# Daring Attack

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## 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
$~n$ $~\lambda_n$ $~h_n$ $~\frac{\partial \lambda_n}{\partial x}$ $~\frac{\partial \lambda_n}{\partial y}$ $~\frac{\partial \lambda_n}{\partial z}$ $~\gamma_{n1}$ $~\gamma_{n2}$ $~\gamma_{n3}$
$~1$ $~(x^2 + q^2 y^2 + p^2 z^2)^{1 / 2}$ $~\lambda_1 \ell_{3D}$ $~\frac{x}{\lambda_1}$ $~\frac{q^2 y}{\lambda_1}$ $~\frac{p^2 z}{\lambda_1}$ $~(x) \ell_{3D}$ $~(q^2 y)\ell_{3D}$ $~(p^2z) \ell_{3D}$
$~2$ --- --- --- --- --- $~\ell_q \ell_{3D} (xp^2z)$ $~\ell_q \ell_{3D} (q^2 y p^2z)$ $~- (x^2 + q^4y^2)\ell_q \ell_{3D}$
$~3$ $~\tan^{-1}\biggl( \frac{y^{1/q^2}}{x} \biggr)$ $~\frac{xq^2 y \ell_q}{\sin\lambda_3 \cos\lambda_3}$ $~-\frac{\sin\lambda_3 \cos\lambda_3}{x}$ $~+\frac{\sin\lambda_3 \cos\lambda_3}{q^2y}$ $~0$ $~-q^2 y \ell_q$ $~x\ell_q$ $~0$
 $~\ell_{3D}$ $~\equiv$ $~[x^2 + q^4 y^2 + p^4 z^2]^{- 1/ 2 }$ $~\ell_q$ $~\equiv$ $~[x^2 + q^4 y^2 ]^{- 1/ 2 }$

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 $~n$ $~\lambda_n$ $~h_n$ $~\frac{\partial \lambda_n}{\partial x}$ $~\frac{\partial \lambda_n}{\partial y}$ $~\frac{\partial \lambda_n}{\partial z}$ $~\gamma_{n1}$ $~\gamma_{n2}$ $~\gamma_{n3}$ $~1$ $~(x^2 + q^2 y^2 + p^2 z^2)^{1 / 2}$ $~\lambda_1 \ell_{3D}$ $~\frac{x}{\lambda_1}$ $~\frac{q^2 y}{\lambda_1}$ $~\frac{p^2 z}{\lambda_1}$ $~(x) \ell_{3D}$ $~(q^2 y)\ell_{3D}$ $~(p^2z) \ell_{3D}$ $~2$ --- --- --- --- --- $~\ell_q \ell_{3D} (xp^2z)$ $~\ell_q \ell_{3D} (q^2 y p^2z)$ $~- (x^2 + q^4y^2)\ell_q \ell_{3D}$ $~3$ $~\frac{y^{1/q^2}}{x}$ $~\frac{xq^2 y \ell_q}{\lambda_3}$ $~-\frac{\lambda_3}{x}$ $~+\frac{\lambda_3}{q^2y}$ $~0$ $~-q^2 y \ell_q$ $~x\ell_q$ $~0$

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,

 $~1$ $~=$ $~\biggl( \frac{x}{a}\biggr)^2 + \biggl( \frac{y}{b}\biggr)^2 + \biggl( \frac{z}{c}\biggr)^2 \, .$

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,

 $~\lambda_2^2$ $~\equiv$ $~1 - \biggl( \frac{z}{c}\biggr)^2 \, ,$

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,

 $~\biggl( \frac{x}{a}\biggr)^2 + \biggl( \frac{y}{b}\biggr)^2$ $~=$ $~ \lambda_2^2$ $~\Rightarrow ~~~ x^2 + q^2 y^2$ $~=$ $~ \lambda_1^2 \lambda_2^2 \, .$

[Note that in the case of spherical coordinates (q2 = p2 = 1), $~\lambda_1 \rightarrow r$, 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,

 $~\lambda_1^2 \lambda_2^2$ $~=$ $~\frac{y^{2/q^2}}{\lambda_3^2} + q^2y^2 \, .$

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,

 $~ 0$ $~=$ $~ 2y^2 + \frac{y}{\lambda_3^2} - \lambda_1^2 \lambda_2^2$ $~ \Rightarrow~~~ y$ $~=$ $~ \frac{1}{4} \biggl\{ -\frac{1}{\lambda_3^2} \pm \biggl[ \frac{1}{\lambda_3^4} + 8 \lambda_1^2 \lambda_2^2 \biggr]^{1 / 2} \biggr\}$ $~=$ $~ \frac{1}{4\lambda_3^2} \biggl\{ \biggl[ 1 + 8 \lambda_1^2 \lambda_2^2 \lambda_3^4\biggr]^{1 / 2} - 1 \biggr\} \, .$

(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),

 $~x^2$ $~=$ $~\frac{y}{\lambda_3^2}$ $~=$ $~ \frac{1}{4\lambda_3^4} \biggl\{ \biggl[ 1 + 8 \lambda_1^2 \lambda_2^2 \lambda_3^4\biggr]^{1 / 2} - 1 \biggr\}$ $~\Rightarrow ~~~ x$ $~=$ $~\pm \frac{1}{2\lambda_3^{2}} \biggl\{ \biggl[ 1 + 8 \lambda_1^2 \lambda_2^2 \lambda_3^4\biggr]^{1 / 2} - 1 \biggr\}^{1 / 2} \, .$

