Editing Apps/RiemannEllipsoidsCompressible (section)

==Example of Riemann S-Type Ellipsoids==
===Preamble===

First, set <math>{\vec{\omega}} = \hat{k}\omega</math> and <math>v_z = 0</math>, and write out the Cartesian components of the vector,
<div align="center">
<math>
\vec{A} \equiv ({\vec{v}}\cdot \nabla){\vec{v}} +2 {\vec{\omega}} \times {\vec{v}} .
</math>
</div>
The components are:
<div align="left">
<math>
~~~~~\hat{i}:~~~~~A_x = v_x \frac{\partial v_x}{\partial x} + v_y \frac{\partial v_x}{\partial y} -2\omega v_y ;
</math><br />
<math>
~~~~~\hat{j}:~~~~~A_y = v_x \frac{\partial v_y}{\partial x} + v_y \frac{\partial v_y}{\partial y} +2\omega v_x ;
</math><br />
<math>
~~~~~\hat{k}:~~~~~A_z = 0 .
</math>
</div>
The curl of <math>\vec{A}</math> (needed in Step #2, above) produces a vector with the following three Cartesian components:
<div align="left">
<math>
~~~~~\hat{i}:~~~~~[\nabla\times\vec{A}]_x = \frac{\partial}{\partial y} \biggl[0 \biggr] - \frac{\partial}{\partial z} \biggl[ v_x \frac{\partial v_y}{\partial x} + v_y \frac{\partial v_y}{\partial y} +2\omega v_x \biggr] ;
</math><br />
<math>
~~~~~\hat{j}:~~~~~[\nabla\times\vec{A}]_y = \frac{\partial}{\partial z} \biggl[ v_x \frac{\partial v_x}{\partial x} + v_y \frac{\partial v_x}{\partial y} -2\omega v_y \biggr] - \frac{\partial}{\partial x} \biggl[0 \biggr] ;
</math><br />
<math>
~~~~~\hat{k}:~~~~~[\nabla\times\vec{A}]_z = \frac{\partial}{\partial x} \biggl[ v_x \frac{\partial v_y}{\partial x} + v_y \frac{\partial v_y}{\partial y} +2\omega v_x \biggr] - \frac{\partial}{\partial y} \biggl[ v_x \frac{\partial v_x}{\partial x} + v_y \frac{\partial v_x}{\partial y} -2\omega v_y  \biggr] .
</math>
</div>
Hence, demanding (as in Step #2) that <math>\nabla\times\vec{A} = 0</math> means that each of these components independently must be zero. This, in turn, implies:
<div align="left">
<math>
~~~~~\hat{i}:~~~~~ v_x \frac{\partial v_y}{\partial x} + v_y \frac{\partial v_y}{\partial y} +2\omega v_x  = C_{z1}(x,y);
</math><br />
<math>
~~~~~\hat{j}:~~~~~ v_x \frac{\partial v_x}{\partial x} + v_y \frac{\partial v_x}{\partial y} -2\omega v_y = C_{z2}(x,y) ;
</math><br />
<math>
~~~~~\hat{k}:~~~~~ \frac{\partial}{\partial x} \biggl[ C_{z1}(x,y) \biggr] = \frac{\partial}{\partial y} \biggl[C_{z2}(x,y)  \biggr] ,
</math>
</div>
where the integration "constants" <math>C_{z1}</math> and <math>C_{z2}</math> may be functions of <math>x</math> and/or <math>y</math> but they must be independent of <math>z</math>. 


Generically, the continuity equation demands,
<div align="center">
<math>
0 = \frac{\partial}{\partial x}\biggl[ \rho v_x \biggr] + \frac{\partial}{\partial y}\biggl[ \rho v_y \biggr] + \frac{\partial}{\partial z}\biggl[ \rho v_z \biggr] . 
</math>
</div>


The divergence of <math>\vec{A}</math> (providing the right-hand-side of the Poisson-like equation in Step #3, above) generates:
<div align="center">
<math>
\nabla\cdot\vec{A} = \frac{\partial}{\partial x} \biggl[ v_x \frac{\partial v_x}{\partial x} + v_y \frac{\partial v_x}{\partial y} -2\omega v_y \biggr] + \frac{\partial}{\partial y} \biggl[ v_x \frac{\partial v_y}{\partial x} + v_y \frac{\partial v_y}{\partial y} +2\omega v_x \biggr] + \frac{\partial}{\partial z} \biggl[ 0 \biggr]
</math><br />
<math>
= \frac{\partial}{\partial x} \biggl[ C_{z2}(x,y) \biggr] + \frac{\partial}{\partial y} \biggl[C_{z1}(x,y)  \biggr]  . 
</math>
</div>

===Riemann Flow-Field===
In Riemann S-Type ellipsoids, the adopted planar flow-field as viewed from the rotating reference frame is,
<div align="center">
<math>
\vec{v} = \hat{i} \biggl( \frac{\lambda a y}{b} \biggr) + \hat{j} \biggl( - \frac{\lambda b x}{a} \biggr) .
</math>
</div>
Hence,
<div align="center">
<math>
