Jaycall/T3Coordinates/SpecialCase

Coordinate Transformations

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

${\lambda }_1$	$\equiv$	${(R^2+2z^2)}^{1/2}$	and	${\lambda }_2$	$\equiv$	$\frac{R^2}{\sqrt{2}z}$
$R^2$	$\equiv$	$\frac{{\lambda }_2}{2}(-{\lambda }_2+\sqrt{4{{\lambda }_1}^2+{{\lambda }_2}^2})=\frac{2{{\lambda }_1}^2}{\mathrm{\Lambda }+1}$	and	$z$	$\equiv$	$\frac{1}{2^{3/2}}(-{\lambda }_2+\sqrt{4{{\lambda }_1}^2+{{\lambda }_2}^2})=\frac{{\lambda }_2}{2^{3/2}}(\mathrm{\Lambda }-1)$

where $\mathrm{\Lambda }\equiv {[1+{\left(\frac{2{\lambda }_1}{{\lambda }_2}\right)}^2]}^{1/2}$ .

From this definition of $\mathrm{\Lambda }$, we can compute both its partials with respect to the T3 coordinates, and its total time derivative.

$\frac{\partial \mathrm{\Lambda }}{\partial {\lambda }_1}=\frac{4{\lambda }_1/{{\lambda }_2}^2}{\mathrm{\Lambda }}=\frac{2{({\mathrm{\Lambda }}^2-1)}^{1/2}}{\mathrm{\Lambda }}\frac{1}{{\lambda }_2}\frac{\partial \mathrm{\Lambda }}{\partial {\lambda }_2}=-\frac{4{{\lambda }_1}^2/{\lambda }_2}{\mathrm{\Lambda }}=-\frac{2{({\mathrm{\Lambda }}^2-1)}^{1/2}}{\mathrm{\Lambda }}{\lambda }_1$

$\dot{\mathrm{\Lambda }}=\frac{2{({\mathrm{\Lambda }}^2-1)}^{1/2}}{\mathrm{\Lambda }}(\frac{\dot{{\lambda }_1}}{{\lambda }_2}-{\lambda }_1\dot{{\lambda }_2})$

Partials of the Coordinates

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

	$\frac{\partial }{\partial R}$	$\frac{\partial }{\partial z}$	$\frac{\partial }{\partial \varphi }$
${\lambda }_1$	$\frac{R}{{\lambda }_1}={\left(\frac{2}{\mathrm{\Lambda }+1}\right)}^{1/2}$	$\frac{2z}{{\lambda }_1}={\left[\frac{2(\mathrm{\Lambda }-1)}{\mathrm{\Lambda }+1}\right]}^{1/2}$	$0$
${\lambda }_2$	$\frac{2{\lambda }_2}{R}=\frac{2^{3/2}}{{(\mathrm{\Lambda }-1)}^{1/2}}$	$-\frac{{\lambda }_2}{z}=-\frac{2^{3/2}}{\mathrm{\Lambda }-1}$	$0$
${\lambda }_3$	$0$	$0$	$1$

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

	$\frac{\partial }{\partial {\lambda }_1}$	$\frac{\partial }{\partial {\lambda }_2}$	$\frac{\partial }{\partial {\lambda }_3}$
$R$	$R{\ell }^2{\lambda }_1=\frac{1}{\mathrm{\Lambda }}{\left(\frac{\mathrm{\Lambda }+1}{2}\right)}^{1/2}$	$2R{z}^2{\ell }^2/{\lambda }_2=\frac{{(\mathrm{\Lambda }-1)}^{3/2}}{2\mathrm{\Lambda }}$	$0$
$z$	$2z{\ell }^2{\lambda }_1=\frac{1}{\mathrm{\Lambda }}{\left(\frac{{\mathrm{\Lambda }}^2-1}{2}\right)}^{1/2}$	$-R^{2}z{\ell }^2/{\lambda }_2=-\frac{\mathrm{\Lambda }-1}{2^{3/2}\mathrm{\Lambda }}$	$0$
$\varphi$	$0$	$0$	$1$

where $\ell \equiv {(R^2+4z^2)}^{-1/2}=\frac{1}{{\lambda }_1}{\left(\frac{\mathrm{\Lambda }+1}{2\mathrm{\Lambda }}\right)}^{1/2}$.

Scale Factors

Furthermore, the scale factors become

$h_1$	$=$	${\lambda }_1\ell ={\left(\frac{\mathrm{\Lambda }+1}{2\mathrm{\Lambda }}\right)}^{1/2}$
$h_2$	$=$	$Rz\ell /{\lambda }_2=\frac{\mathrm{\Lambda }-1}{2{\left(2\mathrm{\Lambda }\right)}^{1/2}}$
$h_3$	$=$	$R={\lambda }_3$

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.

