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User:Jaycall/T3 Coordinates/Special Case

2010-07-17T19:31:39Z

Jaycall: /* Conserved Quantity */ Improved form of integral slightly

==Coordinate Transformations==

If the special case <math>q^2=2</math> is considered, it is possible to invert the coordinate transformations in closed form. The coordinate transformations and their inversions become

<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>\left( R^2+2z^2 \right)^{1/2}</math>
</td>
<td align="center">
      and      
</td>
<td align="right">
<math>
\lambda_2
</math>
</td>
<td align="center">
<math>\equiv</math>
</td>
<td align="left">
<math>\frac{R^2}{\sqrt{2}z}</math>
</td>
</tr>
<tr>
<td align="right">
<math>
R^2
</math>
</td>
<td align="center">
<math>\equiv</math>
</td>
<td align="left">
<math>\frac{\lambda_2}{2} \left( - \lambda_2 + \sqrt{4{\lambda_1}^2 + {\lambda_2}^2} \right) = \frac{2{\lambda_1}^2}{\Lambda+1}</math>
</td>
<td align="center">
      and      
</td>
<td align="right">
<math>
z
</math>
</td>
<td align="center">
<math>\equiv</math>
</td>
<td align="left">
<math>\frac{1}{2^{3/2}} \left( -\lambda_2 + \sqrt{4 {\lambda_1}^2+{\lambda_2}^2} \right) = \frac{\lambda_2}{2^{3/2}} \left( \Lambda -1 \right)</math>
</td>
</tr>
</table>

where <math>\Lambda \equiv \left[ 1 + \left( \frac{2 \lambda_1}{\lambda_2} \right)^2 \right]^{1/2}</math> .

From this definition of <math>\Lambda</math>, we can compute both its partials with respect to the T3 coordinates, and its total time derivative.

<div align="center">
<math>\frac{\partial \Lambda}{\partial \lambda_1} = \frac{4 \lambda_1 / {\lambda_2}^2}{\Lambda} = \frac{2 \left( \Lambda^2 - 1 \right)^{1/2}}{\Lambda} \frac{1}{\lambda_2} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{\partial \Lambda}{\partial \lambda_2} = - \frac{4 {\lambda_1}^2 / \lambda_2}{\Lambda} = - \frac{2 \left( \Lambda^2 - 1 \right)^{1/2}}{\Lambda} \lambda_1</math>
</div>

<div align="center">
<math>\dot{\Lambda} = \frac{2 \left( \Lambda^2 - 1 \right)^{1/2}}{\Lambda} \left( \frac{\dot{\lambda_1}}{\lambda_2} - \lambda_1 \dot{\lambda_2} \right)</math>
</div>

==Partials of the Coordinates==

Partial derivatives of each of the T3 coordinates taken with respect to each of the cylindrical coordinates are:

<table align="center" border="1" cellpadding="5">
<tr>
<td align="center">
 
</td>
<td align="center">
<math>
\frac{\partial}{\partial R}
</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{R}{\lambda_1} = \left( \frac{2}{\Lambda + 1} \right)^{1/2}
</math>
</td>
<td align="center">
<math>
\frac{2z}{\lambda_1} = \left[ \frac{2 \left( \Lambda - 1 \right) }{\Lambda + 1} \right]^{1/2}
</math>
</td>
<td align="center">
<math>
0
</math>
</td>
</tr>

<tr>
<td align="center">
<math>\lambda_2</math>
</td>
<td align="center">
<math>
\frac{2 \lambda_2}{R} = \frac{2^{3/2}}{\left( \Lambda - 1 \right)^{1/2}}
</math>
</td>
<td align="center">
<math>
-\frac{\lambda_2}{z} = - \frac{2^{3/2}}{\Lambda - 1}
</math>
</td>
<td align="center">
<math>0</math>
</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>

And partials of the cylindrical coordinates taken with respect to the T3 coordinates are:

<table align="center" border="1" cellpadding="5">
<tr>
<td align="center">
 
</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>R</math>
</td>
<td align="center">
<math>
R \ell^2 \lambda_1 = \frac{1}{\Lambda} \left( \frac{\Lambda + 1}{2} \right)^{1/2}
</math>
</td>
<td align="center">
<math>
2Rz^2 \ell^2 / \lambda_2 = \frac{\left( \Lambda - 1 \right)^{3/2}}{2 \Lambda}
</math>
</td>
<td align="center">
<math>
0
</math>
</td>
</tr>

<tr>
<td align="center">
<math>z</math>
</td>
<td align="center">
<math>
2z \ell^2 \lambda_1 = \frac{1}{\Lambda} \left( \frac{\Lambda^2 - 1}{2} \right)^{1/2}
</math>
</td>
<td align="center">
<math>
-R^2 z \ell^2 / \lambda_2 = - \frac{\Lambda - 1}{2^{3/2} \Lambda}
</math>
</td>
<td align="center">
<math>0</math>
</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>

where <math>\ell \equiv \left( R^2 + 4z^2 \right)^{-1/2} = \frac{1}{\lambda_1} \left( \frac{\Lambda + 1}{2 \Lambda} \right)^{1/2}</math>.

==Scale Factors==

Furthermore, the scale factors become

<table align="center">
<tr>
<td align="right">
<math>h_1</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>\lambda_1 \ell = \left( \frac{\Lambda+1}{2 \Lambda} \right)^{1/2}</math>
</td>
</tr>

<tr>
<td align="right">
<math>h_2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>Rz \ell / \lambda_2 = \frac{\Lambda-1}{2 \left( 2 \Lambda \right)^{1/2}}</math>
</td>
</tr>

<tr>
<td align="right">
<math>h_3</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>R = \lambda_3</math>
</td>
</tr>
</table>

==Useful Relationships==

In this special case, there are some additional useful relationships between various combinations of cylindrical variables and their T3 equivalents which can be written out.

<table align="center">
<tr>
<td align="right">
<math>R^2 + 2z^2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>{\lambda_1}^2</math>
</td>
</tr>

<tr>
<td align="right">
<math>R^2 + 4z^2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>2 {\lambda_1}^2 + {\lambda_2}^2/2 - \lambda_2 \sqrt{{\lambda_1}^2+{\lambda_2}^2/4} = \ell^{-2}</math>
</td>
</tr>

<tr>
<td align="right">
<math>R^2 + 8z^2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>4 {\lambda_1}^2 + \tfrac{3}{2} {\lambda_2}^2 - 3 \lambda_2 \sqrt{{\lambda_1}^2+{\lambda_2}^2/4} = 3 \ell^{-2} - 2 {\lambda_1}^2</math>
</td>
</tr>

<tr>
<td align="right">
<math>R^2 - 2z^2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>- {\lambda_1}^2 - {\lambda_2}^2 + 2 \lambda_2 \sqrt{{\lambda_1}^2+{\lambda_2}^2/4} = 3 {\lambda_1}^2 -2 \ell^{-2}</math>
</td>
</tr>

<tr>
<td align="right">
<math>Rz</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>\sqrt{\sqrt{2}\lambda_2} \left( -\frac{\lambda_2}{2\sqrt{2}} + \sqrt{\frac{4{\lambda_1}^2+{\lambda_2}^2}{8}} \right)^{3/2} = h_2 \lambda_2 / \ell</math>
</td>
</tr>
</table>

==Additional Partials==

Partials of <math>\ell</math> can be taken with respect to the coordinates of either system. They are:

<table align="center" border="1" cellpadding="5">
<tr>
<td align="center">
 
</td>
<td align="center">
<math>
\frac{\partial}{\partial R}
</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>\ell</math>
</td>
<td align="center">
<math>
-R \ell^3
</math>
</td>
<td align="center">
<math>
-4z \ell^3
</math>
</td>
<td align="center">
<math>
0
</math>
</td>
</tr>
</table>

<table align="center" border="1" cellpadding="5">
<tr>
<td align="center">
 
</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>\ell</math>
</td>
<td align="center">
<math>
- \left( R^2 + 8z^2 \right) \ell^5 \lambda_1 = \ell^3 \lambda_1 \left( 2{h_1}^2 - 3 \right)
</math>
</td>
<td align="center">
<math>
2R^2 z^2 \ell^5 / \lambda_2 = 2 {h_2}^2 \ell^3 \lambda_2
</math>
</td>
<td align="center">
<math>
0
</math>
</td>
</tr>
</table>

Partials of the scale factors taken with respect to the T3 coordinates are:

<table align="center" border="1" cellpadding="5">
<tr>
<td align="center">
 
</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>h_1</math>
</td>
<td align="center">
<math>
\ell \left( 2 {h_1}^4 - 3 {h_1}^2 + 1 \right) = 2 h_2 \lambda_2
</math>
</td>
<td align="center">
<math>
2 {h_2}^2 \ell^3 \lambda_1 \lambda_2
</math>
</td>
<td align="center">
<math>
0
</math>
</td>
</tr>

<tr>
<td align="center">
<math>h_2</math>
</td>
<td align="center">
<math>
2 {h_1}^2 h_2 \ell^2 \lambda_1 = 2 h_2 \ell^4 {\lambda_1}^3
</math>
</td>
<td align="center">
<math>
h_2 \left( 2 {h_2}^2 \ell^2 {\lambda_2}^2 - 3 \ell^2 {\lambda_1}^2 + 1 \right) / \lambda_2
</math>
</td>
<td align="center">
<math>0</math>
</td>
</tr>

<tr>
<td align="center">
<math>h_3</math>
</td>
<td align="center">
<math>
R \ell^2 \lambda_1
</math>
</td>
<td align="center">
<math>
2Rz^2 \ell^2 / \lambda_2
</math>
</td>
<td align="center">
<math>
0
</math>
</td>
</tr>
</table>

==Conserved Quantity==

The conserved quantity associated with the <math>\lambda_2</math> coordinate is

<div align="center">
<math>
m{h_2}^2 \dot{\lambda_2} \exp \int \left[ \left( 4 {\lambda_1}^2 + {\lambda_2}^2 - \lambda_2 \sqrt{4{\lambda_1}^2 + {\lambda_2}^2} \right) \left( \frac{{\lambda_1}^2 \dot{\lambda_2}}{\lambda_2} - \frac{\lambda_2 {\dot{\lambda_1}}^2}{\dot{\lambda_2}} \right) \right] dt .
</math>
</div>

The quantity in brackets needs to be integrated. In terms of <math>\Lambda</math>, it can be written

<div align="center">
<math>{\lambda_2}^2 \Lambda \left( \Lambda - 1 \right) \left[ \frac{\left( \Lambda^2 - 1 \right)^{1/2}}{2} \lambda_1 \dot{\lambda_2} - {\lambda_2}^2 \frac{\dot{\lambda_1}}{\dot{\lambda_2}} \frac{\dot{\lambda_1}}{\lambda_2} \right]</math> .
</div>

Notice that the thing in square brackets looks very closely related to <math>\dot{\Lambda}</math>. Could this be a hint? If only we could figure out what <math>\frac{\dot{\lambda_1}}{\dot{\lambda_2}}</math> is, maybe we could factor out the <math>\lambda_1 \dot{\lambda_2} - \frac{\dot{\lambda_1}}{\lambda_2}</math>, which appears in <math>\dot{\Lambda}</math>...

If, by some miracle, it should turn out that <math>\frac{\left( \Lambda^2 - 1 \right)^{1/2}}{2} = {\lambda_2}^2 \frac{\dot{\lambda_1}}{\dot{\lambda_2}}</math>, factorization would be possible and our integral would read

<div align="center">
<math>-\tfrac{1}{4} \int {\lambda_2}^2 \Lambda^2 \left( \Lambda - 1 \right) \ \dot{\Lambda} \ dt = - \tfrac{1}{4} \int {\lambda_2}^2 \Lambda^2 \left( \Lambda - 1 \right) \ d\Lambda </math> .
</div>

We ought to be able to integrate this, right...? Maybe we could handle the pesky <math>{\lambda_2}^2</math> with integration by parts...

User:Jaycall/T3 Coordinates/Special Case

2010-07-17T19:17:48Z

Jaycall: /* Conserved Quantity */ Expressed in terms of \Lambda

==Coordinate Transformations==

If the special case <math>q^2=2</math> is considered, it is possible to invert the coordinate transformations in closed form. The coordinate transformations and their inversions become

<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>\left( R^2+2z^2 \right)^{1/2}</math>
</td>
<td align="center">
      and      
</td>
<td align="right">
<math>
\lambda_2
</math>
</td>
<td align="center">
<math>\equiv</math>
</td>
<td align="left">
<math>\frac{R^2}{\sqrt{2}z}</math>
</td>
</tr>
<tr>
<td align="right">
<math>
R^2
</math>
</td>
<td align="center">
<math>\equiv</math>
</td>
<td align="left">
<math>\frac{\lambda_2}{2} \left( - \lambda_2 + \sqrt{4{\lambda_1}^2 + {\lambda_2}^2} \right) = \frac{2{\lambda_1}^2}{\Lambda+1}</math>
</td>
<td align="center">
      and      
</td>
<td align="right">
<math>
z
</math>
</td>
<td align="center">
<math>\equiv</math>
</td>
<td align="left">
<math>\frac{1}{2^{3/2}} \left( -\lambda_2 + \sqrt{4 {\lambda_1}^2+{\lambda_2}^2} \right) = \frac{\lambda_2}{2^{3/2}} \left( \Lambda -1 \right)</math>
</td>
</tr>
</table>

where <math>\Lambda \equiv \left[ 1 + \left( \frac{2 \lambda_1}{\lambda_2} \right)^2 \right]^{1/2}</math> .

From this definition of <math>\Lambda</math>, we can compute both its partials with respect to the T3 coordinates, and its total time derivative.

<div align="center">
<math>\frac{\partial \Lambda}{\partial \lambda_1} = \frac{4 \lambda_1 / {\lambda_2}^2}{\Lambda} = \frac{2 \left( \Lambda^2 - 1 \right)^{1/2}}{\Lambda} \frac{1}{\lambda_2} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{\partial \Lambda}{\partial \lambda_2} = - \frac{4 {\lambda_1}^2 / \lambda_2}{\Lambda} = - \frac{2 \left( \Lambda^2 - 1 \right)^{1/2}}{\Lambda} \lambda_1</math>
</div>

<div align="center">
<math>\dot{\Lambda} = \frac{2 \left( \Lambda^2 - 1 \right)^{1/2}}{\Lambda} \left( \frac{\dot{\lambda_1}}{\lambda_2} - \lambda_1 \dot{\lambda_2} \right)</math>
</div>

==Partials of the Coordinates==

Partial derivatives of each of the T3 coordinates taken with respect to each of the cylindrical coordinates are:

<table align="center" border="1" cellpadding="5">
<tr>
<td align="center">
 
</td>
<td align="center">
<math>
\frac{\partial}{\partial R}
</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{R}{\lambda_1} = \left( \frac{2}{\Lambda + 1} \right)^{1/2}
</math>
</td>
<td align="center">
<math>
\frac{2z}{\lambda_1} = \left[ \frac{2 \left( \Lambda - 1 \right) }{\Lambda + 1} \right]^{1/2}
</math>
</td>
<td align="center">
<math>
0
</math>
</td>
</tr>

<tr>
<td align="center">
<math>\lambda_2</math>
</td>
<td align="center">
<math>
\frac{2 \lambda_2}{R} = \frac{2^{3/2}}{\left( \Lambda - 1 \right)^{1/2}}
</math>
</td>
<td align="center">
<math>
-\frac{\lambda_2}{z} = - \frac{2^{3/2}}{\Lambda - 1}
</math>
</td>
<td align="center">
<math>0</math>
</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>

And partials of the cylindrical coordinates taken with respect to the T3 coordinates are:

<table align="center" border="1" cellpadding="5">
<tr>
<td align="center">
 
</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>R</math>
</td>
<td align="center">
<math>
R \ell^2 \lambda_1 = \frac{1}{\Lambda} \left( \frac{\Lambda + 1}{2} \right)^{1/2}
</math>
</td>
<td align="center">
<math>
2Rz^2 \ell^2 / \lambda_2 = \frac{\left( \Lambda - 1 \right)^{3/2}}{2 \Lambda}
</math>
</td>
<td align="center">
<math>
0
</math>
</td>
</tr>

<tr>
<td align="center">
<math>z</math>
</td>
<td align="center">
<math>
2z \ell^2 \lambda_1 = \frac{1}{\Lambda} \left( \frac{\Lambda^2 - 1}{2} \right)^{1/2}
</math>
</td>
<td align="center">
<math>
-R^2 z \ell^2 / \lambda_2 = - \frac{\Lambda - 1}{2^{3/2} \Lambda}
</math>
</td>
<td align="center">
<math>0</math>
</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>

where <math>\ell \equiv \left( R^2 + 4z^2 \right)^{-1/2} = \frac{1}{\lambda_1} \left( \frac{\Lambda + 1}{2 \Lambda} \right)^{1/2}</math>.

==Scale Factors==

Furthermore, the scale factors become

<table align="center">
<tr>
<td align="right">
<math>h_1</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>\lambda_1 \ell = \left( \frac{\Lambda+1}{2 \Lambda} \right)^{1/2}</math>
</td>
</tr>

<tr>
<td align="right">
<math>h_2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>Rz \ell / \lambda_2 = \frac{\Lambda-1}{2 \left( 2 \Lambda \right)^{1/2}}</math>
</td>
</tr>

<tr>
<td align="right">
<math>h_3</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>R = \lambda_3</math>
</td>
</tr>
</table>

==Useful Relationships==

In this special case, there are some additional useful relationships between various combinations of cylindrical variables and their T3 equivalents which can be written out.

<table align="center">
<tr>
<td align="right">
<math>R^2 + 2z^2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>{\lambda_1}^2</math>
</td>
</tr>

<tr>
<td align="right">
<math>R^2 + 4z^2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>2 {\lambda_1}^2 + {\lambda_2}^2/2 - \lambda_2 \sqrt{{\lambda_1}^2+{\lambda_2}^2/4} = \ell^{-2}</math>
</td>
</tr>

<tr>
<td align="right">
<math>R^2 + 8z^2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>4 {\lambda_1}^2 + \tfrac{3}{2} {\lambda_2}^2 - 3 \lambda_2 \sqrt{{\lambda_1}^2+{\lambda_2}^2/4} = 3 \ell^{-2} - 2 {\lambda_1}^2</math>
</td>
</tr>

<tr>
<td align="right">
<math>R^2 - 2z^2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>- {\lambda_1}^2 - {\lambda_2}^2 + 2 \lambda_2 \sqrt{{\lambda_1}^2+{\lambda_2}^2/4} = 3 {\lambda_1}^2 -2 \ell^{-2}</math>
</td>
</tr>

<tr>
<td align="right">
<math>Rz</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>\sqrt{\sqrt{2}\lambda_2} \left( -\frac{\lambda_2}{2\sqrt{2}} + \sqrt{\frac{4{\lambda_1}^2+{\lambda_2}^2}{8}} \right)^{3/2} = h_2 \lambda_2 / \ell</math>
</td>
</tr>
</table>

==Additional Partials==

Partials of <math>\ell</math> can be taken with respect to the coordinates of either system. They are:

<table align="center" border="1" cellpadding="5">
<tr>
<td align="center">
 
</td>
<td align="center">
<math>
\frac{\partial}{\partial R}
</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>\ell</math>
</td>
<td align="center">
<math>
-R \ell^3
</math>
</td>
<td align="center">
<math>
-4z \ell^3
</math>
</td>
<td align="center">
<math>
0
</math>
</td>
</tr>
</table>

<table align="center" border="1" cellpadding="5">
<tr>
<td align="center">
 
</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>\ell</math>
</td>
<td align="center">
<math>
- \left( R^2 + 8z^2 \right) \ell^5 \lambda_1 = \ell^3 \lambda_1 \left( 2{h_1}^2 - 3 \right)
</math>
</td>
<td align="center">
<math>
2R^2 z^2 \ell^5 / \lambda_2 = 2 {h_2}^2 \ell^3 \lambda_2
</math>
</td>
<td align="center">
<math>
0
</math>
</td>
</tr>
</table>

Partials of the scale factors taken with respect to the T3 coordinates are:

<table align="center" border="1" cellpadding="5">
<tr>
<td align="center">
 
</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>h_1</math>
</td>
<td align="center">
<math>
\ell \left( 2 {h_1}^4 - 3 {h_1}^2 + 1 \right) = 2 h_2 \lambda_2
</math>
</td>
<td align="center">
<math>
2 {h_2}^2 \ell^3 \lambda_1 \lambda_2
</math>
</td>
<td align="center">
<math>
0
</math>
</td>
</tr>

<tr>
<td align="center">
<math>h_2</math>
</td>
<td align="center">
<math>
2 {h_1}^2 h_2 \ell^2 \lambda_1 = 2 h_2 \ell^4 {\lambda_1}^3
</math>
</td>
<td align="center">
<math>
h_2 \left( 2 {h_2}^2 \ell^2 {\lambda_2}^2 - 3 \ell^2 {\lambda_1}^2 + 1 \right) / \lambda_2
</math>
</td>
<td align="center">
<math>0</math>
</td>
</tr>

<tr>
<td align="center">
<math>h_3</math>
</td>
<td align="center">
<math>
R \ell^2 \lambda_1
</math>
</td>
<td align="center">
<math>
2Rz^2 \ell^2 / \lambda_2
</math>
</td>
<td align="center">
<math>
0
</math>
</td>
</tr>
</table>

==Conserved Quantity==

The conserved quantity associated with the <math>\lambda_2</math> coordinate is

<div align="center">
<math>
m{h_2}^2 \dot{\lambda_2} \exp \int \left[ \left( 4 {\lambda_1}^2 + {\lambda_2}^2 - \lambda_2 \sqrt{4{\lambda_1}^2 + {\lambda_2}^2} \right) \left( \frac{{\lambda_1}^2 \dot{\lambda_2}}{\lambda_2} - \frac{\lambda_2 {\dot{\lambda_1}}^2}{\dot{\lambda_2}} \right) \right] dt .
</math>
</div>

The quantity in brackets needs to be integrated. In terms of <math>\Lambda</math>, it can be written

<div align="center">
<math>{\lambda_2}^2 \Lambda \left( \Lambda - 1 \right) \left[ \frac{\left( \Lambda^2 - 1 \right)^{1/2}}{2} \lambda_1 \dot{\lambda_2} - {\lambda_2}^2 \frac{\dot{\lambda_1}}{\dot{\lambda_2}} \frac{\dot{\lambda_1}}{\lambda_2} \right]</math> .
</div>

Notice that the thing in square brackets looks very closely related to <math>\dot{\Lambda}</math>. Could this be a hint? If only we could figure out what <math>\frac{\dot{\lambda_1}}{\dot{\lambda_2}}</math> is, maybe we could factor out the <math>\lambda_1 \dot{\lambda_2} - \frac{\dot{\lambda_1}}{\lambda_2}</math>, which appears in <math>\dot{\Lambda}</math>...

If, by some miracle, it should turn out that <math>\frac{\left( \Lambda^2 - 1 \right)^{1/2}}{2} = {\lambda_2}^2 \frac{\dot{\lambda_1}}{\dot{\lambda_2}}</math>, factorization would be possible and our integral would read

<div align="center">
<math>- \int \frac{{\lambda_2}^2 \Lambda^2 \left( \Lambda - 1 \right)}{4} \ \dot{\Lambda} \ dt = - \int \frac{{\lambda_2}^2 \Lambda^2 \left( \Lambda - 1 \right)}{4} \ d\Lambda </math> .
</div>

We ought to be able to integrate this, right...? Maybe we could handle the pesky <math>{\lambda_2}^2</math> with integration by parts...

User:Jaycall/T3 Coordinates/Special Case

2010-07-17T18:32:51Z

Jaycall: /* Coordinate Transformations */ Improved form of R^2 expression

User:Jaycall/T3 Coordinates/Special Case

2010-07-17T18:24:23Z

Jaycall: /* Coordinate Transformations */ Added derivatives of \Lambda

==Coordinate Transformations==

If the special case <math>q^2=2</math> is considered, it is possible to invert the coordinate transformations in closed form. The coordinate transformations and their inversions become

<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>\left( R^2+2z^2 \right)^{1/2}</math>
</td>
<td align="center">
      and      
</td>
<td align="right">
<math>
\lambda_2
</math>
</td>
<td align="center">
<math>\equiv</math>
</td>
<td align="left">
<math>\frac{R^2}{\sqrt{2}z}</math>
</td>
</tr>
<tr>
<td align="right">
<math>
R^2
</math>
</td>
<td align="center">
<math>\equiv</math>
</td>
<td align="left">
<math>-\frac{{\lambda_2}^2}{2} + \lambda_2 \sqrt{{\lambda_1}^2 + {\lambda_2}^2/4} = \frac{2{\lambda_1}^2}{\Lambda+1}</math>
</td>
<td align="center">
      and      
</td>
<td align="right">
<math>
z
</math>
</td>
<td align="center">
<math>\equiv</math>
</td>
<td align="left">
<math>\frac{1}{2^{3/2}} \left[ -\lambda_2 + \sqrt{4 {\lambda_1}^2+{\lambda_2}^2} \right] = \frac{\lambda_2}{2^{3/2}} \left( \Lambda -1 \right)</math>
</td>
</tr>
</table>

where <math>\Lambda \equiv \left[ 1 + \left( \frac{2 \lambda_1}{\lambda_2} \right)^2 \right]^{1/2}</math> .

From this definition of <math>\Lambda</math>, we can compute both its partials with respect to the T3 coordinates, and its total time derivative.

<div align="center">
<math>\frac{\partial \Lambda}{\partial \lambda_1} = \frac{4 \lambda_1 / {\lambda_2}^2}{\Lambda} = \frac{2 \left( \Lambda^2 - 1 \right)^{1/2}}{\Lambda} \frac{1}{\lambda_2} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{\partial \Lambda}{\partial \lambda_2} = - \frac{4 {\lambda_1}^2 / \lambda_2}{\Lambda} = - \frac{2 \left( \Lambda^2 - 1 \right)^{1/2}}{\Lambda} \lambda_1</math>
</div>

<div align="center">
<math>\dot{\Lambda} = \frac{2 \left( \Lambda^2 - 1 \right)^{1/2}}{\Lambda} \left( \frac{\dot{\lambda_1}}{\lambda_2} - \lambda_1 \dot{\lambda_2} \right)</math>
</div>

==Partials of the Coordinates==

Partial derivatives of each of the T3 coordinates taken with respect to each of the cylindrical coordinates are:

<table align="center" border="1" cellpadding="5">
<tr>
<td align="center">
 
</td>
<td align="center">
<math>
\frac{\partial}{\partial R}
</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{R}{\lambda_1} = \left( \frac{2}{\Lambda + 1} \right)^{1/2}
</math>
</td>
<td align="center">
<math>
\frac{2z}{\lambda_1} = \left[ \frac{2 \left( \Lambda - 1 \right) }{\Lambda + 1} \right]^{1/2}
</math>
</td>
<td align="center">
<math>
0
</math>
</td>
</tr>

<tr>
<td align="center">
<math>\lambda_2</math>
</td>
<td align="center">
<math>
\frac{2 \lambda_2}{R} = \frac{2^{3/2}}{\left( \Lambda - 1 \right)^{1/2}}
</math>
</td>
<td align="center">
<math>
-\frac{\lambda_2}{z} = - \frac{2^{3/2}}{\Lambda - 1}
</math>
</td>
<td align="center">
<math>0</math>
</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>

And partials of the cylindrical coordinates taken with respect to the T3 coordinates are:

<table align="center" border="1" cellpadding="5">
<tr>
<td align="center">
 
</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>R</math>
</td>
<td align="center">
<math>
R \ell^2 \lambda_1 = \frac{1}{\Lambda} \left( \frac{\Lambda + 1}{2} \right)^{1/2}
</math>
</td>
<td align="center">
<math>
2Rz^2 \ell^2 / \lambda_2 = \frac{\left( \Lambda - 1 \right)^{3/2}}{2 \Lambda}
</math>
</td>
<td align="center">
<math>
0
</math>
</td>
</tr>

<tr>
<td align="center">
<math>z</math>
</td>
<td align="center">
<math>
2z \ell^2 \lambda_1 = \frac{1}{\Lambda} \left( \frac{\Lambda^2 - 1}{2} \right)^{1/2}
</math>
</td>
<td align="center">
<math>
-R^2 z \ell^2 / \lambda_2 = - \frac{\Lambda - 1}{2^{3/2} \Lambda}
</math>
</td>
<td align="center">
<math>0</math>
</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>

where <math>\ell \equiv \left( R^2 + 4z^2 \right)^{-1/2} = \frac{1}{\lambda_1} \left( \frac{\Lambda + 1}{2 \Lambda} \right)^{1/2}</math>.

==Scale Factors==

Furthermore, the scale factors become

<table align="center">
<tr>
<td align="right">
<math>h_1</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>\lambda_1 \ell = \left( \frac{\Lambda+1}{2 \Lambda} \right)^{1/2}</math>
</td>
</tr>

<tr>
<td align="right">
<math>h_2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>Rz \ell / \lambda_2 = \frac{\Lambda-1}{2 \left( 2 \Lambda \right)^{1/2}}</math>
</td>
</tr>

<tr>
<td align="right">
<math>h_3</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>R = \lambda_3</math>
</td>
</tr>
</table>

==Useful Relationships==

In this special case, there are some additional useful relationships between various combinations of cylindrical variables and their T3 equivalents which can be written out.

<table align="center">
<tr>
<td align="right">
<math>R^2 + 2z^2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>{\lambda_1}^2</math>
</td>
</tr>

<tr>
<td align="right">
<math>R^2 + 4z^2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>2 {\lambda_1}^2 + {\lambda_2}^2/2 - \lambda_2 \sqrt{{\lambda_1}^2+{\lambda_2}^2/4} = \ell^{-2}</math>
</td>
</tr>

<tr>
<td align="right">
<math>R^2 + 8z^2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>4 {\lambda_1}^2 + \tfrac{3}{2} {\lambda_2}^2 - 3 \lambda_2 \sqrt{{\lambda_1}^2+{\lambda_2}^2/4} = 3 \ell^{-2} - 2 {\lambda_1}^2</math>
</td>
</tr>

<tr>
<td align="right">
<math>R^2 - 2z^2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>- {\lambda_1}^2 - {\lambda_2}^2 + 2 \lambda_2 \sqrt{{\lambda_1}^2+{\lambda_2}^2/4} = 3 {\lambda_1}^2 -2 \ell^{-2}</math>
</td>
</tr>

<tr>
<td align="right">
<math>Rz</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>\sqrt{\sqrt{2}\lambda_2} \left( -\frac{\lambda_2}{2\sqrt{2}} + \sqrt{\frac{4{\lambda_1}^2+{\lambda_2}^2}{8}} \right)^{3/2} = h_2 \lambda_2 / \ell</math>
</td>
</tr>
</table>

==Additional Partials==

Partials of <math>\ell</math> can be taken with respect to the coordinates of either system. They are:

<table align="center" border="1" cellpadding="5">
<tr>
<td align="center">
 
</td>
<td align="center">
<math>
\frac{\partial}{\partial R}
</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>\ell</math>
</td>
<td align="center">
<math>
-R \ell^3
</math>
</td>
<td align="center">
<math>
-4z \ell^3
</math>
</td>
<td align="center">
<math>
0
</math>
</td>
</tr>
</table>

<table align="center" border="1" cellpadding="5">
<tr>
<td align="center">
 
</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>\ell</math>
</td>
<td align="center">
<math>
- \left( R^2 + 8z^2 \right) \ell^5 \lambda_1 = \ell^3 \lambda_1 \left( 2{h_1}^2 - 3 \right)
</math>
</td>
<td align="center">
<math>
2R^2 z^2 \ell^5 / \lambda_2 = 2 {h_2}^2 \ell^3 \lambda_2
</math>
</td>
<td align="center">
<math>
0
</math>
</td>
</tr>
</table>

Partials of the scale factors taken with respect to the T3 coordinates are:

<table align="center" border="1" cellpadding="5">
<tr>
<td align="center">
 
</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>h_1</math>
</td>
<td align="center">
<math>
\ell \left( 2 {h_1}^4 - 3 {h_1}^2 + 1 \right) = 2 h_2 \lambda_2
</math>
</td>
<td align="center">
<math>
2 {h_2}^2 \ell^3 \lambda_1 \lambda_2
</math>
</td>
<td align="center">
<math>
0
</math>
</td>
</tr>

<tr>
<td align="center">
<math>h_2</math>
</td>
<td align="center">
<math>
2 {h_1}^2 h_2 \ell^2 \lambda_1 = 2 h_2 \ell^4 {\lambda_1}^3
</math>
</td>
<td align="center">
<math>
h_2 \left( 2 {h_2}^2 \ell^2 {\lambda_2}^2 - 3 \ell^2 {\lambda_1}^2 + 1 \right) / \lambda_2
</math>
</td>
<td align="center">
<math>0</math>
</td>
</tr>

<tr>
<td align="center">
<math>h_3</math>
</td>
<td align="center">
<math>
R \ell^2 \lambda_1
</math>
</td>
<td align="center">
<math>
2Rz^2 \ell^2 / \lambda_2
</math>
</td>
<td align="center">
<math>
0
</math>
</td>
</tr>
</table>

==Conserved Quantity==

The conserved quantity associated with the <math>\lambda_2</math> coordinate is

<div align="center">
<math>
m{h_2}^2 \dot{\lambda_2} \exp \int \left[ \left( 4 {\lambda_1}^2 + {\lambda_2}^2 - \lambda_2 \sqrt{4{\lambda_1}^2 + {\lambda_2}^2} \right) \left( \frac{{\lambda_1}^2 \dot{\lambda_2}}{\lambda_2} - \frac{\lambda_2 {\dot{\lambda_1}}^2}{\dot{\lambda_2}} \right) \right] dt
</math>
</div>

User:Jaycall/T3 Coordinates/Special Case

2010-07-17T18:00:50Z

Jaycall: /* Partials of the Coordinates */ Added definitions in terms of \Lambda

==Coordinate Transformations==

If the special case <math>q^2=2</math> is considered, it is possible to invert the coordinate transformations in closed form. The coordinate transformations and their inversions become

<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>\left( R^2+2z^2 \right)^{1/2}</math>
</td>
<td align="center">
      and      
</td>
<td align="right">
<math>
\lambda_2
</math>
</td>
<td align="center">
<math>\equiv</math>
</td>
<td align="left">
<math>\frac{R^2}{\sqrt{2}z}</math>
</td>
</tr>
<tr>
<td align="right">
<math>
R^2
</math>
</td>
<td align="center">
<math>\equiv</math>
</td>
<td align="left">
<math>-\frac{{\lambda_2}^2}{2} + \lambda_2 \sqrt{{\lambda_1}^2 + {\lambda_2}^2/4} = \frac{2{\lambda_1}^2}{\Lambda+1}</math>
</td>
<td align="center">
      and      
</td>
<td align="right">
<math>
z
</math>
</td>
<td align="center">
<math>\equiv</math>
</td>
<td align="left">
<math>\frac{1}{2^{3/2}} \left[ -\lambda_2 + \sqrt{4 {\lambda_1}^2+{\lambda_2}^2} \right] = \frac{\lambda_2}{2^{3/2}} \left( \Lambda -1 \right)</math>
</td>
</tr>
</table>

where <math>\Lambda \equiv \left[ 1 + \left( \frac{2 \lambda_1}{\lambda_2} \right)^2 \right]^{1/2}</math>

==Partials of the Coordinates==

Partial derivatives of each of the T3 coordinates taken with respect to each of the cylindrical coordinates are:

<table align="center" border="1" cellpadding="5">
<tr>
<td align="center">
 
</td>
<td align="center">
<math>
\frac{\partial}{\partial R}
</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{R}{\lambda_1} = \left( \frac{2}{\Lambda + 1} \right)^{1/2}
</math>
</td>
<td align="center">
<math>
\frac{2z}{\lambda_1} = \left[ \frac{2 \left( \Lambda - 1 \right) }{\Lambda + 1} \right]^{1/2}
</math>
</td>
<td align="center">
<math>
0
</math>
</td>
</tr>

<tr>
<td align="center">
<math>\lambda_2</math>
</td>
<td align="center">
<math>
\frac{2 \lambda_2}{R} = \frac{2^{3/2}}{\left( \Lambda - 1 \right)^{1/2}}
</math>
</td>
<td align="center">
<math>
-\frac{\lambda_2}{z} = - \frac{2^{3/2}}{\Lambda - 1}
</math>
</td>
<td align="center">
<math>0</math>
</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>

And partials of the cylindrical coordinates taken with respect to the T3 coordinates are:

<table align="center" border="1" cellpadding="5">
<tr>
<td align="center">
 
</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>R</math>
</td>
<td align="center">
<math>
R \ell^2 \lambda_1 = \frac{1}{\Lambda} \left( \frac{\Lambda + 1}{2} \right)^{1/2}
</math>
</td>
<td align="center">
<math>
2Rz^2 \ell^2 / \lambda_2 = \frac{\left( \Lambda - 1 \right)^{3/2}}{2 \Lambda}
</math>
</td>
<td align="center">
<math>
0
</math>
</td>
</tr>

<tr>
<td align="center">
<math>z</math>
</td>
<td align="center">
<math>
2z \ell^2 \lambda_1 = \frac{1}{\Lambda} \left( \frac{\Lambda^2 - 1}{2} \right)^{1/2}
</math>
</td>
<td align="center">
<math>
-R^2 z \ell^2 / \lambda_2 = - \frac{\Lambda - 1}{2^{3/2} \Lambda}
</math>
</td>
<td align="center">
<math>0</math>
</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>

where <math>\ell \equiv \left( R^2 + 4z^2 \right)^{-1/2} = \frac{1}{\lambda_1} \left( \frac{\Lambda + 1}{2 \Lambda} \right)^{1/2}</math>.

