Editing Apps/MaclaurinSpheroidSequence

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=Maclaurin Spheroid Sequence=
{| class="MaclaurinSequence" style="float:left; margin-right: 20px; border-style: solid; border-width: 3px border-color: black"
|- 
! style="height: 150px; width: 150px; background-color:maroon;" |[[H_BookTiledMenu#Spheroidal_.26_Spheroidal-Like|<font color="white">Maclaurin<br />Spheroid<br />Sequence</font>]]
|}
==Detailed Force Balance Conditions==

===Equilibrium Angular Velocity===

<table border="0" align="right" cellpadding="3" width="25%">
<tr>
  <td align="center">'''Figure 1'''</td>
</tr>
<tr><td align="center">
[[File:EFE Omega2vsECCwithThomsonTait2.png|center|350px|Maclaurin Spheroid Sequence]]
</td></tr>
<tr>
  <td align="center">
The dark blue circular markers locate 15 of the 18 individual models identified in Table 1. The solid black curve derives from our evaluation of the function, <math>~\omega_0^2(e)\, ;</math> this curve also may be found in:
<div align="center">
Fig. 5 (p. 79) of [<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>];<br />
Fig. 7.2 (p. 173) of [<b>[[Appendix/References#ST83|<font color="red">ST83</font>]]</b>]<br />
</div>
  </td>
</tr>
</table>
The essential structural elements of each Maclaurin spheroid model are uniquely determined once we specify the system's axis ratio, <math>~c/a</math>, or the system's meridional-plane eccentricity, <math>~e</math>, where

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

<tr>
  <td align="right">
<math>~e</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\biggl[1 - \biggl(\frac{c}{a}\biggr)^2\biggr]^{1 / 2} \, ,</math>
  </td>
</tr>
</table>

which varies from ''e = 0'' (spherical structure) to ''e = 1'' (infinitesimally thin disk).  According to our [[Apps/MaclaurinSpheroids#Maclaurin_Spheroids_.28axisymmetric_structure.29|accompanying derivation]], for a given choice of <math>~e</math>, the square of the system's equilibrium angular velocity is,

<table align="center" border="0" cellpadding="5">
<tr>
  <td align="right">
<math>
~ \omega_0^2
</math>
  </td>
  <td align="center">
<math>
~=
</math>
  </td>
  <td align="left">
<math>
2\pi G \rho \biggl[ A_1 - A_3 (1-e^2) \biggr] \, ,
</math>
  </td>
</tr>
<tr>
  <td align="center" colspan="3">
[<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>], &sect;32, p. 77, Eq. (4)<br />
[<b>[[Appendix/References#T78|<font color="red">T78</font>]]</b>], &sect;4.5, p. 86, Eq. (52)<br />
[<b>[[Appendix/References#ST83|<font color="red">ST83</font>]]</b>], &sect;7.3, p. 172, Eq. (7.3.18)
</td>
</tr>
</table> 
where,

<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{1}{e^2} \biggl[\frac{\sin^{-1}e}{e} - (1-e^2)^{1/2}  \biggr](1-e^2)^{1/2} \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>
~A_3
</math>
  </td>
  <td align="center">
<math>
~=
</math>
  </td>
  <td align="left">
<math>
\frac{2}{e^2} \biggl[(1-e^2)^{-1/2} -\frac{\sin^{-1}e}{e} \biggr](1-e^2)^{1/2} \, .
</math>
  </td>
</tr>
<tr>
  <td align="center" colspan="3">
{{ TT1867 }}, &sect;522, p. 392, Eqs. (9) &amp; (7)<br />
[<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>], &sect;17, p. 43, Eq. (36)<br />
[<b>[[Appendix/References#T78|<font color="red">T78</font>]]</b>], &sect;4.5, p. 85, Eqs. (48) &amp; (49)<br />
[<b>[[Appendix/References#ST83|<font color="red">ST83</font>]]</b>], &sect;7.3, p. 170, Eq. (7.3.8)
</td>
</tr>
</table>

<table border="0" cellpadding="10" align="right" width="25%"><tr><td align="right">
<table border="1" cellpadding="5" align="center">
<tr>
<td colspan="5" align="center">
'''Table 1'''<br />Data copied from<br />{{ TT1867 }}, &sect;772, p. 614
</td>
</tr>
<tr>
  <td align="center" width="50%"><math>~e</math></td>
  <td align="center"><math>~\frac{\omega_0^2}{2\pi G \rho}</math></td>
  <td align="center" bgcolor="lightgrey" rowspan="10">&nbsp; &nbsp;</td>
  <td align="center" width="50%"><math>~e</math></td>
  <td align="center"><math>~\frac{\omega_0^2}{2\pi G \rho}</math></td>
</tr>
<tr>
  <td align="center">0.10</td>
  <td align="center">0.0027</td>
  <td align="center">0.91</td>
  <td align="center">0.2225</td>
</tr>
<tr>
  <td align="center">0.20</td>
  <td align="center">0.0107</td>
  <td align="center">0.92</td>
  <td align="center">0.2241</td>
</tr>
<tr>
  <td align="center">0.30</td>
  <td align="center">0.0243</td>
  <td align="center">0.93</td>
  <td align="center">0.2247</td>
</tr>
<tr>
  <td align="center">0.40</td>
  <td align="center">0.0436</td>
  <td align="center">0.94</td>
  <td align="center">0.2239</td>
</tr>
<tr>
  <td align="center">0.50</td>
  <td align="center">0.0690</td>
  <td align="center">0.95</td>
  <td align="center">0.2213</td>
</tr>
<tr>
  <td align="center">0.60</td>
  <td align="center">0.1007</td>
  <td align="center">0.96</td>
  <td align="center">0.2160</td>
</tr>
<tr>
  <td align="center">0.70</td>
  <td align="center">0.1387</td>
  <td align="center">0.97</td>
  <td align="center">0.2063</td>
</tr>
<tr>
  <td align="center">0.80</td>
  <td align="center">0.1816</td>
  <td align="center">0.98</td>
  <td align="center">0.1890</td>
</tr>
<tr>
  <td align="center">0.90</td>
  <td align="center">0.2203</td>
  <td align="center">0.99</td>
  <td align="center">0.1551</td>
</tr>
</table>
</td></tr></table>
<span id="MaclaurinFrequency">In other words,</span>

