Editing ThreeDimensionalConfigurations/RiemannStype (section)

=Feeding a 3D Animation=

==Initial Thoughts==

Let's examine the elliptical trajectory of a Lagrangian particle that is moving in the equatorial plane of a Riemann S-Type ellipsoid. As viewed in a frame that is spinning about the Z-axis at angular frequency, <math>~\Omega</math>, the trajectory is defined by,

<table align="center" border=0 cellpadding="3">
<tr>
  <td align="right">
<math>
r^2
</math>
  </td>
  <td align="center">
<math>
~=
</math>
  </td>
  <td align="left">
<math>~
\biggl(\frac{x}{a} \biggr)^2 + \biggl(\frac{y}{b}\biggr)^2 \, ,
</math>
  </td>
</tr>

</table>
where <math>~0 < r \le 1</math>.  (The surface of the relevant ellipsoid is associated with the value, <math>~r=1</math>.)

Let's choose a pair of axis ratios &#8212; for example, <math>~b/a = 0.28</math> and <math>~c/a = 0.231</math> &#8212; then, from Table 1 of our [[#Models_Examined_by_Ou_.282006.29|above discussion]], draw the associated value of either <math>~\lambda</math> or <math>~\zeta</math> that corresponds to the Jacobi-like equilibrium configuration &#8212; in this example, <math>~\lambda = -0.04714</math> and <math>~\zeta = +0.18156</math>.  Then, for any point <math>~(x,y)</math> inside of the ellipsoid, the fluid's velocity components (as viewed from the rotating frame of reference) are,

<table align="center" border=0 cellpadding="3">
<tr>
  <td align="right">
<math>
v_x  = \frac{dx}{dt}  = \lambda \biggl( \frac{ay}{b} \biggr) = -0.16836 ~y
</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; and, &nbsp; &nbsp; &nbsp;
  </td>
  <td align="left">
<math>~
v_y  = \frac{dy}{dt}  = - \lambda \biggl( \frac{bx}{a} \biggr) = + 0.01320~x \, .
</math>
  </td>
</tr>
</table>

Alternatively, we have,

<table align="center" border=0 cellpadding="3">
<tr>
  <td align="right">
<math>
u_x = \frac{dx}{dt} = Q_1  y = - \biggl[ 1 + \frac{b^2}{a^2} \biggr]^{-1}\zeta ~y = -0.16836 ~y
</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; and, &nbsp; &nbsp; &nbsp;
  </td>
  <td align="left">
<math>~
u_y  = \frac{dy}{dt}= Q_2 x  = + \biggl[ 1 + \frac{a^2}{b^2} \biggr]^{-1}\zeta ~x = + 0.01320~x \, .
</math>
  </td>
</tr>
</table>

Now, each Lagrangian fluid element's motion is oscillatory in both the <math>~x</math> and <math>~y</math> coordinate directions.  So let's see how this plays out.  Suppose,

<table align="center" border=0 cellpadding="3">
<tr>
  <td align="right">
<math>
x = x_\mathrm{max} \cos(\varphi t)
</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; and, &nbsp; &nbsp; &nbsp;
  </td>
  <td align="left">
<math>~
y = y_\mathrm{max} \sin(\varphi t) \, .
</math>
  </td>
</tr>
</table>

Then,

<table align="center" border=0 cellpadding="3">
<tr>
  <td align="right">
<math>
\frac{dx}{dt} = -  x_\mathrm{max}\varphi \sin(\varphi t) = -  \biggl( \frac{x_\mathrm{max}}{y_\mathrm{max}}\biggr) \varphi y = - \varphi \biggl(\frac{ay}{b}\biggr)
</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; and, &nbsp; &nbsp; &nbsp;
  </td>
  <td align="left">
<math>~
\frac{dy}{dt} = y_\mathrm{max} \varphi \cos(\varphi t) = + \biggl( \frac{y_\mathrm{max}}{x_\mathrm{max}}\biggr) \varphi x = + \varphi \biggl(\frac{bx}{a}\biggr) \, .
</math>
  </td>
</tr>
</table>

Hence our functional representation of the time-dependent behavior of both <math>~x</math> and <math>~y</math> works perfectly if, for each orbit inside of or on the surface of the configuration, we set <math>~\varphi = - \lambda</math> and if the ratio <math>~y_\mathrm{max}/x_\mathrm{max} = (b/a)</math>.  Hooray!