Summary (q2 = 2)
 $~z(\lambda_1, \lambda_2, \lambda_3)$ $~=$ $~\frac{\lambda_1(1-\lambda_2^2)^{1 / 2}}{p} \, ,$ $~y(\lambda_1, \lambda_2, \lambda_3)$ $~=$ $~ \frac{1}{4\lambda_3^2} \biggl\{ \biggl[ 1 + 8 \lambda_1^2 \lambda_2^2 \lambda_3^4\biggr]^{1 / 2} - 1 \biggr\} = \frac{(\Lambda - 1)}{4\lambda_3^2}\, ,$ $~x(\lambda_1, \lambda_2, \lambda_3)$ $~=$ $~ \frac{1}{2\lambda_3^{2}} \biggl\{ \biggl[ 1 + 8 \lambda_1^2 \lambda_2^2 \lambda_3^4\biggr]^{1 / 2} - 1 \biggr\}^{1 / 2} = \frac{(\Lambda - 1)^{1 / 2}}{2\lambda_3^{2}} \, .$

For convenience, we have defined,

 $~\Lambda^2$ $~\equiv$ $~1 + 8\lambda_1^2 \lambda_2^2 \lambda_3^4$ $~\Rightarrow ~~~ \lambda_1^2 \lambda_2^2 \lambda_3^4$ $~\equiv$ $~\frac{1}{8}\biggl( \Lambda^2 - 1\biggr) \, .$

Test Example

$~q^2 = 2, p^2=3.15, (x, y, z) = (0.4, 0.63581, 0.1)$

$~(\lambda_1, \lambda_2, \lambda_3) = (1, 0.98412, 1.99344)$

$~\ell_{3D} = 0.730058, ~~ \ell_q = 0.750164$

$~h_1 = 0.730058$

$~\Lambda^2-1 = 122.34879 ~~~\Rightarrow ~~~ \Lambda = 11.10625$

Do we get the correct values of $~(x, y, z)$  ?

 $~z(\lambda_1, \lambda_2, \lambda_3)$ $~=$ $~\frac{\lambda_1(1-\lambda_2^2)^{1 / 2}}{p} = 0.1000000 \, ,$ $~y(\lambda_1, \lambda_2, \lambda_3)$ $~=$ $~ \frac{(\Lambda - 1)}{4\lambda_3^2} = 0.635807\, ,$ $~x(\lambda_1, \lambda_2, \lambda_3)$ $~=$ $~ \frac{(\Lambda - 1)^{1 / 2}}{2\lambda_3^{2}} = 0.400000 \, .$

Evaluate a few partial derivatives …

 $~\frac{\partial z}{\partial \lambda_1}$ $~=$ $~ \frac{(1-\lambda_2^2)^{1 / 2}}{p} = 0.1\, ,$ $~\frac{\partial y}{\partial \lambda_1}$ $~=$ $~ \biggl[ \frac{2\lambda_1 \lambda_2^2 \lambda_3^2}{\Lambda} \biggr] = 0.693054 \, ,$ $~\frac{\partial x}{\partial \lambda_1}$ $~=$ $~ \frac{2\lambda_1 \lambda_2^2 \lambda_3^{2}}{\Lambda (\Lambda-1)^{1 / 2}} = 0.218008 \, .$ $~\Rightarrow ~~~ h_1$ $~=$ $~\biggl[ \biggl(\frac{\partial x}{\partial \lambda_1}\biggr)^2 + \biggl(\frac{\partial y}{\partial \lambda_1}\biggr)^2 + \biggl(\frac{\partial z}{\partial \lambda_1}\biggr)^2 \biggr]^{1 / 2} = 0.733383 \, .$

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 …

$~\frac{\partial z}{\partial \lambda_1} = \biggl[ h_1^2 - \biggl( \frac{\partial x}{\partial \lambda_1} \biggr)^2 - \biggl( \frac{\partial y}{\partial \lambda_1} \biggr)^2 \biggr]^{1 / 2} = 0.071647$.

Alternatively,

 $~\frac{\partial z}{\partial \lambda_1}$ $~=$ $~h_1^2 \biggl( \frac{\partial \lambda_1}{\partial z}\biggr) = (0.730058)^2 \biggl[ \frac{p^2z}{\lambda_1} \biggr]$

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

 $~\Rightarrow~~~ \frac{\partial\Lambda}{\partial \lambda_1}$ $~=$ $~ \frac{1}{2\Lambda}\biggl[16\lambda_1 \lambda_2^2 \lambda_3^4 \biggr] = \frac{(\Lambda^2-1)}{\lambda_1 \Lambda} \, ,$ $~\frac{\partial\Lambda}{\partial \lambda_2}$ $~=$ $~ \frac{1}{2\Lambda}\biggl[16\lambda_1^2 \lambda_2 \lambda_3^4 \biggr] = \frac{(\Lambda^2-1)}{\lambda_2 \Lambda} \, ,$ $~\frac{\partial\Lambda}{\partial \lambda_3}$ $~=$ $~ \frac{1}{2\Lambda}\biggl[32\lambda_1^2 \lambda_2^2 \lambda_3^3 \biggr] = \frac{2(\Lambda^2-1)}{\lambda_3 \Lambda} \, .$

