User:Tohline/Appendix/Ramblings/T1Coordinates - Revision history

Tohline: /* Position Vector */ fix type in position vector definition

2010-05-23T13:48:54Z

Position Vector: fix type in position vector definition

← Older revision		Revision as of 13:48, 23 May 2010
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	<td align="left">		<td align="left">
	<math>		<math>
	\hat{i}x + \hat{j}y + \hat{k}		\hat{i}x + \hat{j}y + \hat{k}z
	</math>		</math>
	</td>		</td>

Tohline: /* Integrals of Motion */ Add link to T3 coordinates

2010-05-23T00:53:10Z

Integrals of Motion: Add link to T3 coordinates

← Older revision		Revision as of 00:53, 23 May 2010
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	==Integrals of Motion==		==Integrals of Motion==
	In [[User:Tohline/Appendix/Ramblings/T2Integrals\|an accompanying chapter]] we begin to examine what the <math>3^\mathrm{rd}</math> integral of motion might be in the T2 coordinate system.		<font color="red">'''T2 Coordinates:'''</font> In [[User:Tohline/Appendix/Ramblings/T2Integrals\|an accompanying chapter]] we begin to examine what the <math>3^\mathrm{rd}</math> integral of motion might be in the T2 coordinate system.


			<font color="red">'''T3 Coordinates:'''</font> In yet [[User:Tohline/Appendix/Ramblings/T3Integrals\|another accompanying chapter]] we define the properties of, and begin to examine what the <math>3^\mathrm{rd}</math> integral of motion might be, in a system we refer to as ''T3 coordinates''.

	=See Also=		=See Also=

Tohline: Insert link to T2Integrals chapter

2010-05-20T01:35:45Z

Insert link to T2Integrals chapter

← Older revision		Revision as of 01:35, 20 May 2010
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	This is the desired inversion which gives <math>\varpi(\chi_1,\chi_2)</math> in the case <math>q^2 = 3</math>.		This is the desired inversion which gives <math>\varpi(\chi_1,\chi_2)</math> in the case <math>q^2 = 3</math>.


			==Integrals of Motion==
			In [[User:Tohline/Appendix/Ramblings/T2Integrals\|an accompanying chapter]] we begin to examine what the <math>3^\mathrm{rd}</math> integral of motion might be in the T2 coordinate system.

	=See Also=		=See Also=

Tohline: Add link to PDF document discussing integrals of motion

2010-05-17T03:06:30Z

Add link to PDF document discussing integrals of motion

← Older revision		Revision as of 03:06, 17 May 2010
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	</div>		</div>
	This is the desired inversion which gives <math>\varpi(\chi_1,\chi_2)</math> in the case <math>q^2 = 3</math>.		This is the desired inversion which gives <math>\varpi(\chi_1,\chi_2)</math> in the case <math>q^2 = 3</math>.


			=See Also=
			* [http://www.phys.lsu.edu/astro/H_Book.current/Appendices/Integrals/integrals.of.motion.pdf Old discussion of ''Integrals of Motion'']

	<br />		<br />
	{{LSU_HBook_footer}}		{{LSU_HBook_footer}}

Tohline: /* Scale Factors */ minor rewrite of h_2 scale factor expression

2010-05-16T23:41:15Z

Scale Factors: minor rewrite of h_2 scale factor expression

← Older revision		Revision as of 23:41, 16 May 2010
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	<td align="left">		<td align="left">
	<math>		<math>
	\frac{\varpi~~^4 \sinh~~^2~~\Zeta~~}{~~q^2~~ \chi_2^2 (\varpi^2 + q^4 z^2)}		\frac{z^2 \varpi^2 }{\chi_2^2 (\varpi^2 + q^4 z^2)}
	</math>		</math>
	</td>		</td>
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	</tr>		</tr>
	</table>		</table>



