Appendix/Ramblings/T6CoordinatesPt2

Concentric Ellipsoidal (T6) Coordinates (Part 2)

Orthogonal Coordinates

Speculation5

Spherical Coordinates

$r\cos \theta$	$=$	$z,$
$r\sin \theta$	$=$	$(x^2+y^2{)}^{1/2},$
$\tan \phi$	$=$	$\frac{y}{x}.$

Use λ₁ Instead of r

Here, as above, we define,

${\lambda }_1^2$

$\equiv$

$x^2+q^2{y}^2+p^2{z}^2$

Using this expression to eliminate "x" (in favor of λ₁) in each of the three spherical-coordinate definitions, we obtain,

$r^2\equiv x^2+y^2+z^2$	$=$	${\lambda }_1^2-y^2(q^2-1)-z^2(p^2-1);$
${\tan }^2\theta \equiv \frac{x^2+y^2}{z^2}$	$=$	$\frac{1}{z^2}[{\lambda }_1^2-y^2(q^2-1)-p^2{z}^2];$
$\frac{1}{{\tan }^2\phi }\equiv \frac{x^2}{y^2}$	$=$	$\frac{{\lambda }_1^2-p^2{z}^2}{y^2}-q^2.$

After a bit of additional algebraic manipulation, we find that,

$\frac{z^2}{{\lambda }_1^2}$	$=$	$\frac{(1+{\tan }^2\phi )}{{{𝒟}}^2},$
$\frac{y^2}{{\lambda }_1^2}$	$=$	$\left[\frac{{{𝒟}}^2{\tan }^2\phi -p^2{\tan }^2\phi (1+{\tan }^2\phi )}{(1+q^2{\tan }^2\phi ){{𝒟}}^2}\right],$
$\frac{x^2}{{\lambda }_1^2}$	$=$	$1-q^2\left(\frac{y^2}{{\lambda }_1^2}\right)-p^2\left(\frac{z^2}{{\lambda }_1^2}\right),$

where,

${{𝒟}}^2$

$\equiv$

$[(1+q^2{\tan }^2\phi )(p^2+{\tan }^2\theta )-p^2(q^2-1){\tan }^2\phi ].$

As a check, let's set $q^2=p^2=1$, which should reduce to the normal spherical coordinate system. ${\lambda }_1^2$ $\to$ $r^2,$       and,       ${{𝒟}}^2$ $\to$ $[(1+{\tan }^2\phi )(1+{\tan }^2\theta )].$ $\Rightarrow \frac{z^2}{{\lambda }_1^2}$ $\to$ $\frac{1}{1+{\tan }^2\theta }={\cos }^2\theta =\frac{z^2}{r^2};$ $\frac{y^2}{{\lambda }_1^2}$ $\to$ $\left[\frac{(1+{\tan }^2\phi )(1+{\tan }^2\theta ){\tan }^2\phi -{\tan }^2\phi (1+{\tan }^2\phi )}{(1+{\tan }^2\phi )(1+{\tan }^2\phi )(1+{\tan }^2\theta )}\right]$   $=$ $\frac{{\tan }^2\phi }{(1+{\tan }^2\phi )}\left[\frac{{\tan }^2\theta }{(1+{\tan }^2\theta )}\right]={\sin }^2\theta {\sin }^2\phi =\frac{y^2}{r^2};$ $\frac{x^2}{{\lambda }_1^2}$ $\to$ $1-\left(\frac{y^2}{{\lambda }_1^2}\right)-\left(\frac{z^2}{{\lambda }_1^2}\right),$   $\to$ $1-{\sin }^2\theta {\sin }^2\phi -{\cos }^2\theta =-{\sin }^2\theta {\sin }^2\phi +{\sin }^2\theta ={\sin }^2\theta {\cos }^2\phi =\frac{x^2}{r^2}.$

Relationship To T3 Coordinates

If we set, $q=1$, but continue to assume that $p>1$, we expect to see a representation that resembles our previously discussed, T3 Coordinates. Note, for example, that the new "radial" coordinate is,

${\lambda }_1^2$

$\to$

$({\varpi }^2+p^2{z}^2),$

and,

${{𝒟}}^2$

$\to$

$[(1+{\tan }^2\phi )(p^2+{\tan }^2\theta )].$

$\Rightarrow \frac{p^2{z}^2}{{\lambda }_1^2}$	$\to$	$\frac{p^2}{(p^2+{\tan }^2\theta )}=\frac{1}{(1+p^{-2}{\tan }^2\theta )},$
$\frac{{\varpi }^2}{{\lambda }_1^2}=\frac{x^2}{{\lambda }_1^2}+\frac{y^2}{{\lambda }_1^2}$	$\to$	$1-p^2\left(\frac{z^2}{{\lambda }_1^2}\right)=[1-\frac{1}{(1+p^{-2}{\tan }^2\theta )}].$

We also see that,

$\frac{{\varpi }^2}{p^2{z}^2}$

$\to$

$(1+p^{-2}{\tan }^2\theta )[1-\frac{1}{(1+p^{-2}{\tan }^2\theta )}]=p^{-2}{\tan }^2\theta .$

Again Consider Full 3D Ellipsoid

Let's try to replace everywhere, $[\varpi /(pz){]}^2=p^{-2}{\tan }^2\theta$ with ${\lambda }_2$. This gives,

