User:Tohline/Appendix/Ramblings/ConcentricEllipsodalDaringAttack

Daring Attack

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^4 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>~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) <math>~z(\lambda_1, \lambda_2, \lambda_3)</math> <math>~=</math> <math>~\frac{\lambda_1(1-\lambda_2^2)^{1 / 2}}{p} \, ,</math> <math>~y(\lambda_1, \lambda_2, \lambda_3)</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\} = \frac{(\Lambda - 1)}{4\lambda_3^2}\, ; </math> <math>~x(\lambda_1, \lambda_2, \lambda_3)</math> <math>~=</math> <math>~ \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} = \frac{(\Lambda - 1)^{1 / 2}}{2\lambda_3^{2}} \, . </math>

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>

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>~\ell_{3D} = 0.730058, ~~ \ell_q = 0.750164</math> <math>~h_1 = 0.730058</math>

<math>~\Lambda^2-1 = 122.34879 ~~~\Rightarrow ~~~ \Lambda = 11.10625</math>

Do we get the correct values of <math>~(x, y, z)</math>   ?

<math>~z(\lambda_1, \lambda_2, \lambda_3)</math>	<math>~=</math>	<math>~\frac{\lambda_1(1-\lambda_2^2)^{1 / 2}}{p} \, ,</math>
<math>~y(\lambda_1, \lambda_2, \lambda_3)</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\} = \frac{(\Lambda - 1)}{4\lambda_3^2}\, ; </math>
<math>~x(\lambda_1, \lambda_2, \lambda_3)</math>	<math>~=</math>	<math>~ \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} = \frac{(\Lambda - 1)^{1 / 2}}{2\lambda_3^{2}} \, . </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

• Direction Cosines

© 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