Editing 3Dconfigurations/DescriptionOfRiemannTypeI (section)

===Vorticity===
Here we examine the expression for the vorticity from several different coordinate-system orientations.

====wrt Rotating Body Frame====

As [[#EFE_Rotating_Cartesian_Frame|provided above]], the steady-state velocity field as viewed from the rotating ellipsoid's body frame is,

<table border="0" align="center" cellpadding="5">
<tr>
  <td align="right"><math>\mathbf{u}_\mathrm{EFE}</math></td>
  <td align="center"><math>=</math></td>
  <td align="right"><math>
\boldsymbol{\hat\imath} \biggl\{ - \biggl[ \frac{a^2}{a^2 + b^2} \biggr] \zeta_3 y + \biggl[ \frac{a^2}{a^2 + c^2} \biggr] \zeta_2 z \biggr\}
+
\boldsymbol{\hat\jmath} \biggl\{ +\biggl[ \frac{b^2}{a^2 + b^2} \biggr] \zeta_3 x \biggr\}
+
\mathbf{\hat{k}} \biggl\{ - \biggl[ \frac{c^2}{a^2 + c^2} \biggr] \zeta_2 x  \biggr\}
\, .</math>
  </td>
</tr>
</table>

Hence, as viewed with respect to the body frame, we find,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\nabla\times \boldsymbol{u}_\mathrm{EFE}</math>
  </td>
  <td align="center">
<math>\equiv</math>
  </td>
  <td align="left">
<math>
\boldsymbol{\hat\imath} \biggl\{ \frac{\partial u_z}{\partial y} - \frac{\partial u_y}{\partial z}  \biggr\}
+
\boldsymbol{\hat\jmath} \biggl\{ \frac{\partial u_x}{\partial z} - \frac{\partial u_z}{\partial x} \biggr\}
+
\boldsymbol{\hat{k}}  \biggl\{ \frac{\partial u_y}{\partial x} - \frac{\partial u_x}{\partial y}  \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\boldsymbol{\hat\jmath} \biggl\{ \biggl[ \frac{a^2}{a^2 + c^2} \biggr] \zeta_2 
+ 
\biggl[ \frac{c^2}{a^2 + c^2} \biggr] \zeta_2 \biggr\}
+
\boldsymbol{\hat{k}}  \biggl\{ \biggl[ \frac{b^2}{a^2 + b^2} \biggr] \zeta_3 
+ 
\biggl[ \frac{a^2}{a^2 + b^2} \biggr] \zeta_3 \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\boldsymbol{\hat\jmath} \zeta_2
+
\boldsymbol{\hat{k}}  \zeta_3 \, .
</math>
  </td>
</tr>
</table>

====Another Choice====

Now let's view the flow from a "tilted plane" in which the vorticity vector aligns with the <math>\boldsymbol{\hat{k}''}</math> axis.

<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^2}{a^2+b^2}\biggr] \zeta_3 (y''\cos\chi - z''\sin\chi) 
+ \biggl[ \frac{a^2}{a^2 + c^2}\biggr] \zeta_2 (z_0 + y''\sin\chi + z''\cos\chi)
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+
\boldsymbol{\hat\jmath''}  \biggl\{ 
\cos\chi \biggl[ \frac{b^2}{a^2 + b^2}\biggr] \zeta_3 
-
\sin\chi\biggl[ \frac{c^2}{a^2 + c^2} \biggr] \zeta_2 
\biggr\}x''
-
\mathbf{\hat{k}''} \biggl\{ 
\sin\chi \biggl[ \frac{b^2}{a^2 + b^2}\biggr] \zeta_3 
+
\cos\chi \biggl[ \frac{c^2}{a^2 + c^2} \biggr] \zeta_2 
\biggr\}x'' \, .
</math>
  </td>
</tr>
</table>

