Appendix/Ramblings/T4Integrals

Integrals of Motion in T4 Coordinates

In an accompanying Wiki document, we have derived the properties of an orthogonal, axisymmetric, T3 coordinate system in which the first coordinate, ${\lambda }_1$, defines a family of concentric oblate-spheroidal surfaces whose (uniform) flattening is defined by a parameter $q\equiv R_{\mathrm{e}\mathrm{q}}/Z_{\mathrm{p}\mathrm{o}\mathrm{l}\mathrm{e}}$. In a separate, but related, Wiki document, we attempt to derive the $3^{\mathrm{r}\mathrm{d}}$ isolating integral of motion for massless particles that move inside a flattened, axisymmetric potential whose equipotential surfaces align with ${\lambda }_1=\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{t}$ surfaces in the special (quadratic) case when $q^2=2$. While examining this special case, we noticed that, in T3 Coordinates, the $h_1$ and $h_2$ scale factors are only a function of the coordinate ratio ${\lambda }_1/{\lambda }_2$. This has led us to wonder whether it might be more fruitful to search for the $3^{\mathrm{r}\mathrm{d}}$ isolating integral using a coordinate system in which one of the coordinates is defined by this T3-coordinate ratio.

It is with this in mind that we explore the development of a new T4 coordinate system. From the very beginning we will restrict the T4-coordinate definition to the special case of $q^2=2$ because, at present, we think that the coordinate T3-coordinate ratio ${\lambda }_1/{\lambda }_2$ is only interesting in the quadratic case. (See, for example, the polynomial root derived to complete the T1-coordinate inversion for the cubic case $q^2=3$; it is another combination of the T3 coordinates that appears to be relevant in the cubic case.)

STOP!

(7/06/2010)

As defined, below, this is not an orthogonal coordinate system.

Definition

In what follows, the coordinates $({\lambda }_1,{\lambda }_2,{\lambda }_3)$ refer to T3 Coordinates. Let's define a set of orthogonal T4 Coordinates for the special (quadratic) case $q^2=2$ such that,

${\xi }_1$	$\equiv$	$({\lambda }_1^2+{\lambda }_2^2{)}^{1/2}$	$=$	$\varpi [1+{\sinh }^2\mathrm{Z}+(\sinh \mathrm{Z}{)}^{2/(1-q^2)}{]}^{1/2}$	$\stackrel{(q^2=2)}{\to }$	$\varpi [1+{\sinh }^2\mathrm{Z}+\frac{1}{{\sinh }^2\mathrm{Z}}{]}^{1/2};$
${\xi }_2$	$\equiv$	$\frac{{\lambda }_2}{{\lambda }_1}$	$=$	$[\frac{(\sinh \mathrm{Z}{)}^{2/(1-q^2)}}{1+{\sinh }^2\mathrm{Z}}{]}^{1/2}$	$\stackrel{(q^2=2)}{\to }$	$[\frac{1}{{\sinh }^2\mathrm{Z}(1+{\sinh }^2\mathrm{Z})}{]}^{1/2};$
$\tan {\xi }_3$	$\equiv$	$\frac{y}{x},$

where,

${\sinh }^2\mathrm{Z}\equiv (\frac{qz}{\varpi }{)}^2\stackrel{(q^2=2)}{\to }\frac{2z^2}{{\varpi }^2}.$

The coordinate inversion — from $({\xi }_1,{\xi }_2,{\xi }_3)$ back to $({\lambda }_1,{\lambda }_2,{\lambda }_3)$ — is straightforward. Specifically,

${\lambda }_1$	$=$	${\xi }_1\cos \left[{\tan }^{}{\xi }_2\right];$
${\lambda }_2$	$=$	${\xi }_1\sin \left[{\tan }^{}{\xi }_2\right];$
${\lambda }_3$	$=$	${\xi }_3.$

Here are some relevant partial derivatives:

$\frac{\partial {\sinh }^2\mathrm{Z}}{\partial \varpi }=-\frac{4z^2}{{\varpi }^3};$

