Appendix/Ramblings/ConcentricEllipsoidalT8Coordinates

Concentric Ellipsoidal (T8) Coordinates

Background

Building on our general introduction to Direction Cosines in the context of orthogonal curvilinear coordinate systems, and on our previous development of T3 (concentric oblate-spheroidal) and T5 (concentric elliptic) coordinate systems, here we explore the creation of a concentric ellipsoidal (T8) coordinate system. This is motivated by our desire to construct a fully analytically prescribable model of a nonuniform-density ellipsoidal configuration that is an analog to Riemann S-Type ellipsoids.

Note that, in a separate but closely related discussion, we made attempts to define this coordinate system, numbering the trials up through "T7." In this "T7" effort, we were able to define a set of three, mutually orthogonal unit vectors that should work to define a fully three-dimensional, concentric ellipsoidal coordinate system. But we were unable to figure out what coordinate function, ${\lambda }_3(x,y,z)$, was associated with the third unit vector. In addition, we found the ${\lambda }_2$ coordinate to be rather strange in that it was not oriented in a manner that resembled the classic spherical coordinate system. Here we begin by redefining the ${\lambda }_2$ coordinate such that its associated ${\hat{e}}_3$ unit vector lies parallel to the x-y plane.

Realigning the Second Coordinate

The first coordinate remains the same as before, namely,

${\lambda }_1^2$

$=$

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

This may be rewritten as,

$1$

$=$

$(\frac{x}{a}{)}^2+(\frac{y}{b}{)}^2+(\frac{z}{c}{)}^2,$

where,

$a={\lambda }_1,$

$b=\frac{{\lambda }_1}{q},$

$c=\frac{{\lambda }_1}{p}.$

By specifying the value of $z=z_0<c$, as well as the value of ${\lambda }_1$, we are picking a plane that lies parallel to, but a distance $z_0$ above, the equatorial plane. The elliptical curve that defines the intersection of the ${\lambda }_1$-constant surface with this plane is defined by the expression,

${\lambda }_1^2-p^2{z}_0^2$	$=$	$x^2+q^2{y}^2$
$\Rightarrow 1$	$=$	$(\frac{x}{a_{2D}}{)}^2+(\frac{y}{b_{2D}}{)}^2,$

where,

$a_{2D}=({\lambda }_1^2-p^2{z}_0^2{)}^{1/2},$

$b_{2D}=\frac{1}{q}({\lambda }_1^2-p^2{z}_0^2{)}^{1/2}.$

At each point along this elliptic curve, the line that is tangent to the curve has a slope that can be determined by simply differentiating the equation that describes the curve, that is,

$0$	$=$	$\frac{2xdx}{a_{2D}^2}+\frac{2ydy}{b_{2D}^2}$
$\Rightarrow \frac{dy}{dx}$	$=$	$-\frac{2x}{a_{2D}^2}\cdot \frac{b_{2D}^2}{2y}=-\frac{x}{q^{2}y}.$
$\Rightarrow \mathrm{\Delta }y$	$=$	$-\left(\frac{x}{q^{2}y}\right)\mathrm{\Delta }x.$

The unit vector that lies tangent to any point on this elliptical curve will be described by the expression,

${\hat{e}}_2$	$=$	$\hat{ı}\left\{\frac{\mathrm{\Delta }x}{[(\mathrm{\Delta }x{)}^2+(\mathrm{\Delta }y{)}^2{]}^{1/2}}\right\}+\hat{ȷ}\left\{\frac{\mathrm{\Delta }y}{[(\mathrm{\Delta }x{)}^2+(\mathrm{\Delta }y{)}^2{]}^{1/2}}\right\}$
	$=$	$\hat{ı}\left\{\frac{1}{[1+x^2/(q^4{y}^2){]}^{1/2}}\right\}-\hat{ȷ}\left\{\frac{x/(q^{2}y)}{[1+x^2/(q^4{y}^2){]}^{1/2}}\right\}$
	$=$	$\hat{ı}\left\{\frac{q^{2}y}{[x^2+q^4{y}^2{]}^{1/2}}\right\}-\hat{ȷ}\left\{\frac{x}{[x^2+q^4{y}^2{]}^{1/2}}\right\}.$

As we have discovered, the coordinate that gives rise to this unit vector is,

${\lambda }_2$

$=$

$\frac{x}{y^{1/q^2}}.$

Other properties that result from this definition of ${\lambda }_2$ are presented in the following table.

