User:Jaycall/T3 Coordinates/Special Case

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...