# Riemann (1826 - 1866)

## Background

Excerpt from p. 539 of R. Baker, C. Christenson & H. Orde (2004):

 "Newton first showed that the departure of the figure of the earth from a sphere is due to its rotation. Jacobi showed in 1834 that gravitational equilibrium of a rotating spheroid is consistent with three distinct axes if the angular momentum exceeds a critical value. Dirichlet had posed and partially analyzed the conditions for a configuration which is an ellipsoid varying with time, such that the motion in an inertial frame, is linear in the coordinates. His results were edited posthumously by Dedekind in 1860. In his published work dated 1861 — five years before his death — Riemann took up this problem of Dirichlet …"

Excerpt from p. 530 of R. Baker, C. Christenson & H. Orde (2004) which, in turn, is taken from R. Dedekind's accounting of The life of Bernhard Riemann:

 "In the Easter vacation of 1860 [Riemann] went on a trip to Paris, where he stayed for a month from 26th March; unfortunately the weather was raw and unfriendly and in the last week of his visit there was one day after another of snow and hail which made it almost impossible to see the sights. However, he was delighted with the friendly reception which he received from the Parisian scholars Serret, Bertrand, Hermite, Puiseux and Briot, with whom he spent a pleasant day in the country at Chatenay, along with Bouquet. In the same year, [Riemann] completed his paper on the motion of a fluid ellipsoid …"

Excerpt from pp. 184-185 of EFE:

 "Riemann's paper 'Ein Beitrag zu den Untersuchungen über die Bewegung eines flüssigen gleichartigen Ellipsoides,' communicated to Der Königlichen Gesellschaft der Wissenschaften zu Göttingen on December 8, 1860, is remarkable for the wealth of new results it contains and for the breadth of its comprehension of the entire range of problems. In the present writer's [S. Chandrasekhar] view this much neglected paper — [for example] there are no references to it in any of the writings of Poincaré, Darwin, or Jeans — deserves to be included among the other great papers of Riemann that are well known. … In view of Riemann's unique place in science, a critical appraisal of this paper is perhaps justified."

## His Published Work on Ellipsoids

Riemann's (1861) work, titled, "Ein Beitrag zu den Untersuchungen über die Bewegung eines flüssigen gleichartigen Ellipsoides" — English translation: "A contribution to the study of the motion of a homogeneous fluid ellipsoid" — can be found in various published collections of his papers:

Our description and detailed analysis of Riemann's (1861) work that follows, draws primarily from the 2004 translation of his collected works.

# Governing Equations

## Step 1

At the beginning of his discussion, Riemann denotes by ${\displaystyle (x,y,z)}$ "the [inertial-frame] coordinates of an element of the fluid body at time ${\displaystyle t}$," and he denotes by ${\displaystyle (\xi ,\eta ,\zeta )}$ "the coordinates of the point ${\displaystyle (x,y,z)}$ with respect to a moving coordinate system, whose axes coincide at each instant with the principal axes of the ellipsoid." Drawing from our accompanying discussion of Euler angles, it seems appropriate to associate ${\displaystyle {\vec {A}}_{XYZ}}$ with the vector that points from the origin to the ${\displaystyle (x,y,z)}$ location of the fluid element as viewed from the inertial reference frame, and to associate ${\displaystyle {\vec {A}}_{\mathrm {body} }}$ with the vector that points from the origin to the same location of the fluid element, but as viewed from Riemann's specified rotating frame. With this in mind, we have the following notation mappings:

 ${\displaystyle {\vec {A}}_{\mathrm {XYZ} }(A_{X},A_{Y},A_{Z})}$ ${\displaystyle \rightarrow }$ ${\displaystyle {\vec {A}}_{\mathrm {inertial} }(x,y,z)={\vec {e}}_{X}(x)+{\vec {e}}_{Y}(y)+{\vec {e}}_{Z}(z)\,,}$ ${\displaystyle {\vec {A}}_{\mathrm {body} }(A_{1},A_{2},A_{3})}$ ${\displaystyle \rightarrow }$ ${\displaystyle {\vec {A}}_{\mathrm {body} }(\xi ,\eta ,\zeta )={\vec {e}}_{1}(\xi )+{\vec {e}}_{2}(\eta )+{\vec {e}}_{3}(\zeta )\,.}$

Again drawing from our accompanying Euler angles discussion, quite generally these two coordinate representations — of the same vector, ${\displaystyle {\vec {A}}}$ — are related to one another via the matrix expression,

${\displaystyle {\vec {A}}_{\mathrm {body} }={\hat {R}}\cdot {\vec {A}}_{\mathrm {inertial} }\,,}$

that is,

 ${\displaystyle {\begin{bmatrix}\xi \\\eta \\\zeta \end{bmatrix}}}$ ${\displaystyle ~=}$ ${\displaystyle {\begin{bmatrix}{\vec {e}}_{1}\cdot {\vec {e}}_{X}&{\vec {e}}_{1}\cdot {\vec {e}}_{Y}&{\vec {e}}_{1}\cdot {\vec {e}}_{Z}\\{\vec {e}}_{2}\cdot {\vec {e}}_{X}&{\vec {e}}_{2}\cdot {\vec {e}}_{Y}&{\vec {e}}_{2}\cdot {\vec {e}}_{Z}\\{\vec {e}}_{3}\cdot {\vec {e}}_{X}&{\vec {e}}_{3}\cdot {\vec {e}}_{Y}&{\vec {e}}_{3}\cdot {\vec {e}}_{Z}\end{bmatrix}}\cdot {\begin{bmatrix}x\\y\\z\end{bmatrix}}={\hat {R}}\cdot {\begin{bmatrix}x\\y\\z\end{bmatrix}}\,.}$

In terms of the trio of Euler angles, the rotation matrix is,

 ${\displaystyle {\hat {R}}(\phi ,\theta ,\psi )}$ ${\displaystyle =}$ ${\displaystyle {\begin{bmatrix}(\cos \psi \cos \phi -\sin \phi \sin \psi \cos \theta )&(\cos \psi \sin \phi +\cos \phi \sin \psi \cos \theta )&(\sin \psi \sin \theta )\\(-\sin \psi \cos \phi -\sin \phi \cos \psi \cos \theta )&(-\sin \phi \sin \psi +\cos \phi \cos \psi \cos \theta )&(\cos \psi \sin \theta )\\(\sin \theta \sin \phi )&(-\sin \theta \cos \phi )&(\cos \theta )\end{bmatrix}}\,.}$

This means that — see our accompanying discussion — in terms of the trio of Euler angles, the following three coordinate mappings hold:

 ${\displaystyle \xi }$ = ${\displaystyle x(\cos \psi \cos \phi -\sin \psi \sin \phi \cos \theta )+y(\sin \phi \cos \psi +\sin \psi \cos \theta \cos \phi )+z(\sin \psi \sin \theta )\,,}$ ${\displaystyle \eta }$ = ${\displaystyle x(-\sin \psi \cos \phi -\sin \phi \cos \theta \cos \psi )+y(-\sin \psi \sin \phi +\cos \psi \cos \theta \cos \phi )+z(\sin \theta \cos \psi )\,,}$ ${\displaystyle \zeta }$ = ${\displaystyle x(\sin \theta \sin \phi )+y(-\sin \theta \cos \phi )+z(\cos \theta )\,.}$

Notice the correspondence between this set of coordinate relations and the set marked by Riemann (1861) as equation (2) of §1.

From §1 of Riemann (1861)

Then ${\displaystyle \xi }$, ${\displaystyle \eta }$, ${\displaystyle \zeta }$ are known … to be linear expressions in ${\displaystyle x}$, ${\displaystyle y}$, ${\displaystyle z}$,

 ${\displaystyle \xi }$ = ${\displaystyle \alpha x+\beta y+\gamma z\,,}$ ${\displaystyle \eta }$ = ${\displaystyle \alpha 'x+\beta 'y+\gamma 'z\,,}$ ${\displaystyle \zeta }$ = ${\displaystyle \alpha ''x+\beta ''y+\gamma ''z\,.}$

The coefficients are the cosines of the angles that the axes of one system form with the axes of the other … For example, ${\displaystyle \alpha ={\vec {e}}_{1}\cdot {\vec {e}}_{X}}$, ${\displaystyle \beta ={\vec {e}}_{1}\cdot {\vec {e}}_{Y}}$, and so on.

Alternatively, given that,

${\displaystyle {\vec {A}}_{\mathrm {inertial} }={\hat {R}}^{-1}\cdot {\vec {A}}_{\mathrm {body} }\,,}$

the following additional three mapping relations also must hold:

 ${\displaystyle x}$ = ${\displaystyle \xi (\cos \psi \cos \phi -\sin \psi \sin \phi \cos \theta )+\eta (-\sin \psi \cos \phi -\sin \phi \cos \theta \cos \psi )+\zeta (\sin \theta \sin \phi )\,,}$ ${\displaystyle y}$ = ${\displaystyle \xi (\sin \phi \cos \psi +\sin \psi \cos \theta \cos \phi )+\eta (-\sin \psi \sin \phi +\cos \psi \cos \theta \cos \phi )+\zeta (-\sin \theta \cos \phi )\,,}$ ${\displaystyle z}$ = ${\displaystyle \xi (\sin \psi \sin \theta )+\eta (\sin \theta \cos \psi )+\zeta (\cos \theta )\,.}$

## Step 2

Given that ${\displaystyle (x,y,z)}$ and ${\displaystyle (\xi ,\eta ,\zeta )}$ are components of the same position vector, ${\displaystyle {\vec {A}}}$, it must in general be the case that the square of the length of the vector is the same for both coordinate representations. That is, as Riemann (1861) states in §1 following equation (2), it must be the case that,

${\displaystyle \xi ^{2}+\eta ^{2}+\zeta ^{2}=x^{2}+y^{2}+z^{2}\,.}$

Let's specifically assess whether or not this holds true when Euler angles are used to relate the components of the two position vector expressions.

