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=
=
\gamma_{22} \, .
\gamma_{22} \, .
</math>
&nbsp; &nbsp; &nbsp; <font color="red">Yes!</font>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\gamma_{31}\gamma_{12} - \gamma_{32}\gamma_{11}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \biggl[ \frac{q^2y}{(x^2 + q^4y^2)^{1 / 2}} \biggr]
\biggl[ q^2y \ell_{3D} \biggr]
-
\biggl[ \frac{x}{(x^2 + q^4y^2)^{1 / 2}} \biggr]
\biggl[ x\ell_{3D} \biggr]
=
-(x^2 + q^4y^2)^{1 / 2}\ell_{3D}
=
\gamma_{23} \, .
</math>
</math>
&nbsp; &nbsp; &nbsp; <font color="red">Yes!</font>
&nbsp; &nbsp; &nbsp; <font color="red">Yes!</font>

Revision as of 03:01, 12 February 2021

Concentric Ellipsoidal (T8) Coordinates

Whitworth's (1981) Isothermal Free-Energy Surface
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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 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, <math>~\lambda_3(x, y, z)</math>, was associated with the third unit vector. In addition, we found the <math>~\lambda_2</math> 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 <math>~\lambda_2</math> coordinate such that its associated <math>~\hat{e}_3</math> unit vector lies parallel to the x-y plane.

Realigning the Second Coordinate

The first coordinate remains the same as before, namely,

<math>~\lambda_1^2</math>

<math>~=</math>

<math>~x^2 + q^2 y^2 + p^2 z^2 \, .</math>

This may be rewritten as,

<math>~1</math>

<math>~=</math>

<math>~\biggl( \frac{x}{a}\biggr)^2 + \biggl( \frac{y}{b}\biggr)^2 + \biggl(\frac{z}{c}\biggr)^2 \, ,</math>

where,

<math>~a = \lambda_1 \, ,</math>

      

<math>~b = \frac{\lambda_1}{q} \, ,</math>

      

<math>~c = \frac{\lambda_1}{p} \, .</math>

By specifying the value of <math>~z = z_0 < c</math>, as well as the value of <math>~\lambda_1</math>, we are picking a plane that lies parallel to, but a distance <math>~z_0</math> above, the equatorial plane. The elliptical curve that defines the intersection of the <math>~\lambda_1</math>-constant surface with this plane is defined by the expression,

<math>~\lambda_1^2 - p^2z_0^2</math>

<math>~=</math>

<math>~x^2 + q^2 y^2 </math>

<math>~\Rightarrow~~~1</math>

<math>~=</math>

<math>~\biggl( \frac{x}{a_{2D}}\biggr)^2 + \biggl( \frac{y}{b_{2D}}\biggr)^2 \, ,</math>

where,

<math>~a_{2D} = \biggl(\lambda_1^2 - p^2z_0^2 \biggr)^{1 / 2} \, ,</math>

      

<math>~b_{2D} = \frac{1}{q} \biggl(\lambda_1^2 - p^2z_0^2 \biggr)^{1 / 2} \, .</math>

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

<math>~0</math>

<math>~=</math>

<math>~\frac{2x dx}{a_{2D}^2} + \frac{2y dy}{b_{2D}^2}</math>

<math>~\Rightarrow~~~\frac{dy}{dx}</math>

<math>~=</math>

<math>~- \frac{2x}{a_{2D}^2} \cdot \frac{b_{2D}^2}{2y} = - \frac{x}{q^2y} \, .</math>

<math>~\Rightarrow~~~\Delta y</math>

<math>~=</math>

<math>~- \biggl( \frac{x}{q^2y} \biggr)\Delta x \, .</math>

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

<math>~\hat{e}_2</math>

<math>~=</math>

<math>~ \hat\imath~ \biggl\{ \frac{\Delta x}{[ (\Delta x)^2 + (\Delta y)^2 ]^{1 / 2}} \biggr\} + \hat\jmath~ \biggl\{ \frac{\Delta y}{[ (\Delta x)^2 + (\Delta y)^2 ]^{1 / 2}} \biggr\} </math>

 

<math>~=</math>

<math>~ \hat\imath~ \biggl\{ \frac{1}{[ 1 + x^2/(q^4y^2) ]^{1 / 2}} \biggr\} - \hat\jmath~ \biggl\{ \frac{x/(q^2y)}{[ 1 + x^2/(q^4y^2) ]^{1 / 2}} \biggr\} </math>

 

<math>~=</math>

<math>~ \hat\imath~ \biggl\{ \frac{q^2y}{[ x^2 + q^4y^2 ]^{1 / 2}} \biggr\} - \hat\jmath~ \biggl\{ \frac{x}{[ x^2 + q^4y^2 ]^{1 / 2}} \biggr\} \, .</math>

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

<math>~\lambda_2</math>

<math>~=</math>

<math>~\frac{x}{y^{1/q^2}} \, .</math>

Other properties that result from this definition of <math>~\lambda_2</math> are presented in the following table.

Direction Cosine Components for T8 Coordinates
<math>~n</math> <math>~\lambda_n</math> <math>~h_n</math> <math>~\frac{\partial \lambda_n}{\partial x}</math> <math>~\frac{\partial \lambda_n}{\partial y}</math> <math>~\frac{\partial \lambda_n}{\partial z}</math> <math>~\gamma_{n1}</math> <math>~\gamma_{n2}</math> <math>~\gamma_{n3}</math>
<math>~1</math> <math>~(x^2 + q^2 y^2 + p^2 z^2)^{1 / 2} </math> <math>~\lambda_1 \ell_{3D}</math> <math>~\frac{x}{\lambda_1}</math> <math>~\frac{q^2 y}{\lambda_1}</math> <math>~\frac{p^2 z}{\lambda_1}</math> <math>~(x) \ell_{3D}</math> <math>~(q^2 y)\ell_{3D}</math> <math>~(p^2z) \ell_{3D}</math>
<math>~2</math> <math>~\frac{x}{ y^{1/q^2}}</math> <math>~\frac{1}{\lambda_2}\biggl[\frac{x q^2 y }{(x^2 + q^4y^2)^{1 / 2}}\biggr] </math> <math>~\frac{\lambda_2}{x}</math> <math>~-\frac{\lambda_2}{q^2 y}</math> <math>~0</math> <math>~\frac{q^2 y }{(x^2 + q^4y^2)^{1 / 2}} </math> <math>~- \frac{x }{(x^2 + q^4y^2)^{1 / 2}} </math> <math>~0</math>
<math>~3</math> --- --- --- --- --- --- --- ---