	====Position Vector====		====Position Vector====

Tohline: /* First Attempt */ Derive scale factors and position vector

2010-05-16T23:26:23Z

First Attempt: Derive scale factors and position vector

Show changes

Tohline: /* Relationship Between the Two (to be checked and perhaps extended by Jay Call) */ Begin presentation of cubic solution

2010-05-16T00:09:51Z

Relationship Between the Two (to be checked and perhaps extended by Jay Call): Begin presentation of cubic solution

← Older revision		Revision as of 00:09, 16 May 2010
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	</math>		</math>
	</div>		</div>
	This is the desired inversion which gives <math>\varpi(\chi_1,\chi_2)</math>. The desired expression for <math>z(\chi_1,\chi_2)</math> is,		This is the desired inversion which gives <math>\varpi(\chi_1,\chi_2)</math> in the case <math>q^2 = 2</math>. The desired expression for <math>z(\chi_1,\chi_2)</math> is,
	<div align="center">		<div align="center">
	<math>		<math>
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	</table>		</table>
	Note that this expression is very similar, but not identical, to the expression just derived for <math>z^2</math>.		Note that this expression is very similar, but not identical, to the expression just derived for <math>z^2</math>.


			===Second Special Case (cubic)===
			When <math>q^2 = 3</math>, the above polynomial becomes a cubic equation and therefore its roots also can be determined analytically. Again setting,
			<div align="center">
			<math>
			x \equiv \varpi^2 ,
			</math>
			</div>
			the relevant cubic equation is,
			<div align="center">
			<math>
			x^3 + \biggl( \frac{A}{\chi_2} \biggr)^2 x - \biggl( \frac{A \chi_1}{B \chi_2} \biggr)^2 = 0 .
			</math>
			</div>


			<table align="center" border="5" cellpadding="20" bordercolor="purple" width="75%">
			<tr><td>
			A cubic equation of the form,
			<div align="center">
			<math>
			x^3 + ax + b = 0 ,
			</math>
			</div>
			has the following real root:
			<div align="center">
			<math>
			x = \alpha + \beta ,
			</math>
			</div>
			where,
			<table align="center" border="0" cellpadding="5">
			<tr>
			<td align="right">
			<math>
			\alpha^3
			</math>
			</td>
			<td align="center">
			<math>
			=
			</math>
			</td>
			<td align="left">
			<math>
			-\frac{b}{2} \biggl\{ 1 - \biggl[ 1 + \biggl( \frac{2}{b} \biggr)^2 \biggl( \frac{a}{3} \biggr)^3 \biggr]^{1/2} \biggr\}
			</math>
			</td>
			</tr>

			<tr>
			<td align="right">
			<math>
			\beta^3
			</math>
			</td>
			<td align="center">
			<math>
			=
			</math>
			</td>
			<td align="left">
			<math>
			-\frac{b}{2} \biggl\{ 1 + \biggl[ 1 + \biggl( \frac{2}{b} \biggr)^2 \biggl( \frac{a}{3} \biggr)^3 \biggr]^{1/2} \biggr\}
			</math>
			</td>
			</tr>

			</table>

			</td></tr>
			</table>


			The root of this function that provides positive values for <math>x</math> (''i.e.,'' for <math>\varpi^2</math>) is,
			<div align="center">
			<math>
			x = \biggl( \frac{A^2 \chi_1^2}{2 B^2\chi_2^2} \biggr)^{1/3} \biggl\{ \biggl[ 1 + \sqrt{1 + \Chi_\mathrm{cube}^2} \biggr]^{1/3} + \biggl[ 1 - \sqrt{1 + \Chi_\mathrm{cube}^2} \biggr]^{1/3} \biggr\} ,
			</math>
			</div>
			where,
			<div align="center">
			<math>
			\Chi_\mathrm{cube}^2 \equiv \frac{2^2 B^4 A^2}{3^3 \chi_1^4 \chi_2^2} .
			</math>
			</div>
			This is the desired inversion which gives <math>\varpi(\chi_1,\chi_2)</math> in the case <math>q^2 = 3</math>.