Direction Cosine Components for T8 Coordinates
$n$	${\lambda }_n$	$h_n$	$\frac{\partial {\lambda }_n}{\partial x}$	$\frac{\partial {\lambda }_n}{\partial y}$	$\frac{\partial {\lambda }_n}{\partial z}$	${\gamma }_{n1}$	${\gamma }_{n2}$	${\gamma }_{n3}$
$1$	$(x^2+q^2{y}^2+p^2{z}^2{)}^{1/2}$	${\lambda }_1{\ell }_{3D}$	$\frac{x}{{\lambda }_1}$	$\frac{q^{2}y}{{\lambda }_1}$	$\frac{p^{2}z}{{\lambda }_1}$	$(x){\ell }_{3D}$	$(q^{2}y){\ell }_{3D}$	$(p^{2}z){\ell }_{3D}$
$2$	$\frac{x}{y^{1/q^2}}$	$\frac{1}{{\lambda }_2}\left[\frac{x{q}^{2}y}{(x^2+q^4{y}^2{)}^{1/2}}\right]$	$\frac{{\lambda }_2}{x}$	$-\frac{{\lambda }_2}{q^{2}y}$	$0$	$\frac{q^{2}y}{(x^2+q^4{y}^2{)}^{1/2}}$	$-\frac{x}{(x^2+q^4{y}^2{)}^{1/2}}$	$0$
$3$	---	---	---	---	---	---	---	---
${\ell }_{3D}$ $\equiv$ $(x^2+q^4{y}^2+p^4{z}^2{)}^{-1/2}$

The associated unit vector is, then,

${\hat{e}}_2$

$=$

$\hat{ı}\left[\frac{q^{2}y}{(x^2+q^4{y}^2{)}^{1/2}}\right]-\hat{ȷ}\left[\frac{x}{(x^2+q^4{y}^2{)}^{1/2}}\right].$

It is easy to see that ${\hat{e}}_2\cdot {\hat{e}}_2=1$. We also see that,

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

$=$

$x{\ell }_{3D}\left[\frac{q^{2}y}{(x^2+q^4{y}^2{)}^{1/2}}\right]-q^{2}y{\ell }_{3D}\left[\frac{x}{(x^2+q^4{y}^2{)}^{1/2}}\right]=0,$

so it is clear that these first two unit vectors are orthogonal to one another.

Search for the Third Coordinate

Cross Product of First Two Unit Vectors

The cross-product of these two unit vectors should give the third, namely,

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

Inserting these component expressions into the last row of the T8 Direction Cosine table gives …

Direction Cosine Components for T8 Coordinates
$n$	${\lambda }_n$	$h_n$	$\frac{\partial {\lambda }_n}{\partial x}$	$\frac{\partial {\lambda }_n}{\partial y}$	$\frac{\partial {\lambda }_n}{\partial z}$	${\gamma }_{n1}$	${\gamma }_{n2}$	${\gamma }_{n3}$
$1$	$(x^2+q^2{y}^2+p^2{z}^2{)}^{1/2}$	${\lambda }_1{\ell }_{3D}$	$\frac{x}{{\lambda }_1}$	$\frac{q^{2}y}{{\lambda }_1}$	$\frac{p^{2}z}{{\lambda }_1}$	$(x){\ell }_{3D}$	$(q^{2}y){\ell }_{3D}$	$(p^{2}z){\ell }_{3D}$
$2$	$\frac{x}{y^{1/q^2}}$	$\frac{1}{{\lambda }_2}\left[\frac{x{q}^{2}y}{(x^2+q^4{y}^2{)}^{1/2}}\right]$	$\frac{{\lambda }_2}{x}$	$-\frac{{\lambda }_2}{q^{2}y}$	$0$	$\frac{q^{2}y}{(x^2+q^4{y}^2{)}^{1/2}}$	$-\frac{x}{(x^2+q^4{y}^2{)}^{1/2}}$	$0$
$3$	---	---	---	---	---	$\frac{x{p}^{2}z{\ell }_{3D}}{(x^2+q^4{y}^2{)}^{1/2}}$	$\frac{q^{2}y{p}^{2}z{\ell }_{3D}}{(x^2+q^4{y}^2{)}^{1/2}}$	$-\frac{(x^2+q^4{y}^2){\ell }_{3D}}{(x^2+q^4{y}^2{)}^{1/2}}$
${\ell }_{3D}$ $\equiv$ $(x^2+q^4{y}^2+p^4{z}^2{)}^{-1/2},$

Associated h₃ Scale Factor

Whiteboard EUREKA moment

After working through various scenarios on my whiteboard today (21 January 2021), I propose that,

$\frac{\partial {\lambda }_3}{\partial x}$

$=$

$\frac{x{p}^{2}z}{(x^2+q^4{y}^2)};$

$\frac{\partial {\lambda }_3}{\partial y}$

$=$

$\frac{q^{2}y{p}^{2}z}{(x^2+q^4{y}^2)};$

and

$\frac{\partial {\lambda }_3}{\partial z}$

$=$

$-1.$

This means that,

$h_3^{-2}$	$=$	$\sum _{i=1}^3(\frac{\partial {\lambda }_3}{\partial x_i}{)}^2$
	$=$	$[\frac{x{p}^{2}z}{(x^2+q^4{y}^2)}{]}^2+[\frac{q^{2}y{p}^{2}z}{(x^2+q^4{y}^2)}{]}^2+[-1{]}^2$
	$=$	$\frac{p^4{z}^2(x^2+q^4{y}^2)}{(x^2+q^4{y}^2{)}^2}+1$
	$=$	$\frac{(x^2+q^4{y}^2+p^4{z}^2)}{(x^2+q^4{y}^2)}$
	$=$	$\frac{1}{{\ell }_{3D}^2(x^2+q^4{y}^2)}$
$\Rightarrow h_3$	$=$	${\ell }_{3D}(x^2+q^4{y}^2{)}^{1/2}.$

This seems to work well because, when combined with the three separate expressions for $\partial {\lambda }_3/\partial x_i$, this single expression for $h_3$ generates all three components of the third unit vector, that is, all three direction cosines, ${\gamma }_{3i}$. All of the elements of this new "EUREKA moment" result have been entered into the following table.

