ThreeDimensionalConfigurations/DescriptionOfRiemannTypeI

Description of Riemann Type I Ellipsoids

Type I Riemann Ellipsoids

A more succinct version of this chapter is titled, 3Dconfigurations/DescriptionOfRiemannTypeI.

Analytic Determination of Equilibrium Model Parameters

Drawing heavily from §47 (pp. 129 - 132) of [EFE] , in a separate chapter we show how the steady-state 2^nd-order tensor virial equations can be used to derive the equilibrium structure of Riemann Ellipsoids of Type I, II, & III. By definition, for these types of Riemann Ellipsoids, the two vectors $\overrightarrow{\mathrm{\Omega }}$ and $\overrightarrow{\zeta }$ are not parallel to any of the principal axes of the ellipsoid, and they are not aligned with each other, but they both lie in the $y-z$ plane — that is to say, $({\mathrm{\Omega }}_1,{\zeta }_1)=(0,0)$. For a given specified density $(\rho )$ and choice of the three semi-axes $(a_1,a_2,a_3)\leftrightarrow (a,b,c)$, all five of the expressions displayed in that chapter's Summary Table must be used in order to determine the equilibrium configuration's associated values of the five unknowns: $\mathrm{\Pi },({\mathrm{\Omega }}_2,{\zeta }_2),({\mathrm{\Omega }}_3,{\zeta }_3)$.

In an effort to simplify the constraint-equation expressions — and following the notation found in [EFE] — we adopt the intermediary parameters:

$\beta$ $=$ $\frac{1}{4a^2}\{(4a^2-b^2+c^2)\mp [(4a^2+c^2-b^2{)}^2-16a^2{c}^2{]}^{1/2}\};$ [ EFE, Chapter 7, §47, Eq. (16) ] $\gamma$ $=$ $\frac{1}{4a^2}\{(4a^2+b^2-c^2)\mp [(4a^2+c^2-b^2{)}^2-16a^2{c}^2{]}^{1/2}\}.$ [ EFE, Chapter 7, §47, Eq. (17) ] As is emphasized in EFE (Chapter 7, §47, p. 131) "… the signs in front of the radicals, in the two expressions, go together. Furthermore, "the two roots … correspond to the fact that, consistent with Dedekind's theorem, two states of internal motions are compatible with the same external figure."

The five relevant constraint equations are,

$2\left[\frac{3\cdot 5}{2^2\pi abc\rho }\right]\mathrm{\Pi }$

$=$

$4\pi G\rho c^2\{A_3+[\frac{a^2(3b^2-4a^2+c^2)B_{23}+b^2(a^2{A}_1-c^2{A}_3)}{4a^4-a^2(b^2+c^2)+b^2{c}^2}\left]\right\}.$

$0$	$=$	${\mathrm{\Omega }}_2^2\beta \left[\frac{c^2-b^2}{c^2}\right]\left[\frac{4a^4-a^2(b^2+c^2)+b^2{c}^2}{2a^2{b}^2}\right]+2\pi G\rho \left[\frac{a^2(3b^2-4a^2+c^2)B_{23}+b^2(a^2{A}_1-c^2{A}_3)}{a^2{b}^2}\right],$
[ EFE, Chapter 7, §51, Eq. (170) ]

${\zeta }_2$	$=$	$-{\mathrm{\Omega }}_2\beta \left[\frac{a^2+c^2}{c^2}\right],$
[ EFE, Chapter 7, §47, Eq. (12) ]

$0$	$=$	$-{\mathrm{\Omega }}_3^2\gamma \left[\frac{c^2-b^2}{b^2}\right]\left[\frac{4a^4-a^2(b^2+c^2)+b^2{c}^2}{2a^2{c}^2}\right]+2\pi G\rho \left[\frac{a^2(b^2+3c^2-4a^2)B_{23}+c^2(a^2{A}_1-b^2{A}_2)}{a^2{c}^2}\right].$
[ EFE, Chapter 7, §51, Eq. (171) ]

${\zeta }_3$	$=$	$-{\mathrm{\Omega }}_3\gamma \left[\frac{a^2+b^2}{b^2}\right],$
[ EFE, Chapter 7, §47, Eq. (12) ]

where,

$B_{23}$	$=$	$\left[\frac{A_2{b}^2-A_3{c}^2}{b^2-c^2}\right].$
[ EFE, Chapter 3, §21, Eqs. (105) & (107) ]

Throughout the EFE discussion of Type I Riemann Ellipsoids, the assumption is made that $b\ge a\ge c$. Presumably then, when evaluating the above set of constraints, we must adopt the associations, $(A_1,a_1)\leftrightarrow (A_m,a_m)$, $(A_2,a_2)\leftrightarrow (A_{\ell },a_{\ell })$, and $(A_3,a_3)\leftrightarrow (A_s,a_s)$. This means that the coefficients, $A_1$, $A_2$, and $A_3$ are defined by the expressions,

