Editing ThreeDimensionalConfigurations/JacobiEllipsoids (section)

==Roots of the Governing Relation==

===Constraint on Axis-Ratio Relationship===
To simplify notation, here we will set,
<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>
in which case the governing relation is,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~f_J</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\chi^2}{1-\chi^2} \biggl[ 2(1-A_1)-A_3\biggr]-\upsilon^2  A_3 =0 \, .</math>
  </td>
</tr>
</table>
</div>

Our plan is to employ the [https://brilliant.org/wiki/newton-raphson-method/ Newton Raphson method] to find the root(s) of the <math>~f_J = 0</math> relation, typically holding <math>~\upsilon</math> fixed and using the Newton-Raphson technique to identify the corresponding "root" value of <math>~\chi</math>.  Using this approach, the [https://brilliant.org/wiki/newton-raphson-method/ Newton Raphson technique] requires specification of, not only the function, <math>~f_J</math>, but also its first derivative,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~f_J^'</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{df_J}{d\chi} \, .</math>
  </td>
</tr>
</table>
</div>

Let's determine the requisite expression, using a prime superscript to indicate differentiation with respect to <math>~\chi</math>.  
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~f_J^'</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ 2(1-A_1)-A_3\biggr]\biggl[ \frac{2\chi}{(1-\chi^2)^2} \biggr]
-\frac{\chi^2}{1-\chi^2} \biggl[ 2A_1^'+A_3^'\biggr]
-\upsilon^2  A_3^' \, ,
</math>
  </td>
</tr>
</table>
</div>
where, given that <math>~\theta</math> does not depend on <math>~\chi</math>,
<div align="center">
<table align="center" border=0 cellpadding="3">
<tr>
  <td align="right">
<math>
~A_1^'
</math>
  </td>
  <td align="center">
<math>
~=
</math>
  </td>
  <td align="left">
<math>~\frac{2\upsilon}{\sin^3\theta}  \cdot \frac{d}{d\chi}\biggl\{ \frac{\chi}{k^2} \biggl[ F(\theta,k) - E(\theta,k) \biggr] \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>
~=
</math>
  </td>
  <td align="left">
<math>~\frac{2\upsilon}{k^3 \sin^3\theta}  \cdot \biggl\{ [ F - E ] [k - 2\chi k^'  ] 
+\chi k [ F^' - E^' ]\biggr\} \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>
~A_3^'
</math>
  </td>
  <td align="center">
<math>
~=
</math>
  </td>
  <td align="left">
<math>
~\frac{2}{\sin^3\theta}  \cdot \frac{d}{d\chi}\biggl\{ \frac{\chi}{(1-k^2)} \biggl[  \chi \sin\theta - \upsilon E(\theta,k)\biggr] \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>
~=
</math>
  </td>
  <td align="left">
<math>
~\frac{2}{(1-k^2)^2\sin^3\theta}  \biggl\{ 
\biggl[  \chi \sin\theta - \upsilon E\biggr]\biggl[ (1-k^2) +2\chi kk^' \biggr] + \chi(1-k^2) \biggl[  \sin\theta - \upsilon E^'\biggr]
\biggr\}\, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~k^'</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{d}{d\chi}\biggl[\frac{1 - \chi^2}{1 - \upsilon^2} \biggr]^{1/2} = \frac{-\chi}{(1 - \chi^2)^{1/2}(1 - \upsilon^2)^{1/2}} \, , 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~F^'</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{\partial F(\theta,k)}{\partial k} \cdot k^' \, , 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~E^'</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{\partial E(\theta,k)}{\partial k} \cdot k^' \, . 
</math>
  </td>
</tr>
</table>
</div>

Now, according to [http://functions.wolfram.com/EllipticIntegrals/EllipticF/introductions/IncompleteEllipticIntegrals/ShowAll.html online WolframResearch documentation] &#8212; see, in particular, the subsection titled, "Representations of Derivatives" &#8212;
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{\partial F(z|m)}{\partial m}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{E(z|m)}{2(1-m)m} - \frac{F(z|m)}{2m} - \frac{\sin(2z)}{4(1-m)\sqrt{1-m\sin^2(z)}} \, ,
</math>
  </td>
</tr>
</table>
</div>
and,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{\partial E(z|m)}{\partial m}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{E(z|m) - F(z|m)}{2m} \, ,</math>
  </td>
</tr>
</table>
</div>
where, <math>~z~\leftrightarrow~\theta</math>, and,
<div align="center">
<math>~m \equiv k^2 ~~~~\Rightarrow~~~~\frac{dm}{dk} = 2k \ .</math>
</div>
Hence, we have,

