User:Tohline/Appendix/Ramblings/PPToriPt2

Stability Analyses of PP Tori (Part 2)

This is a direct extension of our Part 1 discussion. Here we continue our effort to check the validity of the Blaes85 eigenvector. The relevant reference is:

• Blaes (1985), MNRAS, 216, 553 (aka Blaes85) — Oscillations of slender tori.

Start From Scratch

Basic Equations from Blaes85

			Blaes85 Eq. No.
<math>~(\beta\eta)^2</math>	<math>~=</math>	<math>~x^2(1+xb) \, ;</math>	(2.6)
<math>~b</math>	<math>~\equiv</math>	<math>~3\cos\theta - \cos^3\theta \, ;</math>	(2.6)
<math>~f</math>	<math>~=</math>	<math>~1-\eta^2 \, .</math>	(2.5)

			Blaes85 Eq. No.
<math>~LHS \equiv \hat{L}W</math>	<math>~=</math>	<math> ~fx^2 \cdot \frac{\partial^2 W}{\partial x^2} + f \cdot \frac{\partial^2 W}{\partial \theta^2} + \biggl[ \frac{fx(1-2x\cos\theta)}{(1-x\cos\theta)} + nx^2\cdot \frac{\partial f}{\partial x}\biggr]\frac{\partial W}{\partial x} </math>
		<math> + \biggl[ \frac{fx\sin\theta}{(1-x\cos\theta)} + n\cdot \frac{\partial f}{\partial \theta}\biggr]\frac{\partial W}{\partial \theta} + \biggl[ \frac{2nx^2m^2}{\beta^2(1-x\cos\theta)^4} - \frac{m^2 x^2 f}{(1-x\cos\theta)^2} \biggr]W </math>	(4.2)
<math>~RHS</math>	<math>~=</math>	<math> ~-\frac{2nm^2}{\beta^2} \cdot (\beta\eta)^2 \biggl[ M \biggl(\frac{\nu}{m}\biggr)^2 + \frac{N}{m} \biggl(\frac{\nu}{m}\biggr)\biggr] W </math>	(4.1)
	<math>~=</math>	<math> ~-\frac{2nm^2}{\beta^2} \biggl[ x^2 \biggl(\frac{\nu}{m}\biggr)^2 + \frac{2x^2}{(1-x\cos\theta)^2} \biggl(\frac{\nu}{m}\biggr)\biggr] W </math>	(4.2)
<math>~\frac{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~i~\biggl[ \frac{2^3\cdot 3}{(n+1)}\biggr]^{1/2} \eta\cos\theta \biggr\} </math>	(4.13)
<math>~\frac{\nu}{m}</math>	<math>~=</math>	<math> ~-1 ~\pm ~ i~\biggl[ \frac{3}{2(n+1)} \biggr]^{1/2} \beta </math>	(4.14)

Our Manipulation of These Equations

<math>~\Lambda \equiv \frac{2^2(n+1)^2}{m^2}\biggl[\frac{W}{A_{00}}-1\biggr]</math>	<math>~=</math>	<math>~\beta^2 \biggl\{ 2^3(n+1)^2 \eta^2\cos^2\theta - 3\eta^2(n+1)^2 - (4n+1) ~\pm~i~[ 2^7\cdot 3(n+1)^3 ]^{1/2} \eta\cos\theta \biggr\} </math>
	<math>~=</math>	<math>~- (4n+1)\beta^2 + (\beta\eta)^2 (n+1)^2[ 2^3 \cos^2\theta - 3] ~\pm~i~\beta [ 2^7\cdot 3(n+1)^3 ]^{1/2} (\beta\eta) \cos\theta \, ; </math>
<math>~\Rightarrow~~~~\frac{W}{A_{00}} </math>	<math>~=</math>	<math>~1+ \biggl[ \frac{m}{2(n+1)} \biggr]^2 \Lambda </math>

<math>~\frac{LHS}{A_{00}} </math>	<math>~=</math>	<math>~\biggl[ \frac{m}{2(n+1)} \biggr]^2 f ~\biggl[ x^2 \cdot \frac{\partial^2 \Lambda}{\partial x^2} + \frac{\partial^2 \Lambda}{\partial \theta^2}\biggr] + \biggl[ \frac{m}{2(n+1)} \biggr]^2\biggl[ \frac{fx(1-2x\cos\theta)}{(1-x\cos\theta)} + nx^2\cdot \frac{\partial f}{\partial x}\biggr]\frac{\partial \Lambda}{\partial x} </math>
		<math> + \biggl[ \frac{m}{2(n+1)} \biggr]^2\biggl[ \frac{fx\sin\theta}{(1-x\cos\theta)} + n\cdot \frac{\partial f}{\partial \theta}\biggr]\frac{\partial \Lambda}{\partial \theta} + \biggl[ \frac{2nx^2m^2}{\beta^2(1-x\cos\theta)^4} - \frac{m^2 x^2 f}{(1-x\cos\theta)^2} \biggr]\biggl\{1+ \biggl[ \frac{m}{2(n+1)} \biggr]^2 \Lambda\biggr\} </math>
	<math>~=</math>	<math>~\biggl[ \frac{m}{2(n+1)} \biggr]^2 f \biggl\{ ~\biggl[ x^2 \cdot \frac{\partial^2 \Lambda}{\partial x^2} + \frac{\partial^2 \Lambda}{\partial \theta^2}\biggr] + \biggl[ \frac{x(1-2x\cos\theta)}{(1-x\cos\theta)} \biggr]\frac{\partial \Lambda}{\partial x} + \biggl[ \frac{x\sin\theta}{(1-x\cos\theta)} \biggr]\frac{\partial \Lambda}{\partial \theta} - \biggl[ \frac{m^2 x^2 }{(1-x\cos\theta)^2} \biggr] \biggl[ \frac{2^2(n+1)^2}{m^2} + \Lambda\biggr]\biggr\} </math>
		<math> + n\biggl[ \frac{m}{2(n+1)} \biggr]^2 \biggl\{ x^2\cdot \frac{\partial f}{\partial x}\cdot \frac{\partial \Lambda}{\partial x} ~+~ \frac{\partial f}{\partial \theta}\cdot \frac{\partial \Lambda}{\partial \theta} ~+~ \biggl[ \frac{2x^2m^2}{\beta^2(1-x\cos\theta)^4} \biggr]\biggl[ \frac{2^2(n+1)^2}{m^2} + \Lambda\biggr]\biggr\} </math>
	<math>~=</math>	<math>~\frac{x^2 f}{(1-x\cos\theta)^2} \biggl[ \frac{m}{2(n+1)} \biggr]^2 \biggl\{ ~(1-x\cos\theta)^2\biggl[ \frac{\partial^2 \Lambda}{\partial x^2} + \frac{1}{x^2}\cdot \frac{\partial^2 \Lambda}{\partial \theta^2}\biggr] + \frac{(1-x\cos\theta)}{x} \biggl[ (1-2x\cos\theta) \frac{\partial \Lambda}{\partial x} + \sin\theta\cdot \frac{\partial \Lambda}{\partial \theta} \biggr] - [ 2^2(n+1)^2 + m^2\Lambda ]\biggr\} </math>
		<math> + ~\frac{x^2 n}{\beta^2(1-x\cos\theta)^4} \biggl[ \frac{m}{2(n+1)} \biggr]^2 \biggl\{\beta^2 (1-x\cos\theta)^4\biggl[ \frac{\partial f}{\partial x}\cdot \frac{\partial \Lambda}{\partial x} ~+~ \frac{1}{x^2}\cdot \frac{\partial f}{\partial \theta}\cdot \frac{\partial \Lambda}{\partial \theta} \biggr] ~+~ [ 2^3(n+1)^2 + 2m^2\Lambda ]\biggr\} \, . </math>