<math>~\ell_{3D}</math>

<math>~\equiv</math>

<math>~(x^2 + q^4y^2 + p^4 z^2 )^{- 1 / 2} </math>

The associated unit vector is, then,

<math>~\hat{e}_2</math>

<math>~=</math>

<math>~ \hat\imath~\biggl[ \frac{q^2 y }{(x^2 + q^4y^2)^{1 / 2}} \biggr] - \hat\jmath~\biggl[ \frac{x }{(x^2 + q^4y^2)^{1 / 2}} \biggr] \, . </math>

It is easy to see that <math>~\hat{e}_2 \cdot \hat{e}_2 = 1</math>. We also see that,

<math>~\hat{e}_1 \cdot \hat{e}_2</math>

<math>~=</math>

<math>~ x \ell_{3D}\biggl[ \frac{q^2 y }{(x^2 + q^4y^2)^{1 / 2}} \biggr] - q^2y \ell_{3D} \biggl[ \frac{x }{(x^2 + q^4y^2)^{1 / 2}} \biggr] = 0 \, , </math>

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

Search for the Third Coordinate

Cross Product of First Two Unit Vectors

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

<math>~\hat{e}_3 = \hat{e}_1 \times \hat{e}_2</math>

<math>~=</math>

<math>~ \hat\imath~\biggl[ {e}_{1y}{e}_{2z} - {e}_{1z}{e}_{2y} \biggr] + \hat\jmath~\biggl[ {e}_{1z}{e}_{2x} - {e}_{1x}{e}_{2z} \biggr] + \hat{k}~\biggl[ {e}_{1x}{e}_{2y} - {e}_{1y}{e}_{2x} \biggr] </math>

 

<math>~=</math>

<math>~ \hat\imath~\biggl[ \frac{x p^2z\ell_{3D}}{(x^2 + q^4y^2)^{1 / 2}} \biggr] + \hat\jmath~\biggl[ \frac{q^2 y p^2z \ell_{3D}}{(x^2 + q^4y^2)^{1 / 2}} \biggr] - \hat{k}~\biggl[ \frac{x^2\ell_{3D} }{(x^2 + q^4y^2)^{1 / 2}} ~+~ \frac{q^4 y^2\ell_{3D} }{(x^2 + q^4y^2)^{1 / 2}} \biggr] </math>

 

<math>~=</math>

<math>~ \frac{\ell_{3D}}{(x^2 + q^4y^2)^{1 / 2}} \biggl\{~ \hat\imath~( x p^2z ) + \hat\jmath~(q^2 y p^2z ) - \hat{k}~(x^2 + q^4y^2) ~\biggr\} \, . </math>

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

Direction Cosine Components for T8 Coordinates
<math>~n</math> <math>~\lambda_n</math> <math>~h_n</math> <math>~\frac{\partial \lambda_n}{\partial x}</math> <math>~\frac{\partial \lambda_n}{\partial y}</math> <math>~\frac{\partial \lambda_n}{\partial z}</math> <math>~\gamma_{n1}</math> <math>~\gamma_{n2}</math> <math>~\gamma_{n3}</math>
<math>~1</math> <math>~(x^2 + q^2 y^2 + p^2 z^2)^{1 / 2} </math> <math>~\lambda_1 \ell_{3D}</math> <math>~\frac{x}{\lambda_1}</math> <math>~\frac{q^2 y}{\lambda_1}</math> <math>~\frac{p^2 z}{\lambda_1}</math> <math>~(x) \ell_{3D}</math> <math>~(q^2 y)\ell_{3D}</math> <math>~(p^2z) \ell_{3D}</math>
<math>~2</math> <math>~\frac{x}{ y^{1/q^2}}</math> <math>~\frac{1}{\lambda_2}\biggl[\frac{x q^2 y }{(x^2 + q^4y^2)^{1 / 2}}\biggr] </math> <math>~\frac{\lambda_2}{x}</math> <math>~-\frac{\lambda_2}{q^2 y}</math> <math>~0</math> <math>~\frac{q^2 y }{(x^2 + q^4y^2)^{1 / 2}} </math> <math>~- \frac{x }{(x^2 + q^4y^2)^{1 / 2}} </math> <math>~0</math>
<math>~3</math> --- --- --- --- --- <math>~\frac{x p^2 z\ell_{3D}}{(x^2 + q^4y^2)^{1 / 2}} </math> <math>~\frac{q^2 y p^2 z\ell_{3D}}{(x^2 + q^4y^2)^{1 / 2}} </math> <math>~-\frac{(x^2 + q^4 y^2)\ell_{3D}}{(x^2 + q^4y^2)^{1 / 2}} </math>

<math>~\ell_{3D}</math>

<math>~\equiv</math>

<math>~(x^2 + q^4y^2 + p^4 z^2 )^{- 1 / 2} \, ,</math>


Associated h3 Scale Factor

Whiteboard EUREKA moment

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

<math>~\frac{\partial \lambda_3}{\partial x}</math>

<math>~=</math>

<math>~\frac{xp^2z}{(x^2 + q^4y^2)} \, ;</math>

       

