Appendix/Ramblings/EllipticCylinderCoordinates

Elliptic Cylinder Coordinates

Background

Building on our general introduction to Direction Cosines in the context of orthogonal curvilinear coordinate systems, here we detail the properties of Elliptic Cylinder Coordinates. First, we present this coordinate system in the manner described by [MF53]; second, we provide an alternate presentation, obtained from Wikipedia; then, third, we investigate whether or not a related coordinate system based on concentric (rather than confocal) elliptic surfaces can be satisfactorily described.

It is useful to keep in mind various properties of a set of confocal ellipses in which the location of the pair of foci is fixed at, $(x,y)=(\pm c,0)$, and the semi-major axis, $a$, is the parameter. The relevant prescriptive relation is,

$1$

$=$

$\frac{x^2}{a^2}+\frac{y^2}{a^2-c^2}$      for,   $a>c.$

The semi-minor axis length, $b$, and the eccentricity, $e$, of the ellipse are, respectively,

$b$

$=$

$(a^2-c^2{)}^{1/2},$

and,

$e\equiv [1-\frac{b^2}{a^2}{]}^{1/2}$

$=$

$\frac{c}{a}.$

The length, ${\ell }_1$, of the chord that connects one focus to a point, $P(x,y)$, on the ellipse is,

${\ell }_1$

$=$

$a+\left(\frac{c}{a}\right)x;$

and the length, ${\ell }_2$, of the chord that connects the second focus to that same point on the ellipse is,

${\ell }_2$

$=$

$a-\left(\frac{c}{a}\right)x.$

It is easy to see that, for any point on the ellipse, the sum of these two lengths is, $2a$. It is worth noting as well that the associated $y$ coordinate of the relevant point can be obtained from the relation,

${\ell }_1^2$	$=$	$y^2+(c+x{)}^2$
$\Rightarrow (ay{)}^2$	$=$	$(a{\ell }_1{)}^2-(ac+ax{)}^2$
	$=$	$(a^2+cx{)}^2-(ac+ax{)}^2$
	$=$	$(a^4+2a^{2}cx+c^2{x}^2)-(a^2{c}^2+2a^{2}cx+a^2{x}^2)$
	$=$	$(a^4+c^2{x}^2)-(a^2{c}^2+a^2{x}^2)$
	$=$	$(a^2-x^2)(a^2-c^2)$
$\Rightarrow y$	$=$	$\pm \frac{1}{a}[(a^2-x^2)(a^2-c^2){]}^{1/2}.$

MF53

Definition

From MF53's Table of Separable Coordinates in Three Dimensions (see their Chapter 5, beginning on p. 655), we find the following description of Elliptic Cylinder Coordinates (p. 657).

Elliptic Cylindrical Coordinates (MF53 Primary Definition)

$x$ $=$ ${\xi }_1{\xi }_2$ $y$ $=$ $[({\xi }_1^2-d^2)(1-{\xi }_2^2){]}^{1/2}$ $z$ $=$ ${\xi }_3$

Alternate Definition

Making the substitutions, ${\xi }_3\to z$, ${\xi }_2\to \cos \nu$, and ${\xi }_1\to d\cosh \mu$, we equally well obtain: $x$ $=$ $d\cosh \mu \cdot \cos \nu$ $y$ $=$ $d\sinh \mu \cdot \sin \nu$ $z$ $=$ $z$

Notice that,

$\frac{x^2}{d^2{\cosh }^2\mu }+\frac{y^2}{d^2{\sinh }^2\mu }$

$=$

${\cos }^2\nu +{\sin }^2\nu =1.$

Hence, as is pointed out in a related Wikipedia discussion, "… this shows that curves of constant $\mu$ form ellipses." For a given choice of $\mu$ — say, ${\mu }_0$ — let's see how the shape of the resulting ellipse relates to the standard ellipses described in our background discussion, above. The semi-major axis of the selected ellipse must be,

$a=d\cosh {\mu }_0.$

And its eccentricity must be obtainable from the relation,

$a^2-c^2=a^2(1-e^2)$	$=$	$d^2{\sinh }^2{\mu }_0$
	$=$	$a^2{\tanh }^2{\mu }_0=a^2(1-\frac{1}{{\cosh }^2{\mu }_0})$
$\Rightarrow e^2$	$=$	$\frac{1}{{\cosh }^2{\mu }_0}.$

We note, as well, that the x-coordinate location of the focus of the selected ellipse is,

$c^2=a^2{e}^2$

$=$

$d^2.$

This emphasizes a key property of the MF53 Elliptic Cylindrical Coordinate system, viz., the family of ellipses that result from selecting various values of ${\mu }_0$ is a family of confocal ellipses.

Scale Factors

Primary

Appreciating that,

$\frac{\partial y}{\partial {\xi }_1}$	$=$	$+[({\xi }_1^2-d^2)(1-{\xi }_2^2){]}^{-1/2}{\xi }_1(1-{\xi }_2^2),$       and that,
$\frac{\partial y}{\partial {\xi }_2}$	$=$	$-[({\xi }_1^2-d^2)(1-{\xi }_2^2){]}^{-1/2}{\xi }_2({\xi }_1^2-d^2),$

we find that the respective scale factors are given by the expressions,

$h_1^2$	$=$	$(\frac{\partial x}{\partial {\xi }_1}{)}^2+(\frac{\partial y}{\partial {\xi }_1}{)}^2+(\frac{\partial z}{\partial {\xi }_1}{)}^2$
	$=$	${\xi }_2^2+[({\xi }_1^2-d^2)(1-{\xi }_2^2){]}^{-1}{\xi }_1^2(1-{\xi }_2^2{)}^2$
	$=$	$({\xi }_1^2-d^2{)}^{-1}[({\xi }_1^2-d^2){\xi }_2^2+{\xi }_1^2(1-{\xi }_2^2)]$
	$=$	$\left[\frac{{\xi }_1^2-d^2{\xi }_2^2}{{\xi }_1^2-d^2}\right];$
$h_2^2$	$=$	$(\frac{\partial x}{\partial {\xi }_2}{)}^2+(\frac{\partial y}{\partial {\xi }_2}{)}^2+(\frac{\partial z}{\partial {\xi }_2}{)}^2$
	$=$	${\xi }_1^2+[({\xi }_1^2-d^2)(1-{\xi }_2^2){]}^{-1}{\xi }_2^2({\xi }_1^2-d^2{)}^2$
	$=$	$(1-{\xi }_2^2{)}^{-1}[{\xi }_1^2(1-{\xi }_2^2)+{\xi }_2^2({\xi }_1^2-d^2)]$
	$=$	$\left[\frac{{\xi }_1^2-d^2{\xi }_2^2}{1-{\xi }_2^2}\right];$
$h_3^2$	$=$	$(\frac{\partial x}{\partial {\xi }_3}{)}^2+(\frac{\partial y}{\partial {\xi }_3}{)}^2+(\frac{\partial z}{\partial {\xi }_3}{)}^2$
	$=$	$1.$

These match the scale-factor expressions found in MF53.