The vorticity is,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\nabla\times \boldsymbol{u''}_\mathrm{EFE}</math>
  </td>
  <td align="center">
<math>\equiv</math>
  </td>
  <td align="left">
<math>
\boldsymbol{\hat\imath''} \biggl\{ \frac{\partial u_z''}{\partial y''} - \frac{\partial u_y''}{\partial z''}  \biggr\}
+
\boldsymbol{\hat\jmath''} \biggl\{ \frac{\partial u_x''}{\partial z''} - \frac{\partial u_z''}{\partial x''} \biggr\}
+
\boldsymbol{\hat{k}''}  \biggl\{ \frac{\partial u_y''}{\partial x''} - \frac{\partial u_x''}{\partial y''}  \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\boldsymbol{\hat\jmath''} \biggl\{ \biggl[\frac{a^2}{a^2+b^2}\biggr] \zeta_3 (\sin\chi) 
+ \biggl[ \frac{a^2}{a^2 + c^2}\biggr] \zeta_2 (\cos\chi) \biggr\}
+
\boldsymbol{\hat\jmath''} \biggl\{ \sin\chi \biggl[ \frac{b^2}{a^2 + b^2}\biggr] \zeta_3 
+
\cos\chi \biggl[ \frac{c^2}{a^2 + c^2} \biggr] \zeta_2  \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
+
\boldsymbol{\hat{k}''}  \biggl\{ \cos\chi \biggl[ \frac{b^2}{a^2 + b^2}\biggr] \zeta_3 
-
\sin\chi\biggl[ \frac{c^2}{a^2 + c^2} \biggr] \zeta_2  \biggr\}
+
\boldsymbol{\hat{k}''}  \biggl\{ \biggl[\frac{a^2}{a^2+b^2}\biggr] \zeta_3 (\cos\chi ) 
- \biggl[ \frac{a^2}{a^2 + c^2}\biggr] \zeta_2 (\sin\chi )
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\boldsymbol{\hat\jmath''} \biggl\{ \biggl[\frac{a^2}{a^2+b^2} + \frac{b^2}{a^2 + b^2}\biggr] \zeta_3 (\sin\chi) 
+ \biggl[ \frac{a^2}{a^2 + c^2} + \frac{c^2}{a^2 + c^2}\biggr] \zeta_2 (\cos\chi) 
\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{a^2}{a^2+b^2} + \frac{b^2}{a^2 + b^2}\biggr] \zeta_3 (\cos\chi ) 
- \biggl[ \frac{a^2}{a^2 + c^2} + \frac{c^2}{a^2 + c^2}\biggr] \zeta_2 (\sin\chi )
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\boldsymbol{\hat\jmath''} \biggl\{ \zeta_3 (\sin\chi) 
+ \zeta_2 (\cos\chi) 
\biggr\}
+
\boldsymbol{\hat{k}''}  \biggl\{ 
\zeta_3 (\cos\chi ) 
- \zeta_2 (\sin\chi )
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\boldsymbol{\hat\jmath''} \biggl\{ \tan\chi 
+ \frac{\zeta_2}{\zeta_3}  
\biggr\}\zeta_3\cos\chi
+
\boldsymbol{\hat{k}''}  \biggl\{ 
1 
- \frac{\zeta_2}{\zeta_3} \tan\chi 
\biggr\}\zeta_3 \cos\chi \, .
</math>
  </td>
</tr>
</table>
So, in order for the <math>\boldsymbol{\hat{\jmath}''}</math> component to be zero, we choose,

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

<tr>
  <td align="right">
<math>\tan\chi</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
-\frac{\zeta_2}{\zeta_3} \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ \cos\chi</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[1 + \frac{\zeta_2^2}{\zeta_3^2}\biggr]^{-1 / 2} \, ,
</math>
  </td>
</tr>
</table>
in which case we have,

