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

Partial Derivatives & Scale Factors

First Coordinate

$\frac{\partial {\lambda }_1}{\partial x}$	$=$	$\frac{x}{{\lambda }_1},$
$\frac{\partial {\lambda }_1}{\partial y}$	$=$	$\frac{q^{2}y}{{\lambda }_1},$
$\frac{\partial {\lambda }_1}{\partial z}$	$=$	$\frac{p^{2}z}{{\lambda }_1}.$

$h_1^{-2}$	$=$	$(\frac{\partial {\lambda }_1}{\partial x}{)}^2+(\frac{\partial {\lambda }_1}{\partial y}{)}^2+(\frac{\partial {\lambda }_1}{\partial z}{)}^2$	$=$	$(\frac{x}{{\lambda }_1}{)}^2+(\frac{q^{2}y}{{\lambda }_1}{)}^2+(\frac{p^{2}z}{{\lambda }_1}{)}^2.$
$\Rightarrow h_1$	$=$	${\lambda }_1{\ell }_{3D},$

where,

${\ell }_{3D}\equiv (x^2+q^4{y}^2+p^4{z}^2{)}^{-1/2}.$

As a result, the associated unit vector is,

${\hat{e}}_1$	$=$	$\hat{ı}{h}_1\left(\frac{\partial {\lambda }_1}{\partial x}\right)+\hat{ȷ}{h}_1\left(\frac{\partial {\lambda }_1}{\partial y}\right)+\hat{k}{h}_1\left(\frac{\partial {\lambda }_1}{\partial z}\right)$
	$=$	$\hat{ı}x{\ell }_{3D}+\hat{ȷ}{q}^{2}y{\ell }_{3D}+\hat{k}{p}^{2}z{\ell }_{3D}.$

Notice that,

${\hat{e}}_1\cdot {\hat{e}}_1$

$=$

$(x^2+q^4{y}^2+p^4{z}^2){\ell }_{3D}^2=1.$

Second Coordinate (1^st Try)

$\frac{\partial {\lambda }_2}{\partial x}$	$=$	$\frac{1}{pz}\left[\frac{x}{(x^2+y^2{)}^{1/2}}\right],$
$\frac{\partial {\lambda }_2}{\partial y}$	$=$	$\frac{1}{pz}\left[\frac{y}{(x^2+y^2{)}^{1/2}}\right],$
$\frac{\partial {\lambda }_2}{\partial z}$	$=$	$-\frac{(x^2+y^2{)}^{1/2}}{p{z}^2}.$

$h_2^{-2}$	$=$	$(\frac{\partial {\lambda }_2}{\partial x}{)}^2+(\frac{\partial {\lambda }_2}{\partial y}{)}^2+(\frac{\partial {\lambda }_2}{\partial z}{)}^2$
	$=$	$\left\{\frac{1}{pz}\right[\frac{x}{(x^2+y^2{)}^{1/2}}]{\}}^2+\{\frac{1}{pz}\left[\frac{y}{(x^2+y^2{)}^{1/2}}\right]{\}}^2+\{\frac{(x^2+y^2{)}^{1/2}}{p{z}^2}{\}}^2$
	$=$	$\left\{\right[\frac{x^2}{(x^2+y^2)p^2{z}^2}\left]\right\}+\left\{\right[\frac{y^2}{(x^2+y^2)p^2{z}^2}\left]\right\}+\left\{\frac{(x^2+y^2)}{p^2{z}^4}\right\}$
	$=$	$\frac{1}{p^2{z}^2}+\frac{(x^2+y^2)}{p^2{z}^4}=\frac{(x^2+y^2+z^2)}{p^2{z}^4}$
$\Rightarrow h_2$	$=$	$\frac{p{z}^2}{r}$

As a result, the associated unit vector is,

${\hat{e}}_2$	$=$	$\hat{ı}{h}_2\left(\frac{\partial {\lambda }_2}{\partial x}\right)+\hat{ȷ}{h}_2\left(\frac{\partial {\lambda }_2}{\partial y}\right)+\hat{k}{h}_2\left(\frac{\partial {\lambda }_2}{\partial z}\right)$
	$=$	$\hat{ı}\left[\frac{xz}{r(x^2+y^2{)}^{1/2}}\right]+\hat{ȷ}\left[\frac{yz}{r(x^2+y^2{)}^{1/2}}\right]-\hat{k}\left[\frac{(x^2+y^2{)}^{1/2}}{r}\right].$

Notice that,

${\hat{e}}_2\cdot {\hat{e}}_2$

$=$

$\left[\frac{x^2{z}^2}{r^2(x^2+y^2)}\right]+\left[\frac{y^2{z}^2}{r^2(x^2+y^2)}\right]+\left[\frac{(x^2+y^2)}{r^2}\right]=1.$

Let's check to see if this "second" unit vector is orthogonal to the "first."

${\hat{e}}_1\cdot {\hat{e}}_2$	$=$	$x{\ell }_{3D}\left[\frac{xz}{r(x^2+y^2{)}^{1/2}}\right]+q^{2}y{\ell }_{3D}\left[\frac{yz}{r(x^2+y^2{)}^{1/2}}\right]-p^{2}z{\ell }_{3D}\left[\frac{(x^2+y^2{)}^{1/2}}{r}\right]$
	$=$	${\ell }_{3D}\left\{\right[\frac{x^{2}z}{r(x^2+y^2{)}^{1/2}}]+[\frac{q^2{y}^{2}z}{r(x^2+y^2{)}^{1/2}}]-[\frac{p^{2}z(x^2+y^2{)}^{1/2}}{r}\left]\right\}$
	$=$	$\frac{z{\ell }_{3D}}{r(x^2+y^2{)}^{1/2}}\left\{\right[x^2]+[q^2{y}^2]-[p^2(x^2+y^2)\left]\right\}$
	$\ne$	$0.$

Second Coordinate (2^nd Try)

