Appendix/Ramblings/ConcentricEllipsoidalT12Coordinates

Concentric Ellipsoidal (T12) 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.

The 1^st coordinate and its associated unit vector are as follows:

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

where,

${\ell }_{3D}$

$\equiv$

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

Generalized Prescription for 2^nd Coordinate

Default

Let's adopt the following generalized prescription for the 2^nd coordinate:

${\lambda }_2$

$\equiv$

$x^a{y}^b{z}^c,$

in which case,

${\hat{e}}_2$

$=$

$\frac{1}{\mathfrak{𝔏}}\left[\hat{ı}\right(\frac{yz}{bc})+\hat{ȷ}(\frac{xz}{ac})+\hat{k}(\frac{xy}{ab}\left)\right],$

where,

${\mathfrak{𝔏}}^2$

$\equiv$

$\frac{1}{a^2{b}^2{c}^2}[a^2(yz{)}^2+b^2(xz{)}^2+c^2(xy{)}^2].$

Now, to ensure that ${\hat{e}}_2$ is perpendicular to ${\hat{e}}_1$, we need,

${\hat{e}}_1\cdot {\hat{e}}_2$	$=$	$0$
$\Rightarrow 0$	$=$	$\frac{{\ell }_{3D}}{\mathfrak{𝔏}}[\frac{xyz}{bc}+\frac{q^{2}xyz}{ac}+\frac{p^{2}xyz}{ab}]=\frac{{\ell }_{3D}(xyz)}{\mathfrak{𝔏}(abc)}[a+q^{2}b+p^{2}c]$
$\Rightarrow 0$	$=$	$[a+q^{2}b+p^{2}c].$

Henceforth, we will refer to this algebraic relation as the "One-Two Perpendicular Constraint."

Arctangent

Instead, let's try,

${\lambda }_2$

$\equiv$

${\tan }^{-1}[x^a{y}^b{z}^c].$

Then,

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

$=$

$\left[\frac{1}{1+{\tan }^2{\lambda }_2}\right]\frac{\partial }{\partial x_i}\left[x^a{y}^b{z}^c\right]={\cos }^2{\lambda }_2\cdot \frac{\partial }{\partial x_i}\left[x^a{y}^b{z}^c\right];$

that is,

$\frac{\partial {\lambda }_2}{\partial x}$	$=$	${\cos }^2{\lambda }_2\cdot \left[x^a{y}^b{z}^c\right]\frac{a}{x}={\cos }^2{\lambda }_2\cdot [\tan {\lambda }_2]\frac{a}{x}=\sin {\lambda }_2\cos {\lambda }_2\left(\frac{a}{x}\right),$
$\frac{\partial {\lambda }_2}{\partial y}$	$=$	$\sin {\lambda }_2\cos {\lambda }_2\left(\frac{b}{y}\right),$
$\frac{\partial {\lambda }_2}{\partial z}$	$=$	$\sin {\lambda }_2\cos {\lambda }_2\left(\frac{c}{z}\right).$

Hence,

$h_2^{-2}$	$=$	${\sin }^2{\lambda }_2{\cos }^2{\lambda }_2[\frac{a^2}{x^2}+\frac{b^2}{y^2}+\frac{c^2}{z^2}]=\frac{{\sin }^2{\lambda }_2{\cos }^2{\lambda }_2}{(xyz{)}^2}[a^2(yz{)}^2+b^2(xz{)}^2+c^2(xy{)}^2]$
$\Rightarrow h_2$	$=$	$\frac{1}{\sin {\lambda }_2\cos {\lambda }_2}\left[\frac{xyz}{\mathfrak{𝔏}(abc)}\right].$

And the associated unit vector is,

${\hat{e}}_2$

$=$

$\frac{1}{\mathfrak{𝔏}}\left[\hat{ı}\right(\frac{yz}{bc})+\hat{ȷ}(\frac{xz}{ac})+\hat{k}(\frac{xy}{ab}\left)\right],$

which is the same as our default situation.