==Scale Factors==

Furthermore, the scale factors become

<table align="center">
<tr>
<td align="right">
<math>h_1</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>\lambda_1 \ell = \left( \frac{\Lambda+1}{2 \Lambda} \right)^{1/2}</math>
</td>
</tr>

<tr>
<td align="right">
<math>h_2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>Rz \ell / \lambda_2 = \frac{\Lambda-1}{2 \left( 2 \Lambda \right)^{1/2}}</math>
</td>
</tr>

<tr>
<td align="right">
<math>h_3</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>R = \lambda_3</math>
</td>
</tr>
</table>

==Useful Relationships==

In this special case, there are some additional useful relationships between various combinations of cylindrical variables and their T3 equivalents which can be written out.

<table align="center">
<tr>
<td align="right">
<math>R^2 + 2z^2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>{\lambda_1}^2</math>
</td>
</tr>

<tr>
<td align="right">
<math>R^2 + 4z^2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>2 {\lambda_1}^2 + {\lambda_2}^2/2 - \lambda_2 \sqrt{{\lambda_1}^2+{\lambda_2}^2/4} = \ell^{-2}</math>
</td>
</tr>

<tr>
<td align="right">
<math>R^2 + 8z^2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>4 {\lambda_1}^2 + \tfrac{3}{2} {\lambda_2}^2 - 3 \lambda_2 \sqrt{{\lambda_1}^2+{\lambda_2}^2/4} = 3 \ell^{-2} - 2 {\lambda_1}^2</math>
</td>
</tr>

<tr>
<td align="right">
<math>R^2 - 2z^2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>- {\lambda_1}^2 - {\lambda_2}^2 + 2 \lambda_2 \sqrt{{\lambda_1}^2+{\lambda_2}^2/4} = 3 {\lambda_1}^2 -2 \ell^{-2}</math>
</td>
</tr>

<tr>
<td align="right">
<math>Rz</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>\sqrt{\sqrt{2}\lambda_2} \left( -\frac{\lambda_2}{2\sqrt{2}} + \sqrt{\frac{4{\lambda_1}^2+{\lambda_2}^2}{8}} \right)^{3/2} = h_2 \lambda_2 / \ell</math>
</td>
</tr>
</table>

==Additional Partials==

Partials of <math>\ell</math> can be taken with respect to the coordinates of either system. They are:

<table align="center" border="1" cellpadding="5">
<tr>
<td align="center">
 
</td>
<td align="center">
<math>
\frac{\partial}{\partial R}
</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>\ell</math>
</td>
<td align="center">
<math>
-R \ell^3
</math>
</td>
<td align="center">
<math>
-4z \ell^3
</math>
</td>
<td align="center">
<math>
0
</math>
</td>
</tr>
</table>

<table align="center" border="1" cellpadding="5">
<tr>
<td align="center">
 
</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>\ell</math>
</td>
<td align="center">
<math>
- \left( R^2 + 8z^2 \right) \ell^5 \lambda_1 = \ell^3 \lambda_1 \left( 2{h_1}^2 - 3 \right)
</math>
</td>
<td align="center">
<math>
2R^2 z^2 \ell^5 / \lambda_2 = 2 {h_2}^2 \ell^3 \lambda_2
</math>
</td>
<td align="center">
<math>
0
</math>
</td>
</tr>
</table>

Partials of the scale factors taken with respect to the T3 coordinates are:

<table align="center" border="1" cellpadding="5">
<tr>
<td align="center">
 
</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>h_1</math>
</td>
<td align="center">
<math>
\ell \left( 2 {h_1}^4 - 3 {h_1}^2 + 1 \right) = 2 h_2 \lambda_2
</math>
</td>
<td align="center">
<math>
2 {h_2}^2 \ell^3 \lambda_1 \lambda_2
</math>
</td>
<td align="center">
<math>
0
</math>
</td>
</tr>

<tr>
<td align="center">
<math>h_2</math>
</td>
<td align="center">
<math>
2 {h_1}^2 h_2 \ell^2 \lambda_1 = 2 h_2 \ell^4 {\lambda_1}^3
</math>
</td>
<td align="center">
<math>
h_2 \left( 2 {h_2}^2 \ell^2 {\lambda_2}^2 - 3 \ell^2 {\lambda_1}^2 + 1 \right) / \lambda_2
</math>
</td>
<td align="center">
<math>0</math>
</td>
</tr>

<tr>
<td align="center">
<math>h_3</math>
</td>
<td align="center">
<math>
R \ell^2 \lambda_1
</math>
</td>
<td align="center">
<math>
2Rz^2 \ell^2 / \lambda_2
</math>
</td>
<td align="center">
<math>
0
</math>
</td>
</tr>
</table>

==Conserved Quantity==

The conserved quantity associated with the <math>\lambda_2</math> coordinate is

<div align="center">
<math>
m{h_2}^2 \dot{\lambda_2} \exp \int \left[ \left( 4 {\lambda_1}^2 + {\lambda_2}^2 - \lambda_2 \sqrt{4{\lambda_1}^2 + {\lambda_2}^2} \right) \left( \frac{{\lambda_1}^2 \dot{\lambda_2}}{\lambda_2} - \frac{\lambda_2 {\dot{\lambda_1}}^2}{\dot{\lambda_2}} \right) \right] dt
</math>
</div>

User:Jaycall/T3 Coordinates/Special Case

2010-07-17T17:17:53Z

Jaycall: /* Scale Factors */ Added definitions in terms of \Lambda

==Coordinate Transformations==

If the special case <math>q^2=2</math> is considered, it is possible to invert the coordinate transformations in closed form. The coordinate transformations and their inversions become

<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>\left( R^2+2z^2 \right)^{1/2}</math>
</td>
<td align="center">
      and      
</td>
<td align="right">
<math>
\lambda_2
</math>
</td>
<td align="center">
<math>\equiv</math>
</td>
<td align="left">
<math>\frac{R^2}{\sqrt{2}z}</math>
</td>
</tr>
<tr>
<td align="right">
<math>
R^2
</math>
</td>
<td align="center">
<math>\equiv</math>
</td>
<td align="left">
<math>-\frac{{\lambda_2}^2}{2} + \lambda_2 \sqrt{{\lambda_1}^2 + {\lambda_2}^2/4} = \frac{2{\lambda_1}^2}{\Lambda+1}</math>
</td>
<td align="center">
      and      
</td>
<td align="right">
<math>
z
</math>
</td>
<td align="center">
<math>\equiv</math>
</td>
<td align="left">
<math>\frac{1}{2^{3/2}} \left[ -\lambda_2 + \sqrt{4 {\lambda_1}^2+{\lambda_2}^2} \right] = \frac{\lambda_2}{2^{3/2}} \left( \Lambda -1 \right)</math>
</td>
</tr>
</table>

where <math>\Lambda \equiv \left[ 1 + \left( \frac{2 \lambda_1}{\lambda_2} \right)^2 \right]^{1/2}</math>

==Partials of the Coordinates==

Partial derivatives of each of the T3 coordinates taken with respect to each of the cylindrical coordinates are:

<table align="center" border="1" cellpadding="5">
<tr>
<td align="center">
 
</td>
<td align="center">
<math>
\frac{\partial}{\partial R}
</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{R}{\lambda_1}
</math>
</td>
<td align="center">
<math>
\frac{2z}{\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{2 \lambda_2}{R}
</math>
</td>
<td align="center">
<math>
-\frac{\lambda_2}{z}
</math>
</td>
<td align="center">
<math>0</math>
</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>

And partials of the cylindrical coordinates taken with respect to the T3 coordinates are:

<table align="center" border="1" cellpadding="5">
<tr>
<td align="center">
 
</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>R</math>
</td>
<td align="center">
<math>
R \ell^2 \lambda_1
</math>
</td>
<td align="center">
<math>
2Rz^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>
2z \ell^2 \lambda_1
</math>
</td>
<td align="center">
<math>
-R^2 z \ell^2 / \lambda_2
</math>
</td>
<td align="center">
<math>0</math>
</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>

where <math>\ell \equiv \left( R^2 + 4z^2 \right)^{-1/2}</math>.

==Scale Factors==

Furthermore, the scale factors become

<table align="center">
<tr>
<td align="right">
<math>h_1</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>\lambda_1 \ell = \left( \frac{\Lambda+1}{2 \Lambda} \right)^{1/2}</math>
</td>
</tr>

<tr>
<td align="right">
<math>h_2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>Rz \ell / \lambda_2 = \frac{\Lambda-1}{2 \left( 2 \Lambda \right)^{1/2}}</math>
</td>
</tr>

<tr>
<td align="right">
<math>h_3</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>R = \lambda_3</math>
</td>
</tr>
</table>

==Useful Relationships==

In this special case, there are some additional useful relationships between various combinations of cylindrical variables and their T3 equivalents which can be written out.

<table align="center">
<tr>
<td align="right">
<math>R^2 + 2z^2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>{\lambda_1}^2</math>
</td>
</tr>

<tr>
<td align="right">
<math>R^2 + 4z^2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>2 {\lambda_1}^2 + {\lambda_2}^2/2 - \lambda_2 \sqrt{{\lambda_1}^2+{\lambda_2}^2/4} = \ell^{-2}</math>
</td>
</tr>

<tr>
<td align="right">
<math>R^2 + 8z^2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>4 {\lambda_1}^2 + \tfrac{3}{2} {\lambda_2}^2 - 3 \lambda_2 \sqrt{{\lambda_1}^2+{\lambda_2}^2/4} = 3 \ell^{-2} - 2 {\lambda_1}^2</math>
</td>
</tr>

<tr>
<td align="right">
<math>R^2 - 2z^2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>- {\lambda_1}^2 - {\lambda_2}^2 + 2 \lambda_2 \sqrt{{\lambda_1}^2+{\lambda_2}^2/4} = 3 {\lambda_1}^2 -2 \ell^{-2}</math>
</td>
</tr>

<tr>
<td align="right">
<math>Rz</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>\sqrt{\sqrt{2}\lambda_2} \left( -\frac{\lambda_2}{2\sqrt{2}} + \sqrt{\frac{4{\lambda_1}^2+{\lambda_2}^2}{8}} \right)^{3/2} = h_2 \lambda_2 / \ell</math>
</td>
</tr>
</table>

==Additional Partials==

Partials of <math>\ell</math> can be taken with respect to the coordinates of either system. They are:

<table align="center" border="1" cellpadding="5">
<tr>
<td align="center">
 
</td>
<td align="center">
<math>
\frac{\partial}{\partial R}
</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>\ell</math>
</td>
<td align="center">
<math>
-R \ell^3
</math>
</td>
<td align="center">
<math>
-4z \ell^3
</math>
</td>
<td align="center">
<math>
0
</math>
</td>
</tr>
</table>

<table align="center" border="1" cellpadding="5">
<tr>
<td align="center">
 
</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>\ell</math>
</td>
<td align="center">
<math>
- \left( R^2 + 8z^2 \right) \ell^5 \lambda_1 = \ell^3 \lambda_1 \left( 2{h_1}^2 - 3 \right)
</math>
</td>
<td align="center">
<math>
2R^2 z^2 \ell^5 / \lambda_2 = 2 {h_2}^2 \ell^3 \lambda_2
</math>
</td>
<td align="center">
<math>
0
</math>
</td>
</tr>
</table>

Partials of the scale factors taken with respect to the T3 coordinates are:

<table align="center" border="1" cellpadding="5">
<tr>
<td align="center">
 
</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>h_1</math>
</td>
<td align="center">
<math>
\ell \left( 2 {h_1}^4 - 3 {h_1}^2 + 1 \right) = 2 h_2 \lambda_2
</math>
</td>
<td align="center">
<math>
2 {h_2}^2 \ell^3 \lambda_1 \lambda_2
</math>
</td>
<td align="center">
<math>
0
</math>
</td>
</tr>

<tr>
<td align="center">
<math>h_2</math>
</td>
<td align="center">
<math>
2 {h_1}^2 h_2 \ell^2 \lambda_1 = 2 h_2 \ell^4 {\lambda_1}^3
</math>
</td>
<td align="center">
<math>
h_2 \left( 2 {h_2}^2 \ell^2 {\lambda_2}^2 - 3 \ell^2 {\lambda_1}^2 + 1 \right) / \lambda_2
</math>
</td>
<td align="center">
<math>0</math>
</td>
</tr>

<tr>
<td align="center">
<math>h_3</math>
</td>
<td align="center">
<math>
R \ell^2 \lambda_1
</math>
</td>
<td align="center">
<math>
2Rz^2 \ell^2 / \lambda_2
</math>
</td>
<td align="center">
<math>
0
</math>
</td>
</tr>
</table>

==Conserved Quantity==

The conserved quantity associated with the <math>\lambda_2</math> coordinate is

<div align="center">
<math>
m{h_2}^2 \dot{\lambda_2} \exp \int \left[ \left( 4 {\lambda_1}^2 + {\lambda_2}^2 - \lambda_2 \sqrt{4{\lambda_1}^2 + {\lambda_2}^2} \right) \left( \frac{{\lambda_1}^2 \dot{\lambda_2}}{\lambda_2} - \frac{\lambda_2 {\dot{\lambda_1}}^2}{\dot{\lambda_2}} \right) \right] dt
</math>
</div>

User:Jaycall/T3 Coordinates/Special Case

2010-07-17T17:13:40Z

Jaycall: /* Coordinate Transformations */ Added definitions in terms of \Lambda

==Coordinate Transformations==

If the special case <math>q^2=2</math> is considered, it is possible to invert the coordinate transformations in closed form. The coordinate transformations and their inversions become

<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>\left( R^2+2z^2 \right)^{1/2}</math>
</td>
<td align="center">
      and      
</td>
<td align="right">
<math>
\lambda_2
</math>
</td>
<td align="center">
<math>\equiv</math>
</td>
<td align="left">
<math>\frac{R^2}{\sqrt{2}z}</math>
</td>
</tr>
<tr>
<td align="right">
<math>
R^2
</math>
</td>
<td align="center">
<math>\equiv</math>
</td>
<td align="left">
<math>-\frac{{\lambda_2}^2}{2} + \lambda_2 \sqrt{{\lambda_1}^2 + {\lambda_2}^2/4} = \frac{2{\lambda_1}^2}{\Lambda+1}</math>
</td>
<td align="center">
      and      
</td>
<td align="right">
<math>
z
</math>
</td>
<td align="center">
<math>\equiv</math>
</td>
<td align="left">
<math>\frac{1}{2^{3/2}} \left[ -\lambda_2 + \sqrt{4 {\lambda_1}^2+{\lambda_2}^2} \right] = \frac{\lambda_2}{2^{3/2}} \left( \Lambda -1 \right)</math>
</td>
</tr>
</table>

where <math>\Lambda \equiv \left[ 1 + \left( \frac{2 \lambda_1}{\lambda_2} \right)^2 \right]^{1/2}</math>

==Partials of the Coordinates==

Partial derivatives of each of the T3 coordinates taken with respect to each of the cylindrical coordinates are:

<table align="center" border="1" cellpadding="5">
<tr>
<td align="center">
 
</td>
<td align="center">
<math>
\frac{\partial}{\partial R}
</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{R}{\lambda_1}
</math>
</td>
<td align="center">
<math>
\frac{2z}{\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{2 \lambda_2}{R}
</math>
</td>
<td align="center">
<math>
-\frac{\lambda_2}{z}
</math>
</td>
<td align="center">
<math>0</math>
</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>

And partials of the cylindrical coordinates taken with respect to the T3 coordinates are:

<table align="center" border="1" cellpadding="5">
<tr>
<td align="center">
 
</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>R</math>
</td>
<td align="center">
<math>
R \ell^2 \lambda_1
</math>
</td>
<td align="center">
<math>
2Rz^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>
2z \ell^2 \lambda_1
</math>
</td>
<td align="center">
<math>
-R^2 z \ell^2 / \lambda_2
</math>
</td>
<td align="center">
<math>0</math>
</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>

where <math>\ell \equiv \left( R^2 + 4z^2 \right)^{-1/2}</math>.

==Scale Factors==

Furthermore, the scale factors become

<table align="center">
<tr>
<td align="right">
<math>h_1</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>\lambda_1 \ell</math>
</td>
</tr>

<tr>
<td align="right">
<math>h_2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>Rz \ell / \lambda_2</math>
</td>
</tr>

<tr>
<td align="right">
<math>h_3</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>R = \lambda_3</math>
</td>
</tr>
</table>

==Useful Relationships==

In this special case, there are some additional useful relationships between various combinations of cylindrical variables and their T3 equivalents which can be written out.

<table align="center">
<tr>
<td align="right">
<math>R^2 + 2z^2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>{\lambda_1}^2</math>
</td>
</tr>

<tr>
<td align="right">
<math>R^2 + 4z^2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>2 {\lambda_1}^2 + {\lambda_2}^2/2 - \lambda_2 \sqrt{{\lambda_1}^2+{\lambda_2}^2/4} = \ell^{-2}</math>
</td>
</tr>

<tr>
<td align="right">
<math>R^2 + 8z^2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>4 {\lambda_1}^2 + \tfrac{3}{2} {\lambda_2}^2 - 3 \lambda_2 \sqrt{{\lambda_1}^2+{\lambda_2}^2/4} = 3 \ell^{-2} - 2 {\lambda_1}^2</math>
</td>
</tr>

<tr>
<td align="right">
<math>R^2 - 2z^2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>- {\lambda_1}^2 - {\lambda_2}^2 + 2 \lambda_2 \sqrt{{\lambda_1}^2+{\lambda_2}^2/4} = 3 {\lambda_1}^2 -2 \ell^{-2}</math>
</td>
</tr>

<tr>
<td align="right">
<math>Rz</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>\sqrt{\sqrt{2}\lambda_2} \left( -\frac{\lambda_2}{2\sqrt{2}} + \sqrt{\frac{4{\lambda_1}^2+{\lambda_2}^2}{8}} \right)^{3/2} = h_2 \lambda_2 / \ell</math>
</td>
</tr>
</table>

==Additional Partials==

Partials of <math>\ell</math> can be taken with respect to the coordinates of either system. They are:

<table align="center" border="1" cellpadding="5">
<tr>
<td align="center">
 
</td>
<td align="center">
<math>
\frac{\partial}{\partial R}
</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>\ell</math>
</td>
<td align="center">
<math>
-R \ell^3
</math>
</td>
<td align="center">
<math>
-4z \ell^3
</math>
</td>
<td align="center">
<math>
0
</math>
</td>
</tr>
</table>

<table align="center" border="1" cellpadding="5">
<tr>
<td align="center">
 
</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>\ell</math>
</td>
<td align="center">
<math>
- \left( R^2 + 8z^2 \right) \ell^5 \lambda_1 = \ell^3 \lambda_1 \left( 2{h_1}^2 - 3 \right)
</math>
</td>
<td align="center">
<math>
2R^2 z^2 \ell^5 / \lambda_2 = 2 {h_2}^2 \ell^3 \lambda_2
</math>
</td>
<td align="center">
<math>
0
</math>
</td>
</tr>
</table>

Partials of the scale factors taken with respect to the T3 coordinates are:

<table align="center" border="1" cellpadding="5">
<tr>
<td align="center">
 
</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>h_1</math>
</td>
<td align="center">
<math>
\ell \left( 2 {h_1}^4 - 3 {h_1}^2 + 1 \right) = 2 h_2 \lambda_2
</math>
</td>
<td align="center">
<math>
2 {h_2}^2 \ell^3 \lambda_1 \lambda_2
</math>
</td>
<td align="center">
<math>
0
</math>
</td>
</tr>

<tr>
<td align="center">
<math>h_2</math>
</td>
<td align="center">
<math>
2 {h_1}^2 h_2 \ell^2 \lambda_1 = 2 h_2 \ell^4 {\lambda_1}^3
</math>
</td>
<td align="center">
<math>
h_2 \left( 2 {h_2}^2 \ell^2 {\lambda_2}^2 - 3 \ell^2 {\lambda_1}^2 + 1 \right) / \lambda_2
</math>
</td>
<td align="center">
<math>0</math>
</td>
</tr>

<tr>
<td align="center">
<math>h_3</math>
</td>
<td align="center">
<math>
R \ell^2 \lambda_1
</math>
</td>
<td align="center">
<math>
2Rz^2 \ell^2 / \lambda_2
</math>
</td>
<td align="center">
<math>
0
</math>
</td>
</tr>
</table>

==Conserved Quantity==

The conserved quantity associated with the <math>\lambda_2</math> coordinate is

<div align="center">
<math>
m{h_2}^2 \dot{\lambda_2} \exp \int \left[ \left( 4 {\lambda_1}^2 + {\lambda_2}^2 - \lambda_2 \sqrt{4{\lambda_1}^2 + {\lambda_2}^2} \right) \left( \frac{{\lambda_1}^2 \dot{\lambda_2}}{\lambda_2} - \frac{\lambda_2 {\dot{\lambda_1}}^2}{\dot{\lambda_2}} \right) \right] dt
</math>
</div>

User:Jaycall/T3 Coordinates/Special Case

2010-07-17T16:49:49Z

Jaycall: Added headings

==Coordinate Transformations==

If the special case <math>q^2=2</math> is considered, it is possible to invert the coordinate transformations in closed form. The coordinate transformations and their inversions become

<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>\left( R^2+2z^2 \right)^{1/2}</math>
</td>
<td align="center">
      and      
</td>
<td align="right">
<math>
\lambda_2
</math>
</td>
<td align="center">
<math>\equiv</math>
</td>
<td align="left">
<math>\frac{R^2}{\sqrt{2}z}</math>
</td>
</tr>
<tr>
<td align="right">
<math>
R^2
</math>
</td>
<td align="center">
<math>\equiv</math>
</td>
<td align="left">
<math>-\frac{{\lambda_2}^2}{2} + \lambda_2 \sqrt{{\lambda_1}^2 + {\lambda_2}^2/4}</math>
</td>
<td align="center">
      and      
</td>
<td align="right">
<math>
z
</math>
</td>
<td align="center">
<math>\equiv</math>
</td>
<td align="left">
<math>\frac{1}{2^{3/2}} \left[ -\lambda_2 + \sqrt{4 {\lambda_1}^2+{\lambda_2}^2} \right] </math>
</td>
</tr>
</table>

==Partials of the Coordinates==

Partial derivatives of each of the T3 coordinates taken with respect to each of the cylindrical coordinates are:

<table align="center" border="1" cellpadding="5">
<tr>
<td align="center">
 
</td>
<td align="center">
<math>
\frac{\partial}{\partial R}
</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{R}{\lambda_1}
</math>
</td>
<td align="center">
<math>
\frac{2z}{\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{2 \lambda_2}{R}
</math>
</td>
<td align="center">
<math>
-\frac{\lambda_2}{z}
</math>
</td>
<td align="center">
<math>0</math>
</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>

And partials of the cylindrical coordinates taken with respect to the T3 coordinates are:

<table align="center" border="1" cellpadding="5">
<tr>
<td align="center">
 
</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>R</math>
</td>
<td align="center">
<math>
R \ell^2 \lambda_1
</math>
</td>
<td align="center">
<math>
2Rz^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>
2z \ell^2 \lambda_1
</math>
</td>
<td align="center">
<math>
-R^2 z \ell^2 / \lambda_2
</math>
</td>
<td align="center">
<math>0</math>
</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>

where <math>\ell \equiv \left( R^2 + 4z^2 \right)^{-1/2}</math>.

==Scale Factors==

Furthermore, the scale factors become

<table align="center">
<tr>
<td align="right">
<math>h_1</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>\lambda_1 \ell</math>
</td>
</tr>

<tr>
<td align="right">
<math>h_2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>Rz \ell / \lambda_2</math>
</td>
</tr>

<tr>
<td align="right">
<math>h_3</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>R = \lambda_3</math>
</td>
</tr>
</table>

==Useful Relationships==

In this special case, there are some additional useful relationships between various combinations of cylindrical variables and their T3 equivalents which can be written out.

<table align="center">
<tr>
<td align="right">
<math>R^2 + 2z^2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>{\lambda_1}^2</math>
</td>
</tr>

<tr>
<td align="right">
<math>R^2 + 4z^2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>2 {\lambda_1}^2 + {\lambda_2}^2/2 - \lambda_2 \sqrt{{\lambda_1}^2+{\lambda_2}^2/4} = \ell^{-2}</math>
</td>
</tr>

<tr>
<td align="right">
<math>R^2 + 8z^2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>4 {\lambda_1}^2 + \tfrac{3}{2} {\lambda_2}^2 - 3 \lambda_2 \sqrt{{\lambda_1}^2+{\lambda_2}^2/4} = 3 \ell^{-2} - 2 {\lambda_1}^2</math>
</td>
</tr>

<tr>
<td align="right">
<math>R^2 - 2z^2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>- {\lambda_1}^2 - {\lambda_2}^2 + 2 \lambda_2 \sqrt{{\lambda_1}^2+{\lambda_2}^2/4} = 3 {\lambda_1}^2 -2 \ell^{-2}</math>
</td>
</tr>

<tr>
<td align="right">
<math>Rz</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>\sqrt{\sqrt{2}\lambda_2} \left( -\frac{\lambda_2}{2\sqrt{2}} + \sqrt{\frac{4{\lambda_1}^2+{\lambda_2}^2}{8}} \right)^{3/2} = h_2 \lambda_2 / \ell</math>
</td>
</tr>
</table>

==Additional Partials==

Partials of <math>\ell</math> can be taken with respect to the coordinates of either system. They are:

<table align="center" border="1" cellpadding="5">
<tr>
<td align="center">
 
</td>
<td align="center">
<math>
\frac{\partial}{\partial R}
</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>\ell</math>
</td>
<td align="center">
<math>
-R \ell^3
</math>
</td>
<td align="center">
<math>
-4z \ell^3
</math>
</td>
<td align="center">
<math>
0
</math>
</td>
</tr>
</table>

<table align="center" border="1" cellpadding="5">
<tr>
<td align="center">
 
</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>\ell</math>
</td>
<td align="center">
<math>
- \left( R^2 + 8z^2 \right) \ell^5 \lambda_1 = \ell^3 \lambda_1 \left( 2{h_1}^2 - 3 \right)
</math>
</td>
<td align="center">
<math>
2R^2 z^2 \ell^5 / \lambda_2 = 2 {h_2}^2 \ell^3 \lambda_2
</math>
</td>
<td align="center">
<math>
0
</math>
</td>
</tr>
</table>

Partials of the scale factors taken with respect to the T3 coordinates are:

<table align="center" border="1" cellpadding="5">
<tr>
<td align="center">
 
</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>h_1</math>
</td>
<td align="center">
<math>
\ell \left( 2 {h_1}^4 - 3 {h_1}^2 + 1 \right) = 2 h_2 \lambda_2
</math>
</td>
<td align="center">
<math>
2 {h_2}^2 \ell^3 \lambda_1 \lambda_2
</math>
</td>
<td align="center">
<math>
0
</math>
</td>
</tr>

<tr>
<td align="center">
<math>h_2</math>
</td>
<td align="center">
<math>
2 {h_1}^2 h_2 \ell^2 \lambda_1 = 2 h_2 \ell^4 {\lambda_1}^3
</math>
</td>
<td align="center">
<math>
h_2 \left( 2 {h_2}^2 \ell^2 {\lambda_2}^2 - 3 \ell^2 {\lambda_1}^2 + 1 \right) / \lambda_2
</math>
</td>
<td align="center">
<math>0</math>
</td>
</tr>

<tr>
<td align="center">
<math>h_3</math>
</td>
<td align="center">
<math>
R \ell^2 \lambda_1
</math>
</td>
<td align="center">
<math>
2Rz^2 \ell^2 / \lambda_2
</math>
</td>
<td align="center">
<math>
0
</math>
</td>
</tr>
</table>

==Conserved Quantity==

The conserved quantity associated with the <math>\lambda_2</math> coordinate is

<div align="center">
<math>
m{h_2}^2 \dot{\lambda_2} \exp \int \left[ \left( 4 {\lambda_1}^2 + {\lambda_2}^2 - \lambda_2 \sqrt{4{\lambda_1}^2 + {\lambda_2}^2} \right) \left( \frac{{\lambda_1}^2 \dot{\lambda_2}}{\lambda_2} - \frac{\lambda_2 {\dot{\lambda_1}}^2}{\dot{\lambda_2}} \right) \right] dt
</math>
</div>

User:Jaycall/T3 Coordinates/Special Case

2010-07-17T16:44:33Z

Jaycall: Corrected coordinate transformation

If the special case <math>q^2=2</math> is considered, it is possible to invert the coordinate transformations in closed form. The coordinate transformations and their inversions become

<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>\left( R^2+2z^2 \right)^{1/2}</math>
</td>
<td align="center">
      and      
</td>
<td align="right">
<math>
\lambda_2
</math>
</td>
<td align="center">
<math>\equiv</math>
</td>
<td align="left">
<math>\frac{R^2}{\sqrt{2}z}</math>
</td>
</tr>
<tr>
<td align="right">
<math>
R^2
</math>
</td>
<td align="center">
<math>\equiv</math>
</td>
<td align="left">
<math>-\frac{{\lambda_2}^2}{2} + \lambda_2 \sqrt{{\lambda_1}^2 + {\lambda_2}^2/4}</math>
</td>
<td align="center">
      and      
</td>
<td align="right">
<math>
z
</math>
</td>
<td align="center">
<math>\equiv</math>
</td>
<td align="left">
<math>\frac{1}{2^{3/2}} \left[ -\lambda_2 + \sqrt{4 {\lambda_1}^2+{\lambda_2}^2} \right] </math>
</td>
</tr>
</table>

Partial derivatives of each of the T3 coordinates taken with respect to each of the cylindrical coordinates are:

<table align="center" border="1" cellpadding="5">
<tr>
<td align="center">
 
</td>
<td align="center">
<math>
\frac{\partial}{\partial R}
</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{R}{\lambda_1}
</math>
</td>
<td align="center">
<math>
\frac{2z}{\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{2 \lambda_2}{R}
</math>
</td>
<td align="center">
<math>
-\frac{\lambda_2}{z}
</math>
</td>
<td align="center">
<math>0</math>
</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>

And partials of the cylindrical coordinates taken with respect to the T3 coordinates are:

<table align="center" border="1" cellpadding="5">
<tr>
<td align="center">
 
</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>R</math>
</td>
<td align="center">
<math>
R \ell^2 \lambda_1
</math>
</td>
<td align="center">
<math>
2Rz^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>
2z \ell^2 \lambda_1
</math>
</td>
<td align="center">
<math>
-R^2 z \ell^2 / \lambda_2
</math>
</td>
<td align="center">
<math>0</math>
</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>

where <math>\ell \equiv \left( R^2 + 4z^2 \right)^{-1/2}</math>.

Furthermore, the scale factors become

<table align="center">
<tr>
<td align="right">
<math>h_1</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>\lambda_1 \ell</math>
</td>
</tr>

<tr>
<td align="right">
<math>h_2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>Rz \ell / \lambda_2</math>
</td>
</tr>

<tr>
<td align="right">
<math>h_3</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>R = \lambda_3</math>
</td>
</tr>
</table>

In this special case, there are some additional useful relationships between various combinations of cylindrical variables and their T3 equivalents which can be written out.

