Difference between revisions of "User:Tohline/Appendix/Ramblings/ConcentricEllipsodalT12Coordinates"

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Line 764: Line 764:
-  
-  
b^2p^4z^4  
b^2p^4z^4  
\biggr] \, .
\biggr] \, ;
</math>
  </td>
</tr>
</table>
 
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~ \biggl[ \frac{2\mathfrak{A}\mathfrak{B}}{\lambda_3} \biggr] \frac{\partial \lambda_3}{\partial y}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\mathfrak{B} \biggl[ 2q^4y \biggr]
-
\mathfrak{A}\biggl[ 2y(a^2 z^2 + c^2x^2) \biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2y \biggl[
q^4[a^2(yz)^2 + b^2(xz)^2 + c^2(xy)^2]
-
(a^2 z^2 + c^2x^2) (x^2 + q^4y^2 + p^4z^2)
\biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2y \biggl[
a^2q^4(yz)^2 + b^2q^4(xz)^2 + c^2q^4(xy)^2
-
a^2 z^2 (x^2 + q^4y^2 + p^4z^2)
-
c^2x^2 (x^2 + q^4y^2 + p^4z^2)
\biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2y \biggl\{
(yz)^2 \biggl[  a^2q^4 - a^2q^4\biggr] + (xz)^2\biggl[ b^2q^4 -a^2 - c^2p^4\biggr] + (xy)^2 \biggl[ c^2q^4 - c^2q^4 \biggr]
-
a^2 p^4z^4
-
c^2x^4
\biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2y \biggl[
x^2y^2 ( b^2q^4 -a^2 - c^2p^4 )
-
a^2 p^4z^4
-
c^2x^4
\biggr] \, ;
</math>
  </td>
</tr>
</table>
 
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~ \biggl[ \frac{2\mathfrak{A}\mathfrak{B}}{\lambda_3} \biggr] \frac{\partial \lambda_3}{\partial z}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\mathfrak{B} \biggl[ 2p^4z \biggr]
-
\mathfrak{A}\biggl[ 2z(a^2 y^2 + b^2x^2)\biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2z \biggl[
p^4[a^2(yz)^2 + b^2(xz)^2 + c^2(xy)^2]
-
(a^2 y^2 + b^2x^2)(x^2 + q^4y^2 + p^4z^2)
\biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2z \biggl[
a^2p^4(yz)^2 + b^2p^4(xz)^2 + c^2p^4(xy)^2
-
a^2 y^2 (x^2 + q^4y^2 + p^4z^2)
-
b^2x^2(x^2 + q^4y^2 + p^4z^2)
\biggr]
</math>
</math>
   </td>
   </td>

Revision as of 23:20, 8 March 2021

Concentric Ellipsoidal (T12) 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.

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

<math>~\lambda_1</math>

<math>~\equiv</math>

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

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

<math>~=</math>

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

where,

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

<math>~\equiv</math>

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

Generalized Prescription for 2nd Coordinate

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

<math>~\lambda_2</math>

<math>~\equiv</math>

<math>~ x^a y^b z^c \, , </math>

in which case,

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

<math>~=</math>

<math>~ \frac{1}{\mathfrak{L}} \biggl[ \hat\imath \biggl(\frac{yz}{bc}\biggr) + \hat\jmath \biggl(\frac{xz}{ac}\biggr) + \hat{k} \biggl(\frac{xy}{ab}\biggr) \biggr] \, , </math>

where,

<math>~\mathfrak{L}^2</math>

<math>~\equiv</math>

<math>~ \frac{1}{a^2b^2c^2} \biggl[ a^2(yz)^2 + b^2(xz)^2 + c^2(xy)^2 \biggr] \, . </math>

Now, to ensure that <math>~\hat{e}_2</math> is perpendicular to <math>~\hat{e}_1</math>, we need,

