Editing Appendix/Ramblings/T3Integrals

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=Integrals of Motion in T3 Coordinates=
Motivated by the HNM82 derivation, in [[User:Tohline/Appendix/Ramblings/T1Coordinates#T2_Coordinates|an accompanying chapter]] we have introduced a new ''T2 Coordinate System'' and have outlined a few of its properties.  Here we offer a modest redefinition of the second ''radial'' coordinate in an effort to bring even more symmetry to the definition of the position vector, <math>\vec{x}</math>.

==Definition==
By defining the dimensionless angle,
<div align="center">
<math>
\Zeta \equiv \sinh^{-1}\biggl( \frac{qz}{\varpi} \biggr) ,
</math>
</div>
the two key "T3" coordinates will be written as,

<table align="center" border="0" cellpadding="2">
<tr>
  <td align="right">
<math>
\lambda_1
</math>
  </td>
  <td align="center">
<math>\equiv</math>
  </td>
  <td align="left">
<math>\varpi \cosh\Zeta = ( \varpi^2 + q^2z^2 )^{1/2}</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp;
  </td>
  <td align="right">
<math>
\lambda_2
</math>
  </td>
  <td align="center">
<math>\equiv</math>
  </td>
  <td align="left">
<math>\varpi [\sinh\Zeta ]^{1/(1-q^2)} = \biggl[\frac{\varpi^{q^2}}{qz}\biggr]^{1/(q^2-1)}</math>
  </td>
</tr>
</table>

===Relevant Partial Derivatives===
Here are some relevant partial derivatives:  

<table align="center" border="1" cellpadding="5">
<tr>
  <td align="center">
&nbsp;
  </td>
  <td align="center">
<math>
\frac{\partial}{\partial x}
</math>
  </td>
  <td align="center">
<math>
\frac{\partial}{\partial y}
</math>
  </td>
  <td align="center">
<math>
\frac{\partial}{\partial z}
</math>
  </td>
</tr>

<tr>
  <td align="center">
<math>\lambda_1</math>
  </td>
  <td align="center">
<math>
\frac{x}{\lambda_1}
</math>
  </td>
  <td align="center">
<math>
\frac{y}{\lambda_1}
</math>
  </td>
  <td align="center">
<math>
\frac{q^2 z}{\lambda_1}
</math>
  </td>
</tr>

<tr>
  <td align="center">
<math>\lambda_2</math>
  </td>
  <td align="center">
<math>
\frac{1}{(q^2-1)} \biggl[ \frac{\varpi^{q^2-1}}{\sinh\Zeta} \biggr]^{q^2/(q^2-1)} \biggl( \frac{q^3 z}{\varpi^{q^2+2}} \biggr) x
</math><br />
<math>
=\frac{q^2}{(q^2-1)} \biggl[ \frac{\varpi^{q^2}}{qz} \biggr]^{1/(q^2-1)} \biggl( \frac{x}{\varpi^2} \biggr)
</math><br />
<math>
=\frac{q^2}{(q^2-1)} \biggl[ \frac{\varpi}{qz} \biggr]^{1/(q^2-1)} \biggl( \frac{x}{\varpi} \biggr)
</math>
  </td>
  <td align="center">
<math>
\frac{1}{(q^2-1)} \biggl[ \frac{\varpi^{q^2-1}}{\sinh\Zeta} \biggr]^{q^2/(q^2-1)} \biggl( \frac{q^3 z}{\varpi^{q^2+2}} \biggr) y
</math><br />
<math>
=\frac{q^2}{(q^2-1)} \biggl[ \frac{\varpi^{q^2}}{qz} \biggr]^{1/(q^2-1)}  \biggl( \frac{y}{\varpi^2} \biggr)
</math><br />
<math>
=\frac{q^2}{(q^2-1)} \biggl[ \frac{\varpi}{qz} \biggr]^{1/(q^2-1)}  \biggl( \frac{y}{\varpi} \biggr)
</math>
  </td>
  <td align="center">
<math>
- \frac{1}{(q^2-1)} \biggl[ \frac{\varpi^{q^2-1}}{\sinh\zeta} \biggr]^{q^2/(q^2-1)} \frac{q}{\varpi^{q^2}}
</math><br />
<math>
=- \frac{1}{(q^2-1)} \biggl[ \frac{\varpi^{q^2}}{qz} \biggr]^{1/(q^2-1)} \frac{1}{z}
</math>
<br />
<math>
=- \frac{1}{(q^2-1)} \biggl[ \frac{\varpi^{q^2}}{qz^{q^2}} \biggr]^{1/(q^2-1)}
</math>
  </td>
</tr>

<tr>
  <td align="center">
<math>\lambda_3</math>
  </td>
  <td align="center">
<math>
- \frac{y}{\varpi^{2}}
</math>
  </td>
  <td align="center">
<math>
+ \frac{x}{\varpi^{2}}
</math>
  </td>
  <td align="center">
<math>
0
</math>
  </td>
</tr>
</table>

Alternatively, partials can be taken with respect to the cylindrical coordinates, <math>\varpi</math>, <math>z</math> and <math>\phi</math>.  (Incidentally, I have reversed the traditional order of the <math>\phi</math> and <math>z</math> coordinates in an attempt to parallelize structure between cylindrical and T3 coordinates since <math>\lambda_3 \equiv \phi</math>.)