<table align="center" border="0" cellpadding="5">
<tr>
  <td align="right">
<math>
~ \frac{\omega_0^2}{2\pi G \rho }
</math>
  </td>
  <td align="center">
<math>
~=
</math>
  </td>
  <td align="left">
<math>
(3-2e^2)(1-e^2)^{1 / 2} \cdot \frac{\sin^{-1}e}{e^3} - \frac{3(1-e^2)}{e^2}
\, .
</math>
  </td>
</tr>
<tr>
  <td align="center" colspan="3">
{{ TT1867 }}, &sect;771, p. 613, Eq. (1)<br />
[<b>[[Appendix/References#Lamb32|<font color="red">Lamb32</font>]]</b>], 6<sup>th</sup> Ed. (1932), Ch. XII, &sect;374, p. 701, Eq. (6)  &#8212; set <math>~\zeta^2 = (1-e^2)/e^2</math><br />
[https://ui.adsabs.harvard.edu/abs/1886RSPS...41..319D/abstract G. H. Darwin (1886)], p.322, Eq. (14) &#8212; set <math>~\gamma = \sin^{-1}e</math><br />
[https://ui.adsabs.harvard.edu/abs/1928asco.book.....J/abstract J. H. Jeans (1928)], &sect;192, p. 202, Eq. (192.4)<br />
[<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>], &sect;32, p. 78, Eq. (6)<br />
[<b>[[Appendix/References#ST83|<font color="red">ST83</font>]]</b>], &sect;7.3, p. 172, Eq. (7.3.18)
</td>
</tr>
</table> 

Figure 1 shows how the square of the angular velocity varies with eccentricity along the Maclaurin spheroid sequence; given the chosen normalization unit,  <math>~\pi G\rho</math>, it is understood that the density of the configuration is held fixed as the eccentricity is varied.  


<table border="0" cellpadding="5" width="60%" align="center"><tr><td align="left">
Examining the Maclaurin spheroid sequence "<font color="orange">&hellip; we see that the value of <math>~\omega_0^2</math> increases gradually from zero to a maximum as the eccentricity <math>~e</math> rises from zero to about 0.93, and then (more quickly) falls to zero as the eccentricity rises from 0.93 to unity.</font>" &hellip; "<font color="orange">If the angular velocity exceed the value</font> associated with this maximum, "<font color="orange">&hellip; equilibrium is impossible in the form of an ellipsoid of revolution.  If the angular velocity fall short of this limit there are always two ellipsoids of revolution which satisfy the conditions of equilibrium.  In one of these the eccentricity is greater than 0.93, in the other less.</font>"</td></tr>
<tr><td align="right">
--- {{ TT1867 }}, &sect;772, p. 614.
</td></tr></table>


The extremum of the curve occurs where <math>d\omega_0^2/de = 0</math>; that is, it occurs where,
<table border="0" align="center" cellpadding="5">
<tr>
  <td align="right"><math>\frac{\sin^{-1}e}{e}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>(1 - e^2)^{1 / 2} \biggl[\frac{9 - 2e^2}{9 - 8e^2}\biggr] \, .</math></td>
</tr>
</table>

In our Figure 1, the small solid-green square marker identifies the location along the sequence where the system with the maximum angular velocity resides:
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\biggl[ e, \frac{\omega_0^2}{\pi G \rho} \biggr]</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\biggl[ 0.92995, 0.449331 \biggr] \, .</math>
  </td>
</tr>
<tr>
  <td align="center" colspan="3">[<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>], &sect;32, p. 80, Eqs. (9) &amp; (10)</td>
</tr>
</table>

===ASIDE===
Suppose we set,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\lambda</math>
  </td>
  <td align="center">
<math>\equiv</math>
  </td>
  <td align="left">
<math>\sin^{-1}e</math>
  </td>
<td align="center">&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; </td>
  <td align="right">
<math>\Rightarrow~~~ e</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\sin\lambda \, ,</math>
  </td>
</tr>
</table>
valid over the range, <math>0 \le \lambda \le \pi</math>. Then we have,

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

<tr>
  <td align="right">
<math>
A_3
</math>
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
\frac{2}{e^2} \biggl[(1-e^2)^{-1/2} -\frac{\sin^{-1}e}{e} \biggr](1-e^2)^{1/2}
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
<math>
A_1
</math>
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
1 - \tfrac{1}{2}A_3 \, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>
\frac{\omega_0^2}{2\pi G \rho}
</math>
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
A_1 - A_3 (1-e^2) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
1 + (2e^2 - 3)\tfrac{1}{2} A_3
</math>
  </td>
</tr>

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

<table border="1" align="center" cellpadding="8"><tr><td align="left">
[[Appendix/Ramblings/TrigFunctions#Integer-Degree_Angles|Note, for example]], that <math>\lambda = 5\pi/12 ~ \Rightarrow ~ e = (1 + \sqrt{3})/(2\sqrt{2}) \approx 0.965925827</math>, in which case,

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

<tr>
  <td align="right">
<math>
\frac{\omega_0^2}{2\pi G \rho}
</math>
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
1 + \frac{(2e^2 - 3)}{e^2} \biggl[1 - \frac{\lambda \cdot (1 - e^2)^{1 / 2}}{e} \biggr]
\approx
0.210901366\, .
</math>
  </td>
</tr>
</table>

Plugging in the analytic expression for the eccentricity, we find,

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

<tr>
  <td align="right">
<math>
e^2
</math>
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
\biggl[ \frac{(1+\sqrt{3})}{2\sqrt{2}} \biggr]^2
=
\biggl[ \frac{2+\sqrt{3}}{4} \biggr]
=
\biggl[ \frac{1}{2} + \frac{\sqrt{3}}{4} \biggr] \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>
(1 - e^2)^{1 / 2 }
</math>
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
\biggl[ \frac{1}{2} - \frac{\sqrt{3}}{4} \biggr]^{1 / 2}
=
\frac{(2 - \sqrt{3})^{1 / 2}}{2}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>
\Rightarrow ~~~ \frac{\omega_0^2}{2\pi G \rho}
</math>
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
1 + \biggl[ \biggl( \frac{2+\sqrt{3}}{2} \biggr) - 3\biggr] 
\biggl[1 - \frac{5\pi}{12} \cdot \frac{(2 - \sqrt{3})^{1 / 2}}{2}\cdot \biggl( \frac{4}{2+\sqrt{3}} \biggr)^{1 / 2}\biggr]
\biggl[ \frac{4}{2+\sqrt{3}} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
1 + \frac{1}{2}\biggl[ \sqrt{3}-4\biggr] 
\biggl[1 - \frac{5\pi}{12} \cdot \biggl( \frac{2 - \sqrt{3}}{2+\sqrt{3}} \biggr)^{1 / 2}\biggr]
\biggl[ \frac{4}{2+\sqrt{3}} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
1 +  
\frac{1}{(2 + \sqrt{3})^{1 / 2}}\biggl[(2 + \sqrt{3})^{1 / 2} - \frac{5\pi}{12} \cdot ( 2 - \sqrt{3} )^{1 / 2}\biggr]
\biggl[ \frac{2(\sqrt{3}-4)}{2+\sqrt{3}} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
1 +  
\frac{1}{6}\biggl[12(2 + \sqrt{3})^{1 / 2} - 5\pi \cdot ( 2 - \sqrt{3} )^{1 / 2}\biggr]
\biggl[ \frac{(\sqrt{3}-4)}{(2+\sqrt{3})^{3 / 2}} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
0.210901367 \, .
</math>
  </td>
</tr>
</table>

