{x}}^{0}b{)}^{1/2}\}$
	$\approx$	${𝒜}(n+1)\{-2^2(n+1)+m^2{\xcancel{x^2}}^0[[2^3(n+1){\cos }^2\theta -3]-\left(\frac{4n+1}{n+1}\right)\cdot (\frac{\beta }{x}{)}^2\pm i(\frac{\beta }{x}\left)\frac{b_0}{(n+1)}\right]\}$
	$\approx$	$-2^2(n+1)x^2\{(n+1)[8n{\cos }^2\theta +1]-(4n+1)(\frac{\beta }{x}{)}^2\pm i\left(\frac{\beta }{x}\right)\frac{n{b}_0}{2^2(n+1)}[1-2{\xcancel{x}}^0\cos \theta +{\xcancel{x^2}}^0{\cos }^{}\theta +{\xcancel{{𝒪}}}^0(x^3)]\}$
$\Rightarrow \frac{{\mathrm{R}\mathrm{H}\mathrm{S}}_4}{x^4}$	$\approx$	$2^2(n+1)(4n+1)(\frac{\beta }{x}{)}^2-2^2(n+1{)}^2[8n{\cos }^2\theta +1]\pm i(-1)\left(\frac{\beta }{x}\right)n{b}_0.$

This should be compared to,

$\frac{{\mathrm{L}\mathrm{H}\mathrm{S}}_4}{x^2}$	$=$	$\left[\right(\frac{\beta }{x}{)}^2-1-xb](1-x\cos \theta {)}^3\{(1-x\cos \theta )[x^2\frac{{\partial }^2\mathrm{\Lambda }}{\partial x^2}+\frac{{\partial }^2\mathrm{\Lambda }}{\partial {\theta }^2}]+x[(1-2x\cos \theta )\frac{\partial \mathrm{\Lambda }}{\partial x}+\sin \theta \frac{\partial \mathrm{\Lambda }}{\partial \theta }\left]\right\}$
		$-nx(1-x\cos \theta {)}^4[(2+3xb)\frac{\partial \mathrm{\Lambda }}{\partial x}-3{\sin }^{}\theta \frac{\partial \mathrm{\Lambda }}{\partial \theta }].$

Now, from above, we can write,

$[x^2\frac{{\partial }^2\mathrm{\Lambda }}{\partial x^2}+\frac{{\partial }^2\mathrm{\Lambda }}{\partial {\theta }^2}]$	$=$	$x^2\{2(n+1)[2^3(n+1){\cos }^2\theta -3](1+3{\xcancel{x}}^{0}b)\pm i\beta (\frac{b_0}{2^2}\left)\right[\frac{b(4+3xb)}{(1+xb{)}^{3/2}}\left]\right\}+x^2\{2^4(n+1{)}^2({\sin }^2\theta -{\cos }^2\theta )\}$
		$+{\xcancel{x^3}}^0\{-2^4\cdot 3(n+1{)}^2{\cos }^3\theta +2^4(n+1{)}^2{\cos }^5\theta +3^2(n+1)(16n+19){\sin }^2\theta \cos \theta -2^3\cdot 23(n+1{)}^2{\sin }^{}\theta {\cos }^{}\theta \}$
		$\pm i(-1)\beta b_{0}x(1+{\xcancel{x}}^{0}b{)}^{1/2}\{1+\frac{3{\xcancel{x}}^0{\sin }^2\theta (5{\cos }^2\theta -2)}{2(1+xb)\cos \theta }+\frac{3^2{\xcancel{x^2}}^0{\sin }^6\theta }{2^2(1+xb{)}^2}\}$
	$\approx$	$x^2\{2(n+1)[2^3(n+1){\cos }^2\theta -3]+2^4(n+1{)}^2({\sin }^2\theta -{\cos }^2\theta )\}\pm ix\{{\xcancel{x\beta }}^0\left(\frac{b_0}{2^2}\right)\left[\frac{b(4+3xb)}{(1+xb{)}^{3/2}}\right]-\beta b_0\}$
	$\approx$	$2(n+1)x^2\{2^3(n+1){\sin }^2\theta -3\}\pm i{x}^2\{-(\frac{\beta }{x}\left)b_0\right\}.$