<table align="center" border="1" cellpadding="5">
<tr>
  <td align="center">
&nbsp;
  </td>
  <td align="center">
<math>
\frac{\partial}{\partial \varpi}
</math>
  </td>
  <td align="center">
<math>
\frac{\partial}{\partial z}
</math>
  </td>
  <td align="center">
<math>
\frac{\partial}{\partial \phi}
</math>
  </td>
</tr>

<tr>
  <td align="center">
<math>{\lambda_1}</math>
  </td>
  <td align="center">
<math>
\frac{\varpi}{\lambda_1}
</math>
  </td>
  <td align="center">
<math>
\frac{q^2 z}{\lambda_1}
</math>
  </td>
  <td align="center">
<math>
0
</math>
  </td>
</tr>

<tr>
  <td align="center">
<math>\lambda_2</math>
  </td>
  <td align="center">
<math>
\frac{q^2}{q^2-1} \left( \frac{\varpi}{qz} \right)^{1/(q^2-1)}
</math><br />
  </td>
  <td align="center">
<math>
-\frac{1}{q^2-1} \left( \frac{\varpi^{q^2}}{qz^{q^2}} \right)^{1/(q^2-1)}
</math><br />
  </td>
  <td align="center">
<math>
0
</math><br />
  </td>
</tr>

<tr>
  <td align="center">
<math>\lambda_3</math>
  </td>
  <td align="center">
<math>
0
</math>
  </td>
  <td align="center">
<math>
0
</math>
  </td>
  <td align="center">
<math>
1
</math>
  </td>
</tr>
</table>

Furthermore, the inverted partials are

<table align="center" border="1" cellpadding="5">
<tr>
  <td align="center">
&nbsp;
  </td>
  <td align="center">
<math>
\frac{\partial}{\partial \lambda_1}
</math>
  </td>
  <td align="center">
<math>
\frac{\partial}{\partial \lambda_2}
</math>
  </td>
  <td align="center">
<math>
\frac{\partial}{\partial \lambda_3}
</math>
  </td>
</tr>

<tr>
  <td align="center">
<math>{\varpi}</math>
  </td>
  <td align="center">
<math>
\varpi \ell^2 \lambda_1
</math>
  </td>
  <td align="center">
<math>
(q^2-1) q^2 \varpi z^2 \ell^2 / \lambda_2
</math>
  </td>
  <td align="center">
<math>
0
</math>
  </td>
</tr>

<tr>
  <td align="center">
<math>z</math>
  </td>
  <td align="center">
<math>
q^2 z \ell^2 \lambda_1
</math><br />
  </td>
  <td align="center">
<math>
- (q^2-1) \varpi^2 z \ell^2 / \lambda_2
</math><br />
  </td>
  <td align="center">
<math>
0
</math><br />
  </td>
</tr>

<tr>
  <td align="center">
<math>\phi</math>
  </td>
  <td align="center">
<math>
0
</math>
  </td>
  <td align="center">
<math>
0
</math>
  </td>
  <td align="center">
<math>
1
</math>
  </td>
</tr>
</table>

===Scale Factors===
The scale factors are,

<table align="center" border="0" cellpadding="5">
<tr>
  <td align="right">
<math>h_1^2</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[ \biggl( \frac{\partial\lambda_1}{\partial x} \biggr)^2 + \biggl( \frac{\partial\lambda_1}{\partial y} \biggr)^2 + \biggl( \frac{\partial\lambda_1}{\partial z} \biggr)^2 \biggr]^{-1}
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\lambda_1^2 \ell^2 
</math>
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
&nbsp;
  </td>
</tr>

<tr>
  <td align="right">
<math>h_2^2</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[ \biggl( \frac{\partial\lambda_2}{\partial x} \biggr)^2 + \biggl( \frac{\partial\lambda_2}{\partial y} \biggr)^2 + \biggl( \frac{\partial\lambda_2}{\partial z} \biggr)^2 \biggr]^{-1}
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
(q^2-1)^2 \biggl(\frac{\varpi z \ell}{\lambda_2} \biggr)^2 
</math>
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
&nbsp;
  </td>
</tr>

<tr>
  <td align="right">
<math>h_3^2</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[ \biggl( \frac{\partial\lambda_3}{\partial x} \biggr)^2 + \biggl( \frac{\partial\lambda_3}{\partial y} \biggr)^2 + \biggl( \frac{\partial\lambda_3}{\partial z} \biggr)^2 \biggr]^{-1}
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\varpi^2 
</math>
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
&nbsp;
  </td>
</tr>

<tr>
  <td align="left" colspan="7">
where, &nbsp; &nbsp; &nbsp; &nbsp;<math>\ell \equiv (\varpi^2 + q^4 z^2)^{-1/2}</math>.
  </td>
</tr>
</table>

===Direction Cosines===
The following table contains expressions for the [[User:Tohline/Appendix/Ramblings/DirectionCosines#Basic_Definitions_and_Relations|nine direction cosines]].

