Editing ThreeDimensionalConfigurations/JacobiEllipsoids (section)

===Angular Momentum Constraint===

<table border="1" align="center" cellpadding="8" width="80%"><tr><td align="left">
<div align="center"><b>Angular Momentum Determination</b></div>
In the above tables, the square of the angular momentum, <math>L^2</math>, for each equilibrium Jacobi ellipsoid has been determined in the following manner:
<table border="0" align="center" cellpadding="8">

<tr>
  <td align="right"><math>L^2</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>I^2\Omega^2</math></td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>\frac{M^2\Omega^2}{5^2}(a^2 + b^2)^2</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>\frac{M^2}{5^2}\biggl(\frac{\Omega^2}{\pi G \rho}\biggr)\biggl[\frac{3G M}{2^2abc} \biggr](a^2 + b^2)^2</math>
  </td>
</tr>

<tr>
  <td align="right"><math>\Rightarrow ~~~ \frac{L^2}{GM^3 \bar{a}}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>\frac{1}{5^2}\biggl(\frac{\Omega^2}{\pi G \rho}\biggr)\biggl[\frac{3}{2^2abc} \biggr]\frac{(a^2 + b^2)^2}{\bar{a}}</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>\frac{3}{2^2\cdot 5^2}\biggl(\frac{\Omega^2}{\pi G \rho}\biggr)\frac{(a^2 + b^2)^2}{\bar{a}^4} \, .</math>
  </td>
</tr>
<tr><td align="center" colspan="3">[<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>], <font color="#00CC00">&sect;39, p. 103, Eq. (16)</font> </td></tr>
</table>

----

Normalizing in a different manner, we have:
<table border="0" align="center" cellpadding="8">

<tr>
  <td align="right"><math>L^2</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>\frac{M^2}{5^2}\biggl(\frac{\Omega^2}{4\pi G \rho}\biggr)\biggl[4\pi G\rho \biggr](a^2 + b^2)^2</math>
  </td>
</tr>

<tr>
  <td align="right"><math>\Rightarrow ~~~j^2 \equiv \frac{L^2}{4\pi G M^{10/3}\rho^{-1 / 3}}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>\frac{M^{-4/3}}{5^2}\biggl(\frac{\Omega^2}{4\pi G \rho}\biggr)\biggl[\rho \biggr]^{4 / 3}(a^2 + b^2)^2</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>\frac{M^{-4/3}}{5^2}\biggl(\frac{\Omega^2}{4\pi G \rho}\biggr)\biggl[\frac{3M}{4\pi abc} \biggr]^{4 / 3}(a^2 + b^2)^2</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>\frac{1}{2^2\cdot 5^2}\biggl[\frac{3}{4\pi } \biggr]^{4 / 3}\biggl(\frac{\Omega^2}{\pi G \rho}\biggr)\frac{(a^2 + b^2)^2}{\bar{a}^4}</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{1}{3}\biggl[\frac{3}{4\pi } \biggr]^{4 / 3} \biggl[\frac{L^2}{GM^{3}\bar{a}} \biggr] 
= 0.049365924 \biggl[\frac{L^2}{GM^{3}\bar{a}} \biggr] 
\, .
</math>
  </td>
</tr>
</table>

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


Alternatively, let's choose a value for the system's total angular momentum, <math>~L > 4.23296</math>, and solve for the axis-ratio pair that identifies that configuration's location along the Jacobi sequence.  We'll adopt the units used by [http://adsabs.harvard.edu/abs/1995ApJ...446..472C Christodoulou ''et al'' (1995)], that is, <math>~G = 1</math>, <math>~\pi \rho = 1</math> and <math>~M = 5</math>, hence,

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

<tr>
  <td align="right">
<math>~a^3</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{3Ma^2}{4\pi(bc)\rho} = \frac{15}{4}\biggl(\frac{b}{a}\biggr)^{-1}  \biggl(\frac{c}{a}\biggr)^{-1}\, .</math>
  </td>
</tr>
</table>
</div>

