VE/RiemannEllipsoids

Steady-State 2^nd-Order Tensor Virial Equations

Steady-State 2nd-Order Tensor Virial Equations

Drawing from our accompanying discussion of virial equations as viewed from a rotating frame of reference, here we employ the 2^nd-order tensor virial equation (TVE),

$0$

$=$

$2{\mathfrak{𝔗}}_{ij}+{\mathfrak{𝔚}}_{ij}+{\delta }_{ij}\mathrm{\Pi }+{\mathrm{\Omega }}^2{I}_{ij}-{\mathrm{\Omega }}_i{\mathrm{\Omega }}_k{I}_{kj}+2{ϵ}_{ilm}{\mathrm{\Omega }}_m\underset{V}{\int }\rho u_l{x}_{j}dx,$

to determine the equilibrium conditions of uniform-density $(\rho )$ ellipsoids that have semi-axes, $(a_1,a_2,a_3)\leftrightarrow (a,b,c),$ and an internal velocity field, $\overrightarrow{u}$ (as prescribed below), that preserves this specified ellipsoidal shape, as viewed from a frame of reference that is rotating with angular velocity, $\overrightarrow{\mathrm{\Omega }}$. Because each of the indices, $i$ and $j$, run from 1 to 3, inclusive, this TVE appears to provide nine equilibrium constraints; and once the values of the density and the three semi-axes are specified, there appear to be seven unknowns: $\mathrm{\Pi }$ and the three pairs of velocity-field components $({\mathrm{\Omega }}_1,{\zeta }_1)$, $({\mathrm{\Omega }}_2,{\zeta }_2)$, $({\mathrm{\Omega }}_3,{\zeta }_3).$ In practice, however, only five constraints are relevant/independent because, as is encapsulated in …

Riemann's Fundamental Theorem … non-trivial solutions are obtained only if no more than two of the three pairs of velocity-field components are different from zero.

Following EFE, we will set ${\mathrm{\Omega }}_1={\zeta }_1=0$, in which case the only applicable TVE constraint relations are the five identified in the following table of equations.

Indices		Each Associated 2nd-Order TVE Expression
$i$	$j$
$1$	$1$	$0$ $=$ $\left[\frac{3\cdot 5}{2^2\pi abc\rho }\right]\mathrm{\Pi }+\{({\mathrm{\Omega }}_2^2+{\mathrm{\Omega }}_3^2)+2[\frac{b^2}{b^2+a^2}]{\mathrm{\Omega }}_3{\zeta }_3+2[\frac{c^2}{c^2+a^2}]{\mathrm{\Omega }}_2{\zeta }_2-(2\pi G\rho )A_1\}{a}^2+[\frac{a^2}{a^2+b^2}{]}^2{\zeta }_3^2{b}^2+[\frac{a^2}{a^2+c^2}{]}^2{\zeta }_2^2{c}^2$
$2$	$2$	$0$ $=$ $\left[\frac{3\cdot 5}{2^2\pi abc\rho }\right]\mathrm{\Pi }+[\frac{b^2}{b^2+a^2}{]}^2{\zeta }_3^2{a}^2+\{{\mathrm{\Omega }}_3^2+2\left[\frac{a^2}{a^2+b^2}\right]{\mathrm{\Omega }}_3{\zeta }_3-(2\pi G\rho )A_2\}{b}^2$
$3$	$3$	$0$ $=$ $\left[\frac{3\cdot 5}{2^2\pi abc\rho }\right]\mathrm{\Pi }+[\frac{c^2}{c^2+a^2}{]}^2{\zeta }_2^2{a}^2+\{{\mathrm{\Omega }}_2^2+2\left[\frac{a^2}{a^2+c^2}\right]{\mathrm{\Omega }}_2{\zeta }_2-(2\pi G\rho )A_3\}{c}^2$
$2$	$3$	$0$ $=$ $\{1+\frac{{\zeta }_2}{{\mathrm{\Omega }}_2}[\frac{a^2}{a^2+c^2}\left]\right[2+\frac{{\zeta }_3}{{\mathrm{\Omega }}_3}\left(\frac{b^2}{b^2+a^2}\right)\left]\right\}{\mathrm{\Omega }}_2{\mathrm{\Omega }}_3{c}^2$
$3$	$2$	$0$ $=$ $\{1+\frac{{\zeta }_3}{{\mathrm{\Omega }}_3}[\frac{a^2}{a^2+b^2}\left]\right[2+\frac{{\zeta }_2}{{\mathrm{\Omega }}_2}\left(\frac{c^2}{c^2+a^2}\right)\left]\right\}{\mathrm{\Omega }}_2{\mathrm{\Omega }}_3{b}^2$

General Coefficient Expressions

In the context of our discussion of configurations that are triaxial ellipsoids, we begin by adopting the $(\ell ,m,s)$ subscript notation to identify which semi-axis length is the (largest, medium-length, smallest). As has been detailed in an accompanying chapter, the gravitational potential anywhere inside or on the surface of an homogeneous ellipsoid may be given analytically in terms of the following three coefficient expressions:

$\frac{A_{\ell }}{a_{\ell }{a}_m{a}_s}$	$=$	$\frac{2}{a_{\ell }^3}\left[\frac{F(\theta ,k)-E(\theta ,k)}{k^2{\sin }^3\theta }\right],$
$\frac{A_s}{a_{\ell }{a}_m{a}_s}$	$=$	$\frac{2}{a_{\ell }^3}\left[\frac{(a_m/a_s)\sin \theta -E(\theta ,k)}{(1-k^2){\sin }^3\theta }\right],$
$\frac{A_m}{a_{\ell }{a}_m{a}_s}=\frac{2-(A_{\ell }+A_s)}{a_{\ell }{a}_m{a}_s}$	$=$	$\frac{2}{a_{\ell }^3}\left[\frac{E(\theta ,k)-(1-k^2)F(\theta ,k)-(a_s/a_m)k^2\sin \theta }{k^2(1-k^2){\sin }^3\theta }\right],$

where, $F(\theta ,k)$ and $E(\theta ,k)$ are incomplete elliptic integrals of the first and second kind, respectively, with arguments,

$\theta ={\cos }^{-1}\left(\frac{a_s}{a_{\ell }}\right)$

and

$k=[\frac{1-(a_m/a_{\ell }{)}^2}{1-(a_s/a_{\ell }{)}^2}{]}^{1/2}.$

Specific Case of a₁ > a₂ > a₃

When we discuss configurations in which $a_1>a_2>a_3>0$ — such as Jacobi, Dedekind, or most Riemann S-Type ellipsoids — we must adopt the associations, $(A_1,a_1)\leftrightarrow (A_{\ell },a_{\ell })$, $(A_2,a_2)\leftrightarrow (A_m,a_m)$, 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,

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

where, the arguments of the incomplete elliptic integrals are,

$\theta ={\cos }^{-1}\left(\frac{a_3}{a_1}\right)$	and	$k=[\frac{1-(a_2/a_1{)}^2}{1-(a_3/a_1{)}^2}{]}^{1/2}.$
[ EFE, Chapter 3, §17, Eq. (32) ]

Specific Case of a₁ > a₃ > a₂

When we discuss configurations in which $a_1>a_3>a_2>0$ — these are usually referred to in EFE as prolate S-Type Riemann ellipsoids — we must instead adopt the associations, $(A_1,a_1)\leftrightarrow (A_{\ell },a_{\ell })$, $(A_2,a_2)\leftrightarrow (A_s,a_s)$, and $(A_3,a_3)\leftrightarrow (A_m,a_m)$. This means that the coefficients, $A_1$, $A_2$, and $A_3$ are defined by the expressions,

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

where, the arguments of the incomplete elliptic integrals of the first and second kind are,

$\theta ={\cos }^{-1}\left(\frac{a_2}{a_1}\right)$	and	$k=[\frac{1-(a_3/a_1{)}^2}{1-(a_2/a_1{)}^2}{]}^{1/2}.$
[ EFE, Chapter 7, §48d, footnote to Table VII (p. 143) ]

NOTE: All irrotational ellipsoids belong to this category of configurations.

Specific Case of a₂ > a₁ > a₃

When we discuss configurations in which $a_2>a_1>a_3>0$ — for example, most Riemann ellipsoids of Types I, II, & III — we must instead 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

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

Oblate Spheroids [a₂ = a₁ > a₃]

Starting with the case of $a_2>a_1>a_3>0$ and setting $a_2=a_1$, we recognize, first, that $k=0$. Hence, we have,

$A_3$

$=$

$2\left[\frac{\sin \theta -(a_3/a_1)E(\theta ,0)}{{\sin }^3\theta }\right],$

Adopted (Internal) Velocity Field

EFE (p. 130) states that the … kinematical requirement, that the motion $(\overrightarrow{u})$, associated with $\overrightarrow{\zeta }$, preserves the ellipsoidal boundary, leads to the following expressions for its components:

$u_1$	$=$	$-\left[\frac{a_1^2}{a_1^2+a_2^2}\right]{\zeta }_3{x}_2+\left[\frac{a_1^2}{a_1^2+a_3^2}\right]{\zeta }_2{x}_3,$
$u_2$	$=$	$-\left[\frac{a_2^2}{a_2^2+a_3^2}\right]{\zeta }_1{x}_3+\left[\frac{a_2^2}{a_2^2+a_1^2}\right]{\zeta }_3{x}_1,$
$u_3$	$=$	$-\left[\frac{a_3^2}{a_3^2+a_1^2}\right]{\zeta }_2{x}_1+\left[\frac{a_3^2}{a_3^2+a_2^2}\right]{\zeta }_1{x}_2.$
[ EFE, Chapter 7, §47, Eq. (1) ]

Equilibrium Expressions

[EFE §11(b), p. 22] Under conditions of a stationary state, [the tensor virial equation] gives,

$2{\mathfrak{𝔗}}_{ij}+{\mathfrak{𝔚}}_{ij}$

$=$

$-{\delta }_{ij}\mathrm{\Pi }.$

[This] provides six integral relations which must obtain whenever the conditions are stationary.