Necessary 3^rd Coordinate

The unit vector associated with the 3^rd coordinate is obtained from the cross product of the first two unit vectors. That is,

${\hat{e}}_3$	$=$	${\hat{e}}_1\times {\hat{e}}_2$
	$=$	$\hat{ı}[e_{1y}{e}_{2z}-e_{1z}{e}_{2y}]+\hat{ȷ}[e_{1z}{e}_{2x}-e_{1x}{e}_{2z}]+\hat{k}[e_{1x}{e}_{2y}-e_{1y}{e}_{2x}]$
	$=$	$\frac{{\ell }_{3D}}{\mathfrak{𝔏}}\left\{\hat{ı}\right[(q^{2}y)\left(\frac{xy}{ab}\right)-(p^{2}z)\left(\frac{xz}{ac}\right)]+\hat{ȷ}[(p^{2}z)\left(\frac{yz}{bc}\right)-(x)\left(\frac{xy}{ab}\right)]+\hat{k}[(x)\left(\frac{xz}{ac}\right)-(q^{2}y)\left(\frac{yz}{bc}\right)\left]\right\}$
	$=$	$\frac{{\ell }_{3D}}{\mathfrak{𝔏}(abc)}\left\{\hat{ı}\right[(c{q}^2{y}^2)-(b{p}^2{z}^2)]x+\hat{ȷ}[(a{p}^2{z}^2)-(c{x}^2)]y+\hat{k}[(b{x}^2)-(a{q}^2{y}^2)\left]z\right\}$

Old Examples

T6 Coordinates

In the set that we have elsewhere referenced as T6 coordinates, we chose: a = - 1, b = q^-2, c = 0. We note, first, that this set of parameter values satisfies the above-defined One-Two Perpendicular Constraint. In this case, our generalized prescription for the 2^nd coordinate generates a unit vector of the form,

${\hat{e}}_2$	$=$	$\frac{1}{{\mathfrak{𝔏}}_{T6}(abc)}[\hat{ı}(ayz)+\hat{ȷ}(bxz)+\hat{k}(cxy)]=\frac{z}{q^2{\mathfrak{𝔏}}_{T6}(abc)}[-\hat{ı}(q^{2}y)+\hat{ȷ}(x)]$
	$=$	$[-\hat{ı}(q^{2}y)+\hat{ȷ}(x)]{\ell }_q,$

where,

${\ell }_q^{-2}\equiv \left[\frac{q^4{\mathfrak{𝔏}}_{T6}^2(abc{)}^2}{z^2}\right]$

$=$

$\frac{q^4}{z^2}[(yz{)}^2+b^2(xz{)}^2]=[x^2+q^4{y}^2].$

And it implies a unit vector for the 3^rd coordinate of the form,

${\hat{e}}_3$	$=$	$\frac{{\ell }_{3D}}{{\mathfrak{𝔏}}_{T6}(abc)}\left\{\hat{ı}\right[(c{q}^2{y}^2)-(b{p}^2{z}^2)]x+\hat{ȷ}[(a{p}^2{z}^2)-(c{x}^2)]y+\hat{k}[(b{x}^2)-(a{q}^2{y}^2)\left]z\right\}$
	$=$	${\ell }_{3D}\left(\frac{{\ell }_q{q}^2}{z}\right)\{-\hat{ı}[\frac{p^2{z}^2}{q^2}]x-\hat{ȷ}[p^2{z}^2]y+\hat{k}[\frac{x^2}{q^2}+q^2{y}^2\left]z\right\}$
	$=$	${\ell }_q{\ell }_{3D}\{-\hat{ı}(x{p}^{2}z)-\hat{ȷ}(q^{2}y{p}^{2}z)+\hat{k}(x^2+q^4{y}^2)\}.$

T10 Coordinates

In the set that we have elsewhere referenced as T10 coordinates, we chose: a = 1, b = q^-2, c = - 2p^-2. We note, first, that this set of parameter values satisfies the above-defined One-Two Perpendicular Constraint. In this case, our generalized prescription for the 2^nd coordinate generates a unit vector of the form,

${\hat{e}}_2$

$=$

$\frac{1}{{\mathfrak{𝔏}}_{T10}(abc)}[\hat{ı}(ayz)+\hat{ȷ}(bxz)+\hat{k}(cxy)]=\frac{1}{q^2{p}^2{\mathfrak{𝔏}}_{T10}(abc)}[\hat{ı}(q^{2}y{p}^{2}z)+\hat{ȷ}(x{p}^{2}z)-\hat{k}(2x{q}^{2}y)]$

where,

$(abc{)}^2{\mathfrak{𝔏}}_{T10}^2$	$\equiv$	$[a^2(yz{)}^2+b^2(xz{)}^2+c^2(xy{)}^2]=[y^2{z}^2+\frac{x^2{z}^2}{q^4}+\frac{4x^2{y}^2}{p^4}]$
$\Rightarrow {{𝒟}}^2\equiv q^4{p}^4(abc{)}^2{\mathfrak{𝔏}}_{T10}^2$	$=$	$[q^4{y}^2{p}^4{z}^2+x^2{p}^4{z}^2+4x^2{q}^4{y}^2].$