<math>~\frac{\partial \lambda_3}{\partial y}</math>

<math>~=</math>

<math>~\frac{q^2 y p^2z}{(x^2 + q^4y^2)} \, ;</math>

        and        

<math>~\frac{\partial \lambda_3}{\partial z}</math>

<math>~=</math>

<math>~-1 \, .</math>

This means that,

<math>~h_3^{-2}</math>

<math>~=</math>

<math>~\sum_{i=1}^3 \biggl( \frac{\partial \lambda_3}{\partial x_i}\biggr)^2</math>

 

<math>~=</math>

<math>~ \biggl[ \frac{xp^2z}{(x^2 + q^4y^2)} \biggr]^2 + \biggl[ \frac{q^2 y p^2z}{(x^2 + q^4y^2)} \biggr]^2 + \biggl[ -1 \biggr]^2 </math>

 

<math>~=</math>

<math>~ \frac{p^4z^2(x^2 + q^4y^2)}{(x^2 + q^4y^2)^2} + 1 </math>

 

<math>~=</math>

<math>~ \frac{(x^2 + q^4y^2 +p^4z^2)}{(x^2 + q^4y^2)} </math>

 

<math>~=</math>

<math>~ \frac{1}{\ell_{3D}^2 (x^2 + q^4y^2)} </math>

<math>~\Rightarrow~~~ h_3</math>

<math>~=</math>

<math>~ \ell_{3D} (x^2 + q^4y^2)^{1 / 2} \, . </math>

This seems to work well because, when combined with the three separate expressions for <math>~\partial \lambda_3/\partial x_i</math>, this single expression for <math>~h_3</math> generates all three components of the third unit vector, that is, all three direction cosines, <math>~\gamma_{3i}</math>. All of the elements of this new "EUREKA moment" result have been entered into the following table.


Direction Cosine Components for T8 Coordinates
<math>~n</math> <math>~\lambda_n</math> <math>~h_n</math> <math>~\frac{\partial \lambda_n}{\partial x}</math> <math>~\frac{\partial \lambda_n}{\partial y}</math> <math>~\frac{\partial \lambda_n}{\partial z}</math> <math>~\gamma_{n1}</math> <math>~\gamma_{n2}</math> <math>~\gamma_{n3}</math>
<math>~1</math> <math>~(x^2 + q^2 y^2 + p^2 z^2)^{1 / 2} </math> <math>~\lambda_1 \ell_{3D}</math> <math>~\frac{x}{\lambda_1}</math> <math>~\frac{q^2 y}{\lambda_1}</math> <math>~\frac{p^2 z}{\lambda_1}</math> <math>~(x) \ell_{3D}</math> <math>~(q^2 y)\ell_{3D}</math> <math>~(p^2z) \ell_{3D}</math>
<math>~2</math> <math>~\frac{x}{ y^{1/q^2}}</math> <math>~\frac{1}{\lambda_2}\biggl[\frac{x q^2 y }{(x^2 + q^4y^2)^{1 / 2}}\biggr] </math> <math>~\frac{\lambda_2}{x}</math> <math>~-\frac{\lambda_2}{q^2 y}</math> <math>~0</math> <math>~\frac{q^2 y }{(x^2 + q^4y^2)^{1 / 2}} </math> <math>~- \frac{x }{(x^2 + q^4y^2)^{1 / 2}} </math> <math>~0</math>
<math>~3</math> --- <math>~\ell_{3D}(x^2 + q^4 y^2)^{1 / 2}</math> <math>~\frac{xp^2z}{(x^2 + q^4y^2)} </math> <math>~\frac{q^2 y p^2z}{(x^2 + q^4y^2)}</math> <math>~-1</math> <math>~\frac{x p^2 z\ell_{3D}}{(x^2 + q^4y^2)^{1 / 2}} </math> <math>~\frac{q^2 y p^2 z\ell_{3D}}{(x^2 + q^4y^2)^{1 / 2}} </math> <math>~-\frac{(x^2 + q^4 y^2)\ell_{3D}}{(x^2 + q^4y^2)^{1 / 2}} </math>

<math>~\ell_{3D}</math>

<math>~\equiv</math>

<math>~(x^2 + q^4y^2 + p^4 z^2 )^{- 1 / 2} </math>


What is the Third Coordinate Function, λ3

The remaining $64,000 question is, "What is the actual expression for <math>~\lambda_3(x, y, z)</math>  ?  "

Notice that the (partial) derivatives of <math>~\lambda_3</math> with respect to <math>~x</math> and <math>~y</math> may be rewritten, respectively, in the form

<math>~\biggl( \frac{q^2 y}{p^2z} \biggr) \frac{\partial \lambda_3}{\partial x}</math>

<math>~=</math>

<math>~ \frac{q^2 y}{x(1 + \eta^2)} = \frac{\eta}{(1 + \eta^2)} \, , </math>       and,

<math>~\biggl( \frac{x}{p^2z} \biggr) \frac{\partial \lambda_3}{\partial y}</math>

<math>~=</math>

<math>~ \frac{q^2y}{x(1 + \eta^2)} = \frac{\eta}{(1 + \eta^2)} \, , </math>

where,

<math>~\eta</math>

<math>~\equiv</math>

<math>~ \frac{q^2y}{x} ~~~~\Rightarrow~~ \frac{\partial \ln\eta}{\partial \ln x} = -1 \, , ~~~~\frac{\partial \ln\eta}{\partial \ln y} = +1 \, . </math>

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

<math>~\frac{d}{dx_i} \biggl[\frac{1}{\cosh\gamma}\biggr]</math>

<math>~=</math>

<math>~- \biggl[ \frac{\tanh\gamma}{\cosh\gamma} \biggr] \frac{d\gamma}{dx_i} \, .</math>

Hence, it will be useful to adopt the mapping,

<math>~\eta</math>

<math>~~~\rightarrow~~~</math>

<math>~\sinh \gamma \, ,</math>

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

<math>~\frac{\eta}{(1+\eta^2)}</math>

<math>~~~\rightarrow~~~</math>

<math>~\frac{\sinh\gamma}{\cosh^2\gamma} = \frac{\tanh\gamma}{\cosh\gamma} \, .</math>

Guess A

In particular, this suggests that we set,

<math>~\lambda_3</math>

<math>~=</math>

<math>~\frac{A}{\cosh\gamma} \, ,</math>

where,

<math>~\gamma</math>

<math>~\equiv</math>

<math>~\sinh^{-1}\eta = \pm \cosh^{-1}[\eta^2 + 1]^{1 / 2} \, .</math>

In other words,

<math>~\lambda_3</math>

<math>~=</math>

<math>~A[\eta^2 + 1]^{-1 / 2} \, .</math>

Let's check the first and second partial derivatives.

