Difference between revisions of "User:Jaycall/T3 Coordinates/Special Case"

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==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
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


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   </td>
   </td>
   <td align="left">
   <td align="left">
<math>-\frac{{\lambda_2}^2}{2} \pm \sqrt{\frac{{\lambda_2}^2}{4} + {\lambda_1}^2 {\lambda_2}^2}</math>
<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>
   <td align="center">
   <td align="center">
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   </td>
   </td>
   <td align="left">
   <td align="left">
<math>-\frac{\lambda_2}{2\sqrt{2}} \pm \sqrt{\frac{4 {\lambda_1}^2+{\lambda_2}^2}{8}}</math>
<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>
   </td>
</tr>
</tr>
</table>
</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:
Partial derivatives of each of the T3 coordinates taken with respect to each of the cylindrical coordinates are:
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   <td align="center">
   <td align="center">
<math>
<math>
\frac{R}{\lambda_1}
\frac{R}{\lambda_1} = \left( \frac{2}{\Lambda + 1} \right)^{1/2}
</math>
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>
<math>
\frac{2z}{\lambda_1}
\frac{2z}{\lambda_1} = \left[ \frac{2 \left( \Lambda - 1 \right) }{\Lambda + 1} \right]^{1/2}
</math>
</math>
   </td>
   </td>
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   <td align="center">
   <td align="center">
<math>
<math>
\frac{2 \lambda_2}{R}
\frac{2 \lambda_2}{R} = \frac{2^{3/2}}{\left( \Lambda - 1 \right)^{1/2}}
</math>
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>
<math>
-\frac{\lambda_2}{z}
-\frac{\lambda_2}{z} = - \frac{2^{3/2}}{\Lambda - 1}
</math>
</math>
   </td>
   </td>
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   <td align="center">
   <td align="center">
<math>
<math>
R \ell^2 \lambda_1
R \ell^2 \lambda_1 = \frac{1}{\Lambda} \left( \frac{\Lambda + 1}{2} \right)^{1/2}
</math>
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>
<math>
2Rz^2 \ell^2 / \lambda_2
2Rz^2 \ell^2 / \lambda_2 = \frac{\left( \Lambda - 1 \right)^{3/2}}{2 \Lambda}
</math>
</math>
   </td>
   </td>
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   <td align="center">
   <td align="center">
<math>
<math>
2z \ell^2 \lambda_1
2z \ell^2 \lambda_1 = \frac{1}{\Lambda} \left( \frac{\Lambda^2 - 1}{2} \right)^{1/2}
</math>
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>
<math>
-R^2 z \ell^2 / \lambda_2
-R^2 z \ell^2 / \lambda_2 = - \frac{\Lambda - 1}{2^{3/2} \Lambda}
</math>
</math>
   </td>
   </td>
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</table>
</table>


where <math>\ell \equiv \left( R^2 + 4z^2 \right)^{-1/2}</math>.
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
Furthermore, the scale factors become
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</td>
</td>
<td align="left">
<td align="left">
<math>\lambda_1 \ell</math>
<math>\lambda_1 \ell = \left( \frac{\Lambda+1}{2 \Lambda} \right)^{1/2}</math>
</td>
</td>
   </tr>
   </tr>
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</td>
</td>
<td align="left">
<td align="left">
<math>Rz \ell / \lambda_2</math>
<math>Rz \ell / \lambda_2 = \frac{\Lambda-1}{2 \left( 2 \Lambda \right)^{1/2}}</math>
</td>
</td>
   </tr>
   </tr>
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   </tr>
   </tr>
</table>
</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.
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.
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</td>
</td>
<td align="left">
<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>
<math>2 {\lambda_1}^2 + {\lambda_2}^2/2 - \lambda_2 \sqrt{{\lambda_1}^2+{\lambda_2}^2/4} = \ell^{-2}</math>
</td>
</td>
   </tr>
   </tr>
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</td>
</td>
<td align="left">
<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>
<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>
</td>
   </tr>
   </tr>
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</td>
</td>
<td align="left">
<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>
<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>
</td>
   </tr>
   </tr>
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</td>
</td>
<td align="left">
<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>
<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>
</td>
   </tr>
   </tr>
</table>
</table>
==Additional Partials==


Partials of <math>\ell</math> can be taken with respect to the coordinates of either system.  They are:
Partials of <math>\ell</math> can be taken with respect to the coordinates of either system.  They are:
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</tr>
</tr>
</table>
</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">
&nbsp;
  </td>
  <td align="center">
<math>
\frac{\partial}{\partial \lambda_1}
</math>
  </td>
  <td align="center">
<math>
\frac{\partial}{\partial \lambda_2}
</math>
  </td>
  <td align="center">
<math>
\frac{\partial}{\partial \lambda_3}
</math>
  </td>
</tr>
<tr>
  <td align="center">
<math>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...

Latest revision as of 19:31, 17 July 2010

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

<math> \lambda_1 </math>

<math>\equiv</math>

<math>\left( R^2+2z^2 \right)^{1/2}</math>

      and      

<math> \lambda_2 </math>

<math>\equiv</math>

<math>\frac{R^2}{\sqrt{2}z}</math>

<math> R^2 </math>

<math>\equiv</math>

<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>

      and      

<math> z </math>

<math>\equiv</math>

<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>

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.