When viewing the (generally ellipsoidal) configuration from a rotating frame of reference, the 2^nd-order TVE takes on the more general form:

$0$	$=$	$2{\mathfrak{𝔗}}_{ij}+{\mathfrak{𝔚}}_{ij}+{\delta }_{ij}\mathrm{\Pi }+{\mathrm{\Omega }}^2{I}_{ij}-{\mathrm{\Omega }}_i{\mathrm{\Omega }}_k{I}_{kj}+2{ϵ}_{ilm}{\mathrm{\Omega }}_m\underset{V}{\int }\rho u_l{x}_{j}dx.$
[ EFE, Chapter 2, §12, Eq. (64) ]

EFE (p. 57) also shows that … The potential energy tensor … for a homogeneous ellipsoid is given by

$\frac{{\mathfrak{𝔚}}_{ij}}{\pi G\rho }$	$=$	$-2A_i{I}_{ij},$
[ EFE, Chapter 3, §22, Eq. (128) ]

where

$I_{ij}$	$=$	$\frac{1}{5}M{a}_i^2{\delta }_{ij},$
[ EFE, Chapter 3, §22, Eq. (129) ]

is the moment of inertia tensor. Expressions for all nine components of the kinetic energy tensor, ${\mathfrak{𝔗}}_{ij}$ are derived in Appendix E, below; and expressions for each of the six Coriolis components can be found in Appendices B, C, & D.

The Three Diagonal Elements

For $i=j=1$, we have,

$0$	$=$	$2{\mathfrak{𝔗}}_{11}+{\mathfrak{𝔚}}_{11}+\mathrm{\Pi }+{\mathrm{\Omega }}^2{I}_{11}-{\mathrm{\Omega }}_1{\mathrm{\Omega }}_k{I}_{k1}+2{ϵ}_{1lm}{\mathrm{\Omega }}_m\underset{V}{\int }\rho u_l{x}_1{d}^{3}x$
	$=$	$2{\mathfrak{𝔗}}_{11}+{\mathfrak{𝔚}}_{11}+\mathrm{\Pi }+{\mathrm{\Omega }}^2{I}_{11}-{\mathrm{\Omega }}_1^2{I}_{11}+2{\mathrm{\Omega }}_3\underset{V}{\int }\rho u_2{x}_1{d}^{3}x-2{\mathrm{\Omega }}_2\underset{V}{\int }\rho u_3{x}_1{d}^{3}x$
	$=$	$2{\mathfrak{𝔗}}_{11}+{\mathfrak{𝔚}}_{11}+\mathrm{\Pi }+({\mathrm{\Omega }}_2^2+{\mathrm{\Omega }}_3^2)I_{11}+2{\mathrm{\Omega }}_3\rho \underset{V}{\int }{u}_{2}x{d}^{3}x-2{\mathrm{\Omega }}_2\rho \underset{V}{\int }{u}_{3}x{d}^{3}x$
	$=$	$[\frac{a^2}{a^2+b^2}{]}^2{\zeta }_3^2{I}_{22}+[\frac{a^2}{a^2+c^2}{]}^2{\zeta }_2^2{I}_{33}-(2\pi G\rho )A_1{I}_{11}+\mathrm{\Pi }+({\mathrm{\Omega }}_2^2+{\mathrm{\Omega }}_3^2)I_{11}+2\left[\frac{b^2}{b^2+a^2}\right]{\mathrm{\Omega }}_3{\zeta }_3{I}_{11}+2\left[\frac{c^2}{c^2+a^2}\right]{\mathrm{\Omega }}_2{\zeta }_2{I}_{11}$
	$=$	$\mathrm{\Pi }+\{({\mathrm{\Omega }}_2^2+{\mathrm{\Omega }}_3^2)+2[\frac{b^2}{b^2+a^2}]{\mathrm{\Omega }}_3{\zeta }_3+2[\frac{c^2}{c^2+a^2}]{\mathrm{\Omega }}_2{\zeta }_2-(2\pi G\rho )A_1\}{I}_{11}+[\frac{a^2}{a^2+b^2}{]}^2{\zeta }_3^2{I}_{22}+[\frac{a^2}{a^2+c^2}{]}^2{\zeta }_2^2{I}_{33}$
$\Rightarrow -\left[\frac{3\cdot 5}{2^2\pi abc\rho }\right]\mathrm{\Pi }$	$=$	$\{({\mathrm{\Omega }}_2^2+{\mathrm{\Omega }}_3^2)+2[\frac{b^2}{b^2+a^2}]{\mathrm{\Omega }}_3{\zeta }_3+2[\frac{c^2}{c^2+a^2}]{\mathrm{\Omega }}_2{\zeta }_2-(2\pi G\rho )A_1\}{a}^2+[\frac{a^2}{a^2+b^2}{]}^2{\zeta }_3^2{b}^2+[\frac{a^2}{a^2+c^2}{]}^2{\zeta }_2^2{c}^2.$

Once we choose the values of the (semi) axis lengths $(a,b,c)$ of an ellipsoid — from which the value of $A_1$ can be immediately determined — along with a specification of $\rho$, this equation has the following five unknowns: $\mathrm{\Pi },{\mathrm{\Omega }}_2,{\mathrm{\Omega }}_3,{\zeta }_2,{\zeta }_3$. Similarly, for $i=j=2$,

$0$	$=$	$2{\mathfrak{𝔗}}_{22}+{\mathfrak{𝔚}}_{22}+\mathrm{\Pi }+{\mathrm{\Omega }}^2{I}_{22}-{\mathrm{\Omega }}_2{\mathrm{\Omega }}_k{I}_{k2}+2{ϵ}_{2lm}{\mathrm{\Omega }}_m\underset{V}{\int }\rho u_l{x}_2{d}^{3}x$
	$=$	$2{\mathfrak{𝔗}}_{22}+{\mathfrak{𝔚}}_{22}+\mathrm{\Pi }+({\mathrm{\Omega }}_1^2+{\mathrm{\Omega }}_3^2)I_{22}+2{\mathrm{\Omega }}_1\rho \underset{V}{\int }{u}_{3}y{d}^{3}x-2{\mathrm{\Omega }}_3\rho \underset{V}{\int }{u}_{1}y{d}^{3}x$
	$=$	$[\frac{b^2}{b^2+c^2}{]}^2{\zeta }_1^2{I}_{33}+[\frac{b^2}{b^2+a^2}{]}^2{\zeta }_3^2{I}_{11}-(2\pi G\rho )A_2{I}_{22}+\mathrm{\Pi }+({\mathrm{\Omega }}_1^2+{\mathrm{\Omega }}_3^2)I_{22}+2\left[\frac{c^2}{c^2+b^2}\right]{\mathrm{\Omega }}_1{\zeta }_1{I}_{22}+2\left[\frac{a^2}{a^2+b^2}\right]{\mathrm{\Omega }}_3{\zeta }_3{I}_{22}$
	$=$	$\mathrm{\Pi }+[\frac{b^2}{b^2+a^2}{]}^2{\zeta }_3^2{I}_{11}+\{({\mathrm{\Omega }}_1^2+{\mathrm{\Omega }}_3^2)+2\left[\frac{c^2}{c^2+b^2}\right]{\mathrm{\Omega }}_1{\zeta }_1+2\left[\frac{a^2}{a^2+b^2}\right]{\mathrm{\Omega }}_3{\zeta }_3-(2\pi G\rho )A_2\}{I}_{22}+[\frac{b^2}{b^2+c^2}{]}^2{\zeta }_1^2{I}_{33}$
$\Rightarrow -\left[\frac{3\cdot 5}{2^2\pi abc\rho }\right]\mathrm{\Pi }$	$=$	$[\frac{b^2}{b^2+a^2}{]}^2{\zeta }_3^2{a}^2+\{({\mathrm{\Omega }}_1^2+{\mathrm{\Omega }}_3^2)+2\left[\frac{c^2}{c^2+b^2}\right]{\mathrm{\Omega }}_1{\zeta }_1+2\left[\frac{a^2}{a^2+b^2}\right]{\mathrm{\Omega }}_3{\zeta }_3-(2\pi G\rho )A_2\}{b}^2+[\frac{b^2}{b^2+c^2}{]}^2{\zeta }_1^2{c}^2.$

This gives us a second equation, but an additional pair of (for a total of seven) unknowns: ${\mathrm{\Omega }}_1,{\zeta }_1$. For the third diagonal element — that is, for $i=j=3$ — we have,

$0$	$=$	$2{\mathfrak{𝔗}}_{33}+{\mathfrak{𝔚}}_{33}+\mathrm{\Pi }+{\mathrm{\Omega }}^2{I}_{33}-{\mathrm{\Omega }}_3{\mathrm{\Omega }}_k{I}_{k3}+2{ϵ}_{3lm}{\mathrm{\Omega }}_m\underset{V}{\int }\rho u_l{x}_3{d}^{3}x$
	$=$	$2{\mathfrak{𝔗}}_{33}+{\mathfrak{𝔚}}_{33}+\mathrm{\Pi }+({\mathrm{\Omega }}_1^2+{\mathrm{\Omega }}_2^2)I_{33}+2{\mathrm{\Omega }}_2\rho \underset{V}{\int }{u}_{1}z{d}^{3}x-2{\mathrm{\Omega }}_1\rho \underset{V}{\int }{u}_{2}z{d}^{3}x$
	$=$	$[\frac{c^2}{c^2+a^2}{]}^2{\zeta }_2^2{I}_{11}+[\frac{c^2}{c^2+b^2}{]}^2{\zeta }_1^2{I}_{22}-(2\pi G\rho )A_3{I}_{33}+\mathrm{\Pi }+({\mathrm{\Omega }}_1^2+{\mathrm{\Omega }}_2^2)I_{33}+2\left[\frac{a^2}{a^2+c^2}\right]{\mathrm{\Omega }}_2{\zeta }_2{I}_{33}+2\left[\frac{b^2}{b^2+c^2}\right]{\mathrm{\Omega }}_1{\zeta }_1{I}_{33}$
	$=$	$\mathrm{\Pi }+[\frac{c^2}{c^2+a^2}{]}^2{\zeta }_2^2{I}_{11}+[\frac{c^2}{c^2+b^2}{]}^2{\zeta }_1^2{I}_{22}+\{({\mathrm{\Omega }}_1^2+{\mathrm{\Omega }}_2^2)+2[\frac{a^2}{a^2+c^2}]{\mathrm{\Omega }}_2{\zeta }_2+2[\frac{b^2}{b^2+c^2}]{\mathrm{\Omega }}_1{\zeta }_1-(2\pi G\rho )A_3\}{I}_{33}$
$\Rightarrow -\left[\frac{3\cdot 5}{2^2\pi abc\rho }\right]\mathrm{\Pi }$	$=$	$[\frac{c^2}{c^2+a^2}{]}^2{\zeta }_2^2{a}^2+[\frac{c^2}{c^2+b^2}{]}^2{\zeta }_1^2{b}^2+\{({\mathrm{\Omega }}_1^2+{\mathrm{\Omega }}_2^2)+2[\frac{a^2}{a^2+c^2}]{\mathrm{\Omega }}_2{\zeta }_2+2[\frac{b^2}{b^2+c^2}]{\mathrm{\Omega }}_1{\zeta }_1-(2\pi G\rho )A_3\}{c}^2.$

This gives us three equations vs. seven unknowns.

