User:Tohline/Appendix/Ramblings/ConcentricEllipsodalDaringAttack
Daring Attack
|
|---|
| | 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 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> | ||||||
|
||||||||||||||
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.
|
Test Example |
|
<math>~q^2 = 2, p^2=3.15, (x, y, z) = (0.4, 0.63581, 0.1)</math>
<math>~(\lambda_1, \lambda_2, \lambda_3) = (1, 0.98412, 1.99344)</math> |
| <math>~\Lambda^2-1 = 122.34879 ~~~\Rightarrow ~~~ \Lambda = 11.10625</math> |
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>~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> |
[Note that in the case of spherical coordinates (q2 = p2 = 1), <math>~\lambda_1 \rightarrow r</math>, and this "second" coordinate, <math>~\lambda_2</math>, becomes <math>~\sin\theta</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, the exponent of <math>~2q^{-2}</math> that appears in the first term on the right-hand side of this expression prevents us from being able to analytically prescribe the function, <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\{ \biggl[ 1 + 8 \lambda_1^2 \lambda_2^2 \lambda_3^4\biggr]^{1 / 2} - 1 \biggr\} \, . </math> |
(Note that, for reasons of simplicity for the time being, in this last expression we have retained only the "positive" solution.) Again, calling upon the <math>~x - y</math> relationship that is provided through the definition of <math>~\lambda_3</math>, we find (when q2 = 2),
|
<math>~x^2</math> |
<math>~=</math> |
<math>~\frac{y}{\lambda_3^2}</math> |
|
|
<math>~=</math> |
<math>~ \frac{1}{4\lambda_3^4} \biggl\{ \biggl[ 1 + 8 \lambda_1^2 \lambda_2^2 \lambda_3^4\biggr]^{1 / 2} - 1 \biggr\} </math> |
|
<math>~\Rightarrow ~~~ x</math> |
<math>~=</math> |
<math>~\pm \frac{1}{2\lambda_3^{2}} \biggl\{ \biggl[ 1 + 8 \lambda_1^2 \lambda_2^2 \lambda_3^4\biggr]^{1 / 2} - 1 \biggr\}^{1 / 2} \, . </math> |
|
Summary (q2 = 2)
|
For convenience, we have defined,
|
<math>~\Lambda^2</math> |
<math>~\equiv</math> |
<math>~1 + 8\lambda_1^2 \lambda_2^2 \lambda_3^4 </math> |
|
<math>~\Rightarrow ~~~ \lambda_1^2 \lambda_2^2 \lambda_3^4 </math> |
<math>~\equiv</math> |
<math>~\frac{1}{8}\biggl( \Lambda^2 - 1\biggr) \, .</math> |
Next, let's examine all nine partial derivatives, noting at the start that,
|
<math>~\Rightarrow~~~ \frac{\partial\Lambda}{\partial \lambda_1} </math> |
<math>~=</math> |
<math>~ \frac{1}{2\Lambda}\biggl[16\lambda_1 \lambda_2^2 \lambda_3^4 \biggr] = \frac{(\Lambda^2-1)}{\lambda_1 \Lambda} \, , </math> |