 ${\displaystyle \xi ^{2}+\eta ^{2}+\zeta ^{2}}$ = ${\displaystyle {\biggl [}x(\cos \psi \cos \phi -\sin \psi \sin \phi \cos \theta )+y(\sin \phi \cos \psi +\sin \psi \cos \theta \cos \phi )+z(\sin \psi \sin \theta ){\biggr ]}^{2}}$ ${\displaystyle +{\biggl [}x(-\sin \psi \cos \phi -\sin \phi \cos \theta \cos \psi )+y(-\sin \psi \sin \phi +\cos \psi \cos \theta \cos \phi )+z(\sin \theta \cos \psi ){\biggr ]}^{2}}$ ${\displaystyle +{\biggl [}x(\sin \theta \sin \phi )+y(-\sin \theta \cos \phi )+z(\cos \theta ){\biggr ]}^{2}\,.}$

As Riemann (1861) states, between these coefficients, six equations hold …. In particular, when they are written in terms of Euler angles, on the right-hand side, the coefficient of the six various terms is …

 ${\displaystyle x^{2}:}$ ${\displaystyle (\cos \psi \cos \phi -\sin \psi \sin \phi \cos \theta )^{2}+(-\sin \psi \cos \phi -\sin \phi \cos \theta \cos \psi )^{2}+(\sin \theta \sin \phi )^{2}}$ ${\displaystyle =\cos ^{2}\psi \cos ^{2}\phi -2\cos \psi \cos \phi \sin \psi \sin \phi \cos \theta +\sin ^{2}\psi \sin ^{2}\phi \cos ^{2}\theta +\sin ^{2}\psi \cos ^{2}\phi +2\sin \psi \cos \phi \sin \phi \cos \theta \cos \psi +\sin ^{2}\phi \cos ^{2}\theta \cos ^{2}\psi +\sin ^{2}\theta \sin ^{2}\phi }$ ${\displaystyle =(2\sin \psi \cos \phi \sin \phi \cos \theta \cos \psi -2\cos \psi \cos \phi \sin \psi \sin \phi \cos \theta )+(\sin ^{2}\psi +\cos ^{2}\psi )\sin ^{2}\phi \cos ^{2}\theta +(\sin ^{2}+\cos ^{2}\psi )\cos ^{2}\phi +\sin ^{2}\theta \sin ^{2}\phi }$ ${\displaystyle =\sin ^{2}\phi \cos ^{2}\theta +\cos ^{2}\phi +\sin ^{2}\theta \sin ^{2}\phi }$ ${\displaystyle =1\,.}$
 ${\displaystyle y^{2}:}$ ${\displaystyle (\sin \phi \cos \psi +\sin \psi \cos \theta \cos \phi )^{2}+(-\sin \psi \sin \phi +\cos \psi \cos \theta \cos \phi )^{2}+(-\sin \theta \cos \phi )^{2}}$ ${\displaystyle =\sin ^{2}\phi \cos ^{2}\psi +2\sin \phi \cos \psi \sin \psi \cos \theta \cos \phi +\sin ^{2}\psi \cos ^{2}\theta \cos ^{2}\phi +\sin ^{2}\psi \sin ^{2}\phi -2\sin \psi \sin \phi \cos \psi \cos \theta \cos \phi +\cos ^{2}\psi \cos ^{2}\theta \cos ^{2}\phi +\sin ^{2}\theta \cos ^{2}\phi }$ ${\displaystyle =(2\sin \phi \cos \psi \sin \psi \cos \theta \cos \phi -2\sin \psi \sin \phi \cos \psi \cos \theta \cos \phi )+(\cos ^{2}\psi +\sin ^{2}\psi )\sin ^{2}\phi +(\sin ^{2}\psi +\cos ^{2}\psi )\cos ^{2}\theta \cos ^{2}\phi +\sin ^{2}\theta \cos ^{2}\phi }$ ${\displaystyle =\sin ^{2}\phi +\cos ^{2}\theta \cos ^{2}\phi +\sin ^{2}\theta \cos ^{2}\phi }$ ${\displaystyle =1\,.}$
 ${\displaystyle z^{2}:}$ ${\displaystyle (\sin \psi \sin \theta )^{2}+(\sin \theta \cos \psi )^{2}+(\cos \theta )^{2}}$ ${\displaystyle =1\,.}$
 ${\displaystyle 2xy:}$ ${\displaystyle (\cos \psi \cos \phi -\sin \psi \sin \phi \cos \theta )(\sin \phi \cos \psi +\sin \psi \cos \theta \cos \phi )+(-\sin \psi \cos \phi -\sin \phi \cos \theta \cos \psi )(-\sin \psi \sin \phi +\cos \psi \cos \theta \cos \phi )+(\sin \theta \sin \phi )(-\sin \theta \cos \phi )}$ ${\displaystyle =\cos \psi \cos \phi (\sin \phi \cos \psi +\sin \psi \cos \theta \cos \phi )-\sin \psi \sin \phi \cos \theta (\sin \phi \cos \psi +\sin \psi \cos \theta \cos \phi )+(\sin \psi \cos \phi +\sin \phi \cos \theta \cos \psi )(\sin \psi \sin \phi -\cos \psi \cos \theta \cos \phi )-\sin ^{2}\theta \sin \phi \cos \phi }$ ${\displaystyle =\cos ^{2}\psi (\sin \phi \cos \phi )+\sin \psi \cos \psi \cos \theta (-\sin ^{2}\phi +\cos ^{2}\phi )-\sin ^{2}\psi \cos ^{2}\theta (\sin \phi \cos \phi )+\sin \psi \cos \phi (\sin \psi \sin \phi )-\sin \psi \cos \phi (\cos \psi \cos \theta \cos \phi )+\sin \phi \cos \theta \cos \psi (\sin \psi \sin \phi )-\sin \phi \cos \theta \cos \psi (\cos \psi \cos \theta \cos \phi )-\sin ^{2}\theta \sin \phi \cos \phi }$ ${\displaystyle =[\sin \psi (\cos \psi )\cos \theta (\sin ^{2}\phi -\cos ^{2}\phi )+\sin \psi \cos \psi \cos \theta (-\sin ^{2}\phi +\cos ^{2}\phi )]+\sin \phi \cos \phi [\sin ^{2}\psi (1-\cos ^{2}\theta )-\cos ^{2}\theta \cos ^{2}\psi -\sin ^{2}\theta +\cos ^{2}\psi ]}$ ${\displaystyle =\sin \phi \cos \phi [\sin ^{2}\psi (1-\cos ^{2}\theta )-\sin ^{2}\theta \sin ^{2}\psi ]}$ ${\displaystyle =\sin \phi \cos \phi \sin ^{2}\psi [1-\cos ^{2}\theta -\sin ^{2}\theta ]}$ ${\displaystyle =0\,.}$
 ${\displaystyle 2xz:}$ ${\displaystyle (\cos \psi \cos \phi -\sin \psi \sin \phi \cos \theta )(\sin \psi \sin \theta )+(-\sin \psi \cos \phi -\sin \phi \cos \theta \cos \psi )(\sin \theta \cos \psi )+(\sin \theta \sin \phi )\cos \theta }$ ${\displaystyle =\cos \psi \cos \phi (\sin \psi \sin \theta )-\sin \psi \sin \phi \cos \theta (\sin \psi \sin \theta )-\sin \psi \cos \phi (\sin \theta \cos \psi )-\sin \phi \cos \theta \cos \psi (\sin \theta \cos \psi )+(\sin \theta \sin \phi )\cos \theta }$ ${\displaystyle =\cos \psi \sin \psi [\cos \phi \sin \theta -\cos \phi \sin \theta ]-\sin ^{2}\psi \sin \phi \cos \theta \sin \theta +\sin \theta \sin \phi \cos \theta [1-\cos ^{2}\psi ]}$ ${\displaystyle =0\,.}$
 ${\displaystyle 2yz:}$ ${\displaystyle (\sin \phi \cos \psi +\sin \psi \cos \theta \cos \phi )\sin \psi \sin \theta +(-\sin \psi \sin \phi +\cos \psi \cos \theta \cos \phi )\sin \theta \cos \psi +(-\sin \theta \cos \phi )\cos \theta }$ ${\displaystyle =[\sin \psi \cos \psi \sin \phi \sin \theta -\sin \psi \cos \psi \sin \phi \sin \theta ]+\sin ^{2}\psi \cos \theta \cos \phi \sin \theta +(\cos ^{2}\psi -1)\cos \theta \cos \phi \sin \theta }$ ${\displaystyle =0\,.}$

