Appendix/Ramblings/ConcentricEllipsoidalDaringAttack

Daring Attack

Background

Building on our general introduction to Direction Cosines in the context of orthogonal curvilinear coordinate systems, and on our previous development of the so-called T6 (concentric elliptic) coordinate system, here we take a somewhat daring attack on this problem, mixing our approach to identifying the expression for the third curvilinear coordinate. Broadly speaking, this entire study 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.

Direction Cosine Components for T6 Coordinates
$n$	${\lambda }_n$	$h_n$	$\frac{\partial {\lambda }_n}{\partial x}$	$\frac{\partial {\lambda }_n}{\partial y}$	$\frac{\partial {\lambda }_n}{\partial z}$	${\gamma }_{n1}$	${\gamma }_{n2}$	${\gamma }_{n3}$
$1$	$(x^2+q^2{y}^2+p^2{z}^2{)}^{1/2}$	${\lambda }_1{\ell }_{3D}$	$\frac{x}{{\lambda }_1}$	$\frac{q^{2}y}{{\lambda }_1}$	$\frac{p^{2}z}{{\lambda }_1}$	$(x){\ell }_{3D}$	$(q^{2}y){\ell }_{3D}$	$(p^{2}z){\ell }_{3D}$
$2$	---	---	---	---	---	${\ell }_q{\ell }_{3D}(x{p}^{2}z)$	${\ell }_q{\ell }_{3D}(q^{2}y{p}^{2}z)$	$-(x^2+q^4{y}^2){\ell }_q{\ell }_{3D}$
$3$	${\tan }^{-1}\left(\frac{y^{1/q^2}}{x}\right)$	$\frac{x{q}^{2}y{\ell }_q}{\sin {\lambda }_3\cos {\lambda }_3}$	$-\frac{\sin {\lambda }_3\cos {\lambda }_3}{x}$	$+\frac{\sin {\lambda }_3\cos {\lambda }_3}{q^{2}y}$	$0$	$-q^{2}y{\ell }_q$	$x{\ell }_q$	$0$
${\ell }_{3D}$ $\equiv$ $[x^2+q^4{y}^2+p^4{z}^2{]}^{-1/2}$ ${\ell }_q$ $\equiv$ $[x^2+q^4{y}^2{]}^{-1/2}$

As before, let's adopt the first-coordinate expression,

${\lambda }_1$

$\equiv$

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

but for the third-coordinate expression we will abandon the trigonometric expression and instead simply use,

${\lambda }_3$

$\equiv$

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

This modified third-coordinate expression means that the last row of the above table changes, as follows.

Daring Attack
$n$	${\lambda }_n$	$h_n$	$\frac{\partial {\lambda }_n}{\partial x}$	$\frac{\partial {\lambda }_n}{\partial y}$	$\frac{\partial {\lambda }_n}{\partial z}$	${\gamma }_{n1}$	${\gamma }_{n2}$	${\gamma }_{n3}$
$1$	$(x^2+q^2{y}^2+p^2{z}^2{)}^{1/2}$	${\lambda }_1{\ell }_{3D}$	$\frac{x}{{\lambda }_1}$	$\frac{q^{2}y}{{\lambda }_1}$	$\frac{p^{2}z}{{\lambda }_1}$	$(x){\ell }_{3D}$	$(q^{2}y){\ell }_{3D}$	$(p^{2}z){\ell }_{3D}$
$2$	---	---	---	---	---	${\ell }_q{\ell }_{3D}(x{p}^{2}z)$	${\ell }_q{\ell }_{3D}(q^{2}y{p}^{2}z)$	$-(x^2+q^4{y}^2){\ell }_q{\ell }_{3D}$
$3$	$\frac{y^{1/q^2}}{x}$	$\frac{x{q}^{2}y{\ell }_q}{{\lambda }_3}$	$-\frac{{\lambda }_3}{x}$	$+\frac{{\lambda }_3}{q^{2}y}$	$0$	$-q^{2}y{\ell }_q$	$x{\ell }_q$	$0$

Notice that the direction cosine functions for the (as yet, unknown) second-coordinate function remain the same. This is because the direction-cosine functions associated with both ${\lambda }_1$ and ${\lambda }_3$ remain unchanged, so it must be true that the cross product of the first and third unit vectors leads to the same components for the second unit vector.

New Approach

Setup

The surface of an ellipsoid with semi-major axes (a, b, c) is defined by the expression,

$1$

$=$

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

This is identical to our expression for ${\lambda }_1$ if we make the associations,

$a={\lambda }_1,$

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

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

Now, given that ${\lambda }_3$ does not functionally depend on $z$, let's consider that the choice of $z$ is tightly associated with the specification of the second coordinate, ${\lambda }_2$. Specifically, let's adopt the definition,

${\lambda }_2^2$

$\equiv$

$1-(\frac{z}{c}{)}^2,$

in which case, we see that,

$z^2$

$=$

$c^2(1-{\lambda }_2^2)=\frac{{\lambda }_1^2(1-{\lambda }_2^2)}{p^2},$

and,

$(\frac{x}{a}{)}^2+(\frac{y}{b}{)}^2$	$=$	${\lambda }_2^2$
$\Rightarrow x^2+q^2{y}^2$	$=$	${\lambda }_1^2{\lambda }_2^2.$

[Note that in the case of spherical coordinates (q² = p² = 1), ${\lambda }_1\to r$, and this "second" coordinate, ${\lambda }_2$, becomes $\sin \theta$.] Combining this last expression with the $x-y$ relationship that is provided by the definition of ${\lambda }_3$, gives,

${\lambda }_1^2{\lambda }_2^2$

$=$

$\frac{y^{2/q^2}}{{\lambda }_3^2}+q^2{y}^2.$

In general, the exponent of $2q^{-2}$ that appears in the first term on the right-hand side of this expression prevents us from being able to analytically prescribe the function, $y({\lambda }_1,{\lambda }_2,{\lambda }_3)$. But a solution is obtainable for selected values of $q^2>1$.