Off-Diagonal Elements

Notice that the off-diagonal components of both $I_{ij}$ and ${\mathfrak{𝔚}}_{ij}$ are zero. Hence, the equilibrium expression that is dictated by each off-diagonal component of the 2^nd-order TVE is,

$0$

$=$

$2{\mathfrak{𝔗}}_{ij}-{\mathrm{\Omega }}_i{\mathrm{\Omega }}_k{I}_{kj}+2{ϵ}_{ilm}{\mathrm{\Omega }}_m\underset{V}{\int }\rho u_l{x}_j{d}^{3}x.$

For example — as is explicitly illustrated on p. 130 of EFE — for $i=2$ and $j=3$,

$0$	$=$	$2{\mathfrak{𝔗}}_{23}-{\mathrm{\Omega }}_2{\mathrm{\Omega }}_3{I}_{33}+2{\mathrm{\Omega }}_1{\xcancel{\underset{V}{\int }\rho u_3{x}_3{d}^{3}x}}^0-2{\mathrm{\Omega }}_3\underset{V}{\int }\rho u_1{x}_3{d}^{3}x,$
[ EFE, Chapter 7, §47, Eq. (3) ]

whereas for $i=3$ and $j=2$,

$0$	$=$	$2{\mathfrak{𝔗}}_{32}-{\mathrm{\Omega }}_3{\mathrm{\Omega }}_2{I}_{22}+2{\mathrm{\Omega }}_2\underset{V}{\int }\rho u_1{x}_2{d}^{3}x-2{\mathrm{\Omega }}_1{\xcancel{\underset{V}{\int }\rho u_2{x}_2{d}^{3}x}}^0.$
[ EFE, Chapter 7, §47, Eq. (4) ]

Given our adoption of a uniform-density configuration whose surface has a precisely ellipsoidal shape and, along with it, our adoption of the above specific prescription for the internal velocity field, $\overrightarrow{u}$, we recognize that, $\underset{V}{\int }\rho u_i{x}_j{d}^{3}x$ $=$ $0$       if    $i=j.$ [ EFE, Chapter 7, §47, Eq. (5) ] This has allowed us to set to zero one of the integrals in each of these last two expressions. In what follows, we will benefit from recognizing, as well, that, ${\mathfrak{𝔗}}_{32}$ $=$ ${\mathfrak{𝔗}}_{23}$ $=$ $\frac{1}{2}\underset{V}{\int }\rho v_2{v}_3{d}^{3}x.$

Our first off-diagonal element is, then,

$0$	$=$	$2{\mathfrak{𝔗}}_{23}-{\mathrm{\Omega }}_2{\mathrm{\Omega }}_3{I}_{33}-2{\mathrm{\Omega }}_3\rho \underset{V}{\int }{u}_{1}z{d}^{3}x$
	$=$	$-\left[\frac{b^2}{b^2+a^2}\right]\left[\frac{c^2}{c^2+a^2}\right]{\zeta }_2{\zeta }_3{a}^2-{\mathrm{\Omega }}_2{\mathrm{\Omega }}_3{c}^2-2\left[\frac{a^2}{a^2+c^2}\right]{\mathrm{\Omega }}_3{\zeta }_2{c}^2$
	$=$	$\{{\mathrm{\Omega }}_2{\mathrm{\Omega }}_3+[\frac{{\zeta }_2{a}^2}{a^2+c^2}\left]\right[2{\mathrm{\Omega }}_3+\frac{{\zeta }_3{b}^2}{b^2+a^2}\left]\right\}{c}^2$
	$=$	$\{1+\frac{{\zeta }_2}{{\mathrm{\Omega }}_2}[\frac{a^2}{a^2+c^2}\left]\right[2+\frac{{\zeta }_3}{{\mathrm{\Omega }}_3}\left(\frac{b^2}{b^2+a^2}\right)\left]\right\}{\mathrm{\Omega }}_2{\mathrm{\Omega }}_3{c}^2.$

The second is,

$0$	$=$	$2{\mathfrak{𝔗}}_{32}-{\mathrm{\Omega }}_3{\mathrm{\Omega }}_2{I}_{22}+2{\mathrm{\Omega }}_2\rho \underset{V}{\int }{u}_{1}y{d}^{3}x$
	$=$	$-\left[\frac{b^2}{b^2+a^2}\right]\left[\frac{c^2}{c^2+a^2}\right]{\zeta }_2{\zeta }_3{a}^2-{\mathrm{\Omega }}_3{\mathrm{\Omega }}_2{b}^2-2\left[\frac{a^2}{a^2+b^2}\right]{\mathrm{\Omega }}_2{\zeta }_3{b}^2$
	$=$	$\{{\mathrm{\Omega }}_2{\mathrm{\Omega }}_3+[\frac{{\zeta }_3{a}^2}{a^2+b^2}\left]\right[2{\mathrm{\Omega }}_2+\frac{{\zeta }_2{c}^2}{c^2+a^2}\left]\right\}{b}^2$
	$=$	$\{1+\frac{{\zeta }_3}{{\mathrm{\Omega }}_3}[\frac{a^2}{a^2+b^2}\left]\right[2+\frac{{\zeta }_2}{{\mathrm{\Omega }}_2}\left(\frac{c^2}{c^2+a^2}\right)\left]\right\}{\mathrm{\Omega }}_2{\mathrm{\Omega }}_3{b}^2.$

How Solution is Obtained

Adding this pair of governing expressions we obtain,

$0$	$=$	$[2{\mathfrak{𝔗}}_{23}-{\mathrm{\Omega }}_2{\mathrm{\Omega }}_3{I}_{33}-2{\mathrm{\Omega }}_3\underset{V}{\int }\rho u_1{x}_{3}dx]+[2{\mathfrak{𝔗}}_{32}-{\mathrm{\Omega }}_3{\mathrm{\Omega }}_2{I}_{22}+2{\mathrm{\Omega }}_2\underset{V}{\int }\rho u_1{x}_{2}dx]$
	$=$	$4{\mathfrak{𝔗}}_{23}-{\mathrm{\Omega }}_2{\mathrm{\Omega }}_3(I_{22}+I_{33})+2\underset{V}{\int }\rho u_1({\mathrm{\Omega }}_2{x}_2-{\mathrm{\Omega }}_3{x}_3)dx;$
[ EFE, Chapter 7, §47, Eq. (6) ]

and subtracting the pair gives,

$0$	$=$	$[2{\mathfrak{𝔗}}_{23}-{\mathrm{\Omega }}_2{\mathrm{\Omega }}_3{I}_{33}-2{\mathrm{\Omega }}_3\underset{V}{\int }\rho u_1{x}_{3}dx]-[2{\mathfrak{𝔗}}_{32}-{\mathrm{\Omega }}_3{\mathrm{\Omega }}_2{I}_{22}+2{\mathrm{\Omega }}_2\underset{V}{\int }\rho u_1{x}_{2}dx]$
	$=$	${\mathrm{\Omega }}_2{\mathrm{\Omega }}_3(I_{22}-I_{33})-2\underset{V}{\int }\rho u_1({\mathrm{\Omega }}_2{x}_2+{\mathrm{\Omega }}_3{x}_3)dx.$
[ EFE, Chapter 7, §47, Eq. (7) ]

Various Degrees of Simplification

Riemann Ellipsoids of Types I, II, & III

In this, most general, case, 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 our above Summary Table must be used in order to determine the equilibrium configuration's associated values of the five unknowns: $\mathrm{\Pi },({\mathrm{\Omega }}_2,{\zeta }_2),({\mathrm{\Omega }}_3,{\zeta }_3)$. Here we show how these five unknowns can be derived from the five constraint equations, closely following the analysis that is presented in §47 (pp. 129 - 132) of [ EFE ].

Constraints Due to Off-Diagonal Elements

We begin by subtracting the constraint equation provided by the first off-diagonal element $(i,j)=(2,3)$ from the constraint equation provided by the second off-diagonal element $(i,j)=(3,2)$. This gives,

$\{1+\frac{{\zeta }_2}{{\mathrm{\Omega }}_2}[\frac{a^2}{a^2+c^2}\left]\right[2+\frac{{\zeta }_3}{{\mathrm{\Omega }}_3}\left(\frac{b^2}{b^2+a^2}\right)\left]\right\}{\mathrm{\Omega }}_2{\mathrm{\Omega }}_3{c}^2$	$=$	$\{1+\frac{{\zeta }_3}{{\mathrm{\Omega }}_3}[\frac{a^2}{a^2+b^2}\left]\right[2+\frac{{\zeta }_2}{{\mathrm{\Omega }}_2}\left(\frac{c^2}{c^2+a^2}\right)\left]\right\}{\mathrm{\Omega }}_2{\mathrm{\Omega }}_3{b}^2$
$\Rightarrow c^2+\frac{{\zeta }_2}{{\mathrm{\Omega }}_2}\left[\frac{2a^2{c}^2}{a^2+c^2}\right][1+\frac{{\zeta }_3}{2{\mathrm{\Omega }}_3}(\frac{b^2}{b^2+a^2}\left)\right]$	$=$	$b^2+\frac{{\zeta }_3}{{\mathrm{\Omega }}_3}\left[\frac{2a^2{b}^2}{a^2+b^2}\right][1+\frac{{\zeta }_2}{2{\mathrm{\Omega }}_2}(\frac{c^2}{c^2+a^2}\left)\right]$
$\Rightarrow c^2+\frac{{\zeta }_2}{{\mathrm{\Omega }}_2}\left[\frac{2a^2{c}^2}{a^2+c^2}\right]+\frac{{\zeta }_2}{{\mathrm{\Omega }}_2}\cdot \frac{{\zeta }_3}{{\mathrm{\Omega }}_3}\left[\frac{a^2{c}^2}{a^2+c^2}\right]\left[\frac{b^2}{b^2+a^2}\right]$	$=$	$b^2+\frac{{\zeta }_3}{{\mathrm{\Omega }}_3}\left[\frac{2a^2{b}^2}{a^2+b^2}\right]+\frac{{\zeta }_3}{{\mathrm{\Omega }}_3}\cdot \frac{{\zeta }_2}{{\mathrm{\Omega }}_2}\left[\frac{a^2{b}^2}{a^2+b^2}\right]\left[\frac{c^2}{c^2+a^2}\right]$
$\Rightarrow c^2+\frac{{\zeta }_2}{{\mathrm{\Omega }}_2}\left[\frac{2a^2{c}^2}{a^2+c^2}\right]$	$=$	$b^2+\frac{{\zeta }_3}{{\mathrm{\Omega }}_3}\left[\frac{2a^2{b}^2}{a^2+b^2}\right].$
[ EFE, Chapter 7, §47, Eq. (11) ]