Let's try,

${\lambda }_2$	$=$	$\left[\frac{(x^2+q^2{y}^2+\mathfrak{𝔣}\cdot p^2{z}^2{)}^{1/2}}{pz}\right],$
$\Rightarrow \frac{\partial {\lambda }_2}{\partial x}$	$=$	$\frac{x}{pz(x^2+q^2{y}^2+\mathfrak{𝔣}\cdot p^2{z}^2{)}^{1/2}}=\frac{x}{p^2{z}^2{\lambda }_2},$
$\frac{\partial {\lambda }_2}{\partial y}$	$=$	$\frac{q^{2}y}{pz(x^2+q^2{y}^2+\mathfrak{𝔣}\cdot p^2{z}^2{)}^{1/2}}=\frac{q^{2}y}{p^2{z}^2{\lambda }_2},$
$\frac{\partial {\lambda }_2}{\partial z}$	$=$	$\frac{\mathfrak{𝔣}\cdot p^{2}z}{pz(x^2+q^2{y}^2+\mathfrak{𝔣}\cdot p^2{z}^2{)}^{1/2}}-\frac{(x^2+q^2{y}^2+\mathfrak{𝔣}\cdot p^2{z}^2{)}^{1/2}}{p{z}^2}=\frac{1}{p^2{z}^2{\lambda }_2}(\mathfrak{𝔣}\cdot p^{2}z)-\frac{{\lambda }_2}{z}=\frac{1}{p^2{z}^2{\lambda }_2}(\mathfrak{𝔣}\cdot p^{2}z-p^{2}z{\lambda }_2^2).$

Hence,

$h_2^{-2}$	$=$	$(\frac{\partial {\lambda }_2}{\partial x}{)}^2+(\frac{\partial {\lambda }_2}{\partial y}{)}^2+(\frac{\partial {\lambda }_2}{\partial z}{)}^2$
	$=$	$[\frac{x}{p^2{z}^2{\lambda }_2}{]}^2+[\frac{q^{2}y}{p^2{z}^2{\lambda }_2}{]}^2+[\frac{\mathfrak{𝔣}}{z{\lambda }_2}-\frac{{\lambda }_2}{z}{]}^2$
	$=$	$\left[\frac{x^2+q^4{y}^2}{p^4{z}^4{\lambda }_2^2}\right]+\left[\frac{1}{z{\lambda }_2}\right(\mathfrak{𝔣}-{\lambda }_2^2){]}^2$
	$=$	$\frac{1}{p^4{z}^4{\lambda }_2^2}[x^2+q^4{y}^2+p^4{z}^2(\mathfrak{𝔣}-{\lambda }_2^2{)}^2]$
$\Rightarrow h_2$	$=$	$\frac{p^2{z}^2{\lambda }_2}{[x^2+q^4{y}^2+p^4{z}^2(\mathfrak{𝔣}-{\lambda }_2^2{)}^2{]}^{1/2}}.$

So, the associated unit vector is,

${\hat{e}}_2$	$=$	$\hat{ı}{h}_2\left(\frac{\partial {\lambda }_2}{\partial x}\right)+\hat{ȷ}{h}_2\left(\frac{\partial {\lambda }_2}{\partial y}\right)+\hat{k}{h}_2\left(\frac{\partial {\lambda }_2}{\partial z}\right)$
	$=$	$\hat{ı}\left\{\frac{x}{[x^2+q^4{y}^2+p^4{z}^2(\mathfrak{𝔣}-{\lambda }_2^2{)}^2{]}^{1/2}}\right\}+\hat{ȷ}\left\{\frac{q^{2}y}{[x^2+q^4{y}^2+p^4{z}^2(\mathfrak{𝔣}-{\lambda }_2^2{)}^2{]}^{1/2}}\right\}+\hat{k}\left\{\frac{p^{2}z(\mathfrak{𝔣}-{\lambda }_2^2)}{[x^2+q^4{y}^2+p^4{z}^2(\mathfrak{𝔣}-{\lambda }_2^2{)}^2{]}^{1/2}}\right\}.$

Checking orthogonality …

${\hat{e}}_1\cdot {\hat{e}}_2$	$=$	$x{\ell }_{3D}\left\{\frac{x}{[x^2+q^4{y}^2+p^4{z}^2(\mathfrak{𝔣}-{\lambda }_2^2{)}^2{]}^{1/2}}\right\}+q^{2}y{\ell }_{3D}\left\{\frac{q^{2}y}{[x^2+q^4{y}^2+p^4{z}^2(\mathfrak{𝔣}-{\lambda }_2^2{)}^2{]}^{1/2}}\right\}+p^{2}z{\ell }_{3D}\left\{\frac{p^{2}z(\mathfrak{𝔣}-{\lambda }_2^2)}{[x^2+q^4{y}^2+p^4{z}^2(\mathfrak{𝔣}-{\lambda }_2^2{)}^2{]}^{1/2}}\right\}$
	$=$	$\frac{{\ell }_{3D}}{[x^2+q^4{y}^2+p^4{z}^2(\mathfrak{𝔣}-{\lambda }_2^2{)}^2{]}^{1/2}}\{x^2+q^4{y}^2+p^4{z}^2(\mathfrak{𝔣}-{\lambda }_2^2)\}.$
	$=$	$\frac{{\ell }_{3D}}{[x^2+q^4{y}^2+p^4{z}^2(\mathfrak{𝔣}-{\lambda }_2^2{)}^2{]}^{1/2}}\{x^2+q^4{y}^2+p^4{z}^2(\mathfrak{𝔣}-{\lambda }_2^2)\}.$

If $\mathfrak{𝔣}=0$, we have …

$p^{2}z(\mathfrak{𝔣}-{\lambda }_2^2)$

$\to$

$[-p^{2}z{\lambda }_2^2{]}_{\mathfrak{𝔣}=0}=-p^{2}z[\frac{(x^2+q^2{y}^2+{\xcancel{\mathfrak{𝔣}\cdot }}^0{p}^2{z}^2{)}^{1/2}}{pz}{]}^2=-\frac{(x^2+q^2{y}^2)}{z},$

which, in turn, means …

$[x^2+q^4{y}^2+p^4{z}^2({\xcancel{\mathfrak{𝔣}}}^0-{\lambda }_2^2{)}^2{]}^{1/2}$	$=$	$[x^2+q^4{y}^2+p^4{z}^2{\lambda }_2^4{]}^{1/2}$
	$=$	$\{x^2+q^4{y}^2+p^4{z}^2[\frac{(x^2+q^2{y}^2+{\xcancel{\mathfrak{𝔣}\cdot }}^0{p}^2{z}^2{)}^{1/2}}{pz}{]}^4{\}}^{1/2}$
	$=$	$\{x^2+q^4{y}^2+[\frac{(x^2+q^2{y}^2{)}^2}{z^2}]{\}}^{1/2}$
	$=$	$(x^2+q^4{y}^2{)}^{1/2}[1+\frac{(x^2+q^2{y}^2)}{z^2}{]}^{1/2}$
	$=$	$\frac{(x^2+q^4{y}^2{)}^{1/2}}{z}[z^2+(x^2+q^2{y}^2){]}^{1/2},$

and,

${\hat{e}}_1\cdot {\hat{e}}_2$

$=$

$\frac{{\ell }_{3D}}{[x^2+q^4{y}^2+p^4{z}^2(\mathfrak{𝔣}-{\lambda }_2^2{)}^2{]}^{1/2}}\{x^2+q^4{y}^2+p^4{z}^2(\mathfrak{𝔣}-{\lambda }_2^2)\}.$