Also,

<math>~\frac{RHS}{A_{00}}</math>	<math>~=</math>	<math> ~-\frac{2n x^2}{\beta^2(1-x\cos\theta)^2} \biggl[ \frac{m}{2(n+1)} \biggr]^2 \biggl[ (1-x\cos\theta)^2\biggl(\frac{\nu}{m}\biggr)^2 + 2\biggl(\frac{\nu}{m}\biggr)\biggr] [ 2^2(n+1)^2 + m^2\Lambda ] </math>
	<math>~=</math>	<math> ~-\frac{x^2n}{\beta^2(1-x\cos\theta)^4} \biggl[ \frac{m}{2(n+1)} \biggr]^2 \biggl[ (1-x\cos\theta)^4\biggl(\frac{\nu}{m}\biggr)^2 + 2(1-x\cos\theta)^2\biggl(\frac{\nu}{m}\biggr)\biggr] [ 2^3(n+1)^2 + 2m^2\Lambda ] \, . </math>

Putting the two together implies,

Definition of Eigenvalue Problem Associated with the Stability of Slim, Papaloizou-Pringle Tori

<math>~0</math> <math>~=</math> <math>~\frac{1}{x^2}\biggl[\frac{LHS}{A_{00}} - \frac{RHS}{A_{00}}\biggr]\biggl[ \frac{2(n+1)}{m} \biggr]^2 (1-x\cos\theta)^4</math>   <math>~=</math> <math>~f (1-x\cos\theta)^2 \biggl\{ ~(1-x\cos\theta)^2\biggl[ \frac{\partial^2 \Lambda}{\partial x^2} + \frac{1}{x^2}\cdot \frac{\partial^2 \Lambda}{\partial \theta^2}\biggr] + \frac{(1-x\cos\theta)}{x} \biggl[ (1-2x\cos\theta) \frac{\partial \Lambda}{\partial x} + \sin\theta\cdot \frac{\partial \Lambda}{\partial \theta} \biggr] - [ 2^2(n+1)^2 + m^2\Lambda ]\biggr\} </math>     <math> + ~\frac{n}{\beta^2} \biggl\{ (1-x\cos\theta)^4\biggl[ \frac{\partial \Lambda}{\partial x} \cdot \frac{\partial (\beta^2 f)}{\partial x} ~+~ \frac{\partial \Lambda}{\partial \theta} \cdot \frac{\partial (\beta^2 f/x^2)}{\partial \theta} \biggr] ~+~ \biggl[ (1-x\cos\theta)^4\biggl(\frac{\nu}{m}\biggr)^2 + 2(1-x\cos\theta)^2\biggl(\frac{\nu}{m}\biggr)+ 1 \biggr] [ 2^3(n+1)^2 + 2m^2\Lambda ] \biggr\} \, . </math>

The first line of this governing, two-line expression contains the function, <math>~f</math>, as a leading factor, while the leading factor in the second line is the ratio, <math>~n/\beta^2</math>. Presumably the three terms (hereafter, TERM1, TERM2, & TERM3, respectively) inside the curly brackets on the first line must cancel — to a sufficiently high order in <math>~x</math> — and, independently, the two terms (hereafter, TERM4 & Term5, respectively) inside the curly brackets on the second line must cancel. Furthermore, these cancellations must occur separately for the real parts and the imaginary parts of each bracketed expression.

Evaluating various terms using the parameter set,    <math>~(n, \theta, x/\beta) = (1, \tfrac{\pi}{3}, \tfrac{1}{4})</math>    as begun in our "Part 1" analysis, we have:

TERM4	<math>~=</math>	<math>~ -x \ell^4\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 \ell^4\biggl[ (2+3xb)\cdot [~1.515625000~\pm~i~36.23373732 ~\beta] - 3\sin^3\theta \cdot [~-2.388335684~\pm~i~(-1)15.36617018 ~\beta] \biggr] </math>
	<math>~=</math>	<math>~ -x\ell^4 [~9.248046874~\pm~i~139.7753772~\beta] </math>

TERM5 (Case B)	<math>~=</math>	<math>~ \biggl[ \ell^4 [1-0.75\beta^2~\pm~i~(-1)\sqrt{3}\beta] +2\ell^2[ -1~\pm~i~\sqrt{0.75}\beta ] + 1 \biggr] \cdot \biggl[~2^5 + 2\cancelto{1}{m^2}[~- 5\beta + 0.167968750~\pm~i~8.031189202 ~\beta]~\biggr] </math>
	<math>~=</math>	<math>~ \biggl[ \ell^4 [1-0.75\beta^2] - 2\ell^2 + 1 \biggr] \cdot \biggl[~[2^5 - 10\beta + (2)0.167968750]~\pm~i~[(2)8.031189202 ~\beta]~\biggr] </math>
		<math>~ \pm~\sqrt{3}\beta\biggl[ \ell^2-~\ell^4 \biggr] \cdot \biggl[~i~ [2^5 - 10\beta + (2)0.167968750]~-~[(2)8.031189202 ~\beta]~\biggr] </math>
	<math>~=</math>	<math>~ \biggl[ \ell^4 [1-0.75\beta^2] - 2\ell^2 + 1 \biggr] \cdot \biggl[~[2^5 - 10\beta + (2)0.167968750]\biggr] \pm~(-1)\sqrt{3}\beta\biggl[ \ell^2-~\ell^4 \biggr] \cdot \biggl[[(2)8.031189202 ~\beta]~\biggr] </math>
		<math>~ \pm~i~\biggl\{\biggl[ \ell^4 [1-0.75\beta^2] - 2\ell^2 + 1 \biggr] \cdot \biggl[[(2)8.031189202 ~\beta]~\biggr] +~\sqrt{3}\beta\biggl[ \ell^2-~\ell^4 \biggr] \cdot \biggl[~ [2^5 - 10\beta + (2)0.167968750]~\biggr] \biggr\} \, . </math>

Evaluating this TERM5 expression for the case of <math>~\beta = 1</math>, we have,

TERM5 (Case B)	<math>~=</math>	<math>~ \biggl[ 0.25\ell^4 - 2\ell^2 + 1 \biggr] \cdot \biggl[~[2^5 - 10 + (2)0.167968750]\biggr] \pm~(-1)\sqrt{3}\biggl[ \ell^2-~\ell^4 \biggr] \cdot \biggl[[(2)8.031189202 ]~\biggr] </math>
		<math>~ \pm~i~\biggl\{\biggl[ \ell^4 [1-0.75] - 2\ell^2 + 1 \biggr] \cdot \biggl[[(2)8.031189202]~\biggr] +~\sqrt{3}\biggl[ \ell^2-~\ell^4 \biggr] \cdot \biggl[~ [2^5 - 10 + (2)0.167968750]~\biggr] \biggr\} </math>
	<math>~=</math>	<math>~ [ -0.38470459 ] \cdot [22.3359375] \pm~(-1)[ ~0.31080502 ] \cdot [~16.0623784 ~] </math>
		<math>~ \pm~i~\biggl\{[ -0.38470459 ] \cdot [~16.0623784 ~] +~[ ~0.31080502 ] \cdot [22.3359375]\biggr\} </math>
	<math>~=</math>	<math>~[~-13.58500545~] \pm~i~[~0.76285080~] \, . </math>