## Step 3

At the bottom of p. 176 of BCO2004, Riemann summarizes the "geometric significance" of various variables that he has introduced:

1. ${\displaystyle (\xi ',\eta ',\zeta ')}$ are the velocity components of the point ${\displaystyle (x,y,z)}$ of the fluid mass parallel to the axes of ${\displaystyle (\xi ,\eta ,\zeta )}$. Note that, in the terms we have used above, ${\displaystyle (x,y,z)}$ refers to the position of a fluid element as viewed from the inertial reference frame, while ${\displaystyle (\xi ,\eta ,\zeta )}$ refers to the position of that same fluid element as viewed from the body frame. The expressions that Riemann derived for these three components of the (inertial-frame) fluid velocity appear on the same page, as equation (7) of §2, namely,
 ${\displaystyle \xi '}$ = ${\displaystyle {\biggl (}{\frac {\xi }{a}}{\biggr )}{\frac {da}{dt}}+(ar_{1}-br){\frac {\eta }{b}}+(cq-aq_{1}){\frac {\zeta }{c}}\,,}$ ${\displaystyle \eta '}$ = ${\displaystyle {\biggl (}{\frac {\eta }{b}}{\biggr )}{\frac {db}{dt}}+(ar-br_{1}){\frac {\xi }{a}}+(bp_{1}-cp){\frac {\zeta }{c}}\,,}$ ${\displaystyle \zeta '}$ = ${\displaystyle {\biggl (}{\frac {\zeta }{c}}{\biggr )}{\frac {dc}{dt}}+(cq_{1}-aq){\frac {\xi }{a}}+(bp-cp_{1}){\frac {\eta }{b}}\,.}$
2. ${\displaystyle (\partial \xi /\partial t,\partial \eta /\partial t,\partial \zeta /\partial t)}$ are the relative velocities decomposed in the same way, for the coordinate system, ${\displaystyle (\xi ,\eta ,\zeta )}$. The expressions that Riemann derived for these three components of the (body-frame) fluid velocity appear as equation (6) of §2, namely,
 ${\displaystyle {\frac {\partial }{\partial t}}{\biggl (}{\frac {\xi }{a}}{\biggr )}}$ = ${\displaystyle (r_{1}){\frac {\eta }{b}}-(q_{1}){\frac {\zeta }{c}}\,,}$ ${\displaystyle {\frac {\partial }{\partial t}}{\biggl (}{\frac {\eta }{b}}{\biggr )}}$ = ${\displaystyle (p_{1}){\frac {\zeta }{c}}-(r_{1}){\frac {\xi }{a}}\,,}$ ${\displaystyle {\frac {\partial }{\partial t}}{\biggl (}{\frac {\zeta }{c}}{\biggr )}}$ = ${\displaystyle (q_{1}){\frac {\xi }{a}}-(p_{1}){\frac {\eta }{b}}\,.}$
3. ${\displaystyle (p,q,r)}$ are the instantaneous rotations of the coordinate system ${\displaystyle (\xi ,\eta ,\zeta )}$ about its axes.
4. ${\displaystyle (p_{1},q_{1},r_{1})}$ have the same significance for the coordinate system ${\displaystyle (\xi _{1},\eta _{1},\zeta _{1})}$.

Initially we were confused regarding the differences between the two referenced coordinate systems, ${\displaystyle (\xi ,\eta ,\zeta )}$ and ${\displaystyle (\xi _{1},\eta _{1},\zeta _{1})}$. Immediately below equation (1) of §1 (near the top of p. 172 of BCO2004), we find the following definition: "Denote by ${\displaystyle (\xi ,\eta ,\zeta )}$ the coordinates of the point ${\displaystyle (x,y,z)}$ with respect to a moving coordinate system, whose axes coincide at each instant with the principal axes of the ellipsoid. On the other hand, near the top of p. 173 of BCO2004, Riemann refers to a set of coefficients that "… can be treated as the cosines of the angles that the axes of a moving coordinate system ${\displaystyle (\xi _{1},\eta _{1},\zeta _{1})}$ form with the axes of the fixed coordinate system ${\displaystyle (x,y,z)}$."

Adopting Figure 1 along with the notation that has been used in our accompanying discussion of Euler Angles, the distinction seems to be that ${\displaystyle (\xi _{1},\eta _{1},\zeta _{1})}$ stands for the unit vectors ${\displaystyle ({\vec {e}}_{x_{1}},{\vec {e}}_{x_{2}},{\vec {e}}_{x_{3}})}$ associated with a moving coordinate system — in the context of Riemann's work, a "moving" system that is always aligned with the principal axes of the (tumbling) ellipsoid — while ${\displaystyle (\xi ,\eta ,\zeta )}$ gives the coordinates of a single fluid element as referenced to this moving coordinate system — the (red) vector, ${\displaystyle {\vec {A}}}$, extends from the origin to this point in space. If this is the correct interpretation of Riemann's notation, then we would argue that Riemann's explanation of the 3rd item of "geometric significance" is poorly worded.

 Figure 1

If we are interpreting these two velocity expressions correctly, then the difference between the inertial-frame velocity and the rotating-frame velocity should be, ${\displaystyle {\vec {\Omega }}\times {\vec {x}}}$. From our associated discussion of the EFE presentation, we find,

 ${\displaystyle {\vec {\Omega }}\times {\vec {x}}={\boldsymbol {u}}^{(0)}-{\boldsymbol {u}}}$ = ${\displaystyle ({\boldsymbol {\hat {\jmath }}}\Omega _{2}+{\boldsymbol {\hat {k}}}\Omega _{3})\times ({\boldsymbol {\hat {\imath }}}x+{\boldsymbol {\hat {\jmath }}}y+{\boldsymbol {\hat {k}}}z)}$ = ${\displaystyle {\boldsymbol {\hat {\imath }}}(\Omega _{2}z-\Omega _{3}y)+{\boldsymbol {\hat {\jmath }}}(\Omega _{3}x)-{\boldsymbol {\hat {k}}}(\Omega _{2}x)\,.}$

Using Riemann's notation instead — assuming our interpretation of Riemann's description is correct — we have,

 ${\displaystyle {\vec {\Omega }}\times {\vec {x}}}$ = ${\displaystyle {\boldsymbol {\hat {\imath }}}{\biggl [}\xi '-{\frac {\partial \xi }{\partial t}}{\biggr ]}+{\boldsymbol {\hat {\jmath }}}{\biggl [}\eta '-{\frac {\partial \eta }{\partial t}}{\biggr ]}+{\boldsymbol {\hat {k}}}{\biggl [}\zeta '-{\frac {\partial \zeta }{\partial t}}{\biggr ]}}$ = ${\displaystyle {\boldsymbol {\hat {\imath }}}{\biggl \{}{\biggl (}{\frac {\xi }{a}}{\biggr )}{\cancelto {0}{\frac {da}{dt}}}+(ar_{1}-br){\frac {\eta }{b}}+(cq-aq_{1}){\frac {\zeta }{c}}-{\biggl [}(ar_{1}){\frac {\eta }{b}}-(aq_{1}){\frac {\zeta }{c}}{\biggr ]}{\biggr \}}}$ ${\displaystyle +{\boldsymbol {\hat {\jmath }}}{\biggl \{}{\biggl (}{\frac {\eta }{b}}{\biggr )}{\cancelto {0}{\frac {db}{dt}}}+(ar-br_{1}){\frac {\xi }{a}}+(bp_{1}-cp){\frac {\zeta }{c}}-{\biggl [}(bp_{1}){\frac {\zeta }{c}}-(br_{1}){\frac {\xi }{a}}{\biggr ]}{\biggr \}}}$ ${\displaystyle +{\boldsymbol {\hat {k}}}{\biggl \{}{\biggl (}{\frac {\zeta }{c}}{\biggr )}{\cancelto {0}{\frac {dc}{dt}}}+(cq_{1}-aq){\frac {\xi }{a}}+(bp-cp_{1}){\frac {\eta }{b}}-{\biggl [}(cq_{1}){\frac {\xi }{a}}-(cp_{1}){\frac {\eta }{b}}{\biggr ]}{\biggr \}}}$ = ${\displaystyle {\boldsymbol {\hat {\imath }}}{\biggl [}(cq){\frac {\zeta }{c}}-(br){\frac {\eta }{b}}{\biggr ]}+{\boldsymbol {\hat {\jmath }}}{\biggl [}(ar){\frac {\xi }{a}}-(cp){\frac {\zeta }{c}}{\biggr ]}+{\boldsymbol {\hat {k}}}{\biggl [}(bp){\frac {\eta }{b}}-(aq){\frac {\xi }{a}}{\biggr ]}\,.}$