Alternatively

Alternatively, the Wikipedia discussion gives,

$h_{\mu }=h_{\nu }$	$=$	$d\sqrt{{\sinh }^2\mu +{\sin }^2\nu }$
${\mathrm{\nabla }}^2\mathrm{\Phi }$	$=$	$\frac{1}{d^2({\sinh }^2\mu +{\sin }^2\nu )}[\frac{{\partial }^2\mathrm{\Phi }}{\partial {\mu }^2}+\frac{{\partial }^2\mathrm{\Phi }}{\partial {\nu }^2}]+\frac{{\partial }^2\mathrm{\Phi }}{\partial z^2}.$

Inverting Coordinate Mapping

Inverting the original coordinate mappings, we find,

$y^2$	$=$	$({\xi }_1^2-a^2)[1-(\frac{x}{{\xi }_1}{)}^2]$
$\Rightarrow 0$	$=$	$({\xi }_1^2-a^2)({\xi }_1^2-x^2)-{\xi }_1^2{y}^2$
	$=$	$({\xi }_1^2-a^2){\xi }_1^2-({\xi }_1^2-a^2)x^2-{\xi }_1^2{y}^2$
	$=$	${\xi }_1^4-{\xi }_1^2(a^2+x^2+y^2)+a^2{x}^2$
$\Rightarrow {\xi }_1^2$	$=$	$\frac{1}{2}\{(a^2+x^2+y^2)\pm [(a^2+x^2+y^2{)}^2-4a^2{x}^2{]}^{1/2}\}$

Only the superior — that is, only the positive — sign will ensure positive values of ${\xi }_1^2$, so in summary we have,

Coordinate Transformation ${\xi }_1$ $=$ $\frac{1}{\sqrt{2}}\left\{\right[(a^2+x^2+y^2{)}^2-4a^2{x}^2{]}^{1/2}+(a^2+x^2+y^2){\}}^{1/2};$ ${\xi }_2$ $=$ $\frac{x}{{\xi }_1};$ ${\xi }_3$ $=$ $z.$

Alternative Wikipedia Definition

This same MF53 coordinate system — with different variable notation — is referred to in a Wikipedia discussion as an "alternative and geometrically intuitive set of elliptic coordinates." The relevant mapping is, $(d\sigma ,\tau ,z{)}_{\mathrm{W}\mathrm{i}\mathrm{k}\mathrm{i}\mathrm{p}\mathrm{e}\mathrm{d}\mathrm{i}\mathrm{a}}=({\xi }_1,{\xi }_2,{\xi }_3{)}_{\mathrm{M}\mathrm{F}\mathrm{5}\mathrm{3}}$. The identified mapping to Cartesian coordinates is,

$x$	$=$	$(d\sigma )\tau$	$=$	${\xi }_1{\xi }_2;$
$y$	$=$	$d[({\sigma }^2-1)(1-{\tau }^2){]}^{1/2}$	$=$	$[({\xi }_1^2-d^2)(1-{\xi }_2^2){]}^{1/2};$
$z$	$=$	$z$	$=$	${\xi }_3.$

According to the Wikipedia discussion, the three scale factors are,

$h_{\sigma }^2$

$=$

$d^2\left[\frac{{\sigma }^2-{\tau }^2}{{\sigma }^2-1}\right];$

$h_{\tau }^2$

$=$

$d^2\left[\frac{{\sigma }^2-{\tau }^2}{1-{\tau }^2}\right];$

and,

$h_z^2$

$=$

$1.$

Interestingly, the Wikipedia discussion also includes the following expression for the Laplacian in this elliptic cylindrical coordinate system:

${\mathrm{\nabla }}^2\mathrm{\Phi }$

$=$

$\frac{1}{d^2({\sigma }^2-{\tau }^2)}\left[\sqrt{{\sigma }^2-1}\frac{\partial }{\partial \sigma }\right(\sqrt{{\sigma }^2-1}\frac{\partial \mathrm{\Phi }}{\partial \sigma })+\sqrt{1-{\tau }^2}\frac{\partial }{\partial \tau }(\sqrt{1-{\tau }^2}\frac{\partial \mathrm{\Phi }}{\partial \tau }\left)\right]+\frac{{\partial }^2\mathrm{\Phi }}{\partial z^2}.$

T5 Coordinates

Introduction

As has been made clear in our above review of the Elliptic Cylinder Coordinate system $({\xi }_1,{\xi }_2,{\xi }_3)=(d\cosh \mu ,\cos \nu ,z)$, individual curves within a family of confocal ellipses are identified by one's choice of the "radial" coordinate parameter, $\mu$, or, alternatively, ${\xi }_1$. Specifically, while the two foci of every ellipse are positioned along the x-axis at the same points — namely, $(x,y)=(\pm d,0)$ — the length of the semi-major axis is given by, $a={\xi }_1=d\cosh \mu$.

In a separate chapter we have introduced a different orthogonal curvilinear coordinate system that we refer to as, "T3 Coordinates." In this coordinate system, $({\lambda }_1,{\lambda }_2,{\lambda }_3)$, individual surfaces within a family of concentric spheroids are identified by one's choice of a different "radial" coordinate parameter, ${\lambda }_1$. Here we will adopt essentially this same set of orthogonal coordinates, using ${\lambda }_1$ and ${\lambda }_2$ to describe a family of concentric ellipses that is independent of the vertical-coordinate. We will refer to it as the …

T5 Coordinate System

${\lambda }_1$ $\equiv$ $x\cosh \zeta$           $=$ $(x^2+q^2{y}^2{)}^{1/2}$ ${\lambda }_2$ $\equiv$ $x(\sinh \zeta {)}^{1/(1-q^2)}$           $=$ $(\frac{x^{q^2}}{qy}{)}^{1/(q^2-1)}$ ${\lambda }_3$ $=$ $z$           $=$ $z$ where, $\zeta$ $\equiv$ ${\sinh }^{-1}\left(\frac{qy}{x}\right)$ and, $0<q<\mathrm{\infty }$ is the (fixed) parameter used to specify the eccentricity, $e=[(q^2-1{)}^{1/2}/q]$, of every ${\lambda }_1=$ constant curve within the family of concentric ellipses.

