User:Tohline/Appendix/Ramblings/ConcentricEllipsodalDaringAttack
Daring Attack
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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> | ||||||
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As before, let's adopt the first-coordinate expression,
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<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,
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<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,
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<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,
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<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,
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<math>~\lambda_2^2</math> |
<math>~\equiv</math> |
<math>~1 - \biggl( \frac{z}{c}\biggr)^2 \, ,</math> |
in which case, we see that,
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<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,
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<math>~\biggl( \frac{x}{a}\biggr)^2 + \biggl( \frac{y}{b}\biggr)^2 </math> |
<math>~=</math> |
<math>~ \lambda_2^2 </math> |
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<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,
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<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,
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<math>~ 0</math> |
<math>~=</math> |
<math>~ 2y^2 + \frac{y}{\lambda_3^2} - \lambda_1^2 \lambda_2^2 </math> |
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<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> |
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<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),
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<math>~x^2</math> |
<math>~=</math> |
<math>~\frac{y}{\lambda_3^2}</math> |
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<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> |
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<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> |
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Summary (q2 = 2)
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For convenience, we have defined,
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<math>~\Lambda^2</math> |
<math>~\equiv</math> |
<math>~1 + 8\lambda_1^2 \lambda_2^2 \lambda_3^4 </math> |
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<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> |
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Test Example |
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<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>~\ell_{3D} = 0.730058, ~~ \ell_q = 0.750164</math>
<math>~h_1 = 0.730058</math> |
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| <math>~\Lambda^2-1 = 122.34879 ~~~\Rightarrow ~~~ \Lambda = 11.10625</math> | |||||||||||||||
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Do we get the correct values of <math>~(x, y, z)</math> ?
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Evaluate a few partial derivatives …
This matches the numerical value for <math>~h_1</math> as determined below, but it does not match the numerical value obtained previously (0.730058) for <math>~h_1</math>. The most likely piece that needs adjustment is the partial of "z" with respect to λ1. It needs to be … <math>~\frac{\partial z}{\partial \lambda_1} = \biggl[ h_1^2 - \biggl( \frac{\partial x}{\partial \lambda_1} \biggr)^2 - \biggl( \frac{\partial y}{\partial \lambda_1} \biggr)^2 \biggr]^{1 / 2} = 0.071647</math>.
Alternatively,
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Next, let's examine all nine partial derivatives, noting at the start that,
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<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> |
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<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> |
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<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,
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<math>~\frac{\partial z}{\partial \lambda_1}</math> |
<math>~=</math> |
<math>~ \frac{(1-\lambda_2^2)^{1 / 2}}{p} \, , </math> |
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<math>~\frac{\partial z}{\partial \lambda_2}</math> |
<math>~=</math> |
<math>~ -\frac{\lambda_1 \lambda_2}{p(1 - \lambda_2^2)^{1 / 2}} \, , </math> |
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<math>~\frac{\partial z}{\partial \lambda_3}</math> |
<math>~=</math> |
<math>~ 0 \, . </math> |
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<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> |
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<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> |
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<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> |
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<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> |
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<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> |
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<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> |
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<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?