Direction Cosine Components for T8 Coordinates
$n$	${\lambda }_n$	$h_n$	$\frac{\partial {\lambda }_n}{\partial x}$	$\frac{\partial {\lambda }_n}{\partial y}$	$\frac{\partial {\lambda }_n}{\partial z}$	${\gamma }_{n1}$	${\gamma }_{n2}$	${\gamma }_{n3}$
$1$	$(x^2+q^2{y}^2+p^2{z}^2{)}^{1/2}$	${\lambda }_1{\ell }_{3D}$	$\frac{x}{{\lambda }_1}$	$\frac{q^{2}y}{{\lambda }_1}$	$\frac{p^{2}z}{{\lambda }_1}$	$(x){\ell }_{3D}$	$(q^{2}y){\ell }_{3D}$	$(p^{2}z){\ell }_{3D}$
$2$	$\frac{x}{y^{1/q^2}}$	$\frac{1}{{\lambda }_2}\left[\frac{x{q}^{2}y}{(x^2+q^4{y}^2{)}^{1/2}}\right]$	$\frac{{\lambda }_2}{x}$	$-\frac{{\lambda }_2}{q^{2}y}$	$0$	$\frac{q^{2}y}{(x^2+q^4{y}^2{)}^{1/2}}$	$-\frac{x}{(x^2+q^4{y}^2{)}^{1/2}}$	$0$
$3$	---	${\ell }_{3D}(x^2+q^4{y}^2{)}^{1/2}$	$\frac{x{p}^{2}z}{(x^2+q^4{y}^2)}$	$\frac{q^{2}y{p}^{2}z}{(x^2+q^4{y}^2)}$	$-1$	$\frac{x{p}^{2}z{\ell }_{3D}}{(x^2+q^4{y}^2{)}^{1/2}}$	$\frac{q^{2}y{p}^{2}z{\ell }_{3D}}{(x^2+q^4{y}^2{)}^{1/2}}$	$-\frac{(x^2+q^4{y}^2){\ell }_{3D}}{(x^2+q^4{y}^2{)}^{1/2}}$
${\ell }_{3D}$ $\equiv$ $(x^2+q^4{y}^2+p^4{z}^2{)}^{-1/2}$

What is the Third Coordinate Function, λ₃

The remaining $64,000 question is, "What is the actual expression for ${\lambda }_3(x,y,z)$  ?  "

Notice that the (partial) derivatives of ${\lambda }_3$ with respect to $x$ and $y$ may be rewritten, respectively, in the form

$\left(\frac{q^{2}y}{p^{2}z}\right)\frac{\partial {\lambda }_3}{\partial x}$	$=$	$\frac{q^{2}y}{x(1+{\eta }^2)}=\frac{\eta }{(1+{\eta }^2)},$       and,
$\left(\frac{x}{p^{2}z}\right)\frac{\partial {\lambda }_3}{\partial y}$	$=$	$\frac{q^{2}y}{x(1+{\eta }^2)}=\frac{\eta }{(1+{\eta }^2)},$

where,

$\eta$

$\equiv$

$\frac{q^{2}y}{x}\Rightarrow \frac{\partial \ln \eta }{\partial \ln x}=-1,\frac{\partial \ln \eta }{\partial \ln y}=+1.$

Then, after searching through the CRC Mathematical Handbook's pages of familiar derivative expressions, we appreciate that

$\frac{d}{d{x}_i}\left[\frac{1}{\cosh \gamma }\right]$

$=$

$-\left[\frac{\tanh \gamma }{\cosh \gamma }\right]\frac{d\gamma }{d{x}_i}.$

Hence, it will be useful to adopt the mapping,

$\eta$

$\to$

$\sinh \gamma ,$

because the right-hand side of both partial-derivative expressions becomes,

$\frac{\eta }{(1+{\eta }^2)}$

$\to$

$\frac{\sinh \gamma }{{\cosh }^2\gamma }=\frac{\tanh \gamma }{\cosh \gamma }.$

Guess A

In particular, this suggests that we set,

${\lambda }_3$

$=$

$\frac{A}{\cosh \gamma },$

where,

$\gamma$

$\equiv$

${\sinh }^{-1}\eta =\pm {\cosh }^{-1}[{\eta }^2+1{]}^{1/2}.$

In other words,

${\lambda }_3$

$=$

$A[{\eta }^2+1{]}^{-1/2}.$

Let's check the first and second partial derivatives.

