Editing 3Dconfigurations/RiemannEllipsoids (section)

=Steady-State Configurations=
In &sect;6 (pp. 183 - 187) of [http://www.kendrickpress.com/Riemann.htm BCO2004]), Riemann investigates the case in which one of the pairs of quantities &#8212; <math>(u, u')</math>, or <math>(v, v')</math>, or <math>(w, w')</math> &#8212; is zero throughout the time-evolving fluid configuration.  As an example, he chooses, <math>u = u' = 0</math>. According to (the English translation of) Riemann's analysis, <font color="orange">"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."</font>

===Case when u = u' = 0===

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

<tr>
  <td align="right" width="10%">
<math>0</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>\mathrm{mod.01} ~\mathrm{of}~(\alpha.4)</math>
</td></tr></table>
  </td>
</tr>

<tr>
  <td align="right" width="10%">
<math>0</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>\mathrm{mod.01} ~\mathrm{of}~(\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>
0 ~~~\Rightarrow~~~ v(c - a)^2 = \mathrm{constant} \, ,
</math>
  </td>
  <td align="left">
<table border="0" align="right" width="5%" cellpadding="3"><tr><td align="center">
<math>\mathrm{mod.01} ~\mathrm{of}~(\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>
0 ~~~\Rightarrow~~~ v'(c+a)^2 = \mathrm{constant} \, ,
</math>
  </td>
  <td align="left">
<table border="0" align="right" width="5%" cellpadding="3"><tr><td align="center">
<math>\mathrm{mod.01} ~\mathrm{of}~(\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>
0 ~~~\Rightarrow~~~ w(a-b)^2 = \mathrm{constant} \, ,
</math>
  </td>
  <td align="left">
<table border="0" align="right" width="5%" cellpadding="3"><tr><td align="center">
<math>\mathrm{mod.01} ~\mathrm{of}~(\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>
0 ~~~\Rightarrow~~~ w'(a+b)^2 = \mathrm{constant} \, .
</math>
  </td>
  <td align="left">
<table border="0" align="right" width="5%" cellpadding="3"><tr><td align="center">
<math>\mathrm{mod.01} ~\mathrm{of}~(\alpha.9)</math>
</td></tr></table>
  </td>
</tr>
</table>

====''v'' and ''w'' Ratios====

From the first modification (mod.01) of equation <math>(\alpha.4)</math>, we find that,

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

<tr>
  <td align="right">
<math>(b+c+2a)v' w'</math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
-(b+c-2a) vw
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ \frac{ v'}{v}</math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
\frac{(2a - b - c)}{(2a + b + c)}\cdot \frac{ w}{w'} \, .
</math>
  </td>
</tr>
</table>
And from the first modification (mod.01) of equation <math>(\alpha.5)</math>, we find,

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

<tr>
  <td align="right">
<math>(2a - b + c)v' w </math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>(2a + b - c) v w' </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ \frac{ v' w}{vw'}</math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
\frac{(2a + b - c)}{(2a - b + c)} \, .
</math>
  </td>
</tr>
</table>

<span id="vw">Combining these two expressions gives</span>,

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

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

<tr>
  <td align="right">
<math>\Rightarrow ~~~\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>
</table>
and,

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

<tr>
  <td align="right">
<math>\biggl[ \frac{ v'}{v} \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>
</table>

From the first modifications (mod.01) of <math>(\alpha.6)</math> through <math>(\alpha.9)</math>, we also see that,

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

<tr>
  <td align="right">
<math>\biggl[ \frac{ v'}{v} \biggr]^2</math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
\frac{(c+a)^4}{(c-a)^4} \times \mathrm{constant} \, ,
</math>
  </td>
<td align="center">&nbsp; &nbsp; &nbsp; and, &nbsp; &nbsp; &nbsp;
  <td align="right">
<math>\biggl[ \frac{ w'}{w} \biggr]^2</math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
\frac{(a-b)^4}{(a+b)^4} \times \mathrm{constant} \, .
</math>
  </td>
</tr>
</table>

Hence, following Riemann (1861) &#8212; see the bottom of p. 183 of [http://www.kendrickpress.com/Riemann.htm BCO2004] &#8212; we can write,
<table border="1" align="center" cellpadding="5" width="80%"><tr><td align="left">

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

<tr>
  <td align="right">
<math>\biggl[ \frac{ v'}{v} \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)} 
=
\biggl( \frac{c+a}{c-a}\biggr)^4 \times \mathrm{constant} \, ,
</math>
  </td>
</tr>

