Editing 3Dconfigurations/RiemannEllipsoids (section)

=Governing Equations=

== Step 1==
At the beginning of his discussion, Riemann denotes by <math>(x, y, z)</math> <font color="orange">"the [inertial-frame] coordinates of an element of the fluid body at time <math>t</math>,"</font> and he denotes by <math>(\xi, \eta, \zeta)</math> <font color="orange">"the coordinates of the point <math>(x, y, z)</math> with respect to a moving coordinate system, whose axes coincide at each instant with the principal axes of the ellipsoid."</font> Drawing from our [[Appendix/Mathematics/EulerAngles#Euler_Angles|accompanying discussion of Euler angles]], it seems appropriate to associate <math>\vec{A}_{XYZ}</math> with the vector that points from the origin to the <math>(x, y, z)</math> location of the fluid element as viewed from the inertial reference frame, and to associate <math>\vec{A}_\mathrm{body}</math> 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:

<table border="0" align="center" cellpadding="3">
<tr>
  <td align="right">
<math>\vec{A}_\mathrm{XYZ}(A_X, A_Y, A_Z)</math>
  </td>
  <td align="center">&nbsp; <math>\rightarrow</math> &nbsp;</td>
  <td align="left">
<math>\vec{A}_\mathrm{inertial}(x, y, z) = \vec{e}_X(x) + \vec{e}_Y(y) + \vec{e}_Z(z) \, ,</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>\vec{A}_\mathrm{body}(A_1, A_2, A_3)</math>
  </td>
  <td align="center">&nbsp; <math>\rightarrow</math> &nbsp;</td>
  <td align="left">
<math>\vec{A}_\mathrm{body}(\xi, \eta, \zeta) = \vec{e}_1(\xi) + \vec{e}_2(\eta) + \vec{e}_3(\zeta) \, .</math>
  </td>
</tr>
</table>
Again drawing from [[#FormMatrix|our accompanying Euler angles discussion]], quite generally these two coordinate representations  &#8212; of the ''same'' vector, <math>\vec{A}</math> &#8212; are related to one another via the matrix expression,

<div align="center">
<math>\vec{A}_\mathrm{body} = \hat{R} \cdot \vec{A}_\mathrm{inertial} \, ,</math>
</div>

that is,

<table border="0" align="center" cellpadding="8">
<tr>
  <td align="right">
<math>\begin{bmatrix}\xi \\ \eta \\ \zeta \end{bmatrix}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\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} \, .
</math>
  </td>
</tr>
</table>

[[Appendix/Mathematics/EulerAngles#RotationMatrix|In terms of the trio of Euler angles]], the rotation matrix is,

<table border="0" align="center" cellpadding="8">

<tr>
  <td align="right">
<math>\hat{R}(\phi, \theta, \psi)</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\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} \, .
</math>
  </td>
</tr>
</table>

This means that &#8212; see [[Appendix/Mathematics/EulerAngles#Switching_Coordinate_Representations_of_a_Vector|our accompanying discussion]] &#8212; in terms of the trio of Euler angles, the following three coordinate mappings hold:

<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right">
<math>\xi</math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
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) \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\eta</math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
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) \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\zeta</math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
x (\sin\theta\sin\phi)
+ y (-\sin\theta \cos\phi)
+ z (\cos\theta) \, .
</math>
  </td>
</tr>
</table>

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

<table border="1" align="center" width="80%" cellpadding="5">
<tr><td align="center">From &sect;1 of '''Riemann (1861)'''</td></tr>
<tr><td align="left">
<font color="orange">Then <math>\xi</math>, <math>\eta</math>, <math>\zeta</math> are known &hellip; to be linear expressions in <math>x</math>, <math>y</math>, <math>z</math>,</font>