$A_2$	$=$	$2\left(\frac{a_1}{a_2}\right)\left(\frac{a_3}{a_2}\right)\left[\frac{F(\theta ,k)-E(\theta ,k)}{k^2{\sin }^3\theta }\right],$
$A_3$	$=$	$2\left(\frac{a_1}{a_2}\right)\left[\frac{(a_1/a_2)\sin \theta -(a_3/a_2)E(\theta ,k)}{(1-k^2){\sin }^3\theta }\right],$
$A_1=2-(A_2+A_3)$	$=$	$\frac{2a_1{a}_3}{a_2^2}\left[\frac{E(\theta ,k)-(1-k^2)F(\theta ,k)-(a_3/a_1)k^2\sin \theta }{k^2(1-k^2){\sin }^3\theta }\right],$

where, the arguments of the incomplete elliptic integrals are,

$\theta ={\cos }^{-1}\left(\frac{a_3}{a_2}\right)$

and

$k=[\frac{1-(a_1/a_2{)}^2}{1-(a_3/a_2{)}^2}{]}^{1/2}.$

Example Equilibrium Models

Extracted from XXVIII

Here are equilibrium model parameters drawn from 📚 S. Chandrasekhar (1966, ApJ, Vol. 145, pp. 842 - 877) — referred to in [EFE] as Publication XXVIII.

Data Extracted from Table 4 (p. 858) of S. Chandrasekhar (1966) The Equilibrium and the Stability of the Riemann Ellipsoids. II. The Astrophysical Journal, Vol. 145, pp. 842 - 877																Our (reverse-engineered) Determination
The Properties of a Few Riemann Ellipsoids of Type I
$\frac{a_2}{a_1}$	$\frac{a_3}{a_1}$		Direct							Adjoint						A1	A2	A3
			${\mathrm{\Omega }}_2$	${\mathrm{\Omega }}_3$	${\zeta }_2$	${\zeta }_3$	$({\zeta }_2/{\mathrm{\Omega }}_2)$	$({\zeta }_3/{\mathrm{\Omega }}_3)$		${\mathrm{\Omega }}_2^{†}$	${\mathrm{\Omega }}_3^{†}$	${\zeta }_2^{†}$	${\zeta }_3^{†}$	$({\zeta }_2/{\mathrm{\Omega }}_2{)}^{†}$	$({\zeta }_3/{\mathrm{\Omega }}_3{)}^{†}$
1.05263	0.41667		+ 0.14834	+0.73257	-1.41355	-2.61578	-9.52912	-3.57069		+0.50185	+1.30617	-0.41783	-1.46707	-0.83258	-1.12318	0.43008706	0.40190235	1.16801059
1.25000	0.50000		+0.39259	+0.66536	-2.19983	-1.93895	-5.60338	-2.91414		+0.87993	+0.94583	-0.98148	-1.36398	-1.11541	-1.44210	0.50823343	0.37944073	1.11232585
1.44065	0.49273		+0.57179	+0.59896	-2.24560	-1.49425	-3.92732	-2.49474		+0.89032	+0.69996	-1.44219	-1.27866	-1.61986	-1.82676	0.52403947	0.32351421	1.15244632
1.66667	0.33333		+0.71251	+0.52815	-2.37502	-1.19714	-3.33331	-2.26667		+0.71251	+0.52815	-2.37502	-1.19714	-3.33331	-2.26667	0.41805282	0.20718125	1.37476593
1.36444	0.09518		+0.05632	+0.40707	-6.68275	-1.24612	-118.657	-3.06119		+0.63035	+0.59414	-0.59714	-0.85376	-0.94731	-1.43697	0.14374587	0.09152713	1.76472699
1.69351	0.11813		+0.15764	+0.38504	-6.27092	-1.02536	-39.7800	-2.66300		+0.73061	+0.44893	-1.35309	-0.87944	-1.85200	-1.95897	0.18178501	0.08464699	1.73356799
1.52303	0.05315		+0.03311	+0.29600	-9.85239	-0.84580	-297.565	-2.85743		+0.52221	+0.38805	-0.62474	-0.64518	-1.19634	-1.66262	0.08593434	0.04618515	1.86788051
1.78590	0.06233		+0.08952	+0.28558	-9.19424	-0.74657	-102.706	-2.61422		+0.57083	+0.31825	-1.4418	-0.66992	-2.52580	-2.10501	0.10258739	0.04358267	1.85382994
NOTE:   All frequencies are given in the unit of $(\pi G\rho {)}^{1/2}$.

Given the values of ${\mathrm{\Omega }}_2$ and ${\mathrm{\Omega }}_3$ from this table, we have reverse-engineered this problem and determined what numerical values of $A_1$, $A_2$, and $A_3$ were used by 📚 Chandrasekhar (1966; XXVIII) for various models. What follows are the expressions that have been derived via this reverse-engineering effort. The values of $A_1$, $A_2$, and $A_3$ that we have derived in this manner have been recorded in the last three columns of the table.