<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)} 
=
\biggl( \frac{a-b}{a+b} \biggr)^4 \times \mathrm{constant} \, .
</math>
  </td>
</tr>
</table>

</td></tr></table>

As Riemann (1961) points out &#8212; see the top of p. 184 of [http://www.kendrickpress.com/Riemann.htm BCO2004] &#8212; <font color="orange">"Taking this together with</font> the constant volume constraint <math>(\alpha.10)</math> <font color="orange">we see that <math>a, b,  c</math> are constant, and consequently <math>v, v', w, w'</math> are constant."</font>

====''v'' and ''w'' Sums and Differences====

If we combine the just-derived coefficient ratios with equation (2) of &sect;6 (see p. 184 of [http://www.kendrickpress.com/Riemann.htm BCO2004]), we deduce that,
<table border="0" align="center" cellpadding="2">

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

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

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

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

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

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
2^{1 / 2}\biggl\{
(4a^2  - b^2 + c^2)   
\pm [(4a^2  - b^2 + c^2)^2 - 16a^2c^2]^{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>
2^{1 / 2}\biggl\{
(4a^2  - b^2 + c^2)   
\pm [16a^4 - 8a^2b^2  + b^4 - 2b^2c^2 + c^4 - 8a^2c^2]^{1 / 2}\biggr\}^{1 / 2} 
S^{1 / 2} 
</math>
  </td>
</tr>
</table>

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

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

<tr>
  <td align="right">
<math>2^2 a^2 \beta</math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
4a^2 - b^2 +c^2 \pm [4a^2 - (b+c)^2]^{1 / 2}[4a^2 - (b-c)^2]^{1 / 2}
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
4a^2 - b^2 +c^2 \pm [16a^4 - 8a^2b^2 +8a^2bc - 4a^2c^2 +b^4 - 2b^3c + b^2c^2 
-8a^2bc + 2b^3c - 4b^2c^2 + 2bc^3 - 4a^2c^2 + b^2c^2 -2bc^3 + c^4  ]^{1 / 2}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
4a^2 - b^2 +c^2 \pm [16a^4 - 8a^2b^2 - 8a^2c^2 +b^4  
- 2b^2c^2 + c^4  ]^{1 / 2}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ (v' - v)</math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
2^{1 / 2}\biggl\{
2^2 a^2 \beta
\biggr\}^{1 / 2} 
S^{1 / 2} 
=
2a(\beta S)^{1 / 2} \, .
</math>
  </td>
</tr>
</table>


Similarly,

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

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

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

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
(v' - v) \, .
</math>
  </td>
</tr>
</table>

We deduce as well that,
<table border="0" align="center" cellpadding="2">

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

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

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

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

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

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

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

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

And again, similarly,
<table border="0" align="center" cellpadding="2">

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

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

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

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
(w' - w) \, .
</math>
  </td>
</tr>
</table>

====First Three Governing Relation====
Given that <math>a, b,  c</math> are constant and <math>u = u' = 0</math>, the first three governing equations become,

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

<tr>
  <td align="right" width="10%">
<math>\frac{\epsilon A}{2} - \frac{\sigma}{2a^2}</math>
  </td>
  <td align="center" width="3%">=</td>
  <td align="left">
<math>
\frac{1}{2a} \biggl[ (a-c)v^2 + (a+c)(v')^2 + (a-b)w^2 + (a+b)(w')^2 \biggr] \, ,
</math>
  </td>
  <td align="left">
<table border="0" align="right" width="5%" cellpadding="3"><tr><td align="center">
<math>\mathrm{mod.01} ~\mathrm{of}~(\alpha.1)</math>
</td></tr></table>
  </td>
</tr>

<tr>
  <td align="right" width="10%">
<math>\frac{\epsilon B}{2} - \frac{\sigma}{2b^2}</math>
  </td>
  <td align="center" width="3%">=</td>
  <td align="left">
<math>
\frac{1}{2b} \biggl[ (b-a)w^2 + (b+a)(w')^2 \biggr] \, ,
</math>
  </td>
  <td align="left">
<table border="0" align="right" width="5%" cellpadding="3"><tr><td align="center">
<math>\mathrm{mod.01} ~\mathrm{of}~(\alpha.2)</math>
</td></tr></table>
  </td>
</tr>