And it implies a unit vector for the 3^rd coordinate of the form,

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

Develop 3^rd-Coordinate Profile

Setup

Reflecting back on an earlier exploration, let's define the two polynomials,

$\mathfrak{𝔄}\equiv {\ell }_{3D}^{-2}$	$=$	$(x^2+q^4{y}^2+p^4{z}^2),$
$\mathfrak{𝔅}\equiv [\mathfrak{𝔏}(abc){]}^2$	$=$	$[a^2(yz{)}^2+b^2(xz{)}^2+c^2(xy{)}^2].$

$\frac{\partial \mathfrak{𝔄}}{\partial x}$ $=$ $2x,$ $\frac{\partial \mathfrak{𝔄}}{\partial y}$ $=$ $2q^{4}y,$ $\frac{\partial \mathfrak{𝔄}}{\partial z}$ $=$ $2p^{4}z;$ $\frac{\partial \mathfrak{𝔅}}{\partial x}$ $=$ $2x(b^2{z}^2+c^2{y}^2),$ $\frac{\partial \mathfrak{𝔅}}{\partial y}$ $=$ $2y(a^2{z}^2+c^2{x}^2),$ $\frac{\partial \mathfrak{𝔅}}{\partial z}$ $=$ $2z(a^2{y}^2+b^2{x}^2).$

Then the 3^rd unit vector may be written as,

$[{\hat{e}}_3{]}_{\mathrm{n}\mathrm{e}\mathrm{e}\mathrm{d}\mathrm{e}\mathrm{d}}$

$=$

${\mathfrak{𝔄}}^{-1/2}{\mathfrak{𝔅}}^{-1/2}\left\{\hat{ı}\right[(c{q}^2{y}^2)-(b{p}^2{z}^2)]x+\hat{ȷ}[(a{p}^2{z}^2)-(c{x}^2)]y+\hat{k}[(b{x}^2)-(a{q}^2{y}^2)\left]z\right\}.$

Guess Third Coordinate Expression

Let's see what unit vector results if we define,

${\lambda }_3$

$\equiv$

${\mathfrak{𝔄}}^{1/2}{\mathfrak{𝔅}}^{-1/2}.$

Partial Derivatives

$\frac{\partial {\lambda }_3}{\partial x_i}$	$=$	$\left[\frac{1}{2}{\mathfrak{𝔄}}^{-1/2}{\mathfrak{𝔅}}^{-1/2}\right]\frac{\partial \mathfrak{𝔄}}{\partial x_i}-\left[\frac{1}{2}{\mathfrak{𝔄}}^{1/2}{\mathfrak{𝔅}}^{-3/2}\right]\frac{\partial \mathfrak{𝔅}}{\partial x_i}$
$\Rightarrow \left[\frac{2\mathfrak{𝔄}\mathfrak{𝔅}}{{\lambda }_3}\right]\frac{\partial {\lambda }_3}{\partial x_i}$	$=$	$\mathfrak{𝔅}\cdot \frac{\partial \mathfrak{𝔄}}{\partial x_i}-\mathfrak{𝔄}\cdot \frac{\partial \mathfrak{𝔅}}{\partial x_i}.$

First, note that,

$\left[\frac{2\mathfrak{𝔄}\mathfrak{𝔅}}{{\lambda }_3}\right]\frac{\partial {\lambda }_3}{\partial x}$	$=$	$\mathfrak{𝔅}\left[2x\right]-\mathfrak{𝔄}[2x(b^2{z}^2+c^2{y}^2)]$
	$=$	$2x[a^2(yz{)}^2+b^2(xz{)}^2+c^2(xy{)}^2-(x^2+q^4{y}^2+p^4{z}^2)(b^2{z}^2+c^2{y}^2)]$
	$=$	$2x[a^2(yz{)}^2+b^2(xz{)}^2+c^2(xy{)}^2-x^2(b^2{z}^2+c^2{y}^2)-q^4{y}^2(b^2{z}^2+c^2{y}^2)-p^4{z}^2(b^2{z}^2+c^2{y}^2)]$
	$=$	$2x\{(yz{)}^2[a^2-q^4{b}^2-c^2{p}^4]+(xz{)}^2[b^2-b^2]+(xy{)}^2[c^2-c^2]-c^2{q}^4{y}^4-b^2{p}^4{z}^4\}$
	$=$	$2x[y^2{z}^2(a^2-q^4{b}^2-c^2{p}^4)-c^2{q}^4{y}^4-b^2{p}^4{z}^4];$