Speculation6

Determine λ₂

This is very similar to the above, Speculation2. Try,

${\lambda }_2$

$=$

$\frac{x{y}^{1/q^2}}{z^{2/p^2}},$

in which case,

$\frac{\partial {\lambda }_2}{\partial x}$	$=$	$\frac{{\lambda }_2}{x},$
$\frac{\partial {\lambda }_2}{\partial y}$	$=$	$\frac{x}{z^{2/p^2}}\left(\frac{1}{q^2}\right)y^{1/q^2-1}=\frac{{\lambda }_2}{q^{2}y},$
$\frac{\partial {\lambda }_2}{\partial z}$	$=$	$-\frac{2{\lambda }_2}{p^{2}z}.$

The associated scale factor is, then,

$h_2$	$=$	$\left[\right(\frac{\partial {\lambda }_2}{\partial x}{)}^2+(\frac{\partial {\lambda }_2}{\partial y}{)}^2+(\frac{\partial {\lambda }_2}{\partial z}{)}^2{]}^{-1/2}$
	$=$	$\left[\right(\frac{{\lambda }_2}{x}{)}^2+(\frac{{\lambda }_2}{q^{2}y}{)}^2+(-\frac{2{\lambda }_2}{p^{2}z}{)}^2{]}^{-1/2}$
	$=$	$\frac{1}{{\lambda }_2}[\frac{1}{x^2}+\frac{1}{q^4{y}^2}+\frac{4}{p^4{z}^2}{]}^{-1/2}$
	$=$	$\frac{1}{{\lambda }_2}[\frac{(q^4{y}^2{p}^4{z}^2+x^2{p}^4{z}^2+4x^2{q}^4{y}^2}{x^2{q}^4{y}^2{p}^4{z}^2}{]}^{-1/2}$
	$=$	$\frac{1}{{\lambda }_2}\left[\frac{x{q}^{2}y{p}^{2}z}{{𝒟}}\right].$

where,

${𝒟}$

$\equiv$

$(q^4{y}^2{p}^4{z}^2+x^2{p}^4{z}^2+4x^2{q}^4{y}^2{)}^{1/2}.$

The associated unit vector is, then,

${\hat{e}}_2$	$=$	$\hat{ı}{h}_2\left(\frac{\partial {\lambda }_2}{\partial x}\right)+\hat{ȷ}{h}_2\left(\frac{\partial {\lambda }_2}{\partial y}\right)+\hat{k}{h}_2\left(\frac{\partial {\lambda }_2}{\partial z}\right)$
	$=$	$\frac{x{q}^{2}y{p}^{2}z}{{𝒟}}\left\{\hat{ı}\right(\frac{1}{x})+\hat{ȷ}(\frac{1}{q^{2}y})+\hat{k}(-\frac{2}{p^{2}z}\left)\right\}.$

Recalling that the unit vector associated with the "first" coordinate is,

${\hat{e}}_1$

$\equiv$

$\hat{ı}(x{\ell }_{3D})+\hat{ȷ}(q^{2}y{\ell }_{3D})+\hat{k}(p^{2}z{\ell }_{3D}),$

where,

${\ell }_{3D}$

$=$

$(x^2+q^4{y}^2+p^4{z}^2{)}^{-1/2},$

let's check to see whether the "second" unit vector is orthogonal to the "first."

${\hat{e}}_1\cdot {\hat{e}}_2$

$=$

$\frac{(x{q}^{2}y{p}^{2}z){\ell }_{3D}}{{𝒟}}[1+1-2]=0.$

Hooray!

Direction Cosines for Third Unit Vector

Now, what is the unit vector, ${\hat{e}}_3$, that is simultaneously orthogonal to both these "first" and the "second" unit vectors?

${\hat{e}}_3\equiv {\hat{e}}_1\times {\hat{e}}_2$	$=$	$\hat{ı}[(e_{1y})(e_{2z})-(e_{2y})(e_{1z}))]+\hat{ȷ}[(e_{1z})(e_{2x})-(e_{2z})(e_{1x}))]+\hat{k}[(e_{1x})(e_{2y})-(e_{2x})(e_{1y}))]$
	$=$	$\frac{(x{q}^{2}y{p}^{2}z){\ell }_{3D}}{{𝒟}}\left\{\hat{ı}\right[(-\frac{2q^{2}y}{p^{2}z})-\left(\frac{p^{2}z}{q^{2}y}\right)]+\hat{ȷ}[\left(\frac{p^{2}z}{x}\right)-(-\frac{2x}{p^{2}z})]+\hat{k}[\left(\frac{x}{q^{2}y}\right)-\left(\frac{q^{2}y}{x}\right)\left]\right\}$
	$=$	$\frac{(x{q}^{2}y{p}^{2}z){\ell }_{3D}}{{𝒟}}\{-\hat{ı}[\frac{2q^4{y}^2+p^4{z}^2}{q^{2}y{p}^{2}z}]+\hat{ȷ}[\frac{p^4{z}^2+2x^2}{x{p}^{2}z}]+\hat{k}[\frac{x^2-q^4{y}^2}{x{q}^{2}y}\left]\right\}$
	$=$	$\frac{{\ell }_{3D}}{{𝒟}}\{-\hat{ı}[x(2q^4{y}^2+p^4{z}^2)]+\hat{ȷ}[q^{2}y(p^4{z}^2+2x^2)]+\hat{k}[p^{2}z(x^2-q^4{y}^2)\left]\right\}.$

Is this a valid unit vector? First, note that …

$(\frac{{\ell }_{3D}}{{𝒟}}{)}^{-2}$	$=$	$(q^4{y}^2{p}^4{z}^2+x^2{p}^4{z}^2+4x^2{q}^4{y}^2)(x^2+q^4{y}^2+p^4{z}^2)$
	$=$	$(x^2{q}^4{y}^2{p}^4{z}^2+x^4{p}^4{z}^2+4x^4{q}^4{y}^2)+(q^8{y}^4{p}^4{z}^2+x^2{q}^4{y}^2{p}^4{z}^2+4x^2{q}^8{y}^4)+(q^4{y}^2{p}^8{z}^4+x^2{p}^8{z}^4+4x^2{q}^4{y}^2{p}^4{z}^2)$
	$=$	$6x^2{q}^4{y}^2{p}^4{z}^2+x^4(p^4{z}^2+4q^4{y}^2)+q^8{y}^4(p^4{z}^2+4x^2)+p^8{z}^4(x^2+q^4{y}^2).$