<table border="1" align="center" cellpadding="8">
<tr>
  <td align="center" colspan="4">
<font color="darkblue">Direction Cosines for T3 Coordinates</font><br/>
<math>\gamma_{ni} = h_n\frac{\partial\lambda_n}{\partial x_i}</math>
  </td>
</tr>

<tr>
  <td align="center">
&nbsp;
  </td>
  <td align="center" colspan="3">
'''<math>i</math>'''
  </td>
</tr>

<tr>
  <td align="center" valign="middle" rowspan="3">
'''<math>n</math>'''
  </td>
  <td align="center" colspan="1">
<math>\ell x</math>
  </td>
  <td align="center" colspan="1">
<math>\ell y</math>
  </td>
  <td align="center" colspan="1">
<math>q^2 \ell z</math>
  </td>
</tr>

<tr>
  <td align="center" colspan="1">
<math>\frac{q^2 x z \ell}{\varpi}</math>
  </td>
  <td align="center" colspan="1">
<math>\frac{q^2 y z \ell}{\varpi}</math>
  </td>
  <td align="center" colspan="1">
<math>- \varpi \ell</math>
  </td>
</tr>

<tr>
  <td align="center" colspan="1">
<math>- \frac{y}{\varpi}</math>
  </td>
  <td align="center" colspan="1">
<math>+\frac{x}{\varpi}</math>
  </td>
  <td align="center" colspan="1">
<math>0</math>
  </td>
</tr>

<tr>
  <td align="center" colspan="4">
where:
<math>\ell \equiv (\varpi^2 + q^4 z^2)^{-1/2}</math>
  </td>
</tr>
</table>

===Orthogonality Condition===
Next, let's use the example orthogonality condition [[User:Tohline/Appendix/Ramblings/DirectionCosines#Orthogonality|derived elsewhere in connection with our overview of direction cosines]].  Specifically, let's see if [[User:Tohline/Appendix/Ramblings/DirectionCosines#DC.02|Equation DC.02]] is satisfied.
<table align="center" border="0" cellpadding="5">
<tr>
  <td align="right">
<math>
\frac{\partial\lambda_1}{\partial \varpi} \cdot  \frac{\partial\lambda_2}{\partial \varpi} + 
\frac{\partial\lambda_1}{\partial z} \cdot \frac{\partial\lambda_2}{\partial z}
</math>
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
\frac{\varpi}{\lambda_1} \biggl[ \frac{q^2}{q^2-1} \left( \frac{\varpi}{qz} \right)^{1/(q^2-1)}
 \biggr] - 
\frac{q^2z}{\lambda_1} \biggl[ \frac{1}{q^2-1} \left( \frac{\varpi^{q^2}}{qz^{q^2}} \right)^{1/(q^2-1)}
 \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
\frac{q^2}{(q^2-1)\lambda_1} \biggl\{ \varpi\biggl[ \left( \frac{\varpi}{qz} \right)^{1/(q^2-1)}
 \biggr] - 
z \biggl[ \left( \frac{\varpi^{q^2}}{qz^{q^2}} \right)^{1/(q^2-1)}
 \biggr] \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
\frac{q^2}{(q^2-1)\lambda_1} \biggl[ q^{-1/(q^2-1)} \varpi^{1 + 1/(q^2-1)} z^{-1/(q^2-1)} - q^{-1/(q^2-1)} \varpi^{q^2/(q^2-1)} z^{1-q^2/(q^2-1)} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
0 .
</math>
  </td>
</tr>
</table>
Hence, the key orthogonality condition defined by [[User:Tohline/Appendix/Ramblings/DirectionCosines#DC.02|Equation DC.02]] is satisfied. MF53 also gives us relationships that should apply ''between'' the various direction cosines if the coordinate system is orthogonal.  Let's check a few cases to see whether <math>\gamma_{mn} = M_{mn}</math>, where "<math>M_{mn}</math> is the minor of <math>\gamma_{mn}</math> in the determinant <math>|\gamma_{mn}|</math>":
<div align="center">
<math>
M_{11} = \gamma_{22}\gamma_{33} - \gamma_{23}\gamma_{32} = - (-\varpi\ell)\frac{x}{\varpi} = + \ell x .
</math> <br /><br />

<math>
M_{12} = \gamma_{23}\gamma_{31} - \gamma_{21}\gamma_{33} = -\varpi\ell \biggl(-\frac{y}{\varpi}\biggr) = +\ell y .
</math> <br /><br />

<math>
M_{13} = \gamma_{21}\gamma_{32} - \gamma_{22}\gamma_{31} = \biggl( \frac{q^2 x z \ell}{\varpi} \biggr) \frac{x}{\varpi} - \biggl( \frac{q^2 y z \ell}{\varpi} \biggr) \biggl( -\frac{y}{\varpi} \biggr) = q^2 \ell z.
</math> <br /><br />

<math>
M_{31} = \gamma_{12}\gamma_{23} - \gamma_{13}\gamma_{22} = \ell y (-\varpi\ell) - q^2 \ell z \biggl(\frac{q^2 yz\ell}{\varpi}\biggr) = - \frac{y \ell^2}{\varpi} \biggl( \varpi^2 + q^4 z^2 \biggr) = - \frac{y}{\varpi} .
</math> <br /><br />

<math>
M_{33} = \gamma_{11}\gamma_{22} - \gamma_{12}\gamma_{21} = \ell x \biggr(\frac{q^2 y z \ell}{\varpi} \biggr) - \ell y \biggr(\frac{q^2 x z \ell}{\varpi} \biggr) = 0 .
</math>
</div>
All of these beautifully obey the relationship, <math>\gamma_{mn} = M_{mn}</math>.