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

<math>~=</math>

<math>~0</math>

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

<math>~=</math>

<math>~ \frac{\ell_{3D}}{\mathfrak{L}} \biggl[ \frac{xyz}{bc} + \frac{q^2xyz}{ac} + \frac{p^2xyz}{ab} \biggr] = \frac{\ell_{3D} (xyz)}{\mathfrak{L}(abc)} \biggl[ a + q^2b + p^2 c \biggr] </math>

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

<math>~=</math>

<math>~ \biggl[ a + q^2b + p^2 c \biggr]\, . </math>

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

Necessary 3rd Coordinate

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

<math>~\hat{e}_3</math>

<math>~=</math>

<math>~\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>~\frac{\ell_{3D}}{\mathfrak{L}} \biggl\{ \hat\imath \biggl[ (q^2y) \biggl( \frac{xy}{ab} \biggr) - (p^2z) \biggl( \frac{xz}{ac} \biggr) \biggr] + \hat\jmath \biggl[ (p^2z) \biggl( \frac{yz}{bc} \biggr) - (x)\biggl( \frac{xy}{ab} \biggr) \biggr] + \hat{k} \biggl[ (x) \biggl( \frac{xz}{ac} \biggr) - (q^2y) \biggl( \frac{yz}{bc} \biggr) \biggr] \biggr\} </math>

 

<math>~=</math>

<math>~\frac{\ell_{3D}}{\mathfrak{L}(abc)} \biggl\{ \hat\imath \biggl[ (cq^2y^2) - (b p^2z^2) \biggr]x + \hat\jmath \biggl[ (ap^2z^2) - (cx^2) \biggr]y + \hat{k} \biggl[ (bx^2) - (aq^2y^2) \biggr]z \biggr\} </math>

Old Examples

T6 Coordinates

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

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

<math>~=</math>

<math>~ \frac{1}{\mathfrak{L}_{T6} (abc)} \biggl[ \hat\imath (ayz) + \hat\jmath (bxz) + \hat{k} (cxy) \biggr] = \frac{z}{q^2 \mathfrak{L}_{T6} (abc)} \biggl[ -\hat\imath (q^2y) + \hat\jmath (x) \biggr] </math>

 

<math>~=</math>

<math>~ \biggl[ -\hat\imath (q^2y) + \hat\jmath (x) \biggr] \ell_q \, , </math>

where,

<math>~\ell_q^{-2} \equiv \biggl[ \frac{q^4\mathfrak{L}_{T6}^2(abc)^2}{z^2}\biggr]</math>

<math>~=</math>

<math>~ \frac{q^4}{z^2}\biggl[ (yz)^2 + b^2(xz)^2 \biggr] = \biggl[ x^2 + q^4y^2 \biggr] \, . </math>

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

<math>~\hat{e}_3</math>

<math>~=</math>

<math>~\frac{\ell_{3D}}{\mathfrak{L}_{T6} (abc)} \biggl\{ \hat\imath \biggl[ (cq^2y^2) - (b p^2z^2) \biggr]x + \hat\jmath \biggl[ (ap^2z^2) - (cx^2) \biggr]y + \hat{k} \biggl[ (bx^2) - (aq^2y^2) \biggr]z \biggr\} </math>

 

<math>~=</math>

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

 

<math>~=</math>

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

T10 Coordinates

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

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

<math>~=</math>

<math>~ \frac{1}{\mathfrak{L}_{T10} (abc)} \biggl[ \hat\imath (ayz) + \hat\jmath (bxz) + \hat{k} (cxy) \biggr] = \frac{1}{q^2 p^2 \mathfrak{L}_{T10} (abc)} \biggl[ \hat\imath (q^2y p^2z) + \hat\jmath ( x p^2 z ) - \hat{k} ( 2xq^2y) \biggr] </math>

where,

<math>~(abc)^2\mathfrak{L}^2_{T10}</math>

<math>~\equiv</math>

<math>~ \biggl[ a^2(yz)^2 + b^2(xz)^2 + c^2(xy)^2 \biggr] =

\biggl[  

y^2z^2 + \frac{x^2 z^2}{q^4} + \frac{4x^2 y^2}{p^4} \biggr] </math>

<math>~\Rightarrow~~~\mathcal{D}^2 \equiv q^4p^4(abc)^2\mathfrak{L}^2_{T10}</math>

