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   <td align="left">
   <td align="left">
<math>~
<math>~
-\frac{z \eta^{-1}}{(1+\eta^2)}  \biggl[ - \frac{q^2 y}{x^2} \biggr]
-\frac{\eta^{-1}}{(1+\eta^2)}  \biggl[ - \frac{q^2 y}{x^2} \biggr]
=
=
\frac{x}{(x^2 + q^4y^2)}  \, .
\frac{x}{(x^2 + q^4y^2)}  \, ;
</math>
  </td>
</tr>
</table>
and,
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\frac{\partial\lambda_3}{\partial y} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-\frac{\eta^{-1}}{(1+\eta^2)}  \biggl[ \frac{q^2 }{x} \biggr]
=
-\frac{x^2}{y(x^2+ q^4y^2)} 
</math>
</math>
   </td>
   </td>

Revision as of 22:35, 25 January 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>

See Also


Whitworth's (1981) Isothermal Free-Energy Surface

© 2014 - 2021 by Joel E. Tohline
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