First …

$\beta \left(\frac{{\mathrm{\Omega }}_2^2}{\pi G\rho }\right)\left[\frac{b^2-c^2}{2c^2}\right][4a^4-a^2(b^2+c^2)+b^2{c}^2]$	$=$	$[a^2(3b^2-4a^2+c^2)B_{23}+b^2(a^2{A}_1-c^2{A}_3)]$
	$=$	$a^2(3b^2-4a^2+c^2)\left[\frac{A_2{b}^2-A_3{c}^2}{b^2-c^2}\right]+a^2{b}^2[2-(A_2+A_3)]-b^2{c}^2{A}_3$
$\Rightarrow \frac{\beta }{4c^2}\left(\frac{{\mathrm{\Omega }}_2^2}{\pi G\rho }\right)\underset{\mathrm{\Gamma }}{\underbrace{(b^2-c^2{)}^2[4a^4-a^2(b^2+c^2)+b^2{c}^2]}}$	$=$	$a^2(3b^2-4a^2+c^2)[A_2{b}^2-A_3{c}^2]+a^2{b}^2(b^2-c^2)[2-(A_2+A_3)]-b^2{c}^2(b^2-c^2)A_3$
	$=$	$2a^2{b}^2(b^2-c^2)+a^2{b}^2(3b^2-4a^2+c^2)A_2-a^2{b}^2(b^2-c^2)A_2-a^2{c}^2(3b^2-4a^2+c^2)A_3-a^2{b}^2(b^2-c^2)A_3-b^2{c}^2(b^2-c^2)A_3$
	$=$	$2a^2{b}^2(b^2-c^2)+a^2{b}^2[(3b^2-4a^2+c^2)-(b^2-c^2)]A_2-[a^2{c}^2(3b^2-4a^2+c^2)+a^2{b}^2(b^2-c^2)+b^2{c}^2(b^2-c^2)]A_3$
	$=$	$2a^2{b}^2(b^2-c^2)+2a^2{b}^2[b^2-2a^2+c^2]A_2-[(3a^2{b}^2{c}^2-4a^4{c}^2+a^2{c}^4)+(a^2{b}^4-a^2{b}^2{c}^2)+(b^4{c}^2-b^2{c}^4)]A_3$
	$=$	$\underset{{\mathrm{\Lambda }}_{20}}{\underbrace{2a^2{b}^2(b^2-c^2)}}+\underset{{\mathrm{\Lambda }}_{22}}{\underbrace{2a^2{b}^2[b^2-2a^2+c^2]}}{A}_2-\underset{{\mathrm{\Lambda }}_{23}}{\underbrace{[2a^2{b}^2{c}^2-4a^4{c}^2+a^2{c}^4+a^2{b}^4+b^4{c}^2-b^2{c}^4]}}{A}_3.$

Next …

$\gamma \left(\frac{{\mathrm{\Omega }}_3^2}{\pi G\rho }\right)\left[\frac{c^2-b^2}{4b^2}\right][4a^4-a^2(b^2+c^2)+b^2{c}^2]$	$=$	$[a^2(b^2+3c^2-4a^2)B_{23}+c^2(a^2{A}_1-b^2{A}_2)]$
	$=$	$a^2(b^2+3c^2-4a^2)\left[\frac{A_2{b}^2-A_3{c}^2}{b^2-c^2}\right]+a^2{c}^2[2-(A_2+A_3)]-b^2{c}^2{A}_2$
$\Rightarrow -\frac{\gamma }{4b^2}\left(\frac{{\mathrm{\Omega }}_3^2}{\pi G\rho }\right)\underset{\mathrm{\Gamma }}{\underbrace{(b^2-c^2{)}^2[4a^4-a^2(b^2+c^2)+b^2{c}^2]}}$	$=$	$a^2(b^2+3c^2-4a^2)[A_2{b}^2-A_3{c}^2]+a^2{c}^2(b^2-c^2)[2-(A_2+A_3)]-b^2{c}^2(b^2-c^2)A_2$
	$=$	$2a^2{c}^2(b^2-c^2)+a^2{b}^2(b^2+3c^2-4a^2)A_2-a^2{c}^2(b^2-c^2)A_2-b^2{c}^2(b^2-c^2)A_2-a^2{c}^2(b^2+3c^2-4a^2)A_3-a^2{c}^2(b^2-c^2)A_3$
	$=$	$2a^2{c}^2(b^2-c^2)+[a^2{b}^2(b^2+3c^2-4a^2)-a^2{c}^2(b^2-c^2)-b^2{c}^2(b^2-c^2)]A_2-2a^2{c}^2[b^2+c^2-2a^2]A_3$
	$=$	$\underset{{\mathrm{\Lambda }}_{30}}{\underbrace{2a^2{c}^2(b^2-c^2)}}+\underset{{\mathrm{\Lambda }}_{32}}{\underbrace{[2a^2{b}^2{c}^2-4a^4{b}^2+a^2{b}^4+a^2{c}^4-b^4{c}^2+b^2{c}^4]}}{A}_2-\underset{{\mathrm{\Lambda }}_{33}}{\underbrace{2a^2{c}^2[b^2+c^2-2a^2]}}{A}_3.$