Also,

$x[(1-2x\cos \theta )\cdot \frac{\partial \mathrm{\Lambda }}{\partial x}+\sin \theta \cdot \frac{\partial \mathrm{\Lambda }}{\partial \theta }]$	$=$	$x^2(n+1)[-6+2^4(n+1){\cos }^2\theta ]$
		$-{\xcancel{x^3}}^0(n+1)\cos \theta \{[15+2^4(n+1)]-{\cos }^{}\theta [9+2^3\cdot 7(n+1)]+2^3\cdot 3(n+1){\cos }^{}\theta \}$
		$+{\xcancel{x^4}}^0(n+1)\{9-2^2\cdot 3^2(1+2n){\cos }^2\theta -[9+32(n+1)]{\cos }^4\theta +2^3(n+1){\cos }^6\theta \}$
		$\pm i{x}^2\left(\frac{\beta }{x}\right)\left[\frac{2^5\cdot 3(n+1{)}^3}{1+{\xcancel{x}}^0(3\cos \theta -{\cos }^{}\theta )}{]}^{1/2}\right\{2\cos \theta -{\xcancel{x}}^0[2-7{\cos }^{}\theta +3{\cos }^{}\theta ]$
		$-{\xcancel{x^2}}^0\cos \theta [9+4{\cos }^{}\theta -{\cos }^{}\theta ]\}$
	$\approx$	$2(n+1)x^2[2^3(n+1){\cos }^2\theta -3]\pm i{x}^2\left(\frac{\beta }{x}\right)b_0.$

Finally,

$nx[(2+3xb)\cdot \frac{\partial \mathrm{\Lambda }}{\partial x}-3{\sin }^3\theta \cdot \frac{\partial \mathrm{\Lambda }}{\partial \theta }]$	$=$	$x^2\cdot 2^{2}n(n+1)[2^3(n+1){\cos }^2\theta -3]$
		$+n{\xcancel{x^3}}^0\cdot 2^2\cdot 3(n+1)\cos \theta \{[2^{2}n-5]+{\cos }^{}\theta [2^{4}n+19]-2^2(n+1){\cos }^{}\theta \}$
		$+n{\xcancel{x^4}}^0\cdot 3^2(n+1){\cos }^2\theta \{-3^3+2\cdot 3^2{\cos }^2\theta [2^{2}n+5]-99\cdot {\cos }^4\theta +2^3(n+1){\cos }^6\theta \}$
		$+n{\xcancel{x^4}}^0(n+1){\sin }^4\theta \{-3^3+3^2{\cos }^2\theta [2^3\cdot 3n+27]-2^3\cdot 3\cdot 5(n+1){\cos }^4\theta \}$
		$\pm in{x}^2\left(\frac{\beta }{x}\right)\left[\frac{2^5\cdot 3(n+1{)}^3}{1+{\xcancel{x}}^{0}b}{]}^{1/2}\right\{4\cos \theta +6{\xcancel{x}}^0(2b\cos \theta +{\sin }^{}\theta )+3{\xcancel{x^2}}^0(3b^2\cos \theta +2b{\sin }^{}\theta +3{\sin }^{}\theta \cos \theta )\}$
	$\approx$	$x^2\cdot 2^{2}n(n+1)[2^3(n+1){\cos }^2\theta -3]\pm i2n{x}^2\left(\frac{\beta }{x}\right)b_0.$