We have,

 $~\frac{\partial z}{\partial \lambda_1}$ $~=$ $~ \frac{(1-\lambda_2^2)^{1 / 2}}{p} \, ,$ $~\frac{\partial z}{\partial \lambda_2}$ $~=$ $~ -\frac{\lambda_1 \lambda_2}{p(1 - \lambda_2^2)^{1 / 2}} \, ,$ $~\frac{\partial z}{\partial \lambda_3}$ $~=$ $~ 0 \, .$ $~\frac{\partial y}{\partial \lambda_1}$ $~=$ $~ \frac{1}{4\lambda_3^2} \cdot \frac{\partial \Lambda}{\partial \lambda_1} = \biggl[ \frac{(\Lambda^2-1)}{4\lambda_3^2\lambda_1 \Lambda} \biggr] = \biggl[ \frac{8\lambda_1^2 \lambda_2^2 \lambda_3^4}{4\lambda_3^2\lambda_1 \Lambda} \biggr] = \biggl[ \frac{2\lambda_1 \lambda_2^2 \lambda_3^2}{\Lambda} \biggr] \, ,$ $~\frac{\partial y}{\partial \lambda_2}$ $~=$ $~ \frac{1}{4\lambda_3^2} \cdot \frac{\partial \Lambda}{\partial \lambda_2} = \biggl[ \frac{(\Lambda^2-1)}{4\lambda_3^2\lambda_2 \Lambda} \biggr] = \biggl[ \frac{8\lambda_1^2 \lambda_2^2 \lambda_3^4}{4\lambda_3^2\lambda_2 \Lambda} \biggr] = \biggl[ \frac{2\lambda_1^2 \lambda_2 \lambda_3^2}{\Lambda} \biggr] \, ,$ $~\frac{\partial y}{\partial \lambda_3}$ $~=$ $~ \frac{1}{4\lambda_3^2} \cdot \frac{\partial \Lambda}{\partial \lambda_3} - \frac{(\Lambda - 1)}{2\lambda_3^3} = \frac{1}{4\lambda_3^2} \cdot \biggl[ \frac{2(\Lambda^2 - 1)}{\lambda_3\Lambda} \biggr] - \frac{\Lambda(\Lambda - 1)}{2\lambda_3^3 \Lambda} = \biggl[ \frac{(\Lambda^2 - 1) - \Lambda(\Lambda-1)}{2\lambda_3^3\Lambda} \biggr] = \frac{(\Lambda - 1) }{2\lambda_3^3\Lambda} \, .$ $~\frac{\partial x}{\partial \lambda_1}$ $~=$ $~ \frac{1}{4\lambda_3^{2}(\Lambda-1)^{1 / 2}} \cdot \frac{\partial \Lambda}{\partial \lambda_1} = \frac{1}{4\lambda_3^{2}(\Lambda-1)^{1 / 2}} \biggl[ \frac{(\Lambda^2 - 1)}{\lambda_1 \Lambda} \biggr] = \frac{(\Lambda^2 - 1)}{4 \lambda_1 \lambda_3^{2} \Lambda (\Lambda-1)^{1 / 2}} = \frac{2\lambda_1 \lambda_2^2 \lambda_3^{2}}{\Lambda (\Lambda-1)^{1 / 2}} \, ,$ $~\frac{\partial x}{\partial \lambda_2}$ $~=$ $~ \frac{1}{4\lambda_3^{2}(\Lambda-1)^{1 / 2}} \cdot \frac{\partial \Lambda}{\partial \lambda_2} = \frac{1}{4\lambda_3^{2}(\Lambda-1)^{1 / 2}} \biggl[ \frac{(\Lambda^2 - 1)}{\lambda_2 \Lambda} \biggr] = \frac{(\Lambda^2 - 1)}{4 \lambda_2 \lambda_3^{2} \Lambda (\Lambda-1)^{1 / 2}} = \frac{2\lambda_1^2 \lambda_2 \lambda_3^{2}}{\Lambda (\Lambda-1)^{1 / 2}} \, ,$ $~\frac{\partial x}{\partial \lambda_3}$ $~=$ $~ \frac{1}{4\lambda_3^{2}(\Lambda-1)^{1 / 2}} \cdot \frac{\partial \Lambda}{\partial \lambda_3} - \frac{(\Lambda-1)^{1 / 2}}{\lambda_3^{3}} = \frac{1}{4\lambda_3^{2}(\Lambda-1)^{1 / 2}} \biggl[ \frac{2(\Lambda^2 - 1)}{\lambda_3 \Lambda} \biggr] - \frac{(\Lambda-1)^{1 / 2}}{\lambda_3^{3}}$ $~=$ $~ \frac{(\Lambda^2 - 1) -2\Lambda (\Lambda - 1) }{2\lambda_3^{3} \Lambda (\Lambda-1)^{1 / 2}} = - \frac{ (\Lambda - 1)^{3 / 2} }{2\lambda_3^{3} \Lambda} \, .$

What about the derived scale-factors?