Now, given that (see the bottom of p. 177 of BCO2004),

 ${\displaystyle p=(u+u')\,,}$ and, ${\displaystyle q=(v+v')\,,}$ and, ${\displaystyle r=(w+w')\,,}$

this expression becomes,

 ${\displaystyle {\vec {\Omega }}\times {\vec {x}}}$ = ${\displaystyle {\boldsymbol {\hat {\imath }}}{\biggl [}(v+v')\zeta -(w+w')\eta {\biggr ]}+{\boldsymbol {\hat {\jmath }}}{\biggl [}(w+w')\xi -(u+u')\zeta {\biggr ]}+{\boldsymbol {\hat {k}}}{\biggl [}(u+u')\eta -(v+v')\xi {\biggr ]}\,.}$

Given what has been derived below, namely,

 ${\displaystyle {\biggl [}{\frac {w'}{w}}{\biggr ]}^{2}}$ = ${\displaystyle {\frac {(2a-b-c)}{(2a+b+c)}}\cdot {\frac {(2a-b+c)}{(2a+b-c)}}\,,}$ ${\displaystyle {\biggl [}{\frac {v'}{v}}{\biggr ]}^{2}}$ = ${\displaystyle {\biggl [}{\frac {(2a-b-c)}{(2a+b+c)}}\cdot {\frac {(2a+b-c)}{(2a-b+c)}}{\biggr ]}}$

we see that,

 ${\displaystyle (v+v')}$ = ${\displaystyle {\biggl \{}1+{\biggl [}{\frac {(2a-b-c)}{(2a+b+c)}}\cdot {\frac {(2a+b-c)}{(2a-b+c)}}{\biggr ]}^{1/2}{\biggr \}}v}$ = ${\displaystyle {\biggl \{}1+{\biggl [}{\frac {(2a-b-c)}{(2a+b+c)}}\cdot {\frac {(2a+b-c)}{(2a-b+c)}}{\biggr ]}^{1/2}{\biggr \}}(2a+b+c)^{1/2}(2a-b+c)^{1/2}~S^{1/2}}$ = ${\displaystyle {\biggl [}(2a+b+c)^{1/2}(2a-b+c)^{1/2}+(2a-b-c)^{1/2}(2a+b-c)^{1/2}{\biggr ]}~S^{1/2}}$ = ${\displaystyle {\biggl \{}[(2a+c)+b]^{1/2}[(2a+c)-b]^{1/2}+[(2a-c)-b]^{1/2}[(2a-c)+b]^{1/2}{\biggr \}}~S^{1/2}}$ = ${\displaystyle {\biggl \{}[(2a+c)^{2}-b^{2}]^{1/2}+[(2a-c)^{2}-b^{2}]^{1/2}{\biggr \}}~S^{1/2}}$ = ${\displaystyle {\biggl \{}[(4a^{2}+c^{2}-b^{2})+4ac]^{1/2}+[(4a^{2}+c^{2}-b^{2})-4ac]^{1/2}{\biggr \}}~S^{1/2}}$ = ${\displaystyle {\biggl \{}[(4a^{2}+c^{2}-b^{2})+4ac]+[(4a^{2}+c^{2}-b^{2})-4ac]+2[(4a^{2}+c^{2}-b^{2})+4ac]^{1/2}[(4a^{2}+c^{2}-b^{2})-4ac]^{1/2}{\biggr \}}^{1/2}~S^{1/2}}$ = ${\displaystyle {\biggl \{}2(4a^{2}+c^{2}-b^{2})+2[(4a^{2}+c^{2}-b^{2})^{2}-16a^{2}c^{2}]^{1/2}{\biggr \}}^{1/2}~S^{1/2}}$

and,

 ${\displaystyle (w+w')}$ = ${\displaystyle {\biggl \{}1+{\biggl [}{\frac {(2a-b-c)}{(2a+b+c)}}\cdot {\frac {(2a-b+c)}{(2a+b-c)}}{\biggr ]}^{1/2}{\biggr \}}w}$ = ${\displaystyle {\biggl \{}1+{\biggl [}{\frac {(2a-b-c)}{(2a+b+c)}}\cdot {\frac {(2a-b+c)}{(2a+b-c)}}{\biggr ]}^{1/2}{\biggr \}}(2a+b+c)^{1/2}(2a+b-c)^{1/2}~T^{1/2}\,.}$

## Raw Form of Ten Governing Relations

In summary (see p. 178 of BCO2004), Riemann states that the system of equations that the ten unknown functions of time must satisfy becomes …

${\displaystyle {\frac {1}{2}}{\frac {d^{2}a}{dt^{2}}}+\epsilon aA-{\frac {\sigma }{a}}}$

=

${\displaystyle (a-c)v^{2}+(a+c)(v')^{2}+(a-b)w^{2}+(a+b)(w')^{2}\,,}$

 ${\displaystyle (\alpha .1)}$

${\displaystyle {\frac {1}{2}}{\frac {d^{2}b}{dt^{2}}}+\epsilon bB-{\frac {\sigma }{b}}}$

=

${\displaystyle (b-a)w^{2}+(b+a)(w')^{2}+(b-c)u^{2}+(b+c)(u')^{2}\,,}$

 ${\displaystyle (\alpha .2)}$

${\displaystyle {\frac {1}{2}}{\frac {d^{2}c}{dt^{2}}}+\epsilon cC-{\frac {\sigma }{c}}}$

=

${\displaystyle (c-b)u^{2}+(c+b)(u')^{2}+(c-a)v^{2}+(c+a)(v')^{2}\,,}$

 ${\displaystyle (\alpha .3)}$

${\displaystyle {\frac {1}{(b-c)}}{\frac {d}{dt}}{\biggl [}u(b-c)^{2}{\biggr ]}}$

=

${\displaystyle -(b+c-2a)vw-(b+c+2a)v'w'\,,}$

 ${\displaystyle (\alpha .4)}$

${\displaystyle {\frac {1}{(b+c)}}{\frac {d}{dt}}{\biggl [}u'(b+c){\biggr ]}}$

=

${\displaystyle -(b-c+2a)vw'-(b-c-2a)v'w\,,}$

 ${\displaystyle (\alpha .5)}$

${\displaystyle {\frac {1}{(c-a)}}{\frac {d}{dt}}{\biggl [}v(c-a)^{2}{\biggr ]}}$

=

${\displaystyle -(c+a-2b)wu-(c+a+2b)w'u'\,,}$

 ${\displaystyle (\alpha .6)}$

${\displaystyle {\frac {1}{(c+a)}}{\frac {d}{dt}}{\biggl [}v'(c+a)^{2}{\biggr ]}}$

=

${\displaystyle -(c-a+2b)wu'-(c-a-2b)w'u\,,}$

 ${\displaystyle (\alpha .7)}$

${\displaystyle {\frac {1}{(a-b)}}{\frac {d}{dt}}{\biggl [}w(a-b)^{2}{\biggr ]}}$

=

${\displaystyle -(a+b-2c)uv-(a+b+2c)u'v'\,,}$

 ${\displaystyle (\alpha .8)}$

${\displaystyle {\frac {1}{(a+b)}}{\frac {d}{dt}}{\biggl [}w'(a+b)^{2}{\biggr ]}}$

=

${\displaystyle -(a-b+2c)uv'-(a-b-2c)u'v\,,}$

 ${\displaystyle (\alpha .9)}$

${\displaystyle abc}$

=

${\displaystyle a_{0}b_{0}c_{0}\,.}$

 ${\displaystyle (\alpha .10)}$

## Gravitational Potential

For the gravitational potential, ${\displaystyle V}$, (see p. 178 of BCO2004), Riemann adopts the expression,

 ${\displaystyle V}$ = ${\displaystyle H-A\xi ^{2}-B\eta ^{2}-C\zeta ^{2}}$ = ${\displaystyle \pi \int _{0}^{\infty }{\frac {ds}{\Delta _{R}}}{\biggl [}1-{\frac {\xi ^{2}}{(a^{2}+s)}}-{\frac {\eta ^{2}}{(b^{2}+s)}}-{\frac {\zeta ^{2}}{(c^{2}+s)}}{\biggr ]}\,,}$

where,

 ${\displaystyle \Delta _{R}}$ ${\displaystyle \equiv }$ ${\displaystyle {\biggl [}{\biggl (}1+{\frac {s}{a^{2}}}{\biggr )}{\biggl (}1+{\frac {s}{b^{2}}}{\biggr )}{\biggl (}1+{\frac {s}{c^{2}}}{\biggr )}{\biggr ]}^{1/2}\,.}$

From our separate discussion of the gravitational potential of homogeneous ellipsoids, which closely follows the notation used in EFE, we find,

${\displaystyle ~\Phi ({\vec {x}})=-\pi G\rho {\biggl [}I_{\mathrm {BT} }a_{1}^{2}-{\biggl (}A_{1}x^{2}+A_{2}y^{2}+A_{3}z^{2}{\biggr )}{\biggr ]},}$