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<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> |
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<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> |
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<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> |
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<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,
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<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> |
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<math>~=</math> |
<math>~\frac{ (\Lambda+1)\lambda_2^2 }{2 \Lambda } + \frac{z^2}{\lambda_1^2} </math> |
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<math>~=</math> |
<math>~ \frac{1}{\lambda_1^2} \biggl[ \frac{ (\Lambda+1)\lambda_2^2 \lambda_1^2 }{2 \Lambda } + z^2 \biggr] </math> |
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<math>~=</math> |
<math>~ \frac{1}{\lambda_1^2} \biggl[ \frac{ (\Lambda+1)(x^2 + 2y^2) }{2 \Lambda } + z^2 \biggr] \, . </math> |
Note that,
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<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> |
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<math>~\Rightarrow ~~~ \Lambda </math> |
<math>~=</math> |
<math>~\frac{x^2 + 4y^2}{x^2} \, .</math> |
Hence, the scale factor becomes,
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<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> |
|
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<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] = \frac{1.911525}{3.554} = 0.537852 </math> |
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<math>~\Rightarrow ~~~ h_1</math> |
<math>~=</math> |
<math>~ 0.733384 \, . </math> |
Compare this expression with the one derived earlier, namely,
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<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,
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<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,
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<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> |
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<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> |
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<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,
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<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> |
Think Again
Firm Relations
In addition to the functions that are specified in our above Daring Attack Table, we appreciate that,
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<math>~\frac{\partial x}{\partial \lambda_1}</math> |
<math>~=</math> |
<math>~ h_1^2 \biggl( \frac{\partial \lambda_1}{\partial x} \biggr) = \biggl(\lambda_1 \ell_{3D} \biggr)^2 \frac{x}{\lambda_1} = x \lambda_1 \ell_{3D}^2 \, , </math> |
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<math>~\frac{\partial y}{\partial \lambda_1}</math> |
<math>~=</math> |
<math>~ h_1^2 \biggl( \frac{\partial \lambda_1}{\partial y} \biggr) = \biggl(\lambda_1 \ell_{3D} \biggr)^2 \frac{q^2y}{\lambda_1} = q^2 y \lambda_1 \ell_{3D}^2 \, , </math> |
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<math>~\frac{\partial z}{\partial \lambda_1}</math> |
<math>~=</math> |
<math>~ h_1^2 \biggl( \frac{\partial \lambda_1}{\partial z} \biggr) = \biggl(\lambda_1 \ell_{3D} \biggr)^2 \frac{p^2z}{\lambda_1} = p^2 z \lambda_1 \ell_{3D}^2 \, . </math> |
Check …
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<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 = \lambda_1^2 \ell_{3D}^4 \biggl[ x^2 + q^4 y^2 + p^4z^2 \biggr] = \lambda_1^2 \ell_{3D}^2 \, . </math> (Yes!) |
Also,
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<math>~\frac{\partial x}{\partial \lambda_3}</math> |
<math>~=</math> |
<math>~ h_3^2 \biggl( \frac{\partial \lambda_3}{\partial x} \biggr) = \biggl[ \frac{xq^2y \ell_q}{\lambda_3} \biggr]^2 \biggl( - \frac{\lambda_3}{x} \biggr) = - q^4 y^2 \ell_q^2 \biggl( \frac{x}{\lambda_3} \biggr) \, , </math> |
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<math>~\frac{\partial y}{\partial \lambda_3}</math> |
<math>~=</math> |
<math>~ h_3^2 \biggl( \frac{\partial \lambda_3}{\partial y} \biggr) = \biggl[ \frac{xq^2y \ell_q}{\lambda_3} \biggr]^2 \biggl( + \frac{\lambda_3}{q^2y} \biggr) = x^2 \ell_q^2 \biggl( \frac{q^2y} {\lambda_3}\biggr) \, , </math> |
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<math>~\frac{\partial z}{\partial \lambda_3}</math> |
<math>~=</math> |
<math>~ h_3^2 \biggl( \frac{\partial \lambda_3}{\partial z} \biggr) = 0 \, . </math> |
Check …
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<math>~h_3^2</math> |
<math>~=</math> |
<math>~ \biggl( \frac{\partial x}{\partial \lambda_3} \biggr)^2 + \biggl( \frac{\partial y}{\partial \lambda_3} \biggr)^2 + \biggl( \frac{\partial z}{\partial \lambda_3} \biggr)^2 = \frac{\ell_q^4}{\lambda_3^2} \biggl[x^2 q^8 y^4 + x^4 q^4y^2 \biggr] = \frac{x^2 q^4 y^2\ell_q^4}{\lambda_3^2} \biggl[q^4 y^2 + x^2 \biggr] = \frac{x^2 q^4 y^2\ell_q^2}{\lambda_3^2} \, . </math> (Yes!) |