Adding the two instead gives,

$0$	$=$	$\{1+\frac{{\zeta }_2}{{\mathrm{\Omega }}_2}[\frac{a^2}{a^2+c^2}\left]\right[2+\frac{{\zeta }_3}{{\mathrm{\Omega }}_3}\left(\frac{b^2}{b^2+a^2}\right)\left]\right\}{\mathrm{\Omega }}_2{\mathrm{\Omega }}_3{c}^2+\{1+\frac{{\zeta }_3}{{\mathrm{\Omega }}_3}[\frac{a^2}{a^2+b^2}\left]\right[2+\frac{{\zeta }_2}{{\mathrm{\Omega }}_2}\left(\frac{c^2}{c^2+a^2}\right)\left]\right\}{\mathrm{\Omega }}_2{\mathrm{\Omega }}_3{b}^2$
	$=$	$b^2+c^2+\frac{{\zeta }_2}{{\mathrm{\Omega }}_2}\left[\frac{a^2{c}^2}{a^2+c^2}\right][2+\frac{{\zeta }_3}{{\mathrm{\Omega }}_3}(\frac{b^2}{b^2+a^2}\left)\right]+\frac{{\zeta }_3}{{\mathrm{\Omega }}_3}\left[\frac{a^2{b}^2}{a^2+b^2}\right][2+\frac{{\zeta }_2}{{\mathrm{\Omega }}_2}(\frac{c^2}{c^2+a^2}\left)\right]$
	$=$	$b^2+c^2+\frac{{\zeta }_2}{{\mathrm{\Omega }}_2}\left[\frac{2a^2{c}^2}{a^2+c^2}\right]+\frac{{\zeta }_3}{{\mathrm{\Omega }}_3}\left[\frac{2a^2{b}^2}{a^2+b^2}\right]+\frac{{\zeta }_2}{{\mathrm{\Omega }}_2}\cdot \frac{{\zeta }_3}{{\mathrm{\Omega }}_3}\left[\frac{2a^2{b}^2{c}^2}{(a^2+c^2)(b^2+a^2)}\right].$
[ EFE, Chapter 7, §47, Eq. (10) ]

The first of these relations cleanly gives an expression for the frequency ratio, ${\zeta }_3/{\mathrm{\Omega }}_3$, in terms of the other frequency ratio, ${\zeta }_2/{\mathrm{\Omega }}_2$. This allows us to rewrite the second relation in terms of the ratio, ${\zeta }_2/{\mathrm{\Omega }}_2$, alone. We obtain,

$0$	$=$	$b^2+c^2+\frac{{\zeta }_2}{{\mathrm{\Omega }}_2}\left[\frac{2a^2{c}^2}{a^2+c^2}\right]+\left\{\frac{{\zeta }_3}{{\mathrm{\Omega }}_3}\right[\frac{2a^2{b}^2}{a^2+b^2}\left]\right\}+\frac{{\zeta }_2}{{\mathrm{\Omega }}_2}\left[\frac{c^2}{(a^2+c^2)}\right]\cdot \left\{\frac{{\zeta }_3}{{\mathrm{\Omega }}_3}\right[\frac{2a^2{b}^2}{(b^2+a^2)}\left]\right\}$
	$=$	$b^2+c^2+\frac{{\zeta }_2}{{\mathrm{\Omega }}_2}\left[\frac{2a^2{c}^2}{a^2+c^2}\right]+\{c^2-b^2+\frac{{\zeta }_2}{{\mathrm{\Omega }}_2}[\frac{2a^2{c}^2}{a^2+c^2}\left]\right\}+\frac{{\zeta }_2}{{\mathrm{\Omega }}_2}\left[\frac{c^2}{(a^2+c^2)}\right]\cdot \{c^2-b^2+\frac{{\zeta }_2}{{\mathrm{\Omega }}_2}[\frac{2a^2{c}^2}{a^2+c^2}\left]\right\}$
	$=$	$2c^2+\frac{{\zeta }_2}{{\mathrm{\Omega }}_2}\left[\frac{c^2}{a^2+c^2}\right](4a^2+c^2-b^2)+\left\{\frac{{\zeta }_2}{{\mathrm{\Omega }}_2}\right[\frac{c^2}{a^2+c^2}]{\}}^{2}2a^2$

ASIDE:   Alternatively, given that, $\frac{{\zeta }_2}{{\mathrm{\Omega }}_2}\left[\frac{c^2}{a^2+c^2}\right]$ $=$ $\frac{1}{2a^2}[b^2-c^2+\frac{{\zeta }_3}{{\mathrm{\Omega }}_3}(\frac{2a^2{b}^2}{a^2+b^2}\left)\right]$ the quadratic equation that governs the value of the frequency ratio, ${\zeta }_3/{\mathrm{\Omega }}_3$ is … $0$ $=$ $4a^2{c}^2+[b^2-c^2+\frac{{\zeta }_3}{{\mathrm{\Omega }}_3}(\frac{2a^2{b}^2}{a^2+b^2}\left)\right](4a^2+c^2-b^2)+[b^2-c^2+\frac{{\zeta }_3}{{\mathrm{\Omega }}_3}(\frac{2a^2{b}^2}{a^2+b^2}){]}^2$   $=$ $4a^2{c}^2+\left[\frac{{\zeta }_3}{{\mathrm{\Omega }}_3}\right(\frac{2a^2{b}^2}{a^2+b^2}\left)\right](4a^2+c^2-b^2)+(b^2-c^2)(4a^2+c^2-b^2)$     $+(b^2-c^2{)}^2+2(b^2-c^2)[\frac{{\zeta }_3}{{\mathrm{\Omega }}_3}\left(\frac{2a^2{b}^2}{a^2+b^2}\right)]+[\frac{{\zeta }_3}{{\mathrm{\Omega }}_3}\left(\frac{2a^2{b}^2}{a^2+b^2}\right){]}^2$   $=$ $4a^2{c}^2+(b^2-c^2)(4a^2+c^2-b^2)+(b^2-c^2{)}^2+[\frac{{\zeta }_3}{{\mathrm{\Omega }}_3}\left(\frac{2a^2{b}^2}{a^2+b^2}\right)\left]\right[(4a^2+c^2-b^2)+2(b^2-c^2)]+[\frac{{\zeta }_3}{{\mathrm{\Omega }}_3}\left(\frac{2a^2{b}^2}{a^2+b^2}\right){]}^2$   $=$ $\left[\frac{{\zeta }_3}{{\mathrm{\Omega }}_3}\right(\frac{2a^2{b}^2}{a^2+b^2}){]}^2+[\frac{{\zeta }_3}{{\mathrm{\Omega }}_3}\left(\frac{2a^2{b}^2}{a^2+b^2}\right)](4a^2+b^2-c^2)+4a^2{b}^2$ $\Rightarrow 0$ $=$ $\left[\right(\frac{{\zeta }_3}{{\mathrm{\Omega }}_3}{)}^2\frac{a^2{b}^2}{(a^2+b^2{)}^2}]+\frac{1}{2}[\frac{{\zeta }_3}{{\mathrm{\Omega }}_3}\left(\frac{1}{a^2+b^2}\right)](4a^2+b^2-c^2)+1.$ Now, in our discussion of Riemann S-Type ellipsoids, there is also a quadratic equation that governs the equilibrium frequency ratio, $f\equiv {\zeta }_3/{\mathrm{\Omega }}_3$. It is, specifically, $0$ $=$ $\left[\frac{a^2{b}^2}{(a^2+b^2{)}^2}\right]f^2+\left[\frac{2a^2{b}^2{B}_{12}}{c^2{A}_3-a^2{b}^2{A}_{12}}\right]\frac{f}{a^2+b^2}+1.$ [ EFE, §48, Eq. (35) ] Notice that the first and third terms of this quadratic equation exactly match the first and third terms of the quadratic equation, which we have just derived, that governs the same frequency ratio in Riemann ellipsoids of Types I, II & III. Does the second term match? That is, is the coefficient of the linear term the same in both quadratic relations? Well, … $\frac{2a^2{b}^2{B}_{12}}{c^2{A}_3-a^2{b}^2{A}_{12}}$ $=$ $2a^2{b}^2[c^2{A}_3+a^2{b}^2(\frac{A_1-A_2}{a^2-b^2}\left){]}^{-1}\right[A_2+a^2\left(\frac{A_1-A_2}{a^2-b^2}\right)]$   $=$ $2a^2{b}^2[c^2{A}_3(a^2-b^2)+a^2{b}^2(A_1-A_2){]}^{-1}[a^2{A}_1-b^2{A}_2].$ Even appreciating that we can make the substitution, $A_3=(2-A_1-A_2)$, I don't see any way that this coefficient expression can be manipulated to match the associated coefficient in the other expression, namely, $(4a^2+b^2-c^2)/2$.