[ EFE, Chapter 3, Eq. (40)1,2 ]
[ BT87, Chapter 2, Table 2-2 ]

where,

 ${\displaystyle ~A_{i}}$ ${\displaystyle ~\equiv }$ ${\displaystyle ~a_{1}a_{2}a_{3}\int _{0}^{\infty }{\frac {du}{\Delta (a_{i}^{2}+u)}},}$ ${\displaystyle ~I_{\mathrm {BT} }}$ ${\displaystyle ~\equiv }$ ${\displaystyle ~{\frac {a_{2}a_{3}}{a_{1}}}\int _{0}^{\infty }{\frac {du}{\Delta }}=A_{1}+A_{2}{\biggl (}{\frac {a_{2}}{a_{1}}}{\biggr )}^{2}+A_{3}{\biggl (}{\frac {a_{3}}{a_{1}}}{\biggr )}^{2},}$ ${\displaystyle ~\Delta }$ ${\displaystyle ~\equiv }$ ${\displaystyle ~{\biggl [}(a_{1}^{2}+u)(a_{2}^{2}+u)(a_{3}^{2}+u){\biggr ]}^{1/2}.}$

[ EFE, Chapter 3, Eqs. (18), (15 & 22)1, & (8), respectively ]
[ BT87, Chapter 2, Table 2-2 ]

How do these two equations for the potential relate to one another? Multiplying Riemann's expression through by ${\displaystyle (-G\rho )}$, we have,

 ${\displaystyle (-G\rho )V}$ = ${\displaystyle -\pi G\rho \int _{0}^{\infty }{\frac {(abc)ds}{\Delta }}{\biggl [}1-{\frac {\xi ^{2}}{(a^{2}+s)}}-{\frac {\eta ^{2}}{(b^{2}+s)}}-{\frac {\zeta ^{2}}{(c^{2}+s)}}{\biggr ]}}$ = ${\displaystyle -\pi G\rho {\biggl \{}(abc)\int _{0}^{\infty }{\frac {ds}{\Delta }}-(abc)\xi ^{2}\int _{0}^{\infty }{\frac {ds}{\Delta }}{\biggl [}{\frac {1}{(a^{2}+s)}}{\biggr ]}-(abc)\eta ^{2}\int _{0}^{\infty }{\frac {ds}{\Delta }}{\biggl [}{\frac {1}{(b^{2}+s)}}{\biggr ]}-(abc)\zeta ^{2}\int _{0}^{\infty }{\frac {ds}{\Delta }}{\biggl [}{\frac {1}{(c^{2}+s)}}{\biggr ]}{\biggr \}}}$ = ${\displaystyle -\pi G\rho {\biggl \{}I_{\mathrm {BT} }a^{2}-A_{1}\xi ^{2}-A_{2}\eta ^{2}-A_{3}\zeta ^{2}{\biggr \}}\,.}$

We see, therefore, that

 ${\displaystyle (-G\rho )V}$ = ${\displaystyle \Phi ({\vec {x}})\,,}$

and we recognize the following notation associations:

 ${\displaystyle (A,B,C)~~\leftrightarrow ~~(A_{1},A_{2},A_{3})}$ and, ${\displaystyle H~~\leftrightarrow ~~a_{1}^{2}I_{\mathrm {BT} }=a_{1}^{2}A_{1}^{2}+a_{2}^{2}A_{2}^{2}+a_{3}^{2}A_{3}^{2}\,.}$

In §6 (pp. 183 - 187) of BCO2004), Riemann investigates the case in which one of the pairs of quantities — ${\displaystyle (u,u')}$, or ${\displaystyle (v,v')}$, or ${\displaystyle (w,w')}$ — is zero throughout the time-evolving fluid configuration. As an example, he chooses, ${\displaystyle u=u'=0}$. According to (the English translation of) Riemann's analysis, "The geometric significance of this hypothesis is that the principal axis always lies in the invariant plane of the whole moving body, and the instantaneous axis of rotation is perpendicular to this principal axis."

### Case when u = u' = 0

${\displaystyle 0}$

=

${\displaystyle -(b+c-2a)vw-(b+c+2a)v'w'\,,}$

 ${\displaystyle \mathrm {mod.01} ~\mathrm {of} ~(\alpha .4)}$

${\displaystyle 0}$

=

${\displaystyle -(b-c+2a)vw'-(b-c-2a)v'w\,,}$

 ${\displaystyle \mathrm {mod.01} ~\mathrm {of} ~(\alpha .5)}$

${\displaystyle {\frac {1}{(c-a)}}{\frac {d}{dt}}{\biggl [}v(c-a)^{2}{\biggr ]}}$

=

${\displaystyle 0~~~\Rightarrow ~~~v(c-a)^{2}=\mathrm {constant} \,,}$

 ${\displaystyle \mathrm {mod.01} ~\mathrm {of} ~(\alpha .6)}$

${\displaystyle {\frac {1}{(c+a)}}{\frac {d}{dt}}{\biggl [}v'(c+a)^{2}{\biggr ]}}$

=

${\displaystyle 0~~~\Rightarrow ~~~v'(c+a)^{2}=\mathrm {constant} \,,}$

 ${\displaystyle \mathrm {mod.01} ~\mathrm {of} ~(\alpha .7)}$

${\displaystyle {\frac {1}{(a-b)}}{\frac {d}{dt}}{\biggl [}w(a-b)^{2}{\biggr ]}}$

=

${\displaystyle 0~~~\Rightarrow ~~~w(a-b)^{2}=\mathrm {constant} \,,}$

 ${\displaystyle \mathrm {mod.01} ~\mathrm {of} ~(\alpha .8)}$

${\displaystyle {\frac {1}{(a+b)}}{\frac {d}{dt}}{\biggl [}w'(a+b)^{2}{\biggr ]}}$

=

${\displaystyle 0~~~\Rightarrow ~~~w'(a+b)^{2}=\mathrm {constant} \,.}$

 ${\displaystyle \mathrm {mod.01} ~\mathrm {of} ~(\alpha .9)}$

#### v and w Ratios

From the first modification (mod.01) of equation ${\displaystyle (\alpha .4)}$, we find that,

 ${\displaystyle (b+c+2a)v'w'}$ = ${\displaystyle -(b+c-2a)vw}$ ${\displaystyle \Rightarrow ~~~{\frac {v'}{v}}}$ = ${\displaystyle {\frac {(2a-b-c)}{(2a+b+c)}}\cdot {\frac {w}{w'}}\,.}$

And from the first modification (mod.01) of equation ${\displaystyle (\alpha .5)}$, we find,

 ${\displaystyle (2a-b+c)v'w}$ = ${\displaystyle (2a+b-c)vw'}$ ${\displaystyle \Rightarrow ~~~{\frac {v'w}{vw'}}}$ = ${\displaystyle {\frac {(2a+b-c)}{(2a-b+c)}}\,.}$

Combining these two expressions gives,

 ${\displaystyle {\frac {(2a-b-c)}{(2a+b+c)}}\cdot {\frac {w}{w'}}{\biggl [}{\frac {w}{w'}}{\biggr ]}}$ = ${\displaystyle {\biggl [}{\frac {(2a+b-c)}{(2a-b+c)}}{\biggr ]}}$ ${\displaystyle \Rightarrow ~~~{\biggl [}{\frac {w'}{w}}{\biggr ]}^{2}}$ = ${\displaystyle {\frac {(2a-b-c)}{(2a+b+c)}}\cdot {\frac {(2a-b+c)}{(2a+b-c)}}\,;}$

and,

 ${\displaystyle {\biggl [}{\frac {v'}{v}}{\biggr ]}^{2}}$ = ${\displaystyle {\frac {(2a-b-c)}{(2a+b+c)}}\cdot {\frac {(2a+b-c)}{(2a-b+c)}}\,.}$

From the first modifications (mod.01) of ${\displaystyle (\alpha .6)}$ through ${\displaystyle (\alpha .9)}$, we also see that,

 ${\displaystyle {\biggl [}{\frac {v'}{v}}{\biggr ]}^{2}}$ = ${\displaystyle {\frac {(c+a)^{4}}{(c-a)^{4}}}\times \mathrm {constant} \,,}$ and, ${\displaystyle {\biggl [}{\frac {w'}{w}}{\biggr ]}^{2}}$ = ${\displaystyle {\frac {(a-b)^{4}}{(a+b)^{4}}}\times \mathrm {constant} \,.}$

Hence, following Riemann (1861) — see the bottom of p. 183 of BCO2004 — we can write,

 ${\displaystyle {\biggl [}{\frac {v'}{v}}{\biggr ]}^{2}}$ = ${\displaystyle {\frac {(2a-b-c)}{(2a+b+c)}}\cdot {\frac {(2a+b-c)}{(2a-b+c)}}={\biggl (}{\frac {c+a}{c-a}}{\biggr )}^{4}\times \mathrm {constant} \,,}$ ${\displaystyle {\biggl [}{\frac {w'}{w}}{\biggr ]}^{2}}$ = ${\displaystyle {\frac {(2a-b-c)}{(2a+b+c)}}\cdot {\frac {(2a-b+c)}{(2a+b-c)}}={\biggl (}{\frac {a-b}{a+b}}{\biggr )}^{4}\times \mathrm {constant} \,.}$

As Riemann (1961) points out — see the top of p. 184 of BCO2004"Taking this together with the constant volume constraint ${\displaystyle (\alpha .10)}$ we see that ${\displaystyle a,b,c}$ are constant, and consequently ${\displaystyle v,v',w,w'}$ are constant."