<font color="red"><b>Matches!</b></font>

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

===Corresponding Total Angular Momentum===

<table border="0" align="right" cellpadding="3">
<tr>
  <td align="center">'''Figure 2'''</td>
</tr>
<tr><td align="center">
[[File:EFE_AngMomVsEcc.png|center|350px|Maclaurin Spheroid Sequence]]
</td></tr>
<tr>
  <td align="center">
Solid black curve also may be found as:
<div align="center">
Fig. 6 (p. 79) of [<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>];<br />
Fig. 7.3 (p. 174) of [<b>[[Appendix/References#ST83|<font color="red">ST83</font>]]</b>]
</div>
  </td>
</tr>
</table>
The total angular momentum of each uniformly rotating Maclaurin spheroid is given by the expression,
<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>~I \omega_0 \, ,</math>
  </td>
</tr>
</table>
where, the moment of inertia <math>~(I)</math> and the total mass <math>~(M)</math> of a uniform-density spheroid are, respectively,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~I</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl(\frac{2}{5}\biggr) M a^2 \, ,</math>
  </td>
<td align="center">&nbsp; &nbsp; &nbsp; and, &nbsp; &nbsp; &nbsp;</td>
  <td align="right">
<math>~M</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl( \frac{4\pi}{3} \biggr) \rho a^2c \, .</math>
  </td>
</tr>
</table>

Adopting the shorthand notation, <math>\bar{a} \equiv  (a^2 c)^{1 / 3}</math>, we have,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~L^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{2^2 M^2 a^4}{5^2} \biggl[  A_1 - A_3 (1-e^2) \biggr] 2\pi G \biggl[ \frac{3}{2^2\pi} \cdot \frac{M}{a^2c} \biggr]</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{6GM^3 {\bar{a}}}{5^2} \biggl[  A_1 - A_3 (1-e^2) \biggr]\biggl(\frac{a}{c}\biggr)^{4/3} </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{L}{(GM^3\bar{a})^{1 / 2}}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{6^{1 / 2}}{5} \biggl[  A_1 - A_3 (1-e^2) \biggr]^{1 / 2}(1 - e^2)^{-1 / 3} \, .</math>
  </td>
</tr>
<tr>
  <td align="center" colspan="3">
[<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>], &sect;32, p. 78, Eq. (7)<br />
[<b>[[Appendix/References#T78|<font color="red">T78</font>]]</b>], &sect;4.5, p. 86, Eq. (54)
  </td>
</tr>
</table>
<span id="MPT77angmom">This also means,</span>
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>L_*^2 \equiv \frac{L^2}{(GM^3\bar{a})}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\frac{6}{5^2} \biggl[ (3-2e^2)(1-e^2)^{1 / 2} \cdot \frac{\sin^{-1}e}{e^3} - \frac{3(1-e^2)}{e^2}\biggr](1 - e^2)^{-2 / 3}  \, .</math>
  </td>
</tr>
<tr>
  <td align="center" colspan="3">
{{ MPT77 }}, &sect;IVa, p. 591, Eq. (4.2)
  </td>
</tr>
</table>

Figure 2 shows how the system's normalized angular momentum, <math>L_*</math>, varies with eccentricity along the Maclaurin spheroid sequence; given the chosen normalization unit,  <math>~(GM^3\bar{a})^{1 / 2}</math>, it is understood that the mass and the volume &#8212; hence, also the density &#8212; of the configuration are held fixed as the eccentricity is varied.  Strictly speaking, along this sequence the angular momentum asymptotically approaches infinity as <math>~e \rightarrow 1</math>; by limiting the ordinate to a maximum value of 1.2, the plot masks this asymptotic behavior.  The small solid-green square marker identifies the location along this sequence where the system with the maximum angular velocity resides (see Figure 1); this system is not associated with a turning point along this angular-momentum versus eccentricity sequence.

==Alternate Sequence Diagrams==

===Energy Ratio, T/|W|===

<table align="right" border=1 cellpadding="8">
<tr>
  <td colspan="3" align="center">
'''Table 2:''' &nbsp;[[Appendix/Ramblings/PowerSeriesExpressions#Maclaurin_Spheroid_Index_Symbols|Limiting Values]]
  </td>
</tr>
<tr>
  <td align="center">
&nbsp;
  </td>
  <td align="center">
<b><math>e \rightarrow 0</math></b>
  </td>
  <td align="center">
<b><math>\frac{c}{a} \rightarrow 0</math></b>
  </td>
</tr>
<tr>
  <td align="center">
<b><math>A_1</math></b>
  </td>
  <td align="center">
<math>\frac{2}{3}\biggl[1 - \frac{e^2}{5} - \mathcal{O}\biggl(e^4\biggr)\biggr]</math>
  </td>
  <td align="center">
<math>\frac{\pi}{2} \biggl( \frac{c}{a}\biggr) - 2\biggl(\frac{c}{a}\biggr)^2+ \mathcal{O}\biggl(\frac{c^3}{a^3}\biggr)</math>
  </td>
</tr>
<tr>
  <td align="center">
<b><math>A_3</math></b>
  </td>
  <td align="center">
<math>\frac{2}{3}\biggl[1 + \frac{2e^2}{5} + \mathcal{O}\biggl(e^4\biggr)\biggr]</math>
  </td>
  <td align="center">
<math> 2 - \pi \biggl( \frac{c}{a}\biggr) + 4\biggl(\frac{c}{a}\biggr)^2
- \mathcal{O}\biggl(\frac{c^3}{a^3}\biggr)</math>
  </td>
</tr>
<tr>
  <td align="center">
<b><math>~\frac{\sin^{-1}e}{e}</math></b>
  </td>
  <td align="center">
<math>~1 + \frac{e^2}{6} + \mathcal{O}\biggl(e^4\biggr)</math>
  </td>
  <td align="center">
<math>~\frac{\pi}{2} - \biggl(\frac{c}{a}\biggr) +\frac{\pi}{4}\biggl(\frac{c}{a}\biggr)^2
- \mathcal{O}\biggl(\frac{c^3}{a^3}\biggr)</math>
  </td>
</tr>
<tr>
  <td align="center">
<b><math>~\tau \equiv \frac{T_\mathrm{rot}}{|W_\mathrm{grav}|}</math></b>
  </td>
  <td align="center">
<math>~0</math>
  </td>
  <td align="center">
<math>~\frac{1}{2}</math>
  </td>
</tr>
</table>
The rotational kinetic energy of each uniformly rotating Maclaurin spheroid is given by the expression,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~T_\mathrm{rot}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{2}I \omega_0^2
=\frac{Ma^2}{5} \cdot 2\pi G\rho
\biggl[ A_1 - (1-e^2)A_3\biggr]
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{2^3\pi^2}{3\cdot 5} \cdot  G\rho^2 a^5  
\biggl[ \frac{(1-e^2)}{e^3} ~(3 - 2e^2)\sin^{-1}e - \frac{3(1-e^2)^{3 / 2}}{e^2} \biggr] 
\, ;
</math>
  </td>
</tr>
</table>
and the gravitational potential energy of each configuration is,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~W_\mathrm{grav}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \frac{3}{5} \cdot \frac{GM^2}{c} \biggl[ A_1 + \frac{1}{2}(1-e^2)A_3 \biggr] 
=
- \frac{3}{2\cdot 5} \cdot \frac{G}{c} \biggl[ \frac{2^2\pi \rho a^2 c}{3} \biggr]^2 \biggl[ 2A_1 + (1-e^2)A_3 \biggr] 
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \frac{2^4 \pi^2}{3\cdot 5} \cdot G\rho^2 a^5 (1-e^2) \cdot \frac{\sin^{-1}e }{e}  \, .</math>
  </td>
</tr>
</table>