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

<tr>
  <td align="right">
<math>\nabla\times \boldsymbol{u''}_\mathrm{EFE}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\boldsymbol{\hat{k}''}  \biggl\{ 
1 + \frac{\zeta_2^2}{\zeta_3^2}  
\biggr\}\zeta_3 \biggl[1 + \frac{\zeta_2^2}{\zeta_3^2}\biggr]^{-1 / 2}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\boldsymbol{\hat{k}''}  \biggl[ 
1 + \frac{\zeta_2^2}{\zeta_3^2}  
\biggr]^{1 / 2}\zeta_3 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\boldsymbol{\hat{k}''}  (\zeta_2^2 + \zeta_3^2)^{1 / 2} \, .
</math>
  </td>
</tr>
</table>
<font color="red"><b>This makes sense!</b></font>  

Now, can we retrieve the "rotating body frame" expression simply by transforming the coordinates?  Well &hellip;

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

<tr>
  <td align="right">
<math>\boldsymbol{\hat{k}''}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
- \boldsymbol{\hat{\jmath}} \sin\chi + \boldsymbol{\hat{k}} \cos\chi
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ \nabla\times \boldsymbol{u''}_\mathrm{EFE} = \boldsymbol{\hat{k}''}  \zeta_3</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
(\zeta_2^2 + \zeta_3^2)^{1 / 2} \biggl[- \boldsymbol{\hat{\jmath}} \sin\chi + \boldsymbol{\hat{k}} \cos\chi \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
(\zeta_2^2 + \zeta_3^2)^{1 / 2} \biggl[- \boldsymbol{\hat{\jmath}} \tan\chi + \boldsymbol{\hat{k}} \biggr]\cos\chi
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
(\zeta_2^2 + \zeta_3^2)^{1 / 2} \biggl[\boldsymbol{\hat{\jmath}} \biggl( \frac{\zeta_2}{\zeta_3}\biggr) + \boldsymbol{\hat{k}} \biggr]
\biggl[1 + \frac{\zeta_2^2}{\zeta_3^2}\biggr]^{-1 / 2}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\boldsymbol{\hat{\jmath}} \zeta_2 + \boldsymbol{\hat{k}} \zeta_3 \, .
</math>
  </td>
</tr>

</table>
<font color="red"><b>Hooray!</b></font>

====wrt Lagrangian Orbital Planes====

As viewed from the "preferred tilted plane," the steady-state velocity field is,

<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^2}{a^2+b^2}\biggr] \zeta_3 (y'\cos\theta - z'\sin\theta) 
+ \biggl[ \frac{a^2}{a^2 + c^2}\biggr] \zeta_2 (z_0 + y'\sin\theta + z'\cos\theta)
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+
\boldsymbol{\hat\jmath'}  \biggl\{ 
\cos\theta \biggl[ \frac{b^2}{a^2 + b^2}\biggr] \zeta_3 
-
\sin\theta\biggl[ \frac{c^2}{a^2 + c^2} \biggr] \zeta_2 
\biggr\}x'
-
\boldsymbol{\hat{k}'} \cancelto{0}{\biggl\{ 
\tan\theta \biggl[ \frac{b^2}{a^2 + b^2}\biggr] \zeta_3 
+
\biggl[ \frac{c^2}{a^2 + c^2} \biggr] \zeta_2 
\biggr\}} \cos\theta x'
</math>
  </td>
</tr>
</table>
where,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\tan\theta</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
- \frac{\beta \Omega_2}{\gamma \Omega_3} = - \biggl[ \frac{c^2 (a^2 + b^2)}{b^2(a^2 + c^2)}\biggr]\frac{\zeta_2}{\zeta_3} \, .
</math>
  </td>
</tr>
</table>