|
<math>~\frac{\partial\Lambda}{\partial \lambda_2} </math> |
<math>~=</math> |
<math>~ \frac{1}{2\Lambda}\biggl[16\lambda_1^2 \lambda_2 \lambda_3^4 \biggr] = \frac{(\Lambda^2-1)}{\lambda_2 \Lambda} \, , </math> |
|
<math>~\frac{\partial\Lambda}{\partial \lambda_3} </math> |
<math>~=</math> |
<math>~ \frac{1}{2\Lambda}\biggl[32\lambda_1^2 \lambda_2^2 \lambda_3^3 \biggr] = \frac{2(\Lambda^2-1)}{\lambda_3 \Lambda} \, . </math> |
We have,
|
<math>~\frac{\partial z}{\partial \lambda_1}</math> |
<math>~=</math> |
<math>~ \frac{(1-\lambda_2^2)^{1 / 2}}{p} \, , </math> |
|
<math>~\frac{\partial z}{\partial \lambda_2}</math> |
<math>~=</math> |
<math>~ -\frac{\lambda_1 \lambda_2}{p(1 - \lambda_2^2)^{1 / 2}} \, , </math> |
|
<math>~\frac{\partial z}{\partial \lambda_3}</math> |
<math>~=</math> |
<math>~ 0 \, . </math> |
|
<math>~\frac{\partial y}{\partial \lambda_1}</math> |
<math>~=</math> |
<math>~ \frac{1}{4\lambda_3^2} \cdot \frac{\partial \Lambda}{\partial \lambda_1} = \biggl[ \frac{(\Lambda^2-1)}{4\lambda_3^2\lambda_1 \Lambda} \biggr] = \biggl[ \frac{8\lambda_1^2 \lambda_2^2 \lambda_3^4}{4\lambda_3^2\lambda_1 \Lambda} \biggr] = \biggl[ \frac{2\lambda_1 \lambda_2^2 \lambda_3^2}{\Lambda} \biggr] \, , </math> |
|
<math>~\frac{\partial y}{\partial \lambda_2}</math> |
<math>~=</math> |
<math>~ \frac{1}{4\lambda_3^2} \cdot \frac{\partial \Lambda}{\partial \lambda_2} = \biggl[ \frac{(\Lambda^2-1)}{4\lambda_3^2\lambda_2 \Lambda} \biggr] = \biggl[ \frac{8\lambda_1^2 \lambda_2^2 \lambda_3^4}{4\lambda_3^2\lambda_2 \Lambda} \biggr] = \biggl[ \frac{2\lambda_1^2 \lambda_2 \lambda_3^2}{\Lambda} \biggr] \, , </math> |
|
<math>~\frac{\partial y}{\partial \lambda_3}</math> |
<math>~=</math> |
<math>~ \frac{1}{4\lambda_3^2} \cdot \frac{\partial \Lambda}{\partial \lambda_3} - \frac{(\Lambda - 1)}{2\lambda_3^3} = \frac{1}{4\lambda_3^2} \cdot \biggl[ \frac{2(\Lambda^2 - 1)}{\lambda_3\Lambda} \biggr] - \frac{\Lambda(\Lambda - 1)}{2\lambda_3^3 \Lambda} = \biggl[ \frac{(\Lambda^2 - 1) - \Lambda(\Lambda-1)}{2\lambda_3^3\Lambda} \biggr] = \frac{(\Lambda - 1) }{2\lambda_3^3\Lambda} \, . </math> |
|
<math>~\frac{\partial x}{\partial \lambda_1}</math> |
<math>~=</math> |
<math>~ \frac{1}{4\lambda_3^{2}(\Lambda-1)^{1 / 2}} \cdot \frac{\partial \Lambda}{\partial \lambda_1} = \frac{1}{4\lambda_3^{2}(\Lambda-1)^{1 / 2}} \biggl[ \frac{(\Lambda^2 - 1)}{\lambda_1 \Lambda} \biggr] = \frac{(\Lambda^2 - 1)}{4 \lambda_1 \lambda_3^{2} \Lambda (\Lambda-1)^{1 / 2}} = \frac{2\lambda_1 \lambda_2^2 \lambda_3^{2}}{\Lambda (\Lambda-1)^{1 / 2}} \, , </math> |