Testing for Expected Cancellations

Note first that, adopting the shorthand notation,

<math>~\ell</math>	<math>~\equiv</math>	<math>~(1-x\cos\theta)</math>
<math>~\Rightarrow ~~~~\ell^2</math>	<math>~=</math>	<math>~1-2\beta \biggl(\frac{x}{\beta}\biggr)\cos\theta + \beta^2 \biggl(\frac{x}{\beta}\biggr)^2\cos^2\theta + \mathcal{O}(\beta^3) \, ;</math>
<math>~\ell^3</math>	<math>~=</math>	<math>~1-3\beta \biggl(\frac{x}{\beta}\biggr)\cos\theta + 3\beta^2 \biggl(\frac{x}{\beta}\biggr)^2\cos^2\theta + \mathcal{O}(\beta^3) \, ;</math>
<math>~\ell^4</math>	<math>~=</math>	<math>~1-4\beta \biggl(\frac{x}{\beta}\biggr)\cos\theta + 6\beta^2\biggl(\frac{x}{\beta}\biggr)^2\cos^2\theta + \mathcal{O}(\beta^3) \, .</math>

Real Parts

<math>~\mathrm{Re}\biggl[\frac{\mathrm{TERM4}}{\ell^4}\biggr]</math>	<math>~=</math>	<math>~ \biggl\{ (n+1)[2^3(n+1)\cos^2\theta -3]x(2+3xb)\biggr\} \cdot \biggl[ -x(2+3xb) \biggr] </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\} \cdot \biggl[ 3x\sin^3\theta \biggr] </math>
	<math>~=</math>	<math>~ -~(n+1)[2^3(n+1)\cos^2\theta -3]x^2(2+3xb)^2 </math>
		<math>~ -~ 3x^3(n+1)\sin^4\theta \biggl\{ 2^4 (n+1) (1+xb) \cos\theta + 3x \sin^2\theta [2^3(n+1)\cos^2\theta -3] \biggr\} </math>
	<math>~=</math>	<math>~ -~x^2 \cdot 2^2 (n+1)[2^3(n+1)\cos^2\theta -3]\biggl(1+\frac{3xb}{2}\biggr)^2 </math>
		<math>~ -~ x^3 \cdot 2^4\cdot 3(n+1)^2 \cos\theta\sin^4\theta (1+xb) ~-~x^4\cdot 3^2(n+1)\sin^6\theta [2^3(n+1)\cos^2\theta -3] \, .</math>
	<math>~=</math>	<math>~ -x\biggl\{~x[~18.37695315~] + x^2[~72.5625~] + x^3[~7.59375~]~~\biggr\} = -x[~9.24804688~]\, . </math>

<math>~\mathrm{Re}\biggl[\mathrm{TERM5}\biggr]</math>	<math>~=</math>	<math>~ \mathrm{Re}\biggl[ \ell^4\biggl(\frac{\nu}{m}\biggr)^2 + 2\ell^2\biggl(\frac{\nu}{m}\biggr)+ 1 \biggr] \cdot \mathrm{Re}[ 2^3(n+1)^2 + 2m^2\Lambda ] -\mathrm{Im}\biggl[ \ell^4\biggl(\frac{\nu}{m}\biggr)^2 + 2\ell^2\biggl(\frac{\nu}{m}\biggr)+ 1 \biggr] \cdot \mathrm{Im}[ 2^3(n+1)^2 + 2m^2\Lambda ] </math>
Case B:	<math>~=</math>	<math>~ \biggl\{ \ell^4\biggl[1-\frac{3\beta^2}{2(n+1)}\biggr] + 2\ell^2\biggl(-1\biggr)+ 1 \biggr\} \cdot \biggl\{ 2^3(n+1)^2 + 2m^2\biggl[ ~- (4n+1)\beta^2 + (n+1)^2(2^3 \cos^2\theta - 3) x^2(1+xb)\biggr] \biggr\} </math>
		<math>~ -~\biggl\{ \ell^4(-1)\biggl[\frac{2\cdot 3\beta^2}{(n+1)}\biggr]^{1/2} + 2\ell^2\biggl[ \frac{3\beta^2}{2(n+1)}\biggr]^{1/2} \biggr\} \cdot 2m^2\beta [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cos\theta \cdot x(1+xb)^{1/2} </math>
	<math>~=</math>	<math>~ \biggl\{1 - 2\ell^2 + \ell^4-\frac{3\beta^2\ell^4}{2(n+1)} \biggr\} \cdot \biggl\{ \biggl[ 2^3(n+1)^2 - 2m^2(4n+1)\beta^2\biggr] + x^2\cdot 2m^2(n+1)^2(2^3 \cos^2\theta - 3) (1+xb) \biggr\} </math>
		<math>~ -~x\beta^2 \cdot m^2[\ell^2 - \ell^4 ] \cdot [ 2^{10}\cdot 3^2(n+1)^2 ]^{1/2} \cos\theta (1+xb)^{1/2} \, . </math>