$\frac{{{𝒟}}^2}{p^2}$

$\equiv$

$[(1+q^2{\tan }^2\phi ){\lambda }_2-(q^2-1){\tan }^2\phi ].$

which means that,

$\frac{p^2{z}^2}{{\lambda }_1^2}$	$=$	$\frac{(1+{\tan }^2\phi )}{[(1+q^2{\tan }^2\phi ){\lambda }_2-(q^2-1){\tan }^2\phi ]}$
	$=$	$\frac{(1+{\tan }^2\phi )/{\tan }^2\phi }{[q^2{\lambda }_2(1+q^2{\tan }^2\phi )/(q^2{\tan }^2\phi )-(q^2-1)]}=\frac{1/{\sin }^2\phi }{[q^2{\lambda }_2{Q}^2-(q^2-1)]},$
$\frac{q^2{y}^2}{{\lambda }_1^2}$	$=$	$\frac{q^2{\tan }^2\phi }{(1+q^2{\tan }^2\phi )}-\frac{q^2{\tan }^2\phi (1+{\tan }^2\phi )}{(1+q^2{\tan }^2\phi )[(1+q^2{\tan }^2\phi ){\lambda }_2-(q^2-1){\tan }^2\phi ]}$
	$=$	$\frac{q^2{\tan }^2\phi }{(1+q^2{\tan }^2\phi )}\{1-\frac{(1+{\tan }^2\phi )}{[(1+q^2{\tan }^2\phi ){\lambda }_2-(q^2-1){\tan }^2\phi ]}\}$
	$=$	$\frac{q^2{\tan }^2\phi }{(1+q^2{\tan }^2\phi )}\{1-\frac{(1+{\tan }^2\phi )/{\tan }^2\phi }{[q^2{\lambda }_2(1+q^2{\tan }^2\phi )/(q^2{\tan }^2\phi )-(q^2-1)]}\}$
	$=$	$\frac{1}{Q^2}\{1-\frac{1/{\sin }^2\phi }{[q^2{\lambda }_2{Q}^2-(q^2-1)]}\}=\frac{1}{Q^2[]}[[]-\frac{1}{{\sin }^2\phi }],$
$\frac{x^2}{{\lambda }_1^2}$	$=$	$1-\frac{q^2{\tan }^2\phi }{(1+q^2{\tan }^2\phi )}\{1-\frac{(1+{\tan }^2\phi )}{[(1+q^2{\tan }^2\phi ){\lambda }_2-(q^2-1){\tan }^2\phi ]}\}$
		$-\frac{(1+{\tan }^2\phi )}{[(1+q^2{\tan }^2\phi ){\lambda }_2-(q^2-1){\tan }^2\phi ]}$
	$=$	$1-\frac{q^2{\tan }^2\phi }{(1+q^2{\tan }^2\phi )}-\left\{\frac{(1+{\tan }^2\phi )}{[(1+q^2{\tan }^2\phi ){\lambda }_2-(q^2-1){\tan }^2\phi ]}\right\}$
	$=$	$1-\frac{1}{Q^2}-\left\{\frac{1/{\sin }^2\phi }{[q^2{\lambda }_2{Q}^2-(q^2-1)]}\right\}=\frac{1}{Q^2[]}\{Q^2[]-[]-\frac{Q^2}{{\sin }^2\phi }\},$
$\frac{x^2+q^2{y}^2}{{\lambda }_1^2}$	$=$	$1-[1+\frac{q^2{\tan }^2\phi }{(1+q^2{\tan }^2\phi )}]\left\{\frac{(1+{\tan }^2\phi )}{[(1+q^2{\tan }^2\phi ){\lambda }_2-(q^2-1){\tan }^2\phi ]}\right\}.$

Now, notice that,

$\frac{q^2{y}^2{Q}^2}{{\lambda }_1^2}$

$=$

$1-\frac{1}{[]{\sin }^2\phi },$

and,

$\frac{x^2}{{\lambda }_1^2}+\frac{1}{Q^2}$

$=$

$1-\frac{1}{[]{\sin }^2\phi }.$

Hence,

$\frac{x^2}{{\lambda }_1^2}+\frac{1}{Q^2}$	$=$	$\frac{q^2{y}^2{Q}^2}{{\lambda }_1^2}$
$\Rightarrow 0$	$=$	$Q^4\left(\frac{q^2{y}^2}{{\lambda }_1^2}\right)-Q^2\left(\frac{x^2}{{\lambda }_1^2}\right)-1$
	$=$	$Q^4-Q^2\left(\frac{x^2}{q^2{y}^2}\right)-\left(\frac{{\lambda }_1^2}{q^2{y}^2}\right),$

where,

$Q^2$

$\equiv$

$\frac{1+q^2{\tan }^2\phi }{q^2{\tan }^2\phi }.$

Solving the quadratic equation, we have,

$Q^2$	$=$	$\frac{1}{2}\left\{\right(\frac{x^2}{q^2{y}^2})\pm [(\frac{x^2}{q^2{y}^2}{)}^2+4(\frac{{\lambda }_1^2}{q^2{y}^2}\left){]}^{1/2}\right\}$
	$=$	$\left(\frac{x^2}{2q^2{y}^2}\right)\{1\pm [1+4\left(\frac{{\lambda }_1^2{q}^2{y}^2}{x^4}\right){]}^{1/2}\}.$

Tentative Summary ${\lambda }_1$ $\equiv$ $(x^2+q^2{y}^2+p^2{z}^2{)}^{1/2},$ ${\lambda }_2$ $\equiv$ $\frac{(x^2+y^2{)}^{1/2}}{pz},$ ${\lambda }_3=Q^2$ $\equiv$ $\left(\frac{x^2}{2q^2{y}^2}\right)\{1\pm [1+4\left(\frac{{\lambda }_1^2{q}^2{y}^2}{x^4}\right){]}^{1/2}\}.$