<math>~=</math>

<math>~

\biggl[  

q^4y^2 p^4z^2 + x^2 p^4z^2 + 4x^2 q^4y^2 \biggr] \, . </math>


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

<math>~\hat{e}_3</math>

<math>~=</math>

<math>~\frac{\ell_{3D}}{\mathcal{D}} \biggl\{ \hat\imath \biggl[ (cq^2y^2) - (b p^2z^2) \biggr]x + \hat\jmath \biggl[ (ap^2z^2) - (cx^2) \biggr]y + \hat{k} \biggl[ (bx^2) - (aq^2y^2) \biggr]z \biggr\} q^2 p^2 </math>

 

<math>~=</math>

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

Develop 3rd-Coordinate Profile

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

<math>~\mathfrak{A} \equiv \ell_{3D}^{-2}</math>

<math>~=</math>

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

<math>~\mathfrak{B} \equiv [\mathfrak{L}(abc)]^2</math>

<math>~=</math>

<math>~ [a^2(yz)^2 + b^2(xz)^2 + c^2(xy)^2] \, . </math>


<math>~\frac{\partial \mathfrak{A}}{\partial x}</math>

<math>~=</math>

<math>~2x \, ,</math>

<math>~\frac{\partial \mathfrak{A}}{\partial y}</math>

<math>~=</math>

<math>~2q^4 y \, ,</math>

<math>~\frac{\partial \mathfrak{A}}{\partial z}</math>

<math>~=</math>

<math>~2p^4 z \, ;</math>

<math>~\frac{\partial \mathfrak{B}}{\partial x}</math>

<math>~=</math>

<math>~2x(b^2z^2 + c^2y^2) \, ,</math>

<math>~\frac{\partial \mathfrak{B}}{\partial y}</math>

<math>~=</math>

<math>~2y(a^2 z^2 + c^2x^2) \, ,</math>

<math>~\frac{\partial \mathfrak{B}}{\partial z}</math>

<math>~=</math>

<math>~2z(a^2 y^2 + b^2x^2) \, .</math>

Then the 3rd unit vector may be written as,

<math>~\hat{e}_3</math>

<math>~=</math>

<math>~\mathfrak{A}^{-1 / 2} \mathfrak{B}^{-1 / 2} \biggl\{ \hat\imath \biggl[ (cq^2y^2) - (b p^2z^2) \biggr]x + \hat\jmath \biggl[ (ap^2z^2) - (cx^2) \biggr]y + \hat{k} \biggl[ (bx^2) - (aq^2y^2) \biggr]z \biggr\} \, . </math>

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

<math>~\lambda_3</math>

<math>~\equiv</math>

<math>~\mathfrak{A}^{1 / 2} \mathfrak{B}^{-1 / 2} \, .</math>

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

<math>~=</math>

<math>~ \biggl[ \frac{1}{2} \mathfrak{A}^{-1 / 2} \mathfrak{B}^{-1 / 2}\biggr] \frac{\partial \mathfrak{A}}{\partial x_i} - \biggl[ \frac{1}{2} \mathfrak{A}^{1 / 2} \mathfrak{B}^{- 3 / 2} \biggr] \frac{\partial \mathfrak{B}}{\partial x_i} </math>

<math>~\Rightarrow~~~ \biggl[ \frac{2\mathfrak{A}\mathfrak{B}}{\lambda_3} \biggr] \frac{\partial \lambda_3}{\partial x_i}</math>

<math>~=</math>

<math>~ \mathfrak{B}\cdot \frac{\partial \mathfrak{A}}{\partial x_i} - \mathfrak{A}\cdot \frac{\partial \mathfrak{B}}{\partial x_i} \, . </math>

First, note that,

<math>~ \biggl[ \frac{2\mathfrak{A}\mathfrak{B}}{\lambda_3} \biggr] \frac{\partial \lambda_3}{\partial x}</math>