<table align="center">
<tr>
<td align="right">
<math>R^2 + 2z^2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>{\lambda_1}^2</math>
</td>
</tr>

<tr>
<td align="right">
<math>R^2 + 4z^2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>2 {\lambda_1}^2 + {\lambda_2}^2/2 - \lambda_2 \sqrt{{\lambda_1}^2+{\lambda_2}^2/4} = \ell^{-2}</math>
</td>
</tr>

<tr>
<td align="right">
<math>R^2 + 8z^2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>4 {\lambda_1}^2 + \tfrac{3}{2} {\lambda_2}^2 - 3 \lambda_2 \sqrt{{\lambda_1}^2+{\lambda_2}^2/4} = 3 \ell^{-2} - 2 {\lambda_1}^2</math>
</td>
</tr>

<tr>
<td align="right">
<math>R^2 - 2z^2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>- {\lambda_1}^2 - {\lambda_2}^2 + 2 \lambda_2 \sqrt{{\lambda_1}^2+{\lambda_2}^2/4} = 3 {\lambda_1}^2 -2 \ell^{-2}</math>
</td>
</tr>

<tr>
<td align="right">
<math>Rz</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>\sqrt{\sqrt{2}\lambda_2} \left( -\frac{\lambda_2}{2\sqrt{2}} + \sqrt{\frac{4{\lambda_1}^2+{\lambda_2}^2}{8}} \right)^{3/2} = h_2 \lambda_2 / \ell</math>
</td>
</tr>
</table>

Partials of <math>\ell</math> can be taken with respect to the coordinates of either system. They are:

<table align="center" border="1" cellpadding="5">
<tr>
<td align="center">
 
</td>
<td align="center">
<math>
\frac{\partial}{\partial R}
</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>\ell</math>
</td>
<td align="center">
<math>
-R \ell^3
</math>
</td>
<td align="center">
<math>
-4z \ell^3
</math>
</td>
<td align="center">
<math>
0
</math>
</td>
</tr>
</table>

<table align="center" border="1" cellpadding="5">
<tr>
<td align="center">
 
</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>\ell</math>
</td>
<td align="center">
<math>
- \left( R^2 + 8z^2 \right) \ell^5 \lambda_1 = \ell^3 \lambda_1 \left( 2{h_1}^2 - 3 \right)
</math>
</td>
<td align="center">
<math>
2R^2 z^2 \ell^5 / \lambda_2 = 2 {h_2}^2 \ell^3 \lambda_2
</math>
</td>
<td align="center">
<math>
0
</math>
</td>
</tr>
</table>

Partials of the scale factors taken with respect to the T3 coordinates are:

<table align="center" border="1" cellpadding="5">
<tr>
<td align="center">
 
</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>h_1</math>
</td>
<td align="center">
<math>
\ell \left( 2 {h_1}^4 - 3 {h_1}^2 + 1 \right) = 2 h_2 \lambda_2
</math>
</td>
<td align="center">
<math>
2 {h_2}^2 \ell^3 \lambda_1 \lambda_2
</math>
</td>
<td align="center">
<math>
0
</math>
</td>
</tr>

<tr>
<td align="center">
<math>h_2</math>
</td>
<td align="center">
<math>
2 {h_1}^2 h_2 \ell^2 \lambda_1 = 2 h_2 \ell^4 {\lambda_1}^3
</math>
</td>
<td align="center">
<math>
h_2 \left( 2 {h_2}^2 \ell^2 {\lambda_2}^2 - 3 \ell^2 {\lambda_1}^2 + 1 \right) / \lambda_2
</math>
</td>
<td align="center">
<math>0</math>
</td>
</tr>

<tr>
<td align="center">
<math>h_3</math>
</td>
<td align="center">
<math>
R \ell^2 \lambda_1
</math>
</td>
<td align="center">
<math>
2Rz^2 \ell^2 / \lambda_2
</math>
</td>
<td align="center">
<math>
0
</math>
</td>
</tr>
</table>

The conserved quantity associated with the <math>\lambda_2</math> coordinate is

<div align="center">
<math>
m{h_2}^2 \dot{\lambda_2} \exp \int \left[ \left( 4 {\lambda_1}^2 + {\lambda_2}^2 - \lambda_2 \sqrt{4{\lambda_1}^2 + {\lambda_2}^2} \right) \left( \frac{{\lambda_1}^2 \dot{\lambda_2}}{\lambda_2} - \frac{\lambda_2 {\dot{\lambda_1}}^2}{\dot{\lambda_2}} \right) \right] dt
</math>
</div>

User talk:Jaycall

2010-07-15T20:16:02Z

Jaycall: /* Two Views of Equation of Motion */

==Joel's First Talk Message to Jay==
Jay: In the email message that you sent to me today, you indicated that you had added the 3rd component to the dx/dt equation. I don't see that modification in the current T3Coordinates page. --[[User:Tohline|Tohline]] 15:05, 29 May 2010 (MDT)

:Joel: It's at the bottom of the section entitled "Time-Derivative of Position and Velocity Vectors". Under the history tab, can you see the edits that I made? It was the very first edit and was fairly minor. I only added a couple terms. --[[User:Jaycall|Jaycall]] 20:17, 29 May 2010 (MDT)

::The mediaWiki "talk" page recommends that when you reply to a query, you should not start a new subsection but, rather, just indent (using one or more colons) immediately following the query. Also, it recommends that you "sign" each "talk" message. You do this by typing 2 dashes followed by 4 tildes! I'm going to edit your "reply" (and this additional one from me) to put it in this recommended format. But you have to "sign" your own remark. --[[User:Tohline|Tohline]] 16:23, 29 May 2010 (MDT)

::By the way, I ''can'' see the edits that you made to add the 3rd component to the dx/dt equation. I did not spot the changes earlier, but via the history "diff" function, I'm able to see the edits clearly. --[[User:Tohline|Tohline]] 16:41, 29 May 2010 (MDT)

==Killing Vector Approach==
Thanks for writing out all the terms in the expression for <math>d\Xi/dt</math> in your "[[User:Jaycall/KillingVectorApproach|Killing Vector Approach]]" discussion. On the one hand, I'm happy that the term I mentioned cancels with another one because that means we have fewer terms to integrate. On the other hand, it was one term that I thought we might actually be able to manipulate analytically. Of the two terms that remain, one is relatively simple — the one that is proportional to <math>\dot{\lambda}_2</math> — but the first term is a bear! That will take some thinking! --[[User:Tohline|Tohline]] 10:50, 30 May 2010 (MDT)

:I know it seems like the wrong direction, but unless we can express this summary integral entirely in terms of <math>\lambda_1</math>, <math>\lambda_2</math>, <math>\dot{\lambda}_1</math> and <math>\dot{\lambda}_2</math>, I think that in order to make progress we're going to have to write things out in terms of <math>\varpi</math>, <math>z</math>, <math>\dot{\varpi}</math> and <math>\dot{z}</math>. Of course, that will make the expression much messier, but I think it will be very difficult to recognize this quantity as an exact derivative unless everything's in terms of the same coordinates. --[[User:Jaycall|Jaycall]] 12:55, 30 May 2010 (MDT)

::I agree. It is for similar reasons that I am planning on focusing on the case where <math>q^2=2</math>. At least in this special case we can invert the coordinate expressions to give <math>\varpi</math> and <math>z</math> in terms of the <math>\lambda</math>s. Incidentally, I have added a link to your "Killing Vector" page as well as a link to this associated "Talk" page on the [[User:Tohline/Appendix/Ramblings|page where I itemize my various "ramblings"]]. --[[User:Tohline|Tohline]] 16:16, 30 May 2010 (MDT)

:::Good point. That would be a good reason to focus on one specific case. I haven't looked closely at the inverted coordinate transformation for the <math>q^2=2</math> case for several days. Since the relation is quadratic, is it obvious which root should be used?

::The proper physical root was obvious to me when I performed the coordinate inversion in the case of [[User:Tohline/Appendix/Ramblings/T1Coordinates#Coordinate_Inversion|T1 Coordinates]], so I presume it will be obvious for the T3 Coordinate system. But the inversion needs to be redone for the case of T3 Coordinates; do you want to take care of this and type up the result? In the (quadratic) case of <math>q^2=2</math>, it may be as simple as replacing <math>\chi_2</math> with <math>1/\lambda_2</math>. --[[User:Tohline|Tohline]] 11:21, 31 May 2010 (MDT)

:::Sure. --[[User:Jaycall|Jaycall]] 13:34, 31 May 2010 (MDT)

'''<font color="red">Mistake?</font>''' Please check the next to last row in the [[User:Jaycall/KillingVectorApproach#Christoffel_Symbols|tabulated expressions for Christoffel symbols]]; shouldn't the <math>i</math> and <math>j</math> indexes be swapped? --[[User:Tohline|Tohline]] 09:39, 6 June 2010 (MDT)

:You are correct. Good eye.--[[User:Jaycall|Jaycall]] 09:42, 9 June 2010 (MDT)

::Actually, it wasn't my good eye. I discovered this from my inelegant brute-force derivations. In particular, when I was trying to derive the same form of the [[User:Tohline/Appendix/Ramblings/T3CharacteristicVector#Two_Views_of_Equation_of_Motion|equation of motion from two different points of view]], I had to plug in some of your Christoffel symbol expressions. At first I could not get my EOM to match yours; it was while trying to understand this mismatch that I realized that your ''general'' expression for <math>\Gamma^i_{ij}</math> did not match the ''specific'' expression for <math>\Gamma^2_{21}</math> that you typed on the "T3 Coordinates" page. In my effort to get the EOMs to match, I decided that your ''general'' expression was the incorrect one. --[[User:Tohline|Tohline]] 14:58, 9 June 2010 (MDT)

:::For the record, on 10 June 2010, I modified the [[User:Jaycall/KillingVectorApproach#Christoffel_Symbols|table that contains the general Christoffel symbol expressions]] in order to correct the order of these indexes.--[[User:Tohline|Tohline]] 18:55, 11 June 2010 (MDT)

==Logarithmic Derivatives of T3 Scale Factors==

Check out my new subsection (under T3 Coordinates) entitled [[User:Tohline/Appendix/Ramblings/T3Integrals#Logarithmic_Derivatives_of_Scale_Factors|Logarithmic Derivatives of Scale Factors]]. First, see if you agree that there is a mistake (typo) in one of your tables of partial derivatives. Specifically, I think that the correct expression is:
<div align="center">
<math>
\frac{\partial z}{\partial\lambda_2} = - (q^2-1)\frac{\varpi^2 z \ell^2}{\lambda_2} .
</math>
</div>
Second, see if you agree with my derived expressions for <math>\partial\ln h_i/\partial\ln\lambda_j</math>. --[[User:Tohline|Tohline]] 22:28, 31 May 2010 (MDT)

:You're certainly right about the sign error. I have corrected the expression in the relevant table. --[[User:Jaycall|Jaycall]] 14:28, 1 June 2010 (MDT)

:I have not yet been able to confirm your expressions for logarithmic derivatives of the scale factors. I'm not sure what approach you took in deriving them, but since I had already calculated partials of the scale factors (<math>\partial_i h_j</math> and so forth), I derived a little trick to help simplify the work. Do you agree that

<div align="center">
<math>
\frac{\partial \ln h_i}{\partial \ln \lambda_j} \equiv \frac{\lambda_j}{h_i} \partial_j h_i ?
</math>
</div>

::Yes, this is precisely how I defined the logarithmic derivatives. And I used your tables of partial derivatives (e.g., <math>\partial\varpi/\partial\lambda_j</math> and <math>\partial z/\partial\lambda_j</math>) to obtain <math>\partial_i h_j</math> and so forth. --[[User:Tohline|Tohline]] 17:30, 1 June 2010 (MDT)

'''<font color="red">Mistake!</font>''' Jay: I have found a mistake in my original derivation of the logarithmic derivatives of the <math>h_1</math> scale factor. Perhaps now they will match yours. Check it out and let me know! --[[User:Tohline|Tohline]] 12:41, 4 June 2010 (MDT)

::Joel: I am finally able to confirm your derivation of the logarithmic derivatives of the scale factors. I came at them from a different angle with a fresh head, and I am now confident that what you have is correct. Next I will look into the general case. --[[User:Jaycall|Jaycall]] 14:29, 10 July 2010 (MDT)

==Two Views of Equation of Motion==

Jay: I have created a [[User:Tohline/Appendix/Ramblings/T3CharacteristicVector|new wiki page]] to explicitly analyze how your Characteristic Vector formalism applies to the T3 Coordinate system. On this page is a subsection entitled "[[User:Tohline/Appendix/Ramblings/T3CharacteristicVector#Two_Views_of_Equation_of_Motion|Two Views of Equation of Motion]]" in which I have shown that the <math>2^\mathrm{nd}</math> component of the equation of motion can be written in the following form,
<div align="center">
<math>\frac{d(h_2^2 \dot{\lambda}_2)}{dt} = \biggl( h_1 \frac{\partial h_1}{\partial \lambda_2} \biggr) \dot{\lambda}_1^2 + \biggl( h_2 \frac{\partial h_2}{\partial \lambda_2} \biggr) \dot{\lambda}_2^2</math> ,
</div>
assuming that <math>\partial\Phi/\partial\lambda_2 = 0</math>. More to the point, I have shown that I can derive this form of this component of the equation of motion using either (A) your Christoffel symbol formalism, or (B) the more classical formalism that I have been using, which I obtained from Appendix 1.B of Binney & Tremaine (1987) and Morse & Feshbach (1953).

Now, I shouldn't be shocked that both derivations lead to the same form of the equation — after all, we're doing physics aren't we? But here is what surprised me: The Christoffel symbol formalism produces this form of the equation without ever assuming ''a particular coordinate system'' whereas, in order to derive the equation using the more classical formalism, I plugged in some relations between the partial derivatives of the scale factors that ''I thought'' were specific to T3 coordinates. This leads me to ask, "How many of the T3 Coordinate relations [[User:Tohline/Appendix/Ramblings/T3Integrals|derived earlier]] are generic relations that apply to any orthogonal, axisymmetric coordinate system, and how many are unique to the T3 Coordinate System?" I'm particularly interested in knowing how generalizable the relations are that appear inside boxes labeled "T3 Coordinates" under the subsections I have entitled, "[[User:Tohline/Appendix/Ramblings/T3Integrals#Derived_Identity_for_T3_Coordinates|Derived Identity for T3 Corodinates]]" and "[[User:Tohline/Appendix/Ramblings/T3Integrals#Logarithmic_Derivatives_of_Scale_Factors|Logarithmic Derivatives of Scale Factors]]". Can you answer this? --[[User:Tohline|Tohline]] 10:46, 7 June 2010 (MDT)

:Joel: Yes, this is the power of using a covariant formulation. The only assumptions we made were that the coordinates were orthogonal and axisymmetric. I believe that the relation under "[[User:Tohline/Appendix/Ramblings/T3Integrals#Derived_Identity_for_T3_Coordinates|Derived Identity for T3 Coordinates]]" would apply to any coordinate system meeting these criteria. I will double check, though. (How did you determine that the position vector could be written <math>\vec{x} = \hat{e}_1 (h_1 \lambda_1) + \hat{e}_2 (h_2 \lambda_2)</math>?)

:The logarithmic derivatives of the scale factors cannot be general in their current form due to the appearance of factors of <math>q</math>, but I suspect they can be generalized and molded into a form that closely resembles what you have. I will work on this, too. --[[User:Jaycall|Jaycall]] 10:39, 9 June 2010 (MDT)

::Jay: You asked how I determined the expression for the [[User:Tohline/Appendix/Ramblings/T3Integrals#Definition|position vector]]. I don't have my detailed notes with me right now, but I can outline the steps. In Morse & Feshbach's classical presentation of orthogonal curvilinear coordinates, they define a set of so-called ''direction cosines,'' <math>\gamma_{ij}</math>. (I don't have the generalized definition with me at this time, but they are expressible in terms of derivatives of the scale factors with respect to the various coordinates; I'm certain ''direction cosines'' are definable in terms of Christoffel symbols.) Each of the Cartesian unit vectors (<math>\hat{\imath},\hat{\jmath},\hat{k}</math>) can be written in terms of products of these direction cosines and the three unit vectors of your chosen curvilinear coordinate system (<math>\hat{e}_1, \hat{e}_2, \hat{e}_3</math>). According to Morse & Feshbach, for example, <math>\hat{\imath} = (\hat{e}_1\gamma_{11} + \hat{e}_2\gamma_{21} + \hat{e}_3\gamma_{31})</math>. So, all I did to obtain an expression for the position vector in T3 coordinates was define <math>\vec{x} = (\hat\imath x + \hat\jmath y + \hat{k}z)</math>, then replace each Cartesian unit vector by its expression in terms of products of the T3-coordinate unit vectors & the appropriate direct cosine, then gather together and simplify terms. Does this make sense? --[[User:Tohline|Tohline]] 14:59, 10 June 2010 (MDT)

:::Yes, it does. That's very helpful. --[[User:Jaycall|Jaycall]] 09:49, 11 June 2010 (MDT)

::::Jay: For the record, I have put together a wiki page that discusses the definition and utility of [[User:Tohline/Appendix/Ramblings/DirectionCosines|Direction Cosines]].--[[User:Tohline|Tohline]] 18:47, 3 July 2010 (MDT)

Joel, I finally figured out what was going on this afternoon with the strange conserved quantities we were getting, like <math>{h_2}^4 \dot{\lambda_2}^2</math> and, consequently, <math>\lambda_1 \lambda_2</math>. There ''is'' a mistake in the derivation stemming all the way back from Eq. CV.02 on my page on the [[User:Jaycall/KillingVectorApproach|characteristic vector approach]]. My derivation of the brute force condition on <math>C_2</math> was correct, but you can see from above that <math>\Xi</math> does not equal <math>\ln C_2</math>, it equals <math>- \ln C_2</math>. Consequently, the conclusion is no longer that <math>C_2</math> must equal <math>{h_2}^2 \dot{\lambda_2}</math>, but rather <math>\left( {h_2}^2 \dot{\lambda_2} \right)^{-1}</math>. So in fact you ''did'' find a conserved quantity--<math>1</math>! : ) I guess it was a tautology after all; your intuition was right. --[[User:Jaycall|Jaycall]] 18:50, 14 July 2010 (MDT)

:Jay, very good catch! Looking back through the history of that wiki page, it is clear that I am the one who introduced this sign error. I mistakingly set <math>\Xi = \ln C_2</math> in equation [[User:Jaycall/KillingVectorApproach#CV.02|CV.02]] at the same time that I "Added a couple of right-justified equation numbers." I'll march through the relevent wiki pages today and attempt to correct this mistake wherever it appears. --[[User:Tohline|Tohline]] 08:30, 15 July 2010 (MDT)

::Joel, I'm sorry yesterday's excitement didn't pan out. It would have been really neat. Well, at least I learned something. On my drive home yesterday I was thinking about what it meant for there to be a constraint that involved only the coordinates (and not their rates of change). We were uncomfortable with this idea, but now I understand why in general no conserved quantity can take this form. If there were a constraint on only the coordinates, then this would limit a particle to a curve or a surface...a subspace of physical space (as opposed to a subspace of phase space). Yet it would definitely be possible to get a particle to leave the subspace by giving it an initial velocity in any direction normal to the subspace. Any conserved quantity must definitely involve the coordinate velocities (unless you're artificially restricting the motion). --[[User:Jaycall|Jaycall]] 14:15, 15 July 2010 (MDT)

==Special Quadratic Case==
Jay: On a [[User:Tohline/Appendix/Ramblings/T3Integrals/QuadraticCase#Special_Case_.28Quadratic.29|new accompanying wiki page]] I have begun to investigate how our system behaves in the special case of <math>q^2 = 2</math>. I have added a pointer/link to this new page — and a separate pointer/link to the parallel investigation that you have told me you are conducting — on the page containing a table of contents to my [[User:Tohline/Appendix/Ramblings|various research ramblings]]. I have not yet actually gone to the page that details your analysis because, at least for a while, I prefer for our trains of thought to remain independent of one another. However, in an effort to simplify the comparison that we ultimately will need to make between our independent analyses, I strongly recommend that you write your expressions for the scale factors, etc. ''for the special case of <math>q^2=2</math>'' in terms of a variable, <math>\Lambda</math>, defined such that,
<div align="center">
<math>
\Lambda \equiv \biggl[1 + \biggl(\frac{2\lambda_1}{\lambda_2}\biggr)^2\biggr]^{1/2} = \cosh(2\Zeta) .
</math>
</div>
By the way, if you don't agree that <math>\sqrt{1+(2\lambda_1/\lambda_2)^2}</math> equals <math>\cosh(2\Zeta)</math>, please let me know!--[[User:Tohline|Tohline]] 19:49, 11 June 2010 (MDT)

:: Joel: I have checked and agree that in the <math>q^2=2</math> case only, the above relation among <math>\Lambda</math>, <math>Z</math> and the T3 coordinates is correct. --[[User:Jaycall|Jaycall]] 14:58, 10 July 2010 (MDT)

User talk:Jaycall

2010-07-15T00:55:36Z

Jaycall: /* Two Views of Equation of Motion */

==Joel's First Talk Message to Jay==
Jay: In the email message that you sent to me today, you indicated that you had added the 3rd component to the dx/dt equation. I don't see that modification in the current T3Coordinates page. --[[User:Tohline|Tohline]] 15:05, 29 May 2010 (MDT)

:Joel: It's at the bottom of the section entitled "Time-Derivative of Position and Velocity Vectors". Under the history tab, can you see the edits that I made? It was the very first edit and was fairly minor. I only added a couple terms. --[[User:Jaycall|Jaycall]] 20:17, 29 May 2010 (MDT)

::The mediaWiki "talk" page recommends that when you reply to a query, you should not start a new subsection but, rather, just indent (using one or more colons) immediately following the query. Also, it recommends that you "sign" each "talk" message. You do this by typing 2 dashes followed by 4 tildes! I'm going to edit your "reply" (and this additional one from me) to put it in this recommended format. But you have to "sign" your own remark. --[[User:Tohline|Tohline]] 16:23, 29 May 2010 (MDT)

::By the way, I ''can'' see the edits that you made to add the 3rd component to the dx/dt equation. I did not spot the changes earlier, but via the history "diff" function, I'm able to see the edits clearly. --[[User:Tohline|Tohline]] 16:41, 29 May 2010 (MDT)

==Killing Vector Approach==
Thanks for writing out all the terms in the expression for <math>d\Xi/dt</math> in your "[[User:Jaycall/KillingVectorApproach|Killing Vector Approach]]" discussion. On the one hand, I'm happy that the term I mentioned cancels with another one because that means we have fewer terms to integrate. On the other hand, it was one term that I thought we might actually be able to manipulate analytically. Of the two terms that remain, one is relatively simple — the one that is proportional to <math>\dot{\lambda}_2</math> — but the first term is a bear! That will take some thinking! --[[User:Tohline|Tohline]] 10:50, 30 May 2010 (MDT)

:I know it seems like the wrong direction, but unless we can express this summary integral entirely in terms of <math>\lambda_1</math>, <math>\lambda_2</math>, <math>\dot{\lambda}_1</math> and <math>\dot{\lambda}_2</math>, I think that in order to make progress we're going to have to write things out in terms of <math>\varpi</math>, <math>z</math>, <math>\dot{\varpi}</math> and <math>\dot{z}</math>. Of course, that will make the expression much messier, but I think it will be very difficult to recognize this quantity as an exact derivative unless everything's in terms of the same coordinates. --[[User:Jaycall|Jaycall]] 12:55, 30 May 2010 (MDT)

::I agree. It is for similar reasons that I am planning on focusing on the case where <math>q^2=2</math>. At least in this special case we can invert the coordinate expressions to give <math>\varpi</math> and <math>z</math> in terms of the <math>\lambda</math>s. Incidentally, I have added a link to your "Killing Vector" page as well as a link to this associated "Talk" page on the [[User:Tohline/Appendix/Ramblings|page where I itemize my various "ramblings"]]. --[[User:Tohline|Tohline]] 16:16, 30 May 2010 (MDT)

:::Good point. That would be a good reason to focus on one specific case. I haven't looked closely at the inverted coordinate transformation for the <math>q^2=2</math> case for several days. Since the relation is quadratic, is it obvious which root should be used?

::The proper physical root was obvious to me when I performed the coordinate inversion in the case of [[User:Tohline/Appendix/Ramblings/T1Coordinates#Coordinate_Inversion|T1 Coordinates]], so I presume it will be obvious for the T3 Coordinate system. But the inversion needs to be redone for the case of T3 Coordinates; do you want to take care of this and type up the result? In the (quadratic) case of <math>q^2=2</math>, it may be as simple as replacing <math>\chi_2</math> with <math>1/\lambda_2</math>. --[[User:Tohline|Tohline]] 11:21, 31 May 2010 (MDT)

:::Sure. --[[User:Jaycall|Jaycall]] 13:34, 31 May 2010 (MDT)

'''<font color="red">Mistake?</font>''' Please check the next to last row in the [[User:Jaycall/KillingVectorApproach#Christoffel_Symbols|tabulated expressions for Christoffel symbols]]; shouldn't the <math>i</math> and <math>j</math> indexes be swapped? --[[User:Tohline|Tohline]] 09:39, 6 June 2010 (MDT)

:You are correct. Good eye.--[[User:Jaycall|Jaycall]] 09:42, 9 June 2010 (MDT)

::Actually, it wasn't my good eye. I discovered this from my inelegant brute-force derivations. In particular, when I was trying to derive the same form of the [[User:Tohline/Appendix/Ramblings/T3CharacteristicVector#Two_Views_of_Equation_of_Motion|equation of motion from two different points of view]], I had to plug in some of your Christoffel symbol expressions. At first I could not get my EOM to match yours; it was while trying to understand this mismatch that I realized that your ''general'' expression for <math>\Gamma^i_{ij}</math> did not match the ''specific'' expression for <math>\Gamma^2_{21}</math> that you typed on the "T3 Coordinates" page. In my effort to get the EOMs to match, I decided that your ''general'' expression was the incorrect one. --[[User:Tohline|Tohline]] 14:58, 9 June 2010 (MDT)

:::For the record, on 10 June 2010, I modified the [[User:Jaycall/KillingVectorApproach#Christoffel_Symbols|table that contains the general Christoffel symbol expressions]] in order to correct the order of these indexes.--[[User:Tohline|Tohline]] 18:55, 11 June 2010 (MDT)

==Logarithmic Derivatives of T3 Scale Factors==

Check out my new subsection (under T3 Coordinates) entitled [[User:Tohline/Appendix/Ramblings/T3Integrals#Logarithmic_Derivatives_of_Scale_Factors|Logarithmic Derivatives of Scale Factors]]. First, see if you agree that there is a mistake (typo) in one of your tables of partial derivatives. Specifically, I think that the correct expression is:
<div align="center">
<math>
\frac{\partial z}{\partial\lambda_2} = - (q^2-1)\frac{\varpi^2 z \ell^2}{\lambda_2} .
</math>
</div>
Second, see if you agree with my derived expressions for <math>\partial\ln h_i/\partial\ln\lambda_j</math>. --[[User:Tohline|Tohline]] 22:28, 31 May 2010 (MDT)

:You're certainly right about the sign error. I have corrected the expression in the relevant table. --[[User:Jaycall|Jaycall]] 14:28, 1 June 2010 (MDT)

:I have not yet been able to confirm your expressions for logarithmic derivatives of the scale factors. I'm not sure what approach you took in deriving them, but since I had already calculated partials of the scale factors (<math>\partial_i h_j</math> and so forth), I derived a little trick to help simplify the work. Do you agree that

<div align="center">
<math>
\frac{\partial \ln h_i}{\partial \ln \lambda_j} \equiv \frac{\lambda_j}{h_i} \partial_j h_i ?
</math>
</div>

::Yes, this is precisely how I defined the logarithmic derivatives. And I used your tables of partial derivatives (e.g., <math>\partial\varpi/\partial\lambda_j</math> and <math>\partial z/\partial\lambda_j</math>) to obtain <math>\partial_i h_j</math> and so forth. --[[User:Tohline|Tohline]] 17:30, 1 June 2010 (MDT)

'''<font color="red">Mistake!</font>''' Jay: I have found a mistake in my original derivation of the logarithmic derivatives of the <math>h_1</math> scale factor. Perhaps now they will match yours. Check it out and let me know! --[[User:Tohline|Tohline]] 12:41, 4 June 2010 (MDT)

::Joel: I am finally able to confirm your derivation of the logarithmic derivatives of the scale factors. I came at them from a different angle with a fresh head, and I am now confident that what you have is correct. Next I will look into the general case. --[[User:Jaycall|Jaycall]] 14:29, 10 July 2010 (MDT)

==Two Views of Equation of Motion==

Jay: I have created a [[User:Tohline/Appendix/Ramblings/T3CharacteristicVector|new wiki page]] to explicitly analyze how your Characteristic Vector formalism applies to the T3 Coordinate system. On this page is a subsection entitled "[[User:Tohline/Appendix/Ramblings/T3CharacteristicVector#Two_Views_of_Equation_of_Motion|Two Views of Equation of Motion]]" in which I have shown that the <math>2^\mathrm{nd}</math> component of the equation of motion can be written in the following form,
<div align="center">
<math>\frac{d(h_2^2 \dot{\lambda}_2)}{dt} = \biggl( h_1 \frac{\partial h_1}{\partial \lambda_2} \biggr) \dot{\lambda}_1^2 + \biggl( h_2 \frac{\partial h_2}{\partial \lambda_2} \biggr) \dot{\lambda}_2^2</math> ,
</div>
assuming that <math>\partial\Phi/\partial\lambda_2 = 0</math>. More to the point, I have shown that I can derive this form of this component of the equation of motion using either (A) your Christoffel symbol formalism, or (B) the more classical formalism that I have been using, which I obtained from Appendix 1.B of Binney & Tremaine (1987) and Morse & Feshbach (1953).

Now, I shouldn't be shocked that both derivations lead to the same form of the equation — after all, we're doing physics aren't we? But here is what surprised me: The Christoffel symbol formalism produces this form of the equation without ever assuming ''a particular coordinate system'' whereas, in order to derive the equation using the more classical formalism, I plugged in some relations between the partial derivatives of the scale factors that ''I thought'' were specific to T3 coordinates. This leads me to ask, "How many of the T3 Coordinate relations [[User:Tohline/Appendix/Ramblings/T3Integrals|derived earlier]] are generic relations that apply to any orthogonal, axisymmetric coordinate system, and how many are unique to the T3 Coordinate System?" I'm particularly interested in knowing how generalizable the relations are that appear inside boxes labeled "T3 Coordinates" under the subsections I have entitled, "[[User:Tohline/Appendix/Ramblings/T3Integrals#Derived_Identity_for_T3_Coordinates|Derived Identity for T3 Corodinates]]" and "[[User:Tohline/Appendix/Ramblings/T3Integrals#Logarithmic_Derivatives_of_Scale_Factors|Logarithmic Derivatives of Scale Factors]]". Can you answer this? --[[User:Tohline|Tohline]] 10:46, 7 June 2010 (MDT)

:Joel: Yes, this is the power of using a covariant formulation. The only assumptions we made were that the coordinates were orthogonal and axisymmetric. I believe that the relation under "[[User:Tohline/Appendix/Ramblings/T3Integrals#Derived_Identity_for_T3_Coordinates|Derived Identity for T3 Coordinates]]" would apply to any coordinate system meeting these criteria. I will double check, though. (How did you determine that the position vector could be written <math>\vec{x} = \hat{e}_1 (h_1 \lambda_1) + \hat{e}_2 (h_2 \lambda_2)</math>?)