Inserting these three approximate expressions into the LHS_4 ensemble gives,

$\frac{{\mathrm{L}\mathrm{H}\mathrm{S}}_4}{x^2}$	$=$	$\left[\right(\frac{\beta }{x}{)}^2-1-{\xcancel{xb}}^0](1-{\xcancel{x}}^0\cos \theta {)}^3\{(1-{\xcancel{x}}^0\cos \theta )[x^2\frac{{\partial }^2\mathrm{\Lambda }}{\partial x^2}+\frac{{\partial }^2\mathrm{\Lambda }}{\partial {\theta }^2}]+x[(1-2x\cos \theta )\frac{\partial \mathrm{\Lambda }}{\partial x}+\sin \theta \frac{\partial \mathrm{\Lambda }}{\partial \theta }\left]\right\}$
		$-nx(1-{\xcancel{x}}^0\cos \theta {)}^4[(2+3xb)\frac{\partial \mathrm{\Lambda }}{\partial x}-3{\sin }^{}\theta \frac{\partial \mathrm{\Lambda }}{\partial \theta }]$
$$	$\approx$	$\left[\right(\frac{\beta }{x}{)}^2-1\left]\right\{2(n+1)x^2[2^3(n+1){\sin }^2\theta -3]\pm i{x}^2[-(\frac{\beta }{x}\left)b_0\right]$
$$		$+2(n+1)x^2[2^3(n+1){\cos }^2\theta -3]\pm i{x}^2\left(\frac{\beta }{x}\right)b_0\}$
		$-x^2\cdot 2^{2}n(n+1)[2^3(n+1){\cos }^2\theta -3]\pm i2n{x}^2\left(\frac{\beta }{x}\right)b_0$
$\Rightarrow \frac{{\mathrm{L}\mathrm{H}\mathrm{S}}_4}{x^4}$	$\approx$	$\left[\right(\frac{\beta }{x}{)}^2-1\left]\right\{2(n+1)[2^3(n+1){\sin }^2\theta -3]+2(n+1)[2^3(n+1){\cos }^2\theta -3]\}-2^{2}n(n+1)[2^3(n+1){\cos }^2\theta -3]$
$$		$\pm i2n\left(\frac{\beta }{x}\right)b_0$
	$\approx$	$2^2(n+1)(\frac{\beta }{x}{)}^2[2^2(n+1)-3]-(n+1)\{[2^4(n+1)-12]+2^{2}n[2^3(n+1){\cos }^2\theta -3]\}\pm i2n(\frac{\beta }{x})b_0$
	$\approx$	$2^2(n+1)(\frac{\beta }{x}{)}^2[4n+1]-2^2(n+1{)}^2[1+2^{3}n{\cos }^2\theta ]\pm i2n\left(\frac{\beta }{x}\right)b_0.$

Assessment

The good news is that the real part of the ${\mathrm{L}\mathrm{H}\mathrm{S}}_4$ expression exactly matches the real part of the ${\mathrm{R}\mathrm{H}\mathrm{S}}_4$ expression. But the imaginary differ by a factor of 2. So, let's repeat the steps leading to the imaginary parts.

Case B:

$\mathrm{I}\mathrm{m}\left[\frac{\mathrm{R}\mathrm{H}{\mathrm{S}}_{\mathrm{4}}}{x^2}\right]$	$=$	$-2^2(n+1)\cdot \mathrm{I}\mathrm{m}[(n+1){𝒜}]+\frac{m^2}{(n+1)}\{\mathrm{I}\mathrm{m}[(n+1){𝒜}]\cdot \mathrm{R}\mathrm{e}[\mathrm{\Lambda }]+\mathrm{R}\mathrm{e}[(n+1){𝒜}]\cdot \mathrm{I}\mathrm{m}[\mathrm{\Lambda }]\}$
	$=$	$-2^2(n+1)\cdot x\cos \theta (1-x\cos \theta {)}^2[2^3\cdot 3n^2{\beta }^2(n+1){]}^{1/2}$
		$+\frac{m^2}{(n+1)}\{x\cos \theta (1-x\cos \theta {)}^2[2^3\cdot 3n^2{\beta }^2(n+1){]}^{1/2}\}\cdot \{-(4n+1){\beta }^2+(\beta \eta {)}^2(n+1)[2^3(n+1){\cos }^{}\theta -3]\}$
		$+\frac{m^2}{(n+1)}\{2n(n+1)+(n+1)(1-x\cos \theta {)}^2[x^2(1+xb)-{\beta }^2-4n]+(1-x\cos \theta {)}^4[2n(n+1)-3n{\beta }^2]\}\cdot \{\beta \cos \theta [2^7\cdot 3(n+1{)}^3{]}^{1/2}(\beta \eta )\}$
	$=$	$-x^2\left(\frac{\beta }{x}\right)\{(1-x\cos \theta {)}^{2}n{b}_0\}$
		$+m^2{x}^4\left(\frac{\beta }{x}\right)\{(1-x\cos \theta {)}^2\left[\frac{n{b}_0}{2^2(n+1{)}^2}\right]\}\cdot \{-\left[\frac{4n+1}{n+1}\right](\frac{\beta }{x}{)}^2+(1+xb)[2^3(n+1){\cos }^{}\theta -3]\}$
		$+m^2{x}^2\left(\frac{\beta }{x}\right)\{2n+(1-x\cos \theta {)}^2[x^2(1+xb)-{\beta }^2-4n]+(1-x\cos \theta {)}^4\cdot [2n-\frac{3n{\beta }^2}{(n+1)}\left]\right\}\cdot \{(1+xb{)}^{1/2}\}$