$\frac{\partial {\lambda }_3}{\partial x}$

$=$

$-\frac{A}{2}\left[\frac{2\eta }{({\eta }^2+1{)}^{3/2}}\right]\frac{\partial \eta }{\partial x}$

Guess B

What if,

${\lambda }_3$

$=$

$\frac{1}{2}\ln (1+{\eta }^2).$

Then,

$\frac{d{\lambda }_3}{d\eta }$

$=$

$\frac{\eta }{(1+{\eta }^2)},$

in which case we find,

$\frac{\partial {\lambda }_3}{\partial x_i}$

$=$

$\frac{d{\lambda }_3}{d\eta }\cdot \frac{\partial \eta }{\partial x_i}$

which means,

$\frac{\partial {\lambda }_3}{\partial x}$

$=$

$\frac{\eta }{(1+{\eta }^2)}[-\frac{q^{2}y}{x^2}]=-\frac{x}{(x^2+q^4{y}^2)}\left[\frac{q^4{y}^2}{x^2}\right].$

Guess C

What if,

${\lambda }_3$

$=$

$\frac{1}{2}\ln (1+{\eta }^{-2}).$

Then,

$\frac{d{\lambda }_3}{d\eta }$

$=$

$-\frac{{\eta }^{-3}}{(1+{\eta }^{-2})}=-\frac{{\eta }^{-1}}{(1+{\eta }^2)}$

in which case we find,

$\frac{\partial {\lambda }_3}{\partial x_i}$

$=$

$\frac{d{\lambda }_3}{d\eta }\cdot \frac{\partial \eta }{\partial x_i}$

which means,

$\frac{\partial {\lambda }_3}{\partial x}$

$=$

$-\frac{{\eta }^{-1}}{(1+{\eta }^2)}[-\frac{q^{2}y}{x^2}]=\frac{x}{(x^2+q^4{y}^2)};$

and,

$\frac{\partial {\lambda }_3}{\partial y}$

$=$

$-\frac{{\eta }^{-1}}{(1+{\eta }^2)}\left[\frac{q^2}{x}\right]=-\frac{x^2}{y(x^2+q^4{y}^2)}$

Inverting Coordinate Relations

In a Plane Perpendicular to the Z-Axis

General Case

At a fixed value of $z=z_0$, let's invert the ${\lambda }_1(x,y)$ and ${\lambda }_2(x,y)$ relations to obtain expressions for $x({\lambda }_1,{\lambda }_2)$ and $y({\lambda }_1,{\lambda }_2)$. Perhaps this will help us determine what the third coordinate expression should be.

We start with,

${\lambda }_1$	$\equiv$	$(x^2+q^2{y}^2+p^2{z}_0^2{)}^{1/2};$
${\lambda }_2$	$\equiv$	$\frac{x}{y^{1/q^2}}.$

This means that,

$\ln y$

$=$

$q^2(\ln x-\ln {\lambda }_2),$

and,

$q^2{y}^2$	$=$	$({\lambda }_1^2-p^2{z}_0^2)-x^2$
$\Rightarrow 2\ln y$	$=$	$\ln \left\{\frac{1}{q^2}\right[({\lambda }_1^2-p^2{z}_0^2)-x^2\left]\right\}.$

Together, this gives,

$\ln \left\{\frac{1}{q^2}\right[({\lambda }_1^2-p^2{z}_0^2)-x^2\left]\right\}$	$=$	$2q^2(\ln x-\ln {\lambda }_2)$
	$=$	$\ln [\frac{x}{{\lambda }_2}{]}^{2q^2}$
$\Rightarrow ({\lambda }_1^2-p^2{z}_0^2)-x^2$	$=$	$q^2[\frac{x}{{\lambda }_2}{]}^{2q^2}$
$\Rightarrow x^2+q^2[\frac{x}{{\lambda }_2}{]}^{2q^2}+(p^2{z}_0^2-{\lambda }_1^2)$	$=$	$0.$

What if Axisymmetric (q² = 1)

In an axisymmetric configuration, $q^2=1$ and $({\lambda }_1^2-p^2{z}_0^2)={\varpi }^2$, so this general expression for $x$ becomes,

$0$	$=$	$x^2+[\frac{x}{{\lambda }_2}{]}^2-{\varpi }^2$
	$=$	$x^2\left[\frac{1+{\lambda }_2^2}{{\lambda }_2^2}\right]-{\varpi }^2$
$\Rightarrow x$	$=$	$\varpi [\frac{{\lambda }_2^2}{1+{\lambda }_2^2}{]}^{1/2}.$

Given that, for axisymmetric systems,

$x$

$=$

$\varpi \cos \phi ,$

we conclude that when $q^2=1$,

$[\frac{{\lambda }_2^2}{1+{\lambda }_2^2}{]}^{1/2}$	$=$	$\cos \phi$
$\Rightarrow {\lambda }_2^2$	$=$	${\cot }^2\phi$

What if q² = 2

For example, if we choose $q^2=2$, we have a quadratic expression for $x^2$, namely,

$0$	$=$	$x^2+2[\frac{x}{{\lambda }_2}{]}^4+(p^2{z}_0^2-{\lambda }_1^2)$
$\Rightarrow 0$	$=$	$x^4+\frac{1}{2}{\lambda }_2^4{x}^2+\frac{1}{2}{\lambda }_2^4(p^2{z}_0^2-{\lambda }_1^2)$
$\Rightarrow 2x^2$	$=$	$-\frac{{\lambda }_2^4}{2}\pm [\frac{{\lambda }_2^8}{4}-2{\lambda }_2^4(p^2{z}_0^2-{\lambda }_1^2){]}^{1/2}$
	$=$	$\frac{{\lambda }_2^4}{2}\left\{\right[1-\frac{8(p^2{z}_0^2-{\lambda }_1^2)}{{\lambda }_2^4}{]}^{1/2}-1\}$
$\Rightarrow x^2$	$=$	$\frac{{\lambda }_2^4}{4}\left\{\right[1-\frac{8(p^2{z}_0^2-{\lambda }_1^2)}{{\lambda }_2^4}{]}^{1/2}-1\}.$