<span id="EnergyNorm">&nbsp;</span>
<table border="1" align="center" cellpadding="10" width="80%">
<tr><td align="left">
<div align="center">'''Energy Normalization'''</div>
In his tabulation of the properties of Maclaurin Spheroids &#8212; see Appendix D (p. 483) of [<b>[[Appendix/References#T78|<font color="red">T78</font>]]</b>] &#8212; Tassoul adopted the following energy normalization:
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>E_\mathrm{T78}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
(4/3)\pi G \rho M {\bar{a}}^2 \, ,
</math>
  </td>
</tr>
</table>
where, as [[#Corresponding_Total_Angular_Momentum|above]],
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\bar{a}</math>
  </td>
  <td align="center">
<math>\equiv</math>
  </td>
  <td align="left">
<math>
(a^2 c)^{1 / 3}  
= a \biggl(\frac{c}{a}\biggr)^{1 / 3}
= a (1 - e^2)^{1 / 6}
\, .
</math>
  </td>
</tr>
</table>
Given that, <math>M = (4/3)\pi \rho a^2c = (4/3)\pi \rho a^3 (1 - e^2)^{1 / 2}\, ,</math> we can write instead,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>E_\mathrm{T78}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
(4/3)\pi G [\rho (4/3)\pi \rho a^3 (1 - e^2)^{1 / 2}] a^2 (1-e^2)^{1 / 3} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
(2^4 \pi^2/3^2)G \rho^2 a^5 (1 - e^2)^{5 / 6} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl(\frac{4\pi}{3}\biggr)^{1 / 3} G (M^5\rho)^{1 / 3} \, .
</math>
  </td>
</tr>
</table>
After normalization, then, we have,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\frac{T_\mathrm{rot}}{E_\mathrm{T78}}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math> 
\frac{3}{2\cdot 5} 
\biggl[ (3 - 2e^2)\frac{\sin^{-1}e}{e} - 3(1-e^2)^{1 / 2} \biggr] \frac{(1-e^2)^{1 / 6}}{e^2}
\, ;
</math>
  </td>
</tr>
</table>
and,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\frac{W_\mathrm{grav}}{E_\mathrm{T78}}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
- \frac{3}{5}(1-e^2)^{1 / 6} \cdot \frac{\sin^{-1}e }{e}  \, .</math>
  </td>
</tr>
</table>
<b><font color="darkblue">Example</font></b>  &hellip; to be checked against the relevant line of data from Tables D.1 and D.2 of [<b>[[Appendix/References#T78|<font color="red">T78</font>]]</b>]:  &nbsp;If we set <math>e = 0.965646</math>, we find, <math>T_\mathrm{rot}/E_\mathrm{T78} =  0.155578\, ,</math> and <math>W_\mathrm{grav}/E_\mathrm{T78} = -0.518594\, ,</math> which implies that, <math>(T_\mathrm{rot} + W_\mathrm{grav})/E_\mathrm{T78} = -0.363016\, ,</math> and <math>\tau \equiv T_\mathrm{rot}/|W_\mathrm{grav}| = 0.300000\, .</math>

----

Note that {{ Wong74 }} &#8212; see the NOTE appended to his Table 2 (p. 686) &#8212; adopts the normalization,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>E_\mathrm{Wong74}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{3}{5}\biggl(\frac{4\pi}{3}\biggr)^{1 / 2} G (M^5\rho)^{1 / 3} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ \frac{E_\mathrm{T78} }{E_\mathrm{Wong74}}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{5}{3}\biggl(\frac{3}{4\pi}\biggr)^{1 / 6}  \, .
</math>
  </td>
</tr>
</table>

----

Alternatively, in {{ EH85 }} &#8212; see Eq. 7 (p. 291) &#8212; and in {{ CKST95d }}  &#8212; see Eq. 1.3 (p. 511)  &#8212; the energy normalization is,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>E_\mathrm{EH85} = E_\mathrm{CKST95d}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
(4\pi G)^2 M^5 L^{-2} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ \biggl[\frac{E_\mathrm{T78}}{E_\mathrm{EH85}} \biggr]^3 = \biggl[\frac{E_\mathrm{T78}}{E_\mathrm{CKST95d}} \biggr]^3</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{j^6}{3(4\pi)^2} \, . 
</math>
  </td>
</tr>
</table>
</td></tr> 
</table>


<span id="tau">Hence, the energy ratio,</span>
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\tau \equiv \frac{T_\mathrm{rot}}{|W_\mathrm{grav}|}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{ A_1 - (1-e^2)A_3 }{  2A_1 + (1-e^2)A_3 }
</math>
  </td>
</tr>
<tr>
  <td align="center" colspan="3">
[<b>[[Appendix/References#T78|<font color="red">T78</font>]]</b>], &sect;4.5, p. 86, Eq. (53)
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{3}{2e^2}\biggl[ 1 - \frac{e(1-e^2)^{1 / 2}}{\sin^{-1} e}\biggr] - 1 </math>
  </td>
</tr>
<tr>
  <td align="center" colspan="3">
[<b>[[Appendix/References#ST83|<font color="red">ST83</font>]]</b>], &sect;7.3, p. 172, Eq. (7.3.24)<br />
[<b>[[Appendix/References#P00|<font color="red">P00</font>]]</b>], Vol. I, &sect;10.3, p. 489, Eq. (10.54)
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{1}{2e^2\sin^{-1} e}\biggl[ (3-2e^2)\sin^{-1} e - 3e(1-e^2)^{1 / 2}\biggr]  
\, .
</math>
  </td>
</tr>
<tr>
  <td align="center" colspan="3">
{{ MPT77 }}, &sect;IVc, p. 594, Eq. (4.4)
  </td>
</tr>
</table>

Building on an [[Apps/MaclaurinSpheroids#Gravitational_Potential|accompanying discussion of the structure of Maclaurin spheroids]], Table 2 &#8212; shown just above, on the right &#8212; lists the limiting values of several key functions.  Note, in particular, that as the eccentricity varies smoothly from zero (spherical configuration) to unity (infinitesimally thin disk), the energy ratio, <math>~\tau</math>, varies smoothly from zero to one-half.  In his examination of the Maclaurin spheroid sequence, [[Appendix/References#T78|Tassoul (1978)]] chose to use this energy ratio as the ''order parameter'', <span id="Figs3and4">rather than the eccentricity</span>.  