Then we have,

$(\frac{{\ell }_{3D}}{{𝒟}}{)}^{-2}{\hat{e}}_3\cdot {\hat{e}}_3$	$=$	$[x(2q^4{y}^2+p^4{z}^2){]}^2+[q^{2}y(p^4{z}^2+2x^2){]}^2+[p^{2}z(x^2-q^4{y}^2){]}^2$
	$=$	$x^2(4q^8{y}^4+4q^4{y}^2{p}^4{z}^2+p^8{z}^4)+q^4{y}^2(p^8{z}^4+4x^2{p}^4{z}^2+4x^4)+p^4{z}^2(x^4-2x^2{q}^4{y}^2+q^8{y}^4)$
	$=$	$4x^2{q}^8{y}^4+4x^2{q}^4{y}^2{p}^4{z}^2+x^2{p}^8{z}^4+q^4{y}^2{p}^8{z}^4+4x^2{q}^4{y}^2{p}^4{z}^2+4x^4{q}^4{y}^2+x^4{p}^4{z}^2-2x^2{q}^4{y}^2{p}^4{z}^2+q^8{y}^4{p}^4{z}^2$
	$=$	$6x^2{q}^4{y}^2{p}^4{z}^2+p^8{z}^4(x^2+q^4{y}^2)+x^4(4q^4{y}^2+p^4{z}^2)+q^8{y}^4(4x^2+p^4{z}^2)$
	$=$	$(\frac{{\ell }_{3D}}{{𝒟}}{)}^{-2},$

which means that, ${\hat{e}}_3\cdot {\hat{e}}_3=1$.    Hooray! Again (11/11/2020)!

Direction Cosine Components for T6 Coordinates
$n$	${\lambda }_n$	$h_n$	$\frac{\partial {\lambda }_n}{\partial x}$	$\frac{\partial {\lambda }_n}{\partial y}$	$\frac{\partial {\lambda }_n}{\partial z}$	${\gamma }_{n1}$	${\gamma }_{n2}$	${\gamma }_{n3}$
$1$	$(x^2+q^2{y}^2+p^2{z}^2{)}^{1/2}$	${\lambda }_1{\ell }_{3D}$	$\frac{x}{{\lambda }_1}$	$\frac{q^{2}y}{{\lambda }_1}$	$\frac{p^{2}z}{{\lambda }_1}$	$(x){\ell }_{3D}$	$(q^{2}y){\ell }_{3D}$	$(p^{2}z){\ell }_{3D}$
$2$	$\frac{x{y}^{1/q^2}}{z^{2/p^2}}$	$\frac{1}{{\lambda }_2}\left[\frac{x{q}^{2}y{p}^{2}z}{{𝒟}}\right]$	$\frac{{\lambda }_2}{x}$	$\frac{{\lambda }_2}{q^{2}y}$	$-\frac{2{\lambda }_2}{p^{2}z}$	$\frac{x{q}^{2}y{p}^{2}z}{{𝒟}}\left(\frac{1}{x}\right)$	$\frac{x{q}^{2}y{p}^{2}z}{{𝒟}}\left(\frac{1}{q^{2}y}\right)$	$\frac{x{q}^{2}y{p}^{2}z}{{𝒟}}(-\frac{2}{p^{2}z})$
$3$	---	---	---	---	---	$-\frac{{\ell }_{3D}}{{𝒟}}[x(2q^4{y}^2+p^4{z}^2)]$	$\frac{{\ell }_{3D}}{{𝒟}}[q^{2}y(p^4{z}^2+2x^2)]$	$\frac{{\ell }_{3D}}{{𝒟}}[p^{2}z(x^2-q^4{y}^2)]$
${\ell }_{3D}$ $\equiv$ $(x^2+q^4{y}^2+p^4{z}^2{)}^{-1/2},$ ${𝒟}$ $\equiv$ $(q^4{y}^2{p}^4{z}^2+x^2{p}^4{z}^2+4x^2{q}^4{y}^2{)}^{1/2}.$

Let's double-check whether this "third" unit vector is orthogonal to both the "first" and the "second" unit vectors.

${\hat{e}}_1\cdot {\hat{e}}_3$	$=$	$\frac{{\ell }_{3D}^2}{{𝒟}}\{-x[x(2q^4{y}^2+p^4{z}^2)]+q^{2}y[q^{2}y(p^4{z}^2+2x^2)]+p^{2}z[p^{2}z(x^2-q^4{y}^2)\left]\right\}$
	$=$	$\frac{{\ell }_{3D}^2}{{𝒟}}\{-(2x^2{q}^4{y}^2+x^2{p}^4{z}^2)+(q^4{y}^2{p}^4{z}^2+2x^2{q}^4{y}^2)+(x^2{p}^4{z}^2-q^4{y}^2{p}^4{z}^2)\}$
	$=$	$0,$

and,

${\hat{e}}_2\cdot {\hat{e}}_3$	$=$	$\frac{{\ell }_{3D}}{{𝒟}}\cdot \frac{x{q}^{2}y{p}^{2}z}{{𝒟}}\{-[(2q^4{y}^2+p^4{z}^2)]+[(p^4{z}^2+2x^2)]-[2(x^2-q^4{y}^2)\left]\right\}$
	$=$	$0.$

Q. E. D.

Search for Third Coordinate Expression

Let's try …

${\lambda }_3$	$=$	${{𝒟}}^n{\ell }_{3D}^m$
	$=$	$(q^4{y}^2{p}^4{z}^2+x^2{p}^4{z}^2+4x^2{q}^4{y}^2{)}^{n/2}(x^2+q^4{y}^2+p^4{z}^2{)}^{-m/2}$
$\Rightarrow \frac{\partial {\lambda }_3}{\partial x_i}$	$=$	${\ell }_{3D}^m\left[\frac{n}{2}\right(q^4{y}^2{p}^4{z}^2+x^2{p}^4{z}^2+4x^2{q}^4{y}^2{)}^{n/2-1}\left]\frac{\partial }{\partial x_i}\right[(q^4{y}^2{p}^4{z}^2+x^2{p}^4{z}^2+4x^2{q}^4{y}^2)]$
		$+{{𝒟}}^n[-\frac{m}{2}(x^2+q^4{y}^2+p^4{z}^2{)}^{-m/2-1}\left]\frac{\partial }{\partial x_i}\right[(x^2+q^4{y}^2+p^4{z}^2)]$
	$=$	${{𝒟}}^n{\ell }_{3D}^m\left[\frac{n}{2}\right(q^4{y}^2{p}^4{z}^2+x^2{p}^4{z}^2+4x^2{q}^4{y}^2{)}^{-1}\left]\frac{\partial }{\partial x_i}\right[(q^4{y}^2{p}^4{z}^2+x^2{p}^4{z}^2+4x^2{q}^4{y}^2)]$
		$+{{𝒟}}^n{\ell }_{3D}^m[-\frac{m}{2}(x^2+q^4{y}^2+p^4{z}^2{)}^{-1}\left]\frac{\partial }{\partial x_i}\right[(x^2+q^4{y}^2+p^4{z}^2)].$