Given that the relationship between <math>~L</math> and <math>~\Omega</math> in equilibrium Jacobi ellipsoids is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~L</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~a^2\biggl[1 + \biggl(\frac{b}{a}\biggr)^2\biggr]\Omega </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{15}{4}\biggl(\frac{b}{a}\biggr)^{-1}  \biggl(\frac{c}{a}\biggr)^{-1} \biggr]^{2/3}
\biggl[1 + \biggl(\frac{b}{a}\biggr)^2\biggr]\Omega </math>
  </td>
</tr>
</table>
</div>

the [[#JacobiConstraints|constraint on <math>~\Omega^2</math> given above]] implies that,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~L^2 \biggl[ \frac{4}{15}\biggl(\frac{b}{a}\biggr)  \biggl(\frac{c}{a}\biggr) \biggr]^{4/3}
\biggl[1 + \biggl(\frac{b}{a}\biggr)^2\biggr]^{-2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2\biggl\{2 - (A_1+A_3) - \biggl[ \frac{2(1-A_1)-A_3}{1 - (b/a)^2} \biggr] \biggr\} \, .</math>
  </td>
</tr>
</table>
</div>
Or, again adopting the shorthand notation,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\chi \equiv \frac{b}{a}</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp; 
  </td>
  <td align="left">
<math>~\upsilon \equiv \frac{c}{a} \, ,</math>
  </td>
</tr>
</table>
</div>

we seek roots of the function,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~f_L</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~L^2  - \biggl[ \frac{3^4\cdot 5^4}{2^5}  \biggr]^{1/3}\chi^{-4/3} \upsilon^{-4/3}(1 + \chi^2)^{2}
\biggl\{[2 - (A_1+A_3)] - \biggl[ 2(1-A_1)-A_3\biggr](1-\chi^2)^{-1} \biggr\} = 0 \, .</math>
  </td>
</tr>
</table>
</div>

As [[#Constraint_on_Axis-Ratio_Relationship|above]], we will hold <math>~\upsilon</math> fixed and use the Newton-Raphson technique to identify the corresponding "root" value of <math>~\chi</math>.  Hence, we need to specify, not only the function, <math>~f_L</math>, but also its first derivative,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~f_L^'</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\frac{\partial f_L}{\partial \chi} \, .</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- \biggl[ \frac{3^4\cdot 5^4}{2^5}  \biggr]^{1/3}\upsilon^{-4/3} \frac{\partial}{\partial \chi} \biggl\{
\chi^{-4/3} (1 + \chi^2)^{2}
[2 - (A_1+A_3)] 
- \chi^{-4/3} (1 + \chi^2)^{2}(1-\chi^2)^{-1}[ 2(1-A_1)-A_3] 
\biggr\}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- \biggl[ \frac{3^4\cdot 5^4}{2^5}  \biggr]^{1/3}\upsilon^{-4/3} \biggl\{
-\frac{4}{3}\chi^{-7/3} (1 + \chi^2)^{2}[2 - (A_1+A_3)] 
+4\chi^{-1/3} (1 + \chi^2)[2 - (A_1+A_3)] 
-\chi^{-4/3} (1 + \chi^2)^{2}(A_1^'+A_3^') 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \frac{4}{3} \chi^{-7/3} (1 + \chi^2)^{2}(1-\chi^2)^{-1}[ 2(1-A_1)-A_3] 
- 4\chi^{-1/3} (1 + \chi^2)(1-\chi^2)^{-1}[ 2(1-A_1)-A_3] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
- 2 \chi^{-1/3} (1 + \chi^2)^{2}(1-\chi^2)^{-2}[ 2(1-A_1)-A_3] 
- \chi^{-4/3} (1 + \chi^2)^{2}(1-\chi^2)^{-1}[ -2A_1^'-A_3^'] 
\biggr\}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- \biggl[ \frac{3\cdot 5^4}{2^5} \biggr]^{1/3}\upsilon^{-4/3} \chi^{-7/3} (1 + \chi^2)\biggl\{
[12\chi^2-4(1 + \chi^2)][2 - (A_1+A_3)] 
-3\chi (1 + \chi^2)(A_1^'+A_3^') 
+ 3\chi (1 + \chi^2)(1-\chi^2)^{-1}[ 2A_1^' + A_3^'] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ (1-\chi^2)^{-2}\{ 4  (1 + \chi^2)(1-\chi^2)[ 2(1-A_1)-A_3] 
- 12\chi^{2} (1-\chi^2)[ 2(1-A_1)-A_3] 
- 6 \chi^{2} (1 + \chi^2)[ 2(1-A_1)-A_3] \}
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- \biggl[ \frac{3\cdot 5^4}{2^5} \biggr]^{1/3}\upsilon^{-4/3} \chi^{-7/3} (1 + \chi^2)\biggl\{
[8\chi^2-4]A_1
+ 3\chi (1 + \chi^2)(1-\chi^2)^{-1} [ (1+\chi^2)A_1^' + \chi^2A_3^' ]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ 2(1-\chi^2)^{-2} [ 2-2A_1-A_3] [ (4\chi^2-2)(1-\chi^2)^{2} +
2  (1 + \chi^2)(1-\chi^2)  - 6\chi^{2} (1-\chi^2) - 3 \chi^{2} (1 + \chi^2) 
] \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- \biggl[ \frac{3\cdot 5^4}{2^5} \biggr]^{1/3}\upsilon^{-4/3} \chi^{-7/3} (1 + \chi^2)\biggl\{
3\chi (1 + \chi^2)(1-\chi^2)^{-1} [ (1+\chi^2)A_1^' + \chi^2A_3^' ]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~+[8\chi^2-4]A_1
+ 2(1-\chi^2)^{-2} [ 2-2A_1-A_3] [ - \chi^2 - 9\chi^4    + 4\chi^6 ] \biggr\}
</math>
  </td>
</tr>
<!--
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ 4\chi^2 - 8\chi^4 + 4\chi^6 -2+4\chi^2 - 2\chi^4  + 2  - 2\chi^4  - 6\chi^{2}  + 6\chi^4 - 3 \chi^{2} - 3 \chi^4 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~-2 + 2
+ 4\chi^2 +4\chi^2  - 6\chi^{2}  - 3 \chi^{2} - 8\chi^4  - 2\chi^4   - 2\chi^4  + 6\chi^4 - 3 \chi^4 + 4\chi^6  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~- \chi^2 - 9\chi^4    + 4\chi^6  
</math>
  </td>
</tr>
-->
</table>
</div>