#### v and w Sums and Differences

If we combine the just-derived coefficient ratios with equation (2) of §6 (see p. 184 of BCO2004), we deduce that,

 ${\displaystyle v'}$ = ${\displaystyle \pm v{\biggl [}{\frac {(2a-b-c)(2a+b-c)}{(2a+b+c)(2a-b+c)}}{\biggr ]}^{1/2}}$ ${\displaystyle \Rightarrow ~~~(v'-v)}$ = ${\displaystyle -v{\biggl \{}1\mp {\biggl [}{\frac {(2a-b-c)(2a+b-c)}{(2a+b+c)(2a-b+c)}}{\biggr ]}^{1/2}{\biggr \}}}$ = ${\displaystyle -{\biggl \{}1\mp {\biggl [}{\frac {(2a-b-c)(2a+b-c)}{(2a+b+c)(2a-b+c)}}{\biggr ]}^{1/2}{\biggr \}}(2a+b+c)^{1/2}(2a-b+c)^{1/2}{\biggl [}\pm S^{1/2}{\biggr ]}}$ = ${\displaystyle {\biggl \{}\pm {\biggl [}{\frac {(2a-b-c)(2a+b-c)}{(2a+b+c)(2a-b+c)}}{\biggr ]}^{1/2}-1{\biggr \}}(2a+b+c)^{1/2}(2a-b+c)^{1/2}{\biggl [}\pm S^{1/2}{\biggr ]}}$ = ${\displaystyle {\biggl \{}{\biggl [}{\frac {(2a-b-c)(2a+b-c)}{(2a+b+c)(2a-b+c)}}{\biggr ]}^{1/2}\pm 1{\biggr \}}(2a+b+c)^{1/2}(2a-b+c)^{1/2}{\biggl [}S^{1/2}{\biggr ]}}$ = ${\displaystyle {\biggl \{}(2a-b-c)^{1/2}(2a+b-c)^{1/2}\pm (2a+b+c)^{1/2}(2a-b+c)^{1/2}{\biggr \}}S^{1/2}}$ = ${\displaystyle {\biggl \{}[(2a-c)-b]^{1/2}[(2a-c)+b]^{1/2}\pm [(2a+c)+b]^{1/2}[(2a+c)-b]^{1/2}{\biggr \}}S^{1/2}}$ = ${\displaystyle {\biggl \{}[(2a-c)^{2}-b^{2}]^{1/2}\pm [(2a+c)^{2}-b^{2}]^{1/2}{\biggr \}}S^{1/2}}$ = ${\displaystyle {\biggl \{}[(4a^{2}-b^{2}+c^{2})-4ac]^{1/2}\pm [(4a^{2}-b^{2}+c^{2})+4ac]^{1/2}{\biggr \}}S^{1/2}}$ = ${\displaystyle {\biggl \{}(4a^{2}-b^{2}+c^{2})-4ac+(4a^{2}-b^{2}+c^{2})+4ac\pm 2[(4a^{2}-b^{2}+c^{2})-4ac]^{1/2}[(4a^{2}-b^{2}+c^{2})+4ac]^{1/2}{\biggr \}}^{1/2}S^{1/2}}$ = ${\displaystyle 2^{1/2}{\biggl \{}(4a^{2}-b^{2}+c^{2})\pm [(4a^{2}-b^{2}+c^{2})-4ac]^{1/2}[(4a^{2}-b^{2}+c^{2})+4ac]^{1/2}{\biggr \}}^{1/2}S^{1/2}}$ = ${\displaystyle 2^{1/2}{\biggl \{}(4a^{2}-b^{2}+c^{2})\pm [(4a^{2}-b^{2}+c^{2})^{2}-16a^{2}c^{2}]^{1/2}{\biggr \}}^{1/2}S^{1/2}}$ = ${\displaystyle 2^{1/2}{\biggl \{}(4a^{2}-b^{2}+c^{2})\pm [16a^{4}-8a^{2}b^{2}+b^{4}-2b^{2}c^{2}+c^{4}-8a^{2}c^{2}]^{1/2}{\biggr \}}^{1/2}S^{1/2}}$

Note that, from Eq. (16) of EFE's Chapter 7, §47 (p. 131),

 ${\displaystyle 2^{2}a^{2}\beta }$ = ${\displaystyle 4a^{2}-b^{2}+c^{2}\pm [4a^{2}-(b+c)^{2}]^{1/2}[4a^{2}-(b-c)^{2}]^{1/2}}$ = ${\displaystyle 4a^{2}-b^{2}+c^{2}\pm [4a^{2}-b^{2}-2bc-c^{2})]^{1/2}[4a^{2}-b^{2}+2bc-c^{2})]^{1/2}}$ = ${\displaystyle 4a^{2}-b^{2}+c^{2}\pm [16a^{4}-8a^{2}b^{2}+8a^{2}bc-4a^{2}c^{2}+b^{4}-2b^{3}c+b^{2}c^{2}-8a^{2}bc+2b^{3}c-4b^{2}c^{2}+2bc^{3}-4a^{2}c^{2}+b^{2}c^{2}-2bc^{3}+c^{4}]^{1/2}}$ = ${\displaystyle 4a^{2}-b^{2}+c^{2}\pm [16a^{4}-8a^{2}b^{2}-8a^{2}c^{2}+b^{4}-2b^{2}c^{2}+c^{4}]^{1/2}}$ ${\displaystyle \Rightarrow ~~~(v'-v)}$ = ${\displaystyle 2^{1/2}{\biggl \{}2^{2}a^{2}\beta {\biggr \}}^{1/2}S^{1/2}=2a(\beta S)^{1/2}\,.}$

Similarly,

 ${\displaystyle (v'+v)}$ = ${\displaystyle v{\biggl \{}1\pm {\biggl [}{\frac {(2a-b-c)(2a+b-c)}{(2a+b+c)(2a-b+c)}}{\biggr ]}^{1/2}{\biggr \}}}$ = ${\displaystyle {\biggl \{}1\pm {\biggl [}{\frac {(2a-b-c)(2a+b-c)}{(2a+b+c)(2a-b+c)}}{\biggr ]}^{1/2}{\biggr \}}(2a+b+c)^{1/2}(2a-b+c)^{1/2}{\biggl [}\pm S^{1/2}{\biggr ]}}$ = ${\displaystyle {\biggl \{}{\biggl [}{\frac {(2a-b-c)(2a+b-c)}{(2a+b+c)(2a-b+c)}}{\biggr ]}^{1/2}\pm 1{\biggr \}}(2a+b+c)^{1/2}(2a-b+c)^{1/2}{\biggl [}S^{1/2}{\biggr ]}}$ = ${\displaystyle (v'-v)\,.}$

We deduce as well that,

 ${\displaystyle w'}$ = ${\displaystyle \pm w{\biggl [}{\frac {(2a-b-c)(2a-b+c)}{(2a+b+c)(2a+b-c)}}{\biggr ]}^{1/2}}$ ${\displaystyle \Rightarrow ~~~(w'-w)}$ = ${\displaystyle -w{\biggl \{}1\mp {\biggl [}{\frac {(2a-b-c)(2a-b+c)}{(2a+b+c)(2a+b-c)}}{\biggr ]}^{1/2}{\biggr \}}}$ = ${\displaystyle {\biggl \{}\pm {\biggl [}{\frac {(2a-b-c)(2a-b+c)}{(2a+b+c)(2a+b-c)}}{\biggr ]}^{1/2}-1{\biggr \}}(2a+b+c)^{1/2}(2a+b-c)^{1/2}{\biggl [}\pm T^{1/2}{\biggr ]}}$ = ${\displaystyle {\biggl \{}{\biggl [}{\frac {(2a-b-c)(2a-b+c)}{(2a+b+c)(2a+b-c)}}{\biggr ]}^{1/2}\pm 1{\biggr \}}(2a+b+c)^{1/2}(2a+b-c)^{1/2}{\biggl [}T^{1/2}{\biggr ]}}$ = ${\displaystyle {\biggl \{}[(2a-b)-c]^{1/2}[(2a-b)+c]^{1/2}]\pm (2a+b+c)^{1/2}(2a+b-c)^{1/2}{\biggr \}}{\biggl [}T^{1/2}{\biggr ]}}$ = ${\displaystyle {\biggl \{}[(2a-b)^{2}-c^{2}]^{1/2}\pm [(2a+b)^{2}-c^{2}]^{1/2}{\biggr \}}{\biggl [}T^{1/2}{\biggr ]}}$ = ${\displaystyle {\biggl \{}[(2a-b)^{2}-c^{2}]+[(2a+b)^{2}-c^{2}]\pm 2[(2a-b)^{2}-c^{2}]^{1/2}[(2a+b)^{2}-c^{2}]^{1/2}{\biggr \}}^{1/2}{\biggl [}T^{1/2}{\biggr ]}}$ = ${\displaystyle {\biggl \{}[4a^{2}-4ab+b^{2}-c^{2}]+[4a^{2}+4ab+b^{2}-c^{2}]\pm 2[(4a^{2}-4ab+b^{2})-c^{2}]^{1/2}[(4a^{2}+4ab+b^{2})-c^{2}]^{1/2}{\biggr \}}^{1/2}{\biggl [}T^{1/2}{\biggr ]}}$ = ${\displaystyle 2^{1/2}{\biggl \{}[4a^{2}+b^{2}-c^{2}]\pm [(4a^{2}+b^{2}-c^{2})-4ab]^{1/2}[(4a^{2}+b^{2}-c^{2})+4ab]^{1/2}{\biggr \}}^{1/2}{\biggl [}T^{1/2}{\biggr ]}\,.}$