Examine the Case: q² = 2

If we set $q^2=2$, then this last combined expression becomes a quadratic equation for $y$. Specifically, we find,

$0$	$=$	$2y^2+\frac{y}{{\lambda }_3^2}-{\lambda }_1^2{\lambda }_2^2$
$\Rightarrow y$	$=$	$\frac{1}{4}\{-\frac{1}{{\lambda }_3^2}\pm [\frac{1}{{\lambda }_3^4}+8{\lambda }_1^2{\lambda }_2^2{]}^{1/2}\}$
	$=$	$\frac{1}{4{\lambda }_3^2}\left\{\right[1+8{\lambda }_1^2{\lambda }_2^2{\lambda }_3^4{]}^{1/2}-1\}.$

(Note that, for reasons of simplicity for the time being, in this last expression we have retained only the "positive" solution.) Again, calling upon the $x-y$ relationship that is provided through the definition of ${\lambda }_3$, we find (when q² = 2),

$x^2$	$=$	$\frac{y}{{\lambda }_3^2}$
	$=$	$\frac{1}{4{\lambda }_3^4}\left\{\right[1+8{\lambda }_1^2{\lambda }_2^2{\lambda }_3^4{]}^{1/2}-1\}$
$\Rightarrow x$	$=$	$\pm \frac{1}{2{\lambda }_3^2}\left\{\right[1+8{\lambda }_1^2{\lambda }_2^2{\lambda }_3^4{]}^{1/2}-1{\}}^{1/2}.$

Summary (q2 = 2) $z({\lambda }_1,{\lambda }_2,{\lambda }_3)$ $=$ $\frac{{\lambda }_1(1-{\lambda }_2^2{)}^{1/2}}{p},$ $y({\lambda }_1,{\lambda }_2,{\lambda }_3)$ $=$ $\frac{1}{4{\lambda }_3^2}\left\{\right[1+8{\lambda }_1^2{\lambda }_2^2{\lambda }_3^4{]}^{1/2}-1\}=\frac{(\mathrm{\Lambda }-1)}{4{\lambda }_3^2},$ $x({\lambda }_1,{\lambda }_2,{\lambda }_3)$ $=$ $\frac{1}{2{\lambda }_3^2}\left\{\right[1+8{\lambda }_1^2{\lambda }_2^2{\lambda }_3^4{]}^{1/2}-1{\}}^{1/2}=\frac{(\mathrm{\Lambda }-1{)}^{1/2}}{2{\lambda }_3^2}.$

For convenience, we have defined,

${\mathrm{\Lambda }}^2$	$\equiv$	$1+8{\lambda }_1^2{\lambda }_2^2{\lambda }_3^4$
$\Rightarrow {\lambda }_1^2{\lambda }_2^2{\lambda }_3^4$	$\equiv$	$\frac{1}{8}({\mathrm{\Lambda }}^2-1).$

Test Example

$q^2=2,p^2=3.15,(x,y,z)=(0.4,0.63581,0.1)$ $({\lambda }_1,{\lambda }_2,{\lambda }_3)=(1,0.98412,1.99344)$ ${\ell }_{3D}=0.730058,{\ell }_q=0.750164$ $h_1=0.730058$

${\mathrm{\Lambda }}^2-1=122.34879\Rightarrow \mathrm{\Lambda }=11.10625$

Do we get the correct values of $(x,y,z)$  ? $z({\lambda }_1,{\lambda }_2,{\lambda }_3)$ $=$ $\frac{{\lambda }_1(1-{\lambda }_2^2{)}^{1/2}}{p}=0.1000000,$ $y({\lambda }_1,{\lambda }_2,{\lambda }_3)$ $=$ $\frac{(\mathrm{\Lambda }-1)}{4{\lambda }_3^2}=0.635807,$ $x({\lambda }_1,{\lambda }_2,{\lambda }_3)$ $=$ $\frac{(\mathrm{\Lambda }-1{)}^{1/2}}{2{\lambda }_3^2}=0.400000.$

Evaluate a few partial derivatives … $\frac{\partial z}{\partial {\lambda }_1}$ $=$ $\frac{(1-{\lambda }_2^2{)}^{1/2}}{p}=0.1,$ $\frac{\partial y}{\partial {\lambda }_1}$ $=$ $\left[\frac{2{\lambda }_1{\lambda }_2^2{\lambda }_3^2}{\mathrm{\Lambda }}\right]=0.693054,$ $\frac{\partial x}{\partial {\lambda }_1}$ $=$ $\frac{2{\lambda }_1{\lambda }_2^2{\lambda }_3^2}{\mathrm{\Lambda }(\mathrm{\Lambda }-1{)}^{1/2}}=0.218008.$ $\Rightarrow h_1$ $=$ $\left[\right(\frac{\partial x}{\partial {\lambda }_1}{)}^2+(\frac{\partial y}{\partial {\lambda }_1}{)}^2+(\frac{\partial z}{\partial {\lambda }_1}{)}^2{]}^{1/2}=0.733383.$ This matches the numerical value for $h_1$ as determined below, but it does not match the numerical value obtained previously (0.730058) for $h_1$. The most likely piece that needs adjustment is the partial of "z" with respect to λ1. It needs to be … $\frac{\partial z}{\partial {\lambda }_1}=[h_1^2-(\frac{\partial x}{\partial {\lambda }_1}{)}^2-(\frac{\partial y}{\partial {\lambda }_1}{)}^2{]}^{1/2}=0.071647$. Alternatively, $\frac{\partial z}{\partial {\lambda }_1}$ $=$ $h_1^2\left(\frac{\partial {\lambda }_1}{\partial z}\right)=(0.730058{)}^2[\frac{p^{2}z}{{\lambda }_1}]$