And, last …
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<math>~\frac{\partial x}{\partial \lambda_2}</math> |
<math>~=</math> |
<math>~ h_2 \gamma_{21} = h_2 \ell_q \ell_{3D} (xp^2z) \, , </math> |
|
<math>~\frac{\partial y}{\partial \lambda_2}</math> |
<math>~=</math> |
<math>~ h_2 \gamma_{22} = h_2 \ell_q \ell_{3D} (q^2 y p^2 z) \, , </math> |
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<math>~\frac{\partial z}{\partial \lambda_2}</math> |
<math>~=</math> |
<math>~ h_2 \gamma_{23} = - h_2 \ell_q \ell_{3D}(x^2 + q^4y^2) \, . </math> |
Speculation
First
From the direction-cosine expressions for <math>~\partial\lambda_2/\partial x_i</math> that have been summarized in our above Daring Attack Table, it seems reasonable to suggest that,
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<math>~h_2^2</math> |
<math>~=</math> |
<math>~(\ell_q \ell_{3D})^2 = \biggl[ (x^2 + q^4y^2)(x^2 + q^4y^2 + p^4z^2) \biggr]^{-1} \, , </math> |
in which case,
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<math>~\frac{\partial \lambda_2}{\partial x}</math> |
<math>~=</math> |
<math>~xp^2z \, ,</math> |
<math>~\frac{\partial \lambda_2}{\partial y}</math> |
<math>~=</math> |
<math>~q^2yp^2z \, ,</math> |
<math>~\frac{\partial \lambda_2}{\partial z}</math> |
<math>~=</math> |
<math>~-(x^2 + q^4y^2) \, ;</math> |
and,
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<math>~\frac{\partial x}{\partial \lambda_2} = h_2^2 \biggl( \frac{\partial \lambda_2}{\partial x} \biggr)</math> |
<math>~=</math> |
<math>~(\ell_q \ell_{3D})^2 xp^2z \, ,</math> |
<math>~\frac{\partial y}{\partial \lambda_2} = h_2^2 \biggl( \frac{\partial \lambda_2}{\partial y} \biggr)</math> |
<math>~=</math> |
<math>~(\ell_q \ell_{3D})^2 q^2yp^2z \, ,</math> |
<math>~\frac{\partial z}{\partial \lambda_2} = h_2^2 \biggl(\frac{\partial \lambda_2}{\partial z} \biggr)</math> |
<math>~=</math> |
<math>~-(\ell_q \ell_{3D})^2 (x^2 + q^4y^2) \, .</math> |
Second
Alternatively, after examining the direction-cosine expressions for <math>~\partial x_i/\partial \lambda_2</math> that we have just provided, one might suggest that,
|
<math>~h_2^2</math> |
<math>~=</math> |
<math>~(\ell_q \ell_{3D})^{-2} = (x^2 + q^4y^2)(x^2 + q^4y^2 + p^4z^2) = p^4z^2(x^2 + q^4y^2) + (x^2 + q^4y^2)^2 \, , </math> |
in which case, the expressions provided for <math>~\partial \lambda_2/\partial x_i</math> and <math>~\partial x_i/\partial \lambda_2</math> must be swapped relative to our First speculation.
Third
Noticing that <math>~h_1^2</math> is proportional to <math>~\lambda_1^2</math> and that <math>~h_3^2</math> is inversely proportional to <math>~\lambda_3^2</math>, let's consider both as possible behaviors for the 2nd scale factor. Let's try the first of these behaviors. Specifically, what if we assume …
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<math>~\frac{\partial \lambda_2}{\partial x} </math> |
<math>~=</math> |
<math>~\frac{xp^2 z}{\lambda_2} \, ,</math> |
<math>~\frac{\partial \lambda_2}{\partial y} </math> |
<math>~=</math> |
<math>~\frac{q^2y p^2z}{\lambda_2} \, ,</math> |
<math>~\frac{\partial \lambda_2}{\partial z} </math> |
<math>~=</math> |
<math>~-\frac{(x^2 + q^4y^2)}{\lambda_2} \, .</math> |
Then,
|
<math>~h_2^{-2}</math> |
<math>~=</math> |
<math>~ \biggl( \frac{\partial \lambda_2}{\partial x}\biggr)^2 + \biggl( \frac{\partial \lambda_2}{\partial y}\biggr)^2 +\biggl( \frac{\partial \lambda_2}{\partial z}\biggr)^2 = [\lambda_2 \ell_q \ell_{3D} ]^{-2} </math> |
|
<math>~\Rightarrow ~~~ h_2</math> |
<math>~=</math> |
<math>~ \lambda_2 \ell_q \ell_{3D} \, . </math> |
|
Primary implication:
Secondary implication:
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Now, what specifically is the function, <math>~\lambda_2(x, y, z)</math> ? Start by rewriting the three partial derivatives as,
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<math>~\frac{1}{2} \frac{\partial (\lambda_2^2)}{\partial x} </math> |
<math>~=</math> |
<math>~xp^2 z \, ,</math> |
<math>~\frac{1}{2} \frac{\partial (\lambda_2)^2}{\partial y} </math> |
<math>~=</math> |
<math>~q^2y p^2z \, ,</math> |
<math>~\frac{1}{2} \frac{\partial (\lambda_2)^2}{\partial z} </math> |
<math>~=</math> |
<math>~-(x^2 + q^4y^2) \, .</math> |
Suppose that,
|
<math>~\lambda_2^2</math> |
<math>~=</math> |
<math>~(x^2 + q^2y^2)p^2z \, .</math> |
Then we have,
|
<math>~\frac{\partial \lambda_2^2}{\partial x}</math> |
<math>~=</math> |
<math>~2xp^2z \, ,</math> |
and, |
<math>~\frac{\partial \lambda_2^2}{\partial y}</math> |
<math>~=</math> |
<math>~2q^2 yp^2z \, .</math> Great! |
But this cannot be the correct expression for <math>~\lambda_2^2</math> because,
|
<math>~\frac{\partial \lambda_2^2}{\partial z}</math> |
<math>~=</math> |
<math>~(x^2 + q^2y^2)p^2 \, ,</math> |
which does not match the desired partial derivative with respect to <math>~z</math>.
