User:Tohline/Appendix/Ramblings/ConcentricEllipsodalT12Coordinates
Concentric Ellipsoidal (T12) Coordinates
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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:
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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> |
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<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,
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<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:
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<math>~\lambda_2</math> |
<math>~\equiv</math> |
<math>~ x^a y^b z^c \, , </math> |
in which case,
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<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,
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<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,
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<math>~\hat{e}_1 \cdot \hat{e}_2</math> |
<math>~=</math> |
<math>~0</math> |
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<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> |
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<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,
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<math>~\hat{e}_3</math> |
<math>~=</math> |
<math>~\hat{e}_1 \times \hat{e}_2</math> |
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|
<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> |
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|
<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,
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<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> |
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<math>~=</math> |
<math>~ \biggl[ -\hat\imath (q^2y) + \hat\jmath (x) \biggr] \ell_q \, , </math> |
where,
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<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,
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<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> |
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|
<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,
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<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,
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<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> |
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<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,
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<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> |
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|
<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
Setup
Reflecting back on an earlier exploration, let's define the two polynomials,
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<math>~\mathfrak{A} \equiv \ell_{3D}^{-2}</math> |
<math>~=</math> |
<math>~(x^2 + q^4y^2 + p^4z^2) \, ,</math> |
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<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> |
|
Then the 3rd unit vector may be written as,
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<math>~[\hat{e}_3]_\mathrm{needed}</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> |
Guess Third Coordinate Expression
Let's see what unit vector results if we define,
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<math>~\lambda_3</math> |
<math>~\equiv</math> |
<math>~\mathfrak{A}^{1 / 2} \mathfrak{B}^{-1 / 2} \, .</math> |
Partial Derivatives
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<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> |
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<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,
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<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> |
