User:Tohline/Appendix/Ramblings/PPTori - Revision history

Tohline: /* Outer [inferior sign] Solution */

2016-12-20T23:18:36Z

Outer [inferior sign] Solution

← Older revision		Revision as of 23:18, 20 December 2016
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	</div>		</div>

	<div align="center">		<div align="center" id="CubeRootImaginary">
	<table border="1" width="60%" cellpadding="8">		<table border="1" width="60%" cellpadding="8">
	<tr><td align="left">		<tr><td align="left">

Tohline: Type in cautionary introductory paragraph

2016-05-24T20:58:23Z

Type in cautionary introductory paragraph

← Older revision		Revision as of 20:58, 24 May 2016
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	<!-- __NOTOC__ will force TOC off -->		<!-- __NOTOC__ will force TOC off -->
	=Stability Analyses of PP Tori=		=Stability Analyses of PP Tori=

			<font color="red"><b>[Comment by J. E. Tohline on 24 May 2016]</b></font>   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, "[[User:Tohline/Apps/ImamuraHadleyCollaboration#Characteristics_of_Unstable_Eigenvectors_in_Self-Gravitating_Tori\|Characteristics of Unstable Eigenvectors in Self-Gravitating Tori]]," that contains a much more trustworthy analysis of this very interesting problem.

			{{LSU_WorkInProgress}}


	{{LSU_HBook_header}}		{{LSU_HBook_header}}

Tohline: /* Examples */ Append "beta" factor to one imaginary expression

2016-04-30T17:58:36Z

Examples: Append "beta" factor to one imaginary expression

← Older revision		Revision as of 17:58, 30 April 2016
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	<math>~\beta~		<math>~\beta~
	2^2\cdot \sqrt{3} \biggl[\frac{11}{2^3} \biggl(2^2+\frac{3}{2^2}\cdot \frac{11}{2^3} \biggr)\biggr] \biggl[ \frac{2^4}{43}\biggr]^{3/2}		2^2\cdot \sqrt{3} \biggl[\frac{11}{2^3} \biggl(2^2+\frac{3}{2^2}\cdot \frac{11}{2^3} \biggr)\biggr] \biggl[ \frac{2^4}{43}\biggr]^{3/2}
	=\biggl[ \frac{2^3\cdot 3}{43^3}\biggr]^{1/2} \biggl[11\cdot (2^7+33)\biggr]		=\biggl[ \frac{2^3\cdot 3}{43^3}\biggr]^{1/2} \biggl[11\cdot (2^7+33)\biggr] \beta
	</math>		</math>
	</td>		</td>
	<td align="right">		<td align="right">
	30.76957507		30.76957507<math>~\beta</math>
	</td>		</td>
	</tr>		</tr>

Tohline: /* Examples */

2016-04-27T19:01:27Z

Examples

← Older revision		Revision as of 19:01, 27 April 2016
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	</td>		</td>
	<td align="right">		<td align="right">
	<math>~- 5\beta</math> + 0.167968750		<math>~- 5\beta^2</math> + 0.167968750
	</td>		</td>
	</tr>		</tr>
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	</table>		</table>
	</div>		</div>


	=====Step 7=====		=====Step 7=====

Tohline: /* Assessment */

2016-04-15T21:36:11Z

Assessment

← Older revision		Revision as of 21:36, 15 April 2016
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	======Assessment======		======Assessment======

	The good news is that the real part of the <math>~\mathrm{LHS}_4</math> expression exactly matches the real part of the <math>~\mathrm{RHS}_4</math> expression. But the imaginary ~~parts do not agree~~. So, let's repeat the steps leading to the imaginary parts.		The good news is that the real part of the <math>~\mathrm{LHS}_4</math> expression exactly matches the real part of the <math>~\mathrm{RHS}_4</math> expression. But the imaginary differ by a factor of 2. So, let's repeat the steps leading to the imaginary parts.