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

<tr>
  <td align="right">
<math>~F^'</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[\frac{\partial F(z|m)}{\partial m} \cdot \frac{dm}{dk}\biggr] k^' 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{E(\theta,k)}{2(1-k^2)k^2} - \frac{F(\theta,k)}{2k^2} - \frac{\sin(2\theta)}{4(1-k^2)\sqrt{1-k^2\sin^2\theta}} \biggr] 2kk^' \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~E^'</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{\partial E(z|m)}{\partial m} \cdot \frac{dm}{dk}\biggr] k^' 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ E(\theta,k) - F(\theta,k) \biggr] \frac{k^'}{k} \, .
</math>
  </td>
</tr>
</table>
</div>
This, then, gives us all of the expressions necessary to specify the derivative, <math>~f_J^'</math> analytically.



<table border="1" cellpadding="5" align="center">
<tr>
  <th align="center" colspan="1">
<font size="+1">Table 1:&nbsp; Double-Precision Evaluations</font><p></p>
Related to Table IV in [<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>], <font color="#00CC00">Chapter 6, &sect;39 (p. 103)</font>
</th>
</tr>
<tr><td align="left">
<pre>
         b/a      c/a            omega2              angmom              5L/M                fJ              fJderiv

        1.00   0.582724     3.742297785D-01     3.037510987D-01     4.232965627D+00     0.000000000D+00     0.000000000D+00
        0.96   0.570801     3.739782202D-01     3.039551227D-01     4.235808832D+00     1.377942479D-06     1.636908401D-01
        0.92   0.558330     3.731876801D-01     3.046006837D-01     4.244805137D+00    -6.821687132D-07     1.676406830D-01
        0.88   0.545263     3.717835971D-01     3.057488283D-01     4.260805266D+00     8.533280272D-07     1.715558312D-01
        0.84   0.531574     3.696959199D-01     3.074667323D-01     4.284745355D+00    -4.622993727D-08     1.754024874D-01
        0.80   0.517216     3.668370069D-01     3.098368632D-01     4.317774645D+00     2.805300664D-08     1.791408327D-01
        0.76   0.502147     3.631138118D-01     3.129555079D-01     4.361234951D+00     3.221800126D-07     1.827219476D-01
        0.72   0.486322     3.584232032D-01     3.169377270D-01     4.416729718D+00     3.274773094D-08     1.860866255D-01
        0.68   0.469689     3.526490289D-01     3.219229588D-01     4.486202108D+00     1.202999164D-08     1.891636215D-01
        0.64   0.452194     3.456641138D-01     3.280805511D-01     4.572012092D+00     2.681560312D-07     1.918668912D-01
        0.60   0.433781     3.373298891D-01     3.356184007D-01     4.677056841D+00     1.037186290D-08     1.940927000D-01
        0.56   0.414386     3.274928085D-01     3.447962894D-01     4.804956583D+00     1.071021385D-07     1.957166395D-01
        0.52   0.393944     3.159887358D-01     3.559412795D-01     4.960269141D+00     8.098003093D-08     1.965890756D-01
        0.48   0.372384     3.026414267D-01     3.694732246D-01     5.148845443D+00     1.255768368D-07     1.965308751D-01
        0.44   0.349632     2.872670174D-01     3.859399647D-01     5.378319986D+00     1.329168636D-08     1.953277019D-01
        0.40   0.325609     2.696779847D-01     4.060726774D-01     5.658882201D+00    -9.783004411D-08     1.927241063D-01
        0.36   0.300232     2.496925963D-01     4.308722159D-01     6.004479614D+00     1.044268276D-07     1.884168286D-01
        0.32   0.273419     2.271530240D-01     4.617497270D-01     6.434777459D+00    -4.469279448D-08     1.820477545D-01
        0.28   0.245083     2.019461513D-01     5.007767426D-01     6.978643856D+00     7.996820889D-08     1.731984783D-01
        0.24   0.215143     1.740514751D-01     5.511400218D-01     7.680488329D+00     1.099319693D-07     1.613864645D-01
        0.20   0.183524     1.436093757D-01     6.180687545D-01     8.613182979D+00     5.068010978D-08     1.460685065D-01
        0.16   0.150166     1.110438660D-01     7.109267615D-01     9.907218635D+00    -2.170751250D-08     1.266576761D-01
        0.12   0.115038     7.728058393D-02     8.487699974D-01     1.182815219D+01     3.613784147D-09     1.025686850D-01
        0.08   0.078166     4.416740942D-02     1.079303624D+00     1.504078558D+01     3.319018649D-08     7.332782508D-02
        0.04   0.039688     1.541513490D-02     1.582762691D+00     2.205680933D+01    -6.674246644D-09     3.882477311D-02
</pre>
</td></tr>
</table>