<tr>
  <td align="right" width="10%">
<math>\frac{\epsilon C}{2} - \frac{\sigma}{2c^2}</math>
  </td>
  <td align="center" width="3%">=</td>
  <td align="left">
<math>
\frac{1}{2c} \biggl[ (c-a)v^2 + (c+a)(v')^2 \biggr]\, .
</math>
  </td>
  <td align="left">
<table border="0" align="right" width="5%" cellpadding="3"><tr><td align="center">
<math>\mathrm{mod.01} ~\mathrm{of}~(\alpha.3)</math>
</td></tr></table>
  </td>
</tr>
</table>

After adopting a pair of alternate constants, S and T, Riemann shows that these relations can be rewritten in the forms (mod.02),
<table border="0" align="center" cellpadding="2" width="90%">

<tr>
  <td align="right" width="10%">
<math>\frac{\epsilon A}{2} - \frac{\sigma}{2a^2}</math>
  </td>
  <td align="center" width="3%">=</td>
  <td align="left">
<math>
(4a^2 - b^2 -c^2)(T + S) - 2(b^2T + c^2S)
</math>
  </td>
  <td align="left">
<table border="0" align="right" width="5%" cellpadding="3"><tr><td align="center">
<math>\mathrm{mod.02} ~\mathrm{of}~(\alpha.1)</math>
</td></tr></table>
  </td>
</tr>

<tr>
  <td align="right" width="10%">
<math>\frac{\epsilon B}{2} - \frac{\sigma}{2b^2}</math>
  </td>
  <td align="center" width="3%">=</td>
  <td align="left">
<math>
(b^2-c^2)T \, ,
</math>
  </td>
  <td align="left">
<table border="0" align="right" width="5%" cellpadding="3"><tr><td align="center">
<math>\mathrm{mod.02} ~\mathrm{of}~(\alpha.2)</math>
</td></tr></table>
  </td>
</tr>

<tr>
  <td align="right" width="10%">
<math>\frac{\epsilon C}{2} - \frac{\sigma}{2c^2}</math>
  </td>
  <td align="center" width="3%">=</td>
  <td align="left">
<math>
(c^2-b^2)S\, .
</math>
  </td>
  <td align="left">
<table border="0" align="right" width="5%" cellpadding="3"><tr><td align="center">
<math>\mathrm{mod.02} ~\mathrm{of}~(\alpha.3)</math>
</td></tr></table>
  </td>
</tr>
</table>

On p. 158 (Chapter 7, &sect;51) of EFE, at a comparable stage of Chandrasekhar's derivation, we find the following trio of governing relations:
<table border="0" align="center" cellpadding="2" width="90%">

<tr>
  <td align="right" width="10%">
<math>2A_1 - \frac{5\Pi}{Ma_1^2}</math>
  </td>
  <td align="center" width="3%">=</td>
  <td align="left">
<math>
\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) \, ,
</math>
  </td>
  <td align="left">
<table border="0" align="right" width="5%" cellpadding="3"><tr><td align="center">
<math>\mathrm{EFE}, ~\mathrm{Eq.}(164)</math>
</td></tr></table>
  </td>
</tr>

<tr>
  <td align="right" width="10%">
<math>2A_2 - \frac{5\Pi}{Ma_2^2}</math>
  </td>
  <td align="center" width="3%">=</td>
  <td align="left">
<math>
\biggl[ \frac{a_2^2 - a_3^2}{2a_2^2} \biggr]\Omega_3^2 \gamma \, ,
</math>
  </td>
  <td align="left">
<table border="0" align="right" width="5%" cellpadding="3"><tr><td align="center">
<math>\mathrm{EFE}, ~\mathrm{Eq.}(165)</math>
</td></tr></table>
  </td>
</tr>

<tr>
  <td align="right" width="10%">
<math>2A_3 - \frac{5\Pi}{Ma_3^2}</math>
  </td>
  <td align="center" width="3%">=</td>
  <td align="left">
<math>
\biggl[ \frac{a_3^2 - a_2^2}{2a_3^2} \biggr]\Omega_2^2 \beta \, .
</math>
  </td>
  <td align="left">
<table border="0" align="right" width="5%" cellpadding="3"><tr><td align="center">
<math>\mathrm{EFE}, ~\mathrm{Eq.}(166)</math>
</td></tr></table>
  </td>
</tr>
</table>

<span id="Correspondence">Building</span> on our [[#Gravitational_Potential|above discussion of the expression for the gravitational potential]], it appears as though we have the following notational correspondence:

<table border="1" align="center" cellpadding="5">
<tr>
  <td align="center" colspan="2">'''Notational Correspondence'''</td>
</tr>
<tr>
  <td align="center">'''Riemann'''</td>
  <td align="center">'''Chandrasekhar'''</td>
</tr>
<tr>
  <td align="center"><math>(A, B, C)</math></td>
  <td align="center"><math>(A_1, A_2, A_3)</math></td>
</tr>
<tr>
  <td align="center"><math>\sigma</math></td>
  <td align="center"><math>\epsilon \biggl[\frac{5\Pi}{2M}\biggr]</math></td>
</tr>
<tr>
  <td align="center"><math>T</math></td>
  <td align="center"><math>\epsilon \biggl[ \frac{\Omega_3^2 \gamma}{8b^2} \biggr]</math></td>
</tr>
<tr>
  <td align="center"><math>S</math></td>
  <td align="center"><math>\epsilon \biggl[ \frac{\Omega_2^2 \beta}{8c^2} \biggr]</math></td>
</tr>
</table>
As a check, multiplying EFE's equation (164) through by <math>(\epsilon/4)</math> gives,
<table border="0" align="center" cellpadding="2" width="90%">

<tr>
  <td align="right">
<math>\frac{\epsilon A_1}{2} - \epsilon \biggl[ \frac{5\Pi}{4Ma_1^2} \biggr]</math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
\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} \, , 
</math>
  </td>
</tr>
</table>
which, after changing to Riemann's corresponding notation, gives,
<table border="0" align="center" cellpadding="2" width="90%">

<tr>
  <td align="right">
<math>\frac{\epsilon A}{2} - \biggl[ \frac{\sigma}{2a^2} \biggr]</math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
\epsilon (4a^2 - b^2 - c^2)
\biggl(S + T\biggr) 
- 2(c^2 S + b^2 T) \, .
</math>
  </td>
</tr>
</table>
This matches the second modification of Riemann's equation <math>(\alpha.1)</math> precisely.  So we are confident that the entries in our table of notational correspondences are correct.

===Velocities===

If, in our [[#Step_3|above discussion]], we are interpreting Riemann's analysis correctly, then the steady-state velocity components as viewed from the rotating reference frame are,
<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>
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_1 = (u - u') \, ,</math>
  </td>
  <td align="center">&nbsp; &nbsp; &nbsp; and, &nbsp; &nbsp; &nbsp;</td>
  <td align="right">
<math>q_1 = (v - v') \, ,</math>
  </td>
  <td align="center">&nbsp; &nbsp; &nbsp; and, &nbsp; &nbsp; &nbsp;</td>
  <td align="right">
<math>r_1 = (w - w') \, ,</math>
  </td>
</tr>
</table>
these velocity components may be written as,
<table border="0" align="center" cellpadding="2">

<tr>
  <td align="right">
<math>\frac{\partial \xi}{\partial t} </math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
(w - w')\frac{a\eta}{b} - (v - v') \frac{a\zeta}{c} \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\frac{\partial\eta}{\partial t} </math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
(u - u')\frac{b \zeta}{c} - (w - w') \frac{b \xi}{a} \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\frac{\partial \zeta}{\partial t} </math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
(v - v')\frac{c\xi}{a} - (u - u') \frac{c\eta}{b} \, .
</math>
  </td>
</tr>
</table>
Examining the case of <math>u = u' = 0</math>, we have,
<table border="0" align="center" cellpadding="2">

<tr>
  <td align="right">
<math>\frac{\partial \xi}{\partial t} </math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
(w - w')\frac{a\eta}{b} - (v - v') \frac{a\zeta}{c} \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\frac{\partial\eta}{\partial t} </math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
(w' - w) \frac{b \xi}{a} \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\frac{\partial \zeta}{\partial t} </math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
(v - v')\frac{c\xi}{a} \, .
</math>
  </td>
</tr>
</table>

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.
<table border="1" align="center" cellpadding="10">
<tr><td align="center" colspan="3">
EFE, Chapter 7, &sect;51 (p. 156), Eq. (154)
</td></tr>
<tr><td align="left">
<table border="0" align="center" cellpadding="2">

<tr>
  <td align="right">
<math>u_1</math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
(\Omega_3 \gamma) \frac{a^2 x_2}{b^2} - (\Omega_2\beta) \frac{a^2 x_3}{c^2} \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>u_2</math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
- (\Omega_3 \gamma) x_1 \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>u_3</math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
(\Omega_2 \beta) x_1 \, .
</math>
  </td>
</tr>
</table>