Hence, the vorticity is,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\nabla\times \boldsymbol{u'}_\mathrm{EFE}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
- \boldsymbol{\hat\imath'} \biggl\{ \frac{\partial u'_y}{\partial z'}  \biggr\}
+
\boldsymbol{\hat\jmath'} \biggl\{ \frac{\partial u'_x}{\partial z'}  \biggr\}
+
\boldsymbol{\hat{k}'}  \biggl\{ \frac{\partial u'_y}{\partial x'} - \frac{\partial u'_x}{\partial y'}  \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\boldsymbol{\hat\jmath'} \frac{\partial }{\partial z'}\biggl\{ 
\biggl[\frac{a^2}{a^2+b^2}\biggr] \zeta_3 (z'\sin\theta) 
+ \biggl[ \frac{a^2}{a^2 + c^2}\biggr] \zeta_2 (z'\cos\theta)
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
+~
\boldsymbol{\hat{k}'} \frac{\partial }{\partial x'}  \biggl\{ 
x'\cos\theta \biggl[ \frac{b^2}{a^2 + b^2}\biggr] \zeta_3 
-
x'\sin\theta\biggl[ \frac{c^2}{a^2 + c^2} \biggr] \zeta_2 
\biggr\}
+ ~
\boldsymbol{\hat{k}'} \frac{\partial }{\partial y'}  \biggl\{ 
\biggl[\frac{a^2}{a^2+b^2}\biggr] \zeta_3 (y'\cos\theta ) 
- \biggl[ \frac{a^2}{a^2 + c^2}\biggr] \zeta_2 (y'\sin\theta )
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\boldsymbol{\hat\jmath'} \biggl\{ 
\biggl[\frac{a^2}{a^2+b^2}\biggr] \zeta_3 \tan\theta 
+ \biggl[ \frac{a^2}{a^2 + c^2}\biggr] \zeta_2 
\biggr\}\cos\theta
+~
\boldsymbol{\hat{k}'} \biggl\{ 
\biggl[ \frac{b^2}{a^2 + b^2}\biggr] \zeta_3 
-
\tan\theta\biggl[ \frac{c^2}{a^2 + c^2} \biggr] \zeta_2 
\biggr\}\cos\theta
+ ~
\boldsymbol{\hat{k}'} \biggl\{ 
\biggl[\frac{a^2}{a^2+b^2}\biggr] \zeta_3  
- \biggl[ \frac{a^2}{a^2 + c^2}\biggr] \zeta_2 \tan\theta 
\biggr\}\cos\theta
</math>
  </td>
</tr>
</table>


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

<tr>
  <td align="right">
<math>\Rightarrow ~~~ \biggl(\frac{1}{\cos\theta}\biggr)\nabla\times \boldsymbol{u'}_\mathrm{EFE}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\boldsymbol{\hat\jmath'} \biggl\{ 
\biggl[\frac{a^2}{a^2+b^2}\biggr] \zeta_3 \tan\theta 
+ \biggl[ \frac{a^2}{a^2 + c^2}\biggr] \zeta_2 
\biggr\}
+~
\boldsymbol{\hat{k}'} \biggl\{ 
\biggl[ \frac{b^2}{a^2 + b^2}\biggr] \zeta_3 
+ ~
\biggl[\frac{a^2}{a^2+b^2}\biggr] \zeta_3  
-
\biggl[ \frac{c^2}{a^2 + c^2} \biggr] \zeta_2\tan\theta 
- \biggl[ \frac{a^2}{a^2 + c^2}\biggr] \zeta_2 \tan\theta 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\boldsymbol{\hat\jmath'} \biggl[\frac{a^2}{a^2+b^2}\biggr] \biggl\{ 
\zeta_3 \tan\theta 
+ \frac{b^2}{c^2}\biggl[ \frac{c^2(a^2+b^2)}{b^2(a^2 + c^2)}\biggr] \zeta_2 
\biggr\}
+~
\boldsymbol{\hat{k}'} \biggl\{ 
\zeta_3  
-
\zeta_2\tan\theta \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\boldsymbol{\hat\jmath'} \biggl[\frac{a^2}{a^2+b^2}\biggr] \biggl[ 
1 
- \frac{b^2}{c^2} 
\biggr]\zeta_3 \tan\theta
+~
\boldsymbol{\hat{k}'} \biggl\{ 
\zeta_3  
-
\zeta_2\tan\theta \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\boldsymbol{\hat\jmath'} \biggl[ 
\frac{a^2(c^2 - b^2)}{c^2(a^2+b^2)}  
\biggr]\zeta_3\tan\theta
+~
\boldsymbol{\hat{k}'} \biggl\{ 
\zeta_3  
-
\zeta_2\tan\theta \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
-~\boldsymbol{\hat\jmath'} \biggl[ 
\frac{a^2(c^2 - b^2)}{b^2(a^2 + c^2)}  
\biggr]\zeta_2
+~
\boldsymbol{\hat{k}'} \biggl\{ 
\zeta_3  
-
\zeta_2\tan\theta \biggr\} \, .
</math>
  </td>
</tr>
</table>