Checking these expressions, we have,

${\lambda }_2\equiv x(\sinh \zeta {)}^{1/(1-q^2)}$

$=$

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

And,

${\lambda }_1\equiv x\cosh \zeta$

$=$

$x[1+{\sinh }^2\zeta {]}^{1/2}=x[1+(\frac{qy}{x}{)}^2{]}^{1/2}=(x^2+q^2{y}^2{)}^{1/2}.$

Comparing this last expression with the above background description of ellipses, we see that ${\lambda }_1=$ constant — for example, ${\lambda }_0$ — is synonymous with an ellipse having …

• A semi-major axis of length, $a={\lambda }_0$;
• An eccentricity, $e\equiv (1-b^2/a^2{)}^{1/2}=[(q^2-1)/q^2{]}^{1/2}$;
• A pair of foci whose coordinate locations along the major axis are, $(x,y)=(\pm c,0)$, where, $c=ae$.

Invert Coordinate Mapping

Solving for $x({\lambda }_1,{\lambda }_2)$, we find …

${\lambda }_1$	$=$	$(x^2+q^2{y}^2{)}^{1/2}$
$\Rightarrow y^2$	$=$	$\frac{1}{q^2}[{\lambda }_1^2-x^2].$

And,

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

Hence,

$x^{2q^2}{\lambda }_2^{2(1-q^2)}+x^2-{\lambda }_1^2$

$=$

$0.$

Alternatively, solving for $y({\lambda }_1,{\lambda }_2)$, we find …

${\lambda }_1$	$=$	$(x^2+q^2{y}^2{)}^{1/2}$
$\Rightarrow x^2$	$=$	${\lambda }_1^2-q^2{y}^2.$

And,

${\lambda }_2$	$=$	$(\frac{x^{q^2}}{qy}{)}^{1/(q^2-1)}$
$\Rightarrow x$	$=$	$(qy{)}^{1/q^2}{\lambda }_2^{(q^2-1)/q^2}.$

Hence,

$(qy{)}^{2/q^2}{\lambda }_2^{2(q^2-1)/q^2}-{\lambda }_1^2+q^2{y}^2$

$=$

$0.$

 

Summary of Inverted Relations

${\lambda }_2^2(\frac{x}{{\lambda }_2}{)}^{2q^2}+x^2-{\lambda }_1^2$ $=$ $0;$ ${\lambda }_2^2(\frac{qy}{{\lambda }_2}{)}^{2/q^2}+q^2{y}^2-{\lambda }_1^2$ $=$ $0.$

Example:     $q^2=2$ $0$ $=$ $x^4{\lambda }_2^{-2}+x^2-{\lambda }_1^2$ $\leftarrow$   Quadratic Eq. in x2 $\Rightarrow x^2$ $=$ $\frac{{\lambda }_2^2}{2}\left\{\right[1+\frac{4{\lambda }_1^2}{{\lambda }_2^2}{]}^{1/2}-1\}=\frac{{\lambda }_2^2}{2}(\mathrm{\Lambda }-1);$   $0$ $=$ $2y^2+(2^{1/2}{\lambda }_2)y-{\lambda }_1^2$ $\leftarrow$   Quadratic Eq. in y $\Rightarrow y$ $=$ $\frac{1}{4}\{-2^{1/2}{\lambda }_2\pm [2{\lambda }_2^2+8{\lambda }_1^2{]}^{1/2}\}$     $=$ $\frac{{\lambda }_2}{2^{3/2}}\left\{\right[1+\frac{4{\lambda }_1^2}{{\lambda }_2^2}{]}^{1/2}-1\}=\frac{{\lambda }_2}{2^{3/2}}(\mathrm{\Lambda }-1).$   where, $\mathrm{\Lambda }\equiv [1+\frac{4{\lambda }_1^2}{{\lambda }_2^2}{]}^{1/2}.$ Note … $\frac{y}{x^2}$ $=$ $\frac{{\lambda }_2}{2^{3/2}}\cdot \frac{2}{{\lambda }_2^2}=\frac{1}{\sqrt{2}{\lambda }_2};$ and, $\frac{4y^2}{x^2}$ $=$ $[1+\frac{4{\lambda }_1^2}{{\lambda }_2^2}{]}^{1/2}-1$ $\Rightarrow \frac{1}{{\ell }^2}\equiv (x^2+4y^2)$ $=$ $x^2[1+\frac{4{\lambda }_1^2}{{\lambda }_2^2}{]}^{1/2}=x^2\mathrm{\Lambda }=\frac{{\lambda }_2^2}{2}\mathrm{\Lambda }(\mathrm{\Lambda }-1)$      or, $\frac{1}{{\ell }^2}\equiv (x^2+4y^2)$ $=$ $\sqrt{2}{\lambda }_{2}y[1+\frac{4{\lambda }_1^2}{{\lambda }_2^2}{]}^{1/2}=\sqrt{2}{\lambda }_{2}y\mathrm{\Lambda }=\frac{{\lambda }_2^2}{2}\cdot \mathrm{\Lambda }(\mathrm{\Lambda }-1).$ Note as well that, ${\ell }^{-2}=2{\lambda }_1^2\mathrm{\Lambda }/(\mathrm{\Lambda }+1).$

Example:     $q^2=\frac{3}{2}$ $0$ $=$ ${\lambda }_2^2(\frac{x}{{\lambda }_2}{)}^3+x^2-{\lambda }_1^2$        $\leftarrow$   Cubic Eq. in x $0$ $=$ ${\lambda }_2^2(\frac{3y^2}{2{\lambda }_2^2}{)}^{2/3}+\frac{3}{2}{y}^2-{\lambda }_1^2$        $\leftarrow$   Cubic Eq. in y2/3