<math>~=</math>

<math>~ \mathfrak{B} \biggl[ 2x \biggr] - \mathfrak{A}\biggl[ 2x (b^2z^2 + c^2 y^2)\biggr] </math>

 

<math>~=</math>

<math>~2x \biggl[ a^2(yz)^2 + b^2(xz)^2 + c^2(xy)^2 - (x^2 + q^4y^2 + p^4z^2)(b^2z^2 + c^2 y^2) \biggr] </math>

 

<math>~=</math>

<math>~2x \biggl[ a^2(yz)^2 + b^2(xz)^2 + c^2(xy)^2 - x^2 (b^2z^2 + c^2 y^2) - q^4y^2 (b^2z^2 + c^2 y^2) - p^4z^2(b^2z^2 + c^2 y^2)

\biggr] </math>

 

<math>~=</math>

<math>~2x \biggl\{ (yz)^2[a^2 - q^4b^2 - c^2p^4] + (xz)^2[b^2 - b^2] + (xy)^2 [c^2 - c^2] - c^2 q^4y^4 - b^2p^4z^4 \biggr\} </math>

 

<math>~=</math>

<math>~2x \biggl[ y^2z^2(a^2 - q^4b^2 - c^2p^4) - c^2 q^4y^4 - b^2p^4z^4 \biggr] \, ; </math>

<math>~ \biggl[ \frac{2\mathfrak{A}\mathfrak{B}}{\lambda_3} \biggr] \frac{\partial \lambda_3}{\partial y}</math>

<math>~=</math>

<math>~ \mathfrak{B} \biggl[ 2q^4y \biggr] - \mathfrak{A}\biggl[ 2y(a^2 z^2 + c^2x^2) \biggr] </math>

 

<math>~=</math>

<math>~2y \biggl[ q^4[a^2(yz)^2 + b^2(xz)^2 + c^2(xy)^2] - (a^2 z^2 + c^2x^2) (x^2 + q^4y^2 + p^4z^2) \biggr] </math>

 

<math>~=</math>

<math>~2y \biggl[ a^2q^4(yz)^2 + b^2q^4(xz)^2 + c^2q^4(xy)^2 - a^2 z^2 (x^2 + q^4y^2 + p^4z^2) - c^2x^2 (x^2 + q^4y^2 + p^4z^2) \biggr] </math>

 

<math>~=</math>

<math>~2y \biggl\{ (yz)^2 \biggl[ a^2q^4 - a^2q^4\biggr] + (xz)^2\biggl[ b^2q^4 -a^2 - c^2p^4\biggr] + (xy)^2 \biggl[ c^2q^4 - c^2q^4 \biggr] - a^2 p^4z^4 - c^2x^4 \biggr\} </math>

 

<math>~=</math>

<math>~2y \biggl[ x^2y^2 ( b^2q^4 -a^2 - c^2p^4 ) - a^2 p^4z^4 - c^2x^4 \biggr] \, ; </math>

<math>~ \biggl[ \frac{2\mathfrak{A}\mathfrak{B}}{\lambda_3} \biggr] \frac{\partial \lambda_3}{\partial z}</math>

<math>~=</math>

<math>~ \mathfrak{B} \biggl[ 2p^4z \biggr] - \mathfrak{A}\biggl[ 2z(a^2 y^2 + b^2x^2)\biggr] </math>

 

<math>~=</math>

<math>~2z \biggl[ p^4[a^2(yz)^2 + b^2(xz)^2 + c^2(xy)^2] - (a^2 y^2 + b^2x^2)(x^2 + q^4y^2 + p^4z^2) \biggr] </math>

 

<math>~=</math>

<math>~2z \biggl[ a^2p^4(yz)^2 + b^2p^4(xz)^2 + c^2p^4(xy)^2 - a^2 y^2 (x^2 + q^4y^2 + p^4z^2) - b^2x^2(x^2 + q^4y^2 + p^4z^2) \biggr] </math>

See Also


Whitworth's (1981) Isothermal Free-Energy Surface

© 2014 - 2021 by Joel E. Tohline
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