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<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^2z^2 ( b^2q^4 -a^2 - c^2p^4 ) - a^2 p^4z^4 - c^2x^4 \biggr] \, ; </math> |
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<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> |
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|
<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> |
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|
<math>~=</math> |
<math>~2z \biggl\{ (yz)^2 \biggl[ a^2p^4 - a^2p^4 \biggr] + (xz)^2 \biggl[ b^2p^4 - b^2p^4 \biggr] + (xy)^2 \biggl[c^2p^4 - a^2 - b^2q^4 \biggr] - a^2 q^4y^4 - b^2x^4 \biggr\} </math> |
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|
<math>~=</math> |
<math>~2z \biggl[ (xy)^2 ( c^2p^4 - a^2 - b^2q^4 ) - a^2 q^4y^4 - b^2x^4 \biggr] \, . </math> |
After completing a few squares, this last expression may be rewritten as …
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<math>~ \biggl[ \frac{2\mathfrak{A}\mathfrak{B}}{\lambda_3} \biggr] \frac{\partial \lambda_3}{\partial z}</math> |
<math>~=</math> |
<math>~2z \biggl[ (xy)^2 ( c^2p^4 - a^2 - b^2q^4 ) - a^2 q^4y^4 - b^2x^4 \biggr] </math> |
|
|
<math>~=</math> |
<math>~2z \biggl[ (xy)^2 ( c^2p^4 - a^2 - b^2q^4 ) \pm 2abq^2(xy)^2 - (aq^2y^2 \pm bx^2)^2 \biggr] </math> |
|
|
<math>~=</math> |
<math>~2z \biggl[ (xy)^2 ( c^2p^4 - a^2 - b^2q^4 \pm 2abq^2) - (aq^2y^2 \pm bx^2)^2 \biggr] </math> |
|
|
<math>~=</math> |
<math>~2z \biggl[ (xy)^2 [ c^2p^4 -(a \pm bq^2)^2 \pm 4abq^2] - (aq^2y^2 \pm bx^2)^2 \biggr] \, . </math> |
Now, if we choose the superior sign throughout this expression, the above-derived One-Two Perpendicular Constraint can be satisfied by setting, <math>~(a + bq^2) = -cp^2</math>. The expression then becomes,
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<math>~ \biggl[ \frac{2\mathfrak{A}\mathfrak{B}}{\lambda_3} \biggr] \frac{\partial \lambda_3}{\partial z}</math> |
<math>~=</math> |
<math>~-2z \biggl[ (aq^2y^2 + bx^2)^2 - 4abq^2 (xy)^2 \biggr] </math> |
|
|
<math>~=</math> |
<math>~ -2z (aq^2y^2 - bx^2)^2 \, . </math> |
Similarly, we find,
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<math>~ \biggl[ \frac{2\mathfrak{A}\mathfrak{B}}{\lambda_3} \biggr] \frac{\partial \lambda_3}{\partial x}</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>~=</math> |
<math>~2x \biggl[ y^2z^2(a^2 - q^4b^2 - c^2p^4) - (cq^2y^2 \pm bp^2z^2)^2 \pm 2bcq^2y^2p^2z^2 \biggr] </math> |
|
|
<math>~=</math> |
<math>~2x \biggl[ (yz)^2(a^2 - q^4b^2 - c^2p^4 \pm 2bcq^2p^2) - (cq^2y^2 \pm bp^2z^2)^2 \biggr] </math> |
|
|
<math>~=</math> |
<math>~2x \biggl[ (yz)^2[a^2 - (bq^2 \pm cp^2)^2 \pm 4bcq^2p^2] - (cq^2y^2 \pm bp^2z^2)^2 \biggr] \, . </math> |
|
<math>~ \biggl[ \frac{2\mathfrak{A}\mathfrak{B}}{\lambda_3} \biggr] \frac{\partial \lambda_3}{\partial y}</math> |
<math>~=</math> |
<math>~2y \biggl[ x^2z^2 ( b^2q^4 -a^2 - c^2p^4 ) - a^2 p^4z^4 - c^2x^4 \biggr] </math> |
|
|
<math>~=</math> |
<math>~2y \biggl[ x^2z^2 ( b^2q^4 -a^2 - c^2p^4 ) - (ap^2z^2 \pm cx^2)^2 \pm 2acx^2p^2z^2 \biggr] </math> |
|
|
<math>~=</math> |
<math>~2y \biggl[ x^2z^2 ( b^2q^4 -a^2 - c^2p^4 \pm 2acp^2 ) - (ap^2z^2 \pm cx^2)^2 \biggr] </math> |
|
|
<math>~=</math> |
<math>~2y \biggl[ (xz)^2 [ b^2q^4 -(a\pm cp^2)^2 \pm 4acp^2 ] - (ap^2z^2 \pm cx^2)^2 \biggr] \, . </math> |
So, if we again choose the superior sign throughout these expression, the above-derived One-Two Perpendicular Constraint can be satisfied: In the first by setting, <math>~(bq^2 + cp^2) = - a</math>; and in the second by setting, <math>~(a + cp^2) = - bq^2</math>. The expressions then become, respectively,
|
<math>~ \biggl[ \frac{2\mathfrak{A}\mathfrak{B}}{\lambda_3} \biggr] \frac{\partial \lambda_3}{\partial x}</math> |
<math>~=</math> |
<math>~ - 2x \biggl[ (cq^2y^2 + bp^2z^2)^2 - 4bcq^2 y^2 p^2 z^2 \biggr] </math> |
|
|
<math>~=</math> |
<math>~ - 2x (cq^2y^2 - bp^2z^2)^2 \, ; </math> |
|
