Difference between revisions of "User:Tohline/Appendix/Ramblings/PPToriPt2"

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(→‎Imaginary Parts: Finished imaginary part of TERM4)
(→‎Start From Scratch: Finished real part of TERM5)
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===Testing for Expected Cancellations===
===Testing for Expected Cancellations===
Note first that, adopting the shorthand notation,
<div align="center">
<math>~\ell \equiv (1-x\cos\theta) \, ,</math>
</div>
we have (for our <font color="red">Case B</font> frequency ratio),
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\mathrm{TERM5}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \ell^4\biggl(\frac{\nu}{m}\biggr)^2 + 2\ell^2\biggl(\frac{\nu}{m}\biggr)+ 1 \biggr]
[ 2^3(n+1)^2 + 2m^2\Lambda ]
</math>
  </td>
</tr>
</table>
</div>
====Real Parts====
====Real Parts====


Line 440: Line 466:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\mathrm{Re}\biggl[\frac{\mathrm{TERM4}}{(1-x\cos\theta)^4}\biggr]</math>
<math>~\mathrm{Re}\biggl[\frac{\mathrm{TERM4}}{\ell^4}\biggr]</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 523: Line 549:
+~ x^3 \cdot 2^4\cdot 3(n+1)^2 \cos\theta\sin^4\theta (1+xb)   
+~ 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  [3-2^3(n+1)\cos^2\theta] \, .</math>
~-~x^4\cdot 3^2(n+1)\sin^6\theta  [3-2^3(n+1)\cos^2\theta] \, .</math>
  </td>
</tr>
</table>
</div>
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\mathrm{Re}\biggl[\mathrm{TERM5}\biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
  </td>
</tr>
<tr>
  <td align="right">
<b><font color="red" size="+1">Case B:</font></b>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<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>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<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>
   </td>
   </td>
</tr>
</tr>
Line 537: Line 642:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\mathrm{Im}\biggl[\frac{\mathrm{TERM4}}{(1-x\cos\theta)^4}\biggr]</math>
<math>~\mathrm{Im}\biggl[\frac{\mathrm{TERM4}}{\ell^4}\biggr]</math>
   </td>
   </td>
   <td align="center">
   <td align="center">

Revision as of 17:16, 22 April 2016

Stability Analyses of PP Tori (Part 2)

Whitworth's (1981) Isothermal Free-Energy Surface
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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:

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,

<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.

Testing for Expected Cancellations

Note first that, adopting the shorthand notation,

<math>~\ell \equiv (1-x\cos\theta) \, ,</math>

we have (for our Case B frequency ratio),

<math>~\mathrm{TERM5}</math>

<math>~=</math>

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


Imaginary Parts

<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>

See Also


Whitworth's (1981) Isothermal Free-Energy Surface

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