Next, let's examine all nine partial derivatives, noting at the start that,

$\Rightarrow \frac{\partial \mathrm{\Lambda }}{\partial {\lambda }_1}$	$=$	$\frac{1}{2\mathrm{\Lambda }}\left[16{\lambda }_1{\lambda }_2^2{\lambda }_3^4\right]=\frac{({\mathrm{\Lambda }}^2-1)}{{\lambda }_1\mathrm{\Lambda }},$
$\frac{\partial \mathrm{\Lambda }}{\partial {\lambda }_2}$	$=$	$\frac{1}{2\mathrm{\Lambda }}\left[16{\lambda }_1^2{\lambda }_2{\lambda }_3^4\right]=\frac{({\mathrm{\Lambda }}^2-1)}{{\lambda }_2\mathrm{\Lambda }},$
$\frac{\partial \mathrm{\Lambda }}{\partial {\lambda }_3}$	$=$	$\frac{1}{2\mathrm{\Lambda }}\left[32{\lambda }_1^2{\lambda }_2^2{\lambda }_3^3\right]=\frac{2({\mathrm{\Lambda }}^2-1)}{{\lambda }_3\mathrm{\Lambda }}.$

We have,

$\frac{\partial z}{\partial {\lambda }_1}$	$=$	$\frac{(1-{\lambda }_2^2{)}^{1/2}}{p},$
$\frac{\partial z}{\partial {\lambda }_2}$	$=$	$-\frac{{\lambda }_1{\lambda }_2}{p(1-{\lambda }_2^2{)}^{1/2}},$
$\frac{\partial z}{\partial {\lambda }_3}$	$=$	$0.$
$\frac{\partial y}{\partial {\lambda }_1}$	$=$	$\frac{1}{4{\lambda }_3^2}\cdot \frac{\partial \mathrm{\Lambda }}{\partial {\lambda }_1}=\left[\frac{({\mathrm{\Lambda }}^2-1)}{4{\lambda }_3^2{\lambda }_1\mathrm{\Lambda }}\right]=\left[\frac{8{\lambda }_1^2{\lambda }_2^2{\lambda }_3^4}{4{\lambda }_3^2{\lambda }_1\mathrm{\Lambda }}\right]=\left[\frac{2{\lambda }_1{\lambda }_2^2{\lambda }_3^2}{\mathrm{\Lambda }}\right],$
$\frac{\partial y}{\partial {\lambda }_2}$	$=$	$\frac{1}{4{\lambda }_3^2}\cdot \frac{\partial \mathrm{\Lambda }}{\partial {\lambda }_2}=\left[\frac{({\mathrm{\Lambda }}^2-1)}{4{\lambda }_3^2{\lambda }_2\mathrm{\Lambda }}\right]=\left[\frac{8{\lambda }_1^2{\lambda }_2^2{\lambda }_3^4}{4{\lambda }_3^2{\lambda }_2\mathrm{\Lambda }}\right]=\left[\frac{2{\lambda }_1^2{\lambda }_2{\lambda }_3^2}{\mathrm{\Lambda }}\right],$
$\frac{\partial y}{\partial {\lambda }_3}$	$=$	$\frac{1}{4{\lambda }_3^2}\cdot \frac{\partial \mathrm{\Lambda }}{\partial {\lambda }_3}-\frac{(\mathrm{\Lambda }-1)}{2{\lambda }_3^3}=\frac{1}{4{\lambda }_3^2}\cdot \left[\frac{2({\mathrm{\Lambda }}^2-1)}{{\lambda }_3\mathrm{\Lambda }}\right]-\frac{\mathrm{\Lambda }(\mathrm{\Lambda }-1)}{2{\lambda }_3^3\mathrm{\Lambda }}=\left[\frac{({\mathrm{\Lambda }}^2-1)-\mathrm{\Lambda }(\mathrm{\Lambda }-1)}{2{\lambda }_3^3\mathrm{\Lambda }}\right]=\frac{(\mathrm{\Lambda }-1)}{2{\lambda }_3^3\mathrm{\Lambda }}.$
$\frac{\partial x}{\partial {\lambda }_1}$	$=$	$\frac{1}{4{\lambda }_3^2(\mathrm{\Lambda }-1{)}^{1/2}}\cdot \frac{\partial \mathrm{\Lambda }}{\partial {\lambda }_1}=\frac{1}{4{\lambda }_3^2(\mathrm{\Lambda }-1{)}^{1/2}}\left[\frac{({\mathrm{\Lambda }}^2-1)}{{\lambda }_1\mathrm{\Lambda }}\right]=\frac{({\mathrm{\Lambda }}^2-1)}{4{\lambda }_1{\lambda }_3^2\mathrm{\Lambda }(\mathrm{\Lambda }-1{)}^{1/2}}=\frac{2{\lambda }_1{\lambda }_2^2{\lambda }_3^2}{\mathrm{\Lambda }(\mathrm{\Lambda }-1{)}^{1/2}},$
$\frac{\partial x}{\partial {\lambda }_2}$	$=$	$\frac{1}{4{\lambda }_3^2(\mathrm{\Lambda }-1{)}^{1/2}}\cdot \frac{\partial \mathrm{\Lambda }}{\partial {\lambda }_2}=\frac{1}{4{\lambda }_3^2(\mathrm{\Lambda }-1{)}^{1/2}}\left[\frac{({\mathrm{\Lambda }}^2-1)}{{\lambda }_2\mathrm{\Lambda }}\right]=\frac{({\mathrm{\Lambda }}^2-1)}{4{\lambda }_2{\lambda }_3^2\mathrm{\Lambda }(\mathrm{\Lambda }-1{)}^{1/2}}=\frac{2{\lambda }_1^2{\lambda }_2{\lambda }_3^2}{\mathrm{\Lambda }(\mathrm{\Lambda }-1{)}^{1/2}},$
$\frac{\partial x}{\partial {\lambda }_3}$	$=$	$\frac{1}{4{\lambda }_3^2(\mathrm{\Lambda }-1{)}^{1/2}}\cdot \frac{\partial \mathrm{\Lambda }}{\partial {\lambda }_3}-\frac{(\mathrm{\Lambda }-1{)}^{1/2}}{{\lambda }_3^3}=\frac{1}{4{\lambda }_3^2(\mathrm{\Lambda }-1{)}^{1/2}}\left[\frac{2({\mathrm{\Lambda }}^2-1)}{{\lambda }_3\mathrm{\Lambda }}\right]-\frac{(\mathrm{\Lambda }-1{)}^{1/2}}{{\lambda }_3^3}$
	$=$	$\frac{({\mathrm{\Lambda }}^2-1)-2\mathrm{\Lambda }(\mathrm{\Lambda }-1)}{2{\lambda }_3^3\mathrm{\Lambda }(\mathrm{\Lambda }-1{)}^{1/2}}=-\frac{(\mathrm{\Lambda }-1{)}^{3/2}}{2{\lambda }_3^3\mathrm{\Lambda }}.$

What about the derived scale-factors?