<table border="1" align="center"><tr><td align="center">
<table border="0" align="center" cellpadding="3">
<tr>
  <td align="center">'''Figure 3'''</td>
<td align="center" rowspan="3">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</td>
  <td align="center">'''Figure 4'''</td>
</tr>
<tr>
<td align="center">
[[File:T78Fig4.2omega2.png|center|350px|Maclaurin Spheroid Sequence]]
</td>
<td align="center">
[[File:T78Fig4.2angmom.png|center|350px|Maclaurin Spheroid Sequence]]
</td>
</tr>
<tr>
  <td align="center">
Solid black curve also may be found in:
<div align="center">
Fig. 4.2 (p. 88) &amp; Fig. 10.1 (p. 236) of [<b>[[Appendix/References#T78|<font color="red">T78</font>]]</b>]
</div>
  </td>
  <td align="center">
This solid black curve also appears in:
<div align="center">
Fig. 4.2 (p. 88) &amp; Fig. 10.12 (p. 237) of [<b>[[Appendix/References#T78|<font color="red">T78</font>]]</b>]
</div>
  </td>
</tr>
</table>

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

Following Tassoul, our Figure 3 shows how the square of the angular velocity varies with <math>~\tau</math>, and our Figure 4 shows how the system angular momentum varies with <math>~\tau</math>.  In these plots, respectively, the square of the angular velocity has been normalized by <math>~2\pi G \rho</math> &#8212; that is, by a quantity that is a factor of two larger than the normalization adopted in EFE &#8212; while the angular momentum has been normalized to the same quantity used in EFE.  As above, the small solid-green square marker identifies the location along the sequence where the system with the maximum angular velocity resides.

===Angular Velocity or T/|W| ''vs.'' Angular Momentum===

Figures 5 and 6, respectively, show how the square of the angular velocity and how the energy ratio, &tau;, vary with the square of the angular momentum for models along the Maclaurin spheroid sequence.  In generating these plots, following the lead of {{ EH83a }}, we have normalized the square of the angular velocity by <math>~4\pi G \rho</math> &#8212; a factor of four larger than the normalization used in EFE &#8212; and we have adopted a slightly different angular-momentum-squared normalization, namely,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>j^2</math>
  </td>
  <td align="center">
<math>\equiv \frac{L^2}{4\pi G M^{10/3} \rho^{-1 / 3}} = </math>
  </td>
  <td align="left">
<math>
\biggl( \frac{3}{2^8 \pi^4}  \biggr)^{1/3}  \frac{L^2}{(GM^3\bar{a})} \, .
</math>
  </td>
</tr>
</table>

<table border="1" align="center" width="80%" cellpadding="5"><tr><td align="left">
Note that in {{ Wong74 }} &#8212; see the NOTE appended to his Table 2 (p. 686) &#8212; the parameter <math>x</math> provides the measure of the configuration's specific angular momentum; specifically,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>x_\mathrm{Wong74} </math>
  </td>
  <td align="center">
<math>\equiv \frac{25}{12} \biggl(\frac{4\pi}{3}\biggr)^{1 / 3}\frac{L^2\rho^{1 / 3}}{G M^{10/3} } = </math>
  </td>
  <td align="left">
<math>
\frac{5^2}{2^2} \biggl(\frac{4\pi}{3}\biggr)^{4 / 3}  j^2 \, .
</math>
  </td>
</tr>
</table>

----

Alternatively, as has already been [[#MPT77angmom|highlighted above]], {{ MPT77 }} adopt the dimensionless parameter (see their Eq. 4.1),
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>
L_*^2 \equiv 
\biggl(\frac{4\pi}{3}\biggr)^{1 / 3}\frac{L^2}{G M^{10/3} \rho^{-1 / 3}}
=
3 \biggl(\frac{4\pi}{3}\biggr)^{4 / 3} j^2
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{L^2}{(GM^3\bar{a})} \, .
</math>
  </td>
</tr>
</table>
</td></tr></table>

<table border="0" align="center" cellpadding="5"><tr><td align="center">
<table border="0" align="center" cellpadding="3">
<tr>
  <td align="center">'''Figure 5'''</td>
<td align="center" rowspan="3">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</td>
  <td align="center">'''Figure 6'''</td>
</tr>
<tr>
<td align="center">
[[File:EH83Fig3.png|center|350px|Maclaurin Spheroid Sequence]]
</td>
<td align="center">
[[File:EH83Fig4.png|center|350px|Maclaurin Spheroid Sequence]]
</td>
</tr>
<tr>
  <td align="center">
This solid black curve also appears in:
<div align="center">
Fig. 3 (p. 1134) of [https://ui.adsabs.harvard.edu/abs/1983PThPh..69.1131E/abstract Eriguchi &amp; Hachisu (1983)]<br />
Fig. 3 (p. 487) of [https://ui.adsabs.harvard.edu/abs/1986ApJS...61..479H/abstract Hachisu (1986)]<br />
Fig. 4 (p. 4507) of [https://ui.adsabs.harvard.edu/abs/2019MNRAS.487.4504B/abstract Basillais &amp; Hur&eacute; (2019)]
</div>
  </td>
  <td align="center">
This solid black curve also appears in:
<div align="center">
Fig. 4 (p. 487) of [https://ui.adsabs.harvard.edu/abs/1986ApJS...61..479H/abstract Hachisu (1986)]
</div>
  </td>
</tr>
</table>
</td></tr></table>

<span id="OmegaMax">As above,</span> the small solid-green square marker identifies the location along both sequences where the system with the maximum angular velocity resides:
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\biggl[ j^2, \frac{\omega_0^2}{4\pi G \rho}, \tau \biggr]</math>
  </td>
  <td align="center">
<math>\equiv</math>
  </td>
  <td align="left">
<math>\biggl[ 0.010105, 0.112333, 0.237894 \biggr] \, .</math>
  </td>
</tr>
</table>