:The logarithmic derivatives of the scale factors cannot be general in their current form due to the appearance of factors of <math>q</math>, but I suspect they can be generalized and molded into a form that closely resembles what you have. I will work on this, too. --[[User:Jaycall|Jaycall]] 10:39, 9 June 2010 (MDT)

::Jay: You asked how I determined the expression for the [[User:Tohline/Appendix/Ramblings/T3Integrals#Definition|position vector]]. I don't have my detailed notes with me right now, but I can outline the steps. In Morse & Feshbach's classical presentation of orthogonal curvilinear coordinates, they define a set of so-called ''direction cosines,'' <math>\gamma_{ij}</math>. (I don't have the generalized definition with me at this time, but they are expressible in terms of derivatives of the scale factors with respect to the various coordinates; I'm certain ''direction cosines'' are definable in terms of Christoffel symbols.) Each of the Cartesian unit vectors (<math>\hat{\imath},\hat{\jmath},\hat{k}</math>) can be written in terms of products of these direction cosines and the three unit vectors of your chosen curvilinear coordinate system (<math>\hat{e}_1, \hat{e}_2, \hat{e}_3</math>). According to Morse & Feshbach, for example, <math>\hat{\imath} = (\hat{e}_1\gamma_{11} + \hat{e}_2\gamma_{21} + \hat{e}_3\gamma_{31})</math>. So, all I did to obtain an expression for the position vector in T3 coordinates was define <math>\vec{x} = (\hat\imath x + \hat\jmath y + \hat{k}z)</math>, then replace each Cartesian unit vector by its expression in terms of products of the T3-coordinate unit vectors & the appropriate direct cosine, then gather together and simplify terms. Does this make sense? --[[User:Tohline|Tohline]] 14:59, 10 June 2010 (MDT)

:::Yes, it does. That's very helpful. --[[User:Jaycall|Jaycall]] 09:49, 11 June 2010 (MDT)

::::Jay: For the record, I have put together a wiki page that discusses the definition and utility of [[User:Tohline/Appendix/Ramblings/DirectionCosines|Direction Cosines]].--[[User:Tohline|Tohline]] 18:47, 3 July 2010 (MDT)

Joel, I finally figured out what was going on this afternoon with the strange conserved quantities we were getting, like <math>{h_2}^4 \dot{\lambda_2}^2</math> and, consequently, <math>\lambda_1 \lambda_2</math>. There ''is'' a mistake in the derivation stemming all the way back from Eq. CV.02 on my page on the [[User:Jaycall/KillingVectorApproach|characteristic vector approach]]. My derivation of the brute force condition on <math>C_2</math> was correct, but you can see from above that <math>\Xi</math> does not equal <math>\ln C_2</math>, it equals <math>- \ln C_2</math>. Consequently, the conclusion is no longer that <math>C_2</math> must equal <math>{h_2}^2 \dot{\lambda_2}</math>, but rather <math>\left( {h_2}^2 \dot{\lambda_2} \right)^{-1}</math>. So in fact you ''did'' find a conserved quantity--<math>1</math>! : ) I guess it was a tautology after all; your intuition was right. --[[User:Jaycall|Jaycall]] 18:50, 14 July 2010 (MDT)

==Special Quadratic Case==
Jay: On a [[User:Tohline/Appendix/Ramblings/T3Integrals/QuadraticCase#Special_Case_.28Quadratic.29|new accompanying wiki page]] I have begun to investigate how our system behaves in the special case of <math>q^2 = 2</math>. I have added a pointer/link to this new page — and a separate pointer/link to the parallel investigation that you have told me you are conducting — on the page containing a table of contents to my [[User:Tohline/Appendix/Ramblings|various research ramblings]]. I have not yet actually gone to the page that details your analysis because, at least for a while, I prefer for our trains of thought to remain independent of one another. However, in an effort to simplify the comparison that we ultimately will need to make between our independent analyses, I strongly recommend that you write your expressions for the scale factors, etc. ''for the special case of <math>q^2=2</math>'' in terms of a variable, <math>\Lambda</math>, defined such that,
<div align="center">
<math>
\Lambda \equiv \biggl[1 + \biggl(\frac{2\lambda_1}{\lambda_2}\biggr)^2\biggr]^{1/2} = \cosh(2\Zeta) .
</math>
</div>
By the way, if you don't agree that <math>\sqrt{1+(2\lambda_1/\lambda_2)^2}</math> equals <math>\cosh(2\Zeta)</math>, please let me know!--[[User:Tohline|Tohline]] 19:49, 11 June 2010 (MDT)

:: Joel: I have checked and agree that in the <math>q^2=2</math> case only, the above relation among <math>\Lambda</math>, <math>Z</math> and the T3 coordinates is correct. --[[User:Jaycall|Jaycall]] 14:58, 10 July 2010 (MDT)

User talk:Jaycall

2010-07-15T00:50:48Z

Jaycall: /* Two Views of Equation of Motion */

==Joel's First Talk Message to Jay==
Jay: In the email message that you sent to me today, you indicated that you had added the 3rd component to the dx/dt equation. I don't see that modification in the current T3Coordinates page. --[[User:Tohline|Tohline]] 15:05, 29 May 2010 (MDT)

:Joel: It's at the bottom of the section entitled "Time-Derivative of Position and Velocity Vectors". Under the history tab, can you see the edits that I made? It was the very first edit and was fairly minor. I only added a couple terms. --[[User:Jaycall|Jaycall]] 20:17, 29 May 2010 (MDT)

::The mediaWiki "talk" page recommends that when you reply to a query, you should not start a new subsection but, rather, just indent (using one or more colons) immediately following the query. Also, it recommends that you "sign" each "talk" message. You do this by typing 2 dashes followed by 4 tildes! I'm going to edit your "reply" (and this additional one from me) to put it in this recommended format. But you have to "sign" your own remark. --[[User:Tohline|Tohline]] 16:23, 29 May 2010 (MDT)

::By the way, I ''can'' see the edits that you made to add the 3rd component to the dx/dt equation. I did not spot the changes earlier, but via the history "diff" function, I'm able to see the edits clearly. --[[User:Tohline|Tohline]] 16:41, 29 May 2010 (MDT)

==Killing Vector Approach==
Thanks for writing out all the terms in the expression for <math>d\Xi/dt</math> in your "[[User:Jaycall/KillingVectorApproach|Killing Vector Approach]]" discussion. On the one hand, I'm happy that the term I mentioned cancels with another one because that means we have fewer terms to integrate. On the other hand, it was one term that I thought we might actually be able to manipulate analytically. Of the two terms that remain, one is relatively simple — the one that is proportional to <math>\dot{\lambda}_2</math> — but the first term is a bear! That will take some thinking! --[[User:Tohline|Tohline]] 10:50, 30 May 2010 (MDT)

:I know it seems like the wrong direction, but unless we can express this summary integral entirely in terms of <math>\lambda_1</math>, <math>\lambda_2</math>, <math>\dot{\lambda}_1</math> and <math>\dot{\lambda}_2</math>, I think that in order to make progress we're going to have to write things out in terms of <math>\varpi</math>, <math>z</math>, <math>\dot{\varpi}</math> and <math>\dot{z}</math>. Of course, that will make the expression much messier, but I think it will be very difficult to recognize this quantity as an exact derivative unless everything's in terms of the same coordinates. --[[User:Jaycall|Jaycall]] 12:55, 30 May 2010 (MDT)

::I agree. It is for similar reasons that I am planning on focusing on the case where <math>q^2=2</math>. At least in this special case we can invert the coordinate expressions to give <math>\varpi</math> and <math>z</math> in terms of the <math>\lambda</math>s. Incidentally, I have added a link to your "Killing Vector" page as well as a link to this associated "Talk" page on the [[User:Tohline/Appendix/Ramblings|page where I itemize my various "ramblings"]]. --[[User:Tohline|Tohline]] 16:16, 30 May 2010 (MDT)

:::Good point. That would be a good reason to focus on one specific case. I haven't looked closely at the inverted coordinate transformation for the <math>q^2=2</math> case for several days. Since the relation is quadratic, is it obvious which root should be used?

::The proper physical root was obvious to me when I performed the coordinate inversion in the case of [[User:Tohline/Appendix/Ramblings/T1Coordinates#Coordinate_Inversion|T1 Coordinates]], so I presume it will be obvious for the T3 Coordinate system. But the inversion needs to be redone for the case of T3 Coordinates; do you want to take care of this and type up the result? In the (quadratic) case of <math>q^2=2</math>, it may be as simple as replacing <math>\chi_2</math> with <math>1/\lambda_2</math>. --[[User:Tohline|Tohline]] 11:21, 31 May 2010 (MDT)

:::Sure. --[[User:Jaycall|Jaycall]] 13:34, 31 May 2010 (MDT)

'''<font color="red">Mistake?</font>''' Please check the next to last row in the [[User:Jaycall/KillingVectorApproach#Christoffel_Symbols|tabulated expressions for Christoffel symbols]]; shouldn't the <math>i</math> and <math>j</math> indexes be swapped? --[[User:Tohline|Tohline]] 09:39, 6 June 2010 (MDT)

:You are correct. Good eye.--[[User:Jaycall|Jaycall]] 09:42, 9 June 2010 (MDT)

::Actually, it wasn't my good eye. I discovered this from my inelegant brute-force derivations. In particular, when I was trying to derive the same form of the [[User:Tohline/Appendix/Ramblings/T3CharacteristicVector#Two_Views_of_Equation_of_Motion|equation of motion from two different points of view]], I had to plug in some of your Christoffel symbol expressions. At first I could not get my EOM to match yours; it was while trying to understand this mismatch that I realized that your ''general'' expression for <math>\Gamma^i_{ij}</math> did not match the ''specific'' expression for <math>\Gamma^2_{21}</math> that you typed on the "T3 Coordinates" page. In my effort to get the EOMs to match, I decided that your ''general'' expression was the incorrect one. --[[User:Tohline|Tohline]] 14:58, 9 June 2010 (MDT)

:::For the record, on 10 June 2010, I modified the [[User:Jaycall/KillingVectorApproach#Christoffel_Symbols|table that contains the general Christoffel symbol expressions]] in order to correct the order of these indexes.--[[User:Tohline|Tohline]] 18:55, 11 June 2010 (MDT)

==Logarithmic Derivatives of T3 Scale Factors==

Check out my new subsection (under T3 Coordinates) entitled [[User:Tohline/Appendix/Ramblings/T3Integrals#Logarithmic_Derivatives_of_Scale_Factors|Logarithmic Derivatives of Scale Factors]]. First, see if you agree that there is a mistake (typo) in one of your tables of partial derivatives. Specifically, I think that the correct expression is:
<div align="center">
<math>
\frac{\partial z}{\partial\lambda_2} = - (q^2-1)\frac{\varpi^2 z \ell^2}{\lambda_2} .
</math>
</div>
Second, see if you agree with my derived expressions for <math>\partial\ln h_i/\partial\ln\lambda_j</math>. --[[User:Tohline|Tohline]] 22:28, 31 May 2010 (MDT)

:You're certainly right about the sign error. I have corrected the expression in the relevant table. --[[User:Jaycall|Jaycall]] 14:28, 1 June 2010 (MDT)

:I have not yet been able to confirm your expressions for logarithmic derivatives of the scale factors. I'm not sure what approach you took in deriving them, but since I had already calculated partials of the scale factors (<math>\partial_i h_j</math> and so forth), I derived a little trick to help simplify the work. Do you agree that

<div align="center">
<math>
\frac{\partial \ln h_i}{\partial \ln \lambda_j} \equiv \frac{\lambda_j}{h_i} \partial_j h_i ?
</math>
</div>

::Yes, this is precisely how I defined the logarithmic derivatives. And I used your tables of partial derivatives (e.g., <math>\partial\varpi/\partial\lambda_j</math> and <math>\partial z/\partial\lambda_j</math>) to obtain <math>\partial_i h_j</math> and so forth. --[[User:Tohline|Tohline]] 17:30, 1 June 2010 (MDT)

'''<font color="red">Mistake!</font>''' Jay: I have found a mistake in my original derivation of the logarithmic derivatives of the <math>h_1</math> scale factor. Perhaps now they will match yours. Check it out and let me know! --[[User:Tohline|Tohline]] 12:41, 4 June 2010 (MDT)

::Joel: I am finally able to confirm your derivation of the logarithmic derivatives of the scale factors. I came at them from a different angle with a fresh head, and I am now confident that what you have is correct. Next I will look into the general case. --[[User:Jaycall|Jaycall]] 14:29, 10 July 2010 (MDT)

==Two Views of Equation of Motion==

Jay: I have created a [[User:Tohline/Appendix/Ramblings/T3CharacteristicVector|new wiki page]] to explicitly analyze how your Characteristic Vector formalism applies to the T3 Coordinate system. On this page is a subsection entitled "[[User:Tohline/Appendix/Ramblings/T3CharacteristicVector#Two_Views_of_Equation_of_Motion|Two Views of Equation of Motion]]" in which I have shown that the <math>2^\mathrm{nd}</math> component of the equation of motion can be written in the following form,
<div align="center">
<math>\frac{d(h_2^2 \dot{\lambda}_2)}{dt} = \biggl( h_1 \frac{\partial h_1}{\partial \lambda_2} \biggr) \dot{\lambda}_1^2 + \biggl( h_2 \frac{\partial h_2}{\partial \lambda_2} \biggr) \dot{\lambda}_2^2</math> ,
</div>
assuming that <math>\partial\Phi/\partial\lambda_2 = 0</math>. More to the point, I have shown that I can derive this form of this component of the equation of motion using either (A) your Christoffel symbol formalism, or (B) the more classical formalism that I have been using, which I obtained from Appendix 1.B of Binney & Tremaine (1987) and Morse & Feshbach (1953).

Now, I shouldn't be shocked that both derivations lead to the same form of the equation — after all, we're doing physics aren't we? But here is what surprised me: The Christoffel symbol formalism produces this form of the equation without ever assuming ''a particular coordinate system'' whereas, in order to derive the equation using the more classical formalism, I plugged in some relations between the partial derivatives of the scale factors that ''I thought'' were specific to T3 coordinates. This leads me to ask, "How many of the T3 Coordinate relations [[User:Tohline/Appendix/Ramblings/T3Integrals|derived earlier]] are generic relations that apply to any orthogonal, axisymmetric coordinate system, and how many are unique to the T3 Coordinate System?" I'm particularly interested in knowing how generalizable the relations are that appear inside boxes labeled "T3 Coordinates" under the subsections I have entitled, "[[User:Tohline/Appendix/Ramblings/T3Integrals#Derived_Identity_for_T3_Coordinates|Derived Identity for T3 Corodinates]]" and "[[User:Tohline/Appendix/Ramblings/T3Integrals#Logarithmic_Derivatives_of_Scale_Factors|Logarithmic Derivatives of Scale Factors]]". Can you answer this? --[[User:Tohline|Tohline]] 10:46, 7 June 2010 (MDT)

:Joel: Yes, this is the power of using a covariant formulation. The only assumptions we made were that the coordinates were orthogonal and axisymmetric. I believe that the relation under "[[User:Tohline/Appendix/Ramblings/T3Integrals#Derived_Identity_for_T3_Coordinates|Derived Identity for T3 Coordinates]]" would apply to any coordinate system meeting these criteria. I will double check, though. (How did you determine that the position vector could be written <math>\vec{x} = \hat{e}_1 (h_1 \lambda_1) + \hat{e}_2 (h_2 \lambda_2)</math>?)

:The logarithmic derivatives of the scale factors cannot be general in their current form due to the appearance of factors of <math>q</math>, but I suspect they can be generalized and molded into a form that closely resembles what you have. I will work on this, too. --[[User:Jaycall|Jaycall]] 10:39, 9 June 2010 (MDT)

::Jay: You asked how I determined the expression for the [[User:Tohline/Appendix/Ramblings/T3Integrals#Definition|position vector]]. I don't have my detailed notes with me right now, but I can outline the steps. In Morse & Feshbach's classical presentation of orthogonal curvilinear coordinates, they define a set of so-called ''direction cosines,'' <math>\gamma_{ij}</math>. (I don't have the generalized definition with me at this time, but they are expressible in terms of derivatives of the scale factors with respect to the various coordinates; I'm certain ''direction cosines'' are definable in terms of Christoffel symbols.) Each of the Cartesian unit vectors (<math>\hat{\imath},\hat{\jmath},\hat{k}</math>) can be written in terms of products of these direction cosines and the three unit vectors of your chosen curvilinear coordinate system (<math>\hat{e}_1, \hat{e}_2, \hat{e}_3</math>). According to Morse & Feshbach, for example, <math>\hat{\imath} = (\hat{e}_1\gamma_{11} + \hat{e}_2\gamma_{21} + \hat{e}_3\gamma_{31})</math>. So, all I did to obtain an expression for the position vector in T3 coordinates was define <math>\vec{x} = (\hat\imath x + \hat\jmath y + \hat{k}z)</math>, then replace each Cartesian unit vector by its expression in terms of products of the T3-coordinate unit vectors & the appropriate direct cosine, then gather together and simplify terms. Does this make sense? --[[User:Tohline|Tohline]] 14:59, 10 June 2010 (MDT)

:::Yes, it does. That's very helpful. --[[User:Jaycall|Jaycall]] 09:49, 11 June 2010 (MDT)

::::Jay: For the record, I have put together a wiki page that discusses the definition and utility of [[User:Tohline/Appendix/Ramblings/DirectionCosines|Direction Cosines]].--[[User:Tohline|Tohline]] 18:47, 3 July 2010 (MDT)

Joel, I finally figured out what was going on this afternoon with the strange conserved quantities we were getting, like <math>{h_2}^4 \dot{\lambda_2}^2</math> and, consequently, <math>\lambda_1 \lambda_2</math>. There ''is'' a mistake in the derivation stemming all the way back from Eq. CV.02 on my page on the [[User:Jaycall/KillingVectorApproach|characteristic vector approach]]. My derivation of the brute force condition on <math>C_2</math> was correct, but you can see from above that <math>\Xi</math> does not equal <math>\ln C_2</math>, it equals <math>- \ln C_2</math>. Consequently, the conclusion is no longer that <math>C_2</math> must equal <math>{h_2}^2 \dot{\lambda_2}</math>, but rather <math>\left( {h_2}^2 \dot{\lambda_2} \right)^{-1}</math>. So in fact you ''did'' find a conserved quantity--<math>1</math>! : ) I guess it was a tautology after all. --[[User:Jaycall|Jaycall]] 18:50, 14 July 2010 (MDT)

==Special Quadratic Case==
Jay: On a [[User:Tohline/Appendix/Ramblings/T3Integrals/QuadraticCase#Special_Case_.28Quadratic.29|new accompanying wiki page]] I have begun to investigate how our system behaves in the special case of <math>q^2 = 2</math>. I have added a pointer/link to this new page — and a separate pointer/link to the parallel investigation that you have told me you are conducting — on the page containing a table of contents to my [[User:Tohline/Appendix/Ramblings|various research ramblings]]. I have not yet actually gone to the page that details your analysis because, at least for a while, I prefer for our trains of thought to remain independent of one another. However, in an effort to simplify the comparison that we ultimately will need to make between our independent analyses, I strongly recommend that you write your expressions for the scale factors, etc. ''for the special case of <math>q^2=2</math>'' in terms of a variable, <math>\Lambda</math>, defined such that,
<div align="center">
<math>
\Lambda \equiv \biggl[1 + \biggl(\frac{2\lambda_1}{\lambda_2}\biggr)^2\biggr]^{1/2} = \cosh(2\Zeta) .
</math>
</div>
By the way, if you don't agree that <math>\sqrt{1+(2\lambda_1/\lambda_2)^2}</math> equals <math>\cosh(2\Zeta)</math>, please let me know!--[[User:Tohline|Tohline]] 19:49, 11 June 2010 (MDT)

:: Joel: I have checked and agree that in the <math>q^2=2</math> case only, the above relation among <math>\Lambda</math>, <math>Z</math> and the T3 coordinates is correct. --[[User:Jaycall|Jaycall]] 14:58, 10 July 2010 (MDT)

User talk:Jaycall

2010-07-15T00:47:03Z

Jaycall: /* Two Views of Equation of Motion */

==Joel's First Talk Message to Jay==
Jay: In the email message that you sent to me today, you indicated that you had added the 3rd component to the dx/dt equation. I don't see that modification in the current T3Coordinates page. --[[User:Tohline|Tohline]] 15:05, 29 May 2010 (MDT)

:Joel: It's at the bottom of the section entitled "Time-Derivative of Position and Velocity Vectors". Under the history tab, can you see the edits that I made? It was the very first edit and was fairly minor. I only added a couple terms. --[[User:Jaycall|Jaycall]] 20:17, 29 May 2010 (MDT)

::The mediaWiki "talk" page recommends that when you reply to a query, you should not start a new subsection but, rather, just indent (using one or more colons) immediately following the query. Also, it recommends that you "sign" each "talk" message. You do this by typing 2 dashes followed by 4 tildes! I'm going to edit your "reply" (and this additional one from me) to put it in this recommended format. But you have to "sign" your own remark. --[[User:Tohline|Tohline]] 16:23, 29 May 2010 (MDT)

::By the way, I ''can'' see the edits that you made to add the 3rd component to the dx/dt equation. I did not spot the changes earlier, but via the history "diff" function, I'm able to see the edits clearly. --[[User:Tohline|Tohline]] 16:41, 29 May 2010 (MDT)

==Killing Vector Approach==
Thanks for writing out all the terms in the expression for <math>d\Xi/dt</math> in your "[[User:Jaycall/KillingVectorApproach|Killing Vector Approach]]" discussion. On the one hand, I'm happy that the term I mentioned cancels with another one because that means we have fewer terms to integrate. On the other hand, it was one term that I thought we might actually be able to manipulate analytically. Of the two terms that remain, one is relatively simple — the one that is proportional to <math>\dot{\lambda}_2</math> — but the first term is a bear! That will take some thinking! --[[User:Tohline|Tohline]] 10:50, 30 May 2010 (MDT)

:I know it seems like the wrong direction, but unless we can express this summary integral entirely in terms of <math>\lambda_1</math>, <math>\lambda_2</math>, <math>\dot{\lambda}_1</math> and <math>\dot{\lambda}_2</math>, I think that in order to make progress we're going to have to write things out in terms of <math>\varpi</math>, <math>z</math>, <math>\dot{\varpi}</math> and <math>\dot{z}</math>. Of course, that will make the expression much messier, but I think it will be very difficult to recognize this quantity as an exact derivative unless everything's in terms of the same coordinates. --[[User:Jaycall|Jaycall]] 12:55, 30 May 2010 (MDT)

::I agree. It is for similar reasons that I am planning on focusing on the case where <math>q^2=2</math>. At least in this special case we can invert the coordinate expressions to give <math>\varpi</math> and <math>z</math> in terms of the <math>\lambda</math>s. Incidentally, I have added a link to your "Killing Vector" page as well as a link to this associated "Talk" page on the [[User:Tohline/Appendix/Ramblings|page where I itemize my various "ramblings"]]. --[[User:Tohline|Tohline]] 16:16, 30 May 2010 (MDT)

:::Good point. That would be a good reason to focus on one specific case. I haven't looked closely at the inverted coordinate transformation for the <math>q^2=2</math> case for several days. Since the relation is quadratic, is it obvious which root should be used?

::The proper physical root was obvious to me when I performed the coordinate inversion in the case of [[User:Tohline/Appendix/Ramblings/T1Coordinates#Coordinate_Inversion|T1 Coordinates]], so I presume it will be obvious for the T3 Coordinate system. But the inversion needs to be redone for the case of T3 Coordinates; do you want to take care of this and type up the result? In the (quadratic) case of <math>q^2=2</math>, it may be as simple as replacing <math>\chi_2</math> with <math>1/\lambda_2</math>. --[[User:Tohline|Tohline]] 11:21, 31 May 2010 (MDT)

:::Sure. --[[User:Jaycall|Jaycall]] 13:34, 31 May 2010 (MDT)

'''<font color="red">Mistake?</font>''' Please check the next to last row in the [[User:Jaycall/KillingVectorApproach#Christoffel_Symbols|tabulated expressions for Christoffel symbols]]; shouldn't the <math>i</math> and <math>j</math> indexes be swapped? --[[User:Tohline|Tohline]] 09:39, 6 June 2010 (MDT)

:You are correct. Good eye.--[[User:Jaycall|Jaycall]] 09:42, 9 June 2010 (MDT)

::Actually, it wasn't my good eye. I discovered this from my inelegant brute-force derivations. In particular, when I was trying to derive the same form of the [[User:Tohline/Appendix/Ramblings/T3CharacteristicVector#Two_Views_of_Equation_of_Motion|equation of motion from two different points of view]], I had to plug in some of your Christoffel symbol expressions. At first I could not get my EOM to match yours; it was while trying to understand this mismatch that I realized that your ''general'' expression for <math>\Gamma^i_{ij}</math> did not match the ''specific'' expression for <math>\Gamma^2_{21}</math> that you typed on the "T3 Coordinates" page. In my effort to get the EOMs to match, I decided that your ''general'' expression was the incorrect one. --[[User:Tohline|Tohline]] 14:58, 9 June 2010 (MDT)

:::For the record, on 10 June 2010, I modified the [[User:Jaycall/KillingVectorApproach#Christoffel_Symbols|table that contains the general Christoffel symbol expressions]] in order to correct the order of these indexes.--[[User:Tohline|Tohline]] 18:55, 11 June 2010 (MDT)

==Logarithmic Derivatives of T3 Scale Factors==

Check out my new subsection (under T3 Coordinates) entitled [[User:Tohline/Appendix/Ramblings/T3Integrals#Logarithmic_Derivatives_of_Scale_Factors|Logarithmic Derivatives of Scale Factors]]. First, see if you agree that there is a mistake (typo) in one of your tables of partial derivatives. Specifically, I think that the correct expression is:
<div align="center">
<math>
\frac{\partial z}{\partial\lambda_2} = - (q^2-1)\frac{\varpi^2 z \ell^2}{\lambda_2} .
</math>
</div>
Second, see if you agree with my derived expressions for <math>\partial\ln h_i/\partial\ln\lambda_j</math>. --[[User:Tohline|Tohline]] 22:28, 31 May 2010 (MDT)

:You're certainly right about the sign error. I have corrected the expression in the relevant table. --[[User:Jaycall|Jaycall]] 14:28, 1 June 2010 (MDT)

:I have not yet been able to confirm your expressions for logarithmic derivatives of the scale factors. I'm not sure what approach you took in deriving them, but since I had already calculated partials of the scale factors (<math>\partial_i h_j</math> and so forth), I derived a little trick to help simplify the work. Do you agree that

<div align="center">
<math>
\frac{\partial \ln h_i}{\partial \ln \lambda_j} \equiv \frac{\lambda_j}{h_i} \partial_j h_i ?
</math>
</div>

::Yes, this is precisely how I defined the logarithmic derivatives. And I used your tables of partial derivatives (e.g., <math>\partial\varpi/\partial\lambda_j</math> and <math>\partial z/\partial\lambda_j</math>) to obtain <math>\partial_i h_j</math> and so forth. --[[User:Tohline|Tohline]] 17:30, 1 June 2010 (MDT)

'''<font color="red">Mistake!</font>''' Jay: I have found a mistake in my original derivation of the logarithmic derivatives of the <math>h_1</math> scale factor. Perhaps now they will match yours. Check it out and let me know! --[[User:Tohline|Tohline]] 12:41, 4 June 2010 (MDT)

::Joel: I am finally able to confirm your derivation of the logarithmic derivatives of the scale factors. I came at them from a different angle with a fresh head, and I am now confident that what you have is correct. Next I will look into the general case. --[[User:Jaycall|Jaycall]] 14:29, 10 July 2010 (MDT)

==Two Views of Equation of Motion==

Jay: I have created a [[User:Tohline/Appendix/Ramblings/T3CharacteristicVector|new wiki page]] to explicitly analyze how your Characteristic Vector formalism applies to the T3 Coordinate system. On this page is a subsection entitled "[[User:Tohline/Appendix/Ramblings/T3CharacteristicVector#Two_Views_of_Equation_of_Motion|Two Views of Equation of Motion]]" in which I have shown that the <math>2^\mathrm{nd}</math> component of the equation of motion can be written in the following form,
<div align="center">
<math>\frac{d(h_2^2 \dot{\lambda}_2)}{dt} = \biggl( h_1 \frac{\partial h_1}{\partial \lambda_2} \biggr) \dot{\lambda}_1^2 + \biggl( h_2 \frac{\partial h_2}{\partial \lambda_2} \biggr) \dot{\lambda}_2^2</math> ,
</div>
assuming that <math>\partial\Phi/\partial\lambda_2 = 0</math>. More to the point, I have shown that I can derive this form of this component of the equation of motion using either (A) your Christoffel symbol formalism, or (B) the more classical formalism that I have been using, which I obtained from Appendix 1.B of Binney & Tremaine (1987) and Morse & Feshbach (1953).

Now, I shouldn't be shocked that both derivations lead to the same form of the equation — after all, we're doing physics aren't we? But here is what surprised me: The Christoffel symbol formalism produces this form of the equation without ever assuming ''a particular coordinate system'' whereas, in order to derive the equation using the more classical formalism, I plugged in some relations between the partial derivatives of the scale factors that ''I thought'' were specific to T3 coordinates. This leads me to ask, "How many of the T3 Coordinate relations [[User:Tohline/Appendix/Ramblings/T3Integrals|derived earlier]] are generic relations that apply to any orthogonal, axisymmetric coordinate system, and how many are unique to the T3 Coordinate System?" I'm particularly interested in knowing how generalizable the relations are that appear inside boxes labeled "T3 Coordinates" under the subsections I have entitled, "[[User:Tohline/Appendix/Ramblings/T3Integrals#Derived_Identity_for_T3_Coordinates|Derived Identity for T3 Corodinates]]" and "[[User:Tohline/Appendix/Ramblings/T3Integrals#Logarithmic_Derivatives_of_Scale_Factors|Logarithmic Derivatives of Scale Factors]]". Can you answer this? --[[User:Tohline|Tohline]] 10:46, 7 June 2010 (MDT)

:Joel: Yes, this is the power of using a covariant formulation. The only assumptions we made were that the coordinates were orthogonal and axisymmetric. I believe that the relation under "[[User:Tohline/Appendix/Ramblings/T3Integrals#Derived_Identity_for_T3_Coordinates|Derived Identity for T3 Coordinates]]" would apply to any coordinate system meeting these criteria. I will double check, though. (How did you determine that the position vector could be written <math>\vec{x} = \hat{e}_1 (h_1 \lambda_1) + \hat{e}_2 (h_2 \lambda_2)</math>?)

:The logarithmic derivatives of the scale factors cannot be general in their current form due to the appearance of factors of <math>q</math>, but I suspect they can be generalized and molded into a form that closely resembles what you have. I will work on this, too. --[[User:Jaycall|Jaycall]] 10:39, 9 June 2010 (MDT)

::Jay: You asked how I determined the expression for the [[User:Tohline/Appendix/Ramblings/T3Integrals#Definition|position vector]]. I don't have my detailed notes with me right now, but I can outline the steps. In Morse & Feshbach's classical presentation of orthogonal curvilinear coordinates, they define a set of so-called ''direction cosines,'' <math>\gamma_{ij}</math>. (I don't have the generalized definition with me at this time, but they are expressible in terms of derivatives of the scale factors with respect to the various coordinates; I'm certain ''direction cosines'' are definable in terms of Christoffel symbols.) Each of the Cartesian unit vectors (<math>\hat{\imath},\hat{\jmath},\hat{k}</math>) can be written in terms of products of these direction cosines and the three unit vectors of your chosen curvilinear coordinate system (<math>\hat{e}_1, \hat{e}_2, \hat{e}_3</math>). According to Morse & Feshbach, for example, <math>\hat{\imath} = (\hat{e}_1\gamma_{11} + \hat{e}_2\gamma_{21} + \hat{e}_3\gamma_{31})</math>. So, all I did to obtain an expression for the position vector in T3 coordinates was define <math>\vec{x} = (\hat\imath x + \hat\jmath y + \hat{k}z)</math>, then replace each Cartesian unit vector by its expression in terms of products of the T3-coordinate unit vectors & the appropriate direct cosine, then gather together and simplify terms. Does this make sense? --[[User:Tohline|Tohline]] 14:59, 10 June 2010 (MDT)

:::Yes, it does. That's very helpful. --[[User:Jaycall|Jaycall]] 09:49, 11 June 2010 (MDT)

::::Jay: For the record, I have put together a wiki page that discusses the definition and utility of [[User:Tohline/Appendix/Ramblings/DirectionCosines|Direction Cosines]].--[[User:Tohline|Tohline]] 18:47, 3 July 2010 (MDT)

Joel, I finally figured out what was going on this afternoon with the strange conserved quantities we were getting, like <math>{h_2}^4 \dot{\lambda_2}^2</math> and, consequently, <math>\lambda_1 \lambda_2</math>. There ''is'' a mistake in the derivation stemming all the way back from Eq. CV.02 on my page on the [[User:Jaycall/KillingVectorApproach|characteristic vector approach]]. My derivation of the brute force condition on <math>C_2</math> was correct, but you can see from above that <math>\Xi</math> does not equal <math>\ln C_2</math>, it equals <math>- \ln C_2</math>. Consequently, the conclusion is no longer that <math>C_2</math> must equal <math>{h_2}^2 \dot{\lambda_2}</math>, but rather <math>\left( {h_2}^2 \dot{\lambda_2} \right)^{-1}</math>. So in fact you ''did'' find a conserved quantity--<math>1</math>! : ) I guess it was a tautology after all.

==Special Quadratic Case==
Jay: On a [[User:Tohline/Appendix/Ramblings/T3Integrals/QuadraticCase#Special_Case_.28Quadratic.29|new accompanying wiki page]] I have begun to investigate how our system behaves in the special case of <math>q^2 = 2</math>. I have added a pointer/link to this new page — and a separate pointer/link to the parallel investigation that you have told me you are conducting — on the page containing a table of contents to my [[User:Tohline/Appendix/Ramblings|various research ramblings]]. I have not yet actually gone to the page that details your analysis because, at least for a while, I prefer for our trains of thought to remain independent of one another. However, in an effort to simplify the comparison that we ultimately will need to make between our independent analyses, I strongly recommend that you write your expressions for the scale factors, etc. ''for the special case of <math>q^2=2</math>'' in terms of a variable, <math>\Lambda</math>, defined such that,
<div align="center">
<math>
\Lambda \equiv \biggl[1 + \biggl(\frac{2\lambda_1}{\lambda_2}\biggr)^2\biggr]^{1/2} = \cosh(2\Zeta) .
</math>
</div>
By the way, if you don't agree that <math>\sqrt{1+(2\lambda_1/\lambda_2)^2}</math> equals <math>\cosh(2\Zeta)</math>, please let me know!--[[User:Tohline|Tohline]] 19:49, 11 June 2010 (MDT)

:: Joel: I have checked and agree that in the <math>q^2=2</math> case only, the above relation among <math>\Lambda</math>, <math>Z</math> and the T3 coordinates is correct. --[[User:Jaycall|Jaycall]] 14:58, 10 July 2010 (MDT)

User:Tohline/Appendix/Ramblings/T3CharacteristicVector

2010-07-14T17:14:54Z

Jaycall: Thoughts on integration

{{LSU_HBook_header}}

=Characteristic Vector for T3 Coordinates=
Let's apply Jay's [[User:Jaycall/KillingVectorApproach|Characteristic Vector approach]] to Joel's [[User:Tohline/Appendix/Ramblings/T3Integrals|T3 Coordinate System]].

