Editing ThreeDimensionalConfigurations/ChallengesPt4 (section)

==Test Orthogonality==

Let's see if the Riemann fluid velocity vector is everywhere tangent to our identified off-center elliptical (Lagrangian-particle) orbital trajectories.  We can determine this by first deriving an expression for the vector normal to the trajectories, then see if the dot product of this vector and the velocity vector is everywhere zero.

The unit vector that is normal to a trajectory is obtained from the (appropriately normalized) gradient of the function,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~f(x', y')</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\biggl[\frac{x'}{x_\mathrm{max}} \biggr]^2 + \biggl[\frac{y' - y_c}{y_\mathrm{max}} \biggr]^2 - 1\, .</math>
  </td>
</tr>
</table>
We find, first,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\nabla f</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\boldsymbol{\hat\imath'} \biggl[\frac{2x'}{x^2_\mathrm{max}} \biggr] 
+ 
\boldsymbol{\hat\jmath'} \biggl[\frac{2(y' - y_c)}{y^2_\mathrm{max}} \biggr] 
- 
\boldsymbol{\hat{k}'} \biggl[\frac{2(y' - y_c)}{y^2_\mathrm{max}} \biggr] \frac{\partial y_c}{\partial z'} \, ,
</math>
  </td>
</tr>
</table>

where, drawing from [[#Step2|Step #2, above]],
<table border="1" cellpadding="8" align="center" width="60%"><tr><td align="left">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{y_c}{z_0}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- \frac{\sin\theta}{c^2 \kappa^2} \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~z_0^2</math>
  </td>
  <td align="center">
<math>~\le</math>
  </td>
  <td align="left">
<math>~c^2 + b^2\tan^2\theta \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~y_\mathrm{max}^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{\kappa^2}\biggl[ 1 - \frac{z_0^2}{c^2 } - \frac{z_0 \sin\theta}{c^2} \biggr] \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~x_\mathrm{max}^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~a^2 \biggl[ 1 - \frac{z_0^2}{c^2 } - \frac{z_0 \sin\theta}{c^2} \biggr] \, .</math>
  </td>
</tr>
</table>
Note as well that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\biggl( \frac{x_\mathrm{max}}{y_\mathrm{max}}\biggr)^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~a^2\kappa^2 = \frac{a^2}{b^2 c^2} \biggl[ c^2 \cos^2\theta + b^2 \sin^2\theta \biggr] \, .</math>
  </td>
</tr>
</table>

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


Next, after [[#ThetaDef|setting]] <math>~\tan\theta = - (\beta\Omega_2/\gamma\Omega_3)</math>, and pulling from [[#Riemann_Flow|Step #3, above]], we have,

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

<tr>
  <td align="right">
<math>~\boldsymbol{u'}_\mathrm{EFE}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\boldsymbol{\hat\imath'} \biggl[ \biggl(\frac{a}{b}\biggr)^2 \gamma \Omega_3 (  y'\cos\theta + z_0\tan\theta ) 
- \biggl(\frac{a}{c}\biggr)^2 \beta \Omega_2 (  y'\sin\theta ) \biggr]
+
\boldsymbol{\hat\jmath'}  \biggl[\beta \Omega_2  \sin\theta 
- \gamma \Omega_3  \cos\theta \biggr]x' \, .
</math>
  </td>
</tr>
</table>

===Stick With Rotating Frame===
The dot product of these two vectors is,

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

<tr>
  <td align="right">
<math>~\nabla f \cdot \boldsymbol{u'}_\mathrm{EFE}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \biggl(\frac{a}{b}\biggr)^2 \gamma \Omega_3 (  y'\cos\theta + z_0\tan\theta ) 
- \biggl(\frac{a}{c}\biggr)^2 \beta \Omega_2 (  y'\sin\theta ) \biggr]\biggl[\frac{2x'}{x^2_\mathrm{max}} \biggr] 
+
\biggl[\beta \Omega_2  \sin\theta 
- \gamma \Omega_3  \cos\theta \biggr]x' \biggl[\frac{2(y' - y_c)}{y^2_\mathrm{max}} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \biggl(\frac{a}{b}\biggr)^2 \gamma \Omega_3 (  y'\cos\theta ) 
- \biggl(\frac{a}{c}\biggr)^2 \beta \Omega_2 (  y'\sin\theta ) \biggr]\biggl[\frac{2x'}{x^2_\mathrm{max}} \biggr] 
+
\biggl[ \biggl(\frac{a}{b}\biggr)^2 \gamma \Omega_3 (  z_0\tan\theta ) 
\biggr]\biggl[\frac{2x'}{x^2_\mathrm{max}} \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+
\biggl[\beta \Omega_2  \sin\theta 
- \gamma \Omega_3  \cos\theta \biggr]x' \biggl[\frac{2y' }{y^2_\mathrm{max}} \biggr]
-
\biggl[\beta \Omega_2  \sin\theta 
- \gamma \Omega_3  \cos\theta \biggr]x' \biggl[\frac{2y_c }{y^2_\mathrm{max}} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 2x'y' \biggl\{
\biggl[ \biggl(\frac{a}{b}\biggr)^2 \gamma \Omega_3 (  \cos\theta ) 
- \biggl(\frac{a}{c}\biggr)^2 \beta \Omega_2 ( \sin\theta ) \biggr]\biggl[\frac{1}{x^2_\mathrm{max}} \biggr] 
+
\biggl[\beta \Omega_2  \sin\theta 
- \gamma \Omega_3  \cos\theta \biggr] \biggl[\frac{1 }{y^2_\mathrm{max}} \biggr]
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ 2x' \biggl\{
\biggl[ \biggl(\frac{a}{b}\biggr)^2 \gamma \Omega_3 \tan\theta  
\biggr]\biggl[\frac{z_0}{x^2_\mathrm{max}} \biggr] 
-
\biggl[\beta \Omega_2  \sin\theta 
- \gamma \Omega_3  \cos\theta \biggr]\biggl[\frac{y_c }{y^2_\mathrm{max}} \biggr] 
\biggr\} \, .
</math>
  </td>
</tr>
</table>
Now, drawing from the various above "boxed" relations and examining the two terms separately, we first find,