	'''<font color="red" size="+1">Case B:</font>'''		'''<font color="red" size="+1">Case B:</font>'''

Tohline: /* Next Lowest Order */ Simplify imaginary component of LHS_4 by dropping higher order terms in beta

2016-04-15T21:33:48Z

Next Lowest Order: Simplify imaginary component of LHS_4 by dropping higher order terms in beta

← Older revision		Revision as of 21:33, 15 April 2016
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	2(n+1)[2^3(n+1)\cos^2\theta -3]		2(n+1)[2^3(n+1)\cos^2\theta -3]
	+ 2^4(n+1)^2(\sin^2\theta - \cos^2\theta)\biggr\}		+ 2^4(n+1)^2(\sin^2\theta - \cos^2\theta)\biggr\}
	~\pm~~i~x^2\biggl\{ \beta \biggl(\frac{b_0}{2^2}\biggr)\biggl[ \frac{b(4+3xb)}{(1+xb)^{3/2}} \biggr]- ~~\biggl(\frac{~~\beta~~}{x}\biggr)~~ b_0\biggr\}</math>		~\pm~~i~x \biggl\{ \cancelto{0}{x\beta} \biggl(\frac{b_0}{2^2}\biggr)\biggl[ \frac{b(4+3xb)}{(1+xb)^{3/2}} \biggr]- \beta b_0\biggr\}</math>
	</td>		</td>
	</tr>		</tr>
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	<td align="left">		<td align="left">
	<math>~2(n+1)x^2\biggl\{ 2^3(n+1)\sin^2\theta -3 \biggr\}		<math>~2(n+1)x^2\biggl\{ 2^3(n+1)\sin^2\theta -3 \biggr\}
	~\pm~~i~x^2\biggl\{ ~~\beta \biggl(\frac{b_0}{2^2}\biggr)\biggl[ \frac{b(4+3xb)}{(1+xb)^{3/2}} \biggr]~~- \biggl(\frac{\beta}{x}\biggr) b_0\biggr\} \, .</math>		~\pm~~i~x^2\biggl\{ - \biggl(\frac{\beta}{x}\biggr) b_0\biggr\} \, .</math>
	</td>		</td>
	</tr>		</tr>
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	\biggl[ \biggl(\frac{\beta}{x}\biggr)^2 - 1 \biggr] \biggl\{		\biggl[ \biggl(\frac{\beta}{x}\biggr)^2 - 1 \biggr] \biggl\{
	~2(n+1)x^2\biggl[ 2^3(n+1)\sin^2\theta -3 \biggr]		~2(n+1)x^2\biggl[ 2^3(n+1)\sin^2\theta -3 \biggr]
	~\pm~~i~x^2\biggl[ ~~\beta b_0 b~~- \biggl(\frac{\beta}{x}\biggr) b_0\biggr]		~\pm~~i~x^2\biggl[ - \biggl(\frac{\beta}{x}\biggr) b_0\biggr]
	</math>		</math>
	</td>		</td>
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	<td align="left">		<td align="left">
	<math>~		<math>~
	~~~\pm~~i~\biggl[ \biggl(\frac{\beta}{x}\biggr)^2 - 1 \biggr] \biggl\{\beta b_0 b- \biggl(\frac{\beta}{x}\biggr) b_0~+~ \biggl(\frac{\beta}{x}\biggr)b_0~~		~\pm~~i~2n\biggl(\frac{\beta}{x}\biggr)b_0
	~~\biggr\}~~~~\pm~~i~2n\biggl(\frac{\beta}{x}\biggr)b_0
	</math>		</math>
	</td>		</td>
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	-(n+1) \biggl\{ [2^4(n+1) -12]		-(n+1) \biggl\{ [2^4(n+1) -12]
	+~ 2^2n[2^3(n+1)\cos^2\theta -3] \biggr\}		+~ 2^2n[2^3(n+1)\cos^2\theta -3] \biggr\}
	~~</math>~~		~\pm~~i~2n\biggl(\frac{\beta}{x}\biggr)b_0
	~~</td>~~
	~~</tr>~~

	~~<tr>~~
	~~<td align="right">~~
	~~<math>~</math>~~
	~~</td>~~
	~~<td align="center">~~
	~~ ~~
	~~</td>~~
	~~<td align="left">~~
	~~<math>~~~
	~\pm~~i~~~\biggl\{\biggl[ \biggl(\frac{\beta}{x}\biggr)^2 - 1 \biggr] \beta b_0 b ~+~~~2n\biggl(\frac{\beta}{x}\biggr)b_0 ~~\biggr\}~~
	</math>		</math>
	</td>		</td>
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	<math>~		<math>~
	2^2(n+1) \biggl(\frac{\beta}{x}\biggr)^2 [4n+1 ]		2^2(n+1) \biggl(\frac{\beta}{x}\biggr)^2 [4n+1 ]