Hence,

$\frac{\beta }{4c^2}\left(\frac{{\mathrm{\Omega }}_2^2}{\pi G\rho }\right)\mathrm{\Gamma }$	$=$	${\mathrm{\Lambda }}_{20}+{\mathrm{\Lambda }}_{22}{A}_2-{\mathrm{\Lambda }}_{23}{A}_3$
$\Rightarrow A_3$	$=$	$\frac{1}{{\mathrm{\Lambda }}_{23}}[{\mathrm{\Lambda }}_{20}+{\mathrm{\Lambda }}_{22}{A}_2-\frac{\beta }{4c^2}(\frac{{\mathrm{\Omega }}_2^2}{\pi G\rho }\left)\mathrm{\Gamma }\right];$

and,

$-\frac{\gamma }{4b^2}\left(\frac{{\mathrm{\Omega }}_3^2}{\pi G\rho }\right)\mathrm{\Gamma }$	$=$	${\mathrm{\Lambda }}_{30}+{\mathrm{\Lambda }}_{32}{A}_2-{\mathrm{\Lambda }}_{33}{A}_3$
$\Rightarrow A_3$	$=$	$\frac{1}{{\mathrm{\Lambda }}_{33}}[{\mathrm{\Lambda }}_{30}+{\mathrm{\Lambda }}_{32}{A}_2+\frac{\gamma }{4b^2}(\frac{{\mathrm{\Omega }}_3^2}{\pi G\rho }\left)\mathrm{\Gamma }\right].$

Put together, we find,

$\left[\frac{{\mathrm{\Lambda }}_{22}}{{\mathrm{\Lambda }}_{23}}\right]A_2+\frac{1}{{\mathrm{\Lambda }}_{23}}[{\mathrm{\Lambda }}_{20}-\frac{\beta }{4c^2}(\frac{{\mathrm{\Omega }}_2^2}{\pi G\rho }\left)\mathrm{\Gamma }\right]$	$=$	$\left[\frac{{\mathrm{\Lambda }}_{32}}{{\mathrm{\Lambda }}_{33}}\right]A_2+\frac{1}{{\mathrm{\Lambda }}_{33}}[{\mathrm{\Lambda }}_{30}+\frac{\gamma }{4b^2}(\frac{{\mathrm{\Omega }}_3^2}{\pi G\rho }\left)\mathrm{\Gamma }\right]$
$\Rightarrow A_2[\frac{{\mathrm{\Lambda }}_{22}}{{\mathrm{\Lambda }}_{23}}-\frac{{\mathrm{\Lambda }}_{32}}{{\mathrm{\Lambda }}_{33}}]$	$=$	$\frac{1}{{\mathrm{\Lambda }}_{33}}[{\mathrm{\Lambda }}_{30}+\frac{\gamma }{4b^2}(\frac{{\mathrm{\Omega }}_3^2}{\pi G\rho }\left)\mathrm{\Gamma }\right]-\frac{1}{{\mathrm{\Lambda }}_{23}}[{\mathrm{\Lambda }}_{20}-\frac{\beta }{4c^2}(\frac{{\mathrm{\Omega }}_2^2}{\pi G\rho }\left)\mathrm{\Gamma }\right]$

$\frac{a_2}{a_1}$	$\frac{a_3}{a_1}$	$A_1$	$A_2$	$A_3$		Direct					Adjoint
						${\beta }_{+}$	${\gamma }_{+}$	$-{\beta }_{+}(a^2+c^2)/c^2$	$-{\gamma }_{+}(a^2+b^2)/b^2$		${\beta }_{-}$	${\gamma }_{-}$	$-{\beta }_{-}(a^2+c^2)/c^2$	$-{\gamma }_{-}(a^2+b^2)/b^2$
1.25000	0.50000	0.50823343	0.37944073	1.11232585		1.12066896	1.77691896	-5.6033448	-2.9141471		0.22308104	0.87933104	-1.1154052	-1.4421029

Examination of Lagrangian Flow in One Specific Model

This particular set of seven key parameters has been drawn from [EFE] Chapter 7, Table XIII (p. 170). The tabular layout presented here, also appears in a related discussion labeled, Challenges Pt. 2.

$a=a_1=1$

$b=a_2=1.25$

$c=a_3=0.4703$

${\mathrm{\Omega }}_2=0.3639$

${\mathrm{\Omega }}_3=0.6633$

${\zeta }_2=-2.2794$

${\zeta }_3=-1.9637$

As a consequence — see an accompanying discussion (alternatively, ChallengesPt6) for details — the values of other parameters are …

