User:Jaycall/T3 Coordinates/Special Case
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}{2} \pm \sqrt{\frac{{\lambda_2}^2}{4} + {\lambda_1}^2 {\lambda_2}^2}</math> |
and |
<math> z </math> |
<math>\equiv</math> |
<math>-\frac{\lambda_2}{2\sqrt{2}} \pm \sqrt{\frac{4 {\lambda_1}^2+{\lambda_2}^2}{8}}</math> |
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} </math> |
<math> \frac{2z}{\lambda_1} </math> |
<math> 0 </math> |
|
<math>\lambda_2</math> |
<math> \frac{2 \lambda_2}{R} </math> |
<math> -\frac{\lambda_2}{z} </math> |
<math>0</math> |
|
<math>\lambda_3</math> |
<math> 0 </math> |
<math> 0</math> |
<math> 1 </math> |
And partials of the cyldrical 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 </math> |
<math> 2Rz^2 \ell^2 / \lambda_2 </math> |
<math> 0 </math> |
|
<math>z</math> |
<math> 2z \ell^2 \lambda_1 </math> |
<math> -R^2 z \ell^2 / \lambda_2 </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}</math>.
Furthermore, the scale factors become
|
<math>h_1</math> |
<math>=</math> |
<math>\lambda_1 \ell</math> |
|
<math>h_2</math> |
<math>=</math> |
<math>Rz \ell / \lambda_2</math> |
|
<math>h_3</math> |
<math>=</math> |
<math>R = \lambda_3</math> |
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 \mp \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 \mp 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 \pm 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}} \pm \sqrt{\frac{4{\lambda_1}^2+{\lambda_2}^2}{8}} \right)^{3/2} = h_2 \lambda_2 / \ell</math> |
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> |