	- 2^2(n+1)^2 [ 1+~ 2^3n\cos^2\theta ] ~\pm~~i~~~b_0 \biggl\{- \beta b ~+~~~2n\biggl(\frac{\beta}{x}\biggr) ~~+\biggl(\frac{\beta}{x}\biggr)^2 \beta b\biggr\}~~ \, .		- 2^2(n+1)^2 [ 1+~ 2^3n\cos^2\theta ] ~\pm~~i~2n\biggl(\frac{\beta}{x}\biggr)b_0
			\, .
	</math>		</math>
	</td>		</td>

Tohline: /* Assessment */ Finished simplifying lowest-order expression for RHS_4

2016-04-15T18:17:49Z

Assessment: Finished simplifying lowest-order expression for RHS_4

← Older revision		Revision as of 18:17, 15 April 2016
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	<td align="left">		<td align="left">
	<math>~-~(1-x\cos\theta)^2 nb_0		<math>~-~(1-x\cos\theta)^2 nb_0
	+ m^2 \biggl\{ 2n - 4n (1-x\cos\theta)^2 + 2n(1-x\cos\theta)^4~~\cdot~~ \biggr\} \cdot (1+xb)^{1/2}		+ m^2 \biggl\{ 2n - 4n (1-x\cos\theta)^2 + 2n(1-x\cos\theta)^4 \biggr\} \cdot (1+xb)^{1/2}
	</math>		</math>
	</td>		</td>
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	<td align="left">		<td align="left">
	<math>~-~(1-x\cos\theta)^2 nb_0		<math>~-~(1-x\cos\theta)^2 nb_0
	+ m^2 \biggl\{ 2n - 4n (1-x\cos\theta)^2 + 2n(1-x\cos\theta)^4\~~cdot~~ \biggr\} \cdot (1+xb)^{1/2}		+ m^2 \biggl\{ 2n - 4n [1-2x\cos\theta + x^2\cos^2\theta + \mathcal{O}(x^3)]
			+ 2n[ 1-4x\cos\theta + 6x^2\cos^2\theta + \mathcal{O}(x^3)] \biggr\} \cdot (1+xb)^{1/2}
	</math>		</math>
	</td>		</td>
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	<math>~		<math>~
	- m^2 \beta^2 (1-x\cos\theta)^2		- m^2 \beta^2 (1-x\cos\theta)^2
	\biggl\{ nb_0(4n+1) + 2^2(n+1)^2(1+xb)^{1/2} \biggl[ (n+1)		\biggl\{ nb_0(4n+1) + 2^2(n+1)^2(1+xb)^{1/2} [ (n+1)
	+ 3n(1-x\cos\theta)^2 \biggr] \biggr\} \cdot \frac{1}{2^2(n+1)^3}		+ 3n(1-x\cos\theta)^2 ] \biggr\} \cdot \frac{1}{2^2(n+1)^3}
			</math>
			</td>
			</tr>

			<tr>
			<td align="right">

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

			<tr>
			<td align="right">

			</td>
			<td align="center">

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

			<tr>
			<td align="right">

			</td>
			<td align="center">

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

			<tr>
			<td align="right">

			</td>
			<td align="center">
			<math>~\approx</math>
			</td>
			<td align="left">
			<math>~-~ nb_0 [1 -2x\cos\theta ]
			</math>
			</td>
			</tr>

			<tr>
			<td align="right">

			</td>
			<td align="center">

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

			<tr>
			<td align="right">

			</td>
			<td align="center">

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

Tohline: /* Assessment */ More work on RHS_4

2016-04-15T17:41:21Z

Assessment: More work on RHS_4

← Older revision		Revision as of 17:41, 15 April 2016
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	</td>		</td>
	</tr>		</tr>
			</table>
			</div>

			<div align="center">
			<table border="0" cellpadding="5" align="center">

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

			<tr>
			<td align="right">

			</td>