And again, similarly,

 ${\displaystyle w'}$ = ${\displaystyle \pm w{\biggl [}{\frac {(2a-b-c)(2a-b+c)}{(2a+b+c)(2a+b-c)}}{\biggr ]}^{1/2}}$ ${\displaystyle \Rightarrow ~~~(w'+w)}$ = ${\displaystyle w{\biggl \{}1\pm {\biggl [}{\frac {(2a-b-c)(2a-b+c)}{(2a+b+c)(2a+b-c)}}{\biggr ]}^{1/2}{\biggr \}}}$ = ${\displaystyle {\biggl \{}1\pm {\biggl [}{\frac {(2a-b-c)(2a-b+c)}{(2a+b+c)(2a+b-c)}}{\biggr ]}^{1/2}{\biggr \}}(2a+b+c)^{1/2}(2a+b-c)^{1/2}{\biggl [}\pm T^{1/2}{\biggr ]}}$ = ${\displaystyle {\biggl \{}{\biggl [}{\frac {(2a-b-c)(2a-b+c)}{(2a+b+c)(2a+b-c)}}{\biggr ]}^{1/2}\pm 1{\biggr \}}(2a+b+c)^{1/2}(2a+b-c)^{1/2}{\biggl [}T^{1/2}{\biggr ]}}$ = ${\displaystyle (w'-w)\,.}$

#### First Three Governing Relation

Given that ${\displaystyle a,b,c}$ are constant and ${\displaystyle u=u'=0}$, the first three governing equations become,

${\displaystyle {\frac {\epsilon A}{2}}-{\frac {\sigma }{2a^{2}}}}$

=

${\displaystyle {\frac {1}{2a}}{\biggl [}(a-c)v^{2}+(a+c)(v')^{2}+(a-b)w^{2}+(a+b)(w')^{2}{\biggr ]}\,,}$

 ${\displaystyle \mathrm {mod.01} ~\mathrm {of} ~(\alpha .1)}$

${\displaystyle {\frac {\epsilon B}{2}}-{\frac {\sigma }{2b^{2}}}}$

=

${\displaystyle {\frac {1}{2b}}{\biggl [}(b-a)w^{2}+(b+a)(w')^{2}{\biggr ]}\,,}$

 ${\displaystyle \mathrm {mod.01} ~\mathrm {of} ~(\alpha .2)}$

${\displaystyle {\frac {\epsilon C}{2}}-{\frac {\sigma }{2c^{2}}}}$

=

${\displaystyle {\frac {1}{2c}}{\biggl [}(c-a)v^{2}+(c+a)(v')^{2}{\biggr ]}\,.}$

 ${\displaystyle \mathrm {mod.01} ~\mathrm {of} ~(\alpha .3)}$

After adopting a pair of alternate constants, S and T, Riemann shows that these relations can be rewritten in the forms (mod.02),

${\displaystyle {\frac {\epsilon A}{2}}-{\frac {\sigma }{2a^{2}}}}$

=

${\displaystyle (4a^{2}-b^{2}-c^{2})(T+S)-2(b^{2}T+c^{2}S)}$

 ${\displaystyle \mathrm {mod.02} ~\mathrm {of} ~(\alpha .1)}$

${\displaystyle {\frac {\epsilon B}{2}}-{\frac {\sigma }{2b^{2}}}}$

=

${\displaystyle (b^{2}-c^{2})T\,,}$

 ${\displaystyle \mathrm {mod.02} ~\mathrm {of} ~(\alpha .2)}$

${\displaystyle {\frac {\epsilon C}{2}}-{\frac {\sigma }{2c^{2}}}}$

=

${\displaystyle (c^{2}-b^{2})S\,.}$

 ${\displaystyle \mathrm {mod.02} ~\mathrm {of} ~(\alpha .3)}$

On p. 158 (Chapter 7, §51) of EFE, at a comparable stage of Chandrasekhar's derivation, we find the following trio of governing relations:

${\displaystyle 2A_{1}-{\frac {5\Pi }{Ma_{1}^{2}}}}$

=

${\displaystyle {\frac {1}{2}}(4a_{1}^{2}-a_{2}^{2}-a_{3}^{2}){\biggl (}{\frac {\Omega _{2}^{2}\beta }{a_{3}^{2}}}+{\frac {\Omega _{3}^{2}\gamma }{a_{2}^{2}}}{\biggr )}-(\Omega _{2}^{2}\beta +\Omega _{3}^{2}\gamma )\,,}$

 ${\displaystyle \mathrm {EFE} ,~\mathrm {Eq.} (164)}$

${\displaystyle 2A_{2}-{\frac {5\Pi }{Ma_{2}^{2}}}}$

=

${\displaystyle {\biggl [}{\frac {a_{2}^{2}-a_{3}^{2}}{2a_{2}^{2}}}{\biggr ]}\Omega _{3}^{2}\gamma \,,}$

 ${\displaystyle \mathrm {EFE} ,~\mathrm {Eq.} (165)}$

${\displaystyle 2A_{3}-{\frac {5\Pi }{Ma_{3}^{2}}}}$

=

${\displaystyle {\biggl [}{\frac {a_{3}^{2}-a_{2}^{2}}{2a_{3}^{2}}}{\biggr ]}\Omega _{2}^{2}\beta \,.}$

 ${\displaystyle \mathrm {EFE} ,~\mathrm {Eq.} (166)}$

Building on our above discussion of the expression for the gravitational potential, it appears as though we have the following notational correspondence:

 Notational Correspondence Riemann Chandrasekhar ${\displaystyle (A,B,C)}$ ${\displaystyle (A_{1},A_{2},A_{3})}$ ${\displaystyle \sigma }$ ${\displaystyle \epsilon {\biggl [}{\frac {5\Pi }{2M}}{\biggr ]}}$ ${\displaystyle T}$ ${\displaystyle \epsilon {\biggl [}{\frac {\Omega _{3}^{2}\gamma }{8b^{2}}}{\biggr ]}}$ ${\displaystyle S}$ ${\displaystyle \epsilon {\biggl [}{\frac {\Omega _{2}^{2}\beta }{8c^{2}}}{\biggr ]}}$

As a check, multiplying EFE's equation (164) through by ${\displaystyle (\epsilon /4)}$ gives,

 ${\displaystyle {\frac {\epsilon A_{1}}{2}}-\epsilon {\biggl [}{\frac {5\Pi }{4Ma_{1}^{2}}}{\biggr ]}}$ = ${\displaystyle \epsilon (4a_{1}^{2}-a_{2}^{2}-a_{3}^{2}){\biggl (}{\frac {\Omega _{2}^{2}\beta }{8a_{3}^{2}}}+{\frac {\Omega _{3}^{2}\gamma }{8a_{2}^{2}}}{\biggr )}-(\Omega _{2}^{2}\beta +\Omega _{3}^{2}\gamma ){\frac {\epsilon }{4}}\,,}$

which, after changing to Riemann's corresponding notation, gives,

 ${\displaystyle {\frac {\epsilon A}{2}}-{\biggl [}{\frac {\sigma }{2a^{2}}}{\biggr ]}}$ = ${\displaystyle \epsilon (4a^{2}-b^{2}-c^{2}){\biggl (}S+T{\biggr )}-2(c^{2}S+b^{2}T)\,.}$

This matches the second modification of Riemann's equation ${\displaystyle (\alpha .1)}$ precisely. So we are confident that the entries in our table of notational correspondences are correct.

### Velocities

If, in our above discussion, we are interpreting Riemann's analysis correctly, then the steady-state velocity components as viewed from the rotating reference frame are,

 ${\displaystyle {\frac {\partial }{\partial t}}{\biggl (}{\frac {\xi }{a}}{\biggr )}}$ = ${\displaystyle (r_{1}){\frac {\eta }{b}}-(q_{1}){\frac {\zeta }{c}}\,,}$ ${\displaystyle {\frac {\partial }{\partial t}}{\biggl (}{\frac {\eta }{b}}{\biggr )}}$ = ${\displaystyle (p_{1}){\frac {\zeta }{c}}-(r_{1}){\frac {\xi }{a}}\,,}$ ${\displaystyle {\frac {\partial }{\partial t}}{\biggl (}{\frac {\zeta }{c}}{\biggr )}}$ = ${\displaystyle (q_{1}){\frac {\xi }{a}}-(p_{1}){\frac {\eta }{b}}\,.}$

Now, given that (see the bottom of p. 177 of BCO2004),

 ${\displaystyle p_{1}=(u-u')\,,}$ and, ${\displaystyle q_{1}=(v-v')\,,}$ and, ${\displaystyle r_{1}=(w-w')\,,}$

these velocity components may be written as,

 ${\displaystyle {\frac {\partial \xi }{\partial t}}}$ = ${\displaystyle (w-w'){\frac {a\eta }{b}}-(v-v'){\frac {a\zeta }{c}}\,,}$ ${\displaystyle {\frac {\partial \eta }{\partial t}}}$ = ${\displaystyle (u-u'){\frac {b\zeta }{c}}-(w-w'){\frac {b\xi }{a}}\,,}$ ${\displaystyle {\frac {\partial \zeta }{\partial t}}}$ = ${\displaystyle (v-v'){\frac {c\xi }{a}}-(u-u'){\frac {c\eta }{b}}\,.}$

Examining the case of ${\displaystyle u=u'=0}$, we have,

 ${\displaystyle {\frac {\partial \xi }{\partial t}}}$ = ${\displaystyle (w-w'){\frac {a\eta }{b}}-(v-v'){\frac {a\zeta }{c}}\,,}$ ${\displaystyle {\frac {\partial \eta }{\partial t}}}$ = ${\displaystyle (w'-w){\frac {b\xi }{a}}\,,}$ ${\displaystyle {\frac {\partial \zeta }{\partial t}}}$ = ${\displaystyle (v-v'){\frac {c\xi }{a}}\,.}$

It seems reasonable to compare this trio of Riemann expressions with the EFE expressions for the velocity field as viewed from a rotating reference frame.