When added together, we obtain,

<math>~\mathrm{Re}[\mathrm{TERM4} + \mathrm{TERM5}]</math>	<math>~=</math>	<math>~ -~\beta^2 \biggl(\frac{x}{\beta}\biggr)^2 \ell^4\cdot 2^2 (n+1)[2^3(n+1)\cos^2\theta -3 ]\biggl(1+\frac{3xb}{2}\biggr)^2 </math>
		<math>~ -~ \beta^3 \biggl(\frac{x}{\beta}\biggr)^3\ell^4\cdot 2^4\cdot 3(n+1)^2 \cos\theta\sin^4\theta (1+xb) ~-~\beta^4\biggl(\frac{x}{\beta}\biggr)^4 \ell^4\cdot 3^2(n+1)\sin^6\theta [2^3(n+1)\cos^2\theta-3] </math>
		<math>~ +~\biggl\{1 - 2\ell^2 + \ell^4 \biggr\} \cdot \biggl\{ 2^3(n+1)^2 + 2m^2\beta^2\biggr[ - (4n+1) + \biggl(\frac{x}{\beta}\biggr)^2(n+1)^2(2^3 \cos^2\theta - 3) (1+xb) \biggr]\biggr\} </math>
		<math>~ -~\frac{3\beta^2\ell^4}{2(n+1)} \biggl\{ 2^3(n+1)^2 + 2m^2\beta^2\biggr[ - (4n+1) + \biggl(\frac{x}{\beta}\biggr)^2(n+1)^2(2^3 \cos^2\theta - 3) (1+xb) \biggr]\biggr\} </math>
		<math>~ -~\beta^3\biggl(\frac{x}{\beta}\biggr) \cdot m^2[\ell^2 - \ell^4 ] \cdot [ 2^{10}\cdot 3^2(n+1)^2 ]^{1/2} \cos\theta (1+xb)^{1/2} </math>
	<math>~=</math>	<math>~ \beta^0 \cdot 2^3(n+1)^2\biggl\{1 - 2\ell^2 + \ell^4 \biggr\} </math>
		<math>~ -~\beta^2 \cdot 2m^2 [ 1 - 2\ell^2 + \ell^4 ] \cdot \biggr[ (4n+1) - \biggl(\frac{x}{\beta}\biggr)^2(n+1)^2(2^3 \cos^2\theta - 3) (1+xb) \biggr] </math>
		<math>~ -~\beta^2\ell^4 2^2\cdot 3 (n+1) + \beta^2 \biggl(\frac{x}{\beta}\biggr)^2 \ell^4\cdot 2^2 (n+1)[3 - 2^3(n+1)\cos^2\theta ]\biggl(1+\frac{3xb}{2}\biggr)^2 </math>
		<math>~ -~\cancelto{0}{\beta^3}\biggl(\frac{x}{\beta}\biggr) \cdot m^2[\ell^2 - \ell^4 ] \cdot [ 2^{10}\cdot 3^2(n+1)^2 ]^{1/2} \cos\theta (1+xb)^{1/2} </math>
		<math>~ -~ \cancelto{0}{\beta^3} \biggl(\frac{x}{\beta}\biggr)^3\ell^4\cdot 2^4\cdot 3(n+1)^2 \cos\theta\sin^4\theta (1+xb) ~-~\cancelto{0}{\beta^4}\biggl(\frac{x}{\beta}\biggr)^4 \ell^4\cdot 3^2(n+1)\sin^6\theta [2^3(n+1)\cos^2\theta-3] </math>
		<math>~ +~\frac{3\cancelto{0}{\beta^4}\ell^4 m^2}{(n+1)} \biggr[ (4n+1) - \biggl(\frac{x}{\beta}\biggr)^2(n+1)^2(2^3 \cos^2\theta - 3) (1+xb) \biggr] </math>
	<math>~\approx</math>	<math>~ \beta^0 \cdot 2^3(n+1)^2\biggl\{1 - 2\biggl[ 1-2\beta \biggl(\frac{x}{\beta}\biggr)\cos\theta + \beta^2 \biggl(\frac{x}{\beta}\biggr)^2\cos^2\theta + \cancelto{0}{\mathcal{O}(\beta^3)}\biggr] + \biggl[ 1-4\beta \biggl(\frac{x}{\beta}\biggr)\cos\theta + 6\beta^2\biggl(\frac{x}{\beta}\biggr)^2\cos^2\theta + \cancelto{0}{\mathcal{O}(\beta^3)} \biggr] \biggr\} </math>
		<math>~ -~\beta^2 \cdot 2m^2 [ 1 - 2 + 1 ] \cdot \biggr[ (4n+1) - \biggl(\frac{x}{\beta}\biggr)^2(n+1)^2(2^3 \cos^2\theta - 3) (1+\cancelto{0}{x}b) \biggr] </math>
		<math>~ -~\beta^2 2^2\cdot 3 (n+1) + \beta^2 \biggl(\frac{x}{\beta}\biggr)^2 2^2 (n+1)[3 - 2^3(n+1)\cos^2\theta ]\biggl(1+\frac{3\cancelto{0}{x}b}{2}\biggr)^2 </math>
	<math>~\approx</math>	<math>~ \beta^0 \cdot 2^3(n+1)^2\biggl\{1 - 2+ 1 \biggr\} +~\beta^1 \biggl(\frac{x}{\beta}\biggr) \cdot 2^3(n+1)^2\biggl\{4\cos\theta -4\cos\theta \biggr\} </math>
		<math>~ +~\beta^2 \biggl(\frac{x}{\beta}\biggr)^2 \cdot 2^5(n+1)^2 \cos^2\theta </math>
		<math>~ -~\beta^2 \cdot 2m^2 [ 1 - 2 + 1 ] \cdot \biggr[ (4n+1) - \biggl(\frac{x}{\beta}\biggr)^2(n+1)^2(2^3 \cos^2\theta - 3) \biggr] </math>
		<math>~ -~\beta^2 2^2\cdot 3 (n+1) \biggl[1 - \biggl(\frac{x}{\beta}\biggr)^2\biggr] - \beta^2 \biggl(\frac{x}{\beta}\biggr)^2 [2^5(n+1)^2\cos^2\theta ] </math>
	<math>~=</math>	<math>~-~\beta^2 2^2\cdot 3 (n+1) \biggl[1 - \biggl(\frac{x}{\beta}\biggr)^2\biggr] \, .</math>

So we see that the coefficients of the lowest-order <math>(\beta^0 ~\mathrm{and} ~ \beta^1)</math> terms are zero, and the coefficient of the <math>~\beta^2</math> term is almost zero!

Imaginary Parts

TERM4

<math>~\mathrm{Im}\biggl[\frac{\mathrm{TERM4}}{\ell^4}\biggr]</math>	<math>~=</math>	<math>~ \biggl\{ \beta\cos\theta [2^5\cdot 3 (n+1)^3]^{1/2} \cdot \frac{x(2+3xb)}{(\beta\eta)}\biggr\} \cdot \biggl[ -x(2+3xb) \biggr] </math>
		<math>~ -~ \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\} \cdot \biggl[ 3x\sin^3\theta \biggr] </math>
	<math>~=</math>	<math>~ -~x \cdot 2\beta\cos\theta [2^7\cdot 3 (n+1)^3]^{1/2} \cdot (1+xb)^{-1/2}\cdot \biggl(1+\frac{3xb}{2}\biggr)^2 </math>
		<math>~ -~ x^2\cdot 3\beta \sin^4\theta [2^7\cdot 3 (n+1)^3 ]^{1/2} (1+xb)^{1/2} \biggl\{ 1 +\frac{3x}{2}\cdot\biggl[ \frac{\sin^2\theta \cos\theta}{(1+xb)} \biggr]\biggr\} </math>
	<math>~=</math>	<math>~ -x\biggl\{~[~109.8335164~] + x[~119.7674436~]~\biggr\}= -34.94384433 </math>

Alternatively we can write,

<math>~\mathrm{Im}\biggl[\frac{\mathrm{TERM4}}{\ell^4}\biggr]</math>	<math>~=</math>	<math>~ \biggl\{ \beta\cos\theta [2^5\cdot 3 (n+1)^3]^{1/2} \cdot \frac{x(2+3xb)}{(\beta\eta)}\biggr\} \cdot \biggl[ -x(2+3xb) \biggr] </math>
		<math>~ -~ \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\} \cdot \biggl[ 3x\sin^3\theta \biggr] </math>
	<math>~=</math>	<math>~ -2b_0 \beta^2 \biggl(\frac{x}{\beta}\biggr) \biggl(1+\frac{3xb}{2} \biggr)^2 (1+xb)^{-1/2} -~ 3b_0\beta^3 \biggl(\frac{x}{\beta}\biggr)^2 \biggl[\frac{\sin^4\theta}{\cos\theta}\biggr] (1 + xb)^{1/2} </math>
		<math>~ -~ \frac{9b_0}{2} \cdot \beta^4 \biggl(\frac{x}{\beta}\biggr)^3 \sin^6\theta (1 + xb)^{-1/2} </math>