$\Rightarrow \mathrm{I}\mathrm{m}\left[\frac{\mathrm{R}\mathrm{H}{\mathrm{S}}_{\mathrm{4}}}{x^4}\right]\left(\frac{x}{\beta }\right)$	$=$	$-(1-x\cos \theta {)}^{2}n{b}_0$
		$+m^2\{2n+(1-x\cos \theta {)}^2[x^2(1+xb)-{\beta }^2-4n]+(1-x\cos \theta {)}^4\cdot [2n-\frac{3n{\beta }^2}{(n+1)}\left]\right\}\cdot (1+xb{)}^{1/2}$
		$+m^2\{-[\frac{4n+1}{n+1}]{\beta }^2+x^2(1+xb)[2^3(n+1){\cos }^2\theta -3]\}\cdot (1-x\cos \theta {)}^2[\frac{n{b}_0}{2^2(n+1{)}^2}]$
	$=$	$-(1-x\cos \theta {)}^{2}n{b}_0+m^2\{2n-4n(1-x\cos \theta {)}^2+2n(1-x\cos \theta {)}^4\}\cdot (1+xb{)}^{1/2}$
		$+m^2(1-x\cos \theta {)}^2\{x^2(1+xb{)}^{3/2}-{\beta }^2\cdot (1+xb{)}^{1/2}[1+\frac{3n(1-x\cos \theta {)}^2}{(n+1)}]\}$
		$+m^2(1-x\cos \theta {)}^2\{x^2(1+xb)[2^3(n+1){\cos }^{}\theta -3]\cdot \left[\frac{n{b}_0}{2^2(n+1{)}^2}\right]-\left[\frac{n{b}_0(4n+1)}{2^2(n+1{)}^3}\right]{\beta }^2\}$
	$=$	$-(1-x\cos \theta {)}^{2}n{b}_0+m^2\{2n-4n[1-2x\cos \theta +x^2{\cos }^{}\theta +{𝒪}(x^3)]+2n[1-4x\cos \theta +6x^2{\cos }^{}\theta +{𝒪}(x^3)]\}\cdot (1+xb{)}^{1/2}$
		$+m^2{x}^2(1-x\cos \theta {)}^2\{2^2(n+1{)}^2(1+xb{)}^{1/2}+[2^3(n+1){\cos }^{}\theta -3]\cdot n{b}_0\}\cdot \frac{(1+xb)}{2^2(n+1{)}^2}$
		$-m^2{\beta }^2(1-x\cos \theta {)}^2\{n{b}_0(4n+1)+2^2(n+1{)}^2(1+xb{)}^{1/2}[(n+1)+3n(1-x\cos \theta {)}^2]\}\cdot \frac{1}{2^2(n+1{)}^3}$
	$=$	$-n{b}_0[1-2x\cos \theta +x^2{\cos }^{}\theta +{𝒪}(x^3)]+m^2\{8n{x}^2{\cos }^{}\theta +{𝒪}(x^3)\}\cdot (1+{\xcancel{x}}^{0}b{)}^{1/2}$
		$+m^2{x}^2(1-{\xcancel{x}}^0\cos \theta {)}^2\{2^2(n+1{)}^2(1+{\xcancel{x}}^{0}b{)}^{1/2}+[2^3(n+1){\cos }^{}\theta -3]\cdot n{b}_0\}\cdot \frac{(1+{\xcancel{x}}^{0}b)}{2^2(n+1{)}^2}$
		$-m^2{\beta }^2(1-{\xcancel{x}}^0\cos \theta {)}^2\{n{b}_0(4n+1)+2^2(n+1{)}^2(1+{\xcancel{x}}^{0}b{)}^{1/2}[(n+1)+3n(1-{\xcancel{x}}^0\cos \theta {)}^2]\}\cdot \frac{1}{2^2(n+1{)}^3}$
	$\approx$	$-n{b}_0[1-2x\cos \theta ]$
		$-n{b}_0{x}^2{\cos }^2\theta +m^2{x}^2\{2^{5}n(n+1{)}^2{\cos }^2\theta +2^2(n+1{)}^2+[2^3(n+1){\cos }^2\theta -3]\cdot n{b}_0\}\cdot \frac{1}{2^2(n+1{)}^2}$
		$-m^2{\beta }^2\{n{b}_0(4n+1)+2^2(n+1{)}^2[(n+1)+3n]\}\cdot \frac{1}{2^2(n+1{)}^3}$