$\frac{\partial {\sinh }^2\mathrm{Z}}{\partial z}=+\frac{4z}{{\varpi }^2}.$

Partial derivatives with respect to cylindrical coordinates are,

	$\frac{\partial }{\partial \varpi }$	$\frac{\partial }{\partial z}$	$\frac{\partial }{\partial \varphi }$
${\xi }_1$	$\frac{\varpi }{{\xi }_1{z}^2}({\varpi }^2+z^2)$	$\frac{1}{2{\xi }_1{z}^3}(4z^4-{\varpi }^4)$	$0$
${\xi }_2$	$\frac{2{\xi }_2^3{z}^2}{{\varpi }^5}({\varpi }^2+4z^2)$	$-\frac{2{\xi }_2^{3}z}{{\varpi }^4}({\varpi }^2+4z^2)$	$0$
${\xi }_3$	$0$	$0$	$1$

Hence, the partials with respect to Cartesian coordinates are,

	$\frac{\partial }{\partial x}$	$\frac{\partial }{\partial y}$	$\frac{\partial }{\partial z}$
${\xi }_1$	$\frac{x}{(1-q^2){\xi }_1}[1+\frac{q^4{z}^2}{{\varpi }^2}-\frac{q^2{\xi }_1^2}{{\varpi }^2}]$ $\stackrel{(q^2=2)}{\to }\frac{x}{{\xi }_1{z}^2}({\varpi }^2+z^2)$	$\frac{y}{(1-q^2){\xi }_1}[1+\frac{q^4{z}^2}{{\varpi }^2}-\frac{q^2{\xi }_1^2}{{\varpi }^2}]$ $\stackrel{(q^2=2)}{\to }\frac{y}{{\xi }_1{z}^2}({\varpi }^2+z^2)$	$-\frac{{\varpi }^2}{(1-q^2){\xi }_{1}z}[1+\frac{q^4{z}^2}{{\varpi }^2}-\frac{{\xi }_1^2}{{\varpi }^2}]$ $\stackrel{(q^2=2)}{\to }+\frac{1}{2{\xi }_1{z}^3}(4z^4-{\varpi }^4)$
${\xi }_2$	$$	$$	$$
${\xi }_3$	$-\frac{y}{{\varpi }^2}$	$+\frac{x}{{\varpi }^2}$	$0$

The scale factors are,

$h_1^2$	$=$	$\left[\right(\frac{\partial {\xi }_1}{\partial x}{)}^2+(\frac{\partial {\xi }_1}{\partial y}{)}^2+(\frac{\partial {\xi }_1}{\partial z}{)}^2{]}^{-1}$	$=$	$\left[\right(\frac{\partial {\xi }_1}{\partial \varpi }{)}^2+(\frac{\partial {\xi }_1}{\partial z}{)}^2{]}^{-1}$	$=$	$\left[\frac{4{\xi }_1^2{z}^6}{({\varpi }^2+4z^2)({\varpi }^6+4z^6)}\right]$
$h_2^2$	$=$	$\left[\right(\frac{\partial {\xi }_2}{\partial x}{)}^2+(\frac{\partial {\xi }_2}{\partial y}{)}^2+(\frac{\partial {\xi }_2}{\partial z}{)}^2{]}^{-1}$	$=$	$\left[\right(\frac{\partial {\xi }_2}{\partial \varpi }{)}^2+(\frac{\partial {\xi }_2}{\partial z}{)}^2{]}^{-1}$	$=$	$\frac{{\varpi }^{10}}{4{\xi }_2^6{z}^2}\left[\frac{1}{({\varpi }^2+4z^2{)}^2({\varpi }^2+z^2)}\right]$
$h_3^2$	$=$	$\left[\right(\frac{\partial {\xi }_3}{\partial x}{)}^2+(\frac{\partial {\xi }_3}{\partial y}{)}^2+(\frac{\partial {\xi }_3}{\partial z}{)}^2{]}^{-1}$	$=$	${\varpi }^2$
where,        $$.

The position vector is,

$\overrightarrow{x}$

$=$

$\hat{ı}x+\hat{ȷ}y+\hat{k}z$

$=$

$$

See Also

• Old discussion of Integrals of Motion