<math>~\Rightarrow ~~~ \mathrm{Im}\biggl[\frac{\mathrm{TERM4}}{\beta^2}\biggr]</math>	<math>~=</math>	<math>~\biggl\{ -2b_0 \biggl(\frac{x}{\beta}\biggr) \biggl(1+\frac{3xb}{2} \biggr)^2 (1+xb)^{-1/2} -~ 3b_0\beta \biggl(\frac{x}{\beta}\biggr)^2 \biggl[\frac{\sin^4\theta}{\cos\theta}\biggr] (1 + xb)^{1/2} -~ \frac{9b_0}{2} \cdot \beta^2 \biggl(\frac{x}{\beta}\biggr)^3 \sin^6\theta (1 + xb)^{-1/2} \biggr\} </math>
		<math>~ \times \biggl\{ 1 -4\beta\biggl(\frac{x}{\beta}\biggr)\cos\theta + 6\beta^2\biggl(\frac{x}{\beta}\biggr)^2\cos^2\theta -4\beta^3\biggl(\frac{x}{\beta}\biggr)^3\cos^3\theta + \biggl(\frac{x}{\beta}\biggr)^4\cos^4\theta \biggr\} </math>
<math>~</math>	<math>~=</math>	<math>~ \biggl\{ ~-27.45837910~-6.77631589 ~-0.70914934~ \biggr\}\times [~0.58618164~] =\biggl\{ ~-34.94384433~ \biggr\}\times [~0.58618164~] = -20.48343998 </math>
	<math>~\approx</math>	<math>~ -2b_0 \biggl(\frac{x}{\beta}\biggr) \biggl(1+\frac{3xb}{2} \biggr)^2 (1+xb)^{-1/2} -~ 3b_0\beta \biggl(\frac{x}{\beta}\biggr)^2 \biggl[\frac{\sin^4\theta}{\cos\theta}\biggr] (1 + xb)^{1/2} -~ \frac{9b_0}{2} \cdot \beta^2 \biggl(\frac{x}{\beta}\biggr)^3 \sin^6\theta (1 + xb)^{-1/2} </math>
		<math>~+~ 8b_0 \beta \biggl(\frac{x}{\beta}\biggr)^2 \biggl(1+\frac{3xb}{2} \biggr)^2 (1+xb)^{-1/2} \cos\theta +~ 12b_0\beta^2 \biggl(\frac{x}{\beta}\biggr)^3 \biggl[\frac{\sin^4\theta}{\cos\theta}\biggr] (1 + xb)^{1/2} \cos\theta </math>
		<math>~ -12b_0 \beta^2 \biggl(\frac{x}{\beta}\biggr)^3 \biggl(1+\frac{3xb}{2} \biggr)^2 (1+xb)^{-1/2} \cos^2\theta </math>
	<math>~\approx</math>	<math>~ -2b_0 \biggl(\frac{x}{\beta}\biggr) \biggl(1+\frac{3xb}{2} \biggr)^2 (1+xb)^{-1/2} +~b_0 \beta \biggl(\frac{x}{\beta}\biggr)^2 \biggl\{ 8\biggl(1+\frac{3xb}{2} \biggr)^2 (1+xb)^{-1/2} \cos\theta -~ 3 \biggl[\frac{\sin^4\theta}{\cos\theta}\biggr] (1 + xb)^{1/2} \biggr\} </math>
		<math>~+\beta^2 b_0\biggl(\frac{x}{\beta}\biggr)^3\biggl\{ -~ \frac{9}{2} \cdot \sin^6\theta (1 + xb)^{-1/2} +~ 12 \biggl[\frac{\sin^4\theta}{\cos\theta}\biggr] (1 + xb)^{1/2} \cos\theta -12 \biggl(1+\frac{3xb}{2} \biggr)^2 (1+xb)^{-1/2} \cos^2\theta \biggr\} </math>
	<math>~\approx</math>	<math>~ -2b_0 \biggl(\frac{x}{\beta}\biggr) \biggl(1+\frac{3xb}{2} \biggr)^2 (1+xb)^{-1/2} +~b_0 \beta \biggl(\frac{x}{\beta}\biggr)^2 \biggl\{ 8 \cos\theta -~ 3 \biggl[\frac{\sin^4\theta}{\cos\theta}\biggr] \biggr\} \, . </math>

TERM5

<math>~\mathrm{Im}\biggl[\mathrm{TERM5}\biggr]</math>	<math>~=</math>	<math>~ \mathrm{Re}\biggl[ \ell^4\biggl(\frac{\nu}{m}\biggr)^2 + 2\ell^2\biggl(\frac{\nu}{m}\biggr)+ 1 \biggr] \cdot \mathrm{Im}[ 2^3(n+1)^2 + 2m^2\Lambda ] +\mathrm{Im}\biggl[ \ell^4\biggl(\frac{\nu}{m}\biggr)^2 + 2\ell^2\biggl(\frac{\nu}{m}\biggr)+ 1 \biggr] \cdot \mathrm{Re}[ 2^3(n+1)^2 + 2m^2\Lambda ] </math>
Case B:	<math>~=</math>	<math>~ x\cdot 2 \beta m^2 \biggl\{1 - 2\ell^2 + \ell^4 -\frac{3\beta^2\ell^4}{2(n+1)} \biggr\} \cdot [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cos\theta \cdot (1+xb)^{1/2} </math>
		<math>~ +~\beta \biggl[ \frac{2\cdot 3}{(n+1)}\biggr]^{1/2} [\ell^2 -\ell^4] \cdot \biggl\{ \biggl[ 2^3(n+1)^2 ~- 2m^2(4n+1)\beta^2\biggr] + x^2 \cdot 2m^2(n+1)[2^3(n+1) \cos^2\theta - 3] (1+xb) \biggr\} </math>
	<math>~=</math>	<math>~ \cancelto{1}{m^2} \biggl\{1 - 2\ell^2 + \ell^4 -\frac{3\beta^2\ell^4}{2(n+1)} \biggr\} \cdot 2 \beta x[ ~ 32.12475681~] +~\sqrt{3}\beta [\ell^2 -\ell^4] \cdot \biggl\{ \biggl[ 2^5 ~- 10\cancelto{1}{m^2}\beta^2\biggr] + 2m^2x^2 \cdot [ ~2.6875~ ] \biggr\} </math>
	<math>~=</math>	<math>~ \cancelto{1}{m^2} \biggl\{~-0.38470459~\biggr\} \cdot [ ~16.06237841~] +~[~0.31080502~] \cdot \biggl\{ 22.3359375\biggr\}= 0.76285080 \, . </math>

Let's rewrite both of these expressions in terms of a power series in <math>~\beta</math>.