===Position Vector===
The position vector is,

<table align="center" border="0" cellpadding="5">
<tr>
  <td align="right">
<math>\vec{x}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\hat{i}x + \hat{j}y + \hat{k}z
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\hat{e}_1 (h_1 \lambda_1) + \hat{e}_2 (h_2 \lambda_2) .
</math>
  </td>
</tr>
</table>

==Vector Derivatives==

For orthogonal coordinate systems, the time-rate-of-change of the three unit vectors are given by the expressions,
<table align="center" border="0" cellpadding="3">
<tr>
  <td align="right">
<math>
\frac{d}{dt}\hat{e}_1 
</math>
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
\hat{e}_2 A + \hat{e}_3 B
</math>
  </td>
</tr>
<tr>
  <td align="right">  
<math>
\frac{d}{dt}\hat{e}_2 
</math>
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
- \hat{e}_1 A + \hat{e}_3 C
</math>
  </td>
</tr>
<tr>
  <td align="right">  
<math>
\frac{d}{dt}\hat{e}_3 
</math>
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
- \hat{e}_1 B - \hat{e}_2 C
</math>
  </td>
</tr> 
</table>

where,
<table align="center" border="0" cellpadding="3">
<tr>
  <td align="right">
<math>
A
</math>
  </td>
  <td align="center">
<math>
\equiv
</math>
  </td>
  <td align="left">
<math>
\frac{\dot{\lambda}_2}{h_1} \frac{\partial h_2}{\partial \lambda_1} -
\frac{\dot{\lambda}_1}{h_2} \frac{\partial h_1}{\partial \lambda_2}
</math>
  </td>
</tr>
<tr>
  <td align="right">  
<math>
B
</math>
  </td>
  <td align="center">
<math>
\equiv
</math>
  </td>
  <td align="left">
<math>
\frac{\dot{\lambda}_3}{h_1} \frac{\partial h_3}{\partial \lambda_1} -
\frac{\dot{\lambda}_1}{h_3} \frac{\partial h_1}{\partial \lambda_3}
</math>
  </td>
</tr>
<tr>
  <td align="right">  
<math>
C
</math>
  </td>
  <td align="center">
<math>
\equiv
</math>
  </td>
  <td align="left">
<math>
\frac{\dot{\lambda}_3}{h_2} \frac{\partial h_3}{\partial \lambda_2} -
\frac{\dot{\lambda}_2}{h_3} \frac{\partial h_2}{\partial \lambda_3}
</math>
  </td>
</tr> 
</table>

Another way of expressing this involves Christoffel symbols and is most easily written using index notation.
  <div align="center">
<math>
\frac{d}{dt} \hat{e}_c = \frac{h_a}{h_c} \ \Gamma^a_{bc} \dot{\lambda}_b \ \hat{e}_a \ \ (a \ne c)
</math>
  </div>

Here, we have been admittedly slopping with the placement and notation of indices in order to best accommodate the notation we have been using up to here.  The <math>b</math> index is summed over all the coordinates.  The <math>a</math> index is summed over all coordinates EXCEPT the <math>c</math> coordinate.  The <math>c</math> index is NOT summed over because it is a free index, meaning that it can equal any of the coordinates depending on which unit vector you want to differentiate.  

Writing this out for each of the individual unit vectors, and striking through terms that are automatically zero when dealing with an orthogonal coordinate system, produces

<table align="center" border="0" cellpadding="3">
<tr>
  <td align="right">
<math>
\frac{d}{dt} \hat{e}_1
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\overbrace{\frac{h_2}{h_1} \left( \Gamma^2_{11} \dot{\lambda}_1 + \Gamma^2_{21} \dot{\lambda}_2 + \cancel{\Gamma^2_{31} \dot{\lambda}_3} \right)}^{A} \hat{e}_2 + \overbrace{\frac{h_3}{h_1} \left( \Gamma^3_{11} \dot{\lambda}_1 + \cancel{\Gamma^3_{21} \dot{\lambda}_2} + \Gamma^3_{31} \dot{\lambda}_3 \right)}^{B} \hat{e}_3
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>
\frac{d}{dt} \hat{e}_2
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\overbrace{\frac{h_1}{h_2} \left( \Gamma^1_{12} \dot{\lambda}_1 + \Gamma^1_{22} \dot{\lambda}_2 + \cancel{\Gamma^1_{32} \dot{\lambda}_3} \right)}^{-A} \hat{e}_1 + \overbrace{\frac{h_3}{h_2} \left( \cancel{\Gamma^3_{12} \dot{\lambda}_1} + \Gamma^3_{22} \dot{\lambda}_2 + \Gamma^3_{32} \dot{\lambda}_3 \right)}^{C} \hat{e}_3
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>
\frac{d}{dt} \hat{e}_3
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\overbrace{\frac{h_1}{h_3} \left( \Gamma^1_{13} \dot{\lambda}_1 + \cancel{\Gamma^1_{23} \dot{\lambda}_2} + \Gamma^1_{33} \dot{\lambda}_3 \right)}^{-B} \hat{e}_1 + \overbrace{\frac{h_2}{h_3} \left( \cancel{\Gamma^2_{13} \dot{\lambda}_1} + \Gamma^2_{23} \dot{\lambda}_2 + \Gamma^2_{33} \dot{\lambda}_3 \right)}^{-C} \hat{e}_2
</math>
  </td>
</tr>
</table>