Hence,

$\frac{1}{{{𝒟}}^n{\ell }_{3D}^m}\cdot \frac{\partial {\lambda }_3}{\partial x}$	$=$	$\frac{n}{2{{𝒟}}^2}\frac{\partial }{\partial x}[q^4{y}^2{p}^4{z}^2+x^2{p}^4{z}^2+4x^2{q}^4{y}^2]-\frac{m{\ell }_{3D}^2}{2}\frac{\partial }{\partial x}[x^2+q^4{y}^2+p^4{z}^2]$
	$=$	$x\left\{\frac{n}{{{𝒟}}^2}\right[p^4{z}^2+4q^4{y}^2]-m{\ell }_{3D}^2\}$
	$=$	$x\{\frac{n(p^4{z}^2+4q^4{y}^2)}{(q^4{y}^2{p}^4{z}^2+x^2{p}^4{z}^2+4x^2{q}^4{y}^2)}-\frac{m}{(x^2+q^4{y}^2+p^4{z}^2)}\}$

This is overly cluttered! Let's try, instead …

$A\equiv {\ell }_{3D}^{-2}$ $=$ $(x^2+q^4{y}^2+p^4{z}^2),$       and,       $B\equiv {{𝒟}}^2$ $=$ $(q^4{y}^2{p}^4{z}^2+x^2{p}^4{z}^2+4x^2{q}^4{y}^2)$ $\Rightarrow \frac{\partial A}{\partial x}$ $=$ $2x,$ $\frac{\partial A}{\partial y}$ $=$ $2q^{4}y,$ $\frac{\partial A}{\partial z}$ $=$ $2p^{4}z;$ $\frac{\partial B}{\partial x}$ $=$ $2x(4q^4{y}^2+p^4{z}^2),$ $\frac{\partial B}{\partial y}$ $=$ $2q^{4}y(p^4{z}^2+4x^2),$ $\frac{\partial B}{\partial z}$ $=$ $2p^{4}z(q^4{y}^2+x^2).$

Now, let's assume that,

${\lambda }_3$	$\equiv$	$(\frac{A}{B}{)}^{1/2},$
$\Rightarrow \frac{\partial {\lambda }_3}{\partial x_i}$	$=$	$\frac{1}{2(AB{)}^{1/2}}\cdot \frac{\partial A}{\partial x_i}-\frac{A^{1/2}}{2B^{3/2}}\cdot \frac{\partial B}{\partial x_i}$
	$=$	$\frac{{\lambda }_3}{2AB}[B\cdot \frac{\partial A}{\partial x_i}-A\cdot \frac{\partial B}{\partial x_i}].$
$\Rightarrow \left[\frac{2AB}{{\lambda }_3}\right]\frac{\partial {\lambda }_3}{\partial x_i}$	$=$	$(q^4{y}^2{p}^4{z}^2+x^2{p}^4{z}^2+4x^2{q}^4{y}^2)\cdot \frac{\partial A}{\partial x_i}-(x^2+q^4{y}^2+p^4{z}^2)\cdot \frac{\partial B}{\partial x_i}.$

Looking ahead … $h_3^{-2}$ $=$ $\left\{\frac{{\lambda }_3}{2AB}\right[B\cdot \frac{\partial A}{\partial x}-A\cdot \frac{\partial B}{\partial x}]{\}}^2+\{\frac{{\lambda }_3}{2AB}[B\cdot \frac{\partial A}{\partial y}-A\cdot \frac{\partial B}{\partial y}]{\}}^2+\left\{\frac{{\lambda }_3}{2AB}\right[B\cdot \frac{\partial A}{\partial z}-A\cdot \frac{\partial B}{\partial z}]{\}}^2$ $\Rightarrow [\frac{2AB}{{\lambda }_3}{]}^2{h}_3^{-2}$ $=$ $[B\cdot \frac{\partial A}{\partial x}-A\cdot \frac{\partial B}{\partial x}{]}^2+[B\cdot \frac{\partial A}{\partial y}-A\cdot \frac{\partial B}{\partial y}{]}^2+[B\cdot \frac{\partial A}{\partial z}-A\cdot \frac{\partial B}{\partial z}{]}^2$ $\Rightarrow \left[\frac{{\lambda }_3}{2AB}\right]h_3$ $=$ $\left\{\right[B\cdot \frac{\partial A}{\partial x}-A\cdot \frac{\partial B}{\partial x}{]}^2+[B\cdot \frac{\partial A}{\partial y}-A\cdot \frac{\partial B}{\partial y}{]}^2+[B\cdot \frac{\partial A}{\partial z}-A\cdot \frac{\partial B}{\partial z}{]}^2{\}}^{-1/2}$ Then, for example, ${\gamma }_{31}\equiv h_3\left(\frac{\partial {\lambda }_3}{\partial x}\right)$ $=$ $[B\cdot \frac{\partial A}{\partial x}-A\cdot \frac{\partial B}{\partial x}]\left\{\right[B\cdot \frac{\partial A}{\partial x}-A\cdot \frac{\partial B}{\partial x}{]}^2+[B\cdot \frac{\partial A}{\partial y}-A\cdot \frac{\partial B}{\partial y}{]}^2+[B\cdot \frac{\partial A}{\partial z}-A\cdot \frac{\partial B}{\partial z}{]}^2{\}}^{-1/2}$

