Editing Apps/RiemannEllipsoidsCompressible (section)

===No Vertical Motions===
Now we restrict the flow by setting <math>v_z = 0</math>, that is, from here on we will assume that all the motion is planar. Also, following the lead of KP96, we define the vorticity of the fluid,
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<math>
\vec{\zeta} \equiv \nabla\times\vec{v} = \hat{i}\zeta_x + \hat{j}\zeta_y + \hat{k}\zeta_z .
</math>
</div> 
[Note that (unfortunately) KP96 use <math>\omega</math> instead of <math>\zeta</math> to represent the rotating-frame vorticity.]  In terms of the components of the vorticity vector, the steady-state Euler equation therefore becomes,
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<math>
(2\omega + \zeta_z)\hat{k}\times\vec{v} + (\hat{i}\zeta_x + \hat{j}\zeta_y)\times\vec{v} + \nabla \biggl[\frac{1}{2}v^2 + H + \Phi -\frac{1}{2}\omega^2 R^2  \biggr] = 0 .
</math>
</div>
Continuing to follow the lead KP96, we next <font color="red">take the curl of this Euler equation</font>.  Because the curl of a gradient is always zero, this leads us to the same condition discussed above &#8212; but this time written in terms of the components of the vorticity &#8212; namely,
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<math>
\nabla\times\vec{A} = 0 = \nabla\times [(2\omega + \zeta_z)\hat{k}\times\vec{v} + (\hat{i}\zeta_x + \hat{j}\zeta_y)\times\vec{v}]  .
</math>
</div>
Using another <font color="darkgreen">vector identity</font>, namely,
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<math>
\nabla\times(\vec{C} \times \vec{B}) =  (\vec{B}\cdot\nabla)\vec{C} - (\vec{C}\cdot\nabla)\vec{B} + \vec{C}(\nabla\cdot\vec{B}) - \vec{B}(\nabla\cdot\vec{C}),
</math>
</div>
and remembering that we are assuming <math>v_z = 0</math>, we see in this case that the vector condition <math>\nabla\times\vec{A}=0</math> leads to the following three independent scalar constraints:
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<math>
~~~~~\hat{i}:~~~~~ [\nabla\times\vec{A}]_x = - \frac{\partial }{\partial z}\biggl[ (2\omega + \zeta)v_x \biggr] + \frac{\partial}{\partial y} \biggl[ \zeta_x v_y - \zeta_y v_x \biggr] = 0  ;
</math><br />
<math>
~~~~~\hat{j}:~~~~~ [\nabla\times\vec{A}]_y = - \frac{\partial }{\partial z} \biggl[ (2\omega + \zeta)v_y  \biggr] - \frac{\partial}{\partial x} \biggl[ \zeta_x v_y - \zeta_y v_x \biggr] = 0 ;
</math><br />
<math>
~~~~~\hat{k}:~~~~~ [\nabla\times\vec{A}]_z = (2\omega + \zeta)\nabla\cdot\vec{v} + \biggl[ v_x \frac{\partial}{\partial x} + v_y \frac{\partial}{\partial y} \biggr](2\omega + \zeta) = 0 .
</math>
</div>
With the understanding that, by definition,
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<math>
\zeta_x \equiv - \frac{\partial v_y}{\partial z} , ~~~~~
\zeta_y \equiv + \frac{\partial v_x}{\partial z} ,  ~~~~~ \mathrm{and} ~~~~~
\zeta_z \equiv \frac{\partial v_y}{\partial x} - \frac{\partial v_x}{\partial y} ,
</math>
</div>
it can be shown that these three constraints are identical to the ones presented in the preamble, above.