==Preferred Normalizations==

Let's do this again, assuming that <math>~x</math> and <math>~y</math> both have units of length and that <math>~t</math> has the unit of time.  Then, let's use <math>~a</math> to normalize lengths and use <math>~(\pi G \rho)^{-1 / 2}</math> to normalize time. We therefore have,

<table align="center" border=0 cellpadding="3">
<tr>
  <td align="right">
<math>
\frac{x}{a} = \biggl(\frac{ x_\mathrm{max} }{a}\biggr) \cos\biggl[ \frac{\varphi}{(\pi G \rho)^{1 / 2}}  \cdot \frac{t}{(\pi G \rho)^{-1 / 2}} \biggr]
</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; and, &nbsp; &nbsp; &nbsp;
  </td>
  <td align="left">
<math>~
\frac{y}{a} = \biggl(\frac{ y_\mathrm{max} }{a}\biggr) \sin\biggl[ \frac{\varphi}{(\pi G \rho)^{1 / 2}}  \cdot \frac{t}{(\pi G \rho)^{-1 / 2}} \biggr] \, .
</math>
  </td>
</tr>
</table>


<table border="1" cellpadding="10" width="80%" align="center"><tr><td align="left">
<font color="red">'''NOTE:'''</font> &nbsp; When implementing in an xml-based COLLADA (3D animation) file, we associate <math>~\mathrm{TIME} = 4</math> with <math>~t \cdot (\pi G \rho)^{1 / 2} = 2\pi</math>.  Hence we can everywhere replace <math>~t \cdot (\pi G \rho)^{1 / 2}</math> with (in ''radians'') <math>~(\pi/2)\cdot \mathrm{TIME}</math> or (in ''degrees'') <math>~90 \cdot \mathrm{TIME}</math>.

<br />
This also means that, if <math>~\varphi/(\pi G \rho)^{1 / 2} = 1</math>,  each Lagrangian fluid element will move through one complete orbit (as viewed from a frame that is rotating with the ellipsoidal figure) in the time it takes the hand of the wall-mounted clock to complete one cycle.
</td></tr></table>

Next, let's normalize the velocities such that <math>~\rho</math> and the total mass, <math>~M</math>, are both assumed to be the same in every examined Riemann ellipsoid.  In particular, we will normalize to,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~v_0</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~(abc)^{1 / 3}(\pi G \rho)^{1 / 2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~a(\pi G \rho)^{1 / 2} \biggl( \frac{b}{a}\cdot \frac{c}{a} \biggr)^{1 / 3} \, ,</math>
  </td>
</tr>
</table>
in which case we have,

<table align="center" border=0 cellpadding="3">
<tr>
  <td align="right">
<math>
\frac{1}{v_0} \cdot \frac{dx}{dt} =  - \frac{\varphi}{(\pi G \rho)^{1 / 2} } \biggl(\frac{a}{b}\biggr)\biggl( \frac{b}{a}\cdot \frac{c}{a} \biggr)^{-1 / 3} \cdot \biggl(\frac{y}{a}\biggr)
</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; and, &nbsp; &nbsp; &nbsp;
  </td>
  <td align="left">
<math>~
\frac{1}{v_0} \cdot \frac{dy}{dt} =  + \frac{\varphi}{(\pi G \rho)^{1 / 2} }  \biggl(\frac{b}{a}\biggr) \biggl( \frac{b}{a}\cdot \frac{c}{a} \biggr)^{-1 / 3}  \cdot \biggl(\frac{x}{a}\biggr) \, .
</math>
  </td>
</tr>
</table>
Finally, setting, <math>~\varphi/(\pi G\rho)^{1 / 2} \rightarrow -\lambda_\mathrm{EFE}</math> means,

<table align="center" border=0 cellpadding="3">
<tr>
  <td align="right">
<math>
V_x \equiv \frac{1}{v_0} \cdot \frac{dx}{dt} =  \lambda_\mathrm{EFE} \biggl(\frac{a}{b}\biggr)\biggl( \frac{b}{a}\cdot \frac{c}{a} \biggr)^{-1 / 3} \cdot \biggl(\frac{y}{a}\biggr)
</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; and, &nbsp; &nbsp; &nbsp;
  </td>
  <td align="left">
<math>~
V_y \equiv \frac{1}{v_0} \cdot \frac{dy}{dt} = - \lambda_\mathrm{EFE}  \biggl(\frac{b}{a}\biggr) \biggl( \frac{b}{a}\cdot \frac{c}{a} \biggr)^{-1 / 3}  \cdot \biggl(\frac{x}{a}\biggr) \, .
</math>
  </td>
</tr>
</table>