What values of <math>~L</math> should we choose?  In association with our [[ThreeDimensionalConfigurations/EFE_Energies#Conserve_Only_L|discussion of warped free-energy surfaces]], we'd like to specify the eccentricity, <math>~e</math>, of a Maclaurin spheroid and adopt the angular momentum of ''that'' configuration.  According to our [[Apps/MaclaurinSpheroids#Maclaurin_Spheroids_.28axisymmetric_structure.29|accompanying discussion of the properties of Maclaurin spheroids]], 
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~L_\mathrm{Mac}^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2^2a^4\Omega^2</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2^3a^4 [ A_1 -A_3(1-e^2)]_\mathrm{Mac} </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2^3 \biggl[\frac{3\cdot 5}{2^2}(1-e^2)^{-1/2}  \biggr]^{4/3} [ A_1 -A_3(1-e^2)]_\mathrm{Mac} </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ [2\cdot 3^4\cdot 5^4]^{1/3} (1-e^2)^{-2/3}  \biggl\{ \frac{1}{e^2} \biggl[\frac{\sin^{-1}e}{e} - (1-e^2)^{1/2}  \biggr](1-e^2)^{1/2}  
-\frac{2}{e^2} \biggl[(1-e^2)^{-1/2} -\frac{\sin^{-1}e}{e} \biggr](1-e^2)^{3/2}  \biggr\} </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ [2\cdot 3^4\cdot 5^4]^{1/3} (1-e^2)^{-2/3}  \biggl\{ 
\frac{1}{e^2} \biggl[\frac{\sin^{-1}e}{e}  \biggr](1-e^2)^{1/2}  
+\frac{2}{e^2} \biggl[\frac{\sin^{-1}e}{e} \biggr](1-e^2)^{3/2}  
-\frac{1}{e^2} \biggl[ (1-e^2)^{1/2}  \biggr](1-e^2)^{1/2}  
-\frac{2}{e^2} \biggl[(1-e^2)^{-1/2}  \biggr](1-e^2)^{3/2}  
\biggr\} </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ [2\cdot 3^4\cdot 5^4]^{1/3} (1-e^2)^{-2/3}  \biggl\{ 
\frac{1}{e^2} \biggl[\frac{\sin^{-1}e}{e}  \biggr](1-e^2)^{1/2}  \biggl[3-2e^2\biggr]
-\frac{3(1-e^2)}{e^2}   
\biggr\} </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ [2\cdot 3^4\cdot 5^4]^{1/3} \frac{(1-e^2)^{1/3} }{e^2} \biggl\{ 
\biggl[\frac{\sin^{-1}e}{e}  \biggr](1-e^2)^{-1/2}  \biggl[3-2e^2\biggr] - 3   
\biggr\} \, . </math>
  </td>
</tr>
</table>
</div>

Note, for example, that if <math>~e = 0.85</math>, the square-root of this expression gives, <math>~L_\mathrm{Mac} = 4.7148806</math>, which matches the angular momentum that was used by [http://adsabs.harvard.edu/abs/1995ApJ...446..472C Christodoulou ''et al'' (1995)] to generate their Figure 3.