==Brute Force Manipulations==

Starting from '''[[User:Jaycall/KillingVectorApproach#CV.02|Equation CV.02]]''', and plugging in expressions for various [[User:Tohline/Appendix/Ramblings/T3Integrals#Logarithmic_Derivatives_of_Scale_Factors|logarithmic derivatives of the T3 scale factors]], we obtain,

<table align="center" border="0" cellpadding="5">
<tr>
<td align="right">
 
</td>
<td align="right">
<math>
\frac{\dot{C}_2}{C_2} \biggl(\frac{d \ln{\lambda}_2}{dt}\biggr)^{-1}
</math>
</td>
<td align="center">
<math>
=
</math>
</td>
<td align="left">
<math>
\biggl(\frac{h_1 \dot{\lambda}_1}{h_2 \dot{\lambda}_2}\biggr)^2 \frac{\partial \ln h_1}{\partial\ln\lambda_2} + \frac{\partial \ln h_2}{\partial \ln\lambda_2}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="right">
 
</td>
<td align="center">
<math>
=
</math>
</td>
<td align="left">
<math>
\biggl(\frac{h_1 \dot{\lambda}_1}{h_2 \dot{\lambda}_2}\biggr)^2 \biggl( \frac{q h_1 h_2 \lambda_2}{\lambda_1 } \biggr)^2 - ( qh_1^2 )^2
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="right">
 
</td>
<td align="center">
<math>
=
</math>
</td>
<td align="left">
<math>
\biggl[ (h_1 \dot{\lambda}_1)^2 ( q h_1 h_2 \lambda_2 )^2 - (h_2 \dot{\lambda}_2)^{2} ( qh_1^2 \lambda_1 )^2 \biggr](h_2 \lambda_1 \dot{\lambda}_2)^{-2}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="right">
 
</td>
<td align="center">
<math>
=
</math>
</td>
<td align="left">
<math>
\biggl[ \biggl(\frac{\dot{\lambda}_1}{\lambda_1}\biggr)^2 - \biggl( \frac{\dot{\lambda}_2}{\lambda_2} \biggr)^2 \biggr]( q h_1^2 h_2 \lambda_1 \lambda_2 )^2 (h_2 \lambda_1 \dot{\lambda}_2)^{-2}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="right">
 
</td>
<td align="center">
<math>
=
</math>
</td>
<td align="left">
<math>
\biggl[ \frac{\dot{\lambda}_1}{\lambda_1} + \frac{\dot{\lambda}_2}{\lambda_2} \biggr] \biggl[ \frac{\dot{\lambda}_1}{\lambda_1} - \frac{\dot{\lambda}_2}{\lambda_2} \biggr] \biggl( \frac{ q h_1^2 \lambda_2}{\dot{\lambda}_2} \biggr)^2
</math>
</td>
</tr>

<tr>
<td align="right">
<math>\Rightarrow</math>   
</td>
<td align="right">
<math>
\frac{\dot{C}_2}{C_2} \biggl(\frac{d \ln{\lambda}_2}{dt}\biggr)
</math>
</td>
<td align="center">
<math>
=
</math>
</td>
<td align="left">
<math>
\biggl[ \frac{\dot{\lambda}_1}{\lambda_1} + \frac{\dot{\lambda}_2}{\lambda_2} \biggr] \biggl[ \frac{\dot{\lambda}_1}{\lambda_1} - \frac{\dot{\lambda}_2}{\lambda_2} \biggr] ( q h_1^2 )^2
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="right">
 
</td>
<td align="center">
<math>
=
</math>
</td>
<td align="left">
<math>
\biggl[ \frac{\dot{\lambda}_1}{\lambda_1} + \frac{\dot{\lambda}_2}{\lambda_2} \biggr] \frac{d\ln h_2}{dt}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="right">
 
</td>
<td align="center">
<math>
=
</math>
</td>
<td align="left">
<math>
\biggl[ \frac{d\ln(\lambda_1 \lambda_2)}{dt} \biggr] \frac{d\ln h_2}{dt}
</math>
</td>
</tr>



</table>

==Two Views of Equation of Motion==

===Christoffel Symbol Formalism===

The second component of the equation of motion can be obtained by setting <math>i = 2</math> and <math>C_i = 1</math> in '''[[User:Jaycall/KillingVectorApproach#CV.01|Equation CV.01]]''', specifically,

<table border="0" align="center" cellpadding="5">
<tr>
<td align="right">
<math>
\frac{d(h_2^2 \dot{\lambda}_2)}{dt}
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
{h_k}^2 \Gamma^k_{2j} \dot{\lambda}_j \dot{\lambda}_k
</math>
</td>
<td align="left">
<math>
= {h_1}^2 \dot{\lambda}_1 \biggr[ \Gamma^1_{21} \dot{\lambda}_1 + \Gamma^1_{22} \dot{\lambda}_2 \biggl] +
{h_2}^2 \dot{\lambda}_2 \biggr[ \Gamma^2_{21} \dot{\lambda}_1 + \Gamma^2_{22} \dot{\lambda}_2 \biggl]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left" colspan="2">
<math>
{h_1}^2 \dot{\lambda}_1 \biggr[ \biggl( \frac{1}{h_1} \frac{\partial h_1}{\partial\lambda_2} \biggr) \dot{\lambda}_1 - \biggl( \frac{h_2}{h_1^2} \frac{\partial h_2}{\partial\lambda_1} \biggr) \dot{\lambda}_2 \biggl] +
{h_2}^2 \dot{\lambda}_2 \biggr[ \biggl( \frac{1}{h_2} \frac{\partial h_2}{\partial\lambda_1} \biggr) \dot{\lambda}_1 + \biggl( \frac{1}{h_2} \frac{\partial h_2}{\partial\lambda_2} \biggr) \dot{\lambda}_2 \biggl]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left" colspan="2">
<math>
\biggl( h_1 \frac{\partial h_1}{\partial\lambda_2} \biggr) \dot{\lambda}_1^2 +
\biggl( h_2 \frac{\partial h_2}{\partial\lambda_2} \biggr) \dot{\lambda}_2^2
</math>
</td>
</tr>
</table>

===Binney and Tremaine Formalism===

We have also derived the second component of the equation of motion following the formalism outlined by Binney and Tremaine ([[User:Tohline/Appendix/References#BT87|BT87]]). Specifically, in our introductory discussion of the [[User:Tohline/Appendix/Ramblings/T3Integrals|T3 Coordinate System]] our '''[[User:Tohline/Appendix/Ramblings/T3Integrals#EOM.01|Equation EOM.01]]''' has the form,

<table border="0" align="center" cellpadding="5">
<tr>
<td align="right">
<math>
\frac{d(h_2 \dot{\lambda}_2)}{dt}
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\biggl(\frac{\lambda_2 \dot{\lambda}_1}{\lambda_1}\biggr) \frac{dh_2}{dt} .
</math>
</td>
</tr>

</table>

To compare this with the form derived using the Christoffel symbol formalism, we need to multiply through by <math>h_2</math> and bring the scale factor inside the time-derivative on the left-hand-side.

<table border="0" align="center" cellpadding="5">
<tr>
<td align="right">
<math>
\frac{d(h_2^2 \dot{\lambda}_2)}{dt}
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left" colspan="3">
<math>
\biggl[ \biggl(\frac{h_2 \lambda_2 \dot{\lambda}_1}{\lambda_1}\biggr) + (h_2 \dot{\lambda}_2) \biggr]\frac{dh_2}{dt}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\biggl[ \biggl(\frac{h_2 \lambda_2 \dot{\lambda}_1}{\lambda_1}\biggr) + (h_2 \dot{\lambda}_2) \biggr]\biggl[ \frac{\partial h_2}{\partial\lambda_1} \dot{\lambda}_1 + \frac{\partial h_2}{\partial\lambda_2} \dot{\lambda}_2 \biggr]
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>
\biggl[ \biggl(\frac{h_2 \lambda_2 \dot{\lambda}_1}{\lambda_1}\biggr) + (h_2 \dot{\lambda}_2) \biggr]\biggl[ - \frac{\lambda_2}{\lambda_1} \dot{\lambda}_1 + \dot{\lambda}_2 \biggr] \frac{\partial h_2}{\partial\lambda_2}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left" colspan="1">
<math>
\biggl[ \dot{\lambda}_2 + \frac{\lambda_2 }{\lambda_1} \dot{\lambda}_1 \biggr]\biggl[ \dot{\lambda}_2 - \frac{\lambda_2}{\lambda_1} \dot{\lambda}_1 \biggr] h_2 \frac{\partial h_2}{\partial\lambda_2}
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left" colspan="1">
<math>
\biggl[ \dot{\lambda}_2^2 - \biggl( \frac{\lambda_2 }{\lambda_1}\biggr)^2 \dot{\lambda}_1^2 \biggr] h_2 \frac{\partial h_2}{\partial\lambda_2}
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>=</math>
</td>
<td align="left" colspan="1">
<math>
\biggl( h_2 \frac{\partial h_2}{\partial\lambda_2}\biggr) \dot{\lambda}_2^2 - \biggl[\frac{h_2 \lambda_2^2}{\lambda_1^2} \dot{\lambda}_1^2 \biggr] \biggl[- \frac{h_1 \lambda_1^2}{h_2 \lambda_2^2} \frac{\partial h_1}{\partial\lambda_2} \biggr]
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left" colspan="1">
<math>
\biggl( h_2 \frac{\partial h_2}{\partial\lambda_2}\biggr) \dot{\lambda}_2^2 + \biggl( h_1 \frac{\partial h_1}{\partial\lambda_2}\biggr) \dot{\lambda}_1^2
</math>
</td>
</tr>
</table>

===Summary===
So we see that, indeed, the two formalisms produce identical forms of the equation of motion.

==Implications==
Backing up to the expression that began our examination of the Binney and Tremaine formalism, we also can write,

<table border="0" align="center" cellpadding="5">
<tr>
<td align="right" colspan="2">
<math>
\frac{d\ln(h_2^2 \dot{\lambda}_2)}{dt}
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left" colspan="3">
<math>
\frac{\lambda_2}{\dot{\lambda}_2} \biggl[ \frac{\dot{\lambda}_1}{\lambda_1} + \frac{\dot{\lambda}_2}{\lambda_2} \biggr]\frac{d\ln h_2}{dt}
</math>
</td>
</tr>

<tr>
<td align="left">
<math>\Rightarrow</math>    
</td>
<td align="right">
<math>
\frac{d\ln(h_2^2 \dot{\lambda}_2)}{dt} \biggl(\frac{d\ln\lambda_2}{dt}\biggr)
</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left" colspan="3">
<math>
\biggl[ \frac{d\ln(\lambda_1 \lambda_2)}{dt} \biggr]\frac{d\ln h_2}{dt}
</math>
</td>
</tr>
</table>

Comparing this with the ''brute force'' derivation of the condition derived above for the characteristic vector, <math>C_2</math>, we see that the two expressions are the same if we set,
<div align="center">
<math>
C_2 = h_2^2 \dot{\lambda}_2 .
</math>
</div>

<b><font color="darkgreen">
This seems to imply that we have discovered a conserved quantity, namely, <math>(h_2^2 \dot{\lambda}_2)^2</math>. On the other hand, I might just be using a circular argument; I might only be saying that "the equation of motion is the equation of motion!"
</font></b>

Temp note: Joel, I don't quite understand this. Next time we get together, can you explain this page to me?

==Conserved Quantity==

Let's cut to the chase. As shown on the page describing the characteristic vector approach, I can write down the third conserved quantity right now--just not in closed form. Assuming there's no potential variation in the direction of <math>\lambda_2</math>, it is

<div align="center">
<math>
m{h_2}^2 \dot{\lambda_2} \exp \left\{ - \int \frac{{h_k}^2}{{h_2}^2} \Gamma^k_{2j} \frac{\dot{\lambda_j} \dot{\lambda_k}}{\dot{\lambda_2}} \ dt \right\} .
</math>
</div>

In the case of T3 coordinates, this becomes more specific.

<div align="center">
<math>
m {h_2}^2 \dot{\lambda_2} \exp \left\{ - \int 2 {\lambda_1}^2 \ell^4 \left( \frac{\lambda_2 \dot{\lambda_1}^2}{\dot{\lambda_2}} - \frac{\dot{\lambda_2}{\lambda_1}^2}{\lambda_2} \right) dt \right\}
</math>
</div>

Although this is not all that useful in an analytic sense until we can integrate it, I wonder if it can be a guide to building a more accurate numerical model. Certainly this function can be integrated numerically, and that's got to be useful somehow...

==Thoughts on Integrating This Conserved Quantity==

The quantity appearing inside the parentheses has an interesting symmetry. Each variable appearing without a dot in the first term appears in the same place with a dot in the second term, and vice versa. Certainly there must be some differentiation rule that will allow us to express this quantity as a total time derivative.

On the other hand, the factor of <math>\dot{\lambda_2}</math> appearing in the denominator of the first term is troublesome. I can't think of any differentiation rule that puts a derivative in the denominator. Product rule, quotient rule, and chain rule all end up ''multiplying'' by derivatives. So I wonder if there's some way to eliminate the <math>\dot{\lambda_2}</math> in favor of undotted variables. This would require transforming the equation of motion for the <math>\dot{\lambda_2}</math> coordinate into a first-order equation. Right now, the second-order equation reads

<div align="center">
<math>
\ddot{\lambda_2} + \frac{\dot{h_2}}{h_2} \dot{\lambda_2} - \frac{\dot{\lambda_1} \dot{h_2}}{\lambda_1 h_2} \lambda_2 = 0 .
</math>
</div>

The first step in reducing this to a first-order equation is to perform a transformation of variables that eliminates that <math>\dot{\lambda_2}</math> term. I have successfully accomplished this. By defining <math>b \equiv {h_2}^{1/2} \lambda_2</math>, the equation can be written:

<div align="center">
<math>
\ddot{b} + \left( \tfrac{1}{4} \frac{{\dot{h_2}}^2}{h_2} - \tfrac{1}{2} \frac{\ddot{h_2}}{h_2} - \frac{\dot{\lambda_1} \dot{h_2}}{\lambda_1 h_2} \right) b = 0 .
</math>
</div>

 <br />
{{LSU_HBook_footer}}

User:Jaycall/T3 Coordinates/Special Case

2010-07-14T16:40:23Z

Jaycall: Added conserved quantity

If the special case <math>q^2=2</math> is considered, it is possible to invert the coordinate transformations in closed form. The coordinate transformations and their inversions become

<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>\left( R^2+2z^2 \right)^{1/2}</math>
</td>
<td align="center">
      and      
</td>
<td align="right">
<math>
\lambda_2
</math>
</td>
<td align="center">
<math>\equiv</math>
</td>
<td align="left">
<math>\frac{R^2}{\sqrt{2}z}</math>
</td>
</tr>
<tr>
<td align="right">
<math>
R^2
</math>
</td>
<td align="center">
<math>\equiv</math>
</td>
<td align="left">
<math>-\frac{{\lambda_2}^2}{2} + \sqrt{\frac{{\lambda_2}^2}{4} + {\lambda_1}^2 {\lambda_2}^2}</math>
</td>
<td align="center">
      and      
</td>
<td align="right">
<math>
z
</math>
</td>
<td align="center">
<math>\equiv</math>
</td>
<td align="left">
<math>-\frac{\lambda_2}{2\sqrt{2}} + \sqrt{\frac{4 {\lambda_1}^2+{\lambda_2}^2}{8}}</math>
</td>
</tr>
</table>

Partial derivatives of each of the T3 coordinates taken with respect to each of the cylindrical coordinates are:

<table align="center" border="1" cellpadding="5">
<tr>
<td align="center">
 
</td>
<td align="center">
<math>
\frac{\partial}{\partial R}
</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{R}{\lambda_1}
</math>
</td>
<td align="center">
<math>
\frac{2z}{\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{2 \lambda_2}{R}
</math>
</td>
<td align="center">
<math>
-\frac{\lambda_2}{z}
</math>
</td>
<td align="center">
<math>0</math>
</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>

And partials of the cylindrical coordinates taken with respect to the T3 coordinates are:

<table align="center" border="1" cellpadding="5">
<tr>
<td align="center">
 
</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>R</math>
</td>
<td align="center">
<math>
R \ell^2 \lambda_1
</math>
</td>
<td align="center">
<math>
2Rz^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>
2z \ell^2 \lambda_1
</math>
</td>
<td align="center">
<math>
-R^2 z \ell^2 / \lambda_2
</math>
</td>
<td align="center">
<math>0</math>
</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>

where <math>\ell \equiv \left( R^2 + 4z^2 \right)^{-1/2}</math>.

Furthermore, the scale factors become

<table align="center">
<tr>
<td align="right">
<math>h_1</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>\lambda_1 \ell</math>
</td>
</tr>

<tr>
<td align="right">
<math>h_2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>Rz \ell / \lambda_2</math>
</td>
</tr>

<tr>
<td align="right">
<math>h_3</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>R = \lambda_3</math>
</td>
</tr>
</table>

In this special case, there are some additional useful relationships between various combinations of cylindrical variables and their T3 equivalents which can be written out.

<table align="center">
<tr>
<td align="right">
<math>R^2 + 2z^2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>{\lambda_1}^2</math>
</td>
</tr>

<tr>
<td align="right">
<math>R^2 + 4z^2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>2 {\lambda_1}^2 + {\lambda_2}^2/2 - \lambda_2 \sqrt{{\lambda_1}^2+{\lambda_2}^2/4} = \ell^{-2}</math>
</td>
</tr>

<tr>
<td align="right">
<math>R^2 + 8z^2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>4 {\lambda_1}^2 + \tfrac{3}{2} {\lambda_2}^2 - 3 \lambda_2 \sqrt{{\lambda_1}^2+{\lambda_2}^2/4} = 3 \ell^{-2} - 2 {\lambda_1}^2</math>
</td>
</tr>

<tr>
<td align="right">
<math>R^2 - 2z^2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>- {\lambda_1}^2 - {\lambda_2}^2 + 2 \lambda_2 \sqrt{{\lambda_1}^2+{\lambda_2}^2/4} = 3 {\lambda_1}^2 -2 \ell^{-2}</math>
</td>
</tr>

<tr>
<td align="right">
<math>Rz</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>\sqrt{\sqrt{2}\lambda_2} \left( -\frac{\lambda_2}{2\sqrt{2}} + \sqrt{\frac{4{\lambda_1}^2+{\lambda_2}^2}{8}} \right)^{3/2} = h_2 \lambda_2 / \ell</math>
</td>
</tr>
</table>

Partials of <math>\ell</math> can be taken with respect to the coordinates of either system. They are:

<table align="center" border="1" cellpadding="5">
<tr>
<td align="center">
 
</td>
<td align="center">
<math>
\frac{\partial}{\partial R}
</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>\ell</math>
</td>
<td align="center">
<math>
-R \ell^3
</math>
</td>
<td align="center">
<math>
-4z \ell^3
</math>
</td>
<td align="center">
<math>
0
</math>
</td>
</tr>
</table>

<table align="center" border="1" cellpadding="5">
<tr>
<td align="center">
 
</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>\ell</math>
</td>
<td align="center">
<math>
- \left( R^2 + 8z^2 \right) \ell^5 \lambda_1 = \ell^3 \lambda_1 \left( 2{h_1}^2 - 3 \right)
</math>
</td>
<td align="center">
<math>
2R^2 z^2 \ell^5 / \lambda_2 = 2 {h_2}^2 \ell^3 \lambda_2
</math>
</td>
<td align="center">
<math>
0
</math>
</td>
</tr>
</table>

Partials of the scale factors taken with respect to the T3 coordinates are:

<table align="center" border="1" cellpadding="5">
<tr>
<td align="center">
 
</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>h_1</math>
</td>
<td align="center">
<math>
\ell \left( 2 {h_1}^4 - 3 {h_1}^2 + 1 \right) = 2 h_2 \lambda_2
</math>
</td>
<td align="center">
<math>
2 {h_2}^2 \ell^3 \lambda_1 \lambda_2
</math>
</td>
<td align="center">
<math>
0
</math>
</td>
</tr>

<tr>
<td align="center">
<math>h_2</math>
</td>
<td align="center">
<math>
2 {h_1}^2 h_2 \ell^2 \lambda_1 = 2 h_2 \ell^4 {\lambda_1}^3
</math>
</td>
<td align="center">
<math>
h_2 \left( 2 {h_2}^2 \ell^2 {\lambda_2}^2 - 3 \ell^2 {\lambda_1}^2 + 1 \right) / \lambda_2
</math>
</td>
<td align="center">
<math>0</math>
</td>
</tr>

<tr>
<td align="center">
<math>h_3</math>
</td>
<td align="center">
<math>
R \ell^2 \lambda_1
</math>
</td>
<td align="center">
<math>
2Rz^2 \ell^2 / \lambda_2
</math>
</td>
<td align="center">
<math>
0
</math>
</td>
</tr>
</table>

The conserved quantity associated with the <math>\lambda_2</math> coordinate is

<div align="center">
<math>
m{h_2}^2 \dot{\lambda_2} \exp \int \left[ \left( 4 {\lambda_1}^2 + {\lambda_2}^2 - \lambda_2 \sqrt{4{\lambda_1}^2 + {\lambda_2}^2} \right) \left( \frac{{\lambda_1}^2 \dot{\lambda_2}}{\lambda_2} - \frac{\lambda_2 {\dot{\lambda_1}}^2}{\dot{\lambda_2}} \right) \right] dt
</math>
</div>

User:Jaycall/T3 Coordinates/Special Case

2010-07-14T16:10:56Z

Jaycall: Eliminated unphysical root

If the special case <math>q^2=2</math> is considered, it is possible to invert the coordinate transformations in closed form. The coordinate transformations and their inversions become

<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>\left( R^2+2z^2 \right)^{1/2}</math>
</td>
<td align="center">
      and      
</td>
<td align="right">
<math>
\lambda_2
</math>
</td>
<td align="center">
<math>\equiv</math>
</td>
<td align="left">
<math>\frac{R^2}{\sqrt{2}z}</math>
</td>
</tr>
<tr>
<td align="right">
<math>
R^2
</math>
</td>
<td align="center">
<math>\equiv</math>
</td>
<td align="left">
<math>-\frac{{\lambda_2}^2}{2} + \sqrt{\frac{{\lambda_2}^2}{4} + {\lambda_1}^2 {\lambda_2}^2}</math>
</td>
<td align="center">
      and      
</td>
<td align="right">
<math>
z
</math>
</td>
<td align="center">
<math>\equiv</math>
</td>
<td align="left">
<math>-\frac{\lambda_2}{2\sqrt{2}} + \sqrt{\frac{4 {\lambda_1}^2+{\lambda_2}^2}{8}}</math>
</td>
</tr>
</table>

Partial derivatives of each of the T3 coordinates taken with respect to each of the cylindrical coordinates are:

<table align="center" border="1" cellpadding="5">
<tr>
<td align="center">
 
</td>
<td align="center">
<math>
\frac{\partial}{\partial R}
</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{R}{\lambda_1}
</math>
</td>
<td align="center">
<math>
\frac{2z}{\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{2 \lambda_2}{R}
</math>
</td>
<td align="center">
<math>
-\frac{\lambda_2}{z}
</math>
</td>
<td align="center">
<math>0</math>
</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>

And partials of the cylindrical coordinates taken with respect to the T3 coordinates are:

<table align="center" border="1" cellpadding="5">
<tr>
<td align="center">
 
</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>R</math>
</td>
<td align="center">
<math>
R \ell^2 \lambda_1
</math>
</td>
<td align="center">
<math>
2Rz^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>
2z \ell^2 \lambda_1
</math>
</td>
<td align="center">
<math>
-R^2 z \ell^2 / \lambda_2
</math>
</td>
<td align="center">
<math>0</math>
</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>

where <math>\ell \equiv \left( R^2 + 4z^2 \right)^{-1/2}</math>.

Furthermore, the scale factors become

<table align="center">
<tr>
<td align="right">
<math>h_1</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>\lambda_1 \ell</math>
</td>
</tr>

<tr>
<td align="right">
<math>h_2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>Rz \ell / \lambda_2</math>
</td>
</tr>

<tr>
<td align="right">
<math>h_3</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>R = \lambda_3</math>
</td>
</tr>
</table>

In this special case, there are some additional useful relationships between various combinations of cylindrical variables and their T3 equivalents which can be written out.

<table align="center">
<tr>
<td align="right">
<math>R^2 + 2z^2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>{\lambda_1}^2</math>
</td>
</tr>

<tr>
<td align="right">
<math>R^2 + 4z^2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>2 {\lambda_1}^2 + {\lambda_2}^2/2 - \lambda_2 \sqrt{{\lambda_1}^2+{\lambda_2}^2/4} = \ell^{-2}</math>
</td>
</tr>

<tr>
<td align="right">
<math>R^2 + 8z^2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>4 {\lambda_1}^2 + \tfrac{3}{2} {\lambda_2}^2 - 3 \lambda_2 \sqrt{{\lambda_1}^2+{\lambda_2}^2/4} = 3 \ell^{-2} - 2 {\lambda_1}^2</math>
</td>
</tr>

<tr>
<td align="right">
<math>R^2 - 2z^2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>- {\lambda_1}^2 - {\lambda_2}^2 + 2 \lambda_2 \sqrt{{\lambda_1}^2+{\lambda_2}^2/4} = 3 {\lambda_1}^2 -2 \ell^{-2}</math>
</td>
</tr>

<tr>
<td align="right">
<math>Rz</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>\sqrt{\sqrt{2}\lambda_2} \left( -\frac{\lambda_2}{2\sqrt{2}} + \sqrt{\frac{4{\lambda_1}^2+{\lambda_2}^2}{8}} \right)^{3/2} = h_2 \lambda_2 / \ell</math>
</td>
</tr>
</table>

Partials of <math>\ell</math> can be taken with respect to the coordinates of either system. They are:

<table align="center" border="1" cellpadding="5">
<tr>
<td align="center">
 
</td>
<td align="center">
<math>
\frac{\partial}{\partial R}
</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>\ell</math>
</td>
<td align="center">
<math>
-R \ell^3
</math>
</td>
<td align="center">
<math>
-4z \ell^3
</math>
</td>
<td align="center">
<math>
0
</math>
</td>
</tr>
</table>

<table align="center" border="1" cellpadding="5">
<tr>
<td align="center">
 
</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>\ell</math>
</td>
<td align="center">
<math>
- \left( R^2 + 8z^2 \right) \ell^5 \lambda_1 = \ell^3 \lambda_1 \left( 2{h_1}^2 - 3 \right)
</math>
</td>
<td align="center">
<math>
2R^2 z^2 \ell^5 / \lambda_2 = 2 {h_2}^2 \ell^3 \lambda_2
</math>
</td>
<td align="center">
<math>
0
</math>
</td>
</tr>
</table>

Partials of the scale factors taken with respect to the T3 coordinates are:

<table align="center" border="1" cellpadding="5">
<tr>
<td align="center">
 
</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>h_1</math>
</td>
<td align="center">
<math>
\ell \left( 2 {h_1}^4 - 3 {h_1}^2 + 1 \right) = 2 h_2 \lambda_2
</math>
</td>
<td align="center">
<math>
2 {h_2}^2 \ell^3 \lambda_1 \lambda_2
</math>
</td>
<td align="center">
<math>
0
</math>
</td>
</tr>

<tr>
<td align="center">
<math>h_2</math>
</td>
<td align="center">
<math>
2 {h_1}^2 h_2 \ell^2 \lambda_1 = 2 h_2 \ell^4 {\lambda_1}^3
</math>
</td>
<td align="center">
<math>
h_2 \left( 2 {h_2}^2 \ell^2 {\lambda_2}^2 - 3 \ell^2 {\lambda_1}^2 + 1 \right) / \lambda_2
</math>
</td>
<td align="center">
<math>0</math>
</td>
</tr>

<tr>
<td align="center">
<math>h_3</math>
</td>
<td align="center">
<math>
R \ell^2 \lambda_1
</math>
</td>
<td align="center">
<math>
2Rz^2 \ell^2 / \lambda_2
</math>
</td>
<td align="center">
<math>
0
</math>
</td>
</tr>
</table>

User:Jaycall/T3 Coordinates/Special Case

2010-07-14T16:02:20Z

Jaycall: Added partials of scale factors

If the special case <math>q^2=2</math> is considered, it is possible to invert the coordinate transformations in closed form. The coordinate transformations and their inversions become

<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>\left( R^2+2z^2 \right)^{1/2}</math>
</td>
<td align="center">
      and      
</td>
<td align="right">
<math>
\lambda_2
</math>
</td>
<td align="center">
<math>\equiv</math>
</td>
<td align="left">
<math>\frac{R^2}{\sqrt{2}z}</math>
</td>
</tr>
<tr>
<td align="right">
<math>
R^2
</math>
</td>
<td align="center">
<math>\equiv</math>
</td>
<td align="left">
<math>-\frac{{\lambda_2}^2}{2} \pm \sqrt{\frac{{\lambda_2}^2}{4} + {\lambda_1}^2 {\lambda_2}^2}</math>
</td>
<td align="center">
      and      
</td>
<td align="right">
<math>
z
</math>
</td>
<td align="center">
<math>\equiv</math>
</td>
<td align="left">
<math>-\frac{\lambda_2}{2\sqrt{2}} \pm \sqrt{\frac{4 {\lambda_1}^2+{\lambda_2}^2}{8}}</math>
</td>
</tr>
</table>

Partial derivatives of each of the T3 coordinates taken with respect to each of the cylindrical coordinates are:

<table align="center" border="1" cellpadding="5">
<tr>
<td align="center">
 
</td>
<td align="center">
<math>
\frac{\partial}{\partial R}
</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{R}{\lambda_1}
</math>
</td>
<td align="center">
<math>
\frac{2z}{\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{2 \lambda_2}{R}
</math>
</td>
<td align="center">
<math>
-\frac{\lambda_2}{z}
</math>
</td>
<td align="center">
<math>0</math>
</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>

And partials of the cylindrical coordinates taken with respect to the T3 coordinates are:

<table align="center" border="1" cellpadding="5">
<tr>
<td align="center">
 
</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>R</math>
</td>
<td align="center">
<math>
R \ell^2 \lambda_1
</math>
</td>
<td align="center">
<math>
2Rz^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>
2z \ell^2 \lambda_1
</math>
</td>
<td align="center">
<math>
-R^2 z \ell^2 / \lambda_2
</math>
</td>
<td align="center">
<math>0</math>
</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>

where <math>\ell \equiv \left( R^2 + 4z^2 \right)^{-1/2}</math>.

Furthermore, the scale factors become

<table align="center">
<tr>
<td align="right">
<math>h_1</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>\lambda_1 \ell</math>
</td>
</tr>

<tr>
<td align="right">
<math>h_2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>Rz \ell / \lambda_2</math>
</td>
</tr>

<tr>
<td align="right">
<math>h_3</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>R = \lambda_3</math>
</td>
</tr>
</table>

In this special case, there are some additional useful relationships between various combinations of cylindrical variables and their T3 equivalents which can be written out.

<table align="center">
<tr>
<td align="right">
<math>R^2 + 2z^2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>{\lambda_1}^2</math>
</td>
</tr>

<tr>
<td align="right">
<math>R^2 + 4z^2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>2 {\lambda_1}^2 + {\lambda_2}^2/2 \mp \lambda_2 \sqrt{{\lambda_1}^2+{\lambda_2}^2/4} = \ell^{-2}</math>
</td>
</tr>

<tr>
<td align="right">
<math>R^2 + 8z^2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>4 {\lambda_1}^2 + \tfrac{3}{2} {\lambda_2}^2 \mp 3 \lambda_2 \sqrt{{\lambda_1}^2+{\lambda_2}^2/4} = 3 \ell^{-2} - 2 {\lambda_1}^2</math>
</td>
</tr>

<tr>
<td align="right">
<math>R^2 - 2z^2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>- {\lambda_1}^2 - {\lambda_2}^2 \pm 2 \lambda_2 \sqrt{{\lambda_1}^2+{\lambda_2}^2/4} = 3 {\lambda_1}^2 -2 \ell^{-2}</math>
</td>
</tr>

<tr>
<td align="right">
<math>Rz</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>\sqrt{\sqrt{2}\lambda_2} \left( -\frac{\lambda_2}{2\sqrt{2}} \pm \sqrt{\frac{4{\lambda_1}^2+{\lambda_2}^2}{8}} \right)^{3/2} = h_2 \lambda_2 / \ell</math>
</td>
</tr>
</table>

Partials of <math>\ell</math> can be taken with respect to the coordinates of either system. They are:

<table align="center" border="1" cellpadding="5">
<tr>
<td align="center">
 
</td>
<td align="center">
<math>
\frac{\partial}{\partial R}
</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>\ell</math>
</td>
<td align="center">
<math>
-R \ell^3
</math>
</td>
<td align="center">
<math>
-4z \ell^3
</math>
</td>
<td align="center">
<math>
0
</math>
</td>
</tr>
</table>

<table align="center" border="1" cellpadding="5">
<tr>
<td align="center">
 
</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>\ell</math>
</td>
<td align="center">
<math>
- \left( R^2 + 8z^2 \right) \ell^5 \lambda_1 = \ell^3 \lambda_1 \left( 2{h_1}^2 - 3 \right)
</math>
</td>
<td align="center">
<math>
2R^2 z^2 \ell^5 / \lambda_2 = 2 {h_2}^2 \ell^3 \lambda_2
</math>
</td>
<td align="center">
<math>
0
</math>
</td>
</tr>
</table>

Partials of the scale factors taken with respect to the T3 coordinates are:

<table align="center" border="1" cellpadding="5">
<tr>
<td align="center">
 
</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>h_1</math>
</td>
<td align="center">
<math>
\ell \left( 2 {h_1}^4 - 3 {h_1}^2 + 1 \right) = 2 h_2 \lambda_2
</math>
</td>
<td align="center">
<math>
2 {h_2}^2 \ell^3 \lambda_1 \lambda_2
</math>
</td>
<td align="center">
<math>
0
</math>
</td>
</tr>

<tr>
<td align="center">
<math>h_2</math>
</td>
<td align="center">
<math>
2 {h_1}^2 h_2 \ell^2 \lambda_1 = 2 h_2 \ell^4 {\lambda_1}^3
</math>
</td>
<td align="center">
<math>
h_2 \left( 2 {h_2}^2 \ell^2 {\lambda_2}^2 - 3 \ell^2 {\lambda_1}^2 + 1 \right) / \lambda_2
</math>
</td>
<td align="center">
<math>0</math>
</td>
</tr>

<tr>
<td align="center">
<math>h_3</math>
</td>
<td align="center">
<math>
R \ell^2 \lambda_1
</math>
</td>
<td align="center">
<math>
2Rz^2 \ell^2 / \lambda_2
</math>
</td>
<td align="center">
<math>
0
</math>
</td>
</tr>
</table>

User:Jaycall/T3 Coordinates/Special Case

2010-07-14T15:38:41Z

Jaycall: Fixed typo

If the special case <math>q^2=2</math> is considered, it is possible to invert the coordinate transformations in closed form. The coordinate transformations and their inversions become

<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>\left( R^2+2z^2 \right)^{1/2}</math>
</td>
<td align="center">
      and      
</td>
<td align="right">
<math>
\lambda_2
</math>
</td>
<td align="center">
<math>\equiv</math>
</td>
<td align="left">
<math>\frac{R^2}{\sqrt{2}z}</math>
</td>
</tr>
<tr>
<td align="right">
<math>
R^2
</math>
</td>
<td align="center">
<math>\equiv</math>
</td>
<td align="left">
<math>-\frac{{\lambda_2}^2}{2} \pm \sqrt{\frac{{\lambda_2}^2}{4} + {\lambda_1}^2 {\lambda_2}^2}</math>
</td>
<td align="center">
      and      
</td>
<td align="right">
<math>
z
</math>
</td>
<td align="center">
<math>\equiv</math>
</td>
<td align="left">
<math>-\frac{\lambda_2}{2\sqrt{2}} \pm \sqrt{\frac{4 {\lambda_1}^2+{\lambda_2}^2}{8}}</math>
</td>
</tr>
</table>

Partial derivatives of each of the T3 coordinates taken with respect to each of the cylindrical coordinates are:

<table align="center" border="1" cellpadding="5">
<tr>
<td align="center">
 
</td>
<td align="center">
<math>
\frac{\partial}{\partial R}
</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{R}{\lambda_1}
</math>
</td>
<td align="center">
<math>
\frac{2z}{\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{2 \lambda_2}{R}
</math>
</td>
<td align="center">
<math>
-\frac{\lambda_2}{z}
</math>
</td>
<td align="center">
<math>0</math>
</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>

And partials of the cylindrical coordinates taken with respect to the T3 coordinates are:

<table align="center" border="1" cellpadding="5">
<tr>
<td align="center">
 
</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>R</math>
</td>
<td align="center">
<math>
R \ell^2 \lambda_1
</math>
</td>
<td align="center">
<math>
2Rz^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>
2z \ell^2 \lambda_1
</math>
</td>
<td align="center">
<math>
-R^2 z \ell^2 / \lambda_2
</math>
</td>
<td align="center">
<math>0</math>
</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>

where <math>\ell \equiv \left( R^2 + 4z^2 \right)^{-1/2}</math>.