==Oblate Spheroidal Coordinates==
Following the lead of {{ Bardeen71 }}, {{ HE83 }}, and {{ HE84 }} &#8212; also see the succinct summary that is provided in Appendix A (pp. ) of {{ HTE87 }} &#8212; let's shift to oblate-spheroidal coordinates <math>(\xi, \eta, \phi)</math> which are related to Cartesian coordinates via the relations,
<table border="0" align="center" cellpadding="5">
<tr>
  <td align="right"><math>x</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>a_0\biggl[(1+\xi^2)(1 - \eta^2) \biggr]^{1 / 2} \cos\phi \, ,</math></td>
</tr>
<tr>
  <td align="right"><math>y</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>a_0\biggl[(1+\xi^2)(1 - \eta^2) \biggr]^{1 / 2} \sin\phi \, ,</math></td>
</tr>
<tr>
  <td align="right"><math>z</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>a_0\xi\eta \, .</math></td>
</tr>
</table>
For axisymmetric configurations, such as Maclaurin spheroids, we also appreciate that,
<table border="0" align="center" cellpadding="5">
<tr>
  <td align="right"><math>\varpi \equiv (x^2 + y^2)^{1 / 2}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>a_0 \biggl[(1+\xi^2)(1 - \eta^2) \biggr]^{1 / 2}\, .</math></td>
</tr>
<tr>
  <td align="center" colspan="3">
{{ Bardeen71 }}, §IV, p. 429, Eq. (12)<br />
{{ HE83 }}, &sect;A.1, p. 587, Eq. (1)
  </td>
</tr>
</table>
In this coordinate system, the surface of the Maclaurin spheroid is marked by a specific value of the coordinate, <math>\xi</math> &#8212; call it, <math>\xi_s</math> &#8212; and points along the surface (in any meridional plane) are identified by varying <math>\eta</math> from zero (equatorial plane) to unity (the pole).  Given that the eccentricity of the spheroid is <math>e = [1 - c^2/a^2]^{1 / 2}</math>, we understand that,

<table border="0" align="center" cellpadding="5">
<tr>
  <td align="right"><math>a</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>a_0 (1+\xi_s^2)^{1 / 2}\, ,</math></td>
</tr>
<tr>
  <td align="right"><math>c</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>a_0 \xi_s\, ,</math></td>
</tr>
<tr>
  <td align="right"><math>\Rightarrow ~~~ e^2</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
1 - (a_0 \xi_s)^2 \biggl[a_0^2 (1+\xi_s^2)\biggr]^{-1} 
=
1 - \frac{\xi_s^2}{ (1+\xi_s^2)}
=
\frac{1}{ (1+\xi_s^2)}
</math>
  </td>
</tr>
<tr>
  <td align="center" colspan="3">
{{ Bardeen71 }}, §IV, p. 429, Eq. (14)
  </td>
</tr>
<tr>
  <td align="right"><math>\Rightarrow ~~~ \xi_s^2</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{1}{ e^2} - 1
\, .</math>
  </td>
</tr>
</table>
Also, in order for the volume of the spheroid to remain constant &#8212; and equal to that of a sphere of the same total mass and density &#8212; along the sequence of spheroids we understand that,
<table border="0" align="center" cellpadding="8">
<tr>
  <td align="right"><math>\frac{M}{\rho} = \frac{4\pi a^2c}{3}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>\frac{4\pi}{3} a^3 \biggl(\frac{c}{a}\biggr)
=
\frac{4\pi}{3} a^3 \biggl[1 - e^2\biggr]^{1 / 2}
</math>
  </td>
</tr>
<tr>
  <td align="right"><math>\Rightarrow ~~~ \frac{3M}{4\pi\rho} </math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
a_0^3 (1+\xi_s^2)^{3 / 2} \biggl\{
\xi_s^2 \biggl[(1+\xi_s^2)\biggr]^{-1}
\biggr\}^{1 / 2}
</math>
  </td>
</tr>
<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
a_0^3 \xi_s (1+\xi_s^2) 
</math>
  </td>
</tr>
<tr>
  <td align="center" colspan="3">
{{ HE83 }}, &sect;A.2, p. 588, Eq. (10)<br />
{{ HTE87 }}, p. 610, Eq. (A5)
  </td>
</tr>
<tr>
  <td align="right"><math>\Rightarrow ~~~ a_0^3 </math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\biggl(\frac{3M}{4\pi\rho}\biggr) \biggl[ \xi_s (1+\xi_s^2) \biggr]^{-1}
=
\biggl(\frac{3M}{4\pi\rho}\biggr) \frac{e^3}{(1-e^2)^{1 / 2}} \, .
</math>
  </td>
</tr>
</table>


<table border="1" align="center" width="90%" cellpadding="8">
<tr><td align="left">

From Appendix A of {{ HTE87 }} &#8212; hereafter {{ HTE87hereafter }} &#8212; we also appreciate that,
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right"><math>\Omega^2 \equiv \frac{\omega_0^2}{4\pi G \rho}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\xi q_2(\xi) \, ,
</math>
  </td>
</tr>
<tr>
  <td align="center" colspan="3">
{{ HE83 }}, &sect;A.2, p. 588, Eq. (9)<br />
{{ HTE87hereafter }}, p. 610, Eq. (A4)
  </td>
</tr>

<tr>
  <td align="right"><math>L</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\biggl(\frac{8\pi}{15}\biggr) \rho \omega_0 a_0^5 \xi(1+\xi^2)^2 \, ,
</math>
  </td>
</tr>
<tr>
  <td align="center" colspan="3">
{{ HE83 }}, &sect;A.2, p. 588, Eq. (11)<br />
{{ HTE87hereafter }}, p. 610, Eq. (A6)
  </td>
</tr>

<tr>
  <td align="right"><math>T_\mathrm{rot}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\biggl(\frac{4\pi}{15}\biggr) \rho \omega_0^2 a_0^5 \xi(1+\xi^2)^2 \, ,
</math>
  </td>
</tr>
<tr>
  <td align="center" colspan="3">
{{ HTE87hereafter }}, p. 610, Eq. (A7)
  </td>
</tr>

<tr>
  <td align="right"><math>W_\mathrm{grav}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
- \biggl(\frac{16\pi^2}{15}\biggr) G \rho^2 a_0^5 \xi^2 (1+\xi^2)^2 q_0(\xi)\, ,
</math>
  </td>
</tr>
<tr>
  <td align="center" colspan="3">
{{ HTE87hereafter }}, p. 610, Eq. (A8)
  </td>
</tr>

