User:Tohline/Appendix/Ramblings/ConcentricEllipsodalT12Coordinates

Concentric Ellipsoidal (T12) Coordinates

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 1^st 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 2^nd Coordinate

Let's adopt the following generalized prescription for the 2^nd 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 3^rd Coordinate

The unit vector associated with the 3^rd 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 2^nd 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 3^rd 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 2^nd 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>

And it implies a unit vector for the 3^rd coordinate of the form,

<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>
	<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 3^rd-Coordinate Profile

Reflecting back on an earlier exploration, let's define the two polynomials,

<math>~\mathfrak{A} \equiv \ell_{3D}^{-2}</math>	<math>~=</math>	<math>~(x^2 + q^4y^2 + p^4z^2) \, ,</math>
<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>

<math>~\frac{\partial \mathfrak{A}}{\partial x}</math> <math>~=</math> <math>~2x \, ,</math> <math>~\frac{\partial \mathfrak{A}}{\partial y}</math> <math>~=</math> <math>~2q^4 y \, ,</math> <math>~\frac{\partial \mathfrak{A}}{\partial z}</math> <math>~=</math> <math>~2p^4 z \, ;</math> <math>~\frac{\partial \mathfrak{B}}{\partial x}</math> <math>~=</math> <math>~2x(b^2z^2 + c^2y^2) \, ,</math> <math>~\frac{\partial \mathfrak{B}}{\partial y}</math> <math>~=</math> <math>~2y(a^2 z^2 + c^2x^2) \, ,</math> <math>~\frac{\partial \mathfrak{B}}{\partial z}</math> <math>~=</math> <math>~2z(a^2 y^2 + b^2x^2) \, .</math>

Then the 3^rd unit vector may be written as,

<math>~\hat{e}_3</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>

Let's see what unit vector results if we define,

<math>~\lambda_3</math>

<math>~\equiv</math>

<math>~\mathfrak{A}^{1 / 2} \mathfrak{B}^{-1 / 2} \, .</math>

<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>
<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,

<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>

<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^2y^2 ( b^2q^4 -a^2 - c^2p^4 ) - a^2 p^4z^4 - c^2x^4 \biggr] \, ; </math>

<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>
	<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>

See Also

• Direction Cosines
• Trials up through T7 Coordinates

© 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