<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right">
<math>\xi</math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
\alpha x + \beta y + \gamma z \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\eta</math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
\alpha' x + \beta' y + \gamma' z \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\zeta</math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
\alpha'' x + \beta'' y + \gamma'' z \, .
</math>
  </td>
</tr>
</table>
<font color="orange">The coefficients are the cosines of the angles that the axes of one system form with the axes of the other &hellip;</font>  For example, <math>\alpha = \vec{e}_1 \cdot \vec{e}_X</math>, <math>\beta = \vec{e}_1 \cdot \vec{e}_Y</math>, and so on.
</td></tr></table>

Alternatively, given that,
<div align="center">
<math>\vec{A}_\mathrm{inertial} = \hat{R}^{-1} \cdot \vec{A}_\mathrm{body} \, ,</math>
</div>

the following additional three mapping relations also must hold:

<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right">
<math>x</math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
\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) \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>y</math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
\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) \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>z</math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
\xi (\sin\psi \sin\theta)
+ \eta (\sin\theta \cos\psi)
+ \zeta (\cos\theta) \, .
</math>
  </td>
</tr>
</table>

== Step 2==

Given that <math>(x, y, z)</math> and <math>(\xi, \eta, \zeta)</math> are components of the same position vector, <math>\vec{A}</math>, 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 &sect;1 following equation (2), it must be the case that,
<div align="center">
<math> \xi^2 + \eta^2 + \zeta^2 = x^2 + y^2 + z^2 \, .</math>
</div>
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.

<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right">
<math>\xi^2 + \eta^2 + \zeta^2</math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
\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
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">&nbsp;</td>
  <td align="left">
<math>
+
\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
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">&nbsp;</td>
  <td align="left">
<math>
+ \biggl[
x (\sin\theta\sin\phi)
+ y (-\sin\theta \cos\phi)
+ z (\cos\theta) \biggr]^2 \, .
</math>
  </td>
</tr>
</table>

As Riemann (1861) states, <font color="orange">between these coefficients, six equations hold &hellip;</font>. In particular, when they are written in terms of Euler angles, on the right-hand side, the coefficient of ''the six'' various terms is &hellip;

<table border="0" align="center" cellpadding="5" width="90%">

<tr>
  <td align="right" width="10%">
<math>x^2:</math>
  </td>
  <td align="left">
<math>
(\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
</math>
  </td>
</tr>

<tr>
  <td align="right" width="10%">
&nbsp;
  </td>
  <td align="left">
<math>=
\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
</math>
  </td>
</tr>

<tr>
  <td align="right" width="10%">
&nbsp;
  </td>
  <td align="left">
<math>=
( 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
</math>
  </td>
</tr>

<tr>
  <td align="right" width="10%">
&nbsp;
  </td>
  <td align="left">
<math>=
\sin^2\phi \cos^2\theta  + \cos^2\phi + \sin^2\theta \sin^2\phi
</math>
  </td>
</tr>

<tr>
  <td align="right" width="10%">
&nbsp;
  </td>
  <td align="left">
<math>= 1 \, .
</math>
  </td>
</tr>
</table>

<table border="0" align="center" cellpadding="5" width="90%">

<tr>
  <td align="right" width="10%">
<math>y^2:</math>
  </td>
  <td align="left">
<math>
(\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
</math>
  </td>
</tr>

<tr>
  <td align="right" width="10%">
&nbsp;
  </td>
  <td align="left">
<math>=
\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
</math>
  </td>
</tr>

<tr>
  <td align="right" width="10%">
&nbsp;
  </td>
  <td align="left">
<math>=
(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
</math>
  </td>
</tr>

<tr>
  <td align="right" width="10%">
&nbsp;
  </td>
  <td align="left">
<math>=
\sin^2\phi  + \cos^2\theta \cos^2\phi
+ \sin^2\theta \cos^2\phi
</math>
  </td>
</tr>

<tr>
  <td align="right" width="10%">
&nbsp;
  </td>
  <td align="left">
<math>= 1 \, .
</math>
  </td>
</tr>
</table>