C_{z1}(x,y) = \biggl( \frac{\lambda a y}{b} \biggr) \frac{\partial}{\partial x}\biggl( - \frac{\lambda b x}{a} \biggr) + \biggl( - \frac{\lambda b x}{a} \biggr) \frac{\partial}{\partial y}\biggl( - \frac{\lambda b x}{a} \biggr) +2\omega \biggl( \frac{\lambda a y}{b} \biggr) = \biggl[2\omega\biggl( \frac{\lambda a }{b} \biggr)- \lambda^2  \biggr]y,
</math><br />
<math>
C_{z2}(x,y) = \biggl( \frac{\lambda a y}{b} \biggr) \frac{\partial}{\partial x}\biggl( \frac{\lambda a y}{b} \biggr) + \biggl( - \frac{\lambda b x}{a} \biggr) \frac{\partial}{\partial y}\biggl( \frac{\lambda a y}{b} \biggr) -2\omega \biggl( - \frac{\lambda b x}{a} \biggr) = \biggl[2\omega\biggl( \frac{\lambda b }{a} \biggr)   - \lambda^2 \biggr]x .
</math>
</div>
Because <math>C_{z1}</math> is independent of <math>x</math> and <math>C_{z2}</math> is independent of <math>y</math>, we see that <math>[\nabla\times\vec{A}]_z = 0</math>, trivially.  With this specified velocity flow-field and the appreciation that Riemann S-type ellipsoids also have uniform density, the continuity equation is also trivially satisfied; specifically,
<div align="center">
<math>
\frac{\partial}{\partial x}\biggl[ \rho v_x \biggr] + \frac{\partial}{\partial y}\biggl[ \rho v_y \biggr] + \frac{\partial}{\partial z}\biggl[ \rho v_z \biggr] = \rho \biggl[ \frac{\partial v_x}{\partial x} + \frac{\partial v_y}{\partial y} \biggr] = 0 .
</math>
</div>
However, the right-hand-side of our Poisson-like equation is not zero; rather, it is,
<div align="center">
<math>
\nabla\cdot\vec{A} = \biggl[2\omega\biggl( \frac{\lambda b }{a} \biggr)   - \lambda^2 \biggr] + \biggl[2\omega\biggl( \frac{\lambda a }{b} \biggr)- \lambda^2  \biggr] = 2 \biggl[\omega\lambda \biggl( \frac{b}{a} + \frac{a}{b} \biggr)  - \lambda^2 \biggr].
</math>
</div>

===Summary===
What can we learn from the Riemann S-Type ellipsoids?  Well, let's assume that the <i>structure</i> of our equilibrium flow-field will be more complicated than simply "flow along elliptical paths", but let's continue to assume that a solution can be found in which the flow remains <i>planar</i>, that is, <math>v_z=0</math> everywhere.  Also, let's continue to align the rotation axis of the frame with the <math>z</math>-axis of the configuration.  The three steps in our proposed solution strategy become:


====<font color="darkblue">Step #1 (simplified):</font>====
Guess a 3D density distribution where the density goes to zero along the three principal axes at <math>x=a</math>, <math>y = b</math>, and <math>z = c</math>.  Solve the Poisson equation in order to derive values for <math>\Phi</math> everywhere inside and on the surface of the 3D configuration:
<div align="center">
<math>
\nabla^2 \Phi = 4\pi G \rho .
</math>
</div>

====<font color="darkblue">Step #2 (simplified):</font>====
Use the continuity equation and the curl of the Euler equation to numerically derive the <i>structure</i> but not the overall magnitude of the velocity flow-field throughout the 3D configuration, assuming <math>\vec{\omega}=\hat{k}\omega</math> and the 3D density distribution is known.  Here are the relevant equations:
<div align="center">
<math>
\frac{\partial}{\partial x}\biggl[ \rho v_x \biggr] + \frac{\partial}{\partial y}\biggl[ \rho v_y \biggr] = 0 .
</math>
</div>
<div align="left">
<math>
~~~~~\hat{i}:~~~~~ v_x \frac{\partial v_y}{\partial x} + v_y \frac{\partial v_y}{\partial y} +2\omega v_x  = C_{z1}(x,y);
</math><br />
<math>
~~~~~\hat{j}:~~~~~ v_x \frac{\partial v_x}{\partial x} + v_y \frac{\partial v_x}{\partial y} -2\omega v_y = C_{z2}(x,y) ;
</math><br />
<math>
~~~~~\hat{k}:~~~~~ \frac{\partial}{\partial x} \biggl[ C_{z1}(x,y) \biggr] = \frac{\partial}{\partial y} \biggl[C_{z2}(x,y)  \biggr] ,
</math>
</div>