$\left[\frac{2\mathfrak{𝔄}\mathfrak{𝔅}}{{\lambda }_3}\right]\frac{\partial {\lambda }_3}{\partial y}$	$=$	$\mathfrak{𝔅}\left[2q^{4}y\right]-\mathfrak{𝔄}[2y(a^2{z}^2+c^2{x}^2)]$
	$=$	$2y[q^4[a^2(yz{)}^2+b^2(xz{)}^2+c^2(xy{)}^2]-(a^2{z}^2+c^2{x}^2)(x^2+q^4{y}^2+p^4{z}^2)]$
	$=$	$2y[a^2{q}^4(yz{)}^2+b^2{q}^4(xz{)}^2+c^2{q}^4(xy{)}^2-a^2{z}^2(x^2+q^4{y}^2+p^4{z}^2)-c^2{x}^2(x^2+q^4{y}^2+p^4{z}^2)]$
	$=$	$2y\{(yz{)}^2[a^2{q}^4-a^2{q}^4]+(xz{)}^2[b^2{q}^4-a^2-c^2{p}^4]+(xy{)}^2[c^2{q}^4-c^2{q}^4]-a^2{p}^4{z}^4-c^2{x}^4\}$
	$=$	$2y[x^2{z}^2(b^2{q}^4-a^2-c^2{p}^4)-a^2{p}^4{z}^4-c^2{x}^4];$

$\left[\frac{2\mathfrak{𝔄}\mathfrak{𝔅}}{{\lambda }_3}\right]\frac{\partial {\lambda }_3}{\partial z}$	$=$	$\mathfrak{𝔅}\left[2p^{4}z\right]-\mathfrak{𝔄}[2z(a^2{y}^2+b^2{x}^2)]$
	$=$	$2z[p^4[a^2(yz{)}^2+b^2(xz{)}^2+c^2(xy{)}^2]-(a^2{y}^2+b^2{x}^2)(x^2+q^4{y}^2+p^4{z}^2)]$
	$=$	$2z[a^2{p}^4(yz{)}^2+b^2{p}^4(xz{)}^2+c^2{p}^4(xy{)}^2-a^2{y}^2(x^2+q^4{y}^2+p^4{z}^2)-b^2{x}^2(x^2+q^4{y}^2+p^4{z}^2)]$
	$=$	$2z\{(yz{)}^2[a^2{p}^4-a^2{p}^4]+(xz{)}^2[b^2{p}^4-b^2{p}^4]+(xy{)}^2[c^2{p}^4-a^2-b^2{q}^4]-a^2{q}^4{y}^4-b^2{x}^4\}$
	$=$	$2z[(xy{)}^2(c^2{p}^4-a^2-b^2{q}^4)-a^2{q}^4{y}^4-b^2{x}^4].$

After completing a few squares, this last expression may be rewritten as …

$\left[\frac{2\mathfrak{𝔄}\mathfrak{𝔅}}{{\lambda }_3}\right]\frac{\partial {\lambda }_3}{\partial z}$	$=$	$2z[(xy{)}^2(c^2{p}^4-a^2-b^2{q}^4)-a^2{q}^4{y}^4-b^2{x}^4]$
	$=$	$2z[(xy{)}^2(c^2{p}^4-a^2-b^2{q}^4)\pm 2ab{q}^2(xy{)}^2-(a{q}^2{y}^2\pm b{x}^2{)}^2]$
	$=$	$2z[(xy{)}^2(c^2{p}^4-a^2-b^2{q}^4\pm 2ab{q}^2)-(a{q}^2{y}^2\pm b{x}^2{)}^2]$
	$=$	$2z[(xy{)}^2[c^2{p}^4-(a\pm b{q}^2{)}^2\pm 4ab{q}^2]-(a{q}^2{y}^2\pm b{x}^2{)}^2].$

Now, if we choose the superior sign throughout this expression, the above-derived One-Two Perpendicular Constraint can be satisfied by setting, $(a+b{q}^2)=-c{p}^2$. The expression then becomes,

$\left[\frac{2\mathfrak{𝔄}\mathfrak{𝔅}}{{\lambda }_3}\right]\frac{\partial {\lambda }_3}{\partial z}$	$=$	$-2z[(a{q}^2{y}^2+b{x}^2{)}^2-4ab{q}^2(xy{)}^2]$
	$=$	$-2z(a{q}^2{y}^2-b{x}^2{)}^2.$