EFE, Chapter 7, §51 (p. 156), Eq. (154)

 ${\displaystyle u_{1}}$ = ${\displaystyle (\Omega _{3}\gamma ){\frac {a^{2}x_{2}}{b^{2}}}-(\Omega _{2}\beta ){\frac {a^{2}x_{3}}{c^{2}}}\,,}$ ${\displaystyle u_{2}}$ = ${\displaystyle -(\Omega _{3}\gamma )x_{1}\,,}$ ${\displaystyle u_{3}}$ = ${\displaystyle (\Omega _{2}\beta )x_{1}\,.}$

From equation (2) of §6 (see p. 184 of BCO2004), we deduce that,

 ${\displaystyle (v')^{2}}$ = ${\displaystyle (v)^{2}{\biggl [}{\frac {(2a-b-c)(2a+b-c)}{(2a+b+c)(2a-b+c)}}{\biggr ]}}$ ${\displaystyle \Rightarrow ~~~(v'-v)}$ = ${\displaystyle v{\biggl \{}{\biggl [}{\frac {(2a-b-c)(2a+b-c)}{(2a+b+c)(2a-b+c)}}{\biggr ]}^{1/2}-1{\biggr \}}}$ = ${\displaystyle {\biggl \{}{\biggl [}{\frac {(2a-b-c)(2a+b-c)}{(2a+b+c)(2a-b+c)}}{\biggr ]}^{1/2}-1{\biggr \}}(2a+b+c)^{1/2}(2a-b+c)^{1/2}S^{1/2}}$ = ${\displaystyle {\biggl \{}(2a-b-c)^{1/2}(2a+b-c)^{1/2}-(2a+b+c)^{1/2}(2a-b+c)^{1/2}{\biggr \}}S^{1/2}}$

Now, from above, we have surmised that,

 ${\displaystyle S}$ = ${\displaystyle {\frac {\epsilon \Omega _{2}^{2}\beta }{8c^{2}}}\,,}$

in which case,

 ${\displaystyle {\frac {\partial \zeta }{\partial t}}=(v-v'){\frac {c\xi }{a}}}$ = ${\displaystyle {\frac {c\xi }{a}}{\biggl \{}(2a-b-c)^{1/2}(2a+b-c)^{1/2}-(2a+b+c)^{1/2}(2a-b+c)^{1/2}{\biggr \}}{\biggl [}{\frac {\epsilon \Omega _{2}^{2}\beta }{8c^{2}}}{\biggr ]}^{1/2}}$ = ${\displaystyle {\biggl (}{\frac {\epsilon }{2}}{\biggr )}^{1/2}(\Omega _{2}\beta )\xi {\biggl \{}K_{0}{\biggr \}}{\biggl [}{\frac {1}{4a^{2}\beta }}{\biggr ]}^{1/2}\,,}$

where,

 ${\displaystyle K_{0}}$ ${\displaystyle \equiv }$ ${\displaystyle (2a-b-c)^{1/2}(2a+b-c)^{1/2}-(2a+b+c)^{1/2}(2a-b+c)^{1/2}\,.}$

According to EFE — see Chapter 7, §47 (p. 131) …

 ${\displaystyle 4a^{2}\beta }$ = ${\displaystyle 4a^{2}-b^{2}+c^{2}\pm {\biggl \{}[4a^{2}-(b+c)^{2}][4a^{2}-(b-c)^{2}]{\biggr \}}^{1/2}}$
 Test case: ${\displaystyle a=1.0,b=1.2,c=0.33~~~\Rightarrow ~~~K_{0}=-0.83580}$ and (selecting the negative sign),     ${\displaystyle 4a^{2}\beta =2.66890-2.31962=0.34928}$ Hence, ${\displaystyle {\frac {K_{0}}{[4a^{2}\beta ]^{1/2}}}=-1.41421=-{\sqrt {2}}\,.}$ Excellent!   This strongly suggests that in all cases, ${\displaystyle 8a^{2}\beta =K_{0}^{2}\,;}$ and that, as was surmised earlier, ${\displaystyle {\frac {\partial \zeta }{\partial t}}=\epsilon ^{1/2}(\Omega _{2}\beta )\xi \,.}$

Let's check to see if this is indeed a general result. Rewriting ${\displaystyle K_{0}^{2}}$, we find,

 ${\displaystyle K_{0}^{2}}$ = ${\displaystyle {\biggl \{}[2a-b-c]^{1/2}[2a+b-c]^{1/2}-[2a+b+c]^{1/2}[2a-b+c]^{1/2}{\biggr \}}^{2}}$ = ${\displaystyle {\biggl \{}[(2a-c)^{2}-b^{2}]^{1/2}-[(2a+c)^{2}-b^{2}]^{1/2}{\biggr \}}^{2}}$ = ${\displaystyle [(2a-c)^{2}-b^{2}]+[(2a+c)^{2}-b^{2}]-2[(2a+c)^{2}-b^{2}]^{1/2}[(2a-c)^{2}-b^{2}]^{1/2}}$ = ${\displaystyle [4a^{2}-4ac+c^{2}-b^{2}]+[4a^{2}+4ac+c^{2}-b^{2}]-2[(2a+c)^{2}-b^{2}]^{1/2}[(2a-c)^{2}-b^{2}]^{1/2}}$ = ${\displaystyle 2[4a^{2}+c^{2}-b^{2}]-2[(2a+c)^{2}-b^{2}]^{1/2}[(2a-c)^{2}-b^{2}]^{1/2}\,.}$

Hence (selecting the negative sign in the expression for ${\displaystyle 4a^{2}\beta }$),

 ${\displaystyle 4a^{2}\beta -{\frac {K_{0}^{2}}{2}}}$ = ${\displaystyle [b^{2}-4a^{2}-c^{2}]+[(2a+c)^{2}-b^{2}]^{1/2}[(2a-c)^{2}-b^{2}]^{1/2}}$ ${\displaystyle +4a^{2}-b^{2}+c^{2}-[4a^{2}-(b+c)^{2}]^{1/2}[4a^{2}-(b-c)^{2}]^{1/2}}$ = ${\displaystyle {\biggl \{}[(2a+c)^{2}-b^{2}][(2a-c)^{2}-b^{2}]{\biggr \}}^{1/2}-{\biggl \{}[4a^{2}-(b+c)^{2}][4a^{2}-(b-c)^{2}]{\biggr \}}^{1/2}}$ = ${\displaystyle {\biggl \{}(2a+c)^{2}(2a-c)^{2}-b^{2}[(2a+c)^{2}+(2a-c)^{2}]+b^{4}{\biggr \}}^{1/2}}$ ${\displaystyle -{\biggl \{}16a^{4}-4a^{2}[(b-c)^{2}+(b+c)^{2}]+(b+c)^{2}(b-c)^{2}{\biggr \}}^{1/2}}$ = ${\displaystyle {\biggl \{}(4a^{2}-c^{2})^{2}-b^{2}[4a^{2}+4ac+c^{2}+4a^{2}-4ac+c^{2}]+b^{4}{\biggr \}}^{1/2}}$ ${\displaystyle -{\biggl \{}16a^{4}-4a^{2}[b^{2}-2bc+c^{2}+b^{2}+2bc+c^{2}]+(b^{2}-c^{2})^{2}{\biggr \}}^{1/2}}$ = ${\displaystyle {\biggl \{}16a^{4}-8a^{2}c^{2}+c^{4}-8a^{2}b^{2}-2b^{2}c^{2}+b^{4}{\biggr \}}^{1/2}-{\biggl \{}16a^{4}-8a^{2}b^{2}-8a^{2}c^{2}+b^{4}-2b^{2}c^{2}+c^{4}{\biggr \}}^{1/2}}$ = ${\displaystyle 0\,.}$

Terrific!   We have demonstrated that, quite generally, ${\displaystyle 8a^{2}\beta =K_{0}^{2}}$.