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,

$H=H_0{\mathrm{\Theta }}_H$

$=$

$H_0-\frac{H_0}{{\beta }^2}[r^2+r^3(3\cos \theta -{\cos }^{}\theta )+{𝒪}(r^4)].$

[Note:   Here we have replaced the variable name, $x$, as used in Blaes85, with the variable name, $r$, in order (1) to emphasize that the variable represents a dimensionless radial coordinate, and (2) to avoid conflict with the GGN86 variable, $x$, 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,

${\mathrm{\Omega }}_K\equiv \frac{G{M}_{\mathrm{p}\mathrm{t}}}{{\varpi }_0^3},$

we can set,

$H_0$

$=$

$\frac{G{M}_{\mathrm{p}\mathrm{t}}{\beta }^2}{2{\varpi }_0}=\frac{1}{2}{\mathrm{\Omega }}_K^2{\varpi }_0^2{\beta }^2.$

Hence, for the slimmest tori — that is, keeping only the lowest order term in $r$ — the enthalpy distribution becomes,

$H$	$=$	$\frac{1}{2}{\mathrm{\Omega }}_K^2{\varpi }_0^2{\beta }^2-\frac{1}{2}{\mathrm{\Omega }}_K^2{\varpi }_0^2[r^2+{\xcancel{r^3}}^0(3\cos \theta -{\cos }^{}\theta )+{\xcancel{{𝒪}(r^4)}}^0]$
	$\approx$	$\frac{{\mathrm{\Omega }}_K^2}{2}[{\varpi }_0^2{\beta }^2-{\varpi }_0^2{r}^2].$

Following GGN86, the surface of the torus — where the enthalpy drops to zero — occurs at $r=a/{\varpi }_0$. Hence, we recognize that,

$\beta =\frac{a}{{\varpi }_0},$

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

$H$

$=$

$\frac{{\mathrm{\Omega }}_K^2}{2}[a^2-{\varpi }_0^2{r}^2].$

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

$Q_0(x,z)$

$=$

$\frac{{\mathrm{\Omega }}^2}{2}[(2q-3)(a^2-x^2)-z^2],$

once it is appreciated that, in moving from the Blaes85 discussion to the GGN86 discussion, ${\varpi }_0^2{r}^2\to (x^2+z^2)$, and it is recognized that Blaes85 restricted his investigation to tori that have uniform specific angular momentum $(q=2)$.

Additional Notation

$(ky{)}_{\mathrm{G}\mathrm{G}\mathrm{N}}=(\frac{my}{{\varpi }_0}{)}_{\mathrm{G}\mathrm{G}\mathrm{N}}\leftrightarrow (m\varphi {)}_{\mathrm{B}\mathrm{l}\mathrm{a}\mathrm{e}\mathrm{s}}$

${\beta }_{\mathrm{G}\mathrm{G}\mathrm{N}}\equiv (\frac{ma}{{\varpi }_0}{)}_{\mathrm{G}\mathrm{G}\mathrm{N}}\leftrightarrow m{\beta }_{\mathrm{B}\mathrm{l}\mathrm{a}\mathrm{e}\mathrm{s}}$

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

$\psi$

$=$

$1+\frac{1}{4}{k}^2(5x^2-3z^2)\mp 4i(\frac{3}{2}{)}^{1/2}kx{\beta }_{\mathrm{G}\mathrm{G}\mathrm{N}}.$

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

$\mathrm{I}\mathrm{m}(\psi )$	$=$	$\mp 4(\frac{3}{2}{)}^{1/2}kx{\beta }_{\mathrm{G}\mathrm{G}\mathrm{N}}$
	$=$	$\mp 4\left(\frac{3}{2}{)}^{1/2}\right(\frac{m}{{\varpi }_0})[{\varpi }_0(\eta {\beta }_{\mathrm{B}\mathrm{l}\mathrm{a}\mathrm{e}\mathrm{s}})\cos \theta ](m{\beta }_{\mathrm{B}\mathrm{l}\mathrm{a}\mathrm{e}\mathrm{s}})$
	$=$	$\mp 4(\frac{3}{2}{)}^{1/2}{m}^2{\beta }_{\mathrm{B}\mathrm{l}\mathrm{a}\mathrm{e}\mathrm{s}}^2\eta \cos \theta ,$

which exactly matches $\mathrm{I}\mathrm{m}(f_m)$ as derived by Blaes85 and summarized above. Similarly,