<math>~\mathrm{Im}\biggl[\mathrm{TERM5}\biggr]</math>	<math>~=</math>	<math>~ \beta^2\biggl(\frac{x}{\beta}\biggr)\cdot 2 m^2 b_0 \biggl\{1 - 2\biggl[1 -2\beta\biggl(\frac{x}{\beta}\biggr)\cos\theta + \beta^2\biggl(\frac{x}{\beta}\biggr)^2\cos^2\theta + \mathcal{O}(\beta^3) \biggr] </math>
		<math>~ + \biggl[1 -4\beta\biggl(\frac{x}{\beta}\biggr)\cos\theta + 6\beta^2\biggl(\frac{x}{\beta}\biggr)^2\cos^2\theta + \mathcal{O}(\beta^3) \biggr]\biggl[1 -\frac{3\beta^2}{2(n+1)} \biggr]\biggr\} \cdot \biggl\{ 1 +\beta\biggl(\frac{x}{\beta}\biggr)\frac{b}{2} - \beta^2\biggl(\frac{x}{\beta}\biggr)^2\frac{b^2}{8} + \mathcal{O}(\beta^3)\biggr\} </math>
		<math>~ +~\beta \biggl[ \frac{2\cdot 3}{(n+1)}\biggr]^{1/2} \biggl[ \beta^0(1-1) + 2\beta\biggl(\frac{x}{\beta}\biggr)\cos\theta - 5\beta^2\biggl(\frac{x}{\beta}\biggr)^2\cos^2\theta +4\beta^3 \biggl(\frac{x}{\beta}\biggr)^3\cos^3\theta + \mathcal{O}(\beta^4)\biggr] \cdot \biggl\{ 2^3(n+1)^2 \biggr\} </math>
		<math>~ +~\beta \biggl[ \frac{2\cdot 3}{(n+1)}\biggr]^{1/2} \biggl[ \beta^0(1-1) + 2\beta\biggl(\frac{x}{\beta}\biggr)\cos\theta - 5\beta^2\biggl(\frac{x}{\beta}\biggr)^2\cos^2\theta +4\beta^3 \biggl(\frac{x}{\beta}\biggr)^3\cos^3\theta + \mathcal{O}(\beta^4)\biggr] \cdot \biggl\{ ~- 2m^2(4n+1)\beta^2 \biggr\} </math>
		<math>~ +~\beta \biggl[ \frac{2\cdot 3}{(n+1)}\biggr]^{1/2} \biggl[ \beta^0(1-1) + 2\beta\biggl(\frac{x}{\beta}\biggr)\cos\theta - 5\beta^2\biggl(\frac{x}{\beta}\biggr)^2\cos^2\theta +4\beta^3 \biggl(\frac{x}{\beta}\biggr)^3\cos^3\theta + \mathcal{O}(\beta^4)\biggr] \cdot \biggl\{ x^2 \cdot 2m^2(n+1)[2^3(n+1) \cos^2\theta - 3] \biggr\} </math>
		<math>~ +~\beta \biggl[ \frac{2\cdot 3}{(n+1)}\biggr]^{1/2} \biggl[ \beta^0(1-1) + 2\beta\biggl(\frac{x}{\beta}\biggr)\cos\theta - 5\beta^2\biggl(\frac{x}{\beta}\biggr)^2\cos^2\theta +4\beta^3 \biggl(\frac{x}{\beta}\biggr)^3\cos^3\theta + \mathcal{O}(\beta^4)\biggr] \cdot \biggl\{ x^3 b \cdot 2m^2(n+1)[2^3(n+1) \cos^2\theta - 3] \biggr\} </math>

<math>~\Rightarrow~~~\mathrm{Im}\biggl[\frac{\mathrm{TERM5}}{\beta^2}\biggr]</math>	<math>~=</math>	<math>~ \biggl(\frac{x}{\beta}\biggr)\cdot 2 m^2 b_0 \biggl\{\beta^0(1-2+1) +4\beta\biggl(\frac{x}{\beta}\biggr)\cos\theta -2 \beta^2\biggl(\frac{x}{\beta}\biggr)^2\cos^2\theta -4\beta\biggl(\frac{x}{\beta}\biggr)\cos\theta + 6\beta^2\biggl(\frac{x}{\beta}\biggr)^2\cos^2\theta -\frac{3\beta^2}{2(n+1)} + \mathcal{O}(\beta^3) \biggr\} </math>
		<math>~ \times \biggl\{ 1 +\beta\biggl(\frac{x}{\beta}\biggr)\frac{b}{2} - \beta^2\biggl(\frac{x}{\beta}\biggr)^2\frac{b^2}{8} + \mathcal{O}(\beta^3)\biggr\} </math>
		<math>~ +~b_0\biggl[ \frac{(1-1)}{\beta\cos\theta} + 2\beta^0\biggl(\frac{x}{\beta}\biggr) - 5\beta\biggl(\frac{x}{\beta}\biggr)^2\cos\theta +4\beta^2 \biggl(\frac{x}{\beta}\biggr)^3\cos^2\theta + \mathcal{O}(\beta^3)\biggr] </math>
		<math>~ -~ m^2(4n+1)\cdot \biggl[ \frac{2^3\cdot 3}{(n+1)}\biggr]^{1/2} \biggl[ \beta^{1}(1-1) + 2\beta^2\biggl(\frac{x}{\beta}\biggr)\cos\theta - 5\beta^3 \biggl(\frac{x}{\beta}\biggr)^2\cos^2\theta +4\beta^4 \biggl(\frac{x}{\beta}\biggr)^3\cos^3\theta + \mathcal{O}(\beta^5)\biggr] </math>
		<math>~ +~m^2[2^3(n+1) \cos^2\theta - 3] \cdot [ 2^3\cdot 3(n+1) ]^{1/2} \biggl[ \beta^1\biggl( \frac{x}{\beta}\biggr)^2(1-1) + 2\beta^2\biggl( \frac{x}{\beta}\biggr)^3\cos\theta - 5\beta^3\biggl( \frac{x}{\beta}\biggr)^4 \cos^2\theta +4\beta^4\biggl( \frac{x}{\beta}\biggr)^5\cos^3\theta + \mathcal{O}(\beta^3)\biggr] </math>
		<math>~ +~m^2 b [2^3(n+1) \cos^2\theta - 3] \cdot [ 2^3\cdot 3(n+1) ]^{1/2} \biggl[ \beta^2\biggl(\frac{x}{\beta}\biggr)^3 (1-1) + 2\beta^3\biggl(\frac{x}{\beta}\biggr)^4 \cos\theta - 5\beta^4\biggl(\frac{x}{\beta}\biggr)^5 \cos^2\theta +4\beta^5\biggl(\frac{x}{\beta}\biggr)^6 \cos^3\theta + \mathcal{O}(\beta^3)\biggr] </math>

Dropping all terms on the right-hand-side that are <math>~\mathcal{O}(\beta^3)</math> or higher, we have,

