Editing 3Dconfigurations/RiemannEllipsoids (section)

===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.