---

Alternatively, suppose ${\lambda }_3$ $\equiv$ ${\mathfrak{𝔄}}^{m/2}{\mathfrak{𝔅}}^{-n/2}$ $\Rightarrow \left[\frac{2\mathfrak{𝔄}\mathfrak{𝔅}}{{\lambda }_3}\right]\frac{\partial {\lambda }_3}{\partial x_i}$ $=$ $m\mathfrak{𝔅}\cdot \frac{\partial \mathfrak{𝔄}}{\partial x_i}-n\mathfrak{𝔄}\cdot \frac{\partial \mathfrak{𝔅}}{\partial x_i}.$ Then we have, for example, $\frac{1}{2z}\left[\frac{2\mathfrak{𝔄}\mathfrak{𝔅}}{{\lambda }_3}\right]\frac{\partial {\lambda }_3}{\partial z}$ $=$ $m\mathfrak{𝔅}\left[p^4\right]-n\mathfrak{𝔄}[(a^2{y}^2+b^2{x}^2)]$   $=$ $m{p}^4[a^2(yz{)}^2+b^2(xz{)}^2+c^2(xy{)}^2]-n(x^2+q^4{y}^2+p^4{z}^2)(a^2{y}^2+b^2{x}^2)$   $=$ $m{p}^4[a^2(yz{)}^2+b^2(xz{)}^2+c^2(xy{)}^2]-n[(x^2+q^4{y}^2+p^4{z}^2)a^2{y}^2+(x^2+q^4{y}^2+p^4{z}^2)b^2{x}^2]$   $=$ $(m-n)a^2{p}^4(yz{)}^2+(m-n)b^2{p}^4(xz{)}^2+(xy{)}^2[m{c}^2{p}^4-n{a}^2-n{b}^2{q}^4]-n{a}^2{q}^4{y}^4-n{b}^2{x}^4$   $=$ $(m-n)p^4[a^2(yz{)}^2+b^2(xz{)}^2]+(xy{)}^2[m{c}^2{p}^4-n(a^2+b^2{q}^4)]-n[(a{q}^2{y}^2\pm b{x}^2{)}^2\mp 2ab{q}^2(xy{)}^2].$ Choosing the inferior sign then enforcing the above-derived One-Two Perpendicular Constraint by setting, $(a+b{q}^2)=-c{p}^2$, gives, $\frac{1}{2z}\left[\frac{2\mathfrak{𝔄}\mathfrak{𝔅}}{{\lambda }_3}\right]\frac{\partial {\lambda }_3}{\partial z}$ $=$ $(m-n)p^4[a^2(yz{)}^2+b^2(xz{)}^2]+(xy{)}^2[m{c}^2{p}^4-n(a^2+b^2{q}^4+2ab{q}^2)]-n(a{q}^2{y}^2-b{x}^2{)}^2$   $=$ $(m-n)p^4[a^2(yz{)}^2+b^2(xz{)}^2]+(xy{)}^2[m{c}^2{p}^4-n(-c{p}^2{)}^2]-n(a{q}^2{y}^2-b{x}^2{)}^2$   $=$ $(m-n)p^4[a^2(yz{)}^2+b^2(xz{)}^2+c^2(xy{)}^2]-n(b{x}^2-a{q}^2{y}^2{)}^2$   $=$ $(m-n)p^4\mathfrak{𝔅}-n(b{x}^2-a{q}^2{y}^2{)}^2$   $=$ $m{p}^4\mathfrak{𝔅}-n[p^4\mathfrak{𝔅}+(b{x}^2-a{q}^2{y}^2{)}^2].$ Reflecting back on the first line of the "example" derivation, we recognize that, $p^4\mathfrak{𝔅}+(b{x}^2-a{q}^2{y}^2{)}^2$ $=$ $\mathfrak{𝔄}(a^2{y}^2+b^2{x}^2)$ $\Rightarrow (b{x}^2-a{q}^2{y}^2{)}^2$ $=$ $\mathfrak{𝔄}(a^2{y}^2+b^2{x}^2)-p^4\mathfrak{𝔅}$