This is a quadratic equation whose solution gives,

$4a^2\cdot \frac{{\zeta }_2}{{\mathrm{\Omega }}_2}\left[\frac{c^2}{a^2+c^2}\right]$

$=$

$-(4a^2+c^2-b^2)\pm [(4a^2+c^2-b^2{)}^2-16a^2{c}^2{]}^{1/2}.$

For the other frequency ratio we therefore find,

$2\{b^2-c^2+\frac{{\zeta }_3}{{\mathrm{\Omega }}_3}[\frac{2a^2{b}^2}{a^2+b^2}\left]\right\}$	$=$	$2\cdot \frac{{\zeta }_2}{{\mathrm{\Omega }}_2}\left[\frac{2a^2{c}^2}{a^2+c^2}\right]$
	$=$	$-(4a^2+c^2-b^2)\pm [(4a^2+c^2-b^2{)}^2-16a^2{c}^2{]}^{1/2}$
$\Rightarrow 4a^2\cdot \frac{{\zeta }_3}{{\mathrm{\Omega }}_3}\left[\frac{b^2}{a^2+b^2}\right]$	$=$	$-(4a^2+b^2-c^2)\pm [(4a^2+c^2-b^2{)}^2-16a^2{c}^2{]}^{1/2}$

 

SUMMARY:   Riemann Ellipsoids of Types I, II, & III $\beta \equiv -\frac{{\zeta }_2}{{\mathrm{\Omega }}_2}\left[\frac{c^2}{a^2+c^2}\right]$ $=$ $\frac{1}{4a^2}\{(4a^2-b^2+c^2)\mp [(4a^2+c^2-b^2{)}^2-16a^2{c}^2{]}^{1/2}\};$ [ EFE, Chapter 7, §47, Eq. (16) ] $\gamma \equiv -\frac{{\zeta }_3}{{\mathrm{\Omega }}_3}\left[\frac{b^2}{a^2+b^2}\right]$ $=$ $\frac{1}{4a^2}\{(4a^2+b^2-c^2)\mp [(4a^2+c^2-b^2{)}^2-16a^2{c}^2{]}^{1/2}\}.$ [ 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." As has also been pointed out in EFE (Chapter 7, §51, p. 158), from the steps that have led to the development and solution of the above pair of quadratic equations we can demonstrate that the following relations also hold: ${\beta }^2-2\beta +\frac{c^2}{a^2}=\left[\frac{c^2-b^2}{2a^2}\right]\beta ,$         ${\gamma }^2-2\gamma +\frac{b^2}{a^2}=\left[\frac{b^2-c^2}{2a^2}\right]\gamma$ $1-2\beta +\left(\frac{a^2}{c^2}\right){\beta }^2$ $=$ $\left[\frac{4a^2-b^2-3c^2}{2c^2}\right]\beta ,$ $1-2\gamma +\left(\frac{a^2}{b^2}\right){\gamma }^2$ $=$ $\left[\frac{4a^2-c^2-3b^2}{2b^2}\right]\gamma .$ [ EFE, Chapter 7, §51, Eqs. (161) - (163) ]

Constraints Due to Diagonal Elements

Next, to simplify manipulations, let's replace the frequency ratios by these newly defined — and known — parameters, $\beta$ and $\gamma$, in the three diagonal-element expressions that are written out in our above Summary Table.

Indices		Rewritten Diagonal-Element Expressions
$i$	$j$
$1$	$1$	$\left[\frac{3\cdot 5}{2^2\pi abc\rho }\right]\mathrm{\Pi }$ $=$ $-\{({\mathrm{\Omega }}_2^2+{\mathrm{\Omega }}_3^2)+2[\frac{b^2}{b^2+a^2}\left]{\mathrm{\Omega }}_3\right[-\frac{{\mathrm{\Omega }}_3\gamma (a^2+b^2)}{b^2}]+2[\frac{c^2}{c^2+a^2}\left]{\mathrm{\Omega }}_2\right[-\frac{{\mathrm{\Omega }}_2\beta (a^2+c^2)}{c^2}]-(2\pi G\rho )A_1\}{a}^2$     $-\left[\frac{a^2}{a^2+b^2}{]}^2\right[-\frac{{\mathrm{\Omega }}_3\gamma (a^2+b^2)}{b^2}{]}^2{b}^2-\left[\frac{a^2}{a^2+c^2}{]}^2\right[-\frac{{\mathrm{\Omega }}_2\beta (a^2+c^2)}{c^2}{]}^2{c}^2$   $=$ $\{-{\mathrm{\Omega }}_2^2-{\mathrm{\Omega }}_3^2+2{\mathrm{\Omega }}_3^2\gamma +2{\mathrm{\Omega }}_2^2\beta +(2\pi G\rho )A_1\}{a}^2-\left(\frac{a^4}{b^2}\right){\mathrm{\Omega }}_3^2{\gamma }^2-\left(\frac{a^4}{c^2}\right){\mathrm{\Omega }}_2^2{\beta }^2$   $=$ $\left\{{\mathrm{\Omega }}_2^2\right[2\beta -1-\left(\frac{a^2}{c^2}\right){\beta }^2]+{\mathrm{\Omega }}_3^2[2\gamma -1-\left(\frac{a^2}{b^2}\right){\gamma }^2]+(2\pi G\rho )A_1\}{a}^2$ [ EFE, Chapter 7, §51, Eq. (158) ]
$2$	$2$	$\left[\frac{3\cdot 5}{2^2\pi abc\rho }\right]\mathrm{\Pi }$ $=$ $-\left[\frac{b^2}{b^2+a^2}{]}^2\right[-\frac{{\mathrm{\Omega }}_3\gamma (a^2+b^2)}{b^2}{]}^2{a}^2-\{{\mathrm{\Omega }}_3^2+2[\frac{a^2}{a^2+b^2}\left]{\mathrm{\Omega }}_3\right[-\frac{{\mathrm{\Omega }}_3\gamma (a^2+b^2)}{b^2}]-(2\pi G\rho )A_2\}{b}^2$   $=$ $-{\mathrm{\Omega }}_3^2{\gamma }^2{a}^2-{\mathrm{\Omega }}_3^2{b}^2+2a^2{\mathrm{\Omega }}_3^2\gamma +(2\pi G\rho )b^2{A}_2$   $=$ $-a^2{\mathrm{\Omega }}_3^2[{\gamma }^2-2\gamma +(\frac{b^2}{a^2}\left)\right]+(2\pi G\rho )b^2{A}_2$ [ EFE, Chapter 7, §51, Eq. (159) ]
$3$	$3$	$\left[\frac{3\cdot 5}{2^2\pi abc\rho }\right]\mathrm{\Pi }$ $=$ $-\left[\frac{c^2}{c^2+a^2}{]}^2\right[-\frac{{\mathrm{\Omega }}_2\beta (a^2+c^2)}{c^2}{]}^2{a}^2-\{{\mathrm{\Omega }}_2^2+2[\frac{a^2}{a^2+c^2}\left]{\mathrm{\Omega }}_2\right[-\frac{{\mathrm{\Omega }}_2\beta (a^2+c^2)}{c^2}]-(2\pi G\rho )A_3\}{c}^2$   $=$ $-{\mathrm{\Omega }}_2^2{\beta }^2{a}^2-{\mathrm{\Omega }}_2^2{c}^2+2a^2{\mathrm{\Omega }}_2^2\beta +(2\pi G\rho )c^2{A}_3$   $=$ $-a^2{\mathrm{\Omega }}_2^2[{\beta }^2-2\beta +(\frac{c^2}{a^2}\left)\right]+(2\pi G\rho )c^2{A}_3$ [ EFE, Chapter 7, §51, Eq. (160) ]

Using the $(i,j)=(3,3)$ element to preplace $\mathrm{\Pi }$ in the other two expressions, we obtain,

$0$

$=$

${\mathrm{\Omega }}_2^2[{\beta }^2-2\beta +(\frac{c^2}{a^2}\left)\right]+{\mathrm{\Omega }}_2^2[2\beta -1-(\frac{a^2}{c^2}\left){\beta }^2\right]+{\mathrm{\Omega }}_3^2[2\gamma -1-(\frac{a^2}{b^2}\left){\gamma }^2\right]+2\pi G\rho [A_1-(\frac{c^2}{a^2}\left)A_3\right];$

and,

${\mathrm{\Omega }}_2^2[{\beta }^2-2\beta +(\frac{c^2}{a^2}\left)\right]$

$=$

${\mathrm{\Omega }}_3^2[{\gamma }^2-2\gamma +(\frac{b^2}{a^2}\left)\right]+2\pi G\rho \left[\right(\frac{c^2}{a^2})A_3-(\frac{b^2}{a^2}\left)A_2\right].$

Inserting the various relations highlighted above, these two expressions may be rewritten as,

$0$	$=$	${\mathrm{\Omega }}_2^2\left[\frac{c^2-b^2}{2a^2}\right]\beta -{\mathrm{\Omega }}_2^2\left[\frac{4a^2-b^2-3c^2}{2c^2}\right]\beta -{\mathrm{\Omega }}_3^2\left[\frac{4a^2-c^2-3b^2}{2b^2}\right]\gamma +2\pi G\rho [A_1-(\frac{c^2}{a^2}\left)A_3\right]$
	$=$	${\mathrm{\Omega }}_2^2\beta \left[\frac{c^2(c^2-b^2)+a^2(b^2+3c^2-4a^2)}{2a^2{c}^2}\right]-{\mathrm{\Omega }}_3^2\gamma \left[\frac{4a^2-c^2-3b^2}{2b^2}\right]+2\pi G\rho [A_1-(\frac{c^2}{a^2}\left)A_3\right];$

and,

${\mathrm{\Omega }}_2^2\left[\frac{c^2-b^2}{2a^2}\right]\beta$	$=$	${\mathrm{\Omega }}_3^2\left[\frac{b^2-c^2}{2a^2}\right]\gamma +2\pi G\rho \left[\right(\frac{c^2}{a^2})A_3-(\frac{b^2}{a^2}\left)A_2\right]$
$\Rightarrow {\mathrm{\Omega }}_2^2\beta$	$=$	$-{\mathrm{\Omega }}_3^2\gamma +4\pi G\rho \left[\frac{c^2{A}_3-b^2{A}_2}{c^2-b^2}\right]$
$\Rightarrow -{\mathrm{\Omega }}_3^2\gamma$	$=$	${\mathrm{\Omega }}_2^2\beta -4\pi G\rho \left[\frac{c^2{A}_3-b^2{A}_2}{c^2-b^2}\right].$

Together, then,

$0$	$=$	${\mathrm{\Omega }}_2^2\beta \left[\frac{c^2(c^2-b^2)+a^2(b^2+3c^2-4a^2)}{2a^2{c}^2}\right]+2\pi G\rho [A_1-(\frac{c^2}{a^2}\left)A_3\right]+\left[\frac{4a^2-c^2-3b^2}{2b^2}\right]\{{\mathrm{\Omega }}_2^2\beta -4\pi G\rho [\frac{c^2{A}_3-b^2{A}_2}{c^2-b^2}\left]\right\}$
	$=$	${\mathrm{\Omega }}_2^2\beta [\frac{c^2(c^2-b^2)+a^2(b^2+3c^2-4a^2)}{2a^2{c}^2}+\frac{4a^2-c^2-3b^2}{2b^2}]+2\pi G\rho \left[\frac{a^2{A}_1-c^2{A}_3}{a^2}\right]-2\pi G\rho \left[\frac{4a^2-c^2-3b^2}{b^2}\right]\left[\frac{c^2{A}_3-b^2{A}_2}{c^2-b^2}\right]$
	$=$	${\mathrm{\Omega }}_2^2\beta \left\{\frac{b^2[c^2(c^2-b^2)+a^2(b^2+3c^2-4a^2)]+(4a^2-c^2-3b^2)a^2{c}^2}{2a^2{b}^2{c}^2}\right\}+2\pi G\rho \left\{\right[\frac{a^2{A}_1-c^2{A}_3}{a^2}]-[\frac{4a^2-c^2-3b^2}{b^2}\left]\right[\frac{c^2{A}_3-b^2{A}_2}{c^2-b^2}\left]\right\}$
	$=$	${\mathrm{\Omega }}_2^2\beta \left\{\frac{b^2{c}^2(c^2-b^2)+a^2{b}^2(c^2-b^2)+(c^2-b^2)a^2{c}^2+a^2(4a^2-2b^2-2c^2)(c^2-b^2)}{2a^2{b}^2{c}^2}\right\}+2\pi G\rho \left[\frac{b^2(a^2{A}_1-c^2{A}_3)+a^2(3b^2-4a^2+c^2)B_{23}}{a^2{b}^2}\right]$
	$=$	${\mathrm{\Omega }}_2^2\beta \left[\frac{c^2-b^2}{c^2}\right]\left[\frac{4a^4-a^2(b^2+c^2)+b^2{c}^2}{2a^2{b}^2}\right]+2\pi G\rho \left[\frac{a^2(3b^2-4a^2+c^2)B_{23}+b^2(a^2{A}_1-c^2{A}_3)}{a^2{b}^2}\right],$
[ EFE, Chapter 7, §51, Eq. (170) ]