 $~h_1^2$ $~=$ $~ \biggl(\frac{\partial x}{\partial \lambda_1}\biggr)^2 + \biggl(\frac{\partial y}{\partial \lambda_1}\biggr)^2 + \biggl(\frac{\partial z}{\partial \lambda_1}\biggr)^2$ $~=$ $~ \biggl[ \frac{2\lambda_1 \lambda_2^2 \lambda_3^{2}}{\Lambda (\Lambda-1)^{1 / 2}} \biggr]^2 + \biggl[ \frac{2\lambda_1 \lambda_2^2 \lambda_3^2}{\Lambda} \biggr]^2 + \biggl[ \frac{(1-\lambda_2^2)^{1 / 2}}{p} \biggr]^2$ $~=$ $~ \biggl[ \frac{4\lambda_1^2 \lambda_2^4 \lambda_3^{4 }}{\Lambda^2 (\Lambda-1)} \biggr] + \biggl[ \frac{4\lambda_1^2 \lambda_2^4 \lambda_3^4}{\Lambda^2} \biggr] + \biggl[ \frac{(1-\lambda_2^2)}{p^2} \biggr]$ $~=$ $~\frac{1}{p^2 \Lambda^2(\Lambda - 1)} \biggl[ 4 p^2 \lambda_1^2 \lambda_2^4 \lambda_3^4 \Lambda + (1-\lambda_2^2) \Lambda^2(\Lambda - 1) \biggr] \, .$

Written in terms of Cartesian coordinates, this becomes,

 $~h_1^2$ $~=$ $~\frac{ 8 \lambda_1^2 \lambda_2^2 \lambda_3^4 (\lambda_2^2 ) }{2 \Lambda (\Lambda - 1)} + \frac{(1-\lambda_2^2)}{p^2}$ $~=$ $~\frac{ (\Lambda+1)\lambda_2^2 }{2 \Lambda } + \frac{z^2}{\lambda_1^2}$ $~=$ $~ \frac{1}{\lambda_1^2} \biggl[ \frac{ (\Lambda+1)\lambda_2^2 \lambda_1^2 }{2 \Lambda } + z^2 \biggr]$ $~=$ $~ \frac{1}{\lambda_1^2} \biggl[ \frac{ (\Lambda+1)(x^2 + 2y^2) }{2 \Lambda } + z^2 \biggr] \, .$

Note that,

 $~\Lambda -1$ $~=$ $~4x^2\lambda_3^4 = 4x^2 \biggl( \frac{y^2}{x^4} \biggr) = 4\biggl( \frac{y^2}{x^2} \biggr)$ $~\Rightarrow ~~~ \Lambda$ $~=$ $~\frac{x^2 + 4y^2}{x^2} \, .$

Hence, the scale factor becomes,

 $~h_1^2$ $~=$ $~ \frac{1}{2 \lambda_1^2} \biggl[ (x^2 + 2y^2) + \frac{ x^2(x^2 + 2y^2) }{(x^2 + 4y^2) } + 2z^2 \biggr]$ $~=$ $~ \frac{1}{2 \lambda_1^2(x^2 + 4y^2) } \biggl[ (x^2 + 2y^2) (x^2 + 4y^2) + x^2(x^2 + 2y^2) + 2z^2(x^2 + 4y^2) \biggr]$ $~=$ $~ \frac{1}{2 \lambda_1^2(x^2 + 4y^2) } \biggl[ (2x^4 + 8x^2y^2 +8y^4) + 2z^2(x^2 + 4y^2) \biggr] = \frac{1.911525}{3.554} = 0.537852$ $~\Rightarrow ~~~ h_1$ $~=$ $~ 0.733384 \, .$

Compare this expression with the one derived earlier, namely,

 $~h_1^2 \biggr|_{q^2 = 2} = \biggl[\lambda_1^2 \ell_{3D}^2 \biggr]_{q^2 = 2}$ $~=$ $~ \frac{(x^2 + 2y^2 + p^2z^2)}{x^2 + 4y^2 + p^4z^2} \, .$

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

 $~\lambda_1^2 = x^2 + 2y^2 + p^2z^2 \, ,$ $~\lambda_2^2 = \frac{\lambda_1^2 - p^2 z^2}{\lambda_1^2} = \frac{x^2 + 2y^2}{x^2 + 2y^2 + p^2z^2} \, ,$ $~\lambda_3^2 = \frac{y}{x^2} \, .$

Hence,

 $~(\Lambda^2 - 1) = 8\lambda_1^2 \lambda_2^2 \lambda_3^4$ $~=$ $~ 8(x^2 + 2y^2 + p^2z^2)\biggl[ \frac{x^2 + 2y^2}{x^2 + 2y^2 + p^2z^2} \biggr]\frac{y^2}{x^4} = \biggl[ \frac{8y^2(x^2 + 2y^2)}{x^4 } \biggr] \, ,$ $~\Rightarrow ~~~ \Lambda^2$ $~=$ $~ 1 + \biggl[ \frac{8y^2(x^2 + 2y^2)}{x^4 } \biggr] = \frac{1}{x^4}\biggl[x^4 + 8x^2y^2 + 16y^4 \biggr] = \frac{1}{x^4}\biggl[x^2 + 4y^4 \biggr]^2 \, ,$ $~\Rightarrow ~~~ (\Lambda+1)$ $~=$ $~ \frac{(x^2 + 4y^4)}{x^2} + 1 = \frac{2x^2 + 4y^4}{x^2} \, ,$

which means,

 $~h_1^2$ $~=$ $~\frac{1}{8\lambda_1^2 \Lambda^2} \biggl\{ 4\lambda_1^2 \lambda_2^2 \lambda_3 (\Lambda+1) + 4\lambda_1^2 \lambda_2^2 (\Lambda^2 - 1) + 8z^2\Lambda^2 \biggr\}$