			<td align="center">
			<math>~=</math>
			</td>
			<td align="left">
			<math>~-~(1-x\cos\theta)^2 nb_0
			+ m^2 \biggl\{ 2n - 4n (1-x\cos\theta)^2 + 2n(1-x\cos\theta)^4\cdot \biggr\} \cdot (1+xb)^{1/2}
			</math>
			</td>
			</tr>

			<tr>
			<td align="right">

			</td>
			<td align="center">

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

			<tr>
			<td align="right">

			</td>
			<td align="center">

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

Tohline: /* Assessment */ More detailed progress

2016-04-15T16:54:22Z

Assessment: More detailed progress

← Older revision		Revision as of 16:54, 15 April 2016
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	+ (1-x\cos\theta)^4\cdot \biggl[2n - \frac{3n\beta^2}{(n+1)} \biggr]		+ (1-x\cos\theta)^4\cdot \biggl[2n - \frac{3n\beta^2}{(n+1)} \biggr]
	\biggr\} \cdot \biggl\{(1+xb)^{1/2} \biggr\}		\biggr\} \cdot \biggl\{(1+xb)^{1/2} \biggr\}
			</math>
			</td>
			</tr>

			<tr>
			<td align="right">
			<math>~\Rightarrow~~~\mathrm{Im}\biggl[ \frac{\mathrm{RHS_4}}{x^4} \biggr]\biggl(\frac{x}{\beta}\biggr) </math>
			</td>
			<td align="center">
			<math>~=</math>
			</td>
			<td align="left">
			<math>~-~(1-x\cos\theta)^2 nb_0
			</math>
			</td>
			</tr>

			<tr>
			<td align="right">

			</td>
			<td align="center">

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

			<tr>
			<td align="right">

			</td>
			<td align="center">

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

			<tr>
			<td align="right">

			</td>
			<td align="center">
			<math>~=</math>
			</td>
			<td align="left">
			<math>~-~(1-x\cos\theta)^2 nb_0
			+ m^2 \biggl\{ 2n - 4n (1-x\cos\theta)^2 + 2n(1-x\cos\theta)^4\cdot \biggr\} \cdot (1+xb)^{1/2}
			</math>
			</td>
			</tr>

			<tr>
			<td align="right">

			</td>
			<td align="center">

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

			<tr>
			<td align="right">

			</td>
			<td align="center">

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

	</math>		</math>
	</td>		</td>

Tohline: /* Assessment */ More work on RHS_4

2016-04-15T15:47:55Z

Assessment: More work on RHS_4

← Older revision		Revision as of 15:47, 15 April 2016
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	+ (1-x\cos\theta)^4 [2n(n+1) - 3n\beta^2 ]		+ (1-x\cos\theta)^4 [2n(n+1) - 3n\beta^2 ]
	\biggr\} \cdot \biggl\{\beta\cos\theta [2^7\cdot 3 (n+1)^3]^{1/2} (\beta\eta) \biggr\}		\biggr\} \cdot \biggl\{\beta\cos\theta [2^7\cdot 3 (n+1)^3]^{1/2} (\beta\eta) \biggr\}
			</math>
			</td>
			</tr>

			<tr>
			<td align="right">

			</td>
			<td align="center">
			<math>~=</math>
			</td>
			<td align="left">
			<math>~-~x^2 \biggl(\frac{\beta}{x}\biggr) \biggl\{ (1-x\cos\theta)^2 nb_0 \biggr\}
			</math>
			</td>
			</tr>

			<tr>
			<td align="right">

			</td>
			<td align="center">

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

			<tr>
			<td align="right">

			</td>
			<td align="center">

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