					Example Values
$\tan \theta$	$=$	$-\frac{{\zeta }_2}{{\zeta }_3}\left[\frac{a^2+b^2}{a^2+c^2}\right]\frac{c^2}{b^2}=-0.344793$			$\theta =$	$-19.0238^{\circ }$
$\mathrm{\Lambda }$	$\equiv$	$\left[\frac{a^2}{a^2+b^2}\right]{\zeta }_3\cos \theta -\left[\frac{a^2}{a^2+c^2}\right]{\zeta }_2\sin \theta$			$\mathrm{\Lambda }=$	$-1.332892$
$\frac{y_0}{z_0}$	$=$	$\left[\frac{a^2}{a^2+c^2}\right]\frac{{\zeta }_2}{\mathrm{\Lambda }}=-\frac{b^2\sin \theta }{(c^2{\cos }^2\theta +b^2{\sin }^2\theta )}$			$\frac{y_0}{z_0}=$	$+1.400377$
$\frac{x_{\mathrm{m}\mathrm{a}\mathrm{x}}}{y_{\mathrm{m}\mathrm{a}\mathrm{x}}}$	$=$	$\left\{\mathrm{\Lambda }\right[\frac{a^2+b^2}{b^2}]\frac{\cos \theta }{{\zeta }_3}{\}}^{1/2}$			$\frac{x_{\mathrm{m}\mathrm{a}\mathrm{x}}}{y_{\mathrm{m}\mathrm{a}\mathrm{x}}}=$	$+1.025854$
	$=$	$\frac{a(c^2{\cos }^2\theta +b^2{\sin }^2\theta {)}^{1/2}}{bc}$
$|\dot{\phi }|$	$=$	$\left\{\mathrm{\Lambda }\right[\frac{b^2}{a^2+b^2}]\frac{{\zeta }_3}{\cos \theta }{\}}^{1/2}$   Correct!			$|\dot{\phi }|=$	$1.299300$   Correct!
$\dot{\phi }$	$=$	$\frac{{\zeta }_3}{\cos \theta }\left[\frac{abc(c^2{\cos }^2\theta +b^2{\sin }^2\theta {)}^{1/2}}{c^2(a^2+b^2)}\right]$			$\dot{\phi }=$	$-1.299300$

EFE Rotating Cartesian Frame

Concentric triaxial ellipsoids are defined by the expression,

$P$

$=$

$(\frac{x}{a}{)}^2+(\frac{y}{b}{)}^2+(\frac{z}{c}{)}^2,$

where $0\le P\le 1$ is a constant. As viewed from the rotating reference frame, the velocity flow-field everywhere inside $(0\le P<1)$, and on the surface $(P=1)$ of the Type I Riemann ellipsoid is given by the expression — see, for example, an accompanying discussion of the Riemann flow-field,

${\mathbf{𝐮}}_{\mathrm{E}\mathrm{F}\mathrm{E}}$

$=$

$\hat{\mathit{ı}}\{-[\frac{a^2}{a^2+b^2}]{\zeta }_{3}y+[\frac{a^2}{a^2+c^2}\left]{\zeta }_{2}z\right\}+\hat{\mathit{ȷ}}\{+[\frac{b^2}{a^2+b^2}\left]{\zeta }_{3}x\right\}+\hat{\mathbf{k}}\{-[\frac{c^2}{a^2+c^2}\left]{\zeta }_{2}x\right\}.$

In an accompanying discussion, we have shown that,

${\mathbf{𝐮}}_{\mathrm{E}\mathrm{F}\mathrm{E}}\cdot \mathrm{\nabla }P$

$=$

$0,$

which means that, at every location inside and on the surface of the configuration, the velocity vector is orthogonal to a vector that is normal to the ellipsoidal surface at that location.

Tilted Coordinate System

Figure 1:   Tilted Reference Frame
$\hat{\mathit{ı}}$ $\to$ ${\hat{\mathit{ı}}}^{\prime },$ $\hat{\mathit{ȷ}}$ $\to$ ${\hat{\mathit{ȷ}}}^{\prime }\cos \theta -{\hat{\mathit{k}}}^{\prime }\sin \theta ,$ $\hat{\mathit{k}}$ $\to$ ${\hat{\mathit{ȷ}}}^{\prime }\sin \theta +{\hat{\mathit{k}}}^{\prime }\cos \theta .$		$x$ $\to$ $x^{\prime },$ $y$ $\to$ $y^{\prime }\cos \theta -z^{\prime }\sin \theta ,$ $z-z_0$ $\to$ $y^{\prime }\sin \theta +z^{\prime }\cos \theta .$

As we have detailed in our accompanying discussion, as viewed from this "tipped" frame, the concentric ellipsoidal surfaces of a Type I Riemann ellipsoid are defined by the expression,

$P^{\prime }$

$=$

$[\frac{y^{\prime }\cos \theta -z^{\prime }\sin \theta }{b}{]}^2+[\frac{z_0+z^{\prime }\cos \theta +y^{\prime }\sin \theta }{c}{]}^2+(\frac{x^{\prime }}{a}{)}^2.$

and the velocity flow-field is given by the expression,

${{\mathit{u}}^{\prime }}_{\mathrm{E}\mathrm{F}\mathrm{E}}$	$=$	${\hat{\mathit{ı}}}^{\prime }\{-[\frac{a^2}{a^2+b^2}]{\zeta }_3(y^{\prime }\cos \theta -z^{\prime }\sin \theta )+[\frac{a^2}{a^2+c^2}]{\zeta }_2(z_0+y^{\prime }\sin \theta +z^{\prime }\cos \theta )\}$
		$+[{\hat{\mathit{ȷ}}}^{\prime }\cos \theta -{\hat{\mathbf{k}}}^{\prime }\sin \theta ]\left\{\right[\frac{b^2}{a^2+b^2}\left]{\zeta }_3{x}^{\prime }\right\}+[{\hat{\mathit{ȷ}}}^{\prime }\sin \theta +{\hat{\mathbf{k}}}^{\prime }\cos \theta ]\{-[\frac{c^2}{a^2+c^2}\left]{\zeta }_2{x}^{\prime }\right\}.$