Example:     $q^2=3$ $0$ $=$ ${\lambda }_2^2(\frac{x}{{\lambda }_2}{)}^6+x^2-{\lambda }_1^2$        $\leftarrow$   Cubic Eq. in x2 $0$ $=$ ${\lambda }_2^2(\frac{3y^2}{{\lambda }_2^2}{)}^{1/3}+3y^2-{\lambda }_1^2$        $\leftarrow$   Cubic Eq. in y2/3

Example:     $q^2=4$ $0$ $=$ ${\lambda }_2^2(\frac{x}{{\lambda }_2}{)}^8+x^2-{\lambda }_1^2$        $\leftarrow$   Quartic Eq. in x2 $0$ $=$ ${\lambda }_2^2(\frac{2y}{{\lambda }_2}{)}^{1/2}+4y^2-{\lambda }_1^2$        $\leftarrow$   Quartic Eq. in y1/2

 

Relevant Partial Derivatives

Before moving forward, we need to evaluate a number of relevant partial derivatives.

$\frac{\partial {\lambda }_1}{\partial x}$	$=$	$\frac{\partial }{\partial x}[x^2+q^2{y}^2{]}^{1/2}=\frac{1}{2}[x^2+q^2{y}^2{]}^{-1/2}2x=\frac{x}{{\lambda }_1}.$
$\frac{\partial {\lambda }_1}{\partial y}$	$=$	$\frac{\partial }{\partial y}[x^2+q^2{y}^2{]}^{1/2}=\frac{1}{2}[x^2+q^2{y}^2{]}^{-1/2}2q^{2}y=\frac{q^{2}y}{{\lambda }_1}.$
$\frac{\partial {\lambda }_2}{\partial x}$	$=$	$\frac{\partial }{\partial x}[x^{q^2/(q^2-1)}(qy{)}^{-1/(q^2-1)}]=[\frac{q^2}{q^2-1}]\frac{{\lambda }_2}{x}.$
$\frac{\partial {\lambda }_2}{\partial y}$	$=$	$\frac{\partial }{\partial y}[x^{q^2/(q^2-1)}(qy{)}^{-1/(q^2-1)}]=-[\frac{1}{q^2-1}]\frac{{\lambda }_2}{y}.$

We may also need the set of complementary partial derivatives. Even though we are unable to explicitly invert the coordinate mappings, once we have in hand expressions for the three scale factors (see immediately below), we can determine expressions for the set of complementary partial derivatives via the generic relation,

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

$=$

$h_n^2\cdot \frac{\partial {\lambda }_n}{\partial x_i}.$

Example:     $q^2=2$ $y$ $=$ $\frac{{\lambda }_2}{2^{3/2}}\{\mathrm{\Lambda }-1\},$         $x^2$ $=$ $\frac{{\lambda }_2^2}{2}\{\mathrm{\Lambda }-1\},$       where,       $\mathrm{\Lambda }$ $\equiv$ $[1+\frac{4{\lambda }_1^2}{{\lambda }_2^2}{]}^{1/2}.$ Noting that, $\frac{\partial \mathrm{\Lambda }}{\partial {\lambda }_1}$ $=$ $\frac{1}{\mathrm{\Lambda }{\lambda }_1}\left[\frac{4{\lambda }_1^2}{{\lambda }_2^2}\right]$       and,       $\frac{\partial \mathrm{\Lambda }}{\partial {\lambda }_2}$ $=$ $-\frac{1}{\mathrm{\Lambda }{\lambda }_2}\left[\frac{4{\lambda }_1^2}{{\lambda }_2^2}\right],$ we have, $\frac{\partial y}{\partial {\lambda }_1}$ $=$ $\frac{{\lambda }_2}{2^{3/2}}\cdot \frac{\partial \mathrm{\Lambda }}{\partial {\lambda }_1}=\frac{\sqrt{2}{\lambda }_1}{{\lambda }_2}[1+\frac{4{\lambda }_1^2}{{\lambda }_2^2}{]}^{-1/2},$ $\frac{\partial y}{\partial {\lambda }_2}$ $=$ $\frac{1}{2^{3/2}}[\mathrm{\Lambda }-1]+\frac{{\lambda }_2}{2^{3/2}}\cdot \frac{\partial \mathrm{\Lambda }}{\partial {\lambda }_2}=\frac{(\mathrm{\Lambda }-1)}{2^{3/2}}-\frac{\sqrt{2}{\lambda }_1^2}{\mathrm{\Lambda }{\lambda }_2^2}.$   $=$ $\frac{\mathrm{\Lambda }(\mathrm{\Lambda }-1)-4{\lambda }_1^2/{\lambda }_2^2}{2^{3/2}\mathrm{\Lambda }}=\frac{{\mathrm{\Lambda }}^2-\mathrm{\Lambda }-4{\lambda }_1^2/{\lambda }_2^2}{2^{3/2}\mathrm{\Lambda }}$   $=$ $\frac{(1-\mathrm{\Lambda })}{2^{3/2}\mathrm{\Lambda }}.$

Let's compare by drawing from the expressions for ${\ell }^2$, above, and for $h_n^2$ derived below.

$[h_1^2\cdot \frac{\partial {\lambda }_1}{\partial y}{]}_{q^2=2}$	$=$	$\left[{\lambda }_1^2{\ell }^2\right(\frac{q^{2}y}{{\lambda }_1}){]}_{q^2=2}=[2{\lambda }_1{\ell }^{2}y{]}_{q^2=2}$
	$=$	$2{\lambda }_1\left\{\frac{1}{\sqrt{2}{\lambda }_2\mathrm{\Lambda }}\right\}=\frac{\sqrt{2}{\lambda }_1}{{\lambda }_2\mathrm{\Lambda }}.$

Yes! This, indeed matches the just-derived expression for $\partial y/\partial {\lambda }_1$. And we also have,

$[h_2^2\cdot \frac{\partial {\lambda }_2}{\partial y}{]}_{q^2=2}$	$=$	$\{-[\frac{1}{q^2-1}\left]\frac{{\lambda }_2}{y}\right[\frac{(q^2-1)xy\ell }{{\lambda }_2}{]}^2{\}}_{q^2=2}=-[\frac{(q^2-1)x^{2}y{\ell }^2}{{\lambda }_2}{]}_{q^2=2}$
	$=$	$\frac{(1-\mathrm{\Lambda })}{2^{3/2}\mathrm{\Lambda }}.$

Yes, again!