Given that, for $q^2=2$, one of the two defining expression means, ${\lambda }_2=x/\sqrt{y}$, we also have,

$y$

$=$

$\frac{{\lambda }_2^2}{4}\left\{\right[1-\frac{8(p^2{z}_0^2-{\lambda }_1^2)}{{\lambda }_2^4}{]}^{1/2}-1\}.$

New 2^nd Coordinate

Apparently it will be cleaner to define a new "2^nd coordinate," ${\kappa }_2$, such that,

${\kappa }_2^{2q^2}$	$\equiv$	$q^2[\frac{1}{{\lambda }_2}{]}^{2q^2}$
$\Rightarrow {\kappa }_2$	$\equiv$	$q^{1/q^2}\left[\frac{1}{{\lambda }_2}\right]=q^{1/q^2}\left[\frac{y^{1/q^2}}{x}\right]=\frac{(qy{)}^{1/q^2}}{x}.$

(With this new definition, ${\kappa }_2\sim \tan \phi$; it is exactly this when $q^2=1$.) Then we can rewrite the last expression from the above general case as,

$x^2+({\kappa }_{2}x{)}^{2q^2}-({\lambda }_1^2-p^2{z}_0^2)$

$=$

$0.$

When $q=1$ (the axisymmetric case), this gives,

$x^2(1+{\kappa }_2^2)$	$=$	$({\lambda }_1^2-p^2{z}_0^2)$
$\Rightarrow \frac{x^2}{({\lambda }_1^2-p^2{z}_0^2)}$	$=$	$\frac{1}{(1+{\kappa }_2^2)},$

which means that ${\kappa }_2=\tan \phi$. And, for the case of $q^2=2$, after making the substitution,

${\lambda }_2\to (q{)}^{1/q^2}{\kappa }_2^{-1}=\frac{2^{1/4}}{{\kappa }_2},$

we find,

$x^2$	$=$	$\frac{{\lambda }_2^4}{4}\left\{\right[1-\frac{8(p^2{z}_0^2-{\lambda }_1^2)}{{\lambda }_2^4}{]}^{1/2}-1\}.$
$\Rightarrow x^2$	$=$	$\frac{1}{2{\kappa }_2^4}\left\{\right[1+4{\kappa }_2^4({\lambda }_1^2-p^2{z}_0^2){]}^{1/2}-1\},$

and,

$y=\left(\frac{{\kappa }_2^2}{2^{1/2}}\right)x^2$	$=$	$\left(\frac{{\kappa }_2^2}{2^{1/2}}\right)\frac{1}{2{\kappa }_2^4}\left\{\right[1+4{\kappa }_2^4({\lambda }_1^2-p^2{z}_0^2){]}^{1/2}-1\}$
	$=$	$\frac{1}{2^{3/2}{\kappa }_2^2}\left\{\right[1+4{\kappa }_2^4({\lambda }_1^2-p^2{z}_0^2){]}^{1/2}-1\}.$

Angle Between Unit Vectors

We begin by restating that the coordinate, scale factor, and unit vector associated with the normal to our ellipsoidal surface are,

${\lambda }_1$	$\equiv$	$(x^2+q^2{y}^2+p^2{z}^2{)}^{1/2},$
$h_1$	$=$	${\lambda }_1{\ell }_{3D},$
${\hat{e}}_1$	$=$	$\hat{ı}(x{\ell }_{3D})+\hat{ȷ}(q^{2}y{\ell }_{3D})+\hat{k}(p^{2}z{\ell }_{3D}).$

In the Table below titled, "Direction Cosines Components for κ8 Coordinates", there are two fully-formed unit vectors that are each orthogonal to the (first) unit vector that is normal to the ellipsoid's surface. Here we will refer to the coordinates of these two fully-formed unit vectors as,

${\lambda }_2$

$\equiv$

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

and,

${\kappa }_2$

$\equiv$

$\frac{(qy{)}^{1/q^2}}{x}.$

The associated scale factors and unit vectors are given by the following expressions:

$h_{{\lambda }_2}^{-2}$	$=$	$(\frac{\partial {\lambda }_2}{\partial x}{)}^2+(\frac{\partial {\lambda }_2}{\partial y}{)}^2+(\frac{\partial {\lambda }_2}{\partial z}{)}^2$
	$=$	$(\frac{{\lambda }_2}{x}{)}^2+(\frac{{\lambda }_2}{q^{2}y}{)}^2+(-\frac{2{\lambda }_2}{p^{2}z}{)}^2$
	$=$	${\lambda }_2^2\left[\frac{q^4{y}^2{p}^4{z}^2+x^2{p}^4{z}^2+4x^2{q}^4{y}^2}{x^2{q}^4{y}^2{p}^4{z}^2}\right]$
$\Rightarrow h_{{\lambda }_2}$	$=$	$\frac{x{q}^{2}y{p}^{2}z}{{\lambda }_2{𝒟}}$
$\Rightarrow {\hat{e}}_{{\lambda }_2}$	$=$	$\hat{ı}\left[h_{{\lambda }_2}\right(\frac{\partial {\lambda }_2}{\partial x}\left)\right]+\hat{ȷ}\left[h_{{\lambda }_2}\right(\frac{\partial {\lambda }_2}{\partial y}\left)\right]+\hat{k}\left[h_{{\lambda }_2}\right(\frac{\partial {\lambda }_2}{\partial z}\left)\right]$
	$=$	$\hat{ı}\left[\frac{q^{2}y{p}^{2}z}{{𝒟}}\right]+\hat{ȷ}\left[\frac{x{p}^{2}z}{{𝒟}}\right]-\hat{k}\left[\frac{2x{q}^{2}y}{{𝒟}}\right].$

With regard to orthogonality, note that, ${\hat{e}}_1\cdot {\hat{e}}_{{\lambda }_2}$ $=$ $[\hat{ı}(x{\ell }_{3D})+\hat{ȷ}(q^{2}y{\ell }_{3D})+\hat{k}(p^{2}z{\ell }_{3D})]\cdot \left\{\hat{ı}\right[\frac{q^{2}y{p}^{2}z}{{𝒟}}]+\hat{ȷ}[\frac{x{p}^{2}z}{{𝒟}}]-\hat{k}[\frac{2x{q}^{2}y}{{𝒟}}\left]\right\}$   $=$ $(x{\ell }_{3D})\left[\frac{q^{2}y{p}^{2}z}{{𝒟}}\right]+(q^{2}y{\ell }_{3D})\left[\frac{x{p}^{2}z}{{𝒟}}\right]-(p^{2}z{\ell }_{3D})\left[\frac{2x{q}^{2}y}{{𝒟}}\right]$   $=$ $0.$

And,

$h_{{\kappa }_2}^{-2}$	$=$	$(\frac{\partial {\kappa }_2}{\partial x}{)}^2+(\frac{\partial {\kappa }_2}{\partial y}{)}^2+(\frac{\partial {\kappa }_2}{\partial z}{)}^2$
	$=$	$(-\frac{{\kappa }_2}{x}{)}^2+(\frac{{\kappa }_2}{q^{2}y}{)}^2=\frac{{\kappa }_2^2}{x^2{q}^4{y}^2}[x^2+q^4{y}^2]$
$\Rightarrow h_{{\kappa }_2}$	$=$	$\frac{x{q}^{2}y}{{\kappa }_2(x^2+q^4{y}^2{)}^{1/2}}$
$\Rightarrow {\hat{e}}_{{\kappa }_2}$	$=$	$-\hat{ı}\left[\frac{q^{2}y}{(x^2+q^4{y}^2{)}^{1/2}}\right]+\hat{ȷ}\left[\frac{x}{(x^2+q^4{y}^2{)}^{1/2}}\right]+\hat{k}\left[0\right].$

Again, note that with regard to orthogonality, ${\hat{e}}_1\cdot {\hat{e}}_{{\kappa }_2}$ $=$ $[\hat{ı}(x{\ell }_{3D})+\hat{ȷ}(q^{2}y{\ell }_{3D})+\hat{k}(p^{2}z{\ell }_{3D})]\cdot \{-\hat{ı}[\frac{q^{2}y}{(x^2+q^4{y}^2{)}^{1/2}}]+\hat{ȷ}[\frac{x}{(x^2+q^4{y}^2{)}^{1/2}}\left]\right\}$   $=$ $-(x{\ell }_{3D})\left[\frac{q^{2}y}{(x^2+q^4{y}^2{)}^{1/2}}\right]+(q^{2}y{\ell }_{3D})\left[\frac{x}{(x^2+q^4{y}^2{)}^{1/2}}\right]$   $=$ $0.$

From this pair of orthogonality checks, we appreciate that both unit vectors always lie in the plane that is tangent to the surface of our ellipsoid. Next, let's determine the angle, $\alpha$, between these two unit vectors as measured in the relevant tangent-plane.

$\cos \alpha \equiv {\hat{e}}_{{\lambda }_2}\cdot {\hat{e}}_{{\kappa }_2}$	$=$	$\left\{\hat{ı}\right[\frac{q^{2}y{p}^{2}z}{{𝒟}}]+\hat{ȷ}[\frac{x{p}^{2}z}{{𝒟}}]-\hat{k}[\frac{2x{q}^{2}y}{{𝒟}}\left]\right\}\{-\hat{ı}[\frac{q^{2}y}{(x^2+q^4{y}^2{)}^{1/2}}]+\hat{ȷ}[\frac{x}{(x^2+q^4{y}^2{)}^{1/2}}]+\hat{k}[0\left]\right\}$
	$=$	$-\left[\frac{q^{2}y}{(x^2+q^4{y}^2{)}^{1/2}}\right]\left[\frac{q^{2}y{p}^{2}z}{{𝒟}}\right]+\left[\frac{x}{(x^2+q^4{y}^2{)}^{1/2}}\right]\left[\frac{x{p}^{2}z}{{𝒟}}\right]$
	$=$	$\left[\frac{p^{2}z}{{𝒟}(x^2+q^4{y}^2{)}^{1/2}}\right][x^2-q^4{y}^2].$