----

Can we obtain this result by starting from the original, rotating body frame coordinate expression, then transforming the coordinates ?

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

<tr>
  <td align="right">
<math>\nabla\times \boldsymbol{u}_\mathrm{EFE}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\boldsymbol{\hat\jmath} \zeta_2
+
\boldsymbol{\hat{k}}  \zeta_3 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[ \boldsymbol{\hat\jmath'}\cos\theta - \boldsymbol{\hat{k}'}\sin\theta \biggr] \zeta_2
+
\biggl[ \boldsymbol{\hat\jmath'}\sin\theta + \boldsymbol{\hat{k}'}\cos\theta \biggr] \zeta_3 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\boldsymbol{\hat\jmath'}\biggl[ \zeta_2 \cos\theta 
+
\zeta_3 \sin\theta \biggr] 
+
\boldsymbol{\hat{k}'} \biggl[ \zeta_3\cos\theta  
-
\zeta_2 \sin\theta \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ \biggl( \frac{1}{\cos\theta} \biggr) \nabla\times \boldsymbol{u}_\mathrm{EFE}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\boldsymbol{\hat\jmath'}\biggl[ \zeta_2 
+
\zeta_3 \tan\theta \biggr] 
+
\boldsymbol{\hat{k}'} \biggl[ \zeta_3  
-
\zeta_2 \tan\theta \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\boldsymbol{\hat\jmath'}\biggl[ 1 
+
\biggl(\frac{\zeta_3}{\zeta_2}\biggr) \tan\theta \biggr] \zeta_2
+
\boldsymbol{\hat{k}'} \biggl[ \zeta_3  
-
\zeta_2 \tan\theta \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\boldsymbol{\hat\jmath'}\biggl\{ 1 
-
\biggl[ \frac{c^2 (a^2 + b^2)}{b^2(a^2 + c^2)}\biggr] 
\biggr\} \zeta_2
+
\boldsymbol{\hat{k}'} \biggl[ \zeta_3  
-
\zeta_2 \tan\theta \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\boldsymbol{\hat\jmath'}\biggl\{ 
\biggl[ \frac{b^2(a^2 + c^2) - c^2 (a^2 + b^2)}{b^2(a^2 + c^2)}\biggr] 
\biggr\} \zeta_2
+
\boldsymbol{\hat{k}'} \biggl[ \zeta_3  
-
\zeta_2 \tan\theta \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
-~\boldsymbol{\hat\jmath'} 
\biggl[ \frac{a^2 (c^2 - b^2 )}{b^2(a^2 + c^2)}\biggr] 
\zeta_2
+
\boldsymbol{\hat{k}'} \biggl[ \zeta_3  
-
\zeta_2 \tan\theta \biggr]\, .
</math>
  </td>
</tr>
</table>
<font color="red"><b>Q.E.D.</b></font>

----

Let's see if we can rewrite this expression in a more physically insightful way.