<math>~\frac{\partial \lambda_3}{\partial x}</math>

<math>~=</math>

<math>~ - \frac{A}{2} \biggl[ \frac{2\eta}{(\eta^2 + 1)^{3 / 2}} \biggr] \frac{\partial \eta}{\partial x} </math>

Guess B

What if,

<math>~\lambda_3</math>

<math>~=</math>

<math>~\frac{1}{2}\ln(1+\eta^2) \, .</math>

Then,

<math>~\frac{d\lambda_3}{d\eta}</math>

<math>~=</math>

<math>~ \frac{\eta}{(1+\eta^2)} \, , </math>

in which case we find,

<math>~\frac{\partial\lambda_3}{\partial x_i} </math>

<math>~=</math>

<math>~\frac{d\lambda_3}{d\eta} \cdot \frac{\partial\eta}{\partial x_i}</math>

which means,

<math>~\frac{\partial\lambda_3}{\partial x} </math>

<math>~=</math>

<math>~ \frac{\eta}{(1+\eta^2)} \biggl[ - \frac{q^2 y}{x^2} \biggr] = - \frac{x}{(x^2 + q^4y^2)} \biggl[ \frac{q^4 y^2}{x^2} \biggr] \, . </math>

Guess C

What if,

<math>~\lambda_3</math>

<math>~=</math>

<math>~\frac{1}{2} \ln(1+\eta^{-2}) \, .</math>

Then,

<math>~\frac{d\lambda_3}{d\eta}</math>

<math>~=</math>

<math>~- \frac{\eta^{-3}}{(1+\eta^{-2})} = -\frac{ \eta^{-1}}{(1+\eta^{2})} </math>

in which case we find,

<math>~\frac{\partial\lambda_3}{\partial x_i} </math>

<math>~=</math>

<math>~\frac{d\lambda_3}{d\eta} \cdot \frac{\partial\eta}{\partial x_i}</math>

which means,

<math>~\frac{\partial\lambda_3}{\partial x} </math>

<math>~=</math>

<math>~ -\frac{\eta^{-1}}{(1+\eta^2)} \biggl[ - \frac{q^2 y}{x^2} \biggr] = \frac{x}{(x^2 + q^4y^2)} \, ; </math>

and,

<math>~\frac{\partial\lambda_3}{\partial y} </math>

<math>~=</math>

<math>~ -\frac{\eta^{-1}}{(1+\eta^2)} \biggl[ \frac{q^2 }{x} \biggr] = -\frac{x^2}{y(x^2+ q^4y^2)} </math>

Inverting Coordinate Relations

In a Plane Perpendicular to the Z-Axis

General Case

At a fixed value of <math>~z = z_0</math>, let's invert the <math>~\lambda_1(x, y)</math> and <math>~\lambda_2(x, y)</math> relations to obtain expressions for <math>~x(\lambda_1, \lambda_2)</math> and <math>~y(\lambda_1, \lambda_2)</math>. Perhaps this will help us determine what the third coordinate expression should be.

We start with,

<math>~\lambda_1</math>

<math>~\equiv</math>

<math>~(x^2 + q^2y^2 + p^2 z_0^2)^{1 / 2} \, ;</math>

<math>~\lambda_2</math>

<math>~\equiv</math>

<math>~\frac{x}{y^{1/q^2}} \, .</math>

This means that,

<math>~\ln y</math>

<math>~=</math>

<math>~q^2(\ln x - \ln\lambda_2 ) \, ,</math>

and,

<math>~q^2 y^2</math>

<math>~=</math>

<math>~(\lambda_1^2 - p^2z_0^2) - x^2</math>

<math>~\Rightarrow~~~2\ln y</math>

<math>~=</math>

<math>~\ln\biggl\{ \frac{1}{q^2} \biggl[ (\lambda_1^2 - p^2z_0^2) - x^2 \biggr] \biggr\} \, .</math>

Together, this gives,

<math>~\ln\biggl\{ \frac{1}{q^2} \biggl[ (\lambda_1^2 - p^2z_0^2) - x^2 \biggr] \biggr\} </math>

<math>~=</math>

<math>~2q^2(\ln x - \ln\lambda_2 ) </math>

 

<math>~=</math>

<math>~\ln\biggl[\frac{x}{\lambda_2} \biggr]^{2q^2}</math>

<math>~\Rightarrow~~~(\lambda_1^2 - p^2z_0^2) - x^2 </math>

<math>~=</math>

<math>~q^2\biggl[\frac{x}{\lambda_2} \biggr]^{2q^2} </math>

<math>~\Rightarrow~~~ x^2 +q^2\biggl[\frac{x}{\lambda_2} \biggr]^{2q^2} + (p^2z_0^2 - \lambda_1^2) </math>

<math>~=</math>

<math>~0 \, .</math>

What if Axisymmetric (q2 = 1)

In an axisymmetric configuration, <math>~q^2 = 1</math> and <math>~(\lambda_1^2 - p^2z_0^2) = \varpi^2</math>, so this general expression for <math>~x</math> becomes,

<math>~0</math>

<math>~=</math>

<math>~ x^2 + \biggl[\frac{x}{\lambda_2} \biggr]^{2} -\varpi^2 </math>

 