<table border="0" align="center" cellpadding="5" width="90%">

<tr>
  <td align="right" width="10%">
<math>z^2:</math>
  </td>
  <td align="left">
<math>
(\sin\psi \sin\theta )^2 + (\sin\theta \cos\psi )^2 + ( \cos\theta)^2
</math>
  </td>
</tr>

<tr>
  <td align="right" width="10%">
&nbsp;
  </td>
  <td align="left">
<math>= 1 \, .
</math>
  </td>
</tr>
</table>

<table border="0" align="center" cellpadding="5" width="90%">

<tr>
  <td align="right" width="10%">
<math>2xy:</math>
  </td>
  <td align="left">
<math>
(\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)
</math>
  </td>
</tr>

<tr>
  <td align="right" width="10%">
&nbsp;
  </td>
  <td align="left">
<math>=
\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
</math>
  </td>
</tr>

<tr>
  <td align="right" width="10%">
&nbsp;
  </td>
  <td align="left">
<math>=
\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
</math>
  </td>
</tr>

<tr>
  <td align="right" width="10%">
&nbsp;
  </td>
  <td align="left">
<math>=
[\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] 
</math>
  </td>
</tr>

<tr>
  <td align="right" width="10%">
&nbsp;
  </td>
  <td align="left">
<math>=
\sin\phi \cos\phi[\sin^2\psi (1 - \cos^2\theta)- \sin^2\theta \sin^2\psi ] 
</math>
  </td>
</tr>

<tr>
  <td align="right" width="10%">
&nbsp;
  </td>
  <td align="left">
<math>=
\sin\phi \cos\phi\sin^2\psi[ 1 - \cos^2\theta - \sin^2\theta  ] 
</math>
  </td>
</tr>

<tr>
  <td align="right" width="10%">
&nbsp;
  </td>
  <td align="left">
<math>= 0 \, .
</math>
  </td>
</tr>
</table>

<table border="0" align="center" cellpadding="5" width="90%">

<tr>
  <td align="right" width="10%">
<math>2xz:</math>
  </td>
  <td align="left">
<math>
(\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
</math>
  </td>
</tr>

<tr>
  <td align="right" width="10%">
&nbsp;
  </td>
  <td align="left">
<math>=
\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
</math>
  </td>
</tr>

<tr>
  <td align="right" width="10%">
&nbsp;
  </td>
  <td align="left">
<math>=
\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]
</math>
  </td>
</tr>

<tr>
  <td align="right" width="10%">
&nbsp;
  </td>
  <td align="left">
<math>= 0 \, .
</math>
  </td>
</tr>
</table>

<table border="0" align="center" cellpadding="5" width="90%">

<tr>
  <td align="right" width="10%">
<math>2yz:</math>
  </td>
  <td align="left">
<math>
(\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
</math>
  </td>
</tr>

<tr>
  <td align="right" width="10%">
&nbsp;
  </td>
  <td align="left">
<math>=
[\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 
</math>
  </td>
</tr>

<tr>
  <td align="right" width="10%">
&nbsp;
  </td>
  <td align="left">
<math>= 0 \, .
</math>
  </td>
</tr>
</table>

== Step 3==

At the bottom of p. 176 of [http://www.kendrickpress.com/Riemann.htm BCO2004], Riemann summarizes the <font color="orange">"geometric significance"</font> of various variables that he has introduced:
<ol>
  <li>
<font color="orange"><math>(\xi', \eta', \zeta')</math> are the velocity components of the point <math>(x, y, z)</math> of the fluid mass parallel to the axes of <math>(\xi, \eta, \zeta)</math></font>. Note that, in the terms we have used above, <math>(x, y, z)</math> refers to the position of a fluid element as viewed from the ''inertial'' reference frame, while <math>(\xi, \eta, \zeta)</math> 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 &sect;2, namely,
<table border="0" align="center" cellpadding="2">