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

<tr>
  <td align="right">
<math>\biggl[ 1 + \tan^2\theta \biggr]^{1 / 2} \nabla\times \boldsymbol{u}_\mathrm{EFE}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\boldsymbol{\hat\jmath'}\biggl[ \zeta_2  
+
\zeta_3 \tan\theta \biggr] 
+
\boldsymbol{\hat{k}'} \biggl[ \zeta_3  
-
\zeta_2 \tan\theta \biggr] \, ,
</math>
  </td>
</tr>
</table>
where,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\tan\theta</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
- \biggl[ \frac{c^2 (a^2 + b^2)}{b^2(a^2 + c^2)}\biggr]\frac{\zeta_2}{\zeta_3} \, .
</math>
  </td>
</tr>
</table>

====Summary Vorticity Expressions====

<ol><li>
Written in terms of the (unprimed) body-frame coordinates,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\nabla\times \boldsymbol{u}_\mathrm{EFE}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\boldsymbol{\hat\jmath} \zeta_2
+
\boldsymbol{\hat{k}}  \zeta_3 \, .
</math>
  </td>
</tr>
</table>
</li>
<li>
If we view the fluid motion from a (double-primed) frame that is tilted with respect to the (unprimed) body frame by the angle, <math>\chi</math>, such that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\tan\chi</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
-\frac{\zeta_2}{\zeta_3} \, ,
</math>
  </td>
</tr>
</table>
then <math>\boldsymbol{\hat{k}''}</math> will align with the vorticity vector and the vorticity vector will have only one component, namely,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\nabla\times \boldsymbol{u''}_\mathrm{EFE}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\boldsymbol{\hat{k}''}  (\zeta_2^2 + \zeta_3^2)^{1 / 2} \, .
</math>
  </td>
</tr>
</table>

</li>
<li>
If we view the fluid motion from a (single-primed) frame that is tilted with respect to the (unprimed) body frame by an angle, <math>\theta</math>, such that the motion of Lagrangian fluid elements is everywhere parallel to the x'-y' plane &#8212; that is, such that there is no Lagrangian fluid motion in the <math>\boldsymbol{\hat{k}'}</math> direction &#8212; we find,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\tan\theta</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
- \frac{\beta \Omega_2}{\gamma \Omega_3} = - \biggl[ \frac{c^2 (a^2 + b^2)}{b^2(a^2 + c^2)}\biggr]\frac{\zeta_2}{\zeta_3} \, ,
</math>
  </td>
</tr>
</table>
and,
<table border="0" cellpadding="5" align="center">

  <td align="right">
<math>\nabla\times \boldsymbol{u'}_\mathrm{EFE}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\boldsymbol{\hat\jmath'}\overbrace{\biggl[ \zeta_2\cos\theta 
+
\zeta_3 \sin\theta \biggr]}^{\mathrm{due~to~vertical~shear}} 
+
\boldsymbol{\hat{k}'} \underbrace{\biggl[ \zeta_3  \cos\theta
-
\zeta_2 \sin\theta \biggr]}_{\zeta_L} \, .
</math>
  </td>
</table>
</li>

<li>
[[#Vorticity_Implied_by_Lagrangian_Fluid_Motions|From below]], the contribution to the vorticity that is provided by the Lagrangian orbital-element-based description of the motion of the fluid is,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\zeta_\mathrm{L}</math>
  </td>
  <td align="center">
<math>\equiv</math>
  </td>
  <td align="left">
<math>
\biggl\{ \frac{\partial \dot{y}'}{\partial x'} - \frac{\partial \dot{x}'}{\partial y'}  \biggr\}
=
\biggl[ 
\biggl(\frac{x_\mathrm{max}}{y_\mathrm{max}}\biggr)
+
\biggl(\frac{y_\mathrm{max}}{x_\mathrm{max}}\biggr)\biggr] \dot\varphi \, .
</math>
  </td>
</tr>
</table>
<ol type="A">
  <li>
Adopting the parameter (Model001 evaluation in parentheses),
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\Lambda</math>
  </td>
  <td align="center">
<math>\equiv</math>
  </td>
  <td align="left">
<math>
\biggl[ \frac{a^2}{a^2 + b^2} \biggr] \zeta_3  \cos\theta    
-
\biggl[ \frac{a^2}{a^2 + c^2} \biggr] \zeta_2 \sin\theta 
= - 1.332892 \, ,
</math>
  </td>
</tr>
</table>
we have found that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\frac{x_\mathrm{max}}{y_\mathrm{max}}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl\{ \Lambda \biggl[ \frac{a^2 + b^2}{b^2} \biggr] \frac{\cos\theta}{\zeta_3} \biggr\}^{1 / 2} 
= 1.025854
\, ,
</math>
  </td>
  <td align="center">&nbsp; &nbsp; &nbsp; and, &nbsp; &nbsp; &nbsp; </td>
  <td align="right">
<math>\dot\varphi</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl\{ \Lambda \biggl[ \frac{b^2}{a^2 + b^2} \biggr] \frac{\zeta_3 }{\cos\theta} \biggr\}^{1 / 2}
= 1.299300
\, ,
</math>
  </td>
</tr>
</table>
which implies,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\zeta_L\biggr|_\mathrm{Model001}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
2.599446 \, .
</math>
  </td>
</tr>
</table>