We also have explicitly demonstrated that, for any arbitrarily chosen value of the tilt angle, $\theta$,

${{\mathbf{𝐮}}^{\prime }}_{\mathrm{E}\mathrm{F}\mathrm{E}}\cdot \mathrm{\nabla }{P}^{\prime }$

$=$

$0.$

Preferred Tilt

As we discuss elsewhere, if we specifically choose,

$\tan \theta$

$=$

$-\frac{\beta {\mathrm{\Omega }}_2}{\gamma {\mathrm{\Omega }}_3}=-\left[\frac{c^2(a^2+b^2)}{b^2(a^2+c^2)}\right]\frac{{\zeta }_2}{{\zeta }_3}.$

the component of the flow-field in the ${\hat{\mathbf{k}}}^{\prime }$ direction vanishes; that is, in this specific case, as viewed from the tilted reference frame, all of the fluid motion is confined to the x'-y' plane. Notice that this plane is not parallel to any of the three principal planes of the Type I Riemann ellipsoid. I have not seen this fluid-flow behavior previously described in the published literature. Maybe Norman Lebovitz will know.

The three panels of Figure 2, and the text description that follows, have been drawn from a separate discussion.

Figure 2a	Figure 2b
↲
Figure 2c
↲

As has been described in an accompanying discussion of Riemann Type 1 ellipsoids, we have used COLLADA to construct an animated and interactive 3D scene that displays in purple the surface of an example Type I ellipsoid; panels a and b of Figure 2 show what this ellipsoid looks like when viewed from two different perspectives. (As a reminder — see the explanation accompanying Figure 2 of that accompanying discussion — the ellipsoid is tilted about the x-coordinate axis at an angle of 61.25° to the equilibrium spin axis, which is shown in green.) Yellow markers also have been placed in this 3D scene at each of the coordinate locations specified in the table that accompanies that discussion. From the perspective presented in Figure 2b, we can immediately identify three separate, nearly circular trajectories; the largest one corresponds to our choice of z₀ = -0.25, the smallest corresponds to our choice of z₀ = -0.60, and the one of intermediate size correspond to our choice of z₀ = -0.4310. When viewed from the perspective presented in Figure 2a, we see that these three trajectories define three separate planes; each plane is tipped at an angle of θ = -19.02° to the untilted equatorial, x-y plane of the purple ellipsoid.

Lagrangian Fluid Trajectories

Off-Center Ellipse

The yellow dots in Figures 2a and 2b trace three different, nearly circular, closed curves. These curves each show what results from the intersection of the surface of the triaxial ellipsoid and a plane that is tilted with respect to the x = x' axis by the specially chosen angle, $\theta$; the different curves result from different choices of the intersection point, $z_0$. Several additional such curves are displayed in Figure 2c. Each of these curves necessarily also identifies the trajectory that is followed by a fluid element that sits on the surface of the ellipsoid.

We have determined that the $y^{\prime }(x^{\prime })$ function that defines each closed curve is describable analytically by the expression,

$[\frac{y^{\prime }-y{\text{'}}_0}{y{\text{'}}_{\mathrm{m}\mathrm{a}\mathrm{x}}}{]}^2$

$=$

$1-(\frac{x^{\prime }}{x{\text{'}}_{\mathrm{m}\mathrm{a}\mathrm{x}}}{)}^2,$

where (see independent derivations with identical results from ChallengesPt2 and ChallengesPt6),

$x{\text{'}}_{\mathrm{m}\mathrm{a}\mathrm{x}}$	$=$	$a[1-\frac{z_0^2{\cos }^2\theta }{(c^2{\cos }^2\theta +b^2{\sin }^2\theta )}{]}^{1/2}$
	$=$	$a[\frac{(c^2-z_0^2){\cos }^2\theta +b^2{\sin }^2\theta }{(c^2{\cos }^2\theta +b^2{\sin }^2\theta )}{]}^{1/2},$
$y{\text{'}}_{\mathrm{m}\mathrm{a}\mathrm{x}}$	$=$	$bc[(c^2{\cos }^2\theta +b^2{\sin }^2\theta )-z_0^2{\cos }^2\theta {]}^{1/2}(c^2{\cos }^2\theta +b^2{\sin }^2\theta {)}^{-1}$
	$=$	$bc[\frac{(c^2-z_0^2){\cos }^2\theta +b^2{\sin }^2\theta }{(c^2{\cos }^2\theta +b^2{\sin }^2\theta {)}^2}{]}^{1/2},$
$y{\text{'}}_0$	$=$	$-\frac{z_0{b}^2\sin \theta }{(c^2{\cos }^2\theta +b^2{\sin }^2\theta )}.$