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) ]

Similarly, given that (see just above),

${\mathrm{\Omega }}_2^2\beta \left[\frac{c^2-b^2}{c^2}\right]$	$=$	$-{\mathrm{\Omega }}_3^2\gamma \left[\frac{c^2-b^2}{c^2}\right]+4\pi G\rho \left[\frac{c^2{A}_3-b^2{A}_2}{c^2}\right]$
	$=$	$-{\mathrm{\Omega }}_3^2\gamma \left[\frac{c^2-b^2}{c^2}\right]+4\pi G\rho \left[\frac{(c^2-b^2)B_{23}}{c^2}\right],$

we have,

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

Finally, looking back at the $(i,j)=(3,3)$ constraint and recognizing that,

$-{\mathrm{\Omega }}_2^2\beta (c^2-b^2)$

$=$

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

we find,

$2\left[\frac{3\cdot 5}{2^2\pi abc\rho }\right]\mathrm{\Pi }$	$=$	$-2a^2{\mathrm{\Omega }}_2^2[{\beta }^2-2\beta +(\frac{c^2}{a^2}\left)\right]+(4\pi G\rho )c^2{A}_3$
	$=$	$(4\pi G\rho )c^2{A}_3-(c^2-b^2){\mathrm{\Omega }}_2^2\beta$
	$=$	$(4\pi G\rho )c^2{A}_3+4\pi G\rho c^2\left[\frac{a^2(3b^2-4a^2+c^2)B_{23}+b^2(a^2{A}_1-c^2{A}_3)}{4a^4-a^2(b^2+c^2)+b^2{c}^2}\right]$
	$=$	$4\pi G\rho c^2\{A_3+[\frac{a^2(3b^2-4a^2+c^2)B_{23}+b^2(a^2{A}_1-c^2{A}_3)}{4a^4-a^2(b^2+c^2)+b^2{c}^2}\left]\right\}.$

Riemann S-Type Ellipsoids

In this case, we assume that $\overrightarrow{\mathrm{\Omega }}$ and $\overrightarrow{\zeta }$ are aligned with each other and, as well, are aligned with the $z$-axis; that is to say, in addition to setting $({\mathrm{\Omega }}_1,{\zeta }_1)=(0,0)$ we also set $({\mathrm{\Omega }}_2,{\zeta }_2)=(0,0)$. So, there are only three unknowns — $\mathrm{\Pi },({\mathrm{\Omega }}_3,{\zeta }_3)$ — and they can be determined by ignoring off-axis expressions and simultaneously solving the diagonal element expressions displayed in our above Summary Table. Furthermore, two of the three diagonal-element expressions can be simplified because we are setting $({\mathrm{\Omega }}_2,{\zeta }_2)=(0,0)$. The three relevant equilibrium constraints are:

Indices		2nd-Order TVE Expressions that are Relevant to Riemann S-Type Ellipsoids
$i$	$j$
$1$	$1$	$0$ $=$ $\left[\frac{3\cdot 5}{2^2\pi abc\rho }\right]\mathrm{\Pi }+\{{\mathrm{\Omega }}_3^2+2[\frac{b^2}{b^2+a^2}]{\mathrm{\Omega }}_3{\zeta }_3-(2\pi G\rho )A_1\}{a}^2+[\frac{a^2}{a^2+b^2}{]}^2{\zeta }_3^2{b}^2$
$2$	$2$	$0$ $=$ $\left[\frac{3\cdot 5}{2^2\pi abc\rho }\right]\mathrm{\Pi }+[\frac{b^2}{b^2+a^2}{]}^2{\zeta }_3^2{a}^2+\{{\mathrm{\Omega }}_3^2+2\left[\frac{a^2}{a^2+b^2}\right]{\mathrm{\Omega }}_3{\zeta }_3-(2\pi G\rho )A_2\}{b}^2$
$3$	$3$	$0$ $=$ $\left[\frac{3\cdot 5}{2^2\pi abc\rho }\right]\mathrm{\Pi }-(2\pi G\rho )A_3{c}^2$

The $(i,j)=(3,3)$ component expression immediately identifies the value of one of the unknowns, namely,

$\mathrm{\Pi }$

$=$

$\left(\frac{2^3{\pi }^2}{3\cdot 5}\right)G{\rho }^2{A}_{3}ab{c}^3.$

From the remaining pair of diagonal-element expressions, we therefore have,

$0$

$=$

$a^2{\mathrm{\Omega }}_3^2+2\left[\frac{b^2{a}^2}{b^2+a^2}\right]{\mathrm{\Omega }}_3{\zeta }_3+[\frac{a^2}{a^2+b^2}{]}^2{\zeta }_3^2{b}^2+(2\pi G\rho )(A_3{c}^2-A_1{a}^2),$

and,

$0$

$=$

$[\frac{b^2}{b^2+a^2}{]}^2{\zeta }_3^2{a}^2+b^2{\mathrm{\Omega }}_3^2+2[\frac{a^2{b}^2}{a^2+b^2}]{\mathrm{\Omega }}_3{\zeta }_3+(2\pi G\rho )(A_3{c}^2-A_2{b}^2).$

Multiplying the first of these two expressions through by $b^2$ and the second through by $a^2$, then subtracting the second from the first gives,

$0$	$=$	$b^2\left\{2\right[\frac{b^2{a}^2}{b^2+a^2}]{\mathrm{\Omega }}_3{\zeta }_3+[\frac{a^2}{a^2+b^2}{]}^2{\zeta }_3^2{b}^2+(2\pi G\rho )(A_3{c}^2-A_1{a}^2)\}$
		$-a^2\left\{\right[\frac{b^2}{b^2+a^2}{]}^2{\zeta }_3^2{a}^2+2\left[\frac{a^2{b}^2}{a^2+b^2}\right]{\mathrm{\Omega }}_3{\zeta }_3+(2\pi G\rho )(A_3{c}^2-A_2{b}^2)\}$
	$=$	$\left\{2\right[\frac{b^4{a}^2}{b^2+a^2}]{\mathrm{\Omega }}_3{\zeta }_3+(2\pi G\rho )(A_3{c}^2-A_1{a}^2)b^2\}-\left\{2\right[\frac{a^4{b}^2}{a^2+b^2}]{\mathrm{\Omega }}_3{\zeta }_3+(2\pi G\rho )(A_3{c}^2-A_2{b}^2)a^2\}$
$\Rightarrow \left[\frac{b^2{a}^2}{b^2+a^2}\right]{\mathrm{\Omega }}_3{\zeta }_3$	$=$	$\pi G\rho \left[\frac{(A_3{c}^2-A_2{b}^2)a^2-(A_3{c}^2-A_1{a}^2)b^2}{b^2-a^2}\right]$
	$=$	$\pi G\rho [\frac{(A_1-A_2)a^2{b}^2}{b^2-a^2}-A_3{c}^2].$
[ EFE, Chapter 7, §48, Eq. (30) ]

Note that — as EFE has done and as we have recorded in a related discussion — the first term on the right-hand-side of this last expression can be expressed more compactly in terms of the coefficient, $A_{12}$.

Alternatively, dividing the first expression through by $a^2$ and the second by $b^2$, then adding the pair of expressions gives,

$0$	$=$	${\mathrm{\Omega }}_3^2+2\left[\frac{b^2}{b^2+a^2}\right]{\mathrm{\Omega }}_3{\zeta }_3+\left[\frac{a^2{b}^2}{(a^2+b^2{)}^2}\right]{\zeta }_3^2+(2\pi G\rho )(A_3{c}^2-A_1{a}^2)\frac{1}{a^2}$
		$+\left[\frac{a^2{b}^2}{(b^2+a^2{)}^2}\right]{\zeta }_3^2+{\mathrm{\Omega }}_3^2+2\left[\frac{a^2}{a^2+b^2}\right]{\mathrm{\Omega }}_3{\zeta }_3+(2\pi G\rho )(A_3{c}^2-A_2{b}^2)\frac{1}{b^2}$
	$=$	$2{\mathrm{\Omega }}_3^2+2{\mathrm{\Omega }}_3{\zeta }_3+2\left[\frac{a^2{b}^2}{(a^2+b^2{)}^2}\right]{\zeta }_3^2+2\pi G\rho [\frac{A_3{c}^2-A_1{a}^2}{a^2}+\frac{A_3{c}^2-A_2{b}^2}{b^2}].$

If we divide through by 2, then replace the product, ${\mathrm{\Omega }}_3{\zeta }_3$, in this expression by the relation derived immediately above, we have,

${\mathrm{\Omega }}_3^2+\left[\frac{a^2{b}^2}{(a^2+b^2{)}^2}\right]{\zeta }_3^2$	$=$	$-\pi G\rho \left[\frac{b^2(A_3{c}^2-A_1{a}^2)+a^2(A_3{c}^2-A_2{b}^2)}{a^2{b}^2}\right]-\pi G\rho \left[\frac{(A_1-A_2)a^2{b}^2-A_3{c}^2(b^2-a^2)}{b^2-a^2}\right]\left[\frac{b^2+a^2}{b^2{a}^2}\right]$
	$=$	$\frac{\pi G\rho }{a^2{b}^2(a^2-b^2)}\{[b^2(A_3{c}^2-A_1{a}^2)+a^2(A_3{c}^2-A_2{b}^2)](b^2-a^2)+[(A_1-A_2)a^2{b}^2-A_3{c}^2(b^2-a^2)](b^2+a^2)\}$
	$=$	$\frac{\pi G\rho }{a^2{b}^2(a^2-b^2)}\{[-A_1{a}^2{b}^2-A_2{a}^2{b}^2](b^2-a^2)+(A_1-A_2)a^2{b}^2(b^2+a^2)\}$
	$=$	$\frac{2\pi G\rho }{(a^2-b^2)}[A_1{a}^2-A_2{b}^2].$
[ EFE, Chapter 7, §48, Eq. (29) ]