<math>~\mathrm{Im}\biggl[\frac{\mathrm{TERM5}}{\beta^2}\biggr]</math>	<math>~=</math>	<math>~ \biggl(\frac{x}{\beta}\biggr)\cdot 2 m^2 b_0 \biggl\{\beta^0(1-2+1) +(4-4)\beta\biggl(\frac{x}{\beta}\biggr)\cos\theta +4 \beta^2\biggl(\frac{x}{\beta}\biggr)^2\cos^2\theta - \beta^2\biggl[ \frac{3}{2(n+1)}\biggr] + \cancelto{0}{\mathcal{O}(\beta^3)} \biggr\} </math>
		<math>~ \times \biggl\{ 1 +\beta\biggl(\frac{x}{\beta}\biggr)\frac{b}{2} - \beta^2\biggl(\frac{x}{\beta}\biggr)^2\frac{b^2}{8} + \cancelto{0}{\mathcal{O}(\beta^3)}\biggr\} </math>
		<math>~ +~b_0\biggl[ \frac{(1-1)}{\beta\cos\theta} + 2\beta^0\biggl(\frac{x}{\beta}\biggr) - 5\beta\biggl(\frac{x}{\beta}\biggr)^2\cos\theta +4\beta^2 \biggl(\frac{x}{\beta}\biggr)^3\cos^2\theta + \cancelto{0}{\mathcal{O}(\beta^3)}\biggr] </math>
		<math>~ -~ m^2(4n+1)\cdot \biggl[ \frac{2^3\cdot 3}{(n+1)}\biggr]^{1/2} \biggl[ \beta^{1}(1-1) + 2\beta^2\biggl(\frac{x}{\beta}\biggr)\cos\theta + \cancelto{0}{\mathcal{O}(\beta^3)}\biggr] </math>
		<math>~ +~m^2[2^3(n+1) \cos^2\theta - 3] \cdot [ 2^3\cdot 3(n+1) ]^{1/2} \biggl[ \beta^1\biggl( \frac{x}{\beta}\biggr)^2(1-1) + 2\beta^2\biggl( \frac{x}{\beta}\biggr)^3\cos\theta + \cancelto{0}{\mathcal{O}(\beta^3)}\biggr] </math>
		<math>~ +~m^2 b [2^3(n+1) \cos^2\theta - 3] \cdot [ 2^3\cdot 3(n+1) ]^{1/2} \biggl[ \beta^2\biggl(\frac{x}{\beta}\biggr)^3 (1-1) + \cancelto{0}{\mathcal{O}(\beta^3)}\biggr] </math>
	<math>~\approx</math>	<math>~m^2 b_0 \biggl\{- \biggl[ \frac{3}{(n+1)}\biggr]\biggl(\frac{x}{\beta}\biggr) + 8 \biggl(\frac{x}{\beta}\biggr)^3\cos^2\theta \biggr\} \times\biggl\{ \beta^2 +\cancelto{0}{\mathcal{O}(\beta^3)} \biggr\} </math>
		<math>~ +~b_0\biggl[ 2\beta^0\biggl(\frac{x}{\beta}\biggr) - 5\beta\biggl(\frac{x}{\beta}\biggr)^2\cos\theta + 4\beta^2 \biggl(\frac{x}{\beta}\biggr)^3\cos^2\theta \biggr] </math>
		<math>~ -~ m^2(4n+1)\cdot \biggl[ \frac{2^3\cdot 3}{(n+1)}\biggr]^{1/2} \biggl[ 2\beta^2\biggl(\frac{x}{\beta}\biggr)\cos\theta \biggr] </math>
		<math>~ +~m^2[2^3(n+1) \cos^2\theta - 3] \cdot [ 2^3\cdot 3(n+1) ]^{1/2} \biggl[ 2\beta^2\biggl( \frac{x}{\beta}\biggr)^3\cos\theta \biggr] </math>
	<math>~\approx</math>	<math>~2b_0\beta^0\biggl(\frac{x}{\beta}\biggr) - 5b_0\beta\biggl(\frac{x}{\beta}\biggr)^2\cos\theta + 4b_0\beta^2 \biggl(\frac{x}{\beta}\biggr)^3\cos^2\theta </math>
		<math>~+\beta^2 m^2 \biggl\{- \biggl[ \frac{3b_0}{(n+1)}\biggr]\biggl(\frac{x}{\beta}\biggr) + 8 b_0\biggl(\frac{x}{\beta}\biggr)^3\cos^2\theta -~ (4n+1)\cdot \biggl[ \frac{2^3\cdot 3}{(n+1)}\biggr]^{1/2} \biggl[ 2\biggl(\frac{x}{\beta}\biggr)\cos\theta \biggr] +~ [2^3(n+1) \cos^2\theta - 3] \cdot [ 2^3\cdot 3(n+1) ]^{1/2} \biggl[ 2\biggl( \frac{x}{\beta}\biggr)^3\cos\theta \biggr] \biggr\} \, . </math>

Together

Together, then, we have:

<math>~\mathrm{Im}\biggl[\frac{\mathrm{TERM4}+\mathrm{TERM5}}{b_0\beta^2}\biggr]</math>	<math>~\approx</math>	<math>~ -2\biggl(\frac{x}{\beta}\biggr) \biggl(1+\frac{3xb}{2} \biggr)^2 (1+xb)^{-1/2} + \beta \biggl(\frac{x}{\beta}\biggr)^2 \biggl\{ 8 \cos\theta -~ 3 \biggl[\frac{\sin^4\theta}{\cos\theta}\biggr] \biggr\} + 2\beta^0\biggl(\frac{x}{\beta}\biggr) - 5\beta\biggl(\frac{x}{\beta}\biggr)^2\cos\theta </math>
	<math>~\approx</math>	<math>~ -2\biggl(\frac{x}{\beta}\biggr) \biggl(1+3xb \biggr) \biggl(1- \frac{xb}{2} \biggr) + 2\biggl(\frac{x}{\beta}\biggr) + \beta \biggl(\frac{x}{\beta}\biggr)^2 \biggl\{ 3 \cos\theta -~ 3 \biggl[\frac{\sin^4\theta}{\cos\theta}\biggr] \biggr\} </math>
	<math>~\approx</math>	<math>~ -\biggl(\frac{x}{\beta}\biggr) \biggl[2+5bx \biggr] + 2\biggl(\frac{x}{\beta}\biggr) + \frac{3\beta}{\cos\theta} \biggl(\frac{x}{\beta}\biggr)^2 \biggl\{ \cos^2\theta -\sin^4\theta \biggr\} </math>
	<math>~=</math>	<math>~ \biggl(\frac{x}{\beta}\biggr) (-2 + 2) -5\beta\biggl(\frac{x}{\beta}\biggr)^2 [3\cos\theta - \cos^3\theta] + \frac{3\beta}{\cos\theta} \biggl(\frac{x}{\beta}\biggr)^2 \biggl\{ \cos^2\theta -[1-2\cos^2\theta + \cos^4\theta] \biggr\} </math>
	<math>~=</math>	<math>~ \biggl(\frac{x}{\beta}\biggr) (-2 + 2) -\frac{5\beta}{\cos\theta}\biggl(\frac{x}{\beta}\biggr)^2 [3\cos^2\theta - \cos^4\theta] + \frac{3\beta}{\cos\theta} \biggl(\frac{x}{\beta}\biggr)^2 \biggl\{ -1+3\cos^2\theta - \cos^4\theta \biggr\} </math>
	<math>~=</math>	<math>~ \biggl(\frac{x}{\beta}\biggr) (-2 + 2) + \frac{\beta}{\cos\theta} \biggl(\frac{x}{\beta}\biggr)^2 \biggl\{ -3+9\cos^2\theta - 3\cos^4\theta -15\cos^2\theta + 5\cos^4\theta \biggr\} </math>
	<math>~=</math>	<math>~ \biggl(\frac{x}{\beta}\biggr) (-2 + 2) - \frac{\beta}{\cos\theta} \biggl(\frac{x}{\beta}\biggr)^2 \biggl\{ 3 + 6\cos^2\theta - 2\cos^4\theta \biggr\} </math>