<tr>
  <td align="right"><math>\Rightarrow ~~~ \frac{T_\mathrm{rot}}{|W_\mathrm{grav}|}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\biggl(\frac{4\pi}{15}\biggr) \rho \omega_0^2 a_0^5 \xi(1+\xi^2)^2
\biggl[ \biggl(\frac{16\pi^2}{15}\biggr) G \rho^2 a_0^5 \xi^2 (1+\xi^2)^2 q_0(\xi) \biggr]^{-1}
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\biggl[ \frac{\omega_0^2}{4\pi G \rho}\biggr] \frac{1}{\xi q_0(\xi)}
=
\frac{q_2(\xi)}{q_0(\xi)} \, ,
</math>
  </td>
</tr>
</table>
where &#8212; see Eqs. (A15) - (A17) of {{ HTE87hereafter }} and Appendix A (p. 443) of {{ Bardeen71hereafter }}&#8212; the first three spheroidal wave functions of the second kind are,
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right"><math>q_0(\xi)</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\tan^{-1}(1/\xi) \, ,
</math>
  </td>
</tr>
<tr>
  <td align="center" colspan="3">
{{ HTE87hereafter }}, p. 610, Eq. (A15)
  </td>
</tr>

<tr>
  <td align="right"><math>q_1(\xi)</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
-\xi \tan^{-1}(1/\xi) + 1\, ,
</math>
  </td>
</tr>
<tr>
  <td align="center" colspan="3">
{{ HTE87hereafter }}, p. 610, Eq. (A16)
  </td>
</tr>

<tr>
  <td align="right"><math>q_2(\xi)</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{1}{2}\biggl[(3\xi^2 + 1)\tan^{-1}(1/\xi) - 3\xi\biggr] 
\, .
</math>
  </td>
</tr>
<tr>
  <td align="center" colspan="3">
{{ HTE87hereafter }}, p. 610, Eq. (A17)
  </td>
</tr>
</table>

<hr />

'''Check:'''  &nbsp;Given that, <math>\xi^2 = (1-e^2)/e^2</math>, we have,

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

<tr>
  <td align="right"><math>\tan^{-1}(1/\xi)</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\sin^{-1}\biggl[ \frac{1}{\sqrt{\xi^2+1}} \biggr] = \sin^{-1}e\, ,
</math>
  </td>
</tr>
</table>
in which case:

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

<tr>
  <td align="right"><math>\frac{\omega_0^2}{2\pi G\rho}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{(1 - e^2)^{1 / 2}}{e}\biggl\{ \biggl[\frac{3(1-e^2)}{e^2} + 1\biggr]\sin^{-1}e - \frac{3(1-e^2)^{1 / 2}}{e} \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\biggl[(3-2e^2)(1 - e^2)^{1 / 2} \biggr]\cdot \frac{\sin^{-1}e}{e^3} - \frac{3(1-e^2)}{e^2} \, ;
</math>
&nbsp; &nbsp; &nbsp; &nbsp; [[Apps/MaclaurinSpheroidSequence#MaclaurinFrequency|(matches here)]]
  </td>
</tr>

<tr>
  <td align="right">
<math>
L_*^2  \equiv \biggl(\frac{4\pi}{3}\biggr)^{1 / 3}\frac{L^2}{G M^{10/3} \rho^{-1 / 3}}
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl(\frac{4\pi}{3}\biggr)^{1 / 3} G^{-1} M^{-10/3}\rho^{1 / 3}
\biggl(\frac{2^3\pi}{3 \cdot 5}\biggr)^2 \rho^2 a_0^{10} \omega_0^2 
\biggl[ \frac{(1-e^2)}{e^{10}}\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl(\frac{4\pi}{3}\biggr)^{1 / 3} G^{-1} M^{-10/3}\rho^{1 / 3}
\biggl(\frac{2^3\pi}{3 \cdot 5}\biggr)^2 \rho^2 
\biggl[ \frac{(1-e^2)}{e^{10}}\biggr]
\biggl[ \biggl(\frac{3M}{4\pi\rho}\biggr)\frac{e^3}{(1-e^2)^{1 / 2}}\biggr]^{10/3} \omega_0^2
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{6}{25}  
\biggl[\frac{\omega_0^2}{2\pi G\rho}\biggr] (1-e^2)^{- 2/3}\, ;
</math>
&nbsp; &nbsp; &nbsp; &nbsp; [[Apps/MaclaurinToroid#Maclaurin_Spheroid_Reminder|(matches here)]]
  </td>
</tr>

<tr>
  <td align="right"><math>T_\mathrm{rot}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\biggl(\frac{4\pi}{15}\biggr) 2\pi G \rho^2 a_0^5 \biggl[ \frac{\omega_0^2}{2\pi G \rho} \biggr] \xi(1+\xi^2)^2 
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\biggl(\frac{2^3\pi^2}{3 \cdot 5}\biggr) G \rho^2 
(a \cdot e)^5
\biggl[ \frac{\omega_0^2}{2\pi G \rho} \biggr] 
\biggl[\frac{(1-e^2)^{1 / 2}}{e^5}\biggr]  
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\biggl(\frac{2^3\pi^2}{3 \cdot 5}\biggr) G \rho^2 a^5
\biggl[ \frac{\omega_0^2}{2\pi G \rho} \biggr] 
(1-e^2)^{1 / 2} \, ;  
</math>
&nbsp; &nbsp; &nbsp; &nbsp; [[Apps/MaclaurinSpheroidSequence#Alternate_Sequence_Diagrams|(matches here)]]
  </td>
</tr>

<tr>
  <td align="right"><math>W_\mathrm{grav}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
- \biggl(\frac{16\pi^2}{15}\biggr) G \rho^2 (a \cdot e)^5 \biggl[ \frac{(1-e^2)}{e^6} \cdot \sin^{-1}e\biggr]
</math>
  </td>
</tr>
<tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
- \biggl(\frac{16\pi^2}{15}\biggr) G \rho^2 a^5 (1-e^2)\cdot \frac{\sin^{-1}e}{e} \, ;
</math>
&nbsp; &nbsp; &nbsp; &nbsp; [[Apps/MaclaurinSpheroidSequence#Alternate_Sequence_Diagrams|(matches here)]]
  </td>
</tr>
<tr>

<tr>
  <td align="right"><math>\frac{T_\mathrm{rot}}{|W_\mathrm{grav}|} = \frac{q_2(\xi)}{q_0(\xi)}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{1}{2\tan^{-1}(1/\xi)}\biggl[(3\xi^2 + 1)\tan^{-1}(1/\xi) - 3\xi\biggr] 
</math>
  </td>
</tr>

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

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{1}{2e^2\sin^{-1}e}\biggl[(3-2e^2)\sin^{-1}e - 3e(1-e^2)^{1 / 2}\biggr] \, .</math>
&nbsp; &nbsp; &nbsp; &nbsp; [[Apps/MaclaurinSpheroidSequence#tau|(matches here)]]
  </td>
</tr>
</table>

