Editing 3Dconfigurations/RiemannEllipsoids (section)

===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>.