$h_1^2$	$=$	$(\frac{\partial x}{\partial {\lambda }_1}{)}^2+(\frac{\partial y}{\partial {\lambda }_1}{)}^2+(\frac{\partial z}{\partial {\lambda }_1}{)}^2$
	$=$	$[\frac{2{\lambda }_1{\lambda }_2^2{\lambda }_3^2}{\mathrm{\Lambda }(\mathrm{\Lambda }-1{)}^{1/2}}{]}^2+[\frac{2{\lambda }_1{\lambda }_2^2{\lambda }_3^2}{\mathrm{\Lambda }}{]}^2+[\frac{(1-{\lambda }_2^2{)}^{1/2}}{p}{]}^2$
	$=$	$\left[\frac{4{\lambda }_1^2{\lambda }_2^4{\lambda }_3^4}{{\mathrm{\Lambda }}^2(\mathrm{\Lambda }-1)}\right]+\left[\frac{4{\lambda }_1^2{\lambda }_2^4{\lambda }_3^4}{{\mathrm{\Lambda }}^2}\right]+\left[\frac{(1-{\lambda }_2^2)}{p^2}\right]$
	$=$	$\frac{1}{p^2{\mathrm{\Lambda }}^2(\mathrm{\Lambda }-1)}[4p^2{\lambda }_1^2{\lambda }_2^4{\lambda }_3^4\mathrm{\Lambda }+(1-{\lambda }_2^2){\mathrm{\Lambda }}^2(\mathrm{\Lambda }-1)].$

Written in terms of Cartesian coordinates, this becomes,

$h_1^2$	$=$	$\frac{8{\lambda }_1^2{\lambda }_2^2{\lambda }_3^4({\lambda }_2^2)}{2\mathrm{\Lambda }(\mathrm{\Lambda }-1)}+\frac{(1-{\lambda }_2^2)}{p^2}$
	$=$	$\frac{(\mathrm{\Lambda }+1){\lambda }_2^2}{2\mathrm{\Lambda }}+\frac{z^2}{{\lambda }_1^2}$
	$=$	$\frac{1}{{\lambda }_1^2}[\frac{(\mathrm{\Lambda }+1){\lambda }_2^2{\lambda }_1^2}{2\mathrm{\Lambda }}+z^2]$
	$=$	$\frac{1}{{\lambda }_1^2}[\frac{(\mathrm{\Lambda }+1)(x^2+2y^2)}{2\mathrm{\Lambda }}+z^2].$

Note that,

$\mathrm{\Lambda }-1$	$=$	$4x^2{\lambda }_3^4=4x^2\left(\frac{y^2}{x^4}\right)=4\left(\frac{y^2}{x^2}\right)$
$\Rightarrow \mathrm{\Lambda }$	$=$	$\frac{x^2+4y^2}{x^2}.$

Hence, the scale factor becomes,

$h_1^2$	$=$	$\frac{1}{2{\lambda }_1^2}[(x^2+2y^2)+\frac{x^2(x^2+2y^2)}{(x^2+4y^2)}+2z^2]$
	$=$	$\frac{1}{2{\lambda }_1^2(x^2+4y^2)}[(x^2+2y^2)(x^2+4y^2)+x^2(x^2+2y^2)+2z^2(x^2+4y^2)]$
	$=$	$\frac{1}{2{\lambda }_1^2(x^2+4y^2)}[(2x^4+8x^2{y}^2+8y^4)+2z^2(x^2+4y^2)]=\frac{1.911525}{3.554}=0.537852$
$\Rightarrow h_1$	$=$	$0.733384.$

---

Compare this expression with the one derived earlier, namely,

$h_1^2{|}_{q^2=2}=[{\lambda }_1^2{\ell }_{3D}^2{]}_{q^2=2}$

$=$

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

Well … first we recognize that, when q² = 2,

${\lambda }_1^2=x^2+2y^2+p^2{z}^2,$

${\lambda }_2^2=\frac{{\lambda }_1^2-p^2{z}^2}{{\lambda }_1^2}=\frac{x^2+2y^2}{x^2+2y^2+p^2{z}^2},$

${\lambda }_3^2=\frac{y}{x^2}.$

Hence,

$({\mathrm{\Lambda }}^2-1)=8{\lambda }_1^2{\lambda }_2^2{\lambda }_3^4$	$=$	$8(x^2+2y^2+p^2{z}^2)\left[\frac{x^2+2y^2}{x^2+2y^2+p^2{z}^2}\right]\frac{y^2}{x^4}=\left[\frac{8y^2(x^2+2y^2)}{x^4}\right],$
$\Rightarrow {\mathrm{\Lambda }}^2$	$=$	$1+\left[\frac{8y^2(x^2+2y^2)}{x^4}\right]=\frac{1}{x^4}[x^4+8x^2{y}^2+16y^4]=\frac{1}{x^4}[x^2+4y^4{]}^2,$
$\Rightarrow (\mathrm{\Lambda }+1)$	$=$	$\frac{(x^2+4y^4)}{x^2}+1=\frac{2x^2+4y^4}{x^2},$

which means,

$h_1^2$

$=$

$\frac{1}{8{\lambda }_1^2{\mathrm{\Lambda }}^2}\{4{\lambda }_1^2{\lambda }_2^2{\lambda }_3(\mathrm{\Lambda }+1)+4{\lambda }_1^2{\lambda }_2^2({\mathrm{\Lambda }}^2-1)+8z^2{\mathrm{\Lambda }}^2\}$

Think Again

Firm Relations

In addition to the functions that are specified in our above Daring Attack Table, we appreciate that,

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

Check …

$h_1^2$

$=$

$(\frac{\partial x}{\partial {\lambda }_1}{)}^2+(\frac{\partial y}{\partial {\lambda }_1}{)}^2+(\frac{\partial z}{\partial {\lambda }_1}{)}^2={\lambda }_1^2{\ell }_{3D}^4[x^2+q^4{y}^2+p^4{z}^2]={\lambda }_1^2{\ell }_{3D}^2.$       (Yes!)