Furthermore, the scale factors become

<table align="center">
<tr>
<td align="right">
<math>h_1</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>\lambda_1 \ell</math>
</td>
</tr>

<tr>
<td align="right">
<math>h_2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>Rz \ell / \lambda_2</math>
</td>
</tr>

<tr>
<td align="right">
<math>h_3</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>R = \lambda_3</math>
</td>
</tr>
</table>

In this special case, there are some additional useful relationships between various combinations of cylindrical variables and their T3 equivalents which can be written out.

<table align="center">
<tr>
<td align="right">
<math>R^2 + 2z^2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>{\lambda_1}^2</math>
</td>
</tr>

<tr>
<td align="right">
<math>R^2 + 4z^2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>2 {\lambda_1}^2 + {\lambda_2}^2/2 \mp \lambda_2 \sqrt{{\lambda_1}^2+{\lambda_2}^2/4} = \ell^{-2}</math>
</td>
</tr>

<tr>
<td align="right">
<math>R^2 + 8z^2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>4 {\lambda_1}^2 + \tfrac{3}{2} {\lambda_2}^2 \mp 3 \lambda_2 \sqrt{{\lambda_1}^2+{\lambda_2}^2/4} = 3 \ell^{-2} - 2 {\lambda_1}^2</math>
</td>
</tr>

<tr>
<td align="right">
<math>R^2 - 2z^2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>- {\lambda_1}^2 - {\lambda_2}^2 \pm 2 \lambda_2 \sqrt{{\lambda_1}^2+{\lambda_2}^2/4} = 3 {\lambda_1}^2 -2 \ell^{-2}</math>
</td>
</tr>

<tr>
<td align="right">
<math>Rz</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>\sqrt{\sqrt{2}\lambda_2} \left( -\frac{\lambda_2}{2\sqrt{2}} \pm \sqrt{\frac{4{\lambda_1}^2+{\lambda_2}^2}{8}} \right)^{3/2} = h_2 \lambda_2 / \ell</math>
</td>
</tr>
</table>

Partials of <math>\ell</math> can be taken with respect to the coordinates of either system. They are:

<table align="center" border="1" cellpadding="5">
<tr>
<td align="center">
 
</td>
<td align="center">
<math>
\frac{\partial}{\partial R}
</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>\ell</math>
</td>
<td align="center">
<math>
-R \ell^3
</math>
</td>
<td align="center">
<math>
-4z \ell^3
</math>
</td>
<td align="center">
<math>
0
</math>
</td>
</tr>
</table>

<table align="center" border="1" cellpadding="5">
<tr>
<td align="center">
 
</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>\ell</math>
</td>
<td align="center">
<math>
- \left( R^2 + 8z^2 \right) \ell^5 \lambda_1 = \ell^3 \lambda_1 \left( 2{h_1}^2 - 3 \right)
</math>
</td>
<td align="center">
<math>
2R^2 z^2 \ell^5 / \lambda_2 = 2 {h_2}^2 \ell^3 \lambda_2
</math>
</td>
<td align="center">
<math>
0
</math>
</td>
</tr>
</table>

User:Tohline/Appendix/Ramblings/T3CharacteristicVector

2010-07-12T16:11:02Z

Jaycall: Added section on conserved quantity specific to T3 coordinates

User:Jaycall/KillingVectorApproach

2010-07-12T15:27:44Z

Jaycall: /* Killing Vector Approach to Integrals of Motion Problem */

{{LSU_HBook_header}}

=Characteristic Vector Approach to Integrals of Motion Problem=

Motivated by our recent work in relativistic fluid dynamics, one insightful approach to the integrals of motion problem involves rewriting the equations of motion in terms of an arbitrary vector field, which we're currently calling a ''characteristic vector''. The benefit here is that we can produce a mathematical condition that tells us what combintations of variables should be grouped together in order to produce conserved quantities. The details follow.

==Newtonian Equations of Motion==

Using index summation notation (wherein repeated indices are summed over), it is possible to write the Newtonian equations of motion for a test particle in a fixed, static potential in an arbitrary orthogonal coordinate system. The equation describing motion in the <math>i</math>-dimension is:

<div align="center">
<math>
\frac{d}{dt} \left( m \ {h_i}^2 \dot{\lambda}_i \right) = m \ {h_k}^2 \Gamma^k_{ij} \dot{\lambda}_j \dot{\lambda}_k - m \ \partial_i \Phi
</math>
</div>

where <math>m</math> is the mass of the test particle, the <math>\lambda</math>s are the chosen coordinates, and the <math>\dot{\lambda}</math>s are the coordinate velocities (or the total time-derivatives of the coordinates), the <math>h</math>s are the metric scale factors associated with each of the coordinates, <math>\Phi</math> is the gravitational potential, and the <math>\Gamma</math>s are the Christoffel symbols.

==Metric Components, Orthogonality Condition, and Scale Factors==

For any coordinate system it is possible to express the components of the metric in terms of the partial derivatives defining the transformation from any other coordinate system. Written in terms of the transformation from cylindrical coordinates, each metric component is

<div align="center">
<math>
g_{ij} = \left( \partial_i R \right) \left( \partial_j R \right) + \left( \partial_i z \right) \left( \partial_j z \right) + R^2 \left( \partial_i \phi \right) \left( \partial_j \phi \right) .
</math>
</div>

The orthogonality condition is that the metric be diagonal. That is, each off-diagonal component of the metric must equal zero.

<div align="center">
<math>
\left( \partial_i R \right) \left( \partial_j R \right) + \left( \partial_i z \right) \left( \partial_j z \right) + R^2 \left( \partial_i \phi \right) \left( \partial_j \phi \right) = 0_{ij}, \ \ \ \ \ i \neq j .
</math>
</div>

If the coordinates should be orthogonal, then scale factors can also be defined in terms of the diagonal components of the metric such that <math>g_{ii} = {h_i}^2</math>. This results in the following definition of the scale factors if the orthogonality condition is met.

<div align="center">
<math>
h_i = \left[ \left( \partial_i R \right)^2 + \left( \partial_i z \right)^2 + R^2 \left( \partial_i \phi \right)^2 \right]^{1/2} .
</math>
</div>

==Christoffel Symbols==
For an orthogonal coordinate system, the Christoffel symbols can be written (depending on which, if any, indices are repeated) as

<table align="center" border="1" cellpadding="5">
<tr>
<th align="center" colspan="3">
<font color="darkblue">
Christoffel Symbols<br />
</font>
(for orthogonal coordinate systems)
</th>
</tr>
<tr>
<td align="center">
<math>
\Gamma^k_{ij} = 0
</math>
</td>
<td align="center">
for
</td>
<td align="center">
<math>
i \ne j \ne k
</math>
</td>
</tr>
<tr>
<td align="center">
<math>
\Gamma^k_{ii} = - \frac{h_i}{h_k} \frac{\partial_k h_i}{h_k}
</math>
</td>
<td align="center">
for
</td>
<td align="center">
<math>
i \ne k
</math>
</td>
</tr>
<tr>
<td align="center">
<math>
\Gamma^i_{ij} = \Gamma^i_{ji} = \frac{\partial_j h_i}{h_i}
</math>
</td>
<td align="center" colspan="2">
otherwise<br/>
<font color="blue" size="-2">
(Mistake in original expression corrected <br/>06/10/2010, [[User_talk:Jaycall#Killing_Vector_Approach|as discussed]].)
</font>
</td>
</tr>
<tr>
<td align="center">
<math>
\Gamma^i_{ii} = \frac{\partial_i h_i}{h_i}
</math>
</td>
<td align="center" colspan="2">
otherwise
</td>
</tr>
</table>

==Identification of Conserved Quantity==

If, through some miracle, the right-hand side (RHS) of the equation of motion should equal zero, then the combination of variables inside the parentheses must by definition be a conserved quantity. But good luck guessing a coordinate system in which this miracle will occur!

===Introduce Characteristic Vector===
The trick is to introduce an arbitrary vector field, which will be used to build a location-dependent weighted linear combination of the equations of motion. Since the equations of motion associated with each of the coordinates collectively form a single vector equation, we will just form an inner product of our characteristic vector with each side of the vector equation of motion. The result is

<span id="CV.01"><table align="right" border="1" cellpadding="10" width="10%">
<tr><th><font color="darkblue">CV.01</font></th></tr>
</table></span>
<div align="center">
<math>
C_i \frac{d}{dt} \left( m \ {h_i}^2 \dot{\lambda}_i \right) = m \ {h_k}^2 \Gamma^k_{ij} \dot{\lambda}_j \dot{\lambda}_k C_i - m \ C_i \ \partial_i \Phi
</math>
</div>

Next, we bring <math>C_i</math> inside the total time-derivative on the left-hand side (LHS). This produces an additional term, which we promptly move over to the RHS and include as part of what is commonly referred to as the ''source'' (since it's the source of any change in the quantity in parentheses).

<div align="center">
<math>
\frac{d}{dt} \left( m \ {h_i}^2 \dot{\lambda}_i C_i \right) = m \ {h_i}^2 \dot{\lambda}_i \dot{C}_i + m \ {h_k}^2 \Gamma^k_{ij} \dot{\lambda}_j \dot{\lambda}_k C_i - m \ C_i \ \partial_i \Phi
</math>
</div>

By doing this, we have formed a new ''conservative'' quantity; that is, a new quantity in the parentheses which will be conserved if and only if the source is zero. The utility is in the fact that we should be able to force the source associated with this new conservative quantity to go to zero by choosing the right characteristic vector. Once we find the right characteristic vector, we'll be able to use it directly to build the conserved quantity <math>m \ {h_i}^2 \dot{\lambda}_i C_i</math>.

===Strategic Choice of Characteristic Vector===
So we're looking for a vector <math>\vec{C}</math> such that

<div align="center">
<math>
{h_i}^2 \dot{\lambda}_i \dot{C}_i + {h_k}^2 \Gamma^k_{ij} \dot{\lambda}_j \dot{\lambda}_k C_i - C_i \ \partial_i \Phi = 0 .
</math>
</div>

The good news is that this is a single scalar equation constraining the three independent components of <math>\vec{C}</math>, so several families of solutions should exist. Consequently, we can afford to be choosey!

In the problem we're exploring using [[User:Tohline/Appendix/Ramblings/T3Integrals|T3 coordinates]], we already know a conserved quantity associated with the azimuthal coordinate — angular momentum. We're looking for an additional (independent) conserved quantity associated with the two radial coordinates. So it makes sense to look for one of the solutions that has <math>C_3 = 0</math>. Furthermore, we'd like to avoid dealing with the unknown potential function (which varies only with <math>\lambda_1</math>) as much as possible, so as long as we're being choosey, let's look for a solution that has <math>C_1 = 0</math>. We now have three conditions on three components. The condition constraining <math>C_2</math> is

<div align="center">
<math>
{h_2}^2 \dot{\lambda}_2 \dot{C}_2 + {h_k}^2 \Gamma^k_{2j} \dot{\lambda}_j \dot{\lambda}_k C_2 = 0 .
</math>
</div>

This can be rewritten more simply as

<div align="center">
<math>
\dot{C}_2 = - \left( \frac{{h_k}^2}{{h_2}^2} \Gamma^k_{2j} \frac{\dot{\lambda}_j \dot{\lambda}_k}{\dot{\lambda}_2} \right) C_2 .
</math>
</div>

Perhaps the best approach to solving this condition is separation of variables.

<table align="center" border="0" cellpadding="5">
<tr>
<td align="right">
 
</td>
<td align="right">
<math>
\int \frac{dC_2}{C_2}
</math>
</td>
<td align="center">
<math>
=
</math>
</td>
<td align="left">
<math>
- \int \left( \frac{{h_k}^2}{{h_2}^2} \Gamma^k_{2j} \frac{\dot{\lambda}_j \dot{\lambda}_k}{\dot{\lambda}_2} \right) dt
</math>
</td>
</tr>

<tr>
<td align="right">
<math>
\Rightarrow
</math>
</td>
<td align="right">
<math>
\ln C_2
</math>
</td>
<td align="center">
<math>
=
</math>
</td>
<td align="left">
<math>
- \int \left( \frac{{h_k}^2}{{h_2}^2} \Gamma^k_{2j} \frac{\dot{\lambda}_j \dot{\lambda}_k}{\dot{\lambda}_2} \right) dt
</math>
</td>
</tr>

<tr>
<td align="right">
<math>
\Rightarrow
</math>
</td>
<td align="right">
<math>
C_2
</math>
</td>
<td align="center">
<math>
=
</math>
</td>
<td align="left">
<math>
\exp \left\{ - \int \left( \frac{{h_k}^2}{{h_2}^2} \Gamma^k_{2j} \frac{\dot{\lambda}_j \dot{\lambda}_k}{\dot{\lambda}_2} \right) dt \right\} .
</math>
</td>
</tr>

</table>

The final necessary step will be to write the quantity in parentheses as the exact derivative of some other quantity, call it <math>\Xi</math>. The long-sought conserved quantity will then be <math> m {h_2}^2 \dot{\lambda}_2 \exp \left( - \Xi \right) </math> , where

<div align="center">
<math>
\Xi \equiv \int \left( \frac{{h_k}^2}{{h_2}^2} \Gamma^k_{2j} \frac{\dot{\lambda}_j \dot{\lambda}_k}{\dot{\lambda}_2} \right) dt .
</math>
</div>

====Question from Joel====
Is the following logic correct?

Suppose we examine the term in which <math>j=2</math> and <math>k=1</math>. The integral becomes,

<div align="center">
<math>
\Xi\biggr|_{j=2,k=1} = \int \left( \frac{{h_1}^2}{{h_2}^2} \Gamma^1_{22} \frac{\dot{\lambda}_2 \dot{\lambda}_1}{\dot{\lambda}_2} \right) dt = \int \left( \frac{{h_1}^2}{{h_2}^2} \Gamma^1_{22} \right) d\lambda_1
</math>
</div>

Given that the scale factors and the Christoffel symbols are expressible entirely in terms of <math>\lambda_1</math> and <math>\lambda_2</math>, one could imagine a situation — for example in the quadratic case of <math>q^2=2</math> — in which this integral could be completed analytically. (I presume that wherever <math>\lambda_2</math> appears inside this integral, it can be treated as a constant because the two coordinates are independent of one another.)

====Response from Jay====

Yes, I believe you've worked this term out correctly, but as you can see in what follows, it cancels out with the other cross term. Writing out each of the terms in <math>\Xi</math> gives,

<div align="center">
<math>
\frac{d\Xi}{dt} = \frac{{h_1}^2}{{h_2}^2} \Gamma^1_{21} \frac{\dot{\lambda}_1 \dot{\lambda}_1}{\dot{\lambda}_2} + \frac{{h_1}^2}{{h_2}^2} \Gamma^1_{22} \frac{\dot{\lambda}_2 \dot{\lambda}_1}{\dot{\lambda}_2} + \frac{{h_1}^2}{{h_2}^2} \Gamma^1_{23} \frac{\dot{\lambda}_3 \dot{\lambda}_1}{\dot{\lambda}_2} + \frac{{h_2}^2}{{h_2}^2} \Gamma^2_{21} \frac{\dot{\lambda}_1 \dot{\lambda}_2}{\dot{\lambda}_2} + \frac{{h_2}^2}{{h_2}^2} \Gamma^2_{22} \frac{\dot{\lambda}_2 \dot{\lambda}_2}{\dot{\lambda}_2} + \frac{{h_2}^2}{{h_2}^2} \Gamma^2_{23} \frac{\dot{\lambda}_3 \dot{\lambda}_2}{\dot{\lambda}_2} + \frac{{h_3}^2}{{h_2}^2} \Gamma^3_{21} \frac{\dot{\lambda}_1 \dot{\lambda}_3}{\dot{\lambda}_2} + \frac{{h_3}^2}{{h_2}^2} \Gamma^3_{22} \frac{\dot{\lambda}_2 \dot{\lambda}_3}{\dot{\lambda}_2} + \frac{{h_3}^2}{{h_2}^2} \Gamma^3_{23} \frac{\dot{\lambda}_3 \dot{\lambda}_3}{\dot{\lambda}_2} .
</math>
</div>

Plugging in values for the Christoffel symbols leads to the expression

<div align="center">
<math>
\frac{d\Xi}{dt} = \frac{{h_1}^2}{{h_2}^2} \frac{\partial_2 h_1}{h_1} \frac{\dot{\lambda}_1 \dot{\lambda}_1}{\dot{\lambda}_2} - \frac{{h_1}^2}{{h_2}^2} \frac{h_2}{h_1} \frac{\partial_1 h_2}{h_1} \frac{\dot{\lambda}_2 \dot{\lambda}_1}{\dot{\lambda}_2} + \frac{{h_2}^2}{{h_2}^2} \frac{\partial_1 h_2}{h_2} \frac{\dot{\lambda}_1 \dot{\lambda}_2}{\dot{\lambda}_2} + \frac{{h_2}^2}{{h_2}^2} \frac{\partial_2 h_2}{h_2} \frac{\dot{\lambda}_2 \dot{\lambda}_2}{\dot{\lambda}_2} + \frac{{h_3}^2}{{h_2}^2} \frac{\partial_2 h_3}{h_3} \frac{\dot{\lambda}_3 \dot{\lambda}_3}{\dot{\lambda}_2} .
</math>
</div>

And limiting our interest to motion within the meridional plane (setting <math>\dot{\lambda}_3 = 0</math>) and simplifying

<span id="CV.02"><table align="right" border="1" cellpadding="10" width="10%">
<tr><th><font color="darkblue">CV.02</font></th></tr>
</table></span>
<table align="center" border="1" cellpadding="5">
<tr>
<td align="right">
<math>
\frac{d\Xi}{dt} \equiv \frac{d\ln C_2}{dt}
</math>
</td>
<td align="center">
<math>
=
</math>
</td>
<td align="left">
<math>
\frac{h_1}{h_2} \frac{\partial_2 h_1}{h_2} \frac{{\dot{\lambda}_1}^2}{\dot{\lambda}_2} - \cancel{\frac{\partial_1 h_2}{h_2} \dot{\lambda}_1} + \cancel{\frac{\partial_1 h_2}{h_2} \dot{\lambda}_1} + \frac{\partial_2 h_2}{h_2} \dot{\lambda}_2
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>
=
</math>
</td>
<td align="left">
<math>
\biggl[ \biggl(\frac{h_1 \dot{\lambda}_1}{h_2 \dot{\lambda}_2}\biggr)^2 \frac{\partial \ln h_1}{\partial\ln\lambda_2} + \frac{\partial \ln h_2}{\partial \ln\lambda_2}\biggr] \frac{d \ln{\lambda}_2}{dt}
</math>
</td>
</tr>
</table>

=Application to T3 Coordinates=

On a [[User:Tohline/Appendix/Ramblings/T3CharacteristicVector|separate page]], we have applied this "characteristic vector" approach specifically to T3 coordinates.

 <br />
{{LSU_HBook_footer}}

User:Jaycall/KillingVectorApproach

2010-07-12T15:23:39Z

Jaycall:

{{LSU_HBook_header}}

=Killing Vector Approach to Integrals of Motion Problem=

Motivated by our recent work in relativistic fluid dynamics, one insightful approach to the integrals of motion problem involves rewriting the equations of motion in terms of an arbitrary vector field, which we're currently calling a ''characteristic vector''. The benefit here is that we can produce a mathematical condition that tells us what combintations of variables should be grouped together in order to produce conserved quantities. The details follow.

==Newtonian Equations of Motion==

Using index summation notation (wherein repeated indices are summed over), it is possible to write the Newtonian equations of motion for a test particle in a fixed, static potential in an arbitrary orthogonal coordinate system. The equation describing motion in the <math>i</math>-dimension is:

<div align="center">
<math>
\frac{d}{dt} \left( m \ {h_i}^2 \dot{\lambda}_i \right) = m \ {h_k}^2 \Gamma^k_{ij} \dot{\lambda}_j \dot{\lambda}_k - m \ \partial_i \Phi
</math>
</div>

where <math>m</math> is the mass of the test particle, the <math>\lambda</math>s are the chosen coordinates, and the <math>\dot{\lambda}</math>s are the coordinate velocities (or the total time-derivatives of the coordinates), the <math>h</math>s are the metric scale factors associated with each of the coordinates, <math>\Phi</math> is the gravitational potential, and the <math>\Gamma</math>s are the Christoffel symbols.

==Metric Components, Orthogonality Condition, and Scale Factors==

For any coordinate system it is possible to express the components of the metric in terms of the partial derivatives defining the transformation from any other coordinate system. Written in terms of the transformation from cylindrical coordinates, each metric component is

<div align="center">
<math>
g_{ij} = \left( \partial_i R \right) \left( \partial_j R \right) + \left( \partial_i z \right) \left( \partial_j z \right) + R^2 \left( \partial_i \phi \right) \left( \partial_j \phi \right) .
</math>
</div>

The orthogonality condition is that the metric be diagonal. That is, each off-diagonal component of the metric must equal zero.

<div align="center">
<math>
\left( \partial_i R \right) \left( \partial_j R \right) + \left( \partial_i z \right) \left( \partial_j z \right) + R^2 \left( \partial_i \phi \right) \left( \partial_j \phi \right) = 0_{ij}, \ \ \ \ \ i \neq j .
</math>
</div>

If the coordinates should be orthogonal, then scale factors can also be defined in terms of the diagonal components of the metric such that <math>g_{ii} = {h_i}^2</math>. This results in the following definition of the scale factors if the orthogonality condition is met.

<div align="center">
<math>
h_i = \left[ \left( \partial_i R \right)^2 + \left( \partial_i z \right)^2 + R^2 \left( \partial_i \phi \right)^2 \right]^{1/2} .
</math>
</div>

==Christoffel Symbols==
For an orthogonal coordinate system, the Christoffel symbols can be written (depending on which, if any, indices are repeated) as

<table align="center" border="1" cellpadding="5">
<tr>
<th align="center" colspan="3">
<font color="darkblue">
Christoffel Symbols<br />
</font>
(for orthogonal coordinate systems)
</th>
</tr>
<tr>
<td align="center">
<math>
\Gamma^k_{ij} = 0
</math>
</td>
<td align="center">
for
</td>
<td align="center">
<math>
i \ne j \ne k
</math>
</td>
</tr>
<tr>
<td align="center">
<math>
\Gamma^k_{ii} = - \frac{h_i}{h_k} \frac{\partial_k h_i}{h_k}
</math>
</td>
<td align="center">
for
</td>
<td align="center">
<math>
i \ne k
</math>
</td>
</tr>
<tr>
<td align="center">
<math>
\Gamma^i_{ij} = \Gamma^i_{ji} = \frac{\partial_j h_i}{h_i}
</math>
</td>
<td align="center" colspan="2">
otherwise<br/>
<font color="blue" size="-2">
(Mistake in original expression corrected <br/>06/10/2010, [[User_talk:Jaycall#Killing_Vector_Approach|as discussed]].)
</font>
</td>
</tr>
<tr>
<td align="center">
<math>
\Gamma^i_{ii} = \frac{\partial_i h_i}{h_i}
</math>
</td>
<td align="center" colspan="2">
otherwise
</td>
</tr>
</table>

==Identification of Conserved Quantity==

If, through some miracle, the right-hand side (RHS) of the equation of motion should equal zero, then the combination of variables inside the parentheses must by definition be a conserved quantity. But good luck guessing a coordinate system in which this miracle will occur!

===Introduce Characteristic Vector===
The trick is to introduce an arbitrary vector field, which will be used to build a location-dependent weighted linear combination of the equations of motion. Since the equations of motion associated with each of the coordinates collectively form a single vector equation, we will just form an inner product of our characteristic vector with each side of the vector equation of motion. The result is

<span id="CV.01"><table align="right" border="1" cellpadding="10" width="10%">
<tr><th><font color="darkblue">CV.01</font></th></tr>
</table></span>
<div align="center">
<math>
C_i \frac{d}{dt} \left( m \ {h_i}^2 \dot{\lambda}_i \right) = m \ {h_k}^2 \Gamma^k_{ij} \dot{\lambda}_j \dot{\lambda}_k C_i - m \ C_i \ \partial_i \Phi
</math>
</div>

Next, we bring <math>C_i</math> inside the total time-derivative on the left-hand side (LHS). This produces an additional term, which we promptly move over to the RHS and include as part of what is commonly referred to as the ''source'' (since it's the source of any change in the quantity in parentheses).

<div align="center">
<math>
\frac{d}{dt} \left( m \ {h_i}^2 \dot{\lambda}_i C_i \right) = m \ {h_i}^2 \dot{\lambda}_i \dot{C}_i + m \ {h_k}^2 \Gamma^k_{ij} \dot{\lambda}_j \dot{\lambda}_k C_i - m \ C_i \ \partial_i \Phi
</math>
</div>

By doing this, we have formed a new ''conservative'' quantity; that is, a new quantity in the parentheses which will be conserved if and only if the source is zero. The utility is in the fact that we should be able to force the source associated with this new conservative quantity to go to zero by choosing the right characteristic vector. Once we find the right characteristic vector, we'll be able to use it directly to build the conserved quantity <math>m \ {h_i}^2 \dot{\lambda}_i C_i</math>.

===Strategic Choice of Characteristic Vector===
So we're looking for a vector <math>\vec{C}</math> such that

<div align="center">
<math>
{h_i}^2 \dot{\lambda}_i \dot{C}_i + {h_k}^2 \Gamma^k_{ij} \dot{\lambda}_j \dot{\lambda}_k C_i - C_i \ \partial_i \Phi = 0 .
</math>
</div>

The good news is that this is a single scalar equation constraining the three independent components of <math>\vec{C}</math>, so several families of solutions should exist. Consequently, we can afford to be choosey!

In the problem we're exploring using [[User:Tohline/Appendix/Ramblings/T3Integrals|T3 coordinates]], we already know a conserved quantity associated with the azimuthal coordinate — angular momentum. We're looking for an additional (independent) conserved quantity associated with the two radial coordinates. So it makes sense to look for one of the solutions that has <math>C_3 = 0</math>. Furthermore, we'd like to avoid dealing with the unknown potential function (which varies only with <math>\lambda_1</math>) as much as possible, so as long as we're being choosey, let's look for a solution that has <math>C_1 = 0</math>. We now have three conditions on three components. The condition constraining <math>C_2</math> is

<div align="center">
<math>
{h_2}^2 \dot{\lambda}_2 \dot{C}_2 + {h_k}^2 \Gamma^k_{2j} \dot{\lambda}_j \dot{\lambda}_k C_2 = 0 .
</math>
</div>

This can be rewritten more simply as

<div align="center">
<math>
\dot{C}_2 = - \left( \frac{{h_k}^2}{{h_2}^2} \Gamma^k_{2j} \frac{\dot{\lambda}_j \dot{\lambda}_k}{\dot{\lambda}_2} \right) C_2 .
</math>
</div>

Perhaps the best approach to solving this condition is separation of variables.

<table align="center" border="0" cellpadding="5">
<tr>
<td align="right">
 
</td>
<td align="right">
<math>
\int \frac{dC_2}{C_2}
</math>
</td>
<td align="center">
<math>
=
</math>
</td>
<td align="left">
<math>
- \int \left( \frac{{h_k}^2}{{h_2}^2} \Gamma^k_{2j} \frac{\dot{\lambda}_j \dot{\lambda}_k}{\dot{\lambda}_2} \right) dt
</math>
</td>
</tr>

<tr>
<td align="right">
<math>
\Rightarrow
</math>
</td>
<td align="right">
<math>
\ln C_2
</math>
</td>
<td align="center">
<math>
=
</math>
</td>
<td align="left">
<math>
- \int \left( \frac{{h_k}^2}{{h_2}^2} \Gamma^k_{2j} \frac{\dot{\lambda}_j \dot{\lambda}_k}{\dot{\lambda}_2} \right) dt
</math>
</td>
</tr>

<tr>
<td align="right">
<math>
\Rightarrow
</math>
</td>
<td align="right">
<math>
C_2
</math>
</td>
<td align="center">
<math>
=
</math>
</td>
<td align="left">
<math>
\exp \left\{ - \int \left( \frac{{h_k}^2}{{h_2}^2} \Gamma^k_{2j} \frac{\dot{\lambda}_j \dot{\lambda}_k}{\dot{\lambda}_2} \right) dt \right\} .
</math>
</td>
</tr>

</table>

The final necessary step will be to write the quantity in parentheses as the exact derivative of some other quantity, call it <math>\Xi</math>. The long-sought conserved quantity will then be <math> m {h_2}^2 \dot{\lambda}_2 \exp \left( - \Xi \right) </math> , where

<div align="center">
<math>
\Xi \equiv \int \left( \frac{{h_k}^2}{{h_2}^2} \Gamma^k_{2j} \frac{\dot{\lambda}_j \dot{\lambda}_k}{\dot{\lambda}_2} \right) dt .
</math>
</div>

====Question from Joel====
Is the following logic correct?

Suppose we examine the term in which <math>j=2</math> and <math>k=1</math>. The integral becomes,

<div align="center">
<math>
\Xi\biggr|_{j=2,k=1} = \int \left( \frac{{h_1}^2}{{h_2}^2} \Gamma^1_{22} \frac{\dot{\lambda}_2 \dot{\lambda}_1}{\dot{\lambda}_2} \right) dt = \int \left( \frac{{h_1}^2}{{h_2}^2} \Gamma^1_{22} \right) d\lambda_1
</math>
</div>

Given that the scale factors and the Christoffel symbols are expressible entirely in terms of <math>\lambda_1</math> and <math>\lambda_2</math>, one could imagine a situation — for example in the quadratic case of <math>q^2=2</math> — in which this integral could be completed analytically. (I presume that wherever <math>\lambda_2</math> appears inside this integral, it can be treated as a constant because the two coordinates are independent of one another.)

====Response from Jay====

Yes, I believe you've worked this term out correctly, but as you can see in what follows, it cancels out with the other cross term. Writing out each of the terms in <math>\Xi</math> gives,

<div align="center">
<math>
\frac{d\Xi}{dt} = \frac{{h_1}^2}{{h_2}^2} \Gamma^1_{21} \frac{\dot{\lambda}_1 \dot{\lambda}_1}{\dot{\lambda}_2} + \frac{{h_1}^2}{{h_2}^2} \Gamma^1_{22} \frac{\dot{\lambda}_2 \dot{\lambda}_1}{\dot{\lambda}_2} + \frac{{h_1}^2}{{h_2}^2} \Gamma^1_{23} \frac{\dot{\lambda}_3 \dot{\lambda}_1}{\dot{\lambda}_2} + \frac{{h_2}^2}{{h_2}^2} \Gamma^2_{21} \frac{\dot{\lambda}_1 \dot{\lambda}_2}{\dot{\lambda}_2} + \frac{{h_2}^2}{{h_2}^2} \Gamma^2_{22} \frac{\dot{\lambda}_2 \dot{\lambda}_2}{\dot{\lambda}_2} + \frac{{h_2}^2}{{h_2}^2} \Gamma^2_{23} \frac{\dot{\lambda}_3 \dot{\lambda}_2}{\dot{\lambda}_2} + \frac{{h_3}^2}{{h_2}^2} \Gamma^3_{21} \frac{\dot{\lambda}_1 \dot{\lambda}_3}{\dot{\lambda}_2} + \frac{{h_3}^2}{{h_2}^2} \Gamma^3_{22} \frac{\dot{\lambda}_2 \dot{\lambda}_3}{\dot{\lambda}_2} + \frac{{h_3}^2}{{h_2}^2} \Gamma^3_{23} \frac{\dot{\lambda}_3 \dot{\lambda}_3}{\dot{\lambda}_2} .
</math>
</div>

Plugging in values for the Christoffel symbols leads to the expression

<div align="center">
<math>
\frac{d\Xi}{dt} = \frac{{h_1}^2}{{h_2}^2} \frac{\partial_2 h_1}{h_1} \frac{\dot{\lambda}_1 \dot{\lambda}_1}{\dot{\lambda}_2} - \frac{{h_1}^2}{{h_2}^2} \frac{h_2}{h_1} \frac{\partial_1 h_2}{h_1} \frac{\dot{\lambda}_2 \dot{\lambda}_1}{\dot{\lambda}_2} + \frac{{h_2}^2}{{h_2}^2} \frac{\partial_1 h_2}{h_2} \frac{\dot{\lambda}_1 \dot{\lambda}_2}{\dot{\lambda}_2} + \frac{{h_2}^2}{{h_2}^2} \frac{\partial_2 h_2}{h_2} \frac{\dot{\lambda}_2 \dot{\lambda}_2}{\dot{\lambda}_2} + \frac{{h_3}^2}{{h_2}^2} \frac{\partial_2 h_3}{h_3} \frac{\dot{\lambda}_3 \dot{\lambda}_3}{\dot{\lambda}_2} .
</math>
</div>

And limiting our interest to motion within the meridional plane (setting <math>\dot{\lambda}_3 = 0</math>) and simplifying

<span id="CV.02"><table align="right" border="1" cellpadding="10" width="10%">
<tr><th><font color="darkblue">CV.02</font></th></tr>
</table></span>
<table align="center" border="1" cellpadding="5">
<tr>
<td align="right">
<math>
\frac{d\Xi}{dt} \equiv \frac{d\ln C_2}{dt}
</math>
</td>
<td align="center">
<math>
=
</math>
</td>
<td align="left">
<math>
\frac{h_1}{h_2} \frac{\partial_2 h_1}{h_2} \frac{{\dot{\lambda}_1}^2}{\dot{\lambda}_2} - \cancel{\frac{\partial_1 h_2}{h_2} \dot{\lambda}_1} + \cancel{\frac{\partial_1 h_2}{h_2} \dot{\lambda}_1} + \frac{\partial_2 h_2}{h_2} \dot{\lambda}_2
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>
=
</math>
</td>
<td align="left">
<math>
\biggl[ \biggl(\frac{h_1 \dot{\lambda}_1}{h_2 \dot{\lambda}_2}\biggr)^2 \frac{\partial \ln h_1}{\partial\ln\lambda_2} + \frac{\partial \ln h_2}{\partial \ln\lambda_2}\biggr] \frac{d \ln{\lambda}_2}{dt}
</math>
</td>
</tr>
</table>

=Application to T3 Coordinates=

On a [[User:Tohline/Appendix/Ramblings/T3CharacteristicVector|separate page]], we have applied this "characteristic vector" approach specifically to T3 coordinates.