==Example Mach Surface==

Let's try to plot the "Mach surface" for the example model, b41c385, referenced below.  Its relevant parameter values are,
<ul>
  <li><math>~b/a = 0.41</math></li>
  <li><math>~c/a = 0.385</math></li>
  <li><math>~\lambda_\mathrm{EFE} = 0.079886</math></li>
</ul>
Hence, if we set a = 1 then we have,

<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right"><math>~V_x</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>~(0.079886) \biggl(\frac{1}{0.41}\biggr)\biggl( 1.85034 \biggr) \cdot y</math></td>
  <td align="center">&nbsp; &nbsp; &nbsp;and&nbsp; &nbsp; &nbsp;</td>
  <td align="right"><math>~V_y</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>~- (0.079886)  \biggl(0.41\biggr) \biggl( 1.85034 \biggr)  \cdot x</math></td>
</tr>
<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>~0.3605 y</math></td>
  <td align="center">&nbsp; </td>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>~- 0.0606 x</math></td>
</tr>
</table>

Borrowing from an [[Appendix/Ramblings/RiemannB74C692#location_X|accompanying discussion]], we have the following example data set.
<table border="1" align="center" cellpadding="8">
<tr>
  <th align="center" colspan="11">''Direct''</th>
</tr>
<tr>
  <td align="center" rowspan="2">n</td>
  <td align="center" colspan="2">Axisymmetric</td>
  <td align="center" colspan="1">b41c385</td>
  <td align="center" colspan="3" bgcolor="lightblue">Surface</td>
  <td align="center" colspan="3">|V| = 0.1</td>
</tr>
<tr>
  <td align="center">x<sub>0</sub></td>
  <td align="center">y<sub>0</sub></td>
  <td align="center">y = 0.41 &times; y<sub>0</sub></td>
  <td align="center">V<sub>x</sub></td>
  <td align="center">V<sub>y</sub></td>
  <td align="center">|V|</td>
  <td align="center">''factor''</td>
  <td align="center">x</td>
  <td align="center">y</td>
</tr>
<tr>
  <td align="center">1</td>
  <td align="right">1.0000</td>
  <td align="right">0.0000</td>
  <td align="right">0.0000</td>
  <td align="right" bgcolor="lightblue">0.0000</td>
  <td align="right">-0.0606</td>
  <td align="right">0.0606</td>
  <td align="center">---</td>
  <td align="center">---</td>
  <td align="center">---</td>
</tr>
<tr>
  <td align="center">2</td>
  <td align="right">0.9921</td>
  <td align="right">-0.1253</td>
  <td align="right">-0.0514</td>
  <td align="right" bgcolor="lightblue">-0.0185</td>
  <td align="right">-0.0601</td>
  <td align="right">0.0629</td>
  <td align="center">---</td>
  <td align="center">---</td>
  <td align="center">---</td>
</tr>
<tr>
  <td align="center">3</td>
  <td align="right">0.9686</td>
  <td align="right">-0.2487</td>
  <td align="right">-0.1020</td>
  <td align="right" bgcolor="lightblue">-0.0368</td>
  <td align="right">-0.0587</td>
  <td align="right">0.0693</td>
  <td align="center">---</td>
  <td align="center">---</td>
  <td align="center">---</td>
</tr>
<tr>
  <td align="center">4</td>
  <td align="right">0.9298</td>
  <td align="right">-0.3681</td>
  <td align="right">-0.1509</td>
  <td align="right" bgcolor="lightblue">-0.0544</td>
  <td align="right">-0.0563</td>
  <td align="right">0.0783</td>
  <td align="center">---</td>
  <td align="center">---</td>
  <td align="center">---</td>