<tr>
  <td align="right">
<math>\xi'</math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
\biggl( \frac{\xi}{a} \biggr) \frac{da}{dt} + (ar_1 -br)\frac{\eta}{b} + (cq -aq_1) \frac{\zeta}{c} \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\eta'</math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
\biggl( \frac{\eta}{b} \biggr) \frac{db}{dt} + (ar -br_1)\frac{\xi}{a} + (bp_1 - cp) \frac{\zeta}{c} \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\zeta'</math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
\biggl( \frac{\zeta}{c} \biggr) \frac{dc}{dt} + (c q_1 - aq)\frac{\xi}{a} + (bp - cp_1) \frac{\eta}{b} \, .
</math>
  </td>
</tr>
</table>
  </li>

  <li>
<math>(\partial \xi/\partial t, \partial\eta/\partial t, \partial \zeta/\partial t)</math> are <font color="orange">the relative velocities decomposed in the same way, for the coordinate system, <math>(\xi, \eta, \zeta)</math></font>. The expressions that Riemann derived for these three components of the (body-frame) fluid velocity appear as equation (6) of &sect;2, namely,
<table border="0" align="center" cellpadding="2">

<tr>
  <td align="right">
<math>\frac{\partial}{\partial t} \biggl( \frac{\xi}{a} \biggr)</math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
(r_1)\frac{\eta}{b} - (q_1) \frac{\zeta}{c} \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\frac{\partial}{\partial t} \biggl( \frac{\eta}{b} \biggr)</math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
(p_1)\frac{\zeta}{c} - (r_1) \frac{\xi}{a} \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\frac{\partial}{\partial t} \biggl( \frac{\zeta}{c} \biggr)</math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
(q_1)\frac{\xi}{a} - (p_1) \frac{\eta}{b} \, .
</math>
  </td>
</tr>
</table>
  </li>
  <li>
<math>(p, q, r)</math> <font color="orange">are the instantaneous rotations of the coordinate system <math>(\xi, \eta, \zeta)</math> about its axes</font>. 
  </li>
  <li>
<math>(p_1, q_1, r_1)</math> <font color="orange">have the same significance for the coordinate system <math>(\xi_1, \eta_1, \zeta_1)</math></font>. 
  </li>
</ol>
<font color="red">Initially we were confused</font> regarding the differences between the two referenced coordinate systems, <math>(\xi, \eta, \zeta)</math> and <math>(\xi_1, \eta_1, \zeta_1)</math>. Immediately below equation (1) of &sect;1 (near the top of p. 172 of [http://www.kendrickpress.com/Riemann.htm BCO2004]), we find the following definition:  <font color="orange">"Denote by <math>(\xi, \eta, \zeta)</math> the coordinates of the point <math>(x, y, z)</math> with respect to a moving coordinate system, whose axes coincide at each instant with the principal axes of the ellipsoid.</font> On the other hand, near the top of p. 173 of [http://www.kendrickpress.com/Riemann.htm BCO2004], Riemann refers to a set of coefficients that <font color="orange">"&hellip; can be treated as the cosines of the angles that the axes of a moving coordinate system <math>(\xi_1, \eta_1, \zeta_1)</math> form with the axes of the fixed coordinate system <math>(x, y, z)</math>."</font>  

Adopting Figure 1 along with the notation that has been used in our [[Appendix/Mathematics/EulerAngles#Euler_Angles|accompanying discussion of Euler Angles]], the distinction seems to be that <math>(\xi_1, \eta_1, \zeta_1)</math> stands for the unit vectors <math>(\vec{e}_{x_1}, \vec{e}_{x_2}, \vec{e}_{x_3})</math> associated with a moving coordinate system &#8212; in the context of Riemann's work, a "moving" system that is always aligned with the principal axes of the (tumbling) ellipsoid &#8212; while <math>(\xi, \eta, \zeta)</math> gives the coordinates of a single fluid element as referenced to this moving coordinate system &#8212; the (red) vector, <math>\vec{A}</math>, 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 3<sup>rd</sup> item of "geometric significance" is poorly worded.