Scale Factors, Direction Cosines & Unit Vectors

From our accompanying generic discussion of direction cosines, we can write,

$h_1^2$	$=$	$\left[\right(\frac{\partial {\lambda }_1}{\partial x}{)}^2+(\frac{\partial {\lambda }_1}{\partial y}{)}^2+(\frac{\partial {\lambda }_1}{\partial z}{)}^2{]}^{-1}$
	$=$	$\left[\right(\frac{x}{{\lambda }_1}{)}^2+(\frac{q^{2}y}{{\lambda }_1}{)}^2{]}^{-1}={\lambda }_1^2[x^2+q^4{y}^2{]}^{-1}$
	$=$	${\lambda }_1^2{\ell }^2;$
$h_2^2$	$=$	$\left\{\right(\frac{\partial {\lambda }_2}{\partial x}{)}^2+(\frac{\partial {\lambda }_2}{\partial y}{)}^2+(\frac{\partial {\lambda }_2}{\partial z}{)}^2{\}}^{-1}$
	$=$	$\left\{\right[\frac{q^2}{q^2-1}{]}^2(\frac{{\lambda }_2}{x}{)}^2+[\frac{1}{q^2-1}{]}^2(\frac{{\lambda }_2}{y}{)}^2{\}}^{-1}$
	$=$	$[\frac{(q^2-1)xy\ell }{{\lambda }_2}{]}^2;$
$h_3^2$	$=$	$\left\{\right(\frac{\partial {\lambda }_3}{\partial x}{)}^2+(\frac{\partial {\lambda }_3}{\partial y}{)}^2+(\frac{\partial {\lambda }_3}{\partial z}{)}^2{\}}^{-1}=1;$

where,

$\ell \equiv (x^2+q^4{y}^2{)}^{-1/2}.$

Direction Cosines for T5 Coordinates ${\gamma }_{ni}=h_n\left(\frac{\partial {\lambda }_n}{\partial x_i}\right)$
$n$	$i=x,y,z$
$1$	$x\ell$	$q^{2}y\ell$	$0$
$2$	$q^{2}y\ell$	$-x\ell$	$0$
$3$	$0$	$0$	$1$

The unit vectors are,

${\hat{e}}_n$

$=$

$\hat{ı}{\gamma }_{n1}+\hat{ȷ}{\gamma }_{n2}+\hat{k}{\gamma }_{n3},$

that is,

${\hat{e}}_1$	$=$	$\hat{ı}(x\ell )+\hat{ȷ}(q^{2}y\ell ),$
${\hat{e}}_2$	$=$	$\hat{ı}(q^{2}y\ell )-\hat{ȷ}(x\ell ),$
${\hat{e}}_3$	$=$	$\hat{k}.$

Notice that,

${\hat{e}}_1\cdot \widehat{e_2}$

$=$

$q^{2}xy{\ell }^2-q^{2}xy{\ell }^2=0,$

and,

${\hat{e}}_1\cdot \widehat{e_1}$

$=$

$x^2{\ell }^2+q^4{y}^2{\ell }^2={\ell }^2(x^2+q^4{y}^2)=1.$

These are both desired orthogonality conditions. Alternatively,

$\hat{ı}$	$=$	${\hat{e}}_1{\gamma }_{11}+{\hat{e}}_2{\gamma }_{21}+{\hat{e}}_3{\gamma }_{31}={\hat{e}}_1(x\ell )+{\hat{e}}_2(q^{2}y\ell );$
$\hat{ȷ}$	$=$	${\hat{e}}_1{\gamma }_{12}+{\hat{e}}_2{\gamma }_{22}+{\hat{e}}_3{\gamma }_{32}={\hat{e}}_1(q^{2}y\ell )-{\hat{e}}_2(x\ell );$
$\hat{k}$	$=$	${\hat{e}}_1{\gamma }_{13}+{\hat{e}}_2{\gamma }_{23}+{\hat{e}}_3{\gamma }_{33}={\hat{e}}_3.$

Spatial Operators

Summary Reminder

$h_1^2$ $=$ ${\lambda }_1^2{\ell }^2;$       $h_2^2$ $=$ $[\frac{(q^2-1)xy\ell }{{\lambda }_2}{]}^2;$       and,       $h_3^2$ $=$ $1;$

In T5 Coordinates, a couple of relevant operators are:

$\mathrm{\nabla }F$	$=$	${\hat{e}}_1\left[\frac{1}{h_1}\frac{\partial F}{\partial {\lambda }_1}\right]+{\hat{e}}_2\left[\frac{1}{h_2}\frac{\partial F}{\partial {\lambda }_2}\right]+{\hat{e}}_3\left[\frac{1}{h_3}\frac{\partial F}{\partial {\lambda }_3}\right]$
	$=$	${\hat{e}}_1\left(\frac{1}{{\lambda }_1\ell }\right)\frac{\partial F}{\partial {\lambda }_1}+{\hat{e}}_2\left[\frac{{\lambda }_2}{(q^2-1)xy\ell }\right]\frac{\partial F}{\partial {\lambda }_2}+{\hat{e}}_3\frac{\partial F}{\partial {\lambda }_3}.$

${\mathrm{\nabla }}^{2}F$	$=$	$\frac{1}{h_1{h}_2{h}_3}\left[\frac{\partial }{\partial {\lambda }_1}\right(\frac{h_2{h}_3}{h_1}\cdot \frac{\partial F}{\partial {\lambda }_1})+\frac{\partial }{\partial {\lambda }_2}(\frac{h_3{h}_1}{h_2}\cdot \frac{\partial F}{\partial {\lambda }_2})+\frac{\partial }{\partial {\lambda }_3}(\frac{h_1{h}_2}{h_3}\cdot \frac{\partial F}{\partial {\lambda }_3}\left)\right]$
	$=$	$\frac{{\lambda }_2}{{\lambda }_1(q^2-1)xy{\ell }^2}\left\{\frac{\partial }{\partial {\lambda }_1}\right[\frac{(q^2-1)xy}{{\lambda }_1{\lambda }_2}\cdot \frac{\partial F}{\partial {\lambda }_1}]+\frac{\partial }{\partial {\lambda }_2}[\frac{{\lambda }_1{\lambda }_2}{(q^2-1)xy}\cdot \frac{\partial F}{\partial {\lambda }_2}]+\frac{\partial }{\partial {\lambda }_3}[\frac{{\lambda }_1(q^2-1)xy{\ell }^2}{{\lambda }_2}\cdot \frac{\partial F}{\partial {\lambda }_3}\left]\right\}$