## Think Again

### Firm Relations

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

 $~\frac{\partial x}{\partial \lambda_1}$ $~=$ $~ h_1^2 \biggl( \frac{\partial \lambda_1}{\partial x} \biggr) = \biggl(\lambda_1 \ell_{3D} \biggr)^2 \frac{x}{\lambda_1} = x \lambda_1 \ell_{3D}^2 \, ,$ $~\frac{\partial y}{\partial \lambda_1}$ $~=$ $~ h_1^2 \biggl( \frac{\partial \lambda_1}{\partial y} \biggr) = \biggl(\lambda_1 \ell_{3D} \biggr)^2 \frac{q^2y}{\lambda_1} = q^2 y \lambda_1 \ell_{3D}^2 \, ,$ $~\frac{\partial z}{\partial \lambda_1}$ $~=$ $~ h_1^2 \biggl( \frac{\partial \lambda_1}{\partial z} \biggr) = \biggl(\lambda_1 \ell_{3D} \biggr)^2 \frac{p^2z}{\lambda_1} = p^2 z \lambda_1 \ell_{3D}^2 \, .$

Check …

 $~h_1^2$ $~=$ $~ \biggl( \frac{\partial x}{\partial \lambda_1} \biggr)^2 + \biggl( \frac{\partial y}{\partial \lambda_1} \biggr)^2 + \biggl( \frac{\partial z}{\partial \lambda_1} \biggr)^2 = \lambda_1^2 \ell_{3D}^4 \biggl[ x^2 + q^4 y^2 + p^4z^2 \biggr] = \lambda_1^2 \ell_{3D}^2 \, .$       (Yes!)

Also,

 $~\frac{\partial x}{\partial \lambda_3}$ $~=$ $~ h_3^2 \biggl( \frac{\partial \lambda_3}{\partial x} \biggr) = \biggl[ \frac{xq^2y \ell_q}{\lambda_3} \biggr]^2 \biggl( - \frac{\lambda_3}{x} \biggr) = - q^4 y^2 \ell_q^2 \biggl( \frac{x}{\lambda_3} \biggr) \, ,$ $~\frac{\partial y}{\partial \lambda_3}$ $~=$ $~ h_3^2 \biggl( \frac{\partial \lambda_3}{\partial y} \biggr) = \biggl[ \frac{xq^2y \ell_q}{\lambda_3} \biggr]^2 \biggl( + \frac{\lambda_3}{q^2y} \biggr) = x^2 \ell_q^2 \biggl( \frac{q^2y} {\lambda_3}\biggr) \, ,$ $~\frac{\partial z}{\partial \lambda_3}$ $~=$ $~ h_3^2 \biggl( \frac{\partial \lambda_3}{\partial z} \biggr) = 0 \, .$

Check …

 $~h_3^2$ $~=$ $~ \biggl( \frac{\partial x}{\partial \lambda_3} \biggr)^2 + \biggl( \frac{\partial y}{\partial \lambda_3} \biggr)^2 + \biggl( \frac{\partial z}{\partial \lambda_3} \biggr)^2 = \frac{\ell_q^4}{\lambda_3^2} \biggl[x^2 q^8 y^4 + x^4 q^4y^2 \biggr] = \frac{x^2 q^4 y^2\ell_q^4}{\lambda_3^2} \biggl[q^4 y^2 + x^2 \biggr] = \frac{x^2 q^4 y^2\ell_q^2}{\lambda_3^2} \, .$       (Yes!)

And, last …

 $~\frac{\partial x}{\partial \lambda_2}$ $~=$ $~ h_2 \gamma_{21} = h_2 \ell_q \ell_{3D} (xp^2z) \, ,$ $~\frac{\partial y}{\partial \lambda_2}$ $~=$ $~ h_2 \gamma_{22} = h_2 \ell_q \ell_{3D} (q^2 y p^2 z) \, ,$ $~\frac{\partial z}{\partial \lambda_2}$ $~=$ $~ h_2 \gamma_{23} = - h_2 \ell_q \ell_{3D}(x^2 + q^4y^2) \, .$

### 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,

 $~h_2^2$ $~=$ $~(\ell_q \ell_{3D})^2 = \biggl[ (x^2 + q^4y^2)(x^2 + q^4y^2 + p^4z^2) \biggr]^{-1} \, ,$

in which case,

 $~\frac{\partial \lambda_2}{\partial x}$ $~=$ $~xp^2z \, ,$ $~\frac{\partial \lambda_2}{\partial y}$ $~=$ $~q^2yp^2z \, ,$ $~\frac{\partial \lambda_2}{\partial z}$ $~=$ $~-(x^2 + q^4y^2) \, ;$

and,

 $~\frac{\partial x}{\partial \lambda_2} = h_2^2 \biggl( \frac{\partial \lambda_2}{\partial x} \biggr)$ $~=$ $~(\ell_q \ell_{3D})^2 xp^2z \, ,$ $~\frac{\partial y}{\partial \lambda_2} = h_2^2 \biggl( \frac{\partial \lambda_2}{\partial y} \biggr)$ $~=$ $~(\ell_q \ell_{3D})^2 q^2yp^2z \, ,$ $~\frac{\partial z}{\partial \lambda_2} = h_2^2 \biggl(\frac{\partial \lambda_2}{\partial z} \biggr)$ $~=$ $~-(\ell_q \ell_{3D})^2 (x^2 + q^4y^2) \, .$