 <br />
{{LSU_HBook_footer}}

User talk:Jaycall

2010-07-10T20:58:07Z

Jaycall: /* Special Quadratic Case */

==Joel's First Talk Message to Jay==
Jay: In the email message that you sent to me today, you indicated that you had added the 3rd component to the dx/dt equation. I don't see that modification in the current T3Coordinates page. --[[User:Tohline|Tohline]] 15:05, 29 May 2010 (MDT)

:Joel: It's at the bottom of the section entitled "Time-Derivative of Position and Velocity Vectors". Under the history tab, can you see the edits that I made? It was the very first edit and was fairly minor. I only added a couple terms. --[[User:Jaycall|Jaycall]] 20:17, 29 May 2010 (MDT)

::The mediaWiki "talk" page recommends that when you reply to a query, you should not start a new subsection but, rather, just indent (using one or more colons) immediately following the query. Also, it recommends that you "sign" each "talk" message. You do this by typing 2 dashes followed by 4 tildes! I'm going to edit your "reply" (and this additional one from me) to put it in this recommended format. But you have to "sign" your own remark. --[[User:Tohline|Tohline]] 16:23, 29 May 2010 (MDT)

::By the way, I ''can'' see the edits that you made to add the 3rd component to the dx/dt equation. I did not spot the changes earlier, but via the history "diff" function, I'm able to see the edits clearly. --[[User:Tohline|Tohline]] 16:41, 29 May 2010 (MDT)

==Killing Vector Approach==
Thanks for writing out all the terms in the expression for <math>d\Xi/dt</math> in your "[[User:Jaycall/KillingVectorApproach|Killing Vector Approach]]" discussion. On the one hand, I'm happy that the term I mentioned cancels with another one because that means we have fewer terms to integrate. On the other hand, it was one term that I thought we might actually be able to manipulate analytically. Of the two terms that remain, one is relatively simple — the one that is proportional to <math>\dot{\lambda}_2</math> — but the first term is a bear! That will take some thinking! --[[User:Tohline|Tohline]] 10:50, 30 May 2010 (MDT)

:I know it seems like the wrong direction, but unless we can express this summary integral entirely in terms of <math>\lambda_1</math>, <math>\lambda_2</math>, <math>\dot{\lambda}_1</math> and <math>\dot{\lambda}_2</math>, I think that in order to make progress we're going to have to write things out in terms of <math>\varpi</math>, <math>z</math>, <math>\dot{\varpi}</math> and <math>\dot{z}</math>. Of course, that will make the expression much messier, but I think it will be very difficult to recognize this quantity as an exact derivative unless everything's in terms of the same coordinates. --[[User:Jaycall|Jaycall]] 12:55, 30 May 2010 (MDT)

::I agree. It is for similar reasons that I am planning on focusing on the case where <math>q^2=2</math>. At least in this special case we can invert the coordinate expressions to give <math>\varpi</math> and <math>z</math> in terms of the <math>\lambda</math>s. Incidentally, I have added a link to your "Killing Vector" page as well as a link to this associated "Talk" page on the [[User:Tohline/Appendix/Ramblings|page where I itemize my various "ramblings"]]. --[[User:Tohline|Tohline]] 16:16, 30 May 2010 (MDT)

:::Good point. That would be a good reason to focus on one specific case. I haven't looked closely at the inverted coordinate transformation for the <math>q^2=2</math> case for several days. Since the relation is quadratic, is it obvious which root should be used?

::The proper physical root was obvious to me when I performed the coordinate inversion in the case of [[User:Tohline/Appendix/Ramblings/T1Coordinates#Coordinate_Inversion|T1 Coordinates]], so I presume it will be obvious for the T3 Coordinate system. But the inversion needs to be redone for the case of T3 Coordinates; do you want to take care of this and type up the result? In the (quadratic) case of <math>q^2=2</math>, it may be as simple as replacing <math>\chi_2</math> with <math>1/\lambda_2</math>. --[[User:Tohline|Tohline]] 11:21, 31 May 2010 (MDT)

:::Sure. --[[User:Jaycall|Jaycall]] 13:34, 31 May 2010 (MDT)

'''<font color="red">Mistake?</font>''' Please check the next to last row in the [[User:Jaycall/KillingVectorApproach#Christoffel_Symbols|tabulated expressions for Christoffel symbols]]; shouldn't the <math>i</math> and <math>j</math> indexes be swapped? --[[User:Tohline|Tohline]] 09:39, 6 June 2010 (MDT)

:You are correct. Good eye.--[[User:Jaycall|Jaycall]] 09:42, 9 June 2010 (MDT)

::Actually, it wasn't my good eye. I discovered this from my inelegant brute-force derivations. In particular, when I was trying to derive the same form of the [[User:Tohline/Appendix/Ramblings/T3CharacteristicVector#Two_Views_of_Equation_of_Motion|equation of motion from two different points of view]], I had to plug in some of your Christoffel symbol expressions. At first I could not get my EOM to match yours; it was while trying to understand this mismatch that I realized that your ''general'' expression for <math>\Gamma^i_{ij}</math> did not match the ''specific'' expression for <math>\Gamma^2_{21}</math> that you typed on the "T3 Coordinates" page. In my effort to get the EOMs to match, I decided that your ''general'' expression was the incorrect one. --[[User:Tohline|Tohline]] 14:58, 9 June 2010 (MDT)

:::For the record, on 10 June 2010, I modified the [[User:Jaycall/KillingVectorApproach#Christoffel_Symbols|table that contains the general Christoffel symbol expressions]] in order to correct the order of these indexes.--[[User:Tohline|Tohline]] 18:55, 11 June 2010 (MDT)

==Logarithmic Derivatives of T3 Scale Factors==

Check out my new subsection (under T3 Coordinates) entitled [[User:Tohline/Appendix/Ramblings/T3Integrals#Logarithmic_Derivatives_of_Scale_Factors|Logarithmic Derivatives of Scale Factors]]. First, see if you agree that there is a mistake (typo) in one of your tables of partial derivatives. Specifically, I think that the correct expression is:
<div align="center">
<math>
\frac{\partial z}{\partial\lambda_2} = - (q^2-1)\frac{\varpi^2 z \ell^2}{\lambda_2} .
</math>
</div>
Second, see if you agree with my derived expressions for <math>\partial\ln h_i/\partial\ln\lambda_j</math>. --[[User:Tohline|Tohline]] 22:28, 31 May 2010 (MDT)

:You're certainly right about the sign error. I have corrected the expression in the relevant table. --[[User:Jaycall|Jaycall]] 14:28, 1 June 2010 (MDT)

:I have not yet been able to confirm your expressions for logarithmic derivatives of the scale factors. I'm not sure what approach you took in deriving them, but since I had already calculated partials of the scale factors (<math>\partial_i h_j</math> and so forth), I derived a little trick to help simplify the work. Do you agree that

<div align="center">
<math>
\frac{\partial \ln h_i}{\partial \ln \lambda_j} \equiv \frac{\lambda_j}{h_i} \partial_j h_i ?
</math>
</div>

::Yes, this is precisely how I defined the logarithmic derivatives. And I used your tables of partial derivatives (e.g., <math>\partial\varpi/\partial\lambda_j</math> and <math>\partial z/\partial\lambda_j</math>) to obtain <math>\partial_i h_j</math> and so forth. --[[User:Tohline|Tohline]] 17:30, 1 June 2010 (MDT)

'''<font color="red">Mistake!</font>''' Jay: I have found a mistake in my original derivation of the logarithmic derivatives of the <math>h_1</math> scale factor. Perhaps now they will match yours. Check it out and let me know! --[[User:Tohline|Tohline]] 12:41, 4 June 2010 (MDT)

::Joel: I am finally able to confirm your derivation of the logarithmic derivatives of the scale factors. I came at them from a different angle with a fresh head, and I am now confident that what you have is correct. Next I will look into the general case. --[[User:Jaycall|Jaycall]] 14:29, 10 July 2010 (MDT)

==Two Views of Equation of Motion==

Jay: I have created a [[User:Tohline/Appendix/Ramblings/T3CharacteristicVector|new wiki page]] to explicitly analyze how your Characteristic Vector formalism applies to the T3 Coordinate system. On this page is a subsection entitled "[[User:Tohline/Appendix/Ramblings/T3CharacteristicVector#Two_Views_of_Equation_of_Motion|Two Views of Equation of Motion]]" in which I have shown that the <math>2^\mathrm{nd}</math> component of the equation of motion can be written in the following form,
<div align="center">
<math>\frac{d(h_2^2 \dot{\lambda}_2)}{dt} = \biggl( h_1 \frac{\partial h_1}{\partial \lambda_2} \biggr) \dot{\lambda}_1^2 + \biggl( h_2 \frac{\partial h_2}{\partial \lambda_2} \biggr) \dot{\lambda}_2^2</math> ,
</div>
assuming that <math>\partial\Phi/\partial\lambda_2 = 0</math>. More to the point, I have shown that I can derive this form of this component of the equation of motion using either (A) your Christoffel symbol formalism, or (B) the more classical formalism that I have been using, which I obtained from Appendix 1.B of Binney & Tremaine (1987) and Morse & Feshbach (1953).

Now, I shouldn't be shocked that both derivations lead to the same form of the equation — after all, we're doing physics aren't we? But here is what surprised me: The Christoffel symbol formalism produces this form of the equation without ever assuming ''a particular coordinate system'' whereas, in order to derive the equation using the more classical formalism, I plugged in some relations between the partial derivatives of the scale factors that ''I thought'' were specific to T3 coordinates. This leads me to ask, "How many of the T3 Coordinate relations [[User:Tohline/Appendix/Ramblings/T3Integrals|derived earlier]] are generic relations that apply to any orthogonal, axisymmetric coordinate system, and how many are unique to the T3 Coordinate System?" I'm particularly interested in knowing how generalizable the relations are that appear inside boxes labeled "T3 Coordinates" under the subsections I have entitled, "[[User:Tohline/Appendix/Ramblings/T3Integrals#Derived_Identity_for_T3_Coordinates|Derived Identity for T3 Corodinates]]" and "[[User:Tohline/Appendix/Ramblings/T3Integrals#Logarithmic_Derivatives_of_Scale_Factors|Logarithmic Derivatives of Scale Factors]]". Can you answer this? --[[User:Tohline|Tohline]] 10:46, 7 June 2010 (MDT)

:Joel: Yes, this is the power of using a covariant formulation. The only assumptions we made were that the coordinates were orthogonal and axisymmetric. I believe that the relation under "[[User:Tohline/Appendix/Ramblings/T3Integrals#Derived_Identity_for_T3_Coordinates|Derived Identity for T3 Coordinates]]" would apply to any coordinate system meeting these criteria. I will double check, though. (How did you determine that the position vector could be written <math>\vec{x} = \hat{e}_1 (h_1 \lambda_1) + \hat{e}_2 (h_2 \lambda_2)</math>?)

:The logarithmic derivatives of the scale factors cannot be general in their current form due to the appearance of factors of <math>q</math>, but I suspect they can be generalized and molded into a form that closely resembles what you have. I will work on this, too. --[[User:Jaycall|Jaycall]] 10:39, 9 June 2010 (MDT)

::Jay: You asked how I determined the expression for the [[User:Tohline/Appendix/Ramblings/T3Integrals#Definition|position vector]]. I don't have my detailed notes with me right now, but I can outline the steps. In Morse & Feshbach's classical presentation of orthogonal curvilinear coordinates, they define a set of so-called ''direction cosines,'' <math>\gamma_{ij}</math>. (I don't have the generalized definition with me at this time, but they are expressible in terms of derivatives of the scale factors with respect to the various coordinates; I'm certain ''direction cosines'' are definable in terms of Christoffel symbols.) Each of the Cartesian unit vectors (<math>\hat{\imath},\hat{\jmath},\hat{k}</math>) can be written in terms of products of these direction cosines and the three unit vectors of your chosen curvilinear coordinate system (<math>\hat{e}_1, \hat{e}_2, \hat{e}_3</math>). According to Morse & Feshbach, for example, <math>\hat{\imath} = (\hat{e}_1\gamma_{11} + \hat{e}_2\gamma_{21} + \hat{e}_3\gamma_{31})</math>. So, all I did to obtain an expression for the position vector in T3 coordinates was define <math>\vec{x} = (\hat\imath x + \hat\jmath y + \hat{k}z)</math>, then replace each Cartesian unit vector by its expression in terms of products of the T3-coordinate unit vectors & the appropriate direct cosine, then gather together and simplify terms. Does this make sense? --[[User:Tohline|Tohline]] 14:59, 10 June 2010 (MDT)

:::Yes, it does. That's very helpful. --[[User:Jaycall|Jaycall]] 09:49, 11 June 2010 (MDT)

::::Jay: For the record, I have put together a wiki page that discusses the definition and utility of [[User:Tohline/Appendix/Ramblings/DirectionCosines|Direction Cosines]].--[[User:Tohline|Tohline]] 18:47, 3 July 2010 (MDT)

==Special Quadratic Case==
Jay: On a [[User:Tohline/Appendix/Ramblings/T3Integrals/QuadraticCase#Special_Case_.28Quadratic.29|new accompanying wiki page]] I have begun to investigate how our system behaves in the special case of <math>q^2 = 2</math>. I have added a pointer/link to this new page — and a separate pointer/link to the parallel investigation that you have told me you are conducting — on the page containing a table of contents to my [[User:Tohline/Appendix/Ramblings|various research ramblings]]. I have not yet actually gone to the page that details your analysis because, at least for a while, I prefer for our trains of thought to remain independent of one another. However, in an effort to simplify the comparison that we ultimately will need to make between our independent analyses, I strongly recommend that you write your expressions for the scale factors, etc. ''for the special case of <math>q^2=2</math>'' in terms of a variable, <math>\Lambda</math>, defined such that,
<div align="center">
<math>
\Lambda \equiv \biggl[1 + \biggl(\frac{2\lambda_1}{\lambda_2}\biggr)^2\biggr]^{1/2} = \cosh(2\Zeta) .
</math>
</div>
By the way, if you don't agree that <math>\sqrt{1+(2\lambda_1/\lambda_2)^2}</math> equals <math>\cosh(2\Zeta)</math>, please let me know!--[[User:Tohline|Tohline]] 19:49, 11 June 2010 (MDT)

:: Joel: I have checked and agree that in the <math>q^2=2</math> case only, the above relation among <math>\Lambda</math>, <math>Z</math> and the T3 coordinates is correct. --[[User:Jaycall|Jaycall]] 14:58, 10 July 2010 (MDT)

User talk:Jaycall

2010-07-10T20:29:41Z

Jaycall: /* Logarithmic Derivatives of T3 Scale Factors */

==Joel's First Talk Message to Jay==
Jay: In the email message that you sent to me today, you indicated that you had added the 3rd component to the dx/dt equation. I don't see that modification in the current T3Coordinates page. --[[User:Tohline|Tohline]] 15:05, 29 May 2010 (MDT)

:Joel: It's at the bottom of the section entitled "Time-Derivative of Position and Velocity Vectors". Under the history tab, can you see the edits that I made? It was the very first edit and was fairly minor. I only added a couple terms. --[[User:Jaycall|Jaycall]] 20:17, 29 May 2010 (MDT)

::The mediaWiki "talk" page recommends that when you reply to a query, you should not start a new subsection but, rather, just indent (using one or more colons) immediately following the query. Also, it recommends that you "sign" each "talk" message. You do this by typing 2 dashes followed by 4 tildes! I'm going to edit your "reply" (and this additional one from me) to put it in this recommended format. But you have to "sign" your own remark. --[[User:Tohline|Tohline]] 16:23, 29 May 2010 (MDT)

::By the way, I ''can'' see the edits that you made to add the 3rd component to the dx/dt equation. I did not spot the changes earlier, but via the history "diff" function, I'm able to see the edits clearly. --[[User:Tohline|Tohline]] 16:41, 29 May 2010 (MDT)

==Killing Vector Approach==
Thanks for writing out all the terms in the expression for <math>d\Xi/dt</math> in your "[[User:Jaycall/KillingVectorApproach|Killing Vector Approach]]" discussion. On the one hand, I'm happy that the term I mentioned cancels with another one because that means we have fewer terms to integrate. On the other hand, it was one term that I thought we might actually be able to manipulate analytically. Of the two terms that remain, one is relatively simple — the one that is proportional to <math>\dot{\lambda}_2</math> — but the first term is a bear! That will take some thinking! --[[User:Tohline|Tohline]] 10:50, 30 May 2010 (MDT)

:I know it seems like the wrong direction, but unless we can express this summary integral entirely in terms of <math>\lambda_1</math>, <math>\lambda_2</math>, <math>\dot{\lambda}_1</math> and <math>\dot{\lambda}_2</math>, I think that in order to make progress we're going to have to write things out in terms of <math>\varpi</math>, <math>z</math>, <math>\dot{\varpi}</math> and <math>\dot{z}</math>. Of course, that will make the expression much messier, but I think it will be very difficult to recognize this quantity as an exact derivative unless everything's in terms of the same coordinates. --[[User:Jaycall|Jaycall]] 12:55, 30 May 2010 (MDT)

::I agree. It is for similar reasons that I am planning on focusing on the case where <math>q^2=2</math>. At least in this special case we can invert the coordinate expressions to give <math>\varpi</math> and <math>z</math> in terms of the <math>\lambda</math>s. Incidentally, I have added a link to your "Killing Vector" page as well as a link to this associated "Talk" page on the [[User:Tohline/Appendix/Ramblings|page where I itemize my various "ramblings"]]. --[[User:Tohline|Tohline]] 16:16, 30 May 2010 (MDT)

:::Good point. That would be a good reason to focus on one specific case. I haven't looked closely at the inverted coordinate transformation for the <math>q^2=2</math> case for several days. Since the relation is quadratic, is it obvious which root should be used?

::The proper physical root was obvious to me when I performed the coordinate inversion in the case of [[User:Tohline/Appendix/Ramblings/T1Coordinates#Coordinate_Inversion|T1 Coordinates]], so I presume it will be obvious for the T3 Coordinate system. But the inversion needs to be redone for the case of T3 Coordinates; do you want to take care of this and type up the result? In the (quadratic) case of <math>q^2=2</math>, it may be as simple as replacing <math>\chi_2</math> with <math>1/\lambda_2</math>. --[[User:Tohline|Tohline]] 11:21, 31 May 2010 (MDT)

:::Sure. --[[User:Jaycall|Jaycall]] 13:34, 31 May 2010 (MDT)

'''<font color="red">Mistake?</font>''' Please check the next to last row in the [[User:Jaycall/KillingVectorApproach#Christoffel_Symbols|tabulated expressions for Christoffel symbols]]; shouldn't the <math>i</math> and <math>j</math> indexes be swapped? --[[User:Tohline|Tohline]] 09:39, 6 June 2010 (MDT)

:You are correct. Good eye.--[[User:Jaycall|Jaycall]] 09:42, 9 June 2010 (MDT)

::Actually, it wasn't my good eye. I discovered this from my inelegant brute-force derivations. In particular, when I was trying to derive the same form of the [[User:Tohline/Appendix/Ramblings/T3CharacteristicVector#Two_Views_of_Equation_of_Motion|equation of motion from two different points of view]], I had to plug in some of your Christoffel symbol expressions. At first I could not get my EOM to match yours; it was while trying to understand this mismatch that I realized that your ''general'' expression for <math>\Gamma^i_{ij}</math> did not match the ''specific'' expression for <math>\Gamma^2_{21}</math> that you typed on the "T3 Coordinates" page. In my effort to get the EOMs to match, I decided that your ''general'' expression was the incorrect one. --[[User:Tohline|Tohline]] 14:58, 9 June 2010 (MDT)

:::For the record, on 10 June 2010, I modified the [[User:Jaycall/KillingVectorApproach#Christoffel_Symbols|table that contains the general Christoffel symbol expressions]] in order to correct the order of these indexes.--[[User:Tohline|Tohline]] 18:55, 11 June 2010 (MDT)

==Logarithmic Derivatives of T3 Scale Factors==

Check out my new subsection (under T3 Coordinates) entitled [[User:Tohline/Appendix/Ramblings/T3Integrals#Logarithmic_Derivatives_of_Scale_Factors|Logarithmic Derivatives of Scale Factors]]. First, see if you agree that there is a mistake (typo) in one of your tables of partial derivatives. Specifically, I think that the correct expression is:
<div align="center">
<math>
\frac{\partial z}{\partial\lambda_2} = - (q^2-1)\frac{\varpi^2 z \ell^2}{\lambda_2} .
</math>
</div>
Second, see if you agree with my derived expressions for <math>\partial\ln h_i/\partial\ln\lambda_j</math>. --[[User:Tohline|Tohline]] 22:28, 31 May 2010 (MDT)

:You're certainly right about the sign error. I have corrected the expression in the relevant table. --[[User:Jaycall|Jaycall]] 14:28, 1 June 2010 (MDT)

:I have not yet been able to confirm your expressions for logarithmic derivatives of the scale factors. I'm not sure what approach you took in deriving them, but since I had already calculated partials of the scale factors (<math>\partial_i h_j</math> and so forth), I derived a little trick to help simplify the work. Do you agree that

<div align="center">
<math>
\frac{\partial \ln h_i}{\partial \ln \lambda_j} \equiv \frac{\lambda_j}{h_i} \partial_j h_i ?
</math>
</div>

::Yes, this is precisely how I defined the logarithmic derivatives. And I used your tables of partial derivatives (e.g., <math>\partial\varpi/\partial\lambda_j</math> and <math>\partial z/\partial\lambda_j</math>) to obtain <math>\partial_i h_j</math> and so forth. --[[User:Tohline|Tohline]] 17:30, 1 June 2010 (MDT)

'''<font color="red">Mistake!</font>''' Jay: I have found a mistake in my original derivation of the logarithmic derivatives of the <math>h_1</math> scale factor. Perhaps now they will match yours. Check it out and let me know! --[[User:Tohline|Tohline]] 12:41, 4 June 2010 (MDT)

::Joel: I am finally able to confirm your derivation of the logarithmic derivatives of the scale factors. I came at them from a different angle with a fresh head, and I am now confident that what you have is correct. Next I will look into the general case. --[[User:Jaycall|Jaycall]] 14:29, 10 July 2010 (MDT)

==Two Views of Equation of Motion==

Jay: I have created a [[User:Tohline/Appendix/Ramblings/T3CharacteristicVector|new wiki page]] to explicitly analyze how your Characteristic Vector formalism applies to the T3 Coordinate system. On this page is a subsection entitled "[[User:Tohline/Appendix/Ramblings/T3CharacteristicVector#Two_Views_of_Equation_of_Motion|Two Views of Equation of Motion]]" in which I have shown that the <math>2^\mathrm{nd}</math> component of the equation of motion can be written in the following form,
<div align="center">
<math>\frac{d(h_2^2 \dot{\lambda}_2)}{dt} = \biggl( h_1 \frac{\partial h_1}{\partial \lambda_2} \biggr) \dot{\lambda}_1^2 + \biggl( h_2 \frac{\partial h_2}{\partial \lambda_2} \biggr) \dot{\lambda}_2^2</math> ,
</div>
assuming that <math>\partial\Phi/\partial\lambda_2 = 0</math>. More to the point, I have shown that I can derive this form of this component of the equation of motion using either (A) your Christoffel symbol formalism, or (B) the more classical formalism that I have been using, which I obtained from Appendix 1.B of Binney & Tremaine (1987) and Morse & Feshbach (1953).

Now, I shouldn't be shocked that both derivations lead to the same form of the equation — after all, we're doing physics aren't we? But here is what surprised me: The Christoffel symbol formalism produces this form of the equation without ever assuming ''a particular coordinate system'' whereas, in order to derive the equation using the more classical formalism, I plugged in some relations between the partial derivatives of the scale factors that ''I thought'' were specific to T3 coordinates. This leads me to ask, "How many of the T3 Coordinate relations [[User:Tohline/Appendix/Ramblings/T3Integrals|derived earlier]] are generic relations that apply to any orthogonal, axisymmetric coordinate system, and how many are unique to the T3 Coordinate System?" I'm particularly interested in knowing how generalizable the relations are that appear inside boxes labeled "T3 Coordinates" under the subsections I have entitled, "[[User:Tohline/Appendix/Ramblings/T3Integrals#Derived_Identity_for_T3_Coordinates|Derived Identity for T3 Corodinates]]" and "[[User:Tohline/Appendix/Ramblings/T3Integrals#Logarithmic_Derivatives_of_Scale_Factors|Logarithmic Derivatives of Scale Factors]]". Can you answer this? --[[User:Tohline|Tohline]] 10:46, 7 June 2010 (MDT)

:Joel: Yes, this is the power of using a covariant formulation. The only assumptions we made were that the coordinates were orthogonal and axisymmetric. I believe that the relation under "[[User:Tohline/Appendix/Ramblings/T3Integrals#Derived_Identity_for_T3_Coordinates|Derived Identity for T3 Coordinates]]" would apply to any coordinate system meeting these criteria. I will double check, though. (How did you determine that the position vector could be written <math>\vec{x} = \hat{e}_1 (h_1 \lambda_1) + \hat{e}_2 (h_2 \lambda_2)</math>?)

:The logarithmic derivatives of the scale factors cannot be general in their current form due to the appearance of factors of <math>q</math>, but I suspect they can be generalized and molded into a form that closely resembles what you have. I will work on this, too. --[[User:Jaycall|Jaycall]] 10:39, 9 June 2010 (MDT)

::Jay: You asked how I determined the expression for the [[User:Tohline/Appendix/Ramblings/T3Integrals#Definition|position vector]]. I don't have my detailed notes with me right now, but I can outline the steps. In Morse & Feshbach's classical presentation of orthogonal curvilinear coordinates, they define a set of so-called ''direction cosines,'' <math>\gamma_{ij}</math>. (I don't have the generalized definition with me at this time, but they are expressible in terms of derivatives of the scale factors with respect to the various coordinates; I'm certain ''direction cosines'' are definable in terms of Christoffel symbols.) Each of the Cartesian unit vectors (<math>\hat{\imath},\hat{\jmath},\hat{k}</math>) can be written in terms of products of these direction cosines and the three unit vectors of your chosen curvilinear coordinate system (<math>\hat{e}_1, \hat{e}_2, \hat{e}_3</math>). According to Morse & Feshbach, for example, <math>\hat{\imath} = (\hat{e}_1\gamma_{11} + \hat{e}_2\gamma_{21} + \hat{e}_3\gamma_{31})</math>. So, all I did to obtain an expression for the position vector in T3 coordinates was define <math>\vec{x} = (\hat\imath x + \hat\jmath y + \hat{k}z)</math>, then replace each Cartesian unit vector by its expression in terms of products of the T3-coordinate unit vectors & the appropriate direct cosine, then gather together and simplify terms. Does this make sense? --[[User:Tohline|Tohline]] 14:59, 10 June 2010 (MDT)

:::Yes, it does. That's very helpful. --[[User:Jaycall|Jaycall]] 09:49, 11 June 2010 (MDT)

::::Jay: For the record, I have put together a wiki page that discusses the definition and utility of [[User:Tohline/Appendix/Ramblings/DirectionCosines|Direction Cosines]].--[[User:Tohline|Tohline]] 18:47, 3 July 2010 (MDT)

==Special Quadratic Case==
Jay: On a [[User:Tohline/Appendix/Ramblings/T3Integrals/QuadraticCase#Special_Case_.28Quadratic.29|new accompanying wiki page]] I have begun to investigate how our system behaves in the special case of <math>q^2 = 2</math>. I have added a pointer/link to this new page — and a separate pointer/link to the parallel investigation that you have told me you are conducting — on the page containing a table of contents to my [[User:Tohline/Appendix/Ramblings|various research ramblings]]. I have not yet actually gone to the page that details your analysis because, at least for a while, I prefer for our trains of thought to remain independent of one another. However, in an effort to simplify the comparison that we ultimately will need to make between our independent analyses, I strongly recommend that you write your expressions for the scale factors, etc. ''for the special case of <math>q^2=2</math>'' in terms of a variable, <math>\Lambda</math>, defined such that,
<div align="center">
<math>
\Lambda \equiv \biggl[1 + \biggl(\frac{2\lambda_1}{\lambda_2}\biggr)^2\biggr]^{1/2} = \cosh(2\Zeta) .
</math>
</div>
By the way, if you don't agree that <math>\sqrt{1+(2\lambda_1/\lambda_2)^2}</math> equals <math>\cosh(2\Zeta)</math>, please let me know!--[[User:Tohline|Tohline]] 19:49, 11 June 2010 (MDT)

User talk:Jaycall

2010-06-11T15:49:53Z

Jaycall: /* Two Views of Equation of Motion */ Response to explanation of method for determining position vector in T3 coordinates

==Joel's First Talk Message to Jay==
Jay: In the email message that you sent to me today, you indicated that you had added the 3rd component to the dx/dt equation. I don't see that modification in the current T3Coordinates page. --[[User:Tohline|Tohline]] 15:05, 29 May 2010 (MDT)

:Joel: It's at the bottom of the section entitled "Time-Derivative of Position and Velocity Vectors". Under the history tab, can you see the edits that I made? It was the very first edit and was fairly minor. I only added a couple terms. --[[User:Jaycall|Jaycall]] 20:17, 29 May 2010 (MDT)

::The mediaWiki "talk" page recommends that when you reply to a query, you should not start a new subsection but, rather, just indent (using one or more colons) immediately following the query. Also, it recommends that you "sign" each "talk" message. You do this by typing 2 dashes followed by 4 tildes! I'm going to edit your "reply" (and this additional one from me) to put it in this recommended format. But you have to "sign" your own remark. --[[User:Tohline|Tohline]] 16:23, 29 May 2010 (MDT)

::By the way, I ''can'' see the edits that you made to add the 3rd component to the dx/dt equation. I did not spot the changes earlier, but via the history "diff" function, I'm able to see the edits clearly. --[[User:Tohline|Tohline]] 16:41, 29 May 2010 (MDT)

==Killing Vector Approach==
Thanks for writing out all the terms in the expression for <math>d\Xi/dt</math> in your "[[User:Jaycall/KillingVectorApproach|Killing Vector Approach]]" discussion. On the one hand, I'm happy that the term I mentioned cancels with another one because that means we have fewer terms to integrate. On the other hand, it was one term that I thought we might actually be able to manipulate analytically. Of the two terms that remain, one is relatively simple — the one that is proportional to <math>\dot{\lambda}_2</math> — but the first term is a bear! That will take some thinking! --[[User:Tohline|Tohline]] 10:50, 30 May 2010 (MDT)

:I know it seems like the wrong direction, but unless we can express this summary integral entirely in terms of <math>\lambda_1</math>, <math>\lambda_2</math>, <math>\dot{\lambda}_1</math> and <math>\dot{\lambda}_2</math>, I think that in order to make progress we're going to have to write things out in terms of <math>\varpi</math>, <math>z</math>, <math>\dot{\varpi}</math> and <math>\dot{z}</math>. Of course, that will make the expression much messier, but I think it will be very difficult to recognize this quantity as an exact derivative unless everything's in terms of the same coordinates. --[[User:Jaycall|Jaycall]] 12:55, 30 May 2010 (MDT)

::I agree. It is for similar reasons that I am planning on focusing on the case where <math>q^2=2</math>. At least in this special case we can invert the coordinate expressions to give <math>\varpi</math> and <math>z</math> in terms of the <math>\lambda</math>s. Incidentally, I have added a link to your "Killing Vector" page as well as a link to this associated "Talk" page on the [[User:Tohline/Appendix/Ramblings|page where I itemize my various "ramblings"]]. --[[User:Tohline|Tohline]] 16:16, 30 May 2010 (MDT)

:::Good point. That would be a good reason to focus on one specific case. I haven't looked closely at the inverted coordinate transformation for the <math>q^2=2</math> case for several days. Since the relation is quadratic, is it obvious which root should be used?

::The proper physical root was obvious to me when I performed the coordinate inversion in the case of [[User:Tohline/Appendix/Ramblings/T1Coordinates#Coordinate_Inversion|T1 Coordinates]], so I presume it will be obvious for the T3 Coordinate system. But the inversion needs to be redone for the case of T3 Coordinates; do you want to take care of this and type up the result? In the (quadratic) case of <math>q^2=2</math>, it may be as simple as replacing <math>\chi_2</math> with <math>1/\lambda_2</math>. --[[User:Tohline|Tohline]] 11:21, 31 May 2010 (MDT)

:::Sure. --[[User:Jaycall|Jaycall]] 13:34, 31 May 2010 (MDT)

'''<font color="red">Mistake?</font>''' Please check the next to last row in the [[User:Jaycall/KillingVectorApproach#Christoffel_Symbols|tabulated expressions for Christoffel symbols]]; shouldn't the <math>i</math> and <math>j</math> indexes be swapped? --[[User:Tohline|Tohline]] 09:39, 6 June 2010 (MDT)

:You are correct. Good eye.--[[User:Jaycall|Jaycall]] 09:42, 9 June 2010 (MDT)

::Actually, it wasn't my good eye. I discovered this from my inelegant brute-force derivations. In particular, when I was trying to derive the same form of the [[User:Tohline/Appendix/Ramblings/T3CharacteristicVector#Two_Views_of_Equation_of_Motion|equation of motion from two different points of view]], I had to plug in some of your Christoffel symbol expressions. At first I could not get my EOM to match yours; it was while trying to understand this mismatch that I realized that your ''general'' expression for <math>\Gamma^i_{ij}</math> did not match the ''specific'' expression for <math>\Gamma^2_{21}</math> that you typed on the "T3 Coordinates" page. In my effort to get the EOMs to match, I decided that your ''general'' expression was the incorrect one. --[[User:Tohline|Tohline]] 14:58, 9 June 2010 (MDT)

==Logarithmic Derivatives of T3 Scale Factors==

Check out my new subsection (under T3 Coordinates) entitled [[User:Tohline/Appendix/Ramblings/T3Integrals#Logarithmic_Derivatives_of_Scale_Factors|Logarithmic Derivatives of Scale Factors]]. First, see if you agree that there is a mistake (typo) in one of your tables of partial derivatives. Specifically, I think that the correct expression is:
<div align="center">
<math>
\frac{\partial z}{\partial\lambda_2} = - (q^2-1)\frac{\varpi^2 z \ell^2}{\lambda_2} .
</math>
</div>
Second, see if you agree with my derived expressions for <math>\partial\ln h_i/\partial\ln\lambda_j</math>. --[[User:Tohline|Tohline]] 22:28, 31 May 2010 (MDT)

:You're certainly right about the sign error. I have corrected the expression in the relevant table. --[[User:Jaycall|Jaycall]] 14:28, 1 June 2010 (MDT)

:I have not yet been able to confirm your expressions for logarithmic derivatives of the scale factors. I'm not sure what approach you took in deriving them, but since I had already calculated partials of the scale factors (<math>\partial_i h_j</math> and so forth), I derived a little trick to help simplify the work. Do you agree that

<div align="center">
<math>
\frac{\partial \ln h_i}{\partial \ln \lambda_j} \equiv \frac{\lambda_j}{h_i} \partial_j h_i ?
</math>
</div>

::Yes, this is precisely how I defined the logarithmic derivatives. And I used your tables of partial derivatives (e.g., <math>\partial\varpi/\partial\lambda_j</math> and <math>\partial z/\partial\lambda_j</math>) to obtain <math>\partial_i h_j</math> and so forth. --[[User:Tohline|Tohline]] 17:30, 1 June 2010 (MDT)

'''<font color="red">Mistake!</font>''' Jay: I have found a mistake in my original derivation of the logarithmic derivatives of the <math>h_1</math> scale factor. Perhaps now they will match yours. Check it out and let me know! --[[User:Tohline|Tohline]] 12:41, 4 June 2010 (MDT)

==Two Views of Equation of Motion==

Jay: I have created a [[User:Tohline/Appendix/Ramblings/T3CharacteristicVector|new wiki page]] to explicitly analyze how your Characteristic Vector formalism applies to the T3 Coordinate system. On this page is a subsection entitled "[[User:Tohline/Appendix/Ramblings/T3CharacteristicVector#Two_Views_of_Equation_of_Motion|Two Views of Equation of Motion]]" in which I have shown that the <math>2^\mathrm{nd}</math> component of the equation of motion can be written in the following form,
<div align="center">
<math>\frac{d(h_2^2 \dot{\lambda}_2)}{dt} = \biggl( h_1 \frac{\partial h_1}{\partial \lambda_2} \biggr) \dot{\lambda}_1^2 + \biggl( h_2 \frac{\partial h_2}{\partial \lambda_2} \biggr) \dot{\lambda}_2^2</math> ,
</div>
assuming that <math>\partial\Phi/\partial\lambda_2 = 0</math>. More to the point, I have shown that I can derive this form of this component of the equation of motion using either (A) your Christoffel symbol formalism, or (B) the more classical formalism that I have been using, which I obtained from Appendix 1.B of Binney & Tremaine (1987) and Morse & Feshbach (1953).

