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>

If an exact solution, <math>~(\Lambda,\nu/m)</math>, to this eigenvalue problem were plugged into this governing PDE, we would expect that both of the following summations would be exactly zero at all meridional-plane <math>~(x,\theta)</math> locations throughout the torus:

<math>~0</math>	<math>~=</math>	<math>~\mathrm{TERM1} + \mathrm{TERM2} + \mathrm{TERM3} \, ,</math>
<math>~0</math>	<math>~=</math>	<math>~\mathrm{TERM4} + \mathrm{TERM5} \, .</math>

While an exact analytic solution to this eigenvalue problem is not (yet) known, Blaes (1985) has determined that a good approximate solution is an 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>

We have plugged this "Blaes85" approximate eigenvector into the five separate "TERM" expressions — analytically evaluating partial (1^st and 2^nd) derivatives along the way, as appropriate — then have evaluated each of the expressions numerically using an Excel spreadsheet. The appropriate sums of these TERMs are, indeed, nearly zero for slim <math>~(\beta \ll 1)</math> configurations. As the blue diamonds in the following log-log plot show, for example, the real part of the sum, "TERM4 + TERM5," drops by approximately two orders of magnitude for every factor of ten drop in <math>~\beta</math>, dropping a total of roughly eight orders of magnitude over the (displayed) range, <math>~\beta = 1 ~\rightarrow~ 10^{-4}</math>. As the salmon-colored squares in the same plot indicate, the imaginary part of the sum, "TERM4 + TERM5," is even closer to zero, dropping roughly 12 orders of magnitude over the same range of <math>~\beta</math>. This indicates that, with the Blaes85 eigenvector, the real part of the sum of this pair of terms differs from zero by an (error) expression whose leading-order term varies as <math>~\beta^{2}</math> while the corresponding imaginary part of the sum differs from zero by an (error) expression whose leading-order term varies as <math>~\beta^{3}</math>.

As our above analytic analysis shows, when each of the expressions for TERM4 and TERM5 is rewritten as a power series in <math>~\beta</math>, a sum of the two TERMs results in precise cancellation of leading-order terms. For the imaginary component of this sum, the highest order residual term is, precisely,

<math>~Residual_{45}</math>	<math>~\equiv</math>	<math>~\mathrm{Im}[\mathrm{TERM4}+\mathrm{TERM5}]</math>
	<math>~=</math>	<math>~ - \beta^3 \biggl(\frac{x}{\beta}\biggr)^2 [2^7\cdot 3 (n+1)^3]^{1/2}[ 3 + 6\cos^2\theta - 2\cos^4\theta ] + \mathcal{O}(\beta^4) \, . </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.

© 2014 - 2021 by Joel E. Tohline |   H_Book Home   |   YouTube   | Appendices: | Equations | Variables | References | Ramblings | Images | myphys.lsu | ADS | Recommended citation:   Tohline, Joel E. (2021), The Structure, Stability, & Dynamics of Self-Gravitating Fluids, a (MediaWiki-based) Vistrails.org publication, https://www.vistrails.org/index.php/User:Tohline/citation