	<br />		<br />
	{{LSU_HBook_footer}}		{{LSU_HBook_footer}}

Tohline: /* First Special Case */ Correct error in expression for z^2

2010-05-15T03:26:57Z

First Special Case: Correct error in expression for z^2

← Older revision		Revision as of 03:26, 15 May 2010
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	which, in retrospect, could have been derived more directly from the definition of <math>\chi_1</math>.		which, in retrospect, could have been derived more directly from the definition of <math>\chi_1</math>.

	===First Special Case===		===First Special Case (quadratic)===
	When <math>q^2 = 2</math>, the above polynomial becomes a quadratic equation and therefore its roots can be determined analytically. Setting,		When <math>q^2 = 2</math>, the above polynomial becomes a quadratic equation and therefore its roots can be determined analytically. Setting,
	<div align="center">		<div align="center">
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	<div align="center">		<div align="center">
	<math>		<math>
	z^2 = \frac{\chi_1^2}{q^2 B^2}- \frac{A^2}{2q^2 \chi_2^2} \biggl[ \sqrt{1 + \Chi_\mathrm{quad}^2} - 1 \biggr] = \frac{A^2}{4 \chi_2^2}\biggl[ \~~biggl(~~1+ \Chi_\mathrm{quad}^2 ~~\biggr)~~ - \sqrt{1 + \Chi_\mathrm{quad}^2}\biggr] .		z^2 = \frac{\chi_1^2}{q^2 B^2}- \frac{A^2}{2q^2 \chi_2^2} \biggl[ \sqrt{1 + \Chi_\mathrm{quad}^2} - 1 \biggr] = \frac{A^2}{4 \chi_2^2}\biggl[ \frac{1}{2} \Chi_\mathrm{quad}^2 +1 - \sqrt{1 + \Chi_\mathrm{quad}^2}\biggr] .
	</math>		</math>
	</div>		</div>
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	<td align="left">		<td align="left">
	<math>		<math>
	2z^2 .		\frac{A^2}{2\chi_2^2} \biggl[ \Chi_\mathrm{quad}^2 + 1 - \sqrt{ 1 + \Chi_\mathrm{quad}^2} \biggr] .
	</math>		</math>
	</td>		</td>
	</tr>		</tr>
	</table>		</table>
	~~To be honest~~, ~~this result does~~ not ~~make sense! I suspect that there is a mathematical error somewhere~~.		Note that this expression is very similar, but not identical, to the expression just derived for <math>z^2</math>.


	<br />		<br />
	{{LSU_HBook_footer}}		{{LSU_HBook_footer}}

Tohline: /* First Special Case */ Extend expressions for "L"

2010-05-15T02:52:02Z

First Special Case: Extend expressions for "L"

← Older revision		Revision as of 02:52, 15 May 2010
Line 227:		Line 227:
	<div align="center">		<div align="center">
	<math>		<math>
	z^2 = \frac{\chi_1^2}{q^2 B^2}- \frac{A^2}{2q^2 \chi_2^2} \biggl[ \sqrt{1 + \Chi_\mathrm{quad}^2} - 1 \biggr] = \frac{A^2}{~~2q^2~~ \chi_2^2}\biggl[ \biggl(1+ \Chi_\mathrm{quad}^2 \biggr) - \sqrt{1 + \Chi_\mathrm{quad}^2}\biggr] .		z^2 = \frac{\chi_1^2}{q^2 B^2}- \frac{A^2}{2q^2 \chi_2^2} \biggl[ \sqrt{1 + \Chi_\mathrm{quad}^2} - 1 \biggr] = \frac{A^2}{4 \chi_2^2}\biggl[ \biggl(1+ \Chi_\mathrm{quad}^2 \biggr) - \sqrt{1 + \Chi_\mathrm{quad}^2}\biggr] .
	</math>		</math>
	</div>		</div>

