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

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   <td align="left">
   <td align="left">
<math>~
<math>~
q^2y^2 + \frac{y}{\lambda_3^2} - \lambda_1^2 \lambda_2^2
2y^2 + \frac{y}{\lambda_3^2} - \lambda_1^2 \lambda_2^2
</math>
</math>
   </td>
   </td>
Line 333: Line 333:
   <td align="left">
   <td align="left">
<math>~
<math>~
\frac{1}{2q^2} \biggl\{
\frac{1}{4} \biggl\{
-\frac{1}{\lambda_3^2} \pm \biggl[ \frac{1}{\lambda_3^4} + 4q^2 \lambda_1^2 \lambda_2^2 \biggr]^{1 / 2}
-\frac{1}{\lambda_3^2} \pm \biggl[ \frac{1}{\lambda_3^4} + 8 \lambda_1^2 \lambda_2^2 \biggr]^{1 / 2}
\biggr\}
\biggr\}
</math>
</math>
Line 349: Line 349:
   <td align="left">
   <td align="left">
<math>~
<math>~
\frac{1}{2q^2\lambda_3^2} \biggl\{
\frac{1}{4\lambda_3^2} \biggl\{
-1 \pm \biggl[ 1 + 4q^2 \lambda_1^2 \lambda_2^2 \lambda_3^4\biggr]^{1 / 2}
-1 \pm \biggl[ 1 + 8 \lambda_1^2 \lambda_2^2 \lambda_3^4\biggr]^{1 / 2}
\biggr\} \, .
\biggr\} \, .
</math>
</math>

Revision as of 23:06, 17 March 2021

Daring Attack

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 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
<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>~\ell_q \ell_{3D} (xp^2z)</math> <math>~\ell_q \ell_{3D} (q^2 y p^2z) </math> <math>~- (x^2 + q^4y^2)\ell_q \ell_{3D}</math>
<math>~3</math> <math>~\tan^{-1}\biggl( \frac{y^{1/q^2}}{x} \biggr)</math> <math>~\frac{xq^2 y \ell_q}{\sin\lambda_3 \cos\lambda_3}</math> <math>~-\frac{\sin\lambda_3 \cos\lambda_3}{x}</math> <math>~+\frac{\sin\lambda_3 \cos\lambda_3}{q^2y}</math> <math>~0</math> <math>~-q^2 y \ell_q</math> <math>~x\ell_q</math> <math>~0</math>

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

<math>~\equiv</math>

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

<math>~\ell_q</math>

<math>~\equiv</math>

<math>~[x^2 + q^4 y^2 ]^{- 1/ 2 }</math>

As before, let's adopt the first-coordinate expression,

<math>~\lambda_1</math>

<math>~\equiv</math>

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

but for the third-coordinate expression we will abandon the trigonometric expression and instead simply use,

<math>~\lambda_3</math>

<math>~\equiv</math>

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

This modified third-coordinate expression means that the last row of the above table changes, as follows.

Daring Attack
<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>~\ell_q \ell_{3D} (xp^2z)</math> <math>~\ell_q \ell_{3D} (q^2 y p^2z) </math> <math>~- (x^2 + q^4y^2)\ell_q \ell_{3D}</math>
<math>~3</math> <math>~\frac{y^{1/q^2}}{x} </math> <math>~\frac{xq^2 y \ell_q}{\lambda_3}</math> <math>~-\frac{\lambda_3}{x}</math> <math>~+\frac{\lambda_3}{q^2y}</math> <math>~0</math> <math>~-q^2 y \ell_q</math> <math>~x\ell_q</math> <math>~0</math>

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 <math>~\lambda_1</math> and <math>~\lambda_3</math> 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,

<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>

This is identical to our expression for <math>~\lambda_1</math> if we make the associations,

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

     

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

     

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

Now, given that <math>~\lambda_3</math> does not functionally depend on <math>~z</math>, let's consider that the choice of <math>~z</math> is tightly associated with the specification of the second coordinate, <math>~\lambda_2</math>. Specifically, let's adopt the definition,

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

<math>~\equiv</math>

<math>~1 - \biggl( \frac{z}{c}\biggr)^2 \, ,</math>

in which case, we see that,

<math>~\biggl( \frac{x}{a}\biggr)^2 + \biggl( \frac{y}{b}\biggr)^2 = \lambda_2^2 \, ,</math>

      and,      

<math>~z^2 = c^2 (1 - \lambda_2^2) = \frac{\lambda_1^2 (1 - \lambda_2^2)}{p^2} \, .</math>

<math>~z^2</math>

<math>~=</math>

<math>~c^2(1-\lambda_2^2) = \frac{\lambda_1^2(1-\lambda_2^2)}{p^2} \, ,</math>

and,

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

<math>~=</math>

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

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

<math>~=</math>

<math>~ \lambda_1^2 \lambda_2^2 \, .</math>

Combining this last expression with the <math>~x - y</math> relationship that is provided by the definition of <math>~\lambda_3</math>, gives,

<math>~\lambda_1^2 \lambda_2^2</math>

<math>~=</math>

<math>~\frac{y^{2/q^2}}{\lambda_3^2} + q^2y^2 \, .</math>

In general, this expression does not give us an analytical expression for <math>~y(\lambda_1, \lambda_2, \lambda_3)</math>. But a solution is obtainable for selected values of <math>~q^2 > 1</math>.

Examine the Case: q2 = 2

If we set <math>~q^2 = 2</math>, then this last combined expression becomes a quadratic equation for <math>~y</math>. Specifically, we find,

<math>~ 0</math>

<math>~=</math>

<math>~ 2y^2 + \frac{y}{\lambda_3^2} - \lambda_1^2 \lambda_2^2 </math>

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

<math>~=</math>

<math>~ \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\} </math>

 

<math>~=</math>

<math>~ \frac{1}{4\lambda_3^2} \biggl\{ -1 \pm \biggl[ 1 + 8 \lambda_1^2 \lambda_2^2 \lambda_3^4\biggr]^{1 / 2} \biggr\} \, . </math>

See Also


Whitworth's (1981) Isothermal Free-Energy Surface

© 2014 - 2021 by Joel E. Tohline
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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