It has become customary to characterize each Riemann S-Type ellipsoid by the value of its equilibrium frequency ratio,

$f$

$\equiv$

$\frac{{\zeta }_3}{{\mathrm{\Omega }}_3},$

in which case the relevant pair of constraint equations becomes,

$\left[\frac{b^2{a}^2}{b^2+a^2}\right]f{\mathrm{\Omega }}_3^2$ $=$ $\pi G\rho [\frac{(A_1-A_2)a^2{b}^2}{b^2-a^2}-A_3{c}^2];$ [ EFE, Chapter 7, §48, Eq. (34) ] and, ${\mathrm{\Omega }}_3^2\{1+[\frac{a^2{b}^2}{(a^2+b^2{)}^2}\left]f^2\right\}$ $=$ $\frac{2\pi G\rho }{(a^2-b^2)}[A_1{a}^2-A_2{b}^2].$ [ EFE, Chapter 7, §48, Eq. (33) ]

These two equations can straightforwardly be combined to generate a quadratic equation for the frequency ratio, $f$. Then, once the value of $f$ has been determined, either expression can be used to determine the corresponding equilibrium value for ${\mathrm{\Omega }}_3$ in the unit of $(\pi G\rho {)}^{1/2}$. The fact that the value of $f$ is determined from the solution of a quadratic equation underscores the realization that, for a given specification of the ellipsoidal geometry $(a,b,c)$, if an equilibrium exists — i.e., if the solution for $f$ is real rather than imaginary — then two equally valid, and usually different (i.e., non-degenerate), values of $f$ will be realized. This means that two different underlying flows — one direct and the other adjoint — will sustain the shape of the ellipsoidal configuration, as viewed from a frame that is rotating about the $z$-axis with frequency, ${\mathrm{\Omega }}_3$.

Jacobi and Dedekind Ellipsoids

Describe …

Maclaurin Spheroids

Describe …

Appendices:  Various Integrals Over Ellipsoid Volume

Throughout this set of appendices, we work with a uniform-density ellipsoid whose surface is defined by the expression,

$1$

$=$

$\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}.$

Appendix A:  Volume

Here we seek to find the volume of the ellipsoid via the Cartesian integral expression,

$V$

$=$

$\iiint dxdydz.$

Preliminaries

First, we will integrate over $x$ and specify the integration limits via the expression,

$x_{\ell }$

$\equiv$

$a[1-\frac{y^2}{b^2}-\frac{z^2}{c^2}{]}^{1/2};$

second, we will integrate over $z$ and specify the integration limits via the expression,

$z_{\ell }$

$\equiv$

$c[1-\frac{y^2}{b^2}{]}^{1/2};$

third, we will integrate over $y$ and set the limits of integration as $\pm b$.

Carry Out the Integration

Following thestrategy that has just been outlined, we have,

$V$	$=$	$\iint dydz\underset{-x_{\ell }}{\overset{+x_{\ell }}{\int }}dx=\iint dydz[x{]}_{-x_{\ell }}^{+x_{\ell }}=2\int dy\int x_{\ell }dz$
	$=$	$2a\int dy\int [1-\frac{y^2}{b^2}-\frac{z^2}{c^2}{]}^{1/2}dz=\frac{2a}{c}\int dy\underset{-z_{\ell }}{\overset{+z_{\ell }}{\int }}[z_{\ell }^2-z^2{]}^{1/2}dz$
	$=$	$\frac{2a}{c}\int \frac{dy}{2}[z\sqrt{z_{\ell }^2-z^2}+z_{\ell }^2{\sin }^{-1}(\frac{z}{|z_{\ell }|}){]}_{-z_{\ell }}^{+z_{\ell }}$
	$=$	$\frac{2a}{c}\int [z_{\ell }{\xcancel{\sqrt{z_{\ell }^2-z_{\ell }^2}}}^0+z_{\ell }^2{\sin }^{-1}(1\left)\right]dy=\frac{2a}{c}\int \left[\frac{\pi }{2}{z}_{\ell }^2\right]dy$
	$=$	$\pi ac\underset{-b}{\overset{+b}{\int }}(1-\frac{y^2}{b^2})dy=\pi ac[y-\frac{y^3}{3b^2}{]}_{-b}^{+b}$
	$=$	$\frac{4\pi }{3}\cdot abc.$

Appendix B:  Coriolis Component u₁x₂

$\iiint [u_{1}y]dxdydz$	$=$	$\iiint \{-[\frac{a^2}{a^2+b^2}]{\zeta }_{3}y+[\frac{a^2}{a^2+c^2}\left]{\zeta }_{2}z\right\}ydxdydz$
	$=$	$-\left[\frac{a^2}{a^2+b^2}\right]{\zeta }_3\iiint y^{2}dxdydz+\left[\frac{a^2}{a^2+c^2}\right]{\zeta }_2\iiint yzdxdydz$
	$=$	$-\left[\frac{a^2}{a^2+b^2}\right]{\zeta }_3\int y^{2}dy\int dz\underset{-x_{\ell }}{\overset{+x_{\ell }}{\int }}dx+\left[\frac{a^2}{a^2+c^2}\right]{\zeta }_2\int ydy\int zdz\underset{-x_{\ell }}{\overset{+x_{\ell }}{\int }}dx$
	$=$	$-\left[\frac{2a^2}{a^2+b^2}\right]{\zeta }_3\int y^{2}dy\int x_{\ell }dz+\left[\frac{2a^2}{a^2+c^2}\right]{\zeta }_2\int ydy\int z{x}_{\ell }dz$
	$=$	$-\left[\frac{2a^3}{a^2+b^2}\right]{\zeta }_3\int y^{2}dy\int [1-\frac{y^2}{b^2}-\frac{z^2}{c^2}{]}^{1/2}dz+[\frac{2a^3}{a^2+c^2}]{\zeta }_2\int ydy\int z[1-\frac{y^2}{b^2}-\frac{z^2}{c^2}{]}^{1/2}dz$
	$=$	$-\frac{1}{c}\left[\frac{2a^3}{a^2+b^2}\right]{\zeta }_3\int y^{2}dy\underset{-z_{\ell }}{\overset{+z_{\ell }}{\int }}[z_{\ell }^2-z^2{]}^{1/2}dz+\frac{1}{c}[\frac{2a^3}{a^2+c^2}]{\zeta }_2\int ydy\underset{-z_{\ell }}{\overset{+z_{\ell }}{\int }}z[z_{\ell }^2-z^2{]}^{1/2}dz$
	$=$	$-\frac{1}{c}\left[\frac{2a^3}{a^2+b^2}\right]{\zeta }_3\int y^{2}dy\cdot \frac{1}{2}\{z\sqrt{z_{\ell }^2-z^2}+z_{\ell }^2{\sin }^{-1}(\frac{z}{|z_{\ell }|}){\}}_{-z_{\ell }}^{+z_{\ell }}-\frac{1}{c}[\frac{2a^3}{a^2+c^2}]{\zeta }_2\int ydy\cdot \frac{1}{3}\{[z_{\ell }^2-z^2{]}^{3/2}{\}}_{-z_{\ell }}^{+z_{\ell }}$
	$=$	$-\frac{1}{c}\left[\frac{2a^3}{a^2+b^2}\right]{\zeta }_3\int y^{2}dy\cdot \frac{1}{2}\left\{z_{\ell }^2{\sin }^{-1}\right(\frac{z}{|z_{\ell }|}){\}}_{-z_{\ell }}^{+z_{\ell }}=-\pi ac[\frac{a^2}{a^2+b^2}\left]{\zeta }_3\underset{-b}{\overset{b}{\int }}{y}^2\right[1-\frac{y^2}{b^2}]dy$
	$=$	$-\pi ac\left[\frac{a^2}{a^2+b^2}\right]{\zeta }_3[\frac{y^3}{3}-\frac{y^5}{5b^2}{]}_{-b}^{+b}=-2\pi a{b}^{3}c[\frac{a^2}{a^2+b^2}\left]{\zeta }_3\right[\frac{2}{15}]=-\frac{4\pi abc}{3}[\frac{a^2}{a^2+b^2}\left]{\zeta }_3\right[\frac{b^2}{5}]$
	$=$	$-\frac{I_{22}}{\rho }\left[\frac{a^2}{a^2+b^2}\right]{\zeta }_3.$
[ EFE, Chapter 7, §47, p. 130, Eq. (9a) ]

Appendix C:  Coriolis Component u₁x₃

Here we will additionally make use of the integration limits,

$y_{\ell }^2$

$\equiv$

$b^2(1-\frac{z^2}{c^2}).$

Integration over the relevant Coriolis component gives,

$\iiint [u_{1}z]dxdydz$	$=$	$\iiint \{-[\frac{a^2}{a^2+b^2}]{\zeta }_{3}y+[\frac{a^2}{a^2+c^2}\left]{\zeta }_{2}z\right\}zdxdydz$
	$=$	$-\left[\frac{a^2}{a^2+b^2}\right]{\zeta }_3\iiint {\xcancel{yzdxdydz}}^0+\left[\frac{a^2}{a^2+c^2}\right]{\zeta }_2\iiint z^{2}dxdydz$
	$=$	$\left[\frac{a^2}{a^2+c^2}\right]{\zeta }_2\int z^{2}dz\int dy\underset{-x_{\ell }}{\overset{+x_{\ell }}{\int }}dx$
	$=$	$2a\left[\frac{a^2}{a^2+c^2}\right]{\zeta }_2\int z^{2}dz\int dy\left\{\right[1-\frac{y^2}{b^2}-\frac{z^2}{c^2}{]}^{1/2}\}$
	$=$	$\frac{2a}{b}\left[\frac{a^2}{a^2+c^2}\right]{\zeta }_2\int z^{2}dz\underset{-y_{\ell }}{\overset{+y_{\ell }}{\int }}[y_{\ell }^2-y^2{]}^{1/2}dy$
	$=$	$\frac{2a}{b}\left[\frac{a^2}{a^2+c^2}\right]{\zeta }_2\int z^{2}dz\cdot \frac{1}{2}\{y\sqrt{y_{\ell }^2-y^2}+y_{\ell }^2{\sin }^{-1}(\frac{y}{|y_{\ell }|}){\}}_{-y_{\ell }}^{+y_{\ell }}$
	$=$	$\frac{2a}{b}\left[\frac{a^2}{a^2+c^2}\right]{\zeta }_2\underset{-c}{\overset{c}{\int }}{z}^2\left\{\frac{\pi }{2}{y}_{\ell }^2\right\}dz=\pi ab\left[\frac{a^2}{a^2+c^2}\right]{\zeta }_2\underset{-c}{\overset{c}{\int }}{z}^2\{1-\frac{z^2}{c^2}\}dz$
	$=$	$\pi ab\left[\frac{a^2}{a^2+c^2}\right]{\zeta }_2\{\frac{z^3}{3}-\frac{z^5}{5c^2}{\}}_{-c}^{+c}=\pi ab[\frac{a^2}{a^2+c^2}\left]{\zeta }_2\right\{\frac{1}{3}-\frac{1}{5}\}2c^3=\frac{4\pi abc}{3}[\frac{a^2}{a^2+c^2}\left]{\zeta }_2\right\{\frac{c^2}{5}\}$
	$=$	$+\frac{I_{33}}{\rho }\left[\frac{a^2}{a^2+c^2}\right]{\zeta }_2.$
[ EFE, Chapter 7, §47, p. 130, Eq. (9b) ]