$R^2+2z^2$	$=$	${{\lambda }_1}^2$
$R^2+4z^2$	$=$	$2{{\lambda }_1}^2+{{\lambda }_2}^2/2-{\lambda }_2\sqrt{{{\lambda }_1}^2+{{\lambda }_2}^2/4}={\ell }^{-2}$
$R^2+8z^2$	$=$	$4{{\lambda }_1}^2+\frac{3}{2}{{\lambda }_2}^2-3{\lambda }_2\sqrt{{{\lambda }_1}^2+{{\lambda }_2}^2/4}=3{\ell }^{-2}-2{{\lambda }_1}^2$
$R^2-2z^2$	$=$	$-{{\lambda }_1}^2-{{\lambda }_2}^2+2{\lambda }_2\sqrt{{{\lambda }_1}^2+{{\lambda }_2}^2/4}=3{{\lambda }_1}^2-2{\ell }^{-2}$
$Rz$	$=$	$\sqrt{\sqrt{2}{\lambda }_2}{(-\frac{{\lambda }_2}{2\sqrt{2}}+\sqrt{\frac{4{{\lambda }_1}^2+{{\lambda }_2}^2}{8}})}^{3/2}=h_2{\lambda }_2/\ell$

Additional Partials

Partials of $\ell$ can be taken with respect to the coordinates of either system. They are:

	$\frac{\partial }{\partial R}$	$\frac{\partial }{\partial z}$	$\frac{\partial }{\partial \varphi }$
$\ell$	$-R{\ell }^3$	$-4z{\ell }^3$	$0$

	$\frac{\partial }{\partial {\lambda }_1}$	$\frac{\partial }{\partial {\lambda }_2}$	$\frac{\partial }{\partial {\lambda }_3}$
$\ell$	$-(R^2+8z^2){\ell }^5{\lambda }_1={\ell }^3{\lambda }_1(2{h_1}^2-3)$	$2R^2{z}^2{\ell }^5/{\lambda }_2=2{h_2}^2{\ell }^3{\lambda }_2$	$0$

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

	$\frac{\partial }{\partial {\lambda }_1}$	$\frac{\partial }{\partial {\lambda }_2}$	$\frac{\partial }{\partial {\lambda }_3}$
$h_1$	$\ell (2{h_1}^4-3{h_1}^2+1)=2h_2{\lambda }_2$	$2{h_2}^2{\ell }^3{\lambda }_1{\lambda }_2$	$0$
$h_2$	$2{h_1}^2{h}_2{\ell }^2{\lambda }_1=2h_2{\ell }^4{{\lambda }_1}^3$	$h_2(2{h_2}^2{\ell }^2{{\lambda }_2}^2-3{\ell }^2{{\lambda }_1}^2+1)/{\lambda }_2$	$0$
$h_3$	$R{\ell }^2{\lambda }_1$	$2R{z}^2{\ell }^2/{\lambda }_2$	$0$

Conserved Quantity

The conserved quantity associated with the ${\lambda }_2$ coordinate is

$m{h_2}^2\dot{{\lambda }_2}\exp \int \left[(4{{\lambda }_1}^2+{{\lambda }_2}^2-{\lambda }_2\sqrt{4{{\lambda }_1}^2+{{\lambda }_2}^2})(\frac{{{\lambda }_1}^2\dot{{\lambda }_2}}{{\lambda }_2}-\frac{{\lambda }_2{\dot{{\lambda }_1}}^2}{\dot{{\lambda }_2}})\right]dt.$

The quantity in brackets needs to be integrated. In terms of $\mathrm{\Lambda }$, it can be written

${{\lambda }_2}^2\mathrm{\Lambda }(\mathrm{\Lambda }-1)[\frac{{({\mathrm{\Lambda }}^2-1)}^{1/2}}{2}{\lambda }_1\dot{{\lambda }_2}-{{\lambda }_2}^2\frac{\dot{{\lambda }_1}}{\dot{{\lambda }_2}}\frac{\dot{{\lambda }_1}}{{\lambda }_2}]$ .

Notice that the thing in square brackets looks very closely related to $\dot{\mathrm{\Lambda }}$. Could this be a hint? If only we could figure out what $\frac{\dot{{\lambda }_1}}{\dot{{\lambda }_2}}$ is, maybe we could factor out the ${\lambda }_1\dot{{\lambda }_2}-\frac{\dot{{\lambda }_1}}{{\lambda }_2}$, which appears in $\dot{\mathrm{\Lambda }}$...

If, by some miracle, it should turn out that $\frac{{({\mathrm{\Lambda }}^2-1)}^{1/2}}{2}={{\lambda }_2}^2\frac{\dot{{\lambda }_1}}{\dot{{\lambda }_2}}$, factorization would be possible and our integral would read

$-\frac{1}{4}\int {{\lambda }_2}^2{\mathrm{\Lambda }}^2(\mathrm{\Lambda }-1)\dot{\mathrm{\Lambda }}dt=-\frac{1}{4}\int {{\lambda }_2}^2{\mathrm{\Lambda }}^2(\mathrm{\Lambda }-1)d\mathrm{\Lambda }$ .

We ought to be able to integrate this, right...? Maybe we could handle the pesky ${{\lambda }_2}^2$ with integration by parts...