Alternatively, if we assume …
|
<math>~\frac{\partial \lambda_2}{\partial x} </math> |
<math>~=</math> |
<math>~\frac{\lambda_2}{xp^2 z} \, ,</math> |
<math>~\frac{\partial \lambda_2}{\partial y} </math> |
<math>~=</math> |
<math>~\frac{\lambda_2}{q^2y p^2z} \, ,</math> |
<math>~\frac{\partial \lambda_2}{\partial z} </math> |
<math>~=</math> |
<math>~-\frac{\lambda_2}{(x^2 + q^4y^2)} \, ,</math> |
then,
|
<math>~h_2^{-2}</math> |
<math>~=</math> |
<math>~ \biggl( \frac{\partial \lambda_2}{\partial x}\biggr)^2 + \biggl( \frac{\partial \lambda_2}{\partial y}\biggr)^2 +\biggl( \frac{\partial \lambda_2}{\partial z}\biggr)^2 </math> |
|
|
<math>~=</math> |
<math>~ \biggl( \frac{\lambda_2}{xp^2 z} \biggr)^2 + \biggl( \frac{\lambda_2}{q^2y p^2z} \biggr)^2 +\biggl( \frac{\lambda_2}{x^2 + q^4y^2} \biggr)^2 </math> |
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<math>~\Rightarrow ~~~ (h_2 \lambda_2)^{-2}</math> |
<math>~=</math> |
<math>~ \frac{ q^4y^2p^4z^2 (x^2 + q^4y^2)^2 + x^2p^4z^2 (x^2 + q^4y^2)^2 + x^2 q^4y^2 p^8z^4}{x^2 q^4y^2p^8z^4(x^2 + q^4y^2)^2} </math> |
|
<math>~\Rightarrow ~~~ h_2 </math> |
<math>~=</math> |
<math>~\frac{1}{\lambda_2} \biggl[ \frac{x^2 q^4y^2p^8z^4(x^2 + q^4y^2)^2}{ q^4y^2p^4z^2 (x^2 + q^4y^2)^2 + x^2p^4z^2 (x^2 + q^4y^2)^2 + x^2 q^4y^2 p^8z^4} \biggr]^{1 / 2} </math> |
|
|
<math>~=</math> |
<math>~\frac{1}{\lambda_2} \biggl\{ \frac{x q^2y p^2z(x^2 + q^4y^2)}{ [ q^4y^2 (x^2 + q^4y^2)^2 + x^2 (x^2 + q^4y^2)^2 + x^2 q^4y^2 p^4z^2 ]^{1 / 2}} \biggr\} </math> |
|
|
<math>~=</math> |
<math>~\frac{1}{\lambda_2} \biggl\{ \frac{x q^2y p^2z(x^2 + q^4y^2)}{ [ (x^2 + q^4y^2)^3 + x^2 q^4y^2 p^4z^2 ]^{1 / 2}} \biggr\} </math> |
Let's check for consistency with one of the direction-cosines.