</td>
</tr></table>

From equation (2) of &sect;6 (see p. 184 of [http://www.kendrickpress.com/Riemann.htm BCO2004]), we deduce that,
<table border="0" align="center" cellpadding="2">

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

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
\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}
</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>
</table>
Now, from [[#Correspondence|above]], we have surmised that,
<table border="0" align="center" cellpadding="2">

<tr>
  <td align="right">
<math>S</math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
\frac{\epsilon \Omega_2^2 \beta}{8c^2} \, ,
</math>
  </td>
</tr>
</table>
in which case,
<table border="0" align="center" cellpadding="2">

<tr>
  <td align="right">
<math>\frac{\partial \zeta}{\partial t} = (v - v')\frac{c\xi}{a} </math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
\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}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
\biggl(\frac{\epsilon}{2}\biggr)^{1 / 2}(\Omega_2 \beta) \xi\biggl\{
K_0 \biggr\} 
\biggl[\frac{1}{4a^2 \beta}\biggr]^{1 / 2} \, ,
</math>
  </td>
</tr>
</table>

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

<tr>
  <td align="right">
<math>K_0</math>
  </td>
  <td align="center"><math>\equiv</math></td>
  <td align="left">
<math>
(2a-b-c)^{1 / 2}(2a+b-c)^{1 / 2} - (2a + b + c)^{1 / 2} (2a - b +c)^{1 / 2} \, .
</math>
  </td>
</tr>
</table>
According to EFE &#8212; see Chapter 7, &sect;47 (p. 131) &hellip;
<table border="0" align="center" cellpadding="2">

<tr>
  <td align="right">
<math>4a^2\beta </math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
4a^2 - b^2 + c^2 \pm \biggl\{
[4a^2 - (b+c)^2] [4a^2 - (b-c)^2]
\biggr\}^{1 / 2} 
</math>
  </td>
</tr>
</table>

<table border="1" align="center" width="90%" cellpadding="5"><tr><td align="center">
Test case:  <math>a=1.0, b=1.2, c=0.33 ~~~\Rightarrow~~~ K_0 = -0.83580</math><br />
and (selecting the negative sign), &nbsp; &nbsp; <math>4a^2\beta = 2.66890 - 2.31962 = 0.34928</math><br />
Hence, <math>\frac{K_0}{[4a^2\beta]^{1 / 2} } = - 1.41421 = -\sqrt{2} \, .</math><br />
<font color="red">Excellent!</font> &nbsp; This strongly suggests that in all cases, <math>8a^2\beta = K_0^2 \, ;</math> and that, as was surmised earlier,<br />
<math>\frac{\partial\zeta}{\partial t} = \epsilon^{1 / 2} (\Omega_2 \beta)\xi \, .</math>
</td></tr></table>

Let's check to see if this is indeed a ''general'' result.  Rewriting <math>K_0^2</math>, we find,
<table border="0" align="center" cellpadding="2">

<tr>
  <td align="right">
<math>K_0^2 </math>
  </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\}^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\}^2
</math>
  </td>
</tr>

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

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
2[ 4a^2 + c^2 - b^2]  - 2[(2a + c)^2 - b^2]^{1 / 2}[ (2a-c)^2-b^2]^{1 / 2} \, .
</math>
  </td>
</tr>
</table>
Hence (selecting the negative sign in the expression for <math>4a^2\beta</math>),

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

<tr>
  <td align="right">
<math>4a^2 \beta - \frac{K_0^2}{2} </math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
[ b^2 - 4a^2 - c^2 ]  + [(2a + c)^2 - b^2]^{1 / 2}[ (2a-c)^2-b^2]^{1 / 2}
</math>
  </td>
</tr>

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

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

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

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

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
\biggl\{ 16a^4 -8a^2c^2 + c^4
- 8a^2b^2 - 2b^2c^2 + b^4
\biggr\}^{1 / 2}
- \biggl\{ 
16a^4 - 8a^2b^2 - 8a^2c^2  + b^4 -2b^2c^2 + c^4
\biggr\}^{1 / 2}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
0 \, .
</math>
  </td>
</tr>
</table>
<font color="red">Terrific!</font> &nbsp; We have demonstrated that, quite generally, <math>8a^2\beta = K_0^2</math>.