where, quite generally, the 27 Christoffel symbols are,

<table align="center" border="0" cellpadding="3">
<tr>
  <td align="right">
<math>\Gamma^1_{11}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\frac{\partial_1 h_1}{h_1}</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>\Gamma^1_{12} = \Gamma^1_{21}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\frac{\partial_2 h_1}{h_1}</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>\Gamma^1_{13} = \Gamma^1_{31}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>0</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>\Gamma^1_{22}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>-\frac{h_2}{h_1} \frac{\partial_1 h_2}{h_1}</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>\cancel{\Gamma^1_{23}} = \cancel{\Gamma^1_{32}}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>0</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>\Gamma^1_{33}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>-\frac{h_3}{h_1} \frac{\partial_1 h_3}{h_1}</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Gamma^2_{11}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>-\frac{h_1}{h_2} \frac{\partial_2 h_1}{h_2}</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>\Gamma^2_{12} = \Gamma^2_{21}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\frac{\partial_1 h_2}{h_2}</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>\cancel{\Gamma^2_{13}} = \cancel{\Gamma^2_{31}}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>0</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>\Gamma^2_{22}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\frac{\partial_2 h_2}{h_2}</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>\Gamma^2_{23} = \Gamma^2_{32}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>0</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>\Gamma^2_{33}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>-\frac{h_3}{h_2} \frac{\partial_2 h_3}{h_2}</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Gamma^3_{11}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>0</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>\cancel{\Gamma^3_{12}} = \cancel{\Gamma^3_{21}}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>0</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>\Gamma^3_{13} = \Gamma^3_{31}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\frac{\partial_1 h_3}{h_3}</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>\Gamma^3_{22}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>0</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>\Gamma^3_{23} = \Gamma^3_{32}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\frac{\partial_2 h_3}{h_3}</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>\Gamma^3_{33}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>0</math>
  </td>
</tr>
</table>
Notice that it is the Christoffel symbols labeled with all three of the coordinate indices that are automatically zero for orthogonal coordinate systems.

For additional details surrounding the Christoffel symbols (the significant role they play in the field equations, and how they can be calculated) visit this page [[User:Jaycall/KillingVectorApproach|coming soon]].

==Time-Derivative of Position and Velocity Vectors==
In general for an orthogonal coordinate system, the velocity vector can be written as,
<div align="center">
<math>
\vec{v} = \hat{e}_1 (h_1 \dot{\lambda}_1) + \hat{e}_2 (h_2 \dot{\lambda}_2) +\hat{e}_3 (h_3 \dot{\lambda}_3) .
</math>
</div>
So, in general, the time-rate-of-change of the velocity vector is,

<table align="center" border="0" cellpadding="5">
<tr>
  <td align="right">
<math>\frac{d\vec{v}}{dt}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\hat{e}_1 \biggl[\frac{d(h_1 \dot{\lambda}_1)}{dt}\biggr] +  (h_1 \dot{\lambda}_1)\frac{d\hat{e}_1}{dt} +
\hat{e}_2 \biggl[\frac{d(h_2 \dot{\lambda}_2)}{dt}\biggr] +  (h_2 \dot{\lambda}_2)\frac{d\hat{e}_2}{dt} +
\hat{e}_3 \biggl[\frac{d(h_3 \dot{\lambda}_3)}{dt}\biggr] +  (h_3 \dot{\lambda}_3)\frac{d\hat{e}_3}{dt}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\hat{e}_1 \biggl[\frac{d(h_1 \dot{\lambda}_1)}{dt}\biggr] +  (h_1 \dot{\lambda}_1)\biggl[ \hat{e}_2 A + \hat{e}_3 B \biggr] +
\hat{e}_2 \biggl[\frac{d(h_2 \dot{\lambda}_2)}{dt}\biggr] +  (h_2 \dot{\lambda}_2)\biggl[ - \hat{e}_1 A + \hat{e}_3 C \biggr] +
\hat{e}_3 \biggl[\frac{d(h_3 \dot{\lambda}_3)}{dt}\biggr] +  (h_3 \dot{\lambda}_3)\biggl[ - \hat{e}_1 B - \hat{e}_2 C \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\hat{e}_1 \biggl[\frac{d(h_1 \dot{\lambda}_1)}{dt} - A(h_2 \dot{\lambda}_2) - B(h_3 \dot{\lambda}_3) \biggr] + 
\hat{e}_2 \biggl[\frac{d(h_2 \dot{\lambda}_2)}{dt} + A(h_1 \dot{\lambda}_1) - C(h_3 \dot{\lambda}_3) \biggr] + 
\hat{e}_3 \biggl[\frac{d(h_3 \dot{\lambda}_3)}{dt} + B(h_1 \dot{\lambda}_1) + C(h_2 \dot{\lambda}_2) \biggr] 
</math>
  </td>
</tr>
</table>