Now, I shouldn't be shocked that both derivations lead to the same form of the equation — after all, we're doing physics aren't we? But here is what surprised me: The Christoffel symbol formalism produces this form of the equation without ever assuming ''a particular coordinate system'' whereas, in order to derive the equation using the more classical formalism, I plugged in some relations between the partial derivatives of the scale factors that ''I thought'' were specific to T3 coordinates. This leads me to ask, "How many of the T3 Coordinate relations [[User:Tohline/Appendix/Ramblings/T3Integrals|derived earlier]] are generic relations that apply to any orthogonal, axisymmetric coordinate system, and how many are unique to the T3 Coordinate System?" I'm particularly interested in knowing how generalizable the relations are that appear inside boxes labeled "T3 Coordinates" under the subsections I have entitled, "[[User:Tohline/Appendix/Ramblings/T3Integrals#Derived_Identity_for_T3_Coordinates|Derived Identity for T3 Corodinates]]" and "[[User:Tohline/Appendix/Ramblings/T3Integrals#Logarithmic_Derivatives_of_Scale_Factors|Logarithmic Derivatives of Scale Factors]]". Can you answer this? --[[User:Tohline|Tohline]] 10:46, 7 June 2010 (MDT)

:Joel: Yes, this is the power of using a covariant formulation. The only assumptions we made were that the coordinates were orthogonal and axisymmetric. I believe that the relation under "[[User:Tohline/Appendix/Ramblings/T3Integrals#Derived_Identity_for_T3_Coordinates|Derived Identity for T3 Coordinates]]" would apply to any coordinate system meeting these criteria. I will double check, though. (How did you determine that the position vector could be written <math>\vec{x} = \hat{e}_1 (h_1 \lambda_1) + \hat{e}_2 (h_2 \lambda_2)</math>?)

:The logarithmic derivatives of the scale factors cannot be general in their current form due to the appearance of factors of <math>q</math>, but I suspect they can be generalized and molded into a form that closely resembles what you have. I will work on this, too. --[[User:Jaycall|Jaycall]] 10:39, 9 June 2010 (MDT)

::Jay: You asked how I determined the expression for the [[User:Tohline/Appendix/Ramblings/T3Integrals#Definition|position vector]]. I don't have my detailed notes with me right now, but I can outline the steps. In Morse & Feshbach's classical presentation of orthogonal curvilinear coordinates, they define a set of so-called ''direction cosines,'' <math>\gamma_{ij}</math>. (I don't have the generalized definition with me at this time, but they are expressible in terms of derivatives of the scale factors with respect to the various coordinates; I'm certain ''direction cosines'' are definable in terms of Christoffel symbols.) Each of the Cartesian unit vectors (<math>\hat{\imath},\hat{\jmath},\hat{k}</math>) can be written in terms of products of these direction cosines and the three unit vectors of your chosen curvilinear coordinate system (<math>\hat{e}_1, \hat{e}_2, \hat{e}_3</math>). According to Morse & Feshbach, for example, <math>\hat{\imath} = (\hat{e}_1\gamma_{11} + \hat{e}_2\gamma_{21} + \hat{e}_3\gamma_{31})</math>. So, all I did to obtain an expression for the position vector in T3 coordinates was define <math>\vec{x} = (\hat\imath x + \hat\jmath y + \hat{k}z)</math>, then replace each Cartesian unit vector by its expression in terms of products of the T3-coordinate unit vectors & the appropriate direct cosine, then gather together and simplify terms. Does this make sense? --[[User:Tohline|Tohline]] 14:59, 10 June 2010 (MDT)

:::Yes, it does. That's very helpful. --[[User:Jaycall|Jaycall]] 09:49, 11 June 2010 (MDT)

User:Jaycall/T3 Coordinates/Special Case

2010-06-09T18:06:16Z

Jaycall: Created page with 'If the special case <math>q^2=2</math> is considered, it is possible to invert the coordinate transformations in closed form. The coordinate transformations and their inversions…'

If the special case <math>q^2=2</math> is considered, it is possible to invert the coordinate transformations in closed form. The coordinate transformations and their inversions become

<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>\left( R^2+2z^2 \right)^{1/2}</math>
</td>
<td align="center">
      and      
</td>
<td align="right">
<math>
\lambda_2
</math>
</td>
<td align="center">
<math>\equiv</math>
</td>
<td align="left">
<math>\frac{R^2}{\sqrt{2}z}</math>
</td>
</tr>
<tr>
<td align="right">
<math>
R^2
</math>
</td>
<td align="center">
<math>\equiv</math>
</td>
<td align="left">
<math>-\frac{{\lambda_2}^2}{2} \pm \sqrt{\frac{{\lambda_2}^2}{4} + {\lambda_1}^2 {\lambda_2}^2}</math>
</td>
<td align="center">
      and      
</td>
<td align="right">
<math>
z
</math>
</td>
<td align="center">
<math>\equiv</math>
</td>
<td align="left">
<math>-\frac{\lambda_2}{2\sqrt{2}} \pm \sqrt{\frac{4 {\lambda_1}^2+{\lambda_2}^2}{8}}</math>
</td>
</tr>
</table>

Partial derivatives of each of the T3 coordinates taken with respect to each of the cylindrical coordinates are:

<table align="center" border="1" cellpadding="5">
<tr>
<td align="center">
 
</td>
<td align="center">
<math>
\frac{\partial}{\partial R}
</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{R}{\lambda_1}
</math>
</td>
<td align="center">
<math>
\frac{2z}{\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{2 \lambda_2}{R}
</math>
</td>
<td align="center">
<math>
-\frac{\lambda_2}{z}
</math>
</td>
<td align="center">
<math>0</math>
</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>

And partials of the cyldrical coordinates taken with respect to the T3 coordinates are:

<table align="center" border="1" cellpadding="5">
<tr>
<td align="center">
 
</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>R</math>
</td>
<td align="center">
<math>
R \ell^2 \lambda_1
</math>
</td>
<td align="center">
<math>
2Rz^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>
2z \ell^2 \lambda_1
</math>
</td>
<td align="center">
<math>
-R^2 z \ell^2 / \lambda_2
</math>
</td>
<td align="center">
<math>0</math>
</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>

where <math>\ell \equiv \left( R^2 + 4z^2 \right)^{-1/2}</math>.

Furthermore, the scale factors become

<table align="center">
<tr>
<td align="right">
<math>h_1</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>\lambda_1 \ell</math>
</td>
</tr>

<tr>
<td align="right">
<math>h_2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>Rz \ell / \lambda_2</math>
</td>
</tr>

<tr>
<td align="right">
<math>h_3</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>R = \lambda_3</math>
</td>
</tr>
</table>

In this special case, there are some additional useful relationships between various combinations of cylindrical variables and their T3 equivalents which can be written out.

<table align="center">
<tr>
<td align="right">
<math>R^2 + 2z^2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>{\lambda_1}^2</math>
</td>
</tr>

<tr>
<td align="right">
<math>R^2 + 4z^2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>2 {\lambda_1}^2 + {\lambda_2}^2/2 \mp \lambda_2 \sqrt{{\lambda_1}^2+{\lambda_2}^2/4} = \ell^{-2}</math>
</td>
</tr>

<tr>
<td align="right">
<math>R^2 + 8z^2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>4 {\lambda_1}^2 + \tfrac{3}{2} {\lambda_2}^2 \mp 3 \lambda_2 \sqrt{{\lambda_1}^2+{\lambda_2}^2/4} = 3 \ell^{-2} - 2 {\lambda_1}^2</math>
</td>
</tr>

<tr>
<td align="right">
<math>R^2 - 2z^2</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>- {\lambda_1}^2 - {\lambda_2}^2 \pm 2 \lambda_2 \sqrt{{\lambda_1}^2+{\lambda_2}^2/4} = 3 {\lambda_1}^2 -2 \ell^{-2}</math>
</td>
</tr>

<tr>
<td align="right">
<math>Rz</math>
</td>
<td align="center">
<math>=</math>
</td>
<td align="left">
<math>\sqrt{\sqrt{2}\lambda_2} \left( -\frac{\lambda_2}{2\sqrt{2}} \pm \sqrt{\frac{4{\lambda_1}^2+{\lambda_2}^2}{8}} \right)^{3/2} = h_2 \lambda_2 / \ell</math>
</td>
</tr>
</table>

Partials of <math>\ell</math> can be taken with respect to the coordinates of either system. They are:

<table align="center" border="1" cellpadding="5">
<tr>
<td align="center">
 
</td>
<td align="center">
<math>
\frac{\partial}{\partial R}
</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>\ell</math>
</td>
<td align="center">
<math>
-R \ell^3
</math>
</td>
<td align="center">
<math>
-4z \ell^3
</math>
</td>
<td align="center">
<math>
0
</math>
</td>
</tr>
</table>

<table align="center" border="1" cellpadding="5">
<tr>
<td align="center">
 
</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>\ell</math>
</td>
<td align="center">
<math>
- \left( R^2 + 8z^2 \right) \ell^5 \lambda_1 = \ell^3 \lambda_1 \left( 2{h_1}^2 - 3 \right)
</math>
</td>
<td align="center">
<math>
2R^2 z^2 \ell^5 / \lambda_2 = 2 {h_2}^2 \ell^3 \lambda_2
</math>
</td>
<td align="center">
<math>
0
</math>
</td>
</tr>
</table>

User talk:Jaycall

2010-06-09T16:39:14Z

Jaycall: /* Two Views of Equation of Motion */ Response to question about degree of generalized for T3 coordinate relations

User talk:Jaycall

2010-06-09T15:42:23Z

Jaycall: /* Killing Vector Approach */

User:Tohline/Appendix/Ramblings/T3Integrals

2010-06-01T20:28:49Z

Jaycall: /* Logarithmic Derivatives of Scale Factors */ Partial derivative corrected; question about logarithmic derivatives for Joel

{{LSU_HBook_header}}

=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">
      and      
</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>

Here are some relevant partial derivatives:

<table align="center" border="1" cellpadding="5">
<tr>
<td align="center">
 
</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>
</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>
</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>
</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">
 
</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">
 
</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>

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">
 
</td>
<td align="left">
 
</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">
 
</td>
<td align="left">
 
</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">
 
</td>
<td align="left">
 
</td>
</tr>

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

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">
 
</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">
 
</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">
 
</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">
 
</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><br /><br />
or, inserting the relation derived above for <math>A</math> in terms of <math>dh_2/dt</math> for T3 coordinates<br /><br />
<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>
<br /><br />
Or, equivalently (but perhaps more perversely),<br /><br />
<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 one of these last few expressions is equivalent to the <i>simplest form</i> of the conservation that we derived — actually, Jay Call derived it — 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>



==Logarithmic Derivatives of Scale Factors==

<font color="red">
'''NOTE: I think that the sign is incorrect on one of the partial derivatives that has been tabulated, above.'''
</font>
Specifically, I think that the correct expression is:
<div align="center">
<math>
\frac{\partial z}{\partial\lambda_2} = - (q^2-1)\frac{\varpi^2 z \ell^2}{\lambda_2} .
</math>
</div>

<font color="green">
You're certainly right about the sign error. I have corrected the expression in the table above. --[[User:Jaycall|Jaycall]] 14:28, 1 June 2010 (MDT)
</font>

I have not yet been able to confirm your expressions below. I'm not sure what approach you took in deriving them, but since I had already calculated partials of the scale factors (<math>\partial_i h_j</math> and so forth), I derived a little trick to help simplify the work. Do you agree that

<div align="center">
<math>
\frac{\partial \ln h_i}{\partial \ln \lambda_j} \equiv \frac{\lambda_j}{h_i} \partial_j h_i ?
</math>
</div>

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_1 \lambda_2} \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>
- \frac{(q^2+1)}{(q^2-1)} \biggl( \frac{q h_1 h_2}{\lambda_1 \lambda_2} \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} = \biggl[\frac{q^2 + 1}{q^2-1} \biggr] \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>

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

 <br />
{{LSU_HBook_footer}}

User:Tohline/Appendix/Ramblings/T3Integrals

2010-06-01T17:26:32Z

Jaycall: /* Definition */ Corrected sign error on \frac{\partial z}{\partial \lambda_2}

{{LSU_HBook_header}}

=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">
      and      
</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>

Here are some relevant partial derivatives:

<table align="center" border="1" cellpadding="5">
<tr>
<td align="center">
 
</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>
</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>
</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>
</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">
 
</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">
 
</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>

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">
 
</td>
<td align="left">
 
</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">
 
</td>
<td align="left">
 
</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">
 
</td>
<td align="left">
 
</td>
</tr>

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

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">
 
</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">
 
</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">
 
</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">
 
</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><br /><br />
or, inserting the relation derived above for <math>A</math> in terms of <math>dh_2/dt</math> for T3 coordinates<br /><br />
<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>
<br /><br />
Or, equivalently (but perhaps more perversely),<br /><br />
<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 one of these last few expressions is equivalent to the <i>simplest form</i> of the conservation that we derived — actually, Jay Call derived it — 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>



==Logarithmic Derivatives of Scale Factors==

<font color="red">
'''NOTE: I think that the sign is incorrect on one of the partial derivatives that has been tabulated, above.'''
</font>
Specifically, I think that the correct expression is:
<div align="center">
<math>
\frac{\partial z}{\partial\lambda_2} = - (q^2-1)\frac{\varpi^2 z \ell^2}{\lambda_2} .
</math>
</div>

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_1 \lambda_2} \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>
- \frac{(q^2+1)}{(q^2-1)} \biggl( \frac{q h_1 h_2}{\lambda_1 \lambda_2} \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} = \biggl[\frac{q^2 + 1}{q^2-1} \biggr] \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>

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

 <br />
{{LSU_HBook_footer}}

User talk:Jaycall

2010-05-31T19:34:20Z

Jaycall: /* Killing Vector Approach */

User talk:Jaycall

2010-05-31T01:18:53Z

Jaycall: /* Killing Vector Approach */

User talk:Jaycall

2010-05-30T18:55:40Z

Jaycall: Express summary integral in terms of cylindrical coordinates

User talk:Jaycall

2010-05-30T02:17:07Z

Jaycall: Learned how to "sign"

User:Jaycall/KillingVectorApproach

2010-05-30T02:13:11Z

Jaycall: Expanded out summary integral term-by-term and simplified

=Killing Vector Approach to Integrals of Motion Problem=

Motivated by our recent work in relativistic fluid dynamics, one insightful approach to the integrals of motion problem involves rewriting the equations of motion in terms of an arbitrary vector field, which we're currently calling a ''characteristic vector''. The benefit here is that we can produce a mathematical condition that tells us what combintations of variables should be grouped together in order to produce conserved quantities. The details follow.

Using index summation notation (wherein repeated indices are summed over), it is possible to write the Newtonian equations of motion for a test particle in a fixed, static potential in an arbitrary orthogonal coordinate system. The equation describing motion in the <math>i</math>-dimension is:

<div align="center">
<math>
\frac{d}{dt} \left( m \ {h_i}^2 \dot{\lambda}_i \right) = m \ {h_k}^2 \Gamma^k_{ij} \dot{\lambda}_j \dot{\lambda}_k - m \ \partial_i \Phi
</math>
</div>

where <math>m</math> is the mass of the test particle, the <math>\lambda</math>s are the chosen coordinates, and the <math>\dot{\lambda}</math>s are the coordinate velocities (or the total time-derivatives of the coordinates), the <math>h</math>s are the metric scale factors associated with each of the coordinates, <math>\Phi</math> is the gravitational potential, and the <math>\Gamma</math>s are the Christoffel symbols, which for an orthogonal coordinate system can be written (depending on which, if any, indices are repeated) as

<table align="center">
<tr>
<math>
\Gamma^k_{ij} = 0 \ \ (i \ne j \ne k)
</math>
</tr>
<tr>
<math>
\Gamma^k_{ii} = - \frac{h_i}{h_k} \frac{\partial_k h_i}{h_k} \ \ (i \ne k)
</math>
</tr>
<tr>
<math>
\Gamma^i_{ij} = \Gamma^i_{ji} = \frac{\partial_i h_j}{h_j}
</math>
</tr>
<tr>
<math>
\Gamma^i_{ii} = \frac{\partial_i h_i}{h_i}
</math>
</tr>
</table>

If, through some miracle, the right-hand side (RHS) of the equation of motion should equal zero, then the combination of variables inside the parentheses must by definition be a conserved quantity. But good luck guessing a coordinate system in which this miracle will occur! The trick is to introduce an arbitrary vector field, which will be used to build a location-dependent weighted linear combination of the equations of motion. Since the equations of motion associated with each of the coordinates collectively form a single vector equation, we will just form an inner product of our characteristic vector with each side of the vector equation of motion. The result is

<div align="center">
<math>
C_i \frac{d}{dt} \left( m \ {h_i}^2 \dot{\lambda}_i \right) = m \ {h_k}^2 \Gamma^k_{ij} \dot{\lambda}_j \dot{\lambda}_k C_i - m \ C_i \ \partial_i \Phi
</math>
</div>

Next, we bring <math>C_i</math> inside the total time-derivative on the left-hand side (LHS). This produces an additional term, which we promptly move over to the RHS and include as part of what is commonly referred to as the ''source'' (since it's the source of any change in the quantity in parentheses).

<div align="center">
<math>
\frac{d}{dt} \left( m \ {h_i}^2 \dot{\lambda}_i C_i \right) = m \ {h_i}^2 \dot{\lambda}_i \dot{C}_i + m \ {h_k}^2 \Gamma^k_{ij} \dot{\lambda}_j \dot{\lambda}_k C_i - m \ C_i \ \partial_i \Phi
</math>
</div>

By doing this, we have formed a new ''conservative'' quantity; that is, a new quantity in the parentheses which will be conserved if and only if the source is zero. The utility is in the fact that we should be able to force the source associated with this new conservative quantity to go to zero by choosing the right characteristic vector. Once we find the right characteristic vector, we'll be able to use it directly to build the conserved quantity <math>m \ {h_i}^2 \dot{\lambda}_i C_i</math>.

So we're looking for a vector <math>\vec{C}</math> such that

<div align="center">
<math>
{h_i}^2 \dot{\lambda}_i \dot{C}_i + {h_k}^2 \Gamma^k_{ij} \dot{\lambda}_j \dot{\lambda}_k C_i - C_i \ \partial_i \Phi = 0 .
</math>
</div>

The good news is that this is a single scalar equation constraining the three independent components of <math>\vec{C}</math>, so several families of solutions should exist. Consequently, we can afford to be choosey! In the problem we're exploring using T3 coordinates, we already know a conserved quantity associated with the azimuthal coordinate--angular momentum. We're looking for an additional (independent) conserved quantity associated with the two radial coordinates. So it makes sense to look for one of the solutions that has <math>C_3 = 0</math>. Furthermore, we'd like to avoid dealing with the unknown potential function (which varies only with <math>\lambda_1</math>) as much as possible, so as long as we're being choosey, let's look for a solution that has <math>C_1 = 0</math>. We now have three conditions on three components. The condition constraining <math>C_2</math> is

<div align="center">
<math>
{h_2}^2 \dot{\lambda}_2 \dot{C}_2 + {h_k}^2 \Gamma^k_{2j} \dot{\lambda}_j \dot{\lambda}_k C_2 = 0 .
</math>
</div>

This can be rewritten more simply as

<div align="center">
<math>
\dot{C}_2 = - \left( \frac{{h_k}^2}{{h_2}^2} \Gamma^k_{2j} \frac{\dot{\lambda}_j \dot{\lambda}_k}{\dot{\lambda}_2} \right) C_2 .
</math>
</div>

Perhaps the best approach to solving this condition is separation of variables.

<div align="center">
<math>
\int \frac{dC_2}{C_2} = - \int \left( \frac{{h_k}^2}{{h_2}^2} \Gamma^k_{2j} \frac{\dot{\lambda}_j \dot{\lambda}_k}{\dot{\lambda}_2} \right) dt .
</math>
</div>

<div align="center">
<math>
\Longrightarrow \ln C_2 = - \int \left( \frac{{h_k}^2}{{h_2}^2} \Gamma^k_{2j} \frac{\dot{\lambda}_j \dot{\lambda}_k}{\dot{\lambda}_2} \right) dt .
</math>
</div>

<div align="center">
<math>
\Longrightarrow C_2 = \exp \left\{ - \int \left( \frac{{h_k}^2}{{h_2}^2} \Gamma^k_{2j} \frac{\dot{\lambda}_j \dot{\lambda}_k}{\dot{\lambda}_2} \right) dt \right\} .
</math>
</div>

The final necessary step will be to write the quantity in parentheses as the exact derivative of some other quantity, call it <math>\Xi</math>. The long-sought conserved quantity will then be <math> m {h_2}^2 \dot{\lambda}_2 \exp \left( - \Xi \right) </math> , where

<div align="center">
<math>
\Xi \equiv \int \left( \frac{{h_k}^2}{{h_2}^2} \Gamma^k_{2j} \frac{\dot{\lambda}_j \dot{\lambda}_k}{\dot{\lambda}_2} \right) dt .
</math>
</div>

Writing out each of the terms in <math>\Xi</math>, this becomes

<div align="center">
<math>
\frac{d\Xi}{dt} = \frac{{h_1}^2}{{h_2}^2} \Gamma^1_{21} \frac{\dot{\lambda}_1 \dot{\lambda}_1}{\dot{\lambda}_2} + \frac{{h_1}^2}{{h_2}^2} \Gamma^1_{22} \frac{\dot{\lambda}_2 \dot{\lambda}_1}{\dot{\lambda}_2} + \frac{{h_1}^2}{{h_2}^2} \Gamma^1_{23} \frac{\dot{\lambda}_3 \dot{\lambda}_1}{\dot{\lambda}_2} + \frac{{h_2}^2}{{h_2}^2} \Gamma^2_{21} \frac{\dot{\lambda}_1 \dot{\lambda}_2}{\dot{\lambda}_2} + \frac{{h_2}^2}{{h_2}^2} \Gamma^2_{22} \frac{\dot{\lambda}_2 \dot{\lambda}_2}{\dot{\lambda}_2} + \frac{{h_2}^2}{{h_2}^2} \Gamma^2_{23} \frac{\dot{\lambda}_3 \dot{\lambda}_2}{\dot{\lambda}_2} + \frac{{h_3}^2}{{h_2}^2} \Gamma^3_{21} \frac{\dot{\lambda}_1 \dot{\lambda}_3}{\dot{\lambda}_2} + \frac{{h_3}^2}{{h_2}^2} \Gamma^3_{22} \frac{\dot{\lambda}_2 \dot{\lambda}_3}{\dot{\lambda}_2} + \frac{{h_3}^2}{{h_2}^2} \Gamma^3_{23} \frac{\dot{\lambda}_3 \dot{\lambda}_3}{\dot{\lambda}_2} .
</math>
</div>

Plugging in values for the Christoffel symbols leads to the expression

<div align="center">
<math>
\frac{d\Xi}{dt} = \frac{{h_1}^2}{{h_2}^2} \frac{\partial_2 h_1}{h_1} \frac{\dot{\lambda}_1 \dot{\lambda}_1}{\dot{\lambda}_2} - \frac{{h_1}^2}{{h_2}^2} \frac{h_2}{h_1} \frac{\partial_1 h_2}{h_1} \frac{\dot{\lambda}_2 \dot{\lambda}_1}{\dot{\lambda}_2} + \frac{{h_2}^2}{{h_2}^2} \frac{\partial_1 h_2}{h_2} \frac{\dot{\lambda}_1 \dot{\lambda}_2}{\dot{\lambda}_2} + \frac{{h_2}^2}{{h_2}^2} \frac{\partial_2 h_2}{h_2} \frac{\dot{\lambda}_2 \dot{\lambda}_2}{\dot{\lambda}_2} + \frac{{h_3}^2}{{h_2}^2} \frac{\partial_2 h_3}{h_3} \frac{\dot{\lambda}_3 \dot{\lambda}_3}{\dot{\lambda}_2} .
</math>
</div>

And limiting our interest to motion within the meridianal plane (setting <math>\dot{\lambda}_3 = 0</math>) and simplifying

<div align="center">
<math>
\frac{d\Xi}{dt} = \frac{h_1}{h_2} \frac{\partial_2 h_1}{h_2} \frac{{\dot{\lambda}_1}^2}{\dot{\lambda}_2} - \cancel{\frac{\partial_1 h_2}{h_2} \dot{\lambda}_1} + \cancel{\frac{\partial_1 h_2}{h_2} \dot{\lambda}_1} + \frac{\partial_2 h_2}{h_2} \dot{\lambda}_2 .
</math>
</div>

==Question from Joel==
Is the following logic correct?

Suppose we examine the term in which <math>j=2</math> and <math>k=1</math>. The integral becomes,

<div align="center">
<math>
\Xi\biggr|_{j=2,k=1} = \int \left( \frac{{h_1}^2}{{h_2}^2} \Gamma^1_{22} \frac{\dot{\lambda}_2 \dot{\lambda}_1}{\dot{\lambda}_2} \right) dt = \int \left( \frac{{h_1}^2}{{h_2}^2} \Gamma^1_{22} \right) d\lambda_1
</math>
</div>

Given that the scale factors and the Christoffel symbols are expressible entirely in terms of <math>\lambda_1</math> and <math>\lambda_2</math>, one could imagine a situation — for example in the quadratic case of <math>q^2=2</math> — in which this integral could be completed analytically. (I presume that wherever <math>\lambda_2</math> appears inside this integral, it can be treated as a constant because the two coordinates are independent of one another.)

==Response from Jay==

Yes, I believe you've worked this term out correctly, but as you can see above it cancels out with the other cross term.

User talk:Jaycall

2010-05-29T21:16:20Z

Jaycall: /* Jay's Reply */ Fixed typo

User talk:Jaycall

2010-05-29T21:15:44Z

Jaycall: Location of third component of dx/dt

User:Jaycall/KillingVectorApproach

2010-05-29T18:43:56Z

Jaycall: Added definition of \Xi

User:Jaycall/KillingVectorApproach

2010-05-29T18:27:40Z

Jaycall: Corrected final form of conserved quantity

User:Jaycall/KillingVectorApproach

2010-05-29T18:22:50Z

Jaycall: Furthered solution of C_2 constraint

User:Jaycall/KillingVectorApproach

2010-05-29T18:06:53Z

Jaycall: Improved appearance of equation constraining C_2

User:Jaycall/KillingVectorApproach

2010-05-29T18:00:57Z

Jaycall: Minor corrections to Christoffel symbols

User:Jaycall/KillingVectorApproach

2010-05-29T17:55:38Z

Jaycall: Created page

User:Tohline/Appendix/Ramblings/T3Integrals

2010-05-29T15:41:00Z

Jaycall: /* Vector Derivatives */ Corrected link to KillingVectorApproach

{{LSU_HBook_header}}

=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">
      and      
</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>

Here are some relevant partial derivatives:

<table align="center" border="1" cellpadding="5">
<tr>
<td align="center">
 
</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>
</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>
</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>
</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">
 
</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">
 
</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>

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">
 
</td>
<td align="left">
 
</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">
 
</td>
<td align="left">
 
</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">
 
</td>
<td align="left">
 
</td>
</tr>

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

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 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>
\frac{h_2}{h_1} \left( \Gamma^2_{11} \dot{\lambda}_1 + \Gamma^2_{21} \dot{\lambda}_2 + \Gamma^2_{31} \dot{\lambda}_3 \right) \hat{e}_2 + \frac{h_3}{h_1} \left( \Gamma^3_{11} \dot{\lambda}_1 + \Gamma^3_{21} \dot{\lambda}_2 + \Gamma^3_{31} \dot{\lambda}_3 \right) \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>
\frac{h_1}{h_2} \left( \Gamma^1_{12} \dot{\lambda}_1 + \Gamma^1_{22} \dot{\lambda}_2 + \Gamma^1_{32} \dot{\lambda}_3 \right) \hat{e}_1 + \frac{h_3}{h_2} \left( \Gamma^3_{12} \dot{\lambda}_1 + \Gamma^3_{22} \dot{\lambda}_2 + \Gamma^3_{32} \dot{\lambda}_3 \right) \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>
\frac{h_1}{h_3} \left( \Gamma^1_{13} \dot{\lambda}_1 + \Gamma^1_{23} \dot{\lambda}_2 + \Gamma^1_{33} \dot{\lambda}_3 \right) \hat{e}_1 + \frac{h_2}{h_3} \left( \Gamma^2_{13} \dot{\lambda}_1 + \Gamma^2_{23} \dot{\lambda}_2 + \Gamma^2_{33} \dot{\lambda}_3 \right) \hat{e}_2
</math>
</td>
</tr>
</table>

where the 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>\Gamma^1_{23} = \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>\Gamma^2_{13} = \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>\Gamma^3_{12} = \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>

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">
 
</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">
 
</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">
 
</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">
 
</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><br /><br />
or, inserting the relation derived above for <math>A</math> in terms of <math>dh_2/dt</math> for T3 coordinates<br /><br />
<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>
<br /><br />
Or, equivalently (but perhaps more perversely),<br /><br />
<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 one of these last few expressions is equivalent to the <i>simplest form</i> of the conservation that we derived — actually, Jay Call derived it — 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>



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

 <br />
{{LSU_HBook_footer}}

User:Tohline/Appendix/Ramblings/T3Integrals

2010-05-29T15:21:13Z

Jaycall: /* Vector Derivatives */ Added Christoffel symbols

{{LSU_HBook_header}}

=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">
      and      
</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>

Here are some relevant partial derivatives:

<table align="center" border="1" cellpadding="5">
<tr>
<td align="center">
 
</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>
</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>
</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>
</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">
 
</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">
 
</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>

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">
 
</td>
<td align="left">
 
</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">
 
</td>
<td align="left">
 
</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">
 
</td>
<td align="left">
 
</td>
</tr>

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

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 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>
\frac{h_2}{h_1} \left( \Gamma^2_{11} \dot{\lambda}_1 + \Gamma^2_{21} \dot{\lambda}_2 + \Gamma^2_{31} \dot{\lambda}_3 \right) \hat{e}_2 + \frac{h_3}{h_1} \left( \Gamma^3_{11} \dot{\lambda}_1 + \Gamma^3_{21} \dot{\lambda}_2 + \Gamma^3_{31} \dot{\lambda}_3 \right) \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>
\frac{h_1}{h_2} \left( \Gamma^1_{12} \dot{\lambda}_1 + \Gamma^1_{22} \dot{\lambda}_2 + \Gamma^1_{32} \dot{\lambda}_3 \right) \hat{e}_1 + \frac{h_3}{h_2} \left( \Gamma^3_{12} \dot{\lambda}_1 + \Gamma^3_{22} \dot{\lambda}_2 + \Gamma^3_{32} \dot{\lambda}_3 \right) \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>
\frac{h_1}{h_3} \left( \Gamma^1_{13} \dot{\lambda}_1 + \Gamma^1_{23} \dot{\lambda}_2 + \Gamma^1_{33} \dot{\lambda}_3 \right) \hat{e}_1 + \frac{h_2}{h_3} \left( \Gamma^2_{13} \dot{\lambda}_1 + \Gamma^2_{23} \dot{\lambda}_2 + \Gamma^2_{33} \dot{\lambda}_3 \right) \hat{e}_2
</math>
</td>
</tr>
</table>

where the 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>\Gamma^1_{23} = \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>\Gamma^2_{13} = \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>\Gamma^3_{12} = \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>

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:Call/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">
 
</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">
 
</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">
 
</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">
 
</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><br /><br />
or, inserting the relation derived above for <math>A</math> in terms of <math>dh_2/dt</math> for T3 coordinates<br /><br />
<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>
<br /><br />
Or, equivalently (but perhaps more perversely),<br /><br />
<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 one of these last few expressions is equivalent to the <i>simplest form</i> of the conservation that we derived — actually, Jay Call derived it — 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>



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

 <br />
{{LSU_HBook_footer}}

User:Tohline/Appendix/Ramblings/T3Integrals

2010-05-29T14:12:19Z

Jaycall: /* Definition */ Added partials of T3 coordinates with respect to cylindrical coordinates and vice versa

User:Tohline/Appendix/Ramblings/T3Integrals

2010-05-29T13:27:35Z

Jaycall: /* Time-Derivative of Position and Velocity Vectors */ Added third component to dx/dt