<math>~=</math>

<math>~ x^2\biggl[\frac{1 + \lambda_2^2}{\lambda_2^2}\biggr] -\varpi^2 </math>

<math>~\Rightarrow~~~x</math>

<math>~=</math>

<math>~ \varpi \biggl[\frac{\lambda_2^2}{1 + \lambda_2^2}\biggr]^{1 / 2} \, . </math>

Given that, for axisymmetric systems,

<math>~x</math>

<math>~=</math>

<math>~\varpi \cos\varphi \, ,</math>

we conclude that when <math>~q^2 = 1</math>,

<math>~\biggl[\frac{\lambda_2^2}{1 + \lambda_2^2}\biggr]^{1 / 2}</math>

<math>~=</math>

<math>~\cos\varphi </math>

<math>~\Rightarrow~~~\lambda_2^2</math>

<math>~=</math>

<math>~\cot^2\varphi </math>

What if q2 = 2

For example, if we choose <math>~q^2 = 2</math>, we have a quadratic expression for <math>~x^2</math>, namely,

<math>~0 </math>

<math>~=</math>

<math>~x^2 + 2\biggl[\frac{x}{\lambda_2} \biggr]^{4} + (p^2z_0^2 - \lambda_1^2) </math>

<math>~\Rightarrow ~~~ 0 </math>

<math>~=</math>

<math>~x^{4} + \frac{1}{2} \lambda_2^4~ x^2 + \frac{1}{2}\lambda_2^4(p^2z_0^2 - \lambda_1^2) </math>

<math>~\Rightarrow ~~~ 2x^2 </math>

<math>~=</math>

<math>~ - \frac{\lambda_2^4}{2} \pm \biggl[ \frac{\lambda_2^8}{4} - 2\lambda_2^4(p^2z_0^2 - \lambda_1^2) \biggr]^{1 / 2} </math>

 

<math>~=</math>

<math>~ \frac{\lambda_2^4}{2} \biggl\{ \biggl[ 1 - \frac{8(p^2z_0^2 - \lambda_1^2)}{\lambda_2^4} \biggr]^{1 / 2} - 1 \biggr\} </math>

<math>~\Rightarrow ~~~ x^2 </math>

<math>~=</math>

<math>~ \frac{\lambda_2^4}{4} \biggl\{ \biggl[ 1 - \frac{8(p^2z_0^2 - \lambda_1^2)}{\lambda_2^4} \biggr]^{1 / 2} - 1 \biggr\} \, . </math>

Given that, for <math>~q^2 = 2</math>, one of the two defining expression means, <math>~\lambda_2 = x/\sqrt{y}</math>, we also have,

<math>~y </math>

<math>~=</math>

<math>~ \frac{\lambda_2^2}{4} \biggl\{ \biggl[ 1 - \frac{8(p^2z_0^2 - \lambda_1^2)}{\lambda_2^4} \biggr]^{1 / 2} - 1 \biggr\} \, . </math>

New 2nd Coordinate

Apparently it will be cleaner to define a new "2nd coordinate," <math>~\kappa_2</math>, such that,

<math>~\kappa_2^{2q^2}</math>

<math>~\equiv</math>

<math>~ q^2\biggl[\frac{1}{\lambda_2} \biggr]^{2q^2} </math>

<math>~\Rightarrow~~~\kappa_2</math>

<math>~\equiv</math>

<math>~ q^{1/q^2}\biggl[\frac{1}{\lambda_2} \biggr] = q^{1/q^2}\biggl[\frac{y^{1/q^2}}{x} \biggr] = \frac{(qy)^{1/q^2}}{x} \, . </math>

(With this new definition, <math>~\kappa_2 \sim \tan\varphi</math>; it is exactly this when <math>~q^2 = 1</math>.) Then we can rewrite the last expression from the above general case as,

<math>~x^2 +(\kappa_2 x)^{2q^2} - (\lambda_1^2 - p^2z_0^2 ) </math>

<math>~=</math>

<math>~0 \, .</math>

When <math>~q=1</math> (the axisymmetric case), this gives,

<math>~x^2(1+\kappa_2^2)</math>

<math>~=</math>

<math>~ (\lambda_1^2 - p^2z_0^2 ) </math>

<math>~\Rightarrow~~~\frac{x^2}{(\lambda_1^2 - p^2z_0^2 ) }</math>

<math>~=</math>

<math>~ \frac{1}{(1+\kappa_2^2)} \, ,</math>

which means that <math>~\kappa_2 = \tan\varphi</math>. And, for the case of <math>~q^2 = 2</math>, after making the substitution,

<math>~\lambda_2 \rightarrow (q)^{1/q^2}\kappa_2^{-1} = \frac{2^{1 / 4}}{\kappa_2} \, ,</math>

we find,

<math>~x^2 </math>

<math>~=</math>

<math>~ \frac{\lambda_2^4}{4} \biggl\{ \biggl[ 1 - \frac{8(p^2z_0^2 - \lambda_1^2)}{\lambda_2^4} \biggr]^{1 / 2} - 1 \biggr\} \, . </math>

<math>~\Rightarrow~~~ x^2 </math>

<math>~=</math>

<math>~ \frac{1}{2\kappa_2^4} \biggl\{ \biggl[ 1 + 4\kappa_2^4(\lambda_1^2 - p^2z_0^2 ) \biggr]^{1 / 2} - 1 \biggr\} \, , </math>

and,

<math>~y = \biggl(\frac{\kappa_2^2}{2^{1 / 2}} \biggr) x^2 </math>

<math>~=</math>

<math>~\biggl(\frac{\kappa_2^2}{2^{1 / 2}} \biggr) \frac{1}{2\kappa_2^4} \biggl\{ \biggl[ 1 + 4\kappa_2^4(\lambda_1^2 - p^2z_0^2 ) \biggr]^{1 / 2} - 1 \biggr\} </math>

 

<math>~=</math>

<math>~ \frac{1}{2^{3 / 2}\kappa_2^2} \biggl\{ \biggl[ 1 + 4\kappa_2^4(\lambda_1^2 - p^2z_0^2 ) \biggr]^{1 / 2} - 1 \biggr\} \, . </math>