<span id="Table2"><b>With regard to our Table 1 (immediately above):</b></span>   Given each pair of axis ratios, <math>~(\tfrac{b}{a},\tfrac{c}{a})</math> &#8212; copied from Table IV of [<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>] (see columns 1 and 2 of our Table 1) &#8212; and the corresponding coefficient values, <math>~A_1</math>, <math>~A_2</math>, and <math>~A_3</math>, as tabulated in [[ThreeDimensionalConfigurations/HomogeneousEllipsoids#Table2|Table 2 of our accompanying discussion]], we calculated corresponding values of <math>~\Omega^2</math> (column 3) and total angular momentum (column 4) in the units used in Table IV of [<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>], as well as  (column 5) the total angular momentum in units used by [http://adsabs.harvard.edu/abs/1995ApJ...446..472C Christodoulou, ''et al.'' (1995, ApJ, 446, 472)] &#8212; see [[ThreeDimensionalConfigurations/HomogeneousEllipsoids#Example_Evaluations|our related discussion of these physical quantities]].  We also have tabulated associated values of the function, <math>~f_J</math>, (column 6) and its first derivative, <math>~f_J^'</math>, (column 7) as defined immediately above.  Notice that <math>~f_J</math> is very nearly zero in all cases, which indicates that each axis-ratio pair indeed identifies a configuration that lies along the Jacobi sequence.

<table border="1" cellpadding="5" align="center">
<tr>
  <th align="center" colspan="1">
<font size="+1">Table 2:&nbsp; Jacobi Sequence</font>
</th>
</tr>
<tr><td align="left">
<pre>
   b/a       c/a        A1        A2        A3      omega2       a       5L/M

  0.990699  0.580000  0.512818  0.518962  0.968220  0.374217  1.868761  4.233113
  0.901558  0.552381  0.481786  0.549836  0.968378  0.372621  1.960046  4.251259
  0.820783  0.524762  0.450993  0.580215  0.968792  0.368424  2.057217  4.299402
  0.747135  0.497143  0.420459  0.610088  0.969452  0.361716  2.161309  4.377683
  0.679613  0.469524  0.390210  0.639442  0.970348  0.352587  2.273548  4.486951
  0.617393  0.441905  0.360273  0.668258  0.971469  0.341129  2.395412  4.628802
  0.559798  0.414286  0.330684  0.696516  0.972800  0.327439  2.528716  4.805667
  0.506257  0.386667  0.301483  0.724187  0.974329  0.311620  2.675723  5.020964
  0.456291  0.359048  0.272719  0.751241  0.976040  0.293786  2.839307  5.279337
  0.409492  0.331429  0.244450  0.777636  0.977914  0.274062  3.023190  5.587020
  0.365507  0.303810  0.216744  0.803324  0.979931  0.252593  3.232298  5.952388
  0.324034  0.276190  0.189686  0.828246  0.982067  0.229546  3.473314  6.386811
  0.284807  0.248571  0.163376  0.852329  0.984295  0.205118  3.755577  6.906010
  0.247591  0.220952  0.137939  0.875480  0.986581  0.179549  4.092599  7.532311
  0.212179  0.193333  0.113527  0.897587  0.988885  0.153130  4.504785  8.298565
  0.178382  0.165714  0.090333  0.918505  0.991162  0.126229  5.024664  9.255452
  0.146026  0.138095  0.068601  0.938044  0.993355  0.099316  5.707871 10.486253
  0.114948  0.110476  0.048654  0.955953  0.995393  0.073010  6.659169 12.140357
  0.084989  0.082857  0.030927  0.971879  0.997194  0.048162  8.105501 14.522397
  0.055982  0.055238  0.016051  0.985298  0.998651  0.026008 10.663879 18.396951
  0.027738  0.027619  0.005032  0.995331  0.999637  0.008539 16.979084 26.660547
</pre>
</td></tr>
</table>

<b>With regard to our Table 2 (immediately above):</b>   Here we specified twenty-one values of the axis ratio, <math>~\tfrac{c}{a}</math>, (column 2) and used our Newton-Raphson-based root finder to identify corresponding values of the companion axis ratio, <math>~\tfrac{b}{a}</math>, (column 1) that satisfies the governing relation, <math>~f_J = 0</math>.

===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.