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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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When added together, we obtain,

<math>~\mathrm{Im}[\mathrm{TERM4} + \mathrm{TERM5}]</math>	<math>~=</math>	<math>~ -~\beta^2 \biggl(\frac{x}{\beta}\biggr) \ell^4 \cos\theta [2^9\cdot 3 (n+1)^3]^{1/2} \cdot (1+xb)^{-1/2}\cdot \biggl(1+\frac{3xb}{2}\biggr)^2 </math>
		<math>~ -~\cancelto{0}{\beta^3} \biggl(\frac{x}{\beta}\biggr)^2\cdot 3 \ell^4 \sin^4\theta [2^7\cdot 3 (n+1)^3 ]^{1/2} (1+xb)^{1/2} \biggl\{ 1 +\frac{3x}{2}\cdot\biggl[ \frac{\sin^2\theta \cos\theta}{(1+xb)} \biggr]\biggr\} </math>
		<math>~+\beta^2 \biggl(\frac{x}{\beta}\biggr)\cdot 2 m^2 [1 - 2\ell^2 + \ell^4 ] \cdot [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cos\theta \cdot (1+xb)^{1/2} </math>
		<math>~-\cancelto{0}{\beta^4} \biggl(\frac{x}{\beta}\biggr) \biggl[\frac{3 m^2\ell^4}{(n+1)} \biggr] \cdot [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cos\theta \cdot (1+xb)^{1/2} </math>
		<math>~ -~\beta [ 2^7\cdot 3 (n+1)^3]^{1/2} [\ell^2 -\ell^4] </math>
		<math>~ +~\cancelto{0}{\beta^3} \biggl[ \frac{2\cdot 3}{(n+1)}\biggr]^{1/2} [\ell^2 -\ell^4] \cdot \biggl[ 2m^2(4n+1) - \biggl(\frac{x}{\beta}\biggr)^2 2m^2(n+1)^2(2^3 \cos^2\theta - 3) (1+xb) \biggr] </math>
	<math>~\approx</math>	<math>~ -~\beta^1 [ 2^7\cdot 3 (n+1)^3]^{1/2} \biggl\{ \biggl[ 1-2\beta \biggl(\frac{x}{\beta}\biggr)\cos\theta + \cancelto{0}{\mathcal{O}(\beta^2)} \biggr] - \biggl[ 1-4\beta \biggl(\frac{x}{\beta}\biggr)\cos\theta + \cancelto{0}{\mathcal{O}(\beta^2)} \biggr] \biggr\} </math>
		<math>~ -~\beta^2 \biggl(\frac{x}{\beta}\biggr) \cos\theta [2^9\cdot 3 (n+1)^3]^{1/2} \cdot (1+\cancelto{0}{x}b)^{-1/2}\cdot \biggl(1+\frac{3\cancelto{0}{x}b}{2}\biggr)^2 </math>
		<math>~+\beta^2 \biggl(\frac{x}{\beta}\biggr)\cdot 2 m^2 [1 - 2 + 1 ] \cdot [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cos\theta \cdot (1+\cancelto{0}{x}b)^{1/2} </math>
	<math>~\approx</math>	<math>~ -~\beta^1 [ 2^7\cdot 3 (n+1)^3]^{1/2} [1 - 1] </math>
		<math>~ -~\beta^2 \biggl(\frac{x}{\beta}\biggr)\cos\theta [ 2^9\cdot 3 (n+1)^3]^{1/2} -~\beta^2 \biggl(\frac{x}{\beta}\biggr) \cos\theta [2^9\cdot 3 (n+1)^3]^{1/2} </math>
		<math>~+\beta^2 \biggl(\frac{x}{\beta}\biggr)\cdot 2 m^2 [1 - 2 + 1 ] \cdot [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cos\theta </math>

Summary

As stated above, the eigenvalue problem that must be solved in order to identify the eigenfunction, <math>~\Lambda(x,\theta)</math>, and eigenfrequency, <math>~(\nu/m)</math>, of unstable (as well as stable) nonaxisymmetric modes in slim <math>~(\beta \ll 1)</math>, polytropic <math>~(n)</math> PP tori with uniform specific angular momentum is defined by the following two-dimensional <math>~(x,\theta)</math>, 2^nd-order PDE:

<math>~0</math>

<math>~=</math>

<math>~f (1-x\cos\theta)^2 \biggl\{ ~\mathrm{TERM1} + \mathrm{TERM2} + \mathrm{TERM3} \biggr\} + ~\frac{n}{\beta^2} \biggl\{ \mathrm{TERM4} ~+~ \mathrm{TERM5}\biggr\} \, , </math>

where, <math>~f(x,\theta)</math> is the unperturbed enthalpy distribution in the unperturbed, axisymmetric torus,

<math>~\mathrm{TERM1}</math>	<math>~\equiv</math>	<math>~(1-x\cos\theta)^2\biggl[ \frac{\partial^2 \Lambda}{\partial x^2} + \frac{1}{x^2}\cdot \frac{\partial^2 \Lambda}{\partial \theta^2}\biggr] \, ,</math>
<math>~\mathrm{TERM2}</math>	<math>~\equiv</math>	<math>~\frac{(1-x\cos\theta)}{x} \biggl[ (1-2x\cos\theta) \frac{\partial \Lambda}{\partial x} + \sin\theta\cdot \frac{\partial \Lambda}{\partial \theta} \biggr] \, ,</math>
<math>~\mathrm{TERM3}</math>	<math>~\equiv</math>	<math>~- [ 2^2(n+1)^2 + m^2\Lambda ] \, ,</math>
<math>~\mathrm{TERM4}</math>	<math>~\equiv</math>	<math>~(1-x\cos\theta)^4\biggl[ \frac{\partial \Lambda}{\partial x} \cdot \frac{\partial (\beta^2 f)}{\partial x} ~+~ \frac{\partial \Lambda}{\partial \theta} \cdot \frac{\partial (\beta^2 f/x^2)}{\partial \theta} \biggr] \, ,</math>
<math>~\mathrm{TERM5}</math>	<math>~\equiv</math>	<math>~\biggl[ (1-x\cos\theta)^4\biggl(\frac{\nu}{m}\biggr)^2 + 2(1-x\cos\theta)^2\biggl(\frac{\nu}{m}\biggr)+ 1 \biggr] [ 2^3(n+1)^2 + 2m^2\Lambda ] \, .</math>

Blaes (1985) has determined that a good approximate solution to this eigenvalue problem is the eigenvector defined by the imaginary eigenfrequency,

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

<math>~=</math>

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

and, simultaneously, the imaginary eigenfunction,

<math>~\Lambda</math>

<math>~=</math>

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

where,

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

<math>~=</math>

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

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, <math>~M_*/M_d = 0.0</math>.
  • 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 <math>~0.0 \le M_*/M_d \le 10^3</math> and inner to outer edge aspect ratios of <math>~0.1 < r_-/r_+ < 0.75</math>."
  • 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.

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