</tr>
<tr>
  <td align="center">5</td>
  <td align="right">0.8763</td>
  <td align="right">-0.4818</td>
  <td align="right">-0.1975</td>
  <td align="right" bgcolor="lightblue">-0.0712</td>
  <td align="right">-0.0531</td>
  <td align="right">0.0888</td>
  <td align="center">---</td>
  <td align="center">---</td>
  <td align="center">---</td>
</tr>
<tr>
  <td align="center">6</td>
  <td align="right">0.8090</td>
  <td align="right">-0.5878</td>
  <td align="right">-0.2410</td>
  <td align="right" bgcolor="lightblue">-0.0869</td>
  <td align="right">-0.0490</td>
  <td align="right">0.0998</td>
  <td align="center">1.002</td>
  <td align="center">0.8090</td>
  <td align="center">-0.2410</td>
</tr>
<tr>
  <td align="center">7</td>
  <td align="right">0.7290</td>
  <td align="right">-0.6845</td>
  <td align="right">-0.2807</td>
  <td align="right" bgcolor="lightblue">-0.1012</td>
  <td align="right">-0.0442</td>
  <td align="right">0.1104</td>
  <td align="center">0.9058</td>
  <td align="center">0.6603</td>
  <td align="center">-0.2543</td>
</tr>
<tr>
  <td align="center">8</td>
  <td align="right">0.6374</td>
  <td align="right">-0.7705</td>
  <td align="right">-0.3159</td>
  <td align="right" bgcolor="lightblue">-0.1139</td>
  <td align="right">-0.0386</td>
  <td align="right">0.1203</td>
  <td align="center">0.8313</td>
  <td align="center">0.5298</td>
  <td align="center">-0.2626</td>
</tr>
<tr>
  <td align="center">9</td>
  <td align="right">0.5358</td>
  <td align="right">-0.8443</td>
  <td align="right">-0.3462</td>
  <td align="right" bgcolor="lightblue">-0.1248</td>
  <td align="right">-0.0325</td>
  <td align="right">0.1290</td>
  <td align="center">0.7752</td>
  <td align="center">0.4153</td>
  <td align="center">-0.2684</td>
</tr>
<tr>
  <td align="center">10</td>
  <td align="right">0.4258</td>
  <td align="right">-0.9048</td>
  <td align="right">-0.3710</td>
  <td align="right" bgcolor="lightblue">-0.1337</td>
  <td align="right">-0.0258</td>
  <td align="right">0.1362</td>
  <td align="center">0.7342</td>
  <td align="center">0.3126</td>
  <td align="center">-0.2724</td>
</tr>
<tr>
  <td align="center">11</td>
  <td align="right">0.3090</td>
  <td align="right">-0.9511</td>
  <td align="right">-0.3899</td>
  <td align="right" bgcolor="lightblue">-0.1406</td>
  <td align="right">-0.0187</td>
  <td align="right">0.1418</td>
  <td align="center">0.7052</td>
  <td align="center">0.2179</td>
  <td align="center">-0.2750</td>
</tr>
<tr>
  <td align="center">12</td>
  <td align="right">0.1874</td>
  <td align="right">-0.9823</td>
  <td align="right">-0.4027</td>
  <td align="right" bgcolor="lightblue">-0.1452</td>
  <td align="right">-0.0114</td>
  <td align="right">0.1456</td>
  <td align="center">0.6868</td>
  <td align="center">0.1287</td>
  <td align="center">-0.2766</td>
</tr>
<tr>
  <td align="center">13</td>
  <td align="right">0.0628</td>
  <td align="right">-0.9980</td>
  <td align="right">-0.4092</td>
  <td align="right" bgcolor="lightblue">-0.1475</td>
  <td align="right">-0.0038</td>
  <td align="right">0.1476</td>
  <td align="center">0.6775</td>
  <td align="center">0.0425</td>
  <td align="center">-0.2772</td>
</tr>
</table>