	We note as well that in this special case, where <math>q^2 = 2</math>,		We note as well that in this special case, where <math>q^2 = 2</math>,
	<~~div~~ align="center">		<table align="center" border="0" cellpadding="5">
			<tr>
			<td align="right">
	<math>		<math>
	4 L^2 ~~= \varpi^2 \biggl[1 + 2 \sinh^2 \Zeta \biggr] = \varpi^2 \cosh(2\Zeta) ;~~		4 L^2
	</math>		</math>
	</~~div~~>		</td>
	~~that is,~~		<td align="center">
	<~~div~~ align="~~center~~">		<math>
			=
			</math>
			</td>
			<td align="left">
	<math>		<math>
	L = \~~frac{~~\varpi}{2} \biggl[\~~cosh~~(2\Zeta)\biggr]^{1/2} .		\varpi^2 \biggl[1 + 2 \sinh^2 \Zeta \biggr] = \varpi^2 \cosh(2\Zeta) = \varpi^2 \biggl[ 1+\sinh^2(2\Zeta) \biggr]^{1/2}
	</math>		</math>
	</~~div~~>		</td>
			</tr>

			<tr>
			<td align="right">

			</td>
			<td align="center">
			<math>
			=
			</math>
			</td>
			<td align="left">
			<math>
			\frac{A^2}{2\chi_2^2} \biggl[ \sqrt{1 + \Chi_\mathrm{quad}^2} - 1 \biggr] \sqrt{ 1 + \Chi_\mathrm{quad}^2}
			</math>
			</td>
			</tr>

			<tr>
			<td align="right">

			</td>
			<td align="center">
			<math>
			=
			</math>
			</td>
			<td align="left">
			<math>
			2z^2 .
			</math>
			</td>
			</tr>
			</table>
			To be honest, this result does not make sense! I suspect that there is a mathematical error somewhere.


	<br />		<br />
	{{LSU_HBook_footer}}		{{LSU_HBook_footer}}

Tohline: /* Relationship Between the Two (to be checked and perhaps extended by Jay Call) */ Introduce function "L"

2010-05-15T02:28:22Z

Relationship Between the Two (to be checked and perhaps extended by Jay Call): Introduce function "L"

← Older revision		Revision as of 02:28, 15 May 2010
Line 150:		Line 150:
	</div>		</div>
	While the relationship between <math>\chi_2</math> and <math>\xi_2</math> looks rather convoluted, there is at least a rational reason (provided by HNM82) for specifying the form of <math>\chi_2</math>, whereas the original specification of <math>\xi_2</math> was chosen rather arbitrarily in an effort to look like spherical coordinates.		While the relationship between <math>\chi_2</math> and <math>\xi_2</math> looks rather convoluted, there is at least a rational reason (provided by HNM82) for specifying the form of <math>\chi_2</math>, whereas the original specification of <math>\xi_2</math> was chosen rather arbitrarily in an effort to look like spherical coordinates.

			We should probably address as well the structure of the awkward function that appears in the denominator of the T1-coordinate scale factors, namely,
			<div align="center">
			<math>
			L \equiv \biggl[ z^2 + \frac{\varpi^2}{q^4} \biggr]^{1/2} .
			</math>
			</div>
			In terms of our newly introduced hyperbolic functions, this may be written as,
			<div align="center">
			<math>
			q^4 L^2 = \varpi^2 + q^4 z^2 = \varpi^2 \biggl[1 + q^2 \sinh^2 \Zeta \biggr] .
			</math>
			</div>



	===Coordinate Inversion===		===Coordinate Inversion===
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	</math>		</math>
	</div>		</div>

			We note as well that in this special case, where <math>q^2 = 2</math>,
			<div align="center">
			<math>
			4 L^2 = \varpi^2 \biggl[1 + 2 \sinh^2 \Zeta \biggr] = \varpi^2 \cosh(2\Zeta) ;
			</math>
			</div>
			that is,
			<div align="center">
			<math>
			L = \frac{\varpi}{2} \biggl[\cosh(2\Zeta)\biggr]^{1/2} .
			</math>
			</div>