Also,

$\frac{\partial x}{\partial {\lambda }_3}$	$=$	$h_3^2\left(\frac{\partial {\lambda }_3}{\partial x}\right)=\left[\frac{x{q}^{2}y{\ell }_q}{{\lambda }_3}{]}^2\right(-\frac{{\lambda }_3}{x})=-q^4{y}^2{\ell }_q^2(\frac{x}{{\lambda }_3}),$
$\frac{\partial y}{\partial {\lambda }_3}$	$=$	$h_3^2\left(\frac{\partial {\lambda }_3}{\partial y}\right)=\left[\frac{x{q}^{2}y{\ell }_q}{{\lambda }_3}{]}^2\right(+\frac{{\lambda }_3}{q^{2}y})=x^2{\ell }_q^2(\frac{q^{2}y}{{\lambda }_3}),$
$\frac{\partial z}{\partial {\lambda }_3}$	$=$	$h_3^2\left(\frac{\partial {\lambda }_3}{\partial z}\right)=0.$

Check …

$h_3^2$

$=$

$(\frac{\partial x}{\partial {\lambda }_3}{)}^2+(\frac{\partial y}{\partial {\lambda }_3}{)}^2+(\frac{\partial z}{\partial {\lambda }_3}{)}^2=\frac{{\ell }_q^4}{{\lambda }_3^2}[x^2{q}^8{y}^4+x^4{q}^4{y}^2]=\frac{x^2{q}^4{y}^2{\ell }_q^4}{{\lambda }_3^2}[q^4{y}^2+x^2]=\frac{x^2{q}^4{y}^2{\ell }_q^2}{{\lambda }_3^2}.$       (Yes!)

And, last …

$\frac{\partial x}{\partial {\lambda }_2}$	$=$	$h_2{\gamma }_{21}=h_2{\ell }_q{\ell }_{3D}(x{p}^{2}z),$
$\frac{\partial y}{\partial {\lambda }_2}$	$=$	$h_2{\gamma }_{22}=h_2{\ell }_q{\ell }_{3D}(q^{2}y{p}^{2}z),$
$\frac{\partial z}{\partial {\lambda }_2}$	$=$	$h_2{\gamma }_{23}=-h_2{\ell }_q{\ell }_{3D}(x^2+q^4{y}^2).$

Speculation

First

From the direction-cosine expressions for $\partial {\lambda }_2/\partial x_i$ that have been summarized in our above Daring Attack Table, it seems reasonable to suggest that,

$h_2^2$

$=$

$({\ell }_q{\ell }_{3D}{)}^2=[(x^2+q^4{y}^2)(x^2+q^4{y}^2+p^4{z}^2){]}^{-1},$

in which case,

$\frac{\partial {\lambda }_2}{\partial x}$

$=$

$x{p}^{2}z,$

$\frac{\partial {\lambda }_2}{\partial y}$

$=$

$q^{2}y{p}^{2}z,$

$\frac{\partial {\lambda }_2}{\partial z}$

$=$

$-(x^2+q^4{y}^2);$

and,

$\frac{\partial x}{\partial {\lambda }_2}=h_2^2\left(\frac{\partial {\lambda }_2}{\partial x}\right)$

$=$

$({\ell }_q{\ell }_{3D}{)}^{2}x{p}^{2}z,$

$\frac{\partial y}{\partial {\lambda }_2}=h_2^2\left(\frac{\partial {\lambda }_2}{\partial y}\right)$

$=$

$({\ell }_q{\ell }_{3D}{)}^2{q}^{2}y{p}^{2}z,$

$\frac{\partial z}{\partial {\lambda }_2}=h_2^2\left(\frac{\partial {\lambda }_2}{\partial z}\right)$

$=$

$-({\ell }_q{\ell }_{3D}{)}^2(x^2+q^4{y}^2).$

Second

Alternatively, after examining the direction-cosine expressions for $\partial x_i/\partial {\lambda }_2$ that we have just provided, one might suggest that,

$h_2^2$

$=$

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

in which case, the expressions provided for $\partial {\lambda }_2/\partial x_i$ and $\partial x_i/\partial {\lambda }_2$ must be swapped relative to our First speculation.

Third

Noticing that $h_1^2$ is proportional to ${\lambda }_1^2$ and that $h_3^2$ is inversely proportional to ${\lambda }_3^2$, let's consider both as possible behaviors for the 2^nd scale factor. Let's try the first of these behaviors. Specifically, what if we assume …

$\frac{\partial {\lambda }_2}{\partial x}$

$=$

$\frac{x{p}^{2}z}{{\lambda }_2},$

$\frac{\partial {\lambda }_2}{\partial y}$

$=$

$\frac{q^{2}y{p}^{2}z}{{\lambda }_2},$

$\frac{\partial {\lambda }_2}{\partial z}$

$=$

$-\frac{(x^2+q^4{y}^2)}{{\lambda }_2}.$

Then,

$h_2^{-2}$	$=$	$(\frac{\partial {\lambda }_2}{\partial x}{)}^2+(\frac{\partial {\lambda }_2}{\partial y}{)}^2+(\frac{\partial {\lambda }_2}{\partial z}{)}^2=[{\lambda }_2{\ell }_q{\ell }_{3D}{]}^{-2}$
$\Rightarrow h_2$	$=$	${\lambda }_2{\ell }_q{\ell }_{3D}.$

