Editing 3Dconfigurations/RiemannEllipsoids (section)

== 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>