<table border="1" align="center" cellpadding="3">
<tr><td align="center" colspan="2">'''Figure 1'''</td></tr>
<tr>
  <td align="center">
[[File:BerciuFig1a.png|250px|Berciu Figure 1a]]
  </td>
  <td align="center">
[[File:BerciuFig1bAgain.png|250px|Berciu Figure 1b]]
  </td>
</tr>
</table>

If we are interpreting these two velocity expressions correctly, then the difference between the inertial-frame velocity and the rotating-frame velocity should be, <math>\vec\Omega \times \vec{x}</math>.  From our [https://www.vistrails.org/index.php/User:Tohline/ThreeDimensionalConfigurations/ChallengesPt2#Inertial-Frame_Expressions associated discussion of the EFE presentation], we find,
<table border="0" align="center" cellpadding="2">

<tr>
  <td align="right">
<math>\vec\Omega \times \vec{x} = \boldsymbol{u}^{(0)} - \boldsymbol{u}</math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
(\boldsymbol{\hat\jmath} \Omega_2 + \boldsymbol{\hat{k}} \Omega_3)
\times
(\boldsymbol{\hat\imath} x + \boldsymbol{\hat\jmath}y + \boldsymbol{\hat{k}}z)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
\boldsymbol{\hat\imath}(\Omega_2 z - \Omega_3 y) + \boldsymbol{\hat\jmath}(\Omega_3 x) - \boldsymbol{\hat{k}}(\Omega_2 x)\, .
</math>
  </td>
</tr>
</table>
Using Riemann's notation instead &#8212; assuming our interpretation of Riemann's description is correct &#8212; we have,
<table border="0" align="center" cellpadding="2">

<tr>
  <td align="right">
<math>\vec\Omega \times \vec{x} </math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
\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]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
\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\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">&nbsp;</td>
  <td align="left">
<math>
+
\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\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">&nbsp;</td>
  <td align="left">
<math>
+
\boldsymbol{\hat{k}} \biggl\{
\biggl( \frac{\zeta}{c} \biggr) \cancelto{0}{\frac{dc}{dt}} + (c q_1 - aq)\frac{\xi}{a} + (bp - cp_1) \frac{\eta}{b} 
- \biggl[ (cq_1)\frac{\xi}{a} - (cp_1) \frac{\eta}{b} \biggr]
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
\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] \, .
</math>
  </td>
</tr>
</table>
Now, given that (see the bottom of p. 177 of [http://www.kendrickpress.com/Riemann.htm BCO2004]),
<table border="0" align="center" cellpadding="2">

<tr>
  <td align="right">
<math>p = (u + u') \, ,</math>
  </td>
  <td align="center">&nbsp; &nbsp; &nbsp; and, &nbsp; &nbsp; &nbsp;</td>
  <td align="right">
<math>q = (v + v') \, ,</math>
  </td>
  <td align="center">&nbsp; &nbsp; &nbsp; and, &nbsp; &nbsp; &nbsp;</td>
  <td align="right">
<math>r = (w + w') \, ,</math>
  </td>
</tr>
</table>
this expression becomes,
<table border="0" align="center" cellpadding="2">

<tr>
  <td align="right">
<math>\vec\Omega \times \vec{x} </math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
\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] \, .
</math>
  </td>
</tr>
</table>
Given what has been derived [[#vw|below]], namely,

<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right">
<math>\biggl[\frac{ w'}{w} \biggr]^2 </math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
\frac{(2a - b - c)}{(2a + b + c)} \cdot \frac{(2a - b + c)}{(2a + b - c)} \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\biggl[ \frac{ v'}{v} \biggr]^2</math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
\biggl[
\frac{(2a - b - c)}{(2a + b + c)} \cdot \frac{(2a + b - c)}{(2a - b + c)}
\biggr] 
</math>
  </td>
</tr>
</table>
we see that,