This is the equation that describes a closed ellipse with semi-axes, $(x{\text{'}}_{\mathrm{m}\mathrm{a}\mathrm{x}},y{\text{'}}_{\mathrm{m}\mathrm{a}\mathrm{x}})$, that is offset from the z'-axis along the y'-axis by a distance, $y{\text{'}}_0$. Notice that the degree of flattening,

$[\frac{x^{\prime }}{y^{\prime }}{]}_{\mathrm{m}\mathrm{a}\mathrm{x}}$

$=$

$\frac{a(c^2{\cos }^2\theta +b^2{\sin }^2\theta {)}^{1/2}}{bc},$

is independent of $z_0$; that is to say, the degree of flattening of all of the elliptical trajectories is identical! Notice, as well, that the y-offset, $y{\text{'}}_0$, is linearly proportional to $z_0$.

In a separate discussion, we have demonstrated that the compact version of the tilted flow-field is everywhere orthogonal to the elliptical trajectory whose analytic definition is given by the off-set ellipse equation.

Associated Lagrangian Velocities

Let's presume that, as a function of time, the x'-y' coordinates and associated velocity components of each Lagrangian fluid element are given by the expressions,

$x^{\prime }$	$=$	$x_{\mathrm{m}\mathrm{a}\mathrm{x}}\cos (\dot{\phi }t)$	and,	$y^{\prime }-y_0$	$=$	$y_{\mathrm{m}\mathrm{a}\mathrm{x}}\sin (\dot{\phi }t),$
${\dot{x}}^{\prime }$	$=$	$-x_{\mathrm{m}\mathrm{a}\mathrm{x}}\dot{\phi }\cdot \sin (\dot{\phi }t)=(y_0-y^{\prime })\left[\frac{x_{\mathrm{m}\mathrm{a}\mathrm{x}}}{y_{\mathrm{m}\mathrm{a}\mathrm{x}}}\right]\dot{\phi }$	and,	${\dot{y}}^{\prime }$	$=$	$y_{\mathrm{m}\mathrm{a}\mathrm{x}}\dot{\phi }\cdot \cos (\dot{\phi }t)=x^{\prime }\left[\frac{y_{\mathrm{m}\mathrm{a}\mathrm{x}}}{x_{\mathrm{m}\mathrm{a}\mathrm{x}}}\right]\dot{\phi }.$

If this is the correct description of the Lagrangian motion in a $z^{\prime }=0$ plane of motion, then the velocity components, ${\dot{x}}^{\prime }$ and ${\dot{y}}^{\prime }$, must match the respective components of the Riemann flow-field, namely,

${{\mathit{u}}^{\prime }}_{\mathrm{E}\mathrm{F}\mathrm{E}}{|}_{z^{\prime }=0}$

$=$

${\hat{\mathit{ı}}}^{\prime }\left\{\right[\frac{a^2}{a^2+c^2}\left]{\zeta }_2{z}_0\right\}+{\hat{\mathit{ı}}}^{\prime }\{{\xcancel{z^{\prime }(c^2-b^2)\tan \theta }}^0-y^{\prime }[c^2+b^2{\tan }^2\theta ]\}\frac{{\zeta }_3\cos \theta }{c^2}\left[\frac{a^2}{a^2+b^2}\right]+{\hat{\mathit{ȷ}}}^{\prime }\left[\frac{b^2}{a^2+b^2}\right]\frac{{\zeta }_3{x}^{\prime }}{\cos \theta }.$

First, let's compare the ${\hat{\mathit{ȷ}}}^{\prime }$ components.

$x^{\prime }[\frac{y^{\prime }}{x^{\prime }}{]}_{\mathrm{m}\mathrm{a}\mathrm{x}}\dot{\phi }$	$\leftrightarrow$	$x^{\prime }\left[\frac{b^2}{a^2+b^2}\right]\frac{{\zeta }_3}{\cos \theta }$
$\Rightarrow \dot{\phi }$	$=$	$\left[\frac{b^2}{a^2+b^2}\right]\frac{{\zeta }_3}{\cos \theta }[\frac{x^{\prime }}{y^{\prime }}{]}_{\mathrm{m}\mathrm{a}\mathrm{x}}$
	$=$	$\left[\frac{b^2}{a^2+b^2}\right]\frac{{\zeta }_3}{\cos \theta }\left[\frac{a(c^2{\cos }^2\theta +b^2{\sin }^2\theta {)}^{1/2}}{bc}\right]$
	$=$	$\frac{{\zeta }_3}{\cos \theta }\left[\frac{abc(c^2{\cos }^2\theta +b^2{\sin }^2\theta {)}^{1/2}}{c^2(a^2+b^2)}\right].$

Now let's compare the ${\hat{\mathit{ı}}}^{\prime }$ components.