<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>

<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>

Partials of the Coordinates

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

 

<math> \frac{\partial}{\partial R} </math>

<math> \frac{\partial}{\partial z} </math>

<math> \frac{\partial}{\partial \phi} </math>

<math>\lambda_1</math>

<math> \frac{R}{\lambda_1} = \left( \frac{2}{\Lambda + 1} \right)^{1/2} </math>

<math> \frac{2z}{\lambda_1} = \left[ \frac{2 \left( \Lambda - 1 \right) }{\Lambda + 1} \right]^{1/2} </math>

<math> 0 </math>

<math>\lambda_2</math>

<math> \frac{2 \lambda_2}{R} = \frac{2^{3/2}}{\left( \Lambda - 1 \right)^{1/2}} </math>

<math> -\frac{\lambda_2}{z} = - \frac{2^{3/2}}{\Lambda - 1} </math>

<math>0</math>

<math>\lambda_3</math>

<math> 0 </math>

<math> 0</math>

<math> 1 </math>

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

 

<math> \frac{\partial}{\partial \lambda_1} </math>

<math> \frac{\partial}{\partial \lambda_2} </math>

<math> \frac{\partial}{\partial \lambda_3} </math>

<math>R</math>

<math> R \ell^2 \lambda_1 = \frac{1}{\Lambda} \left( \frac{\Lambda + 1}{2} \right)^{1/2} </math>

<math> 2Rz^2 \ell^2 / \lambda_2 = \frac{\left( \Lambda - 1 \right)^{3/2}}{2 \Lambda} </math>

<math> 0 </math>

<math>z</math>

<math> 2z \ell^2 \lambda_1 = \frac{1}{\Lambda} \left( \frac{\Lambda^2 - 1}{2} \right)^{1/2} </math>

<math> -R^2 z \ell^2 / \lambda_2 = - \frac{\Lambda - 1}{2^{3/2} \Lambda} </math>

<math>0</math>

<math>\phi</math>

<math> 0 </math>

<math> 0</math>

<math> 1 </math>

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

<math>h_1</math>

<math>=</math>

<math>\lambda_1 \ell = \left( \frac{\Lambda+1}{2 \Lambda} \right)^{1/2}</math>

<math>h_2</math>

<math>=</math>

<math>Rz \ell / \lambda_2 = \frac{\Lambda-1}{2 \left( 2 \Lambda \right)^{1/2}}</math>

<math>h_3</math>

<math>=</math>

<math>R = \lambda_3</math>

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.

<math>R^2 + 2z^2</math>

<math>=</math>

<math>{\lambda_1}^2</math>

<math>R^2 + 4z^2</math>

<math>=</math>

<math>2 {\lambda_1}^2 + {\lambda_2}^2/2 - \lambda_2 \sqrt{{\lambda_1}^2+{\lambda_2}^2/4} = \ell^{-2}</math>

<math>R^2 + 8z^2</math>

<math>=</math>

<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>

<math>R^2 - 2z^2</math>

<math>=</math>

<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>

<math>Rz</math>

<math>=</math>

<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>

Additional Partials

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

 

<math> \frac{\partial}{\partial R} </math>

<math> \frac{\partial}{\partial z} </math>

<math> \frac{\partial}{\partial \phi} </math>

<math>\ell</math>

<math> -R \ell^3 </math>

<math> -4z \ell^3 </math>

<math> 0 </math>

 

<math> \frac{\partial}{\partial \lambda_1} </math>

<math> \frac{\partial}{\partial \lambda_2} </math>

<math> \frac{\partial}{\partial \lambda_3} </math>

<math>\ell</math>

<math> - \left( R^2 + 8z^2 \right) \ell^5 \lambda_1 = \ell^3 \lambda_1 \left( 2{h_1}^2 - 3 \right) </math>

<math> 2R^2 z^2 \ell^5 / \lambda_2 = 2 {h_2}^2 \ell^3 \lambda_2 </math>

<math> 0 </math>

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

 

<math> \frac{\partial}{\partial \lambda_1} </math>

<math> \frac{\partial}{\partial \lambda_2} </math>

<math> \frac{\partial}{\partial \lambda_3} </math>

<math>h_1</math>

<math> \ell \left( 2 {h_1}^4 - 3 {h_1}^2 + 1 \right) = 2 h_2 \lambda_2 </math>

<math> 2 {h_2}^2 \ell^3 \lambda_1 \lambda_2 </math>

<math> 0 </math>

<math>h_2</math>

<math> 2 {h_1}^2 h_2 \ell^2 \lambda_1 = 2 h_2 \ell^4 {\lambda_1}^3 </math>

<math> h_2 \left( 2 {h_2}^2 \ell^2 {\lambda_2}^2 - 3 \ell^2 {\lambda_1}^2 + 1 \right) / \lambda_2 </math>

<math>0</math>

<math>h_3</math>

<math> R \ell^2 \lambda_1 </math>

<math> 2Rz^2 \ell^2 / \lambda_2 </math>

<math> 0 </math>

Conserved Quantity

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

<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>

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

<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> .

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

<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> .

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