Now, for the adopted flow-field of the Riemann ellipsoids, <math>C_{z1}(x,y)</math> is only a function of <math>y</math> and <math>C_{z2}(x,y)</math> is only a function of <math>x</math> &#8212; that is, <math>C_{z1}(x,y) \rightarrow C_{z1}(y)</math> and <math>C_{z2}(x,y) \rightarrow C_{z2}(x)</math>.  Hence, the third (<math>\hat{k}</math>) condition is automatically satisfied.  I don't know whether or not we will find that our more general velocity fields exhibit the same nice character.

====<font color="darkblue">Step #3 (simplified):</font>====
Take the divergence of the Euler equation and use it to solve for <math>H</math> throughout the configuration, given the structure of the flow-field obtained in Step #2.  The relevant "Poisson"-like equation is:

<div align="center">
<math>
\nabla^2 \biggl[H + \Phi -\frac{1}{2}\omega^2 (x^2 + y^2) \biggr] = - \frac{\partial}{\partial x} \biggl[ C_{z2}(x,y) \biggr] - \frac{\partial}{\partial y} \biggl[C_{z1}(x,y)  \biggr] .
</math>
</div>
In the Riemann ellipsoids, the RHS of this expression is a constant.  More importantly, in the Riemann case, it is possible to bring the constants from Step #2 inside the spatial operator on the LHS and establish the following <i>algebraic</i> condition:
<div align="center">
<math>
H + \Phi -\frac{1}{2}\omega^2 (x^2 + y^2) + f(x,y) = C_\mathrm{Bernoulli} ,
</math>
</div>
where the function <math>f(x,y)</math> contains only quadratic terms in <math>x</math> and <math>y</math>. Specifically [see Eq. (6) of Ou (2006)],
<div align="center">
<math>
f(x,y) = x^2 \biggl[\omega\lambda \frac{b}{a}-\frac{\lambda^2}{2}  \biggr] + y^2 \biggl[ \omega\lambda \frac{a}{b}-\frac{\lambda^2}{2} \biggr] .
</math>
</div>
This makes sense because in the Riemann case the equipotential surfaces were perfect ellipsoids, just like the shape of the centrifugal potential term.  So the remaining "source" terms could adopt the same "shape".  But in our more general case, surfaces of constant <math>\Phi</math> will have more complicated shapes, so it will be necessary for the function <math>f(x,y)</math> (or equivalent) to incorporate higher order terms in <math>x</math> and <math>y</math>.  Is it possible that equilibrium configurations can be found by including additional terms just in <math>x^4</math> and <math>y^4</math>?