In an [[Apps/MaclaurinSpheroids#Equilibrium_Structure|accompanying discussion]] of axisymmetric configurations, we have recognized that, at any point inside the configuration, the square of the sound speed is given approximately by the enthalpy where,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~
c^2 \sim H(x, y, z)
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{P(x, y, z)}{\rho} =
C_B - \Phi_\mathrm{eff}(x, y, z)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
C_B - \biggl[
\Phi_\mathrm{grav} - \frac{1}{2} \Omega_f^2(x^2 + y^2) 
- \frac{1}{2} \lambda^2(x^2 + y^2)  
+ \Omega_f \lambda \biggl(\frac{b}{a}x^2 + \frac{a}{b}y^2 \biggr)  
\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
C_B + 
\pi G \rho \biggl[ I_\mathrm{BT} a^2 - \biggl(A_1 x^2 + A_2 y^2 +A_3 z^2 \biggr) \biggr]
+ \frac{1}{2} \Omega_f^2(x^2 + y^2) 
+ \frac{1}{2} \lambda^2(x^2 + y^2)  
- \Omega_f \lambda \biggl(\frac{b}{a}x^2 + \frac{a}{b}y^2 \biggr)  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-\pi G \rho \biggl[ I_\mathrm{BT} a^2 - A_3 c^2  \biggr] + 
\pi G \rho \biggl[ I_\mathrm{BT} a^2 - \biggl(A_1 x^2 + A_2 y^2 +A_3 z^2 \biggr) \biggr]
+ \frac{1}{2} \Omega_f^2(x^2 + y^2) 
+ \frac{1}{2} \lambda^2(x^2 + y^2)  
- \Omega_f \lambda \biggl(\frac{b}{a}x^2 + \frac{a}{b}y^2 \biggr)  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\pi G \rho \biggl[A_3 c^2   - \biggl(A_1 x^2 + A_2 y^2 +A_3 z^2 \biggr) \biggr]
+ \frac{ (\Omega_f^2+\lambda^2) }{2} \biggl[ x^2 + y^2\biggr] 
- \Omega_f \lambda \biggl[ \biggl(\frac{b}{a}\biggr) x^2 + \biggl( \frac{a}{b} \biggr)y^2 \biggr]  
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~ \frac{c^2}{a^2 (\pi G\rho)}</math>
  </td>
  <td align="center">
<math>~\sim</math>
  </td>
  <td align="left">
<math>~
\frac{ (\Omega_f^2+\lambda^2) }{2(\pi G \rho)} \biggl[ \biggl(\frac{x}{a}\biggr)^2 + \biggl( \frac{y}{a}\biggr)^2\biggr] 
- \frac{\Omega_f \lambda}{(\pi G \rho)} \biggl[ \biggl(\frac{b}{a}\biggr) \biggl( \frac{x}{a}\biggr)^2 + \biggl( \frac{a}{b} \biggr) \biggl( \frac{y}{a}\biggr)^2 \biggr]  
- \biggl[A_1 \biggl( \frac{x}{a}\biggr)^2 + A_2 \biggl(\frac{y}{a}\biggr)^2 +A_3 \biggl( \frac{z^2 - c^2}{a^2}\biggr)  \biggr] \, .
</math>
  </td>
</tr>
</table>

Drawing from equation (7) of {{ Ou2006 }} and from a [[ThreeDimensionalConfigurations/HomogeneousEllipsoids#Gravitational_Potential|separate discussion of gravitational potential of homogeneous ellipsoids]], we see that the effective potential is,

<div align="center">
<math>
~\Phi_\mathrm{eff}(\vec{x})\equiv \Phi + \Psi = -\pi G \rho \biggl[ I_\mathrm{BT} a^2 - \biggl(A_1 x^2 + A_2 y^2 +A_3 z^2 \biggr) \biggr] 
+ \omega\lambda \biggl( \frac{b}{a}x^2 + \frac{a}{b} y^2\biggr) - \frac{\omega^2}{2}\biggl( x^2 + y^2\biggr) -\frac{\lambda^2}{2}\biggl(x^2 + y^2\biggr) \, ,
</math><br />
</div>

where,
<table align="center" border=0 cellpadding="3">

<tr>
  <td align="right">
<math>
~I_\mathrm{BT}
</math>
  </td>
  <td align="center">
<math>
~\equiv
</math>
  </td>
  <td align="left">
<math>
~A_1 + A_2\biggl(\frac{b}{a}\biggr)^2+ A_3\biggl(\frac{c}{a}\biggr)^2 \, .
</math>
  </td>
</tr>
</table>

Setting <math>~\pi G \rho = 1</math>, let's use the test case from above &#8212; that is, (b/a, c/a) = (0.9, 0.641) &#8212; and see if we get the same value of the Bernoulli constant on the surface at each of the three principal axes.  First, let's set x = y = 0 and z = c.  In this case <math>~I_\mathrm{BT} = 1.36658564</math> we find,
<table border="0" align="center" cellpadding="8">
<tr>
<td align="right">
<math>~\Phi_\mathrm{eff}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\biggl[1.36658564 - 0.36298  \biggr] = 1.00350567 \, .</math>
</td>
</tr>
</table>

Next, let's set y = z = 0 and x = 1.  In this case,
<table border="0" align="center" cellpadding="8">
<tr>
<td align="right">
<math>~\Phi_\mathrm{eff}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~\biggl[1.36658564 - 0.52145027  \biggr] + 0.10151682 -0.64000 - 0.01136605 = 0.29518175 \, .</math>
</td>
</tr>
</table>