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

<tr>
  <td align="right">
TERM1
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{2x'y'}{x_\mathrm{max}^2} \biggl( \frac{a^2}{b^2 c^2} \biggr) \biggl\{
\biggl[ c^2 \gamma \Omega_3 (  \cos\theta ) 
- b^2 \beta \Omega_2 ( \sin\theta ) \biggr] 
+
\biggl[\beta \Omega_2  \sin\theta 
- \gamma \Omega_3  \cos\theta \biggr] \biggl[\frac{x_\mathrm{max}^2 }{y^2_\mathrm{max}} \biggr]\biggl( \frac{b^2 c^2}{a^2} \biggr)
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{2x'y'}{x_\mathrm{max}^2} \biggl( \frac{a^2}{b^2 c^2} \biggr) \biggl\{
\biggl[ c^2 (  \cos\theta ) 
+ b^2 \tan\theta ( \sin\theta ) \biggr] 
-
\biggl[ \tan\theta  \sin\theta 
+  \cos\theta \biggr] \biggl[ c^2 \cos^2\theta +  b^2\sin^2\theta \biggr]
\biggr\} \gamma \Omega_3
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{2x'y'}{x_\mathrm{max}^2} \biggl( \frac{a^2}{b^2 c^2} \biggr) \biggl\{
\biggl[ c^2 + b^2 \tan^2\theta\biggr] 
-
\biggl[ \tan^2\theta  +  1 \biggr] \biggl[ c^2 \cos^2\theta +  b^2\sin^2\theta \biggr]
\biggr\} \gamma \Omega_3 \cos\theta
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{2x'y'}{x_\mathrm{max}^2} \biggl( \frac{a^2}{b^2 c^2} \biggr) \biggl\{
c^2 + b^2 \tan^2\theta
-
\biggl[ \tan^2\theta  +  1 \biggr] \biggl[ c^2 \cos^2\theta \biggr]
-
\biggl[ \tan^2\theta  +  1 \biggr] \biggl[  b^2\sin^2\theta \biggr]
\biggr\} \gamma \Omega_3 \cos\theta
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{2x'y'}{x_\mathrm{max}^2} \biggl( \frac{a^2}{b^2 c^2} \biggr) \biggl\{
c^2 + b^2 \tan^2\theta
- c^2 - b^2\tan^2\theta 
\biggr\} \gamma \Omega_3 \cos\theta
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 0 \, .</math>
  </td>
</tr>
</table>

And, in order for the second term to go to zero as well, we need &hellip;

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

<tr>
  <td align="right">
<math>~0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{2x'}{ x^2_\mathrm{max} } \biggl\{
\biggl[ \biggl(\frac{a}{b}\biggr)^2 \gamma \Omega_3 \tan\theta  
\biggr]z_0 
-
y_c \biggl[\beta \Omega_2  \sin\theta 
- \gamma \Omega_3  \cos\theta \biggr]\biggl[\frac{ x^2_\mathrm{max} }{y^2_\mathrm{max}} \biggr] 
\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{2x'}{ x^2_\mathrm{max} } \biggl\{
\biggl[ c^2  \tan\theta  
\biggr]z_0 
+
y_c \biggl[(\tan\theta) \sin\theta 
+  \cos\theta \biggr]( c^2 \cos^2\theta + b^2 \sin^2\theta ) 
\biggr\} \biggl( \frac{a^2}{b^2 c^2}  \biggr)\gamma \Omega_3
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{2x'}{ x^2_\mathrm{max} } \biggl\{
\biggl[ c^2  \tan\theta  
\biggr]z_0 
+
y_c \cos\theta( c^2 + b^2 \tan^2\theta ) 
\biggr\} \biggl( \frac{a^2}{b^2 c^2}  \biggr)\gamma \Omega_3
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~y_c \cos\theta( c^2 + b^2 \tan^2\theta )</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-\biggl[ c^2  \tan\theta  
\biggr]z_0 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~\frac{y_c}{z_0} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-  \frac{c^2  \tan\theta}{\cos\theta( c^2 + b^2 \tan^2\theta )}
=
-  \frac{c^2  \tan^2\theta}{\sin\theta( c^2 + b^2 \tan^2\theta )}
=
-  \frac{c^2  \sin\theta}{( c^2\cos^2\theta + b^2 \sin^2\theta )}
=
-  \frac{\sin\theta}{b^2  \kappa^2} \, .
</math>
  </td>
</tr>
</table>