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

=Bifurcation Points Along Maclaurin-Spheroid Sequence=

==The Perturbed Configuration==

Referencing the [[AxisymmetricConfigurations/HSCF#Constructing_Two-Dimensional,_Axisymmetric_Structures|Hachisu Self-Consistent Field (HSCF) technique]], our objective is to solve an ''algebraic'' expression for hydrostatic balance,
<div align="center">
<math>~H + \Phi + \Psi = C_0</math> ,
</div>
in conjunction with the Poisson equation in a form that is appropriate for two-dimensional, axisymmetric systems &#8212; written in cylindrical coordinates, for example,
<div align="center">
<math>~
\frac{1}{\varpi} \frac{\partial }{\partial\varpi} \biggl[ \varpi \frac{\partial \Phi}{\partial\varpi} \biggr] + \frac{\partial^2 \Phi}{\partial z^2} = 4\pi G \rho .
</math>
</div>
In both of these expressions, <math>\Phi</math> is the gravitational potential.  In the algebraic expression, <math>C_0</math> is a constant throughout the volume, and on the surface, of the equilibrium configuration.  Here, we seek a uniform-density (incompressible) configuration, in which [[SR#Time-Independent_Problems|the enthalpy]], <math>H = P/\rho</math>, goes to zero at all points across the surface.  And the centrifugal potential, <math>\Psi</math>, [[AxisymmetricConfigurations/SolutionStrategies#Simple_Rotation_Profile_and_Centrifugal_Potential|is given by the expression]]

<table border="0" align="center" cellpadding="5">
<tr>
  <td align="right" width="40%"><math>\Psi</math></td>
  <td align="center"><math>\equiv</math></td>
  <td align="left"><math>- ~\int \frac{h^2(\varpi)}{\varpi^3} d\varpi \, ,</math></td>
</tr>
<tr>
  <td align="center" colspan="3">
{{ OM68 }}, &sect;IIId (p. 1084), eq. (44)<br />
{{ MPT77 }}, &sect;III (p. 590), eq. (3.4)
</td>
</tr>
</table>
where, the (cylindrical) radial distribution of the specific angular momentum,
<div align="center">
<math>h(\varpi) = \varpi^2 \dot\varphi(\varpi) \, ,</math>
</div>
is to be specified according to the physical problem in hand &#8212; usually chosen from a familiar set of "[[AxisymmetricConfigurations/SolutionStrategies#SRPtable|simple rotation profiles]]."  Therefore, across the surface of each equilibrium configuration, the algebraic expression for hydrostatic balance takes the form,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right"><math>\Phi - \int \frac{h^2(\varpi)}{\varpi^3} d\varpi</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>C_0 \, .</math></td>
</tr>
<tr>
  <td align="center" colspan="3">
{{ EH85 }}, &sect;2.1 (p. 290), Eq. (5)
  </td>
</tr>
</table>

===Uniform Rotation===

{{ HE83 }} sought to find bifurcation points along the Maclaurin spheroid sequence where the associated, deformed equilibrium configuration is uniformly rotating.  From the set of familiar [[AxisymmetricConfigurations/SolutionStrategies#SRPtable|simple rotation profiles]], therefore, they set
<div align="center">
<math>
\Psi = - \frac{1}{2} \varpi^2 \omega_0^2 \, ,
</math>
</div>
in which case, in their investigation, the condition (along the surface) for hydrostatic balance is,

<div align="center">
<math>\Phi - \frac{1}{2} \varpi^2 \omega_0^2 = C_0</math> .
</div>
Replacing <math>\varpi^2</math> with its equivalent expression in terms of oblate spheroidal coordinates gives,
<table border="0" align="center" cellpadding="5">
<tr>
  <td align="right">
<math>- ~\Phi + \frac{\omega_0^2}{2} a_0^2 (1 + \xi^2)(1 - \eta^2) </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>- ~C_0 \, ,</math>
  </td>
</tr>
<tr>
  <td align="center" colspan="3">
{{ HE83 }}, &sect;A.1 (p. 587), Eq. (6)
  </td>
</tr>
</table>
which is the same as their Appendix (&sect;A.1) Eq. (6), except they chose a different sign when defining the constant, <math>C_0</math>.


===n' = 0 Configurations===

{{ EH85 }} sought to find bifurcation points along the Maclaurin spheroid sequence where the associated, deformed equilibrium configuration has the same radial distribution of specific angular momentum &#8212; as a function of the integrated mass fraction &#8212; as does a uniformly rotating, uniform density sphere.  That is, inside the integral that defines the centrifugal potential, <math>\Psi</math>, they set,

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

<tr>
  <td align="right">
<math>h(\varpi) = \varpi^2 \dot\varphi(\varpi)</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{5J}{2M}\biggl\{ 1 - [1 - m(\varpi) ]^{2 / 3} \biggr\} \, ,
</math>
  </td>
</tr>
<tr>
  <td align="center" colspan="3">
{{ Stoeckly65 }}, &sect;II.c, eq. (12)<br />
{{ EH85 }}, &sect;2.1 (p. 290), Eq. (1)
  </td>
</tr>
</table>

where, the mass fraction,
<div align="center">
<math>m(\varpi) \equiv \frac{M_\varpi(\varpi)}{M} \, .</math>
</div>

From our example set of familiar [[AxisymmetricConfigurations/SolutionStrategies#SRPtable|simple rotation profiles]], these might reasonably be referred to as [[AxisymmetricConfigurations/SolutionStrategies#Uniform-Density_Initially_(n'_=_0)|<math>n' = 0</math> configurations]].  Instead, {{ EH85 }} label their deformed equilibrium configurations as follows:  "The Maclaurin spheroidal sequence bifurcates into a ''concave hamburger like'' configuration and reaches &#8212; as originally discovered and labeled by {{ MPT77 }} &#8212; the ''Maclaurin toroidal'' sequence."

==Particularly Interesting Models Along the Maclaurin Spheroid Sequence==

<table border="1" cellpadding="8" align="center" width="60%"><tr><td align="center" bgcolor="lightblue">
Go to our associated discussion of [[Appendix/Ramblings/MacSphCriticalPoints|Critical Points along the Maclaurin Spheroid Sequence]].
</td></tr></table>

=See Also=
* [[Apps/MaclaurinSpheroids#Maclaurin_Spheroids_.28axisymmetric_structure.29|Properties of Maclaurin Spheroids]]
* [[Apps/MaclaurinSpheroids/GoogleBooks#Excerpts_from_A_Treatise_of_Fluxions|Excerpts from Maclaurin's (1742) ''A Treatise of Fluxions'']]
* [[ThreeDimensionalConfigurations/EFE_Energies#Properties_of_Homogeneous_Ellipsoids_.282.29|Properties of Homogeneous Ellipsoids]]
* Equilibrium Configurations and Sequences Generated by [[Apps/EriguchiHachisu/Models|Eriguchi, Hachisu, and their various colleagues]]


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