#### Second

Alternatively, after examining the direction-cosine expressions for $~\partial x_i/\partial \lambda_2$ that we have just provided, one might suggest that,

 $~h_2^2$ $~=$ $~(\ell_q \ell_{3D})^{-2} = (x^2 + q^4y^2)(x^2 + q^4y^2 + p^4z^2) = p^4z^2(x^2 + q^4y^2) + (x^2 + q^4y^2)^2 \, ,$

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 2nd scale factor. Let's try the first of these behaviors. Specifically, what if we assume …

 $~\frac{\partial \lambda_2}{\partial x}$ $~=$ $~\frac{xp^2 z}{\lambda_2} \, ,$ $~\frac{\partial \lambda_2}{\partial y}$ $~=$ $~\frac{q^2y p^2z}{\lambda_2} \, ,$ $~\frac{\partial \lambda_2}{\partial z}$ $~=$ $~-\frac{(x^2 + q^4y^2)}{\lambda_2} \, .$

Then,

 $~h_2^{-2}$ $~=$ $~ \biggl( \frac{\partial \lambda_2}{\partial x}\biggr)^2 + \biggl( \frac{\partial \lambda_2}{\partial y}\biggr)^2 +\biggl( \frac{\partial \lambda_2}{\partial z}\biggr)^2 = [\lambda_2 \ell_q \ell_{3D} ]^{-2}$ $~\Rightarrow ~~~ h_2$ $~=$ $~ \lambda_2 \ell_q \ell_{3D} \, .$

Primary implication:

 $~\gamma_{21} = h_2 \biggl(\frac{\partial \lambda_2}{\partial x} \biggr)$ $~=$ $~(xp^2 z) \ell_q \ell_{3D} \ ,$ $~\gamma_{22} = h_2 \biggl(\frac{\partial \lambda_2}{\partial y} \biggr)$ $~=$ $~(q^2 y p^2 z) \ell_q \ell_{3D} \ ,$ $~\gamma_{23} = h_2 \biggl(\frac{\partial \lambda_2}{\partial z} \biggr)$ $~=$ $~-(x^2 + q^4 y^2) \ell_q \ell_{3D} \ .$ 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:

 $~\frac{\partial x}{\partial \lambda_2}$ $~=$ $~ h_2 \gamma_{21} = \lambda_2 \ell_q^2 \ell_{3D}^2 (xp^2z) \, ,$ $~\frac{\partial y}{\partial \lambda_2}$ $~=$ $~ h_2 \gamma_{22} = \lambda_2 \ell_q^2 \ell_{3D}^2 (q^2 y p^2 z) \, ,$ $~\frac{\partial z}{\partial \lambda_2}$ $~=$ $~ h_2 \gamma_{23} = - \lambda_2 \ell_q^2 \ell_{3D}^2(x^2 + q^4y^2) \, .$

Now, what specifically is the function, $~\lambda_2(x, y, z)$ ? Start by rewriting the three partial derivatives as,

 $~\frac{1}{2} \frac{\partial (\lambda_2^2)}{\partial x}$ $~=$ $~xp^2 z \, ,$ $~\frac{1}{2} \frac{\partial (\lambda_2)^2}{\partial y}$ $~=$ $~q^2y p^2z \, ,$ $~\frac{1}{2} \frac{\partial (\lambda_2)^2}{\partial z}$ $~=$ $~-(x^2 + q^4y^2) \, .$

Suppose that,

 $~\lambda_2^2$ $~=$ $~(x^2 + q^2y^2)p^2z \, .$

Then we have,

 $~\frac{\partial \lambda_2^2}{\partial x}$ $~=$ $~2xp^2z \, ,$ and, $~\frac{\partial \lambda_2^2}{\partial y}$ $~=$ $~2q^2 yp^2z \, .$      Great!

But this cannot be the correct expression for $~\lambda_2^2$ because,

 $~\frac{\partial \lambda_2^2}{\partial z}$ $~=$ $~(x^2 + q^2y^2)p^2 \, ,$

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

#### Fourth

Alternatively, if we assume …

 $~\frac{\partial \lambda_2}{\partial x}$ $~=$ $~\frac{\lambda_2}{xp^2 z} \, ,$ $~\frac{\partial \lambda_2}{\partial y}$ $~=$ $~\frac{\lambda_2}{q^2y p^2z} \, ,$ $~\frac{\partial \lambda_2}{\partial z}$ $~=$ $~-\frac{\lambda_2}{(x^2 + q^4y^2)} \, ,$