Kappa (κ8) Coordinates

<math>~\kappa_1</math>

<math>~\equiv</math>

<math>~(x^2 + q^2y^2 + p^2 z^2)^{1 / 2} \, ;</math>

<math>~\kappa_3</math>

<math>~\equiv</math>

<math>~\tan^{-1}\biggl[ \frac{(qy)^{1/q^2}}{x} \biggr] \, .</math>

<math>~\frac{\partial\kappa_3}{\partial x}</math>

<math>~=</math>

<math>~ \biggl[1 + \frac{(qy)^{2/q^2}}{x^2} \biggr]^{-1} \biggl[- \frac{(qy)^{1/q^2}}{x^2} \biggr] </math>

 

<math>~=</math>

<math>~ -\frac{\sin^2\kappa_3}{(qy)^{1/q^2}} = -\frac{\sin^2\kappa_3}{x\tan\kappa_3} </math>

 

<math>~=</math>

<math>~ -\frac{\sin\kappa_3 \cos\kappa_3}{x}\, . </math>

<math>~\frac{\partial\kappa_3}{\partial y}</math>

<math>~=</math>

<math>~ \biggl[1 + \frac{(qy)^{2/q^2}}{x^2} \biggr]^{-1} \biggl[\frac{q^{1/q^2} y^{1/q^2}}{q^2 x y} \biggr] </math>

 

<math>~=</math>

<math>~ \biggl[1 + \frac{(qy)^{2/q^2}}{x^2} \biggr]^{-1} \biggl[\frac{(qy)^{1/q^2}}{x } \biggr] \frac{1}{q^2y} </math>

 

<math>~=</math>

<math>~

\frac{1}{q^2y}\biggl[\frac{\tan\kappa_3}{1 + \tan^2\kappa_3 }\biggr]

</math>

 

<math>~=</math>

<math>~ + \frac{\sin\kappa_3 \cos\kappa_3}{q^2y} \, . </math>

Therefore,

<math>~h_3^{-2}</math>

<math>~=</math>

<math>~ \biggl[-\frac{\sin\kappa_3 \cos\kappa_3}{x} \biggr]^2 + \biggl[\frac{\sin\kappa_3 \cos\kappa_3}{q^2y} \biggr]^2 </math>

 

<math>~=</math>

<math>~(x^2 + q^4y^2) \biggl[\frac{\sin\kappa_3 \cos\kappa_3}{xq^2y} \biggr]^2 </math>

<math>~\Rightarrow~~~ h_3</math>

<math>~=</math>

<math>~(x^2 + q^4y^2)^{-1 / 2} \biggl[\frac{xq^2y}{\sin\kappa_3 \cos\kappa_3} \biggr] \, . </math>

Direction Cosine Components for κ8 Coordinates
<math>~n</math> <math>~\kappa_n</math> <math>~h_n</math> <math>~\frac{\partial \kappa_n}{\partial x}</math> <math>~\frac{\partial \kappa_n}{\partial y}</math> <math>~\frac{\partial \kappa_n}{\partial z}</math> <math>~\gamma_{n1}</math> <math>~\gamma_{n2}</math> <math>~\gamma_{n3}</math>
<math>~1</math> <math>~(x^2 + q^2 y^2 + p^2 z^2)^{1 / 2} </math> <math>~\kappa_1 \ell_{3D}</math> <math>~\frac{x}{\kappa_1}</math> <math>~\frac{q^2 y}{\kappa_1}</math> <math>~\frac{p^2 z}{\kappa_1}</math> <math>~(x) \ell_{3D}</math> <math>~(q^2 y)\ell_{3D}</math> <math>~(p^2z) \ell_{3D}</math>
<math>~2</math> --- <math>~\ell_{3D}(x^2 + q^4 y^2)^{1 / 2}</math> <math>~\frac{xp^2z}{(x^2 + q^4y^2)} </math> <math>~\frac{q^2 y p^2z}{(x^2 + q^4y^2)}</math> <math>~-1</math> <math>~\frac{x p^2 z\ell_{3D}}{(x^2 + q^4y^2)^{1 / 2}} </math> <math>~\frac{q^2 y p^2 z\ell_{3D}}{(x^2 + q^4y^2)^{1 / 2}} </math> <math>~-\frac{(x^2 + q^4 y^2)\ell_{3D}}{(x^2 + q^4y^2)^{1 / 2}} </math>
<math>~3</math> <math>~\tan^{-1}\biggl[ \frac{(qy)^{1/q^2}}{x} \biggr]</math> <math>~\frac{1}{\sin\kappa_3 \cos\kappa_3}\biggl[\frac{x q^2 y }{(x^2 + q^4y^2)^{1 / 2}}\biggr] </math> <math>~-\frac{\sin\kappa_3 \cos\kappa_3}{x}</math> <math>~\frac{\sin\kappa_3 \cos\kappa_3}{q^2 y}</math> <math>~0</math> <math>~- \frac{q^2 y }{(x^2 + q^4y^2)^{1 / 2}} </math> <math>~\frac{x }{(x^2 + q^4y^2)^{1 / 2}} </math> <math>~0</math>

<math>~\ell_{3D}</math>

<math>~\equiv</math>

<math>~(x^2 + q^4y^2 + p^4 z^2 )^{- 1 / 2} </math>

<math>~4</math> <math>~\frac{x y^{1/q^2}}{ z^{2/p^2}}</math> <math>~\frac{1}{\kappa_4}\biggl[\frac{x q^2 y p^2 z}{ \mathcal{D}}\biggr] </math> <math>~\frac{\kappa_4}{x}</math> <math>~\frac{\kappa_4}{q^2 y}</math> <math>~-\frac{2\kappa_4}{p^2 z}</math> <math>~\frac{x q^2 y p^2 z}{\mathcal{D}} \biggl(\frac{1}{x}\biggr)</math> <math>~ \frac{x q^2 y p^2 z}{\mathcal{D}} \biggl(\frac{1}{q^2y}\biggr)</math> <math>~\frac{x q^2 y p^2 z}{\mathcal{D}} \biggl(-\frac{2}{p^2z}\biggr)</math>
<math>~5</math> --- --- --- --- --- <math>~-\frac{\ell_{3D}}{\mathcal{D}}\biggl[ x(2 q^4y^2 + p^4z^2 ) \biggl]</math> <math>~\frac{\ell_{3D}}{\mathcal{D}}\biggl[ q^2 y(p^4z^2 + 2x^2 ) \biggl]</math> <math>~\frac{\ell_{3D}}{\mathcal{D}}\biggl[ p^2z( x^2 - q^4y^2 ) \biggl]</math>