Primary implication: ${\gamma }_{21}=h_2\left(\frac{\partial {\lambda }_2}{\partial x}\right)$ $=$ $(x{p}^{2}z){\ell }_q{\ell }_{3D},$ ${\gamma }_{22}=h_2\left(\frac{\partial {\lambda }_2}{\partial y}\right)$ $=$ $(q^{2}y{p}^{2}z){\ell }_q{\ell }_{3D},$ ${\gamma }_{23}=h_2\left(\frac{\partial {\lambda }_2}{\partial z}\right)$ $=$ $-(x^2+q^4{y}^2){\ell }_q{\ell }_{3D}.$ These perfectly match the direction-cosine expressions (${\gamma }_{2i}$ for i = 1, 3) that have been summarized in our above Daring Attack Table. Secondary implication: $\frac{\partial x}{\partial {\lambda }_2}$ $=$ $h_2{\gamma }_{21}={\lambda }_2{\ell }_q^2{\ell }_{3D}^2(x{p}^{2}z),$ $\frac{\partial y}{\partial {\lambda }_2}$ $=$ $h_2{\gamma }_{22}={\lambda }_2{\ell }_q^2{\ell }_{3D}^2(q^{2}y{p}^{2}z),$ $\frac{\partial z}{\partial {\lambda }_2}$ $=$ $h_2{\gamma }_{23}=-{\lambda }_2{\ell }_q^2{\ell }_{3D}^2(x^2+q^4{y}^2).$

Now, what specifically is the function, ${\lambda }_2(x,y,z)$ ? Start by rewriting the three partial derivatives as,

$\frac{1}{2}\frac{\partial ({\lambda }_2^2)}{\partial x}$

$=$

$x{p}^{2}z,$

$\frac{1}{2}\frac{\partial ({\lambda }_2{)}^2}{\partial y}$

$=$

$q^{2}y{p}^{2}z,$

$\frac{1}{2}\frac{\partial ({\lambda }_2{)}^2}{\partial z}$

$=$

$-(x^2+q^4{y}^2).$

Suppose that,

${\lambda }_2^2$

$=$

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

Then we have,

$\frac{\partial {\lambda }_2^2}{\partial x}$

$=$

$2x{p}^{2}z,$

and,

$\frac{\partial {\lambda }_2^2}{\partial y}$

$=$

$2q^{2}y{p}^{2}z.$      Great!

But this cannot be the correct expression for ${\lambda }_2^2$ because,

$\frac{\partial {\lambda }_2^2}{\partial z}$

$=$

$(x^2+q^2{y}^2)p^2,$

which does not match the desired partial derivative with respect to $z$.

Fourth

Alternatively, if we assume …

$\frac{\partial {\lambda }_2}{\partial x}$

$=$

$\frac{{\lambda }_2}{x{p}^{2}z},$

$\frac{\partial {\lambda }_2}{\partial y}$

$=$

$\frac{{\lambda }_2}{q^{2}y{p}^{2}z},$

$\frac{\partial {\lambda }_2}{\partial z}$

$=$

$-\frac{{\lambda }_2}{(x^2+q^4{y}^2)},$

then,

$h_2^{-2}$	$=$	$(\frac{\partial {\lambda }_2}{\partial x}{)}^2+(\frac{\partial {\lambda }_2}{\partial y}{)}^2+(\frac{\partial {\lambda }_2}{\partial z}{)}^2$
	$=$	$(\frac{{\lambda }_2}{x{p}^{2}z}{)}^2+(\frac{{\lambda }_2}{q^{2}y{p}^{2}z}{)}^2+(\frac{{\lambda }_2}{x^2+q^4{y}^2}{)}^2$
$\Rightarrow (h_2{\lambda }_2{)}^{-2}$	$=$	$\frac{q^4{y}^2{p}^4{z}^2(x^2+q^4{y}^2{)}^2+x^2{p}^4{z}^2(x^2+q^4{y}^2{)}^2+x^2{q}^4{y}^2{p}^8{z}^4}{x^2{q}^4{y}^2{p}^8{z}^4(x^2+q^4{y}^2{)}^2}$
$\Rightarrow h_2$	$=$	$\frac{1}{{\lambda }_2}[\frac{x^2{q}^4{y}^2{p}^8{z}^4(x^2+q^4{y}^2{)}^2}{q^4{y}^2{p}^4{z}^2(x^2+q^4{y}^2{)}^2+x^2{p}^4{z}^2(x^2+q^4{y}^2{)}^2+x^2{q}^4{y}^2{p}^8{z}^4}{]}^{1/2}$
	$=$	$\frac{1}{{\lambda }_2}\left\{\frac{x{q}^{2}y{p}^{2}z(x^2+q^4{y}^2)}{[q^4{y}^2(x^2+q^4{y}^2{)}^2+x^2(x^2+q^4{y}^2{)}^2+x^2{q}^4{y}^2{p}^4{z}^2{]}^{1/2}}\right\}$
	$=$	$\frac{1}{{\lambda }_2}\left\{\frac{x{q}^{2}y{p}^{2}z(x^2+q^4{y}^2)}{[(x^2+q^4{y}^2{)}^3+x^2{q}^4{y}^2{p}^4{z}^2{]}^{1/2}}\right\}$

Let's check for consistency with one of the direction-cosines.