---

Similarly, we find,

$\left[\frac{2\mathfrak{𝔄}\mathfrak{𝔅}}{{\lambda }_3}\right]\frac{\partial {\lambda }_3}{\partial x}$	$=$	$2x[y^2{z}^2(a^2-q^4{b}^2-c^2{p}^4)-c^2{q}^4{y}^4-b^2{p}^4{z}^4]$
	$=$	$2x[y^2{z}^2(a^2-q^4{b}^2-c^2{p}^4)-(c{q}^2{y}^2\pm b{p}^2{z}^2{)}^2\pm 2bc{q}^2{y}^2{p}^2{z}^2]$
	$=$	$2x[(yz{)}^2(a^2-q^4{b}^2-c^2{p}^4\pm 2bc{q}^2{p}^2)-(c{q}^2{y}^2\pm b{p}^2{z}^2{)}^2]$
	$=$	$2x[(yz{)}^2[a^2-(b{q}^2\pm c{p}^2{)}^2\pm 4bc{q}^2{p}^2]-(c{q}^2{y}^2\pm b{p}^2{z}^2{)}^2].$

$\left[\frac{2\mathfrak{𝔄}\mathfrak{𝔅}}{{\lambda }_3}\right]\frac{\partial {\lambda }_3}{\partial y}$	$=$	$2y[x^2{z}^2(b^2{q}^4-a^2-c^2{p}^4)-a^2{p}^4{z}^4-c^2{x}^4]$
	$=$	$2y[x^2{z}^2(b^2{q}^4-a^2-c^2{p}^4)-(a{p}^2{z}^2\pm c{x}^2{)}^2\pm 2ac{x}^2{p}^2{z}^2]$
	$=$	$2y[x^2{z}^2(b^2{q}^4-a^2-c^2{p}^4\pm 2ac{p}^2)-(a{p}^2{z}^2\pm c{x}^2{)}^2]$
	$=$	$2y[(xz{)}^2[b^2{q}^4-(a\pm c{p}^2{)}^2\pm 4ac{p}^2]-(a{p}^2{z}^2\pm c{x}^2{)}^2].$

So, if we again choose the superior sign throughout these expression, the above-derived One-Two Perpendicular Constraint can be satisfied:   In the first by setting, $(b{q}^2+c{p}^2)=-a$; and in the second by setting, $(a+c{p}^2)=-b{q}^2$. The expressions then become, respectively,

$\left[\frac{2\mathfrak{𝔄}\mathfrak{𝔅}}{{\lambda }_3}\right]\frac{\partial {\lambda }_3}{\partial x}$	$=$	$-2x[(c{q}^2{y}^2+b{p}^2{z}^2{)}^2-4bc{q}^2{y}^2{p}^2{z}^2]$
	$=$	$-2x(c{q}^2{y}^2-b{p}^2{z}^2{)}^2;$
$\left[\frac{2\mathfrak{𝔄}\mathfrak{𝔅}}{{\lambda }_3}\right]\frac{\partial {\lambda }_3}{\partial y}$	$=$	$-2y[(a{p}^2{z}^2+c{x}^2{)}^2-4ac{x}^2{p}^2{z}^2]$
	$=$	$-2y(a{p}^2{z}^2-c{x}^2{)}^2.$

Summary Given, ${\lambda }_3$ $\equiv$ ${\mathfrak{𝔄}}^{1/2}{\mathfrak{𝔅}}^{-1/2}=[\frac{(x^2+q^4{y}^2+p^4{z}^2)}{a^2(yz{)}^2+b^2(xz{)}^2+c^2(xy{)}^2}{]}^{1/2}$ the three relevant partial derivatives are: $\left[\frac{\mathfrak{𝔄}\mathfrak{𝔅}}{{\lambda }_3}\right]\frac{\partial {\lambda }_3}{\partial x}$ $=$ $-x(c{q}^2{y}^2-b{p}^2{z}^2{)}^2,$ $\left[\frac{\mathfrak{𝔄}\mathfrak{𝔅}}{{\lambda }_3}\right]\frac{\partial {\lambda }_3}{\partial y}$ $=$ $-y(a{p}^2{z}^2-c{x}^2{)}^2,$ $\left[\frac{\mathfrak{𝔄}\mathfrak{𝔅}}{{\lambda }_3}\right]\frac{\partial {\lambda }_3}{\partial z}$ $=$ $-z(a{q}^2{y}^2-b{x}^2{)}^2.$