then,

 $~h_2^{-2}$ $~=$ $~ \biggl( \frac{\partial \lambda_2}{\partial x}\biggr)^2 + \biggl( \frac{\partial \lambda_2}{\partial y}\biggr)^2 +\biggl( \frac{\partial \lambda_2}{\partial z}\biggr)^2$ $~=$ $~ \biggl( \frac{\lambda_2}{xp^2 z} \biggr)^2 + \biggl( \frac{\lambda_2}{q^2y p^2z} \biggr)^2 +\biggl( \frac{\lambda_2}{x^2 + q^4y^2} \biggr)^2$ $~\Rightarrow ~~~ (h_2 \lambda_2)^{-2}$ $~=$ $~ \frac{ q^4y^2p^4z^2 (x^2 + q^4y^2)^2 + x^2p^4z^2 (x^2 + q^4y^2)^2 + x^2 q^4y^2 p^8z^4}{x^2 q^4y^2p^8z^4(x^2 + q^4y^2)^2}$ $~\Rightarrow ~~~ h_2$ $~=$ $~\frac{1}{\lambda_2} \biggl[ \frac{x^2 q^4y^2p^8z^4(x^2 + q^4y^2)^2}{ q^4y^2p^4z^2 (x^2 + q^4y^2)^2 + x^2p^4z^2 (x^2 + q^4y^2)^2 + x^2 q^4y^2 p^8z^4} \biggr]^{1 / 2}$ $~=$ $~\frac{1}{\lambda_2} \biggl\{ \frac{x q^2y p^2z(x^2 + q^4y^2)}{ [ q^4y^2 (x^2 + q^4y^2)^2 + x^2 (x^2 + q^4y^2)^2 + x^2 q^4y^2 p^4z^2 ]^{1 / 2}} \biggr\}$ $~=$ $~\frac{1}{\lambda_2} \biggl\{ \frac{x q^2y p^2z(x^2 + q^4y^2)}{ [ (x^2 + q^4y^2)^3 + x^2 q^4y^2 p^4z^2 ]^{1 / 2}} \biggr\}$

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

 $~\gamma_{21} = h_2 \biggl( \frac{\partial \lambda_2}{\partial x} \biggr)$ $~=$ $~\biggl\{ \frac{q^2y (x^2 + q^4y^2)}{ [ (x^2 + q^4y^2)^3 + x^2 q^4y^2 p^4z^2 ]^{1 / 2}} \biggr\}$ $~\Rightarrow ~~~ \frac{ \gamma_{21} }{\ell_q(xp^2z) }$ $~=$ $~\frac{(x^2 + q^4y^2)^{1 / 2}}{xp^2z} \biggl\{ \frac{q^2y (x^2 + q^4y^2)}{ [ (x^2 + q^4y^2)^3 + x^2 q^4y^2 p^4z^2 ]^{1 / 2}} \biggr\}$ $~=$ $~\frac{q^2y}{xp^2z} \biggl\{ \frac{(x^2 + q^4y^2)^{3 / 2}}{ [ (x^2 + q^4y^2)^3 + x^2 q^4y^2 p^4z^2 ]^{1 / 2}} \biggr\}$ $~=$ $~\frac{q^2y}{xp^2z} \biggl[1 + \frac{x^2q^4y^2p^4z^2}{(x^2 + q^4y^2)^3} \biggr]^{-1 / 2} \, .$

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,

 $~\gamma_{21}$ $~=$ $~\ell_q \ell_{3D} (xp^2z) \, ,$ $~\gamma_{22}$ $~=$ $~\ell_q \ell_{3D} (q^2 y p^2z) \, ,$ $~\gamma_{23}$ $~=$ $~-\ell_q \ell_{3D} (x^2 + q^4 y^2) \, .$

It is easy to see that the desired orthogonality relationship,

 $~\sum_{i=1}^3 (\gamma_{2i})^2$ $~=$ $~1 \, ,$

is satisfied because,

 $~(xp^2z)^2 + (q^2y p^2z)^2 + (x^2 + q^4y^2)^2$ $~=$ $~(x^2 + q^4y^2)(x^2 + q^4y^2 + p^4z^2) = ( \ell_q \ell_{3D} )^{-2} \, .$

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,

 $~\frac{\partial \lambda_2}{\partial x}$ $~=$ $~A(x, y, z) (xp^2z) \, ,$ $~\frac{\partial \lambda_2}{\partial y}$ $~=$ $~A(x, y, z) (q^2 y p^2z) \, ,$ $~\frac{\partial \lambda_2}{\partial z}$ $~=$ $~-A(x, y, z) (x^2 + q^4 y^2) \, ,$

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,

 $~\frac{\partial \lambda_2}{\partial x}$ $~=$ $~\frac{xp^2 z}{\lambda_2} \, ,$ $~\frac{\partial \lambda_2}{\partial y}$ $~=$ $~\frac{q^2y p^2z}{\lambda_2} \, ,$ $~\frac{\partial \lambda_2}{\partial z}$ $~=$ $~-\frac{(x^2 + q^4y^2)}{\lambda_2} \, .$

## 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 2nd 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:

 $~x$ $~~\rightarrow~~$ $~\frac{y^{1/q^2}}{\lambda_3} \, ,$ $~z$ $~~\rightarrow~~$ $~ \frac{1}{p}\biggl[ \lambda_1^2 - q^2y^2 - x^2 \biggr]^{1 / 2}$ $~~\rightarrow~~$ $~ \frac{1}{p}\biggl[ \lambda_1^2 - q^2y^2 - \frac{y^{2/q^2}}{\lambda_3^2} \biggr]^{1 / 2} = \frac{1}{p\lambda_3}\biggl[ \lambda_1^2 \lambda_3^2 - (qy\lambda_3)^2 - y^{2/q^2} \biggr]^{1 / 2} \, .$