And if $F$ is a function only of ${\lambda }_1$, then,

${\mathrm{\nabla }}^{2}F$	$=$	$\frac{{\lambda }_2}{{\lambda }_1(q^2-1)xy{\ell }^2}\left\{\frac{\partial }{\partial {\lambda }_1}\right[\frac{(q^2-1)xy}{{\lambda }_1{\lambda }_2}\cdot \frac{\partial F}{\partial {\lambda }_1}\left]\right\}$
	$=$	$\frac{{\lambda }_2}{{\lambda }_1(q^2-1)xy{\ell }^2}\left\{\right[\frac{(q^2-1)xy}{{\lambda }_1{\lambda }_2}\left]\right[\frac{{\partial }^{2}F}{\partial {\lambda }_1^2}]+\frac{\partial F}{\partial {\lambda }_1}\cdot \frac{\partial }{\partial {\lambda }_1}[\frac{(q^2-1)xy}{{\lambda }_1{\lambda }_2}\left]\right\}$
	$=$	$\left[\frac{1}{{\lambda }_1^2{\ell }^2}\right]\left[\frac{{\partial }^{2}F}{\partial {\lambda }_1^2}\right]+\left[\frac{1}{{\lambda }_{1}xy{\ell }^2}\right]\frac{\partial F}{\partial {\lambda }_1}\cdot \frac{\partial }{\partial {\lambda }_1}\left[\frac{xy}{{\lambda }_1}\right].$

In order to complete this evaluation, we need a couple of "complementary partial derivatives." Referencing the relation provided above, we find,

$\frac{\partial }{\partial {\lambda }_1}\left[\frac{xy}{{\lambda }_1}\right]$	$=$	$xy\left[\frac{\partial }{\partial {\lambda }_1}\right({\lambda }_1^{-1}\left)\right]+\frac{y}{{\lambda }_1}\left[\frac{\partial x}{\partial {\lambda }_1}\right]+\frac{x}{{\lambda }_1}\left[\frac{\partial y}{\partial {\lambda }_1}\right]$
	$=$	$-\frac{xy}{{\lambda }_1^2}+\frac{y}{{\lambda }_1}\left[h_1^2\frac{\partial {\lambda }_1}{\partial x}\right]+\frac{x}{{\lambda }_1}\left[h_1^2\frac{\partial {\lambda }_1}{\partial y}\right]$
	$=$	$-\frac{xy}{{\lambda }_1^2}+\frac{h_1^2}{{\lambda }_1}[\frac{xy}{{\lambda }_1}+\frac{q^{2}xy}{{\lambda }_1}]$
	$=$	$\frac{xy}{{\lambda }_1^2}[{\lambda }_1^2{\ell }^2(1+q^2)-1].$

Hence,

${\mathrm{\nabla }}^{2}F$	$=$	$\left[\frac{1}{{\lambda }_1^2{\ell }^2}\right]\left[\frac{{\partial }^{2}F}{\partial {\lambda }_1^2}\right]+\frac{xy}{{\lambda }_1^2}[{\lambda }_1^2{\ell }^2(1+q^2)-1]\left[\frac{1}{{\lambda }_{1}xy{\ell }^2}\right]\frac{\partial F}{\partial {\lambda }_1}$
	$=$	$\left[\frac{1}{{\lambda }_1^2{\ell }^2}\right]\left[\frac{{\partial }^{2}F}{\partial {\lambda }_1^2}\right]+[{\lambda }_1^2{\ell }^2(1+q^2)-1]\left[\frac{1}{{\lambda }_1^3{\ell }^2}\right]\frac{\partial F}{\partial {\lambda }_1}$
	$=$	$\left[\frac{1}{{\lambda }_1^2{\ell }^2}\right]\left[\frac{{\partial }^{2}F}{\partial {\lambda }_1^2}\right]-\left[\frac{1}{{\lambda }_1^3{\ell }^2}\right]\frac{\partial F}{\partial {\lambda }_1}+\left[\frac{(1+q^2)}{{\lambda }_1}\right]\frac{\partial F}{\partial {\lambda }_1}.$

Example (q² = 2) Poisson Equation

Setup

Let's see if we can solve the,

Poisson Equation User:Tohline/Math/EQ Poisson01

obtaining an analytic expression for the gravitational potential in the case where, independent of the coordinate, $z$,

$\rho ={\rho }_c\sigma$	$=$	${\rho }_c[1-(\frac{x^2}{a^2}+\frac{y^2}{b^2}\left)\right]$
	$=$	${\rho }_c[1-\frac{1}{a^2}(x^2+q^2{y}^2\left)\right].$

Given that the density distribution is independent of $z$, we expect the potential to be independent of $z$ as well. So, in terms of T5-Coordinates, the Poisson equation may be written as,

$\frac{{\lambda }_2}{{\lambda }_1(q^2-1)xy{\ell }^2}\left\{\frac{\partial }{\partial {\lambda }_1}\right[\frac{(q^2-1)xy}{{\lambda }_1{\lambda }_2}\cdot \frac{\partial \mathrm{\Phi }}{\partial {\lambda }_1}]+\frac{\partial }{\partial {\lambda }_2}[\frac{{\lambda }_1{\lambda }_2}{(q^2-1)xy}\cdot \frac{\partial \mathrm{\Phi }}{\partial {\lambda }_2}\left]\right\}$

$=$

$4\pi G{\rho }_c[1-\frac{{\lambda }_1^2}{a^2}]$

If we specifically consider the case where $q^2=a^2/b^2=2$, this can be rewritten as,