Now, for the T3 coordinate system the position vector has a similar form, specifically,
<div align="center">
<math>
\vec{x} = \hat{e}_1 (h_1 {\lambda}_1) + \hat{e}_2 (h_2 {\lambda}_2) .
</math>
</div>
By analogy, then, the time-rate-of-change of the position vector is,

<table align="center" border="0" cellpadding="5">
<tr>
  <td align="right">
<math>\frac{d\vec{x}}{dt}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\hat{e}_1 \biggl[\frac{d(h_1 {\lambda}_1)}{dt}\biggr] +  (h_1 {\lambda}_1)\frac{d\hat{e}_1}{dt} +
\hat{e}_2 \biggl[\frac{d(h_2 {\lambda}_2)}{dt}\biggr] +  (h_2 {\lambda}_2)\frac{d\hat{e}_2}{dt} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\hat{e}_1 \biggl[\frac{d(h_1 {\lambda}_1)}{dt}\biggr] +  (h_1 {\lambda}_1)\biggl[ \hat{e}_2 A + \hat{e}_3 B \biggr] +
\hat{e}_2 \biggl[\frac{d(h_2 {\lambda}_2)}{dt}\biggr] +  (h_2 {\lambda}_2)\biggl[ - \hat{e}_1 A + \hat{e}_3 C \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\hat{e}_1 \biggl[\frac{d(h_1 {\lambda}_1)}{dt} - A(h_2 {\lambda}_2) \biggr] + 
\hat{e}_2 \biggl[\frac{d(h_2 {\lambda}_2)}{dt} + A(h_1 {\lambda}_1) \biggr] +
\hat{e}_3 \biggl[B(h_1 {\lambda}_1) + C(h_2 {\lambda}_2) \biggr]
</math>
  </td>
</tr>
</table>

==Derived Identity for T3 Coordinates==

Looking at the "<math>\hat{e}_2</math>" component of this last expression, we have,

<div align="center">
<math>
\hat{e}_2 \cdot \frac{d\vec{x}}{dt} = \frac{d(h_2 {\lambda}_2)}{dt} + A(h_1 {\lambda}_1) .
</math>
</div>

But we also know that,

<div align="center">
<math>
\hat{e}_2 \cdot \vec{v} = h_2 \dot{\lambda}_2 .
</math>
</div>

Hence, it must be true that,

<table align="center" border="1" cellpadding="10" width="50%">
<tr>
  <td align="center">
<font color="darkblue">
For T3 Coordinates
</font>
  </td>
</tr>
<tr>
  <td align="center">
<math>
\lambda_2 \frac{d h_2}{dt} = - A(h_1 {\lambda}_1) 
</math><br /><br />
or, equivalently,<br /><br />
<math>
A  = - \frac{\lambda_2}{h_1 \lambda_1} \frac{d h_2}{dt}
</math>
  </td>
</tr>
</table>

At some point, this identity needs to be checked by taking various partial derivatives of the scale factors and plugging them into the generic definition of <math>A</math>, given above.  (<font color="red">Actually, this shouldn't be necessary because in January, 2009, we derived the same equation-of-motion result shown below while using the uglier expression for <math>A</math>.  So this must be a correct identity in the context of T3 coordinates.</font>)

==Implications of Equation of Motion==

Looking now at the "<math>\hat{e}_2</math>" component of the acceleration (which we will set equal to zero), and assuming no motion in the <math>3^\mathrm{rd}</math> component direction, we have,

<div align="center">
<math>
\hat{e}_2 \cdot \frac{d\vec{v}}{dt} = \frac{d(h_2 \dot{\lambda}_2)}{dt} + A(h_1 \dot{\lambda}_1) =0 
</math><br /><br />
<math>
\Rightarrow ~~~~~ \frac{d(h_2 \dot{\lambda}_2)}{dt} = -  A(h_1 \dot{\lambda}_1) 
</math>
</div>

or, inserting the relation derived above for <math>A</math> in terms of <math>dh_2/dt</math> for T3 coordinates,

<span id="EOM.01"><table align="right" border="1" cellpadding="10" width="10%">
<tr><th><font color="darkblue">EOM.01</font></th></tr>
</table></span>
<div align="center">
<math>
\Rightarrow ~~~~~ \frac{d(h_2 \dot{\lambda}_2)}{dt} =  \biggl(\frac{\lambda_2 \dot{\lambda}_1}{\lambda_1}\biggr) \frac{dh_2}{dt} 
</math>
<br /><br />
<math>
\Rightarrow ~~~~~ \frac{d(h_2 \lambda_1 \dot{\lambda}_2)}{dt} =  \dot{\lambda}_1 \frac{d(h_2 \lambda_2)}{dt} .
</math>
</div>