As a result, we have,

$\Rightarrow \left[\frac{2AB}{{\lambda }_3}\right]\frac{\partial {\lambda }_3}{\partial x}$	$=$	$2x[(q^4{y}^2{p}^4{z}^2+x^2{p}^4{z}^2+4x^2{q}^4{y}^2)-(x^2+q^4{y}^2+p^4{z}^2)(4q^4{y}^2+p^4{z}^2)]$
	$=$	$2x[(q^4{y}^2{p}^4{z}^2+x^2{p}^4{z}^2+4x^2{q}^4{y}^2)-(4x^2{q}^4{y}^2+x^2{p}^4{z}^2+4q^8{y}^4+q^4{y}^2{p}^4{z}^2+4q^4{y}^2{p}^4{z}^2+p^8{z}^4)]$
	$=$	$2x[-(4q^8{y}^4+4q^4{y}^2{p}^4{z}^2+p^8{z}^4)]$
	$=$	$-2x(2q^4{y}^2+p^4{z}^2{)}^2$
	$=$	$-8x(q^4{y}^2+\frac{p^4{z}^2}{2}{)}^2$
$\Rightarrow \left[AB\right]\frac{\partial \ln {\lambda }_3}{\partial \ln x}$	$=$	$-\left[2x\right(q^4{y}^2+\frac{p^4{z}^2}{2}){]}^2;$

and,

$\Rightarrow \left[\frac{2AB}{{\lambda }_3}\right]\frac{\partial {\lambda }_3}{\partial y}$	$=$	$2q^{4}y[(q^4{y}^2{p}^4{z}^2+x^2{p}^4{z}^2+4x^2{q}^4{y}^2)-(x^2+q^4{y}^2+p^4{z}^2)(p^4{z}^2+4x^2)]$
	$=$	$2q^{4}y[(q^4{y}^2{p}^4{z}^2+x^2{p}^4{z}^2+4x^2{q}^4{y}^2)-(x^2{p}^4{z}^2+4x^4+q^4{y}^2{p}^4{z}^2+4x^2{q}^4{y}^2+p^8{z}^4+4x^2{p}^4{z}^2)]$
	$=$	$-2q^{4}y(4x^4+p^8{z}^4+4x^2{p}^4{z}^2)$
	$=$	$-2q^{4}y(2x^2+p^4{z}^2{)}^2$
$\Rightarrow \left[AB\right]\frac{\partial \ln {\lambda }_3}{\partial \ln y}$	$=$	$-\left[2q^{2}y\right(x^2+\frac{p^4{z}^2}{2}){]}^2;$

and,

$\Rightarrow \left[\frac{2AB}{{\lambda }_3}\right]\frac{\partial {\lambda }_3}{\partial z}$	$=$	$2p^{4}z[(q^4{y}^2{p}^4{z}^2+x^2{p}^4{z}^2+4x^2{q}^4{y}^2)-(x^2+q^4{y}^2+p^4{z}^2)(q^4{y}^2+x^2)]$
	$=$	$2p^{4}z[(q^4{y}^2{p}^4{z}^2+x^2{p}^4{z}^2+4x^2{q}^4{y}^2)-(x^2{q}^4{y}^2+x^4+q^8{y}^4+x^2{q}^4{y}^2+q^4{y}^2{p}^4{z}^2+x^2{p}^4{z}^2)]$
	$=$	$2p^{4}z[(2x^2{q}^4{y}^2)-(x^4+q^8{y}^4)]$
	$=$	$-2p^{4}z[x^4+q^8{y}^4-2x^2{q}^4{y}^2]$
	$=$	$-2p^{4}z(x^2-q^4{y}^2{)}^2$
$\Rightarrow \left[AB\right]\frac{\partial \ln {\lambda }_3}{\partial \ln z}$	$=$	$-4\left[\right(\frac{p^4{z}^2}{4})(x^2-q^4{y}^2{)}^2]$
	$=$	$-\left[2\right(\frac{p^{2}z}{2})(x^2-q^4{y}^2){]}^2.$

Wow!   Really close! (13 November 2020)

Just for fun, let's see what we get for $h_3$. It is given by the expression,

$h_3^{-2}$	$=$	$(\frac{\partial {\lambda }_3}{\partial x}{)}^2+(\frac{\partial {\lambda }_3}{\partial y}{)}^2+(\frac{\partial {\lambda }_3}{\partial z}{)}^2$
	$=$	$\left\{\frac{{\lambda }_3}{ABx}\right[2x(q^4{y}^2+\frac{p^4{z}^2}{2}){]}^2{\}}^2+\left\{\frac{{\lambda }_3}{ABy}\right[2q^{2}y(x^2+\frac{p^4{z}^2}{2}){]}^2{\}}^2+\left\{\frac{{\lambda }_3}{ABz}\right[2\left(\frac{p^{2}z}{2}\right)(x^2-q^4{y}^2){]}^2{\}}^2$
$\Rightarrow [\frac{AB}{{\lambda }_3}{]}^2{h}_3^{-2}$	$=$	$\left\{\frac{1}{x^2}\right[2x(q^4{y}^2+\frac{p^4{z}^2}{2}){]}^4\}+\{\frac{1}{y^2}\left[2q^{2}y\right(x^2+\frac{p^4{z}^2}{2}\left){]}^4\right\}+\left\{\frac{1}{z^2}\right[2\left(\frac{p^{2}z}{2}\right)(x^2-q^4{y}^2){]}^4\}$

Fiddle Around

Let …

${{ℒ}}_x$	$\equiv$	$-[B\cdot \frac{\partial A}{\partial x}-A\cdot \frac{\partial B}{\partial x}]$	$=$	$8x(q^4{y}^2+\frac{p^4{z}^2}{2}{)}^2$	$=$	$\frac{2}{x}\left[2x\right(q^4{y}^2+\frac{p^4{z}^2}{2}){]}^2$	$=$	$8x{\mathfrak{𝔉}}_x(y,z)$
${{ℒ}}_y$	$\equiv$	$-[B\cdot \frac{\partial A}{\partial y}-A\cdot \frac{\partial B}{\partial y}]$	$=$	$8q^{4}y(x^2+\frac{p^4{z}^2}{2}{)}^2$	$=$	$\frac{2}{y}\left[2q^{2}y\right(x^2+\frac{p^4{z}^2}{2}){]}^2$	$=$	$8y{\mathfrak{𝔉}}_y(x,z)$
${{ℒ}}_z$	$\equiv$	$-[B\cdot \frac{\partial A}{\partial z}-A\cdot \frac{\partial B}{\partial z}]$	$=$	$2p^{4}z(x^2-q^4{y}^2{)}^2$	$=$	$\frac{2}{z}\left[p^{2}z\right(x^2-q^4{y}^2){]}^2$	$=$	$8z{\mathfrak{𝔉}}_z(x,y)$