<math>~\mathcal{D}</math>

<math>~\equiv</math>

<math>~(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2)^{1 / 2} \, .</math>

Also note …

<math>~AB \equiv \biggl( \frac{\mathcal{D}}{\ell_{3D}} \biggr)^{2} </math>

<math>~=</math>

<math>~ \biggl[ x(2 q^4y^2 + p^4z^2 ) \biggl]^2 + \biggl[ q^2 y(p^4z^2 + 2x^2 ) \biggl]^2 + \biggl[ p^2z( x^2 - q^4y^2 ) \biggl]^2 \, , </math>

and the partial derivatives of <math>~A</math> and <math>~B</math> are detailed in an accompanying discussion.

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

<math>~\hat{e}_2 = \hat{e}_3 \times \hat{e}_1</math>

<math>~=</math>

<math>~ \hat\imath \biggl[ e_{3y}e_{1z} - e_{3z} e_{1y} \biggr] + \hat\jmath \biggl[ e_{3z}e_{1x} - e_{3x} e_{1z} \biggr] + \hat{k} \biggl[ e_{3x}e_{1y} - e_{3y} e_{1x} \biggr] </math>

 

<math>~=</math>

<math>~ \hat\imath \biggl[ \frac{x (p^2z) \ell_{3D} }{(x^2 + q^4y^2)^{1 / 2}} \biggr] + \hat\jmath \biggl[ \frac{q^2 y (p^2z) \ell_{3D}}{(x^2 + q^4y^2)^{1 / 2}}\biggr] - \hat{k} \biggl[ \frac{(x^2 + q^4y^2) \ell_{3D} }{(x^2 + q^4y^2)^{1 / 2}} \biggr] \, . </math>

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

Attempt 1

Let's try …

<math>~\kappa_2</math>

<math>~\equiv</math>

<math>~\tan^{-1}\biggl[ \frac{x(qy)^{1/q^2}}{(pz)^{1/p^2}} \biggr] \, ,</math>

which leads to,

<math>~\frac{\partial\kappa_3}{\partial x}</math>

<math>~=</math>

<math>~ \biggl[1 + \frac{x^2(qy)^{2/q^2}}{(pz)^{2/p^2}} \biggr]^{-1} \biggl[ \frac{(qy)^{1/q^2}}{(pz)^{1/p^2}} \biggr] </math>

 

<math>~=</math>

<math>~ \frac{1}{x}\biggl[1+\tan^2\kappa_2\biggr]^{-1} \tan\kappa_2 = \frac{\sin\kappa_2 \cos\kappa_2}{x} \, ; </math>

<math>~\frac{\partial\kappa_3}{\partial y}</math>

<math>~=</math>

<math>~ \biggl[1 + \frac{x^2(qy)^{2/q^2}}{(pz)^{2/p^2}} \biggr]^{-1} \biggl[ \frac{xq^{1/q^2}(y)^{1/q^2}}{q^2 y(pz)^{1/p^2}} \biggr] </math>

 

<math>~=</math>

<math>~ \frac{1}{q^2y}\biggl[1+\tan^2\kappa_2\biggr]^{-1} \tan\kappa_2 = \frac{\sin\kappa_2 \cos\kappa_2}{q^2y} \, ; </math>

<math>~\frac{\partial\kappa_3}{\partial z}</math>

<math>~=</math>

<math>~ - \frac{\sin\kappa_2 \cos\kappa_2}{p^2z} \, . </math>

Hence,

<math>~h_2^{-2}</math>

<math>~=</math>

<math>~\sin^2\kappa_2 \cos^2\kappa_2 \biggl[ \frac{1}{x^2} + \frac{1}{q^4y^2} + \frac{1}{p^4z^2} \biggr]</math>

 

<math>~=</math>

<math>~ \sin^2\kappa_2 \cos^2\kappa_2 \biggl[ \frac{x^2 + q^4 y^2 + p^4 z^2}{x^2 q^4y^2 p^4 z^2} \biggr] = \biggl[ \frac{\sin^2\kappa_2 \cos^2\kappa_2 }{x^2 q^4y^2 p^4 z^2 \ell_{3D}^2} \biggr] </math>

<math>~\Rightarrow~~~ h_2</math>

<math>~=</math>

<math>~ \biggl[ \frac{x q^2y p^2 z \ell_{3D}}{\sin\kappa_2 \cos\kappa_2 } \biggr] \, . </math>

The three direction-cosines are, then,

<math>~\gamma_{21} = h_2 \biggl(\frac{\partial \kappa_2}{\partial x}\biggr)</math>

<math>~=</math>

<math>~ \biggl[ \frac{x q^2y p^2 z \ell_{3D}}{\sin\kappa_2 \cos\kappa_2 } \biggr] \frac{\sin\kappa_2 \cos\kappa_2}{x} = q^2y p^2z \ell_{3D} \, . </math>

Complete Orthogonality Check

If the set of unit vectors is indeed orthogonal, then we must find that,

<math>~\gamma_{mn}</math>

<math>~=</math>

<math>~M_{mn} \, ,</math>

where the quantity <math>~M_{mn}</math> is the minor of <math>~\gamma_{mn}</math> in the determinant, <math>~|\gamma_{mn}|</math>. (Note: This last expression is true only for right-handed coordinate systems. If the coordinate system is left-handed, we should find, <math>~\gamma_{mn} = - M_{mn}</math>.) More specifically, for any right-handed, orthogonal curvilinear coordinate system we should find:

<math>~\gamma_{11}</math>

<math>~=</math>

<math>~\gamma_{22}\gamma_{33} - \gamma_{23}\gamma_{32} \, ,</math>

<math>~\gamma_{12}</math>

<math>~=</math>

<math>~\gamma_{23}\gamma_{31} - \gamma_{21}\gamma_{33} \, ,</math>

<math>~\gamma_{13}</math>

<math>~=</math>

<math>~\gamma_{21}\gamma_{32} - \gamma_{22}\gamma_{31} \, ,</math>

   

<math>~\gamma_{21}</math>

<math>~=</math>

<math>~\gamma_{32}\gamma_{13} - \gamma_{33}\gamma_{12} \, ,</math>

<math>~\gamma_{22}</math>

<math>~=</math>

<math>~\gamma_{33}\gamma_{11} - \gamma_{31}\gamma_{13} \, ,</math>

<math>~\gamma_{23}</math>

<math>~=</math>

<math>~\gamma_{31}\gamma_{12} - \gamma_{32}\gamma_{11} \, ,</math>

   

<math>~\gamma_{31}</math>

<math>~=</math>

<math>~\gamma_{12}\gamma_{23} - \gamma_{13}\gamma_{22} \, ,</math>

<math>~\gamma_{32}</math>

<math>~=</math>

<math>~\gamma_{13}\gamma_{21} - \gamma_{11}\gamma_{23} \, ,</math>

<math>~\gamma_{33}</math>

<math>~=</math>

<math>~\gamma_{11}\gamma_{22} - \gamma_{12}\gamma_{21} \, .</math>

For (κ1, κ2, κ3)

<math>~\gamma_{22}\gamma_{33} - \gamma_{23}\gamma_{32}</math>

<math>~=</math>

<math>~ \biggl[ \frac{q^2y p^2z \ell_{3D}}{(x^2 + q^4y^2)^{1 / 2}} \biggr] \biggl[ 0 \biggr] + \biggl[ \frac{( x^2 + q^4y^2 ) \ell_{3D}}{(x^2 + q^4y^2)^{1 / 2}} \biggr] \biggl[ \frac{x}{(x^2 + q^4y^2)^{1 / 2}} \biggr] = x\ell_{3D} = \gamma_{11} \, . </math>       Yes!

<math>~\gamma_{23}\gamma_{31} - \gamma_{21}\gamma_{33}</math>

<math>~=</math>

<math>~ \bigg[ (x^2 + q^4y^2)^{1 / 2} \ell_{3D} \biggr] \biggl[ \frac{q^2y}{(x^2 + q^4y^2)^{1 / 2}} \biggr] - \biggl[ \frac{xp^2 z \ell_{3D}}{(x^2 + q^4y^2)^{1 / 2}} \biggr] \biggl[ 0 \biggr] = q^2 y \ell_{3D} = \gamma_{12} \, . </math>       Yes!

<math>~\gamma_{21}\gamma_{32} - \gamma_{22}\gamma_{31}</math>

<math>~=</math>

<math>~ \biggl[ \frac{xp^2 z \ell_{3D}}{(x^2 + q^4y^2)^{1 / 2}} \biggr] \biggl[ \frac{x}{(x^2 + q^4y^2)^{1 / 2}} \biggr] + \biggl[ \frac{q^2 y p^2 z \ell_{3D}}{(x^2 + q^4y^2)^{1 / 2}} \biggr] \biggl[ \frac{q^2y}{(x^2 + q^4y^2)^{1 / 2}} \biggr] </math>

 

<math>~=</math>

<math>~ \biggl[ \frac{x^2p^2 z \ell_{3D}}{(x^2 + q^4y^2)} \biggr] + \biggl[ \frac{q^4 y^2 p^2 z \ell_{3D}}{(x^2 + q^4y^2)} \biggr] = p^2 z \ell_{3D} = \gamma_{13} \, . </math>       Yes!

<math>~\gamma_{32}\gamma_{13} - \gamma_{33}\gamma_{12}</math>

<math>~=</math>

<math>~ \biggl[ \frac{x}{(x^2 + q^4y^2)^{1 / 2}} \biggr] \biggl[ p^2z \ell_{3D} \biggr] - \biggl[ 0\biggr] \biggl[ q^2y \ell_{3D} \biggr] = \biggl[ \frac{xp^2 z\ell_{3D}}{(x^2 + q^4y^2)^{1 / 2}} \biggr] = \gamma_{21} \, . </math>       Yes!

<math>~\gamma_{33}\gamma_{11} - \gamma_{31}\gamma_{13}</math>

<math>~=</math>

<math>~ \biggl[ 0\biggr] \biggl[ x\ell_{3D} \biggr] + \biggl[ \frac{q^2y}{(x^2 + q^4y^2)^{1 / 2}} \biggr] \biggl[ p^2z \ell_{3D} \biggr] = \biggl[ \frac{q^2y p^2z \ell_{3D}}{(x^2 + q^4y^2)^{1 / 2}} \biggr] = \gamma_{22} \, . </math>       Yes!

<math>~\gamma_{31}\gamma_{12} - \gamma_{32}\gamma_{11}</math>

<math>~=</math>

<math>~ - \biggl[ \frac{q^2y}{(x^2 + q^4y^2)^{1 / 2}} \biggr] \biggl[ q^2y \ell_{3D} \biggr] - \biggl[ \frac{x}{(x^2 + q^4y^2)^{1 / 2}} \biggr] \biggl[ x\ell_{3D} \biggr] = -(x^2 + q^4y^2)^{1 / 2}\ell_{3D} = \gamma_{23} \, . </math>       Yes!

See Also


Whitworth's (1981) Isothermal Free-Energy Surface

© 2014 - 2021 by Joel E. Tohline
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