---

Let's again visit the unit vector that we know lies in the tangent-plane and is always orthogonal to ${\lambda }_2$, namely,

${\hat{e}}_{{\lambda }_3}$

$=$

$-\hat{ı}[x(2q^4{y}^2+p^4{z}^2)]\frac{{\ell }_{3D}}{{𝒟}}+\hat{ȷ}[q^{2}y(p^4{z}^2+2x^2)]\frac{{\ell }_{3D}}{{𝒟}}+\hat{k}[p^{2}z(x^2-q^4{y}^2)]\frac{{\ell }_{3D}}{{𝒟}}.$

We acknowledge that,

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

Kappa (κ8) Coordinates

${\kappa }_1$	$\equiv$	$(x^2+q^2{y}^2+p^2{z}^2{)}^{1/2};$
${\kappa }_3$	$\equiv$	${\tan }^{-1}\left[\frac{(qy{)}^{1/q^2}}{x}\right].$

$\frac{\partial {\kappa }_3}{\partial x}$	$=$	$[1+\frac{(qy{)}^{2/q^2}}{x^2}{]}^{-1}[-\frac{(qy{)}^{1/q^2}}{x^2}]$
	$=$	$-\frac{{\sin }^2{\kappa }_3}{(qy{)}^{1/q^2}}=-\frac{{\sin }^2{\kappa }_3}{x\tan {\kappa }_3}$
	$=$	$-\frac{\sin {\kappa }_3\cos {\kappa }_3}{x}.$
$\frac{\partial {\kappa }_3}{\partial y}$	$=$	$[1+\frac{(qy{)}^{2/q^2}}{x^2}{]}^{-1}[\frac{q^{1/q^2}{y}^{1/q^2}}{q^{2}xy}]$
	$=$	$[1+\frac{(qy{)}^{2/q^2}}{x^2}{]}^{-1}[\frac{(qy{)}^{1/q^2}}{x}]\frac{1}{q^{2}y}$
	$=$	$\frac{1}{q^{2}y}\left[\frac{\tan {\kappa }_3}{1+{\tan }^2{\kappa }_3}\right]$
	$=$	$+\frac{\sin {\kappa }_3\cos {\kappa }_3}{q^{2}y}.$

Therefore,

$h_3^{-2}$	$=$	$[-\frac{\sin {\kappa }_3\cos {\kappa }_3}{x}{]}^2+[\frac{\sin {\kappa }_3\cos {\kappa }_3}{q^{2}y}{]}^2$
	$=$	$(x^2+q^4{y}^2)[\frac{\sin {\kappa }_3\cos {\kappa }_3}{x{q}^{2}y}{]}^2$
$\Rightarrow h_3$	$=$	$(x^2+q^4{y}^2{)}^{-1/2}[\frac{x{q}^{2}y}{\sin {\kappa }_3\cos {\kappa }_3}].$

 

Direction Cosine Components for κ8 Coordinates
$n$	${\kappa }_n$	$h_n$	$\frac{\partial {\kappa }_n}{\partial x}$	$\frac{\partial {\kappa }_n}{\partial y}$	$\frac{\partial {\kappa }_n}{\partial z}$	${\gamma }_{n1}$	${\gamma }_{n2}$	${\gamma }_{n3}$
$1$	$(x^2+q^2{y}^2+p^2{z}^2{)}^{1/2}$	${\kappa }_1{\ell }_{3D}$	$\frac{x}{{\kappa }_1}$	$\frac{q^{2}y}{{\kappa }_1}$	$\frac{p^{2}z}{{\kappa }_1}$	$(x){\ell }_{3D}$	$(q^{2}y){\ell }_{3D}$	$(p^{2}z){\ell }_{3D}$
$2$	---	${\ell }_{3D}(x^2+q^4{y}^2{)}^{1/2}$	$\frac{x{p}^{2}z}{(x^2+q^4{y}^2)}$	$\frac{q^{2}y{p}^{2}z}{(x^2+q^4{y}^2)}$	$-1$	$\frac{x{p}^{2}z{\ell }_{3D}}{(x^2+q^4{y}^2{)}^{1/2}}$	$\frac{q^{2}y{p}^{2}z{\ell }_{3D}}{(x^2+q^4{y}^2{)}^{1/2}}$	$-\frac{(x^2+q^4{y}^2){\ell }_{3D}}{(x^2+q^4{y}^2{)}^{1/2}}$
$3$	${\tan }^{-1}\left[\frac{(qy{)}^{1/q^2}}{x}\right]$	$\frac{1}{\sin {\kappa }_3\cos {\kappa }_3}\left[\frac{x{q}^{2}y}{(x^2+q^4{y}^2{)}^{1/2}}\right]$	$-\frac{\sin {\kappa }_3\cos {\kappa }_3}{x}$	$\frac{\sin {\kappa }_3\cos {\kappa }_3}{q^{2}y}$	$0$	$-\frac{q^{2}y}{(x^2+q^4{y}^2{)}^{1/2}}$	$\frac{x}{(x^2+q^4{y}^2{)}^{1/2}}$	$0$
${\ell }_{3D}$ $\equiv$ $(x^2+q^4{y}^2+p^4{z}^2{)}^{-1/2}$
$4$	$\frac{x{y}^{1/q^2}}{z^{2/p^2}}$	$\frac{1}{{\kappa }_4}\left[\frac{x{q}^{2}y{p}^{2}z}{{𝒟}}\right]$	$\frac{{\kappa }_4}{x}$	$\frac{{\kappa }_4}{q^{2}y}$	$-\frac{2{\kappa }_4}{p^{2}z}$	$\frac{x{q}^{2}y{p}^{2}z}{{𝒟}}\left(\frac{1}{x}\right)$	$\frac{x{q}^{2}y{p}^{2}z}{{𝒟}}\left(\frac{1}{q^{2}y}\right)$	$\frac{x{q}^{2}y{p}^{2}z}{{𝒟}}(-\frac{2}{p^{2}z})$
$5$	---	---	---	---	---	$-\frac{{\ell }_{3D}}{{𝒟}}[x(2q^4{y}^2+p^4{z}^2)]$	$\frac{{\ell }_{3D}}{{𝒟}}[q^{2}y(p^4{z}^2+2x^2)]$	$\frac{{\ell }_{3D}}{{𝒟}}[p^{2}z(x^2-q^4{y}^2)]$
${𝒟}$ $\equiv$ $(q^4{y}^2{p}^4{z}^2+x^2{p}^4{z}^2+4x^2{q}^4{y}^2{)}^{1/2}.$ Also note … $AB\equiv (\frac{{𝒟}}{{\ell }_{3D}}{)}^2$ $=$ $[x(2q^4{y}^2+p^4{z}^2){]}^2+[q^{2}y(p^4{z}^2+2x^2){]}^2+[p^{2}z(x^2-q^4{y}^2){]}^2,$ and the partial derivatives of $A$ and $B$ are detailed in an accompanying discussion.