With this shorthand in place, we can write,

${\hat{e}}_3$	$=$	$\frac{{\ell }_{3D}}{{𝒟}}\{-\hat{ı}[x(2q^4{y}^2+p^4{z}^2)]+\hat{ȷ}[q^{2}y(p^4{z}^2+2x^2)]+\hat{k}[p^{2}z(x^2-q^4{y}^2)\left]\right\}$
	$=$	$\frac{1}{(AB{)}^{1/2}}\{-\hat{ı}[\frac{x{{ℒ}}_x}{2}{]}^{1/2}+\hat{ȷ}[\frac{y{{ℒ}}_y}{2}{]}^{1/2}+\hat{k}[\frac{z{{ℒ}}_z}{2}{]}^{1/2}\}.$

We therefore also recognize that,

$h_3\left(\frac{\partial {\lambda }_3}{\partial x}\right)$	$=$	$-\frac{1}{(AB{)}^{1/2}}[\frac{x{{ℒ}}_x}{2}{]}^{1/2}$	$=$	$-\frac{1}{(AB{)}^{1/2}}[4x^2{\mathfrak{𝔉}}_x{]}^{1/2},$
$h_3\left(\frac{\partial {\lambda }_3}{\partial y}\right)$	$=$	$\frac{1}{(AB{)}^{1/2}}[\frac{y{{ℒ}}_y}{2}{]}^{1/2}$	$=$	$\frac{1}{(AB{)}^{1/2}}[4y^2{\mathfrak{𝔉}}_y{]}^{1/2},$
$h_3\left(\frac{\partial {\lambda }_3}{\partial z}\right)$	$=$	$\frac{1}{(AB{)}^{1/2}}[\frac{z{{ℒ}}_z}{2}{]}^{1/2}$	$=$	$\frac{1}{(AB{)}^{1/2}}[4z^2{\mathfrak{𝔉}}_z{]}^{1/2}.$

Now, if — and it is a BIG "if" — $h_3=h_0(AB{)}^{-1/2}$, then we have,

$h_0\left(\frac{\partial {\lambda }_3}{\partial x}\right)$	$=$	$-[4x^2{\mathfrak{𝔉}}_x{]}^{1/2}$	$=$	$-2x[{\mathfrak{𝔉}}_x{]}^{1/2},$
$h_0\left(\frac{\partial {\lambda }_3}{\partial y}\right)$	$=$	$[4y^2{\mathfrak{𝔉}}_y{]}^{1/2}$	$=$	$2y[{\mathfrak{𝔉}}_y{]}^{1/2},$
$h_0\left(\frac{\partial {\lambda }_3}{\partial z}\right)$	$=$	$[4z^2{\mathfrak{𝔉}}_z{]}^{1/2}$	$=$	$2z[{\mathfrak{𝔉}}_z{]}^{1/2},$

$\Rightarrow h_0{\lambda }_3$

$=$

$-x^2[{\mathfrak{𝔉}}_x{]}^{1/2}+y^2[{\mathfrak{𝔉}}_y{]}^{1/2}+z^2[{\mathfrak{𝔉}}_z{]}^{1/2}.$

But if this is the correct expression for ${\lambda }_3$ and its three partial derivatives, then it must be true that,

$h_3^{-2}$	$=$	$(\frac{\partial {\lambda }_3}{\partial x}{)}^2+(\frac{\partial {\lambda }_3}{\partial y}{)}^2+(\frac{\partial {\lambda }_3}{\partial z}{)}^2$
$\Rightarrow (\frac{h_3}{h_0}{)}^{-2}$	$=$	$4x^2\left[{\mathfrak{𝔉}}_x\right]+4y^2\left[{\mathfrak{𝔉}}_y\right]+4z^2\left[{\mathfrak{𝔉}}_z\right]$
	$=$	$4x^2\left[\right(q^4{y}^2+\frac{p^4{z}^2}{2}{)}^2]+4y^2[q^4(x^2+\frac{p^4{z}^2}{2}{)}^2]+4z^2\left[\frac{p^4}{4}\right(x^2-q^4{y}^2{)}^2]$
	$=$	$x^2(2q^4{y}^2+p^4{z}^2{)}^2+q^4{y}^2(2x^2+p^4{z}^2{)}^2+p^4{z}^2(x^2-q^4{y}^2{)}^2$

Well … the right-hand side of this expression is identical to the right-hand side of the above expression, where we showed that it equals $({\ell }_{3D}/{𝒟}{)}^{-2}$. That is to say, we are now showing that,

$(\frac{h_3}{h_0}{)}^{-2}$	$=$	$(\frac{{\ell }_{3D}}{{𝒟}}{)}^{-2}=[AB]$
$\Rightarrow \frac{h_3}{h_0}$	$=$	$(AB{)}^{-1/2}.$

And this is precisely what, just a few lines above, we hypothesized the functional expression for $h_3$ ought to be. EUREKA!

Summary

In summary, then …

${\lambda }_3$	$\equiv$	$-x^2[{\mathfrak{𝔉}}_x{]}^{1/2}+y^2[{\mathfrak{𝔉}}_y{]}^{1/2}+z^2[{\mathfrak{𝔉}}_z{]}^{1/2}$
	$=$	$-x^2\left[\right(q^4{y}^2+\frac{p^4{z}^2}{2}{)}^2{]}^{1/2}+y^2\left[q^4\right(x^2+\frac{p^4{z}^2}{2}{)}^2{]}^{1/2}+z^2\left[\frac{p^4}{4}\right(x^2-q^4{y}^2{)}^2{]}^{1/2}$
	$=$	$-x^2(q^4{y}^2+\frac{p^4{z}^2}{2})+q^2{y}^2(x^2+\frac{p^4{z}^2}{2})+\frac{p^2{z}^2}{2}(x^2-q^4{y}^2),$

and,

$h_3$	$=$	$(AB{)}^{-1/2}$
	$=$	$[(x^2+q^4{y}^2+p^4{z}^2)(q^4{y}^2{p}^4{z}^2+x^2{p}^4{z}^2+4x^2{q}^4{y}^2){]}^{-1/2}.$

No! Once again this does not work. The direction cosines — and, hence, the components of the ${\hat{e}}_3$ unit vector — are not correct!