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<math>~\gamma_{21} = h_2 \biggl( \frac{\partial \lambda_2}{\partial x} \biggr)</math> |
<math>~=</math> |
<math>~\biggl\{ \frac{q^2y (x^2 + q^4y^2)}{ [ (x^2 + q^4y^2)^3 + x^2 q^4y^2 p^4z^2 ]^{1 / 2}} \biggr\} </math> |
|
<math>~\Rightarrow ~~~ \frac{ \gamma_{21} }{\ell_q(xp^2z) }</math> |
<math>~=</math> |
<math>~\frac{(x^2 + q^4y^2)^{1 / 2}}{xp^2z} \biggl\{ \frac{q^2y (x^2 + q^4y^2)}{ [ (x^2 + q^4y^2)^3 + x^2 q^4y^2 p^4z^2 ]^{1 / 2}} \biggr\} </math> |
|
|
<math>~=</math> |
<math>~\frac{q^2y}{xp^2z} \biggl\{ \frac{(x^2 + q^4y^2)^{3 / 2}}{ [ (x^2 + q^4y^2)^3 + x^2 q^4y^2 p^4z^2 ]^{1 / 2}} \biggr\} </math> |
|
|
<math>~=</math> |
<math>~\frac{q^2y}{xp^2z} \biggl[1 + \frac{x^2q^4y^2p^4z^2}{(x^2 + q^4y^2)^3} \biggr]^{-1 / 2} \, . </math> |
This does not match the term in the expression for <math>~\gamma_{21}</math> — namely, <math>~\ell_{3D}</math> — that is expected from the original tabulation.
Study the Functional Forms
We know the functional forms of two of the desired curvilinear coordinates, namely,
|
<math>~\lambda_1(x, y, z)</math> |
<math>~\equiv</math> |
<math>~(x^2 + q^2 y^2 + p^2 z^2)^{1 / 2} \, ,</math> |
|
<math>~\lambda_3(x, y, z)</math> |
<math>~\equiv</math> |
<math>~\frac{y^{1/q^2}}{x} \, ,</math> |
but we do not yet have a valid expression for the 2nd coordinate, <math>~\lambda_2(x, y, z)</math>. Nevertheless, let's see if we can guess the functional forms for <math>~x_i(\lambda_1, \lambda_2, \lambda_3)</math>, by inverting the two known curvilinear-coordinate functions. As a starting point, let's impose the following mappings:
|
<math>~x</math> |
<math>~~\rightarrow~~</math> |
<math>~\frac{y^{1/q^2}}{\lambda_3} \, ,</math> |
||
|
<math>~z</math> |
<math>~~\rightarrow~~</math> |
<math>~ \frac{1}{p}\biggl[ \lambda_1^2 - q^2y^2 - x^2 \biggr]^{1 / 2} </math> |
<math>~~\rightarrow~~</math> |
<math>~ \frac{1}{p}\biggl[ \lambda_1^2 - q^2y^2 - \frac{y^{2/q^2}}{\lambda_3^2} \biggr]^{1 / 2} \, . </math> |
This means, for example, that,
|
<math>~\ell_q^{-2}</math> |
<math>~~\rightarrow~~</math> |
<math>~ q^4y^2 + \frac{y^{2/q^2}}{\lambda_3^2} = \lambda_3^{-2} \biggl[ q^2(qy\lambda_3)^2 + y^{2/q^2} \biggr] \, , </math> |
|
<math>~\ell_{3D}^{-2}</math> |
<math>~~\rightarrow~~</math> |
<math>~ q^4y^2 + \frac{y^{2/q^2}}{\lambda_3^2} + p^2\biggl[ \lambda_1^2 - q^2y^2 - \frac{y^{2/q^2}}{\lambda_3^2} \biggr] = (q^2-p^2)q^2y^2 + \frac{(1-p^2)y^{2/q^2}}{\lambda_3^2} + p^2\lambda_1^2 \, . </math> |
Derivatives of x
|
<math>~\frac{\partial x}{\partial \lambda_1}</math> |
<math>~=</math> |
<math>~ h_1^2 \biggl( \frac{\partial \lambda_1}{\partial x} \biggr) = \biggl(\lambda_1 \ell_{3D} \biggr)^2 \frac{x}{\lambda_1} = x \lambda_1 \ell_{3D}^2 \, , </math> |
See Also
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© 2014 - 2021 by Joel E. Tohline |