  </li>
  <li>
Alternatively, we have found that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\frac{x_\mathrm{max}}{y_\mathrm{max}}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{a(c^2\cos^2\theta + b^2\sin^2\theta)^{1 / 2}}{bc}
= 1.025854
\, ,
</math>
  </td>
  <td align="center">&nbsp; &nbsp; &nbsp; and, &nbsp; &nbsp; &nbsp; </td>
  <td align="right">
<math>\dot\varphi</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{\zeta_3}{\cos\theta}
\biggl[\frac{ abc ( c^2 \cos^2\theta + b^2 \sin^2\theta )^{1 / 2}}{c^2(a^2 + b^2)}  \biggr] 
= -1.299300
\, ,
</math>
  </td>
</tr>
</table>
which implies,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\zeta_L\biggr|_\mathrm{Model001}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
-2.599446 \, .
</math>
  </td>
</tr>
</table>
  </li>
  <li>
In step #3, immediately above, we have determined that the <math>\boldsymbol{\hat{k}'}</math> component of the fluid vorticity is given by the expression,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\zeta_L</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
(\zeta_3\cos\theta - \zeta_2\sin\theta) 
= -2.599446 \, .
</math>
  </td>
</tr>
</table>

  </li>
</ol>
</li>
</ol>

<table border="1" width="80%" align="center" cellpadding="5"><tr><td align="left">
<font color="red">It appears as though we have separately derived three expressions for the quantity, <math>\zeta_L</math>.  It would be great if we could demonstrate analytically that the three expressions are, indeed, identical.</font>

Keep in mind that the definition of <math>\tan\theta</math> establishes the relation,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>b^2(a^2 + c^2)\zeta_3\sin\theta</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
- c^2 (a^2 + b^2) \zeta_2 \cos\theta
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ \frac{\zeta_3}{c^2(a^2 + b^2) \cos\theta}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
- \frac{\zeta_2}{b^2(a^2 + c^2)  \sin\theta}
</math>
  </td>
</tr>
</table>

----

Let,
<table border="0" cellpadding="8" align="center">

<tr>
  <td align="right">
<math>\Upsilon</math>
  </td>
  <td align="center">
<math>\equiv</math>
  </td>
  <td align="left">
<math>
(c^2\cos^2\theta + b^2\sin^2\theta)^{1 / 2} \, .
</math>
  </td>
</tr>
</table>

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

<tr>
  <td align="right">
<math>\zeta_\mathrm{L}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[ 
\biggl(\frac{x_\mathrm{max}}{y_\mathrm{max}}\biggr)
+
\biggl(\frac{y_\mathrm{max}}{x_\mathrm{max}}\biggr)\biggr] \dot\varphi 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[ 
\frac{a\Upsilon}{bc} + \frac{bc}{a\Upsilon}
\biggr] 
\frac{\zeta_3}{\cos\theta}
\biggl[\frac{ abc \Upsilon}{c^2(a^2 + b^2)}  \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{\zeta_3}{\cos\theta}\biggl\{
\biggl[ \frac{a\Upsilon}{bc} \biggr] 
\biggl[\frac{ abc \Upsilon}{c^2(a^2 + b^2)}  \biggr] 
+
\biggl[ \frac{bc}{a\Upsilon} \biggr] 
\biggl[\frac{ abc \Upsilon}{c^2(a^2 + b^2)}  \biggr] 