Scale Factor

Hence, the associated scale factor is,

$h_3^{-2}$	$\equiv$	$(\frac{\partial {\lambda }_3}{\partial x}{)}^2+(\frac{\partial {\lambda }_3}{\partial y}{)}^2+(\frac{\partial {\lambda }_3}{\partial z}{)}^2$
$$	$=$	$\left[\frac{{\lambda }_3}{\mathfrak{𝔄}\mathfrak{𝔅}}{]}^2\right\{[-x(c{q}^2{y}^2-b{p}^2{z}^2{)}^2{]}^2+[-y(a{p}^2{z}^2-c{x}^2{)}^2{]}^2+[-z(a{q}^2{y}^2-b{x}^2{)}^2{]}^2\}$
$$	$=$	$\left[\frac{{\lambda }_3}{\mathfrak{𝔄}\mathfrak{𝔅}}{]}^2\right\{x^2(c{q}^2{y}^2-b{p}^2{z}^2{)}^4+y^2(a{p}^2{z}^2-c{x}^2{)}^4+z^2(a{q}^2{y}^2-b{x}^2{)}^4\}$
$\Rightarrow h_3$	$=$	$\left[\frac{\mathfrak{𝔄}\mathfrak{𝔅}}{{\lambda }_3}\right]\frac{1}{\mathfrak{ℭ}},$

where,

$\mathfrak{ℭ}$

$\equiv$

$[x^2(c{q}^2{y}^2-b{p}^2{z}^2{)}^4+y^2(a{p}^2{z}^2-c{x}^2{)}^4+z^2(a{q}^2{y}^2-b{x}^2{)}^4{]}^{1/2}.$

Direction Cosines and Unit Vector

And the associated triplet of direction cosines is:

${\gamma }_{31}\equiv h_3\left(\frac{\partial {\lambda }_3}{\partial x}\right)$	$=$	$-x(c{q}^2{y}^2-b{p}^2{z}^2{)}^2[\frac{1}{\mathfrak{ℭ}}],$
${\gamma }_{32}\equiv h_3\left(\frac{\partial {\lambda }_3}{\partial y}\right)$	$=$	$-y(a{p}^2{z}^2-c{x}^2{)}^2[\frac{1}{\mathfrak{ℭ}}],$
${\gamma }_{33}\equiv h_3\left(\frac{\partial {\lambda }_3}{\partial y}\right)$	$=$	$-z(a{q}^2{y}^2-b{x}^2{)}^2[\frac{1}{\mathfrak{ℭ}}].$

This means that, for our particular guess of the 3^rd coordinate, the relevant unit vector is,

$[{\hat{e}}_3{]}_{\mathrm{g}\mathrm{u}\mathrm{e}\mathrm{s}\mathrm{s}}$

$=$

$\frac{1}{\mathfrak{ℭ}}\{-\hat{ı}[x(c{q}^2{y}^2-b{p}^2{z}^2{)}^2]-\hat{ȷ}[y(a{p}^2{z}^2-c{x}^2{)}^2]-\hat{k}[z(a{q}^2{y}^2-b{x}^2{)}^2]\}.$

Contrast

The unit vector resulting (just derived) from our guess of the third-coordinate expression should be compared with the needed unit vector as described above, namely,

$[{\hat{e}}_3{]}_{\mathrm{n}\mathrm{e}\mathrm{e}\mathrm{d}\mathrm{e}\mathrm{d}}$

$=$

${\mathfrak{𝔄}}^{-1/2}{\mathfrak{𝔅}}^{-1/2}\left\{\hat{ı}\right[x(c{q}^2{y}^2-b{p}^2{z}^2)]+\hat{ȷ}[y(a{p}^2{z}^2-c{x}^2)]+\hat{k}[z(b{x}^2-a{q}^2{y}^2)\left]\right\}.$

At first glance, apart from the difference in signs, the three terms inside the curly braces appear to be identical in these two separate unit-vector expressions. But they are not! Relative to the needed expression, key components of each term are squared in the guessed expression. Very close … but no cigar!

ASIDE Note that, $[{\hat{e}}_3{]}_{\mathrm{n}\mathrm{e}\mathrm{e}\mathrm{d}\mathrm{e}\mathrm{d}}\cdot [{\hat{e}}_3{]}_{\mathrm{n}\mathrm{e}\mathrm{e}\mathrm{d}\mathrm{e}\mathrm{d}}=1$ implies that, $\mathfrak{𝔄}\mathfrak{𝔅}=[\frac{(abc)\mathfrak{𝔏}}{{\ell }_{3D}}{]}^2$ $=$ $[x(c{q}^2{y}^2-b{p}^2{z}^2){]}^2+[y(a{p}^2{z}^2-c{x}^2){]}^2+[z(b{x}^2-a{q}^2{y}^2){]}^2.$ But we also know that (see, for example, immediately below), $\mathfrak{𝔄}\mathfrak{𝔅}$ $=$ $(x^2+q^4{y}^2+p^4{z}^2)[a^2(yz{)}^2+b^2(xz{)}^2+c^2(xy{)}^2].$