$\mathrm{R}\mathrm{e}(\psi )$	$=$	$1+\frac{1}{4}{k}^2(5x^2-3z^2)$
	$=$	$1+\frac{1}{4}(\frac{m}{{\varpi }_0}{)}^2[{\varpi }_0^2{r}^2(5{\cos }^2\theta -3{\sin }^2\theta )]$
	$=$	$1+\frac{1}{4}{\eta }^2{m}^2{\beta }_{\mathrm{B}\mathrm{l}\mathrm{a}\mathrm{e}\mathrm{s}}^2[8{\cos }^2\theta -3]$
	$=$	$1+m^2{\beta }_{\mathrm{B}\mathrm{l}\mathrm{a}\mathrm{e}\mathrm{s}}^2[2{\eta }^2{\cos }^2\theta -\frac{3{\eta }^2}{4}].$

This exactly matches $\mathrm{R}\mathrm{e}(f_m)$ 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:

$f_m(\eta ,\theta )+\frac{1}{4}{\beta }^2{m}^2$

$=$

${\beta }^2{m}^2[2{\eta }^2{\cos }^2\theta -\frac{3{\eta }^2}{4}\pm 4i(\frac{3}{2}{)}^{1/2}\eta \cos \theta ].$

By comparison, from GGN86 we obtain:

$\mathrm{\Psi }(\eta ,\theta )-1$

$=$

$m^2{\beta }_{\mathrm{B}\mathrm{l}\mathrm{a}\mathrm{e}\mathrm{s}}^2[2{\eta }^2{\cos }^2\theta -\frac{3{\eta }^2}{4}\mp 4i(\frac{3}{2}{)}^{1/2}\eta \cos \theta ].$

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),

$F(\eta ,\theta )$	$=$	$\left\{\right[2{\eta }^2{\cos }^2\theta -\frac{3{\eta }^2}{4}{]}^2+\left[4\right(\frac{3}{2}{)}^{1/2}\eta \cos \theta {]}^2{\}}^{1/2}$
	$=$	$\left\{{\eta }^2\right[4^2\left(\frac{3}{2}\right){\cos }^2\theta ]+\frac{1}{4^2}{\eta }^4[8{\cos }^2\theta -3{]}^2{\}}^{1/2}$
	$=$	$\left\{\right(\frac{\eta }{2}{)}^2[2^5\cdot 3{\cos }^2\theta ]+\left(\frac{\eta }{2}{)}^4\right[8{\cos }^2\theta -3{]}^2{\}}^{1/2};$

and the associate phase function is,

$m{\varphi }_m$

$=$

${\tan }^{-1}\left[\frac{2^7\cdot 3{\cos }^2\theta }{{\eta }^2(8{\cos }^2\theta -3{)}^2}\right]+k\theta .$

See Also

• Imamura & Hadley collaboration:
  • Paper I: K. Hadley & J. N. Imamura (2011, Astrophysics and Space Science, 334, 1-26), "Nonaxisymmetric instabilities in self-gravitating disks.   I. Toroids" — In this paper, Hadley & Imamura perform linear stability analyses on fully self-gravitating toroids; that is, there is no central point-like stellar object and, hence, $M_{*}/M_d=0.0$.
  • Paper II: K. Z. Hadley, P. Fernandez, J. N. Imamura, E. Keever, R. Tumblin, & W. Dumas (2014, Astrophysics and Space Science, 353, 191-222), "Nonaxisymmetric instabilities in self-gravitating disks.   II. Linear and quasi-linear analyses" — In this paper, the Imamura & Hadley collaboration performs "an extensive study of nonaxisymmetric global instabilities in thick, self-gravitating star-disk systems creating a large catalog of star/disk systems … for star masses of $0.0\le M_{*}/M_d\le 10^3$ and inner to outer edge aspect ratios of $0.1<r_{-}/r_{+}<0.75$."
  • Paper III: K. Z. Hadley, W. Dumas, J. N. Imamura, E. Keever, & R. Tumblin (2015, Astrophysics and Space Science, 359, article id. 10, 23 pp.), "Nonaxisymmetric instabilities in self-gravitating disks.   III. Angular momentum transport" — In this paper, the Imamura & Hadley collaboration carries out nonlinear simulations of nonaxisymmetric instabilities found in self-gravitating star/disk systems and compares these results with the linear and quasi-linear modeling results presented in Papers I and II.