<math>~ \biggl[ \frac{2\mathfrak{A}\mathfrak{B}}{\lambda_3} \biggr] \frac{\partial \lambda_3}{\partial y}</math> |
<math>~=</math> |
<math>~- 2y \biggl[ (ap^2z^2 + cx^2)^2 - 4acx^2 p^2 z^2 \biggr] </math> |
|
|
<math>~=</math> |
<math>~ - 2y(ap^2z^2 - cx^2)^2 \, . </math> |
|
Summary
Given,
the three relevant partial derivatives are:
|
Scale Factor
Hence, the associated scale factor is,
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<math>~h_3^{-2}</math> |
<math>~\equiv</math> |
<math>~ \biggl( \frac{\partial \lambda_3}{\partial x} \biggr)^2 + \biggl( \frac{\partial \lambda_3}{\partial y} \biggr)^2 + \biggl( \frac{\partial \lambda_3}{\partial z} \biggr)^2 </math> |
|
<math>~</math> |
<math>~=</math> |
<math>~ \biggl[ \frac{\lambda_3}{\mathfrak{A}\mathfrak{B}} \biggr]^2 \biggl\{ \biggl[ - x (cq^2y^2 - bp^2z^2)^2 \biggr]^2 + \biggl[ - y(ap^2z^2 - cx^2)^2 \biggr]^2 + \biggl[ -z (aq^2y^2 - bx^2)^2 \biggr]^2 \biggr\} </math> |
|
<math>~</math> |
<math>~=</math> |
<math>~ \biggl[ \frac{\lambda_3}{\mathfrak{A}\mathfrak{B}} \biggr]^2 \biggl\{ x^2 (cq^2y^2 - bp^2z^2)^4 + y^2 (ap^2z^2 - cx^2)^4 + z^2 (aq^2y^2 - bx^2)^4 \biggr\} </math> |
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<math>~\Rightarrow ~~~h_3</math> |
<math>~=</math> |
<math>~ \biggl[ \frac{\mathfrak{A}\mathfrak{B}}{\lambda_3} \biggr]\frac{1}{\mathfrak{C}} \, , </math> |
where,
|
<math>~\mathfrak{C}</math> |
<math>~\equiv</math> |
<math>~ \biggl[ x^2 (cq^2y^2 - bp^2z^2)^4 + y^2 (ap^2z^2 - cx^2)^4 + z^2 (aq^2y^2 - bx^2)^4 \biggr]^{1 / 2} \, . </math> |
Direction Cosines and Unit Vector
And the associated triplet of direction cosines is:
|
<math>~\gamma_{31} \equiv h_3 \biggl( \frac{\partial \lambda_3}{\partial x} \biggr)</math> |
<math>~=</math> |
<math>~ - x (cq^2y^2 - bp^2z^2)^2\biggl[ \frac{1}{\mathfrak{C}} \biggr] \, , </math> |
|
<math>~\gamma_{32} \equiv h_3 \biggl( \frac{\partial \lambda_3}{\partial y} \biggr)</math> |
<math>~=</math> |
<math>~ - y(ap^2z^2 - cx^2)^2 \biggl[ \frac{1}{\mathfrak{C}} \biggr] \, , </math> |
|
<math>~ \gamma_{33} \equiv h_3 \biggl( \frac{\partial \lambda_3}{\partial y} \biggr)</math> |
<math>~=</math> |
<math>~ -z (aq^2y^2 - bx^2)^2 \biggl[ \frac{1}{\mathfrak{C}} \biggr] \, . </math> |
This means that, for our particular guess of the 3rd coordinate, the relevant unit vector is,
|
<math>~[\hat{e}_3]_\mathrm{guess}</math> |
<math>~=</math> |
<math>~\frac{1}{\mathfrak{C}} \biggl\{ - \hat\imath \biggl[ x (cq^2y^2 - bp^2z^2)^2 \biggr] - \hat\jmath \biggl[ y(ap^2z^2 - cx^2)^2 \biggr] - \hat{k} \biggl[ z (aq^2y^2 - bx^2)^2 \biggr] \biggr\} \, . </math> |
Contrast
The unit vector resulting (just derived) from our guess of the third-coordinate expression should be compared with the needed unit vector as described above, namely,
|
<math>~[\hat{e}_3]_\mathrm{needed}</math> |
<math>~=</math> |
<math>~\mathfrak{A}^{-1 / 2} \mathfrak{B}^{-1 / 2} \biggl\{ \hat\imath \biggl[ x(cq^2y^2 - b p^2z^2) \biggr] + \hat\jmath \biggl[ y(ap^2z^2 - cx^2) \biggr] + \hat{k} \biggl[ z(bx^2 - aq^2y^2) \biggr] \biggr\} \, . </math> |
At first glance, apart from the difference in signs, the three terms inside the curly braces appear to be identical in these two separate unit-vector expressions. But they are not! Relative to the needed expression, key components of each term are squared in the guessed expression. Very close … but no cigar!
What about the overall leading coefficient? Well, given that,
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<math>~\mathfrak{A} </math> |
<math>~=</math> |
<math>~(x^2 + q^4y^2 + p^4z^2) \, ,</math> |
|
<math>~\mathfrak{B} </math> |
<math>~=</math> |
<math>~ [a^2(yz)^2 + b^2(xz)^2 + c^2(xy)^2] \, , </math> |
we can write,
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<math>~\mathfrak{A} \mathfrak{B}</math> |
<math>~=</math> |
<math>~ (x^2 + q^4y^2 + p^4z^2)[a^2(yz)^2 + b^2(xz)^2 + c^2(xy)^2] </math> |
See Also
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© 2014 - 2021 by Joel E. Tohline |