Speculation7

Direction Cosine Components for T6 Coordinates
$n$	${\lambda }_n$	$h_n$	$\frac{\partial {\lambda }_n}{\partial x}$	$\frac{\partial {\lambda }_n}{\partial y}$	$\frac{\partial {\lambda }_n}{\partial z}$	${\gamma }_{n1}$	${\gamma }_{n2}$	${\gamma }_{n3}$
$1$	$(x^2+q^2{y}^2+p^2{z}^2{)}^{1/2}$	${\lambda }_1{\ell }_{3D}$	$\frac{x}{{\lambda }_1}$	$\frac{q^{2}y}{{\lambda }_1}$	$\frac{p^{2}z}{{\lambda }_1}$	$(x){\ell }_{3D}$	$(q^{2}y){\ell }_{3D}$	$(p^{2}z){\ell }_{3D}$
$2$	$\frac{x{y}^{1/q^2}}{z^{2/p^2}}$	$\frac{1}{{\lambda }_2}\left[\frac{x{q}^{2}y{p}^{2}z}{{𝒟}}\right]$	$\frac{{\lambda }_2}{x}$	$\frac{{\lambda }_2}{q^{2}y}$	$-\frac{2{\lambda }_2}{p^{2}z}$	$\frac{x{q}^{2}y{p}^{2}z}{{𝒟}}\left(\frac{1}{x}\right)$	$\frac{x{q}^{2}y{p}^{2}z}{{𝒟}}\left(\frac{1}{q^{2}y}\right)$	$\frac{x{q}^{2}y{p}^{2}z}{{𝒟}}(-\frac{2}{p^{2}z})$
$3$	---	---	---	---	---	$-\frac{{\ell }_{3D}}{{𝒟}}[x(2q^4{y}^2+p^4{z}^2)]$	$\frac{{\ell }_{3D}}{{𝒟}}[q^{2}y(p^4{z}^2+2x^2)]$	$\frac{{\ell }_{3D}}{{𝒟}}[p^{2}z(x^2-q^4{y}^2)]$
${\ell }_{3D}$ $\equiv$ $(x^2+q^4{y}^2+p^4{z}^2{)}^{-1/2},$ ${𝒟}$ $\equiv$ $(q^4{y}^2{p}^4{z}^2+x^2{p}^4{z}^2+4x^2{q}^4{y}^2{)}^{1/2}.$

On my white-board I have shown that, if

${\lambda }_3\equiv {\ell }_{3D}{𝒟},$

then everything will work out as long as,

${ℒ}$

$=$

$(\frac{{𝒟}}{{\ell }_{3D}}{)}^2\frac{1}{{\ell }_{3D}^4},$

where,

${ℒ}$	$\equiv$	$x^2(2q^4{y}^2+p^4{z}^2{)}^4+q^8{y}^2(2x^2+p^4{z}^2{)}^4+p^8{z}^2(x^2-q^4{y}^2{)}^4$
	$=$	$x^2(4q^8{y}^4+4q^4{y}^2{p}^4{z}^2+p^8{z}^4{)}^2+q^8{y}^2(4x^4+4x^2{p}^4{z}^2+p^8{z}^4{)}^2+p^8{z}^2(x^4-2x^2{q}^4{y}^2+q^8{y}^4{)}^2$
	$=$	$x^2[16q^{16}{y}^8+16q^{12}{y}^6{p}^4{z}^2+4q^8{y}^4{p}^8{z}^4+16q^{12}{y}^6{p}^4{z}^2+16q^8{y}^4{p}^8{z}^4+4q^4{y}^2{p}^{12}{z}^6+4q^8{y}^4{p}^8{z}^4+4q^4{y}^2{p}^{12}{z}^6+p^{16}{z}^8]$
		$+q^8{y}^2[16x^8+16x^6{p}^4{z}^2+4x^4{p}^8{z}^4+16x^6{p}^4{z}^2+16x^4{p}^8{z}^4+4x^2{p}^{12}{z}^6+4x^4{p}^8{z}^4+4x^2{p}^{12}{z}^6+p^{16}{z}^8]$
		$+p^8{z}^2[x^8-2x^6{q}^4{y}^2+x^4{q}^8{y}^4-2x^6{q}^4{y}^2+4x^4{q}^8{y}^4-2x^2{q}^{12}{y}^6+x^4{q}^8{y}^4-2x^2{q}^{12}{y}^6+q^{16}{y}^8]$
	$=$	$x^2[16q^{16}{y}^8+32q^{12}{y}^6{p}^4{z}^2+24q^8{y}^4{p}^8{z}^4+8q^4{y}^2{p}^{12}{z}^6+p^{16}{z}^8]$
		$+q^8{y}^2[16x^8+32x^6{p}^4{z}^2+24x^4{p}^8{z}^4+8x^2{p}^{12}{z}^6+p^{16}{z}^8]$
		$+p^8{z}^2[x^8-4x^6{q}^4{y}^2+6x^4{q}^8{y}^4-4x^2{q}^{12}{y}^6+q^{16}{y}^8]$

Let's check this out.

$\mathrm{R}\mathrm{H}\mathrm{S}\equiv (\frac{{𝒟}}{{\ell }_{3D}}{)}^2\frac{1}{{\ell }_{3D}^4}$	$=$	$(q^4{y}^2{p}^4{z}^2+x^2{p}^4{z}^2+4x^2{q}^4{y}^2)(x^2+q^4{y}^2+p^4{z}^2{)}^3$
	$=$	$(x^2+q^4{y}^2+p^4{z}^2)(q^4{y}^2{p}^4{z}^2+x^2{p}^4{z}^2+4x^2{q}^4{y}^2)[x^4+2x^2{q}^4{y}^2+2x^2{p}^4{z}^2+q^8{y}^4+2q^4{y}^2{p}^4{z}^2+p^8{z}^4]$
	$=$	$[(6x^2{q}^4{y}^2{p}^4{z}^2+x^4{p}^4{z}^2+4x^4{q}^4{y}^2)+(q^8{y}^4{p}^4{z}^2+4x^2{q}^8{y}^4)+(q^4{y}^2{p}^8{z}^4+x^2{p}^8{z}^4)]$
		$\times [x^4+2x^2{q}^4{y}^2+2x^2{p}^4{z}^2+q^8{y}^4+2q^4{y}^2{p}^4{z}^2+p^8{z}^4]$
	$=$	$[6x^2{q}^4{y}^2{p}^4{z}^2+x^4(p^4{z}^2+4q^4{y}^2)+q^8{y}^4(p^4{z}^2+4x^2)+p^8{z}^4(q^4{y}^2+x^2)]$
		$\times [x^4+2x^2{q}^4{y}^2+2x^2{p}^4{z}^2+q^8{y}^4+2q^4{y}^2{p}^4{z}^2+p^8{z}^4]$