${\gamma }_{21}=h_2\left(\frac{\partial {\lambda }_2}{\partial x}\right)$	$=$	$\left\{\frac{q^{2}y(x^2+q^4{y}^2)}{[(x^2+q^4{y}^2{)}^3+x^2{q}^4{y}^2{p}^4{z}^2{]}^{1/2}}\right\}$
$\Rightarrow \frac{{\gamma }_{21}}{{\ell }_q(x{p}^{2}z)}$	$=$	$\frac{(x^2+q^4{y}^2{)}^{1/2}}{x{p}^{2}z}\left\{\frac{q^{2}y(x^2+q^4{y}^2)}{[(x^2+q^4{y}^2{)}^3+x^2{q}^4{y}^2{p}^4{z}^2{]}^{1/2}}\right\}$
	$=$	$\frac{q^{2}y}{x{p}^{2}z}\left\{\frac{(x^2+q^4{y}^2{)}^{3/2}}{[(x^2+q^4{y}^2{)}^3+x^2{q}^4{y}^2{p}^4{z}^2{]}^{1/2}}\right\}$
	$=$	$\frac{q^{2}y}{x{p}^{2}z}[1+\frac{x^2{q}^4{y}^2{p}^4{z}^2}{(x^2+q^4{y}^2{)}^3}{]}^{-1/2}.$

This does not match the term in the expression for ${\gamma }_{21}$ — namely, ${\ell }_{3D}$ — that is expected from the original tabulation.

Better Organized

From our above Daring Attack Table, we appreciate that the three direction cosines associated with the (as yet unknown) second curvilinear coordinate are,

${\gamma }_{21}$

$=$

${\ell }_q{\ell }_{3D}(x{p}^{2}z),$

${\gamma }_{22}$

$=$

${\ell }_q{\ell }_{3D}(q^{2}y{p}^{2}z),$

${\gamma }_{23}$

$=$

$-{\ell }_q{\ell }_{3D}(x^2+q^4{y}^2).$

It is easy to see that the desired orthogonality relationship,

$\sum _{i=1}^3({\gamma }_{2i}{)}^2$

$=$

$1,$

is satisfied because,

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

$=$

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

Now, as we attempt to determine the functional form of the second curvilinear coordinate, ${\lambda }_2(x,y,z)$, a seemingly useful intermediate step is to determine the functional form of each of the three partial derivatives of this key coordinate function, namely, $\partial {\lambda }_2/\partial x_i$, for i = 1, 3. Here, we will accomplish this intermediate step by guessing the functional form of the second scale factor, $h_2(x,y,z)$, then applying the relation,

$\frac{\partial {\lambda }_2}{\partial x_i}$

$=$

$\frac{{\gamma }_{2i}}{h_2}.$

Notice that, without violating the above-state orthogonality relationship, we can adopt virtually any functional form for $h_2(x,y,z)$ and deduce that,

$\frac{\partial {\lambda }_2}{\partial x}$

$=$

$A(x,y,z)(x{p}^{2}z),$

$\frac{\partial {\lambda }_2}{\partial y}$

$=$

$A(x,y,z)(q^{2}y{p}^{2}z),$

$\frac{\partial {\lambda }_2}{\partial z}$

$=$

$-A(x,y,z)(x^2+q^4{y}^2),$

as long as,

$A(x,y,z)$

$\equiv$

$\frac{{\ell }_q{\ell }_{3D}}{h_2}.$

This key, leading coefficient function is unity — and, hence, is independent of position — if, as in our First speculation above, we guess that $h_2^2=({\ell }_q{\ell }_{3D}{)}^2$. If, as in our Second speculation above, we guess that $h_2^2=({\ell }_q{\ell }_{3D}{)}^{-2}$, we find that, $A=({\ell }_q{\ell }_{3D}{)}^2$. Our above Third speculation is replicated if we guess that $h_2^2=({\lambda }_2{\ell }_q{\ell }_{3D}{)}^2$; we immediately see that, in this Third case,

$\frac{\partial {\lambda }_2}{\partial x}$

$=$

$\frac{x{p}^{2}z}{{\lambda }_2},$

$\frac{\partial {\lambda }_2}{\partial y}$

$=$

$\frac{q^{2}y{p}^{2}z}{{\lambda }_2},$

$\frac{\partial {\lambda }_2}{\partial z}$

$=$

$-\frac{(x^2+q^4{y}^2)}{{\lambda }_2}.$

Study the Functional Forms

We know the functional forms of two of the desired curvilinear coordinates, namely,

${\lambda }_1(x,y,z)$	$\equiv$	$(x^2+q^2{y}^2+p^2{z}^2{)}^{1/2},$
${\lambda }_3(x,y,z)$	$\equiv$	$\frac{y^{1/q^2}}{x},$

but we do not yet have a valid expression for the 2^nd coordinate, ${\lambda }_2(x,y,z)$. Nevertheless, let's see if we can guess the functional forms for $x_i({\lambda }_1,{\lambda }_2,{\lambda }_3)$, by inverting the two known curvilinear-coordinate functions. As a starting point, let's impose the following mappings:

$x$	$\to$	$\frac{y^{1/q^2}}{{\lambda }_3},$
$z$	$\to$	$\frac{1}{p}[{\lambda }_1^2-q^2{y}^2-x^2{]}^{1/2}$	$\to$	$\frac{1}{p}[{\lambda }_1^2-q^2{y}^2-\frac{y^{2/q^2}}{{\lambda }_3^2}{]}^{1/2}=\frac{1}{p{\lambda }_3}[{\lambda }_1^2{\lambda }_3^2-(qy{\lambda }_3{)}^2-y^{2/q^2}{]}^{1/2}.$

This means, for example, that,

${\ell }_q^{-2}$	$\to$	$q^4{y}^2+\frac{y^{2/q^2}}{{\lambda }_3^2}={\lambda }_3^{-2}[q^2(qy{\lambda }_3{)}^2+y^{2/q^2}],$
${\ell }_{3D}^{-2}$	$\to$	$q^4{y}^2+\frac{y^{2/q^2}}{{\lambda }_3^2}+p^2[{\lambda }_1^2-q^2{y}^2-\frac{y^{2/q^2}}{{\lambda }_3^2}]={\lambda }_3^{-2}[(q^2-p^2)(qy{\lambda }_3{)}^2+(1-p^2)y^{2/q^2}+p^2{\lambda }_1^2{\lambda }_3^2].$