===Try Shifting to (Quasi-) Inertial Frame===

Let's adjust both remaining components of <math>~\boldsymbol{u'}_\mathrm{EFE}</math> to see what their equivalent rotating-frame expressions are.  The relevant shift is,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\boldsymbol{v'}_\mathrm{shift} \equiv \boldsymbol{\Omega'} \times \boldsymbol{x'}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\boldsymbol{\hat{k}} \Omega' \times \biggl[ \boldsymbol{\hat\imath'}x' +  \boldsymbol{\hat\jmath'}y' \biggr]
=
- \boldsymbol{\hat\imath'} \Omega' y' + \boldsymbol{\hat\jmath'} \Omega' x' \, ,
</math>
  </td>
</tr>
</table>
where,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\Omega'</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \Omega_2^2 + \Omega_3^2 \biggr]^{1 / 2} \, .</math>
  </td>
</tr>
</table>
We have, then,

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

<tr>
  <td align="right">
<math>~\boldsymbol{u'}_\mathrm{inertial} \equiv \boldsymbol{u'}_\mathrm{EFE} + \boldsymbol{v'}_\mathrm{shift}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\boldsymbol{\hat\imath'} \biggl[ \biggl(\frac{a}{b}\biggr)^2 \gamma \Omega_3 (  y'\cos\theta + z_0\tan\theta ) 
- \biggl(\frac{a}{c}\biggr)^2 \beta \Omega_2 (  y'\sin\theta )  - \Omega' y'\biggr]
+
\boldsymbol{\hat\jmath'}  \biggl[\beta \Omega_2  \sin\theta 
- \gamma \Omega_3  \cos\theta + \Omega'\biggr]x' 
</math>
  </td>
</tr>
</table>
The dot product of this expression with <math>~\nabla f</math> is,

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

<tr>
  <td align="right">
<math>~\nabla f \cdot \boldsymbol{u'}_\mathrm{inertial}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \biggl(\frac{a}{b}\biggr)^2 \gamma \Omega_3 (  y'\cos\theta + z_0\tan\theta ) 
- \biggl(\frac{a}{c}\biggr)^2 \beta \Omega_2 (  y'\sin\theta ) - \Omega' y' \biggr]\biggl[\frac{2x'}{x^2_\mathrm{max}} \biggr] 
+
\biggl[\beta \Omega_2  \sin\theta 
- \gamma \Omega_3  \cos\theta + \Omega'\biggr]x' \biggl[\frac{2(y' - y_c)}{y^2_\mathrm{max}} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \biggl(\frac{a}{b}\biggr)^2 \gamma \Omega_3 (  \cos\theta ) 
- \biggl(\frac{a}{c}\biggr)^2 \beta \Omega_2 (  \sin\theta ) - \Omega' \biggr]\biggl[\frac{2x' y'}{x^2_\mathrm{max}} \biggr] 
+
\biggl[ \biggl(\frac{a}{b}\biggr)^2 \gamma \Omega_3 (  z_0\tan\theta ) 
\biggr]\biggl[\frac{2x'}{x^2_\mathrm{max}} \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+
\biggl[\beta \Omega_2  \sin\theta 
- \gamma \Omega_3  \cos\theta  + \Omega'\biggr]x' \biggl[\frac{2y' }{y^2_\mathrm{max}} \biggr]
-
\biggl[\beta \Omega_2  \sin\theta 
- \gamma \Omega_3  \cos\theta + \Omega' \biggr]x' \biggl[\frac{2y_c }{y^2_\mathrm{max}} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 2x'y' \biggl\{
\biggl[ \biggl(\frac{a}{b}\biggr)^2 \gamma \Omega_3 (  \cos\theta ) 
- \biggl(\frac{a}{c}\biggr)^2 \beta \Omega_2 ( \sin\theta ) - \Omega'\biggr]\biggl[\frac{1}{x^2_\mathrm{max}} \biggr] 
+
\biggl[\beta \Omega_2  \sin\theta 
- \gamma \Omega_3  \cos\theta + \Omega' \biggr] \biggl[\frac{1 }{y^2_\mathrm{max}} \biggr]
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ 2x' \biggl\{
\biggl[ \biggl(\frac{a}{b}\biggr)^2 \gamma \Omega_3 \tan\theta  
\biggr]\biggl[\frac{z_0}{x^2_\mathrm{max}} \biggr] 
-
\biggl[\beta \Omega_2  \sin\theta 
- \gamma \Omega_3  \cos\theta + \Omega' \biggr]\biggl[\frac{y_c }{y^2_\mathrm{max}} \biggr] 
\biggr\} \, .
</math>
  </td>
</tr>
</table>
Now, drawing from the various above "boxed" relations and examining the two terms separately, we first find,