What about the overall leading coefficient? That is, does $\mathfrak{𝔄}\mathfrak{𝔅}={\mathfrak{ℭ}}^2$  ?  Well, given that,

$\mathfrak{𝔄}={\ell }_{3D}^{-2}$

$=$

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

and,

$\mathfrak{𝔅}=(abc{)}^2{\mathfrak{𝔏}}^2$

$=$

$[a^2(yz{)}^2+b^2(xz{)}^2+c^2(xy{)}^2],$

we have,

$\mathfrak{𝔄}\mathfrak{𝔅}$	$=$	$(x^2+q^4{y}^2+p^4{z}^2)[a^2(yz{)}^2+b^2(xz{)}^2+c^2(xy{)}^2]$
	$=$	$x^2[a^2(yz{)}^2+b^2(xz{)}^2+c^2(xy{)}^2]+q^4{y}^2[a^2(yz{)}^2+b^2(xz{)}^2+c^2(xy{)}^2]+p^4{z}^2[a^2(yz{)}^2+b^2(xz{)}^2+c^2(xy{)}^2]$
	$=$	$x^2{y}^2{z}^2[a^2+b^2{q}^4+c^2{p}^4]+x^4[b^2{z}^2+c^2{y}^2]+q^4{y}^4[a^2{z}^2+c^2{x}^2]+p^4{z}^4[a^2{y}^2+b^2{x}^2]$

On the other hand,

${\mathfrak{ℭ}}^2$

$\equiv$

$x^2(c{q}^2{y}^2-b{p}^2{z}^2{)}^4+y^2(a{p}^2{z}^2-c{x}^2{)}^4+z^2(a{q}^2{y}^2-b{x}^2{)}^4.$

Dot With 1^st Unit Vector

Is $[{\hat{e}}_3{]}_{\mathrm{g}\mathrm{u}\mathrm{e}\mathrm{s}\mathrm{s}}$ orthogonal to ${\hat{e}}_1$? Let's take their dot product to see; note that, for simplicity, we will flip the sign on $[{\hat{e}}_3{]}_{\mathrm{g}\mathrm{u}\mathrm{e}\mathrm{s}\mathrm{s}}$.

$-{\hat{e}}_1\cdot [{\hat{e}}_3{]}_{\mathrm{g}\mathrm{u}\mathrm{e}\mathrm{s}\mathrm{s}}[\frac{\mathfrak{ℭ}}{{\ell }_{3D}}]$	$=$	$\{\hat{ı}(x)+\hat{ȷ}(q^{2}y)+\hat{k}(p^{2}z)\}\cdot \left\{\hat{ı}\right[x(c{q}^2{y}^2-b{p}^2{z}^2{)}^2]+\hat{ȷ}[y(a{p}^2{z}^2-c{x}^2{)}^2]+\hat{k}[z(a{q}^2{y}^2-b{x}^2{)}^2]\}$
	$=$	$x^2(c{q}^2{y}^2-b{p}^2{z}^2{)}^2+q^2{y}^2(a{p}^2{z}^2-c{x}^2{)}^2+p^2{z}^2(a{q}^2{y}^2-b{x}^2{)}^2$
	$=$	$x^2(c^2{q}^4{y}^4-2c{q}^2{y}^{2}b{p}^2{z}^2+b^2{p}^4{z}^4)+q^2{y}^2(a^2{p}^4{z}^4-2ac{x}^2{p}^2{z}^2+c^2{x}^4)+p^2{z}^2(a^2{q}^4{y}^4-2a{q}^2{y}^{2}b{x}^2+b^2{x}^4)$
	$=$	$-2x^2{y}^2{z}^2[c{q}^{2}b{p}^2+a{q}^{2}c{p}^2+a{q}^{2}b{p}^2]+x^2(c^2{q}^4{y}^4+b^2{p}^4{z}^4)+q^2{y}^2(a^2{p}^4{z}^4+c^2{x}^4)+p^2{z}^2(a^2{q}^4{y}^4+b^2{x}^4).$

It does not appear as though the RHS of this expression can be zero for all values of the Cartesian coordinates, (x, y, z). Hence $[{\hat{e}}_3{]}_{\mathrm{g}\mathrm{u}\mathrm{e}\mathrm{s}\mathrm{s}}$ is not orthogonal to ${\hat{e}}_1$.

See Also

• Direction Cosines
• Trials up through T7 Coordinates