Derivatives of x

$\frac{\partial x}{\partial {\lambda }_1}$	$=$	$x{\lambda }_1{\ell }_{3D}^2=y^{1/q^2}{\lambda }_1{\lambda }_3[(q^2-p^2)(qy{\lambda }_3{)}^2+(1-p^2)y^{2/q^2}+p^2{\lambda }_1^2{\lambda }_3^2{]}^{-1},$
$\frac{\partial x}{\partial {\lambda }_3}$	$=$	$-q^4{y}^2{\ell }_q^2\left(\frac{x}{{\lambda }_3}\right)=-q^2(qy{)}^2(y^{1/q^2}\left){\lambda }_3\right[q^2(qy{\lambda }_3{)}^2+y^{2/q^2}{]}^{-1},$
$\frac{\partial x}{\partial {\lambda }_2}$	$=$	$h_2{\ell }_q{\ell }_{3D}(x{p}^{2}z)=\left[h_2{\ell }_q{\ell }_{3D}\right]\left(y^{1/q^2}\right)\frac{p}{{\lambda }_3^2}[{\lambda }_1^2{\lambda }_3^2-(qy{\lambda }_3{)}^2-y^{2/q^2}{]}^{1/2}$
	$=$	$h_2\left(y^{1/q^2}\right)p[{\lambda }_1^2{\lambda }_3^2-(qy{\lambda }_3{)}^2-y^{2/q^2}{]}^{1/2}[q^2(qy{\lambda }_3{)}^2+y^{2/q^2}{]}^{-1/2}[(q^2-p^2)(qy{\lambda }_3{)}^2+(1-p^2)y^{2/q^2}+p^2{\lambda }_1^2{\lambda }_3^2{]}^{-1/2}$

Struggling

I have noticed that, in this last set of expressions, there are recurring terms of the form, $(qy{\lambda }_3)$ and $(y^{2/q^2})$. So, while keeping the same definition of the ccordinate, ${\lambda }_1$, let's replace ${\lambda }_2$ and ${\lambda }_3$ with a pair of coordinates defined as follows:

${\lambda }_4\equiv y{\lambda }_3=\frac{y^{(q^2+1)/q^2}}{x},$

and,

${\lambda }_5\equiv y^{2/q^2}.$

This means that,

$y$	$=$	${\lambda }_5^{q^2/2},$
$x$	$=$	$\frac{y^{(q^2-1)/q^2}}{{\lambda }_4}={\lambda }_4^{-1}{\lambda }_5^{(q^2-1)/2},$
$z^2$	$=$	$\frac{1}{p^2}[{\lambda }_1^2-x^2-q^2{y}^2]=\frac{1}{p^2{\lambda }_4^2}[{\lambda }_1^2{\lambda }_4^2-{\lambda }_5^{q^2-1}-q^2{\lambda }_4^2{\lambda }_5^{q^2}].$

Is this a set of orthogonal coordinates? Well …       No!

New Insight

Following the development of our above, Better Organized discussion, we reverted to several hours of pen & paper derivations, primarily investigating whether it will help us to rewrite various expressions using the [MF53] Direction-Cosine Relations. We discovered that if we set,

$h_2^2$

$=$

$[(x{q}^{2}y){\ell }_q{\ell }_{3D}{]}^2,$

then,

$\frac{\partial {\lambda }_2}{\partial x}$

$=$

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

$\frac{\partial {\lambda }_2}{\partial y}$

$=$

$\frac{p^{2}z}{x},$

$\frac{\partial {\lambda }_2}{\partial z}$

$=$

$-[\frac{x}{q^{2}y}+\frac{q^{2}y}{x}].$

This seems to be a promising method of attack because — in all three cases, i = 1,3 — the derivative of ${\lambda }_2$ with respect to $x_i$ does not depend on $x_i$. Perhaps this simplification will help us identify the function that defines ${\lambda }_2$. This proposed prescription for $h_2(x,y,z)$ and some of its implications are reflected in the following "New Insight" table. (Keep in mind that, although the expressions for ${\gamma }_{21},{\gamma }_{22},\mathrm{a}\mathrm{n}\mathrm{d}{\gamma }_{23}$ remain correct, the tabulated expression is a guess for $h_2$ and, hence, the tabulated expressions for all three $\partial {\lambda }_2/\partial x_i$ are pure speculation.)

New Insight
$n$	${\lambda }_n$	$h_n$	$\frac{\partial {\lambda }_n}{\partial x}$	$\frac{\partial {\lambda }_n}{\partial y}$	$\frac{\partial {\lambda }_n}{\partial z}$	${\gamma }_{n1}$	${\gamma }_{n2}$	${\gamma }_{n3}$
$1$	$(x^2+q^2{y}^2+p^2{z}^2{)}^{1/2}$	${\lambda }_1{\ell }_{3D}$	$\frac{x}{{\lambda }_1}$	$\frac{q^{2}y}{{\lambda }_1}$	$\frac{p^{2}z}{{\lambda }_1}$	$(x){\ell }_{3D}$	$(q^{2}y){\ell }_{3D}$	$(p^{2}z){\ell }_{3D}$
$2$	---	${\ell }_q{\ell }_{3D}(x{q}^{2}y)$	$\frac{p^{2}z}{q^{2}y}$	$\frac{p^{2}z}{x}$	$-[\frac{x}{q^{2}y}+\frac{q^{2}y}{x}]$	${\ell }_q{\ell }_{3D}(x{q}^{2}y)\left[\frac{p^{2}z}{q^{2}y}\right]$	${\ell }_q{\ell }_{3D}(x{q}^{2}y)\left[\frac{p^{2}z}{x}\right]$	$-{\ell }_q{\ell }_{3D}(x{q}^{2}y)[\frac{x}{q^{2}y}+\frac{q^{2}y}{x}]$
$3$	$\frac{y^{1/q^2}}{x}$	$\frac{x{q}^{2}y{\ell }_q}{{\lambda }_3}$	$-\frac{{\lambda }_3}{x}$	$+\frac{{\lambda }_3}{q^{2}y}$	$0$	$-q^{2}y{\ell }_q$	$x{\ell }_q$	$0$

See Also

• Direction Cosines