The direction-cosines of the second unit vector — as has already been inserted into the "κ8 coordinates" table — should be obtainable from the first and third unit vectors via the cross product,

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

The other boxes in the n = 2 row have been drawn from our accompanying EUREKA! moment and the n = 3 row of the table that details "Direction Cosine Components for T8 Coordinates."

Attempt 1

Let's try …

${\kappa }_2$

$\equiv$

${\tan }^{-1}\left[\frac{x(qy{)}^{1/q^2}}{(pz{)}^{1/p^2}}\right],$

which leads to,

$\frac{\partial {\kappa }_3}{\partial x}$	$=$	$[1+\frac{x^2(qy{)}^{2/q^2}}{(pz{)}^{2/p^2}}{]}^{-1}[\frac{(qy{)}^{1/q^2}}{(pz{)}^{1/p^2}}]$
	$=$	$\frac{1}{x}[1+{\tan }^2{\kappa }_2{]}^{-1}\tan {\kappa }_2=\frac{\sin {\kappa }_2\cos {\kappa }_2}{x};$
$\frac{\partial {\kappa }_3}{\partial y}$	$=$	$[1+\frac{x^2(qy{)}^{2/q^2}}{(pz{)}^{2/p^2}}{]}^{-1}[\frac{x{q}^{1/q^2}(y{)}^{1/q^2}}{q^{2}y(pz{)}^{1/p^2}}]$
	$=$	$\frac{1}{q^{2}y}[1+{\tan }^2{\kappa }_2{]}^{-1}\tan {\kappa }_2=\frac{\sin {\kappa }_2\cos {\kappa }_2}{q^{2}y};$
$\frac{\partial {\kappa }_3}{\partial z}$	$=$	$-\frac{\sin {\kappa }_2\cos {\kappa }_2}{p^{2}z}.$

Hence,

$h_2^{-2}$	$=$	${\sin }^2{\kappa }_2{\cos }^2{\kappa }_2[\frac{1}{x^2}+\frac{1}{q^4{y}^2}+\frac{1}{p^4{z}^2}]$
	$=$	${\sin }^2{\kappa }_2{\cos }^2{\kappa }_2\left[\frac{x^2+q^4{y}^2+p^4{z}^2}{x^2{q}^4{y}^2{p}^4{z}^2}\right]=\left[\frac{{\sin }^2{\kappa }_2{\cos }^2{\kappa }_2}{x^2{q}^4{y}^2{p}^4{z}^2{\ell }_{3D}^2}\right]$
$\Rightarrow h_2$	$=$	$\left[\frac{x{q}^{2}y{p}^{2}z{\ell }_{3D}}{\sin {\kappa }_2\cos {\kappa }_2}\right].$

The three direction-cosines are, then,

${\gamma }_{21}=h_2\left(\frac{\partial {\kappa }_2}{\partial x}\right)$

$=$

$\left[\frac{x{q}^{2}y{p}^{2}z{\ell }_{3D}}{\sin {\kappa }_2\cos {\kappa }_2}\right]\frac{\sin {\kappa }_2\cos {\kappa }_2}{x}=q^{2}y{p}^{2}z{\ell }_{3D}.$

See Also

• Direction Cosines
• Trials up through T7 Coordinates