This means, for example, that,

 $~\ell_q^{-2}$ $~~\rightarrow~~$ $~ q^4y^2 + \frac{y^{2/q^2}}{\lambda_3^2} = \lambda_3^{-2} \biggl[ q^2(qy\lambda_3)^2 + y^{2/q^2} \biggr] \, ,$ $~\ell_{3D}^{-2}$ $~~\rightarrow~~$ $~ q^4y^2 + \frac{y^{2/q^2}}{\lambda_3^2} + p^2\biggl[ \lambda_1^2 - q^2y^2 - \frac{y^{2/q^2}}{\lambda_3^2} \biggr] = \lambda_3^{-2} \biggl[ (q^2-p^2)(qy \lambda_3)^2 + (1-p^2)y^{2/q^2} + p^2\lambda_1^2\lambda_3^2 \biggr] \, .$

### Derivatives of x

 $~\frac{\partial x}{\partial \lambda_1}$ $~=$ $~ x \lambda_1 \ell_{3D}^2 = y^{1/q^2} \lambda_1 \lambda_3 \biggl[ (q^2-p^2)(qy \lambda_3)^2 + (1-p^2)y^{2/q^2} + p^2\lambda_1^2\lambda_3^2 \biggr]^{-1} \, , $  $~\frac{\partial x}{\partial \lambda_3}$ $~=$ $~ - q^4 y^2 \ell_q^2 \biggl( \frac{x}{\lambda_3} \biggr) = - q^2 (qy)^2 \biggl( y^{1/q^2} \biggr) \lambda_3 \biggl[ q^2(qy\lambda_3)^2 + y^{2/q^2} \biggr]^{-1} \, ,$ $~\frac{\partial x}{\partial \lambda_2}$ $~=$ $~ h_2 \ell_q \ell_{3D} (xp^2z) = \biggl[ h_2 \ell_q \ell_{3D}\biggr] \biggl(y^{1/q^2} \biggr) \frac{p}{\lambda_3^2}\biggl[ \lambda_1^2 \lambda_3^2 - (qy\lambda_3)^2 - y^{2/q^2} \biggr]^{1 / 2}$ $~=$ $~ h_2 \biggl( y^{1/q^2} \biggr) p \biggl[ \lambda_1^2 \lambda_3^2 - (qy \lambda_3)^2 - y^{2/q^2} \biggr]^{1 / 2} \biggl[ q^2(qy\lambda_3)^2 + y^{2/q^2} \biggr]^{-1 / 2} \biggl[ (q^2-p^2)(qy \lambda_3)^2 + (1-p^2)y^{2/q^2} + p^2\lambda_1^2\lambda_3^2 \biggr]^{-1 / 2}$

### 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,

 $~y$ $~=$ $~\lambda_5^{q^2/2} \, ,$ $~x$ $~=$ $~ \frac{y^{(q^2-1)/q^2}}{\lambda_4} = \lambda_4^{-1} \lambda_5^{ (q^2-1)/2} \, ,$ $~z^2$ $~=$ $~ \frac{1}{p^2} \biggl[\lambda_1^2 - x^2 - q^2y^2 \biggr] = \frac{1}{p^2 \lambda_4^2} \biggl[ \lambda_1^2 \lambda_4^2 - \lambda_5^{q^2-1} - q^2 \lambda_4^2\lambda_5^{q^2} \biggr] \, .$

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,

 $~h_2^2$ $~=$ $~[(xq^2y)\ell_q \ell_{3D}]^2 \, ,$

then,

 $~\frac{\partial \lambda_2}{\partial x}$ $~=$ $~\frac{p^2 z}{q^2y} \, ,$ $~\frac{\partial \lambda_2}{\partial y}$ $~=$ $~\frac{p^2z}{x} \, ,$ $~\frac{\partial \lambda_2}{\partial z}$ $~=$ $~- \biggl[ \frac{x}{q^2y} + \frac{q^2y}{x} \biggr] \, .$

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 $~\gamma_{21}, \gamma_{22}, ~\mathrm{and}~ \gamma_{23}$ 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 $~n$ $~\lambda_n$ $~h_n$ $~\frac{\partial \lambda_n}{\partial x}$ $~\frac{\partial \lambda_n}{\partial y}$ $~\frac{\partial \lambda_n}{\partial z}$ $~\gamma_{n1}$ $~\gamma_{n2}$ $~\gamma_{n3}$ $~1$ $~(x^2 + q^2 y^2 + p^2 z^2)^{1 / 2}$ $~\lambda_1 \ell_{3D}$ $~\frac{x}{\lambda_1}$ $~\frac{q^2 y}{\lambda_1}$ $~\frac{p^2 z}{\lambda_1}$ $~(x) \ell_{3D}$ $~(q^2 y)\ell_{3D}$ $~(p^2z) \ell_{3D}$ $~2$ --- $~\ell_q \ell_{3D} (xq^2y)$ $~\frac{p^2z}{q^2y}$ $~\frac{p^2z}{x}$ $~-\biggl[ \frac{x}{q^2y} + \frac{q^2y}{x} \biggr]$ $~\ell_q \ell_{3D} (xq^2y) \biggl[ \frac{p^2z}{q^2y} \biggr]$ $~\ell_q \ell_{3D} (xq^2y) \biggl[ \frac{ p^2z}{x} \biggr]$ $~- \ell_q \ell_{3D} (xq^2y) \biggl[ \frac{x}{q^2y} + \frac{q^2y}{x} \biggr]$ $~3$ $~\frac{y^{1/q^2}}{x}$ $~\frac{xq^2 y \ell_q}{\lambda_3}$ $~-\frac{\lambda_3}{x}$ $~+\frac{\lambda_3}{q^2y}$ $~0$ $~-q^2 y \ell_q$ $~x\ell_q$ $~0$