|
<math>~\frac{\partial x}{\partial \lambda_2}</math> |
<math>~=</math> |
<math>~ \frac{1}{4\lambda_3^{2}(\Lambda-1)^{1 / 2}} \cdot \frac{\partial \Lambda}{\partial \lambda_2} = \frac{1}{4\lambda_3^{2}(\Lambda-1)^{1 / 2}} \biggl[ \frac{(\Lambda^2 - 1)}{\lambda_2 \Lambda} \biggr] = \frac{(\Lambda^2 - 1)}{4 \lambda_2 \lambda_3^{2} \Lambda (\Lambda-1)^{1 / 2}} = \frac{2\lambda_1^2 \lambda_2 \lambda_3^{2}}{\Lambda (\Lambda-1)^{1 / 2}} \, , </math> |
|
<math>~\frac{\partial x}{\partial \lambda_3}</math> |
<math>~=</math> |
<math>~ \frac{1}{4\lambda_3^{2}(\Lambda-1)^{1 / 2}} \cdot \frac{\partial \Lambda}{\partial \lambda_3} - \frac{(\Lambda-1)^{1 / 2}}{\lambda_3^{3}} = \frac{1}{4\lambda_3^{2}(\Lambda-1)^{1 / 2}} \biggl[ \frac{2(\Lambda^2 - 1)}{\lambda_3 \Lambda} \biggr] - \frac{(\Lambda-1)^{1 / 2}}{\lambda_3^{3}} </math> |
|
|
<math>~=</math> |
<math>~ \frac{(\Lambda^2 - 1) -2\Lambda (\Lambda - 1) }{2\lambda_3^{3} \Lambda (\Lambda-1)^{1 / 2}} = - \frac{ (\Lambda - 1)^{3 / 2} }{2\lambda_3^{3} \Lambda} \, . </math> |
What about the derived scale-factors?
|
<math>~h_1^2</math> |
<math>~=</math> |
<math>~ \biggl(\frac{\partial x}{\partial \lambda_1}\biggr)^2 + \biggl(\frac{\partial y}{\partial \lambda_1}\biggr)^2 + \biggl(\frac{\partial z}{\partial \lambda_1}\biggr)^2 </math> |
|
|
<math>~=</math> |
<math>~ \biggl[ \frac{2\lambda_1 \lambda_2^2 \lambda_3^{2}}{\Lambda (\Lambda-1)^{1 / 2}} \biggr]^2 + \biggl[ \frac{2\lambda_1 \lambda_2^2 \lambda_3^2}{\Lambda} \biggr]^2 + \biggl[ \frac{(1-\lambda_2^2)^{1 / 2}}{p} \biggr]^2 </math> |
|
|
<math>~=</math> |
<math>~ \biggl[ \frac{4\lambda_1^2 \lambda_2^4 \lambda_3^{4 }}{\Lambda^2 (\Lambda-1)} \biggr] + \biggl[ \frac{4\lambda_1^2 \lambda_2^4 \lambda_3^4}{\Lambda^2} \biggr] + \biggl[ \frac{(1-\lambda_2^2)}{p^2} \biggr] </math> |
|
|
<math>~=</math> |
<math>~\frac{1}{p^2 \Lambda^2(\Lambda - 1)} \biggl[ 4 p^2 \lambda_1^2 \lambda_2^4 \lambda_3^4 \Lambda + (1-\lambda_2^2) \Lambda^2(\Lambda - 1) \biggr] \, . </math> |
Written in terms of Cartesian coordinates, this becomes,
|
<math>~h_1^2</math> |
<math>~=</math> |
<math>~\frac{ 8 \lambda_1^2 \lambda_2^2 \lambda_3^4 (\lambda_2^2 ) }{2 \Lambda (\Lambda - 1)} + \frac{(1-\lambda_2^2)}{p^2} </math> |
|
|
<math>~=</math> |
<math>~\frac{ (\Lambda+1)\lambda_2^2 }{2 \Lambda } + \frac{z^2}{\lambda_1^2} </math> |
|