<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right">
<math>(v + v')</math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
\biggl\{ 1 + \biggl[
\frac{(2a - b - c)}{(2a + b + c)} \cdot \frac{(2a + b - c)}{(2a - b + c)}
\biggr]^{1 / 2} \biggr\}v
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
\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} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
\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}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
\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}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
\biggl\{ [(2a + c)^2 - b^2]^{1 / 2} +[(2a - c)^2 - b^2]^{1 / 2} 
\biggr\}~S^{1 / 2}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
\biggl\{ [(4a^2 + c^2 - b^2) + 4ac]^{1 / 2} +[(4a^2 + c^2 - b^2) - 4ac]^{1 / 2} 
\biggr\}~S^{1 / 2}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
\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}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
\biggl\{ 2(4a^2 + c^2 - b^2)  
+
2[(4a^2 + c^2 - b^2)^2 - 16a^2c^2]^{1 / 2}
\biggr\}^{1 / 2} ~S^{1 / 2}
</math>
  </td>
</tr>
</table>
and,

<table border="0" align="center" cellpadding="5">
<tr>
  <td align="right">
<math>(w + w')  </math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
\biggl\{1 + \biggl[
\frac{(2a - b - c)}{(2a + b + c)} \cdot \frac{(2a - b + c)}{(2a + b - c)}
\biggr]^{1 / 2} \biggr\} w
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
\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}
\, .
</math>
  </td>
</tr>
</table>

==Raw Form of Ten Governing Relations==
In summary (see p. 178 of [http://www.kendrickpress.com/Riemann.htm BCO2004]), Riemann states that <font color="orange">the system of equations that the ten unknown functions of time must satisfy becomes &hellip;</font>

<table border="0" align="center" cellpadding="2" width="90%">

<tr>
  <td align="right" width="10%">
<math>\frac{1}{2} \frac{d^2a}{dt^2} + \epsilon a A - \frac{\sigma}{a}</math>
  </td>
  <td align="center" width="3%">=</td>
  <td align="left">
<math>
(a-c)v^2 + (a+c)(v')^2 + (a-b)w^2 + (a+b)(w')^2 \, ,
</math>
  </td>
  <td align="left">
<table border="0" align="right" width="5%" cellpadding="3"><tr><td align="center">
<math>(\alpha.1)</math>
</td></tr></table>
  </td>
</tr>

<tr>
  <td align="right" width="10%">
<math>\frac{1}{2} \frac{d^2b}{dt^2} + \epsilon b B - \frac{\sigma}{b}</math>
  </td>
  <td align="center" width="3%">=</td>
  <td align="left">
<math>
(b-a)w^2 + (b+a)(w')^2 + (b-c)u^2 + (b+c)(u')^2 \, ,
</math>
  </td>
  <td align="left">
<table border="0" align="right" width="5%" cellpadding="3"><tr><td align="center">
<math>(\alpha.2)</math>
</td></tr></table>
  </td>
</tr>

<tr>
  <td align="right" width="10%">
<math>\frac{1}{2} \frac{d^2c}{dt^2} + \epsilon c C - \frac{\sigma}{c}</math>
  </td>
  <td align="center" width="3%">=</td>
  <td align="left">
<math>
(c-b)u^2 + (c+b)(u')^2 + (c-a)v^2 + (c+a)(v')^2 \, ,
</math>
  </td>
  <td align="left">
<table border="0" align="right" width="5%" cellpadding="3"><tr><td align="center">
<math>(\alpha.3)</math>
</td></tr></table>
  </td>
</tr>

<tr>
  <td align="right" width="10%">
<math>\frac{1}{(b-c)}\frac{d}{dt} \biggl[ u(b - c)^2 \biggr]</math>
  </td>
  <td align="center" width="3%">=</td>
  <td align="left">
<math>
- (b+c-2a) v w - (b+c+2a)v' w' \, ,
</math>
  </td>
  <td align="left">
<table border="0" align="right" width="5%" cellpadding="3"><tr><td align="center">
<math>(\alpha.4)</math>
</td></tr></table>
  </td>
</tr>