Or, equivalently (but perhaps more perversely),

<span id="EOM.02"><table align="right" border="1" cellpadding="10" width="10%">
<tr><th><font color="darkblue">EOM.02</font></th></tr>
</table></span>
<div align="center">
<math>
\frac{\dot{\lambda}_2}{\lambda_2} \biggl[ \frac{1}{h_2 \dot{\lambda}_2} \frac{d(h_2 \dot{\lambda}_2)}{dt} \biggr] =  \frac{\dot{\lambda}_1}{\lambda_1} \biggl[ \frac{1}{h_2} \frac{dh_2}{dt} \biggr] 
</math>
<br /><br />
<math>
\Rightarrow ~~~~~ \frac{d \ln\lambda_2}{dt} \biggl[ \frac{d\ln(h_2 \dot{\lambda}_2)}{dt} \biggr] =  \frac{d\ln\lambda_1}{dt} \biggl[\frac{d\ln h_2}{dt} \biggr] 
</math>

</div><br />

<font color="red">'''Note that Equation '''EOM.01''' is equivalent to the <i>simplest form</i> of the conservation that we derived &#8212; actually, Jay Call derived it &#8212; back in January, 2009.'''</font>  Specifically,
<table align="center" border="1" cellpadding="10" width="30%">
<tr>
  <td align="center">
<math>
\lambda_1 \frac{d(h_2 \dot{\lambda}_2)}{dt} =  \lambda_2 \dot{\lambda}_1 \frac{dh_2}{dt} 
</math>
  </td>
</tr>
</table>

<font color="green">'''The 64-thousand dollar question is, "Can we turn any of these expressions into a form which states that the total time-derivative of some function equals zero?" '''</font>

<!--
Another option is to look at the equivalent angular momentum element, <math>h_2^2 \dot{\lambda}_2</math>:


<div align="center">
<math>
\hat{e}_2 \cdot \frac{d\vec{v}}{dt} = \frac{d(h_2 \dot{\lambda}_2)}{dt} + A(h_1 \dot{\lambda}_1) =0 
</math><br /><br />
<math>
\Rightarrow ~~~~~ h_2 \frac{d(h_2 \dot{\lambda}_2)}{dt} = -  A(h_1 h_2 \dot{\lambda}_1) 
</math><br /><br />
<math>
\Rightarrow ~~~~~ \frac{d(h_2^2 \dot{\lambda}_2)}{dt} - h_2 \dot{\lambda}_2 \frac{dh_2}{dt} = -  A(h_1 h_2 \dot{\lambda}_1) 
</math><br /><br />
<math>
\Rightarrow ~~~~~ \frac{d(h_2^2 \dot{\lambda}_2)}{dt} + h_2 \dot{\lambda}_2 \biggl[A h_1 \biggl( \frac{\lambda_1}{\lambda_2} \biggr)  \biggr] = -  A(h_1 h_2 \dot{\lambda}_1) 
</math><br /><br />
<math>
\Rightarrow ~~~~~ \frac{d(h_2^2 \dot{\lambda}_2)}{dt} = -  A h_1 h_2 \lambda_1 \biggl[ \frac{\dot{\lambda}_1}{\lambda_1} + 
\frac{\dot{\lambda}_2}{\lambda_2} \biggr] = -   h_1 h_2 \lambda_1 \biggl[ \frac{\dot{\lambda}_1}{\lambda_1} + 
\frac{\dot{\lambda}_2}{\lambda_2} \biggr]\biggl[\frac{\dot{\lambda}_2}{h_1} \frac{\partial h_2}{\partial \lambda_1} -
\frac{\dot{\lambda}_1}{h_2} \frac{\partial h_1}{\partial \lambda_2}\biggr] .
</math><br /><br />

</div>

<font color="green">'''I'm not yet quite sure what to do with all of this.'''</font>
-->

Let's look at the <math>1^\mathrm{st}</math> component of the equation of motion.  

<div align="center">
<math>
\hat{e}_1 \cdot \frac{d\vec{v}}{dt} = \frac{d(h_1 \dot{\lambda}_1)}{dt} - A(h_2 \dot{\lambda}_2) = - \frac{1}{h_1}\frac{\partial\Phi}{\partial\lambda_1} 
</math><br /><br />
<math>
\Rightarrow ~~~~~ \frac{d(h_1 \dot{\lambda}_1)}{dt} = A(h_2 \dot{\lambda}_2) - \frac{1}{h_1}\frac{\partial\Phi}{\partial\lambda_1} = - \biggl[ \frac{h_2 \lambda_2 }{h_1\lambda_1} \biggr]\dot{\lambda}_2 \frac{dh_2}{dt} - \frac{1}{h_1}\frac{\partial\Phi}{\partial\lambda_1}
</math>
</div>

==Logarithmic Derivatives of Scale Factors==

Given the collection of expressions detailed in our derivation up to this point, the logarithmic derivatives of the scale factors are:

<table align="center" border="1" cellpadding="5">
<tr>
  <td align="right">
<math>
\frac{\partial\ln h_1}{\partial\ln\lambda_1}
</math>
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
- \biggl( \frac{q h_1 h_2 \lambda_2}{\lambda_1 }  \biggr)^2
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>
\frac{\partial\ln h_1}{\partial\ln\lambda_2}
</math>
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
+ \biggl( \frac{q h_1 h_2 \lambda_2}{\lambda_1 }  \biggr)^2
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>
\frac{\partial\ln h_2}{\partial\ln\lambda_1}
</math>
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
+ (q h_1^2)^2
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>
\frac{\partial\ln h_2}{\partial\ln\lambda_2}
</math>
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
- ( qh_1^2 )^2
</math>
  </td>
</tr>
</table>

This means, for example, that,
<div align="center">
<math>
\frac{\partial\ln h_1}{\partial\ln\lambda_2} = - \frac{\partial\ln h_1}{\partial\ln\lambda_1} ,
</math>
</div>

and,
<div align="center">
<math>
\frac{\partial\ln h_2}{\partial\ln\lambda_2} = - \frac{\partial\ln h_2}{\partial\ln\lambda_1} .
</math>
</div>

It also means that,
<div align="center">
<math>
\frac{d\ln h_2}{dt} = \biggl(\frac{\partial\ln h_2}{\partial\ln\lambda_1} \biggr) \frac{\dot{\lambda}_1}{\lambda_1} + \biggl(\frac{\partial\ln h_2}{\partial\ln\lambda_2} \biggr) \frac{\dot{\lambda}_2}{\lambda_2} = ( q h_1^2)^2 \biggl[\frac{\dot{\lambda}_1}{\lambda_1} - \frac{\dot{\lambda}_2}{\lambda_2} \biggr] 
</math><br />

<math>
\Rightarrow ~~~~~ \frac{1}{h_1^4}\frac{d\ln h_2}{dt} = \frac{d\ln(\lambda_1/\lambda_2)^{q^2}}{dt}  ;
</math>
</div>

and,
<div align="center">
<math>
\frac{d\ln h_1}{dt} = \biggl(\frac{\partial\ln h_1}{\partial\ln\lambda_1} \biggr) \frac{\dot{\lambda}_1}{\lambda_1} + \biggl(\frac{\partial\ln h_1}{\partial\ln\lambda_2} \biggr) \frac{\dot{\lambda}_2}{\lambda_2} = - \biggl( \frac{q h_1 h_2 \lambda_2}{\lambda_1}  \biggr)^2 \biggl[\frac{\dot{\lambda}_1}{\lambda_1}- \frac{\dot{\lambda}_2}{\lambda_2} \biggr] 
</math><br />

<math>
\Rightarrow ~~~~~ - \biggl( \frac{\lambda_1}{h_1 h_2 \lambda_2}  \biggr)^2 \frac{d\ln h_1}{dt} =  \frac{d\ln(\lambda_1/\lambda_2)^{q^2}}{dt}.
</math>

</div>

Hence, the logarithmic time-derivatives of the two key scale factors can be related to one another via the expression,
<table border="1" align="center" cellpadding="10" width="40%">
<tr>
  <th align="center">
<font color="darkblue">T3 Coordinates</font>
  </th>
</tr>

<tr><td align="center">
<math>
(h_1 \lambda_1)^2 \frac{d\ln h_1}{dt} + (h_2 \lambda_2)^2 \frac{d\ln h_2}{dt}  = 0 .
</math>
</td></tr>
</table>

Also, plugging the relevant logarithmic derivatives of the key scale factors directly into the definition of <math>A</math> given in our discussion of vector derivatives, above, we can confirm the relationship derived earlier between <math>A</math> and <math>dh_2/dt</math>.  We also can relate <math>A</math> directly to <math>dh_1/dt</math>.  Specifically,

<table border="1" align="center" cellpadding="10" width="40%">
<tr>
  <th align="center">
<font color="darkblue">T3 Coordinates</font>
  </th>
</tr>

<tr><td align="center">
<math>
A = - \biggl( \frac{h_2 \lambda_2}{h_1 \lambda_1} \biggr) \frac{d\ln h_2}{dt} = + \biggl( \frac{h_1 \lambda_1}{h_2 \lambda_2} \biggr) \frac{d\ln h_1}{dt} .
</math>
</td></tr>
</table>

It must therefore also be true that,

<table border="1" align="center" cellpadding="10" width="40%">
<tr>
  <th align="center">
<font color="darkblue">T3 Coordinates</font>
  </th>
</tr>

<tr><td align="center">
<math>
(h_1\lambda_1)^2 \frac{\partial \ln h_1}{\partial\lambda_j} = - (h_2\lambda_2)^2 \frac{\partial \ln h_2}{\partial\lambda_j}
</math>
</td></tr>
</table>

=Special Case (quadratic)=

On a [[User:Tohline/Appendix/Ramblings/T3Integrals/QuadraticCase|separate page]] we examine the structure of the second component of the equation of motion in the special "quadratic" case of <math>q^2 = 2</math>.  Jay Call is independently investigating this same special case [[User:Jaycall/T3 Coordinates/Special_Case|on yet another wiki page]] so that we can ultimately compare and check each others' results.

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

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