$(y_0-y^{\prime })[\frac{x^{\prime }}{y^{\prime }}{]}_{\mathrm{m}\mathrm{a}\mathrm{x}}\dot{\phi }$	$\leftrightarrow$	$\left[\frac{a^2}{a^2+c^2}\right]{\zeta }_2{z}_0-y^{\prime }[c^2+b^2{\tan }^2\theta ]\frac{{\zeta }_3\cos \theta }{c^2}\left[\frac{a^2}{a^2+b^2}\right]$
$\Rightarrow (y_0-y^{\prime })\dot{\phi }$	$=$	$\left\{\right[\frac{a^2}{a^2+c^2}]{\zeta }_2{z}_0-y^{\prime }[c^2+b^2{\tan }^2\theta ]\frac{{\zeta }_3\cos \theta }{c^2}[\frac{a^2}{a^2+b^2}\left]\right\}[\frac{y^{\prime }}{x^{\prime }}{]}_{\mathrm{m}\mathrm{a}\mathrm{x}}$
	$=$	$\left\{\right[\frac{a^2}{a^2+c^2}]{\zeta }_2{z}_0-y^{\prime }[c^2+b^2{\tan }^2\theta \left]\frac{{\zeta }_3\cos \theta }{c^2}\right[\frac{a^2}{a^2+b^2}\left]\right\}\left[\frac{bc}{a(c^2{\cos }^2\theta +b^2{\sin }^2\theta {)}^{1/2}}\right].$

Inserting the just-derived expression for $\dot{\phi }$ into this last expression gives,

$(y_0-y^{\prime })$	$=$	$\left\{\right[\frac{a^2}{a^2+c^2}]{\zeta }_2{z}_0-y^{\prime }[c^2+b^2{\tan }^2\theta \left]\frac{{\zeta }_3\cos \theta }{c^2}\right[\frac{a^2}{a^2+b^2}\left]\right\}\left[\frac{bc}{a(c^2{\cos }^2\theta +b^2{\sin }^2\theta {)}^{1/2}}\right]\left[\frac{a^2+b^2}{b^2}\right]\frac{\cos \theta }{{\zeta }_3}\left[\frac{bc}{a(c^2{\cos }^2\theta +b^2{\sin }^2\theta {)}^{1/2}}\right]$
	$=$	$\left\{\right[\frac{a^2}{a^2+c^2}]{\zeta }_2{z}_0-y^{\prime }[c^2+b^2{\tan }^2\theta \left]\frac{{\zeta }_3\cos \theta }{c^2}\right[\frac{a^2}{a^2+b^2}\left]\right\}\frac{\cos \theta }{{\zeta }_3}\left[\frac{c^2(a^2+b^2)}{a^2(c^2{\cos }^2\theta +b^2{\sin }^2\theta )}\right]$
	$=$	$\left\{\right[\frac{a^2}{a^2+c^2}\left]{\zeta }_2{z}_0\right\}\frac{\cos \theta }{{\zeta }_3}\left[\frac{c^2(a^2+b^2)}{a^2(c^2{\cos }^2\theta +b^2{\sin }^2\theta )}\right]+\{-y^{\prime }[c^2+b^2{\tan }^2\theta \left]\frac{{\zeta }_3\cos \theta }{c^2}\right[\frac{a^2}{a^2+b^2}\left]\right\}\frac{\cos \theta }{{\zeta }_3}\left[\frac{c^2(a^2+b^2)}{a^2(c^2{\cos }^2\theta +b^2{\sin }^2\theta )}\right]$
	$=$	$\left\{\right[\frac{c^2(a^2+b^2)}{b^2(a^2+c^2)}\left]\frac{{\zeta }_2}{{\zeta }_3}\right\}\left[\frac{b^2{z}_0\cos \theta }{(c^2{\cos }^2\theta +b^2{\sin }^2\theta )}\right]-y^{\prime }.$

But,

$\tan \theta$

$=$

$-\left[\frac{c^2(a^2+b^2)}{b^2(a^2+c^2)}\right]\frac{{\zeta }_2}{{\zeta }_3}.$

Hence,

$(y_0-y^{\prime })$	$=$	$\left[\frac{-b^2{z}_0\sin \theta }{(c^2{\cos }^2\theta +b^2{\sin }^2\theta )}\right]-y^{\prime }.$
$\Rightarrow y_0$	$=$	$\left[\frac{-b^2{z}_0\sin \theta }{(c^2{\cos }^2\theta +b^2{\sin }^2\theta )}\right].$

SUCCESS !!!

See Also

• Description of Riemann Type I Ellipsoids (best)
• Description of Riemann Type I Ellipsoids (older introduction)
• Riemann Type 1 Ellipsoids (oldest introduction)
• Construction Challenges (Pt. 1)
• Construction Challenges (Pt. 2)
• Construction Challenges (Pt. 3)
• Construction Challenges (Pt. 4)
• Construction Challenges (Pt. 5)
• Construction Challenges (Pt. 6)

• Related discussions of models viewed from a rotating reference frame:
  • PGE
  • NOTE to Eric Hirschmann & David Neilsen... I have moved the earlier contents of this page to a new Wiki location called Compressible Riemann Ellipsoids.