<tr>
  <td align="right" width="10%">
<math>\frac{1}{(b+c)}\frac{d}{dt} \biggl[ u' (b+c)\biggr]</math>
  </td>
  <td align="center" width="3%">=</td>
  <td align="left">
<math>
- (b-c+2a) v w' - (b-c-2a)v' w \, ,
</math>
  </td>
  <td align="left">
<table border="0" align="right" width="5%" cellpadding="3"><tr><td align="center">
<math>(\alpha.5)</math>
</td></tr></table>
  </td>
</tr>

<tr>
  <td align="right" width="10%">
<math>\frac{1}{(c-a)}\frac{d}{dt}\biggl[v(c - a)^2\biggr]</math>
  </td>
  <td align="center" width="3%">=</td>
  <td align="left">
<math>
- (c + a - 2b) wu - (c + a + 2b)w' u' \, ,
</math>
  </td>
  <td align="left">
<table border="0" align="right" width="5%" cellpadding="3"><tr><td align="center">
<math>(\alpha.6)</math>
</td></tr></table>
  </td>
</tr>

<tr>
  <td align="right" width="10%">
<math>\frac{1}{(c+a)}\frac{d}{dt}\biggl[v' (c + a)^2\biggr]</math>
  </td>
  <td align="center" width="3%">=</td>
  <td align="left">
<math>
- (c - a + 2b) wu' - (c - a - 2b)w' u \, ,
</math>
  </td>
  <td align="left">
<table border="0" align="right" width="5%" cellpadding="3"><tr><td align="center">
<math>(\alpha.7)</math>
</td></tr></table>
  </td>
</tr>

<tr>
  <td align="right" width="10%">
<math>\frac{1}{(a-b)}\frac{d}{dt}\biggl[w(a-b)^2\biggr]</math>
  </td>
  <td align="center" width="3%">=</td>
  <td align="left">
<math>
- (a + b - 2c) uv - (a + b + 2c)u'v' \, ,
</math>
  </td>
  <td align="left">
<table border="0" align="right" width="5%" cellpadding="3"><tr><td align="center">
<math>(\alpha.8)</math>
</td></tr></table>
  </td>
</tr>

<tr>
  <td align="right" width="10%">
<math>\frac{1}{(a+b)}\frac{d}{dt}\biggl[w'(a+b)^2\biggr]</math>
  </td>
  <td align="center" width="3%">=</td>
  <td align="left">
<math>
- (a - b + 2c) uv' - (a - b - 2c)u'v \, ,
</math>
  </td>
  <td align="left">
<table border="0" align="right" width="5%" cellpadding="3"><tr><td align="center">
<math>(\alpha.9)</math>
</td></tr></table>
  </td>
</tr>

<tr>
  <td align="right" width="10%">
<math>abc</math>
  </td>
  <td align="center" width="3%">=</td>
  <td align="left">
<math>
a_0 b_0 c_0 \, .
</math>
  </td>
  <td align="left">
<table border="0" align="right" width="5%" cellpadding="3"><tr><td align="center">
<math>(\alpha.10)</math>
</td></tr></table>
  </td>
</tr>
</table>

==Gravitational Potential==
For the gravitational potential, <math>V</math>, (see p. 178 of [http://www.kendrickpress.com/Riemann.htm BCO2004]), Riemann adopts the expression,

<table border="0" align="center" cellpadding="2" width="90%">

<tr>
  <td align="right" width="10%">
<math>V</math>
  </td>
  <td align="center" width="3%">=</td>
  <td align="left">
<math>
H - A\xi^2 -B\eta^2 - C\zeta^2
</math>
  </td>
</tr>

<tr>
  <td align="right" width="10%">
&nbsp;
  </td>
  <td align="center" width="3%">=</td>
  <td align="left">
<math>
\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] \, ,
</math>
  </td>
</tr>
</table>
where,