$4\pi G{\rho }_c[1-\frac{{\lambda }_1^2}{a^2}]$	$=$	$\frac{{\lambda }_2}{{\lambda }_1(q^2-1)xy{\ell }^2}\left\{\frac{\partial }{\partial {\lambda }_1}\right[\frac{(q^2-1)xy}{{\lambda }_1{\lambda }_2}\cdot \frac{\partial \mathrm{\Phi }}{\partial {\lambda }_1}]+\frac{\partial }{\partial {\lambda }_2}[\frac{{\lambda }_1{\lambda }_2}{(q^2-1)xy}\cdot \frac{\partial \mathrm{\Phi }}{\partial {\lambda }_2}\left]\right\}$
	$=$	$\frac{{\lambda }_2}{{\lambda }_1}\cdot (\mathrm{\Lambda }-1{)}^{-3/2}\frac{2^2}{{\lambda }_2^2}\cdot \frac{{\lambda }_2^2}{2}(\mathrm{\Lambda }-1)\mathrm{\Lambda }\{\frac{\partial }{\partial {\lambda }_1}[\frac{(\mathrm{\Lambda }-1)}{2}\cdot \frac{\partial \mathrm{\Phi }}{\partial {\lambda }_1}]+\frac{\partial }{\partial {\lambda }_2}[\frac{2}{(\mathrm{\Lambda }-1)}\cdot \frac{\partial \mathrm{\Phi }}{\partial {\lambda }_2}]\}$
	$=$	$\frac{4\mathrm{\Lambda }}{(\mathrm{\Lambda }-1)}\left\{\frac{\partial }{\partial {\lambda }_1}\right[\frac{(\mathrm{\Lambda }-1)}{2}\cdot \frac{\partial \mathrm{\Phi }}{\partial {\lambda }_1}]+\frac{\partial }{\partial {\lambda }_2}[\frac{2}{(\mathrm{\Lambda }-1)}\cdot \frac{\partial \mathrm{\Phi }}{\partial {\lambda }_2}\left]\right\}$

where we have used the following expressions derived above:

$x^2{y}^2$	$=$	$(\mathrm{\Lambda }-1)[\frac{{\lambda }_2^2}{2^2}(\mathrm{\Lambda }-1){]}^2,$
$\frac{1}{{\ell }^2}$	$=$	$\frac{{\lambda }_2^2}{2}(\mathrm{\Lambda }-1)\mathrm{\Lambda }=\frac{2{\lambda }_1^2\mathrm{\Lambda }}{(\mathrm{\Lambda }+1)},$
$\mathrm{\Lambda }$	$\equiv$	$[1+\frac{4{\lambda }_1^2}{{\lambda }_2^2}{]}^{1/2}\Rightarrow \frac{1}{2}({\mathrm{\Lambda }}^2-1{)}^{1/2}=\frac{{\lambda }_1}{{\lambda }_2}.$
$\frac{xy}{{\lambda }_1{\lambda }_2}$	$=$	$\frac{1}{2^2}[\frac{{\lambda }_2}{{\lambda }_1}(\mathrm{\Lambda }-1{)}^{3/2}]=\frac{1}{2}(\mathrm{\Lambda }-1).$

Now,

$\frac{\partial (\mathrm{\Lambda }-1)}{\partial {\lambda }_1}$	$=$	$\frac{1}{2\mathrm{\Lambda }}\left(\frac{8{\lambda }_1}{{\lambda }_2^2}\right)=\frac{4}{{\lambda }_1\mathrm{\Lambda }}\left(\frac{{\lambda }_1^2}{{\lambda }_2^2}\right)=\frac{({\mathrm{\Lambda }}^2-1)}{{\lambda }_1\mathrm{\Lambda }};$
$\frac{\partial (\mathrm{\Lambda }-1{)}^{-1}}{\partial {\lambda }_2}$	$=$	$-\frac{1}{(\mathrm{\Lambda }-1{)}^2}\left[\frac{1}{2\mathrm{\Lambda }}\right](-\frac{8{\lambda }_1^2}{{\lambda }_2^3})=\frac{4}{(\mathrm{\Lambda }-1{)}^2}\left[\frac{1}{{\lambda }_2\mathrm{\Lambda }}\right]\left(\frac{{\lambda }_1^2}{{\lambda }_2^2}\right)=\frac{\mathrm{\Lambda }+1}{(\mathrm{\Lambda }-1)}\left[\frac{1}{{\lambda }_2\mathrm{\Lambda }}\right].$

Hence,

$4\pi G{\rho }_c[1-\frac{{\lambda }_1^2}{a^2}]$	$=$	$\frac{4\mathrm{\Lambda }}{(\mathrm{\Lambda }-1)}\left\{\frac{(\mathrm{\Lambda }-1)}{2}\right[\frac{{\partial }^2\mathrm{\Phi }}{\partial {\lambda }_1^2}]+\frac{\partial \mathrm{\Phi }}{\partial {\lambda }_1}\cdot \frac{\partial }{\partial {\lambda }_1}[\frac{(\mathrm{\Lambda }-1)}{2}]+\frac{2}{(\mathrm{\Lambda }-1)}[\frac{{\partial }^2\mathrm{\Phi }}{\partial {\lambda }_2^2}]+\frac{\partial \mathrm{\Phi }}{\partial {\lambda }_2}\cdot \frac{\partial }{\partial {\lambda }_2}[\frac{2}{(\mathrm{\Lambda }-1)}\left]\right\}$
	$=$	$\frac{4\mathrm{\Lambda }}{(\mathrm{\Lambda }-1)}\left\{\frac{(\mathrm{\Lambda }-1)}{2}\right[\frac{{\partial }^2\mathrm{\Phi }}{\partial {\lambda }_1^2}]+[\frac{({\mathrm{\Lambda }}^2-1)}{2{\lambda }_1\mathrm{\Lambda }}]\frac{\partial \mathrm{\Phi }}{\partial {\lambda }_1}+\frac{2}{(\mathrm{\Lambda }-1)}[\frac{{\partial }^2\mathrm{\Phi }}{\partial {\lambda }_2^2}]+[\frac{2(\mathrm{\Lambda }+1)}{(\mathrm{\Lambda }-1){\lambda }_2\mathrm{\Lambda }}\left]\frac{\partial \mathrm{\Phi }}{\partial {\lambda }_2}\right\}$
	$=$	$2\mathrm{\Lambda }\left[\frac{{\partial }^2\mathrm{\Phi }}{\partial {\lambda }_1^2}\right]+\left[\frac{2(\mathrm{\Lambda }+1)}{{\lambda }_1}\right]\frac{\partial \mathrm{\Phi }}{\partial {\lambda }_1}+\frac{8\mathrm{\Lambda }}{(\mathrm{\Lambda }-1{)}^2}\left[\frac{{\partial }^2\mathrm{\Phi }}{\partial {\lambda }_2^2}\right]+\left[\frac{8(\mathrm{\Lambda }+1)}{(\mathrm{\Lambda }-1{)}^2{\lambda }_2}\right]\frac{\partial \mathrm{\Phi }}{\partial {\lambda }_2}$
	$=$	$\frac{2}{{\lambda }_1}\left\{\mathrm{\Lambda }{\lambda }_1\right[\frac{{\partial }^2\mathrm{\Phi }}{\partial {\lambda }_1^2}]+(\mathrm{\Lambda }+1)\frac{\partial \mathrm{\Phi }}{\partial {\lambda }_1}\}+\frac{8}{{\lambda }_2(\mathrm{\Lambda }-1{)}^2}\left\{\mathrm{\Lambda }{\lambda }_2\right[\frac{{\partial }^2\mathrm{\Phi }}{\partial {\lambda }_2^2}]+(\mathrm{\Lambda }+1)\frac{\partial \mathrm{\Phi }}{\partial {\lambda }_2}\}.$