Appendix D:   The Other Four Coriolis Components

It follows that,

$\iiint [u_{2}x]dxdydz$	$=$	$\iiint \{-{\xcancel{\left[\frac{a_2^2}{a_2^2+a_3^2}\right]{\zeta }_{1}z}}^0+[\frac{a_2^2}{a_2^2+a_1^2}\left]{\zeta }_{3}x\right\}xdxdydz$
	$=$	$+\frac{I_{11}}{\rho }\left[\frac{b^2}{b^2+a^2}\right]{\zeta }_3;$
$\iiint [u_{2}z]dxdydz$	$=$	$\iiint \{-[\frac{a_2^2}{a_2^2+a_3^2}]{\zeta }_{1}z+{\xcancel{\left[\frac{a_2^2}{a_2^2+a_1^2}\right]{\zeta }_{3}x}}^0\}zdxdydz$
	$=$	$-\frac{I_{33}}{\rho }\left[\frac{b^2}{b^2+c^2}\right]{\zeta }_1;$
$\iiint [u_{3}x]dxdydz$	$=$	$\iiint \{-[\frac{a_3^2}{a_3^2+a_1^2}]{\zeta }_{2}x+{\xcancel{\left[\frac{a_3^2}{a_3^2+a_2^2}\right]{\zeta }_{1}y}}^0\}xdxdydz$
	$=$	$-\frac{I_{11}}{\rho }\left[\frac{c^2}{c^2+a^2}\right]{\zeta }_2;$
$\iiint [u_{3}y]dxdydz$	$=$	$\iiint \{-{\xcancel{\left[\frac{a_3^2}{a_3^2+a_1^2}\right]{\zeta }_{2}x}}^0+[\frac{a_3^2}{a_3^2+a_2^2}\left]{\zeta }_{1}y\right\}ydxdydz$
	$=$	$+\frac{I_{22}}{\rho }\left[\frac{c^2}{c^2+b^2}\right]{\zeta }_1.$

Appendix E:   Kinetic Energy Components

Diagonal Elements

$\left(\frac{2}{\rho }\right){\mathfrak{𝔗}}_{11}=\underset{V}{\int }{u}_1{u}_1{d}^{3}x$	$=$	$\iiint [u_1^2]dxdydz$
	$=$	$\iiint \{-[\frac{a^2}{a^2+b^2}]{\zeta }_{3}y+[\frac{a^2}{a^2+c^2}]{\zeta }_{2}z{\}}^{2}dxdydz$
	$=$	$\iiint \left\{\right[\frac{a^2}{a^2+b^2}{]}^2{\zeta }_3^2{y}^2-2{\xcancel{\left[\frac{a^2}{a^2+b^2}\right]\left[\frac{a^2}{a^2+c^2}\right]{\zeta }_2{\zeta }_3}}^{0}yz+\left[\frac{a^2}{a^2+c^2}{]}^2{\zeta }_2^2{z}^2\right\}dxdydz$
	$=$	$[\frac{a^2}{a^2+b^2}{]}^2{\zeta }_3^2\iiint y^{2}dxdydz+[\frac{a^2}{a^2+c^2}{]}^2{\zeta }_2^2\iiint z^{2}dxdydz$
	$=$	$\left[\frac{a^2}{a^2+b^2}{]}^2{\zeta }_3^2\right[\frac{I_{22}}{\rho }]+[\frac{a^2}{a^2+c^2}{]}^2{\zeta }_2^2\left[\frac{I_{33}}{\rho }\right].$

Similarly,

$\left(\frac{2}{\rho }\right){\mathfrak{𝔗}}_{22}=\underset{V}{\int }{u}_2{u}_2{d}^{3}x$	$=$	$\iiint [u_2^2]dxdydz$
	$=$	$\iiint \{-[\frac{b^2}{b^2+c^2}]{\zeta }_{1}z+[\frac{b^2}{b^2+a^2}]{\zeta }_{3}x{\}}^{2}dxdydz$
	$=$	$\left[\frac{b^2}{b^2+c^2}{]}^2{\zeta }_1^2\right[\frac{I_{33}}{\rho }]+[\frac{b^2}{b^2+a^2}{]}^2{\zeta }_3^2\left[\frac{I_{11}}{\rho }\right];$
$\left(\frac{2}{\rho }\right){\mathfrak{𝔗}}_{33}=\underset{V}{\int }{u}_3{u}_3{d}^{3}x$	$=$	$\iiint [u_2^2]dxdydz$
	$=$	$\iiint \{-[\frac{c^2}{c^2+a^2}]{\zeta }_{2}x+[\frac{c^2}{c^2+b^2}]{\zeta }_{1}y{\}}^{2}dxdydz$
	$=$	$\left[\frac{c^2}{c^2+a^2}{]}^2{\zeta }_2^2\right[\frac{I_{11}}{\rho }]+[\frac{c^2}{c^2+b^2}{]}^2{\zeta }_1^2\left[\frac{I_{22}}{\rho }\right].$

Off-Diagonal Elements

$\left(\frac{2}{\rho }\right){\mathfrak{𝔗}}_{23}=\underset{V}{\int }{u}_2{u}_3{d}^{3}x$	$=$	$\iiint [u_2{u}_3]dxdydz$
	$=$	$\iiint \{-[\frac{b^2}{b^2+c^2}]{\zeta }_{1}z+[\frac{b^2}{b^2+a^2}\left]{\zeta }_{3}x\right\}\{-[\frac{c^2}{c^2+a^2}]{\zeta }_{2}x+[\frac{c^2}{c^2+b^2}\left]{\zeta }_{1}y\right\}dxdydz$
	$=$	$\iiint \{-[\frac{b^2}{b^2+c^2}\left]{\zeta }_{1}z\right\}\{-[\frac{c^2}{c^2+a^2}]{\zeta }_{2}x+[\frac{c^2}{c^2+b^2}\left]{\zeta }_{1}y\right\}dxdydz$
		$+\iiint \left\{\right[\frac{b^2}{b^2+a^2}\left]{\zeta }_{3}x\right\}\{-[\frac{c^2}{c^2+a^2}]{\zeta }_{2}x+[\frac{c^2}{c^2+b^2}\left]{\zeta }_{1}y\right\}dxdydz$
	$=$	$-\iiint \left\{\right[\frac{b^2}{b^2+a^2}\left]\right[\frac{c^2}{c^2+a^2}\left]{\zeta }_2{\zeta }_3\right\}{x}^{2}dxdydz$
	$=$	$-\frac{I_{11}}{\rho }\left[\frac{b^2}{b^2+a^2}\right]\left[\frac{c^2}{c^2+a^2}\right]{\zeta }_2{\zeta }_3$
[ EFE, Chapter 7, §47, p. 130, Eq. (8) ]

Similarly,

$\left(\frac{2}{\rho }\right){\mathfrak{𝔗}}_{12}=\underset{V}{\int }{u}_1{u}_2{d}^{3}x$	$=$	$\iiint [u_1{u}_2]dxdydz$
	$=$	$\iiint \{-[\frac{a^2}{a^2+b^2}]{\zeta }_{3}y+[\frac{a^2}{a^2+c^2}\left]{\zeta }_{2}z\right\}\{-[\frac{b^2}{b^2+c^2}]{\zeta }_{1}z+[\frac{b^2}{b^2+a^2}\left]{\zeta }_{3}x\right\}dxdydz$
	$=$	$-\left[\frac{a^2}{a^2+c^2}\right]\left[\frac{b^2}{b^2+c^2}\right]{\zeta }_1{\zeta }_2\iiint z^{2}dxdydz$
	$=$	$-\frac{I_{33}}{\rho }\left[\frac{a^2}{a^2+c^2}\right]\left[\frac{b^2}{b^2+c^2}\right]{\zeta }_1{\zeta }_2;$
$\left(\frac{2}{\rho }\right){\mathfrak{𝔗}}_{31}=\underset{V}{\int }{u}_3{u}_1{d}^{3}x$	$=$	$\iiint [u_3{u}_1]dxdydz$
	$=$	$\iiint \{-[\frac{c^2}{c^2+a^2}]{\zeta }_{2}x+[\frac{c^2}{c^2+b^2}\left]{\zeta }_{1}y\right\}\{-[\frac{a^2}{a^2+b^2}]{\zeta }_{3}y+[\frac{a^2}{a^2+c^2}\left]{\zeta }_{2}z\right\}dxdydz$
	$=$	$-\left[\frac{c^2}{c^2+b^2}\right]\left[\frac{a^2}{a^2+b^2}\right]{\zeta }_1{\zeta }_3\iiint y^{2}dxdydz$
	$=$	$-\frac{I_{22}}{\rho }\left[\frac{c^2}{c^2+b^2}\right]\left[\frac{a^2}{a^2+b^2}\right]{\zeta }_1{\zeta }_3.$

And, finally,

${\mathfrak{𝔗}}_{32}$

$=$

${\mathfrak{𝔗}}_{23};$

${\mathfrak{𝔗}}_{21}$

$=$

${\mathfrak{𝔗}}_{12};$

and,

${\mathfrak{𝔗}}_{13}$

$=$

${\mathfrak{𝔗}}_{31}.$

See Also

• Properties of Maclaurin Spheroids
• Excerpts from Maclaurin's (1742) A Treatise of Fluxions