User:Jaycall/T3 Coordinates/Special Case

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Revision as of 16:44, 17 July 2010 by Jaycall (talk | contribs) (Corrected coordinate transformation)
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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} + \lambda_2 \sqrt{{\lambda_1}^2 + {\lambda_2}^2/4}</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] </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 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 </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 - \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>

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>

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>