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

From VistrailsWiki
Jump to navigation Jump to search
Line 399: Line 399:
\biggr]  
\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>
</math>
   </td>
   </td>

Revision as of 20:03, 8 March 2021

Concentric Ellipsoidal (T12) Coordinates

Whitworth's (1981) Isothermal Free-Energy Surface
|   Tiled Menu   |   Tables of Content   |  Banner Video   |  Tohline Home Page   |

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>

See Also


Whitworth's (1981) Isothermal Free-Energy Surface

© 2014 - 2021 by Joel E. Tohline
|   H_Book Home   |   YouTube   |
Appendices: | Equations | Variables | References | Ramblings | Images | myphys.lsu | ADS |
Recommended citation:   Tohline, Joel E. (2021), The Structure, Stability, & Dynamics of Self-Gravitating Fluids, a (MediaWiki-based) Vistrails.org publication, https://www.vistrails.org/index.php/User:Tohline/citation