\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{b^2\zeta_3}{(a^2 + b^2)\cos\theta}\biggl\{
\biggl[\frac{ a^2 \Upsilon^2}{b^2c^2}  \biggr] 
+
1 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{\zeta_3}{c^2(a^2 + b^2)\cos\theta}\biggl\{
\biggl[a^2 (c^2\cos^2\theta + b^2\sin^2\theta) \biggr] 
+
b^2c^2 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{\zeta_3}{c^2(a^2 + b^2)\cos\theta}\biggl\{
a^2 c^2\cos^2\theta + a^2b^2\sin^2\theta 
+
b^2c^2(\sin^2\theta + \cos^2\theta) 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{\zeta_3}{c^2(a^2 + b^2)\cos\theta}\biggl\{
c^2(a^2 + b^2)\cos^2\theta  
\biggr\}
+
\frac{\zeta_3}{c^2(a^2 + b^2)\cos\theta}\biggl\{
b^2(a^2 + c^2)\sin^2\theta 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\zeta_3\cos\theta - \zeta_2\sin\theta \, .
</math>
  </td>
</tr>
</table>
<font color="red"><b>Q.E.D.</b></font>

----

Alternatively, pulling from the expressions that have been derived in terms of the parameter, <math>\Lambda</math>, we find,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\zeta_\mathrm{L}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[ 
\biggl(\frac{x_\mathrm{max}}{y_\mathrm{max}}\biggr)
+
\biggl(\frac{y_\mathrm{max}}{x_\mathrm{max}}\biggr)\biggr] \dot\varphi 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl\{ \Lambda \biggl[ \frac{a^2 + b^2}{b^2} \biggr] \frac{\cos\theta}{\zeta_3} \biggr\}^{1 / 2}
\biggl\{ \Lambda \biggl[ \frac{b^2}{a^2 + b^2} \biggr] \frac{\zeta_3 }{\cos\theta} \biggr\}^{1 / 2}
+
\biggl\{ \Lambda^{-1} \biggl[ \frac{b^2}{a^2 + b^2} \biggr] \frac{\zeta_3}{\cos\theta} \biggr\}^{1 / 2}
\biggl\{ \Lambda \biggl[ \frac{b^2}{a^2 + b^2} \biggr] \frac{\zeta_3 }{\cos\theta} \biggr\}^{1 / 2}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\Lambda
+
\biggl[ \frac{b^2}{a^2 + b^2} \biggr] \frac{\zeta_3}{\cos\theta} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[ \frac{a^2}{a^2 + b^2} \biggr] \zeta_3  \cos\theta    
-
\biggl[ \frac{a^2}{a^2 + c^2} \biggr] \zeta_2 \sin\theta 
+
\biggl[ \frac{b^2}{a^2 + b^2} \biggr] \frac{\zeta_3}{\cos\theta}\biggl\{\sin^2\theta + \cos^2\theta\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\zeta_3  \cos\theta    
-
\biggl[ \frac{a^2}{a^2 + c^2} \biggr] \zeta_2 \sin\theta 
+
\biggl[ b^2c^2 \biggr] \frac{\zeta_3}{c^2(a^2 + b^2)\cos\theta}\biggl\{\sin^2\theta \biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\zeta_3  \cos\theta    
-
\biggl[ \frac{a^2}{a^2 + c^2} \biggr] \zeta_2 \sin\theta 
-
\biggl[ b^2c^2 \biggr] \frac{\zeta_2}{b^2(a^2 + c^2)  \sin\theta}\biggl\{\sin^2\theta \biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\zeta_3  \cos\theta    
-
\zeta_2\sin\theta \, .
</math>
  </td>
</tr>
</table>
<font color="red"><b>Q.E.D.</b></font>
</td></tr></table>