|
<math>~=</math> |
<math>~ \frac{1}{\lambda_1^2} \biggl[ \frac{ (\Lambda+1)\lambda_2^2 \lambda_1^2 }{2 \Lambda } + z^2 \biggr] </math> |
|
|
<math>~=</math> |
<math>~ \frac{1}{\lambda_1^2} \biggl[ \frac{ (\Lambda+1)(x^2 + 2y^2) }{2 \Lambda } + z^2 \biggr] \, . </math> |
Note that,
|
<math>~\Lambda -1</math> |
<math>~=</math> |
<math>~4x^2\lambda_3^4 = 4x^2 \biggl( \frac{y^2}{x^4} \biggr) = 4\biggl( \frac{y^2}{x^2} \biggr) </math> |
|
<math>~\Rightarrow ~~~ \Lambda </math> |
<math>~=</math> |
<math>~\frac{x^2 + 4y^2}{x^2} \, .</math> |
Hence, the scale factor becomes,
|
<math>~h_1^2</math> |
<math>~=</math> |
<math>~ \frac{1}{2 \lambda_1^2} \biggl[ (x^2 + 2y^2) + \frac{ x^2(x^2 + 2y^2) }{(x^2 + 4y^2) } + 2z^2 \biggr] </math> |
|
|
<math>~=</math> |
<math>~ \frac{1}{2 \lambda_1^2(x^2 + 4y^2) } \biggl[ (x^2 + 2y^2) (x^2 + 4y^2) + x^2(x^2 + 2y^2) + 2z^2(x^2 + 4y^2) \biggr] </math> |
|
|
<math>~=</math> |
<math>~ \frac{1}{2 \lambda_1^2(x^2 + 4y^2) } \biggl[ (2x^4 + 8x^2y^2 +8y^4) + 2z^2(x^2 + 4y^2) \biggr] \, . </math> |
Compare this expression with the one derived earlier, namely,
|
<math>~h_1^2 \biggr|_{q^2 = 2} = \biggl[\lambda_1^2 \ell_{3D}^2 \biggr]_{q^2 = 2}</math> |
<math>~=</math> |
<math>~ \frac{(x^2 + 2y^2 + p^2z^2)}{x^2 + 4y^2 + p^4z^2} \, . </math> |
Well … first we recognize that, when q2 = 2,
|
<math>~\lambda_1^2 = x^2 + 2y^2 + p^2z^2 \, ,</math> |
<math>~\lambda_2^2 = \frac{\lambda_1^2 - p^2 z^2}{\lambda_1^2} = \frac{x^2 + 2y^2}{x^2 + 2y^2 + p^2z^2} \, , </math> |
<math>~\lambda_3^2 = \frac{y}{x^2} \, .</math> |
Hence,
|
<math>~(\Lambda^2 - 1) = 8\lambda_1^2 \lambda_2^2 \lambda_3^4</math> |
<math>~=</math> |
<math>~ 8(x^2 + 2y^2 + p^2z^2)\biggl[ \frac{x^2 + 2y^2}{x^2 + 2y^2 + p^2z^2} \biggr]\frac{y^2}{x^4} = \biggl[ \frac{8y^2(x^2 + 2y^2)}{x^4 } \biggr] \, , </math> |
|
<math>~\Rightarrow ~~~ \Lambda^2 </math> |
<math>~=</math> |
<math>~ 1 + \biggl[ \frac{8y^2(x^2 + 2y^2)}{x^4 } \biggr] = \frac{1}{x^4}\biggl[x^4 + 8x^2y^2 + 16y^4 \biggr] = \frac{1}{x^4}\biggl[x^2 + 4y^4 \biggr]^2 \, , </math> |
|
<math>~\Rightarrow ~~~ (\Lambda+1) </math> |
<math>~=</math> |
<math>~ \frac{(x^2 + 4y^4)}{x^2} + 1 = \frac{2x^2 + 4y^4}{x^2} \, , </math> |
which means,
|
<math>~h_1^2</math> |
<math>~=</math> |
<math>~\frac{1}{8\lambda_1^2 \Lambda^2} \biggl\{ 4\lambda_1^2 \lambda_2^2 \lambda_3 (\Lambda+1) + 4\lambda_1^2 \lambda_2^2 (\Lambda^2 - 1) + 8z^2\Lambda^2 \biggr\} </math> |
See Also
|
|
|---|
|
© 2014 - 2021 by Joel E. Tohline |