Trials

Try,

$\mathrm{\Phi }$	$=$	$A{\lambda }_1^{\alpha }+B{\lambda }_2^{\beta }$
$\Rightarrow \frac{\partial \mathrm{\Phi }}{\partial {\lambda }_1}$	$=$	$A\alpha {\lambda }_1^{\alpha -1},$	and,	$\frac{\partial \mathrm{\Phi }}{\partial {\lambda }_2}$	$=$	$B\beta {\lambda }_2^{\beta -1}.$

In this case we find,

$4\pi G{\rho }_c[1-\frac{{\lambda }_1^2}{a^2}]$	$=$	$\frac{2}{{\lambda }_1}\left\{\mathrm{\Lambda }{\lambda }_1\frac{\partial }{\partial {\lambda }_1}\right[\frac{\partial \mathrm{\Phi }}{\partial {\lambda }_1}]+(\mathrm{\Lambda }+1)[\frac{\partial \mathrm{\Phi }}{\partial {\lambda }_1}\left]\right\}+\frac{8}{{\lambda }_2(\mathrm{\Lambda }-1{)}^2}\left\{\mathrm{\Lambda }{\lambda }_2\frac{\partial }{\partial {\lambda }_2}\right[\frac{\partial \mathrm{\Phi }}{\partial {\lambda }_2}]+(\mathrm{\Lambda }+1)[\frac{\partial \mathrm{\Phi }}{\partial {\lambda }_2}\left]\right\}$
	$=$	$\frac{2A}{{\lambda }_1}\left\{\mathrm{\Lambda }{\lambda }_1\frac{\partial }{\partial {\lambda }_1}\right[\alpha {\lambda }_1^{\alpha -1}]+(\mathrm{\Lambda }+1)[\alpha {\lambda }_1^{\alpha -1}\left]\right\}+\frac{8B}{{\lambda }_2(\mathrm{\Lambda }-1{)}^2}\left\{\mathrm{\Lambda }{\lambda }_2\frac{\partial }{\partial {\lambda }_2}\right[\beta {\lambda }_2^{\beta -1}]+(\mathrm{\Lambda }+1)[\beta {\lambda }_2^{\beta -1}\left]\right\}$
	$=$	$\frac{2A\alpha }{{\lambda }_1}\{\mathrm{\Lambda }(\alpha -1){\lambda }_1^{\alpha -1}+(\mathrm{\Lambda }+1){\lambda }_1^{\alpha -1}\}+\frac{8B\beta }{{\lambda }_2(\mathrm{\Lambda }-1{)}^2}\{\mathrm{\Lambda }(\beta -1){\lambda }_2^{\beta -1}+(\mathrm{\Lambda }+1){\lambda }_2^{\beta -1}\}$
	$=$	$2A\alpha {\lambda }_1^{\alpha -2}\{\mathrm{\Lambda }(\alpha -1)+(\mathrm{\Lambda }+1)\}+\frac{8B\beta {\lambda }_2^{\beta -2}}{(\mathrm{\Lambda }-1{)}^2}\{\mathrm{\Lambda }(\beta -1)+(\mathrm{\Lambda }+1)\}$
	$=$	$2A\alpha {\lambda }_1^{\alpha -2}\{\alpha \mathrm{\Lambda }+1\}+\frac{8B\beta {\lambda }_2^{\beta -2}}{(\mathrm{\Lambda }-1{)}^2}\{\beta \mathrm{\Lambda }+1\}.$

If $\alpha =4$,

$\frac{8B\beta {\lambda }_2^{\beta -2}}{(\mathrm{\Lambda }-1{)}^2}[\beta \mathrm{\Lambda }+1]$	$=$	$4\pi G{\rho }_c[1-\frac{{\lambda }_1^2}{a^2}]-8A{\lambda }_1^2-32A{\lambda }_1^2\mathrm{\Lambda }$
$\Rightarrow 8B\beta {\lambda }_2^{\beta -2}[\beta \mathrm{\Lambda }+1]$	$=$	$\{4\pi G{\rho }_c-32A{\lambda }_1^2\mathrm{\Lambda }-{\lambda }_1^2[\frac{4\pi G{\rho }_c}{a^2}+8A\left]\right\}({\mathrm{\Lambda }}^2-2\mathrm{\Lambda }+1).$

If, then, $8A{a}^2=-4\pi G{\rho }_c$,

$\Rightarrow 8B\beta {\lambda }_2^{\beta -2}[\beta \mathrm{\Lambda }+1]$

$=$

$4\pi G{\rho }_c\{1+[\frac{4{\lambda }_1^2}{a^2}\left]\mathrm{\Lambda }\right\}({\mathrm{\Lambda }}^2-2\mathrm{\Lambda }+1).$

But, we also know that, ${\lambda }_1^2={\lambda }_2^2({\mathrm{\Lambda }}^2-1)/4$, so …

$8a^{2}B\beta {\lambda }_2^{\beta -2}[\beta \mathrm{\Lambda }+1]$

$=$

$4\pi G{\rho }_c\{a^2+{\lambda }_2^2({\mathrm{\Lambda }}^2-1)\mathrm{\Lambda }\}({\mathrm{\Lambda }}^2-2\mathrm{\Lambda }+1).$

(25 October 2020) I give up … for now.

See Also

• Direction Cosines