<table border="0" align="center" cellpadding="2" width="90%">

<tr>
  <td align="right" width="10%">
<math>\Delta_R</math>
  </td>
  <td align="center" width="3%"><math>\equiv</math></td>
  <td align="left">
<math>
\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}\, .
</math>
  </td>
</tr>
</table>
From our [https://www.vistrails.org/index.php/User:Tohline/ThreeDimensionalConfigurations/HomogeneousEllipsoids#Gravitational_Potential separate discussion of the gravitational potential of homogeneous ellipsoids], which closely follows the notation used in EFE, we find,

<div align="center">
<math>
~\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],
</math><br />

[ [[User:Tohline/Appendix/References#EFE|EFE]], <font color="#00CC00">Chapter 3, Eq. (40)</font><sup>1,2</sup> ]<br />
[ [[User:Tohline/Appendix/References#BT87|BT87]], <font color="#00CC00">Chapter 2, Table 2-2</font> ]
</div>

<!--
[[File:CommentButton02.png|right|100px|Comment by J. E. Tohline on 15 August 2020:  This integral definition of A_i also appears as Eq. (5) of &sect;10.2 (p. 234) of T78, but it contains an error &#8212; in the denominator on the right-hand-side, a_1 appears instead of a_i.]]
-->
where,

<table align="center" border=0 cellpadding="3">
<tr>
  <td align="right">
<math>
~A_i
</math>
  </td>
  <td align="center">
<math>
~\equiv
</math>
  </td>
  <td align="left">
<math>
~a_1 a_2 a_3 \int_0^\infty \frac{du}{\Delta (a_i^2 + u )} ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>
~I_\mathrm{BT}
</math>
  </td>
  <td align="center">
<math>
~\equiv
</math>
  </td>
  <td align="left">
<math>
~\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 ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>
~\Delta
</math>
  </td>
  <td align="center">
<math>
~\equiv
</math>
  </td>
  <td align="left">
<math>
~\biggl[ (a_1^2 + u)(a_2^2 + u)(a_3^2 + u)  \biggr]^{1/2} .
</math>
  </td>
</tr>
</table>

<div align="center">
[ [[User:Tohline/Appendix/References#EFE|EFE]], <font color="#00CC00">Chapter 3, Eqs. (18), (15 &amp; 22)</font><sup>1</sup><font color="#00CC00">, &amp; (8)</font>, respectively ]<br />
[ [[User:Tohline/Appendix/References#BT87|BT87]], <font color="#00CC00">Chapter 2, Table 2-2</font> ]
</div>
How do these two equations for the potential relate to one another?  Multiplying Riemann's expression through by <math>(-G\rho)</math>, we have,

<table border="0" align="center" cellpadding="2" width="90%">

<tr>
  <td align="right" width="10%">
<math>(-G\rho)V</math>
  </td>
  <td align="center" width="3%">=</td>
  <td align="left">
<math>
- \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]
</math>
  </td>
</tr>

<tr>
  <td align="right" width="10%">
&nbsp;
  </td>
  <td align="center" width="3%">=</td>
  <td align="left">
<math>
- \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\}
</math>
  </td>
</tr>

<tr>
  <td align="right" width="10%">
&nbsp;
  </td>
  <td align="center" width="3%">=</td>
  <td align="left">
<math>
- \pi G\rho \biggl\{ 
I_\mathrm{BT}a^2 
- A_1\xi^2 - A_2\eta^2 - A_3 \zeta^2
\biggr\} \, .
</math>
  </td>
</tr>
</table>
We see, therefore, that 

<table border="0" align="center" cellpadding="2">

<tr>
  <td align="right">
<math>(-G\rho)V</math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
\Phi(\vec{x}) \, ,
</math>
  </td>
</tr>
</table>
and we recognize the following notation associations:

<table border="0" align="center" cellpadding="2">

<tr>
  <td align="right">
<math>(A, B, C)~~ \leftrightarrow ~~ (A_1, A_2, A_3)</math>
  </td>
  <td align="center">&nbsp; &nbsp; &nbsp; and, &nbsp; &nbsp; &nbsp; </td>
  <td align="left">
<math>
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 \, .
</math>
  </td>
</tr>
</table>