Editing Apps/MaclaurinToroid (section)

==Maclaurin Spheroid Reminder==

<span id="L*">As has been demonstrated</span> in our [[Apps/MaclaurinSpheroidSequence#Corresponding_Total_Angular_Momentum|accompanying discussion of the Maclaurin spheroid sequence]], the (square of the) normalized angular momentum that is associated with a spheroid of eccentricity, <math>e \equiv (1 - c^2/a^2)^{1 / 2}</math>, is,

<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>
In that [[Apps/MaclaurinSpheroidSequence#tau|same discussion]], we have demonstrated that the corresponding ratio of rotational to gravitational potential energy is given by the expression,
<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{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>

[[Apps/MaclaurinSpheroidSequence#Figs3and4|Figure 4 from this accompanying discussion]] shows how <math>L_*</math> varies with <math>\tau</math> along the Maclaurin Spheroid sequence.  In an effort to conform to {{ MPT77hereafter }}'s presentation, our Figure 1 (immediately below) displays the same information as displayed in Figure 4 of this separate chapter, but the axes have been swapped and the maximum displayed value of <math>L_*</math> has been extended from 1 to 1.5.

<table border="1" align="center"><tr><td align="center">
<table border="0" align="center" cellpadding="0">
<tr>
<td align="center" rowspan="2">
<b>EFE Diagram</b><br />
[[File:OurEFEannotated.png|300px|OurEFE]]
</td>
<td align="center">
&nbsp;<br /><b>Figure 1</b><br />
[[File:MPT77fiveModified.png|300px|MPT77five]]
</td>
<td align="center">
&nbsp;<br /><b>Figure 2</b><br />
[[File:MPT77sixModified.png|300px|MPT77six]]
</td>
</tr>
<tr>
  <td align="center" colspan="2">
The multicolor curve that appears here in Figures 1 and 2 also appears as a solid black curve in, respectively,
<br />Fig. 5 (p. 594) and Fig. 4 (p. 593) of {{ MPT77 }}
  </td>
</tr>
</table>

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

{{ MPT77hereafter }} also evaluate the normalized total energy, <math>E_\mathrm{tot}/|E_0|</math>, of each of their constructed equilibrium configurations, where
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>E_\mathrm{tot}</math>
  </td>
  <td align="center">
<math>\equiv</math>
  </td>
  <td align="left">
<math>
T_\mathrm{rot} + W_\mathrm{grav} \, ,
</math>
  </td>
</tr>
</table>
and, according to the caption of their Figure 4, <math>E_0</math> is <font color="darkgreen">"&hellip; the energy of a nonrotating sphere of equal mass and volume."</font>  Drawing from our [[Apps/MaclaurinSpheroidSequence#EnergyNorm|separate discussion of the Maclaurin spheroid sequence]], it would be reasonable to assume that the energy normalization adopted by {{ MPT77hereafter }} is the same as the normalization used by [<b>[[Appendix/References#T78|<font color="red">T78</font>]]</b>], namely,
<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>
\biggl(\frac{4\pi}{3}\biggr)^{1 / 3}G (M^5 \rho)^{1 / 3} \, .
</math>
  </td>
</tr>
</table>
For models along the Maclaurin spheroid sequence, this normalization leads to expressions for the two key energy terms of the form,
<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>

<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>
in which case, in the limit of a nonrotating sphere, 
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\lim_{e\rightarrow 0}\biggl[ \frac{T_\mathrm{rot} + W_\mathrm{grav}}{E_\mathrm{T78}}\biggr]</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math> 
- \frac{3}{5} \, .
</math>
  </td>
</tr>
</table>
But in Figure 4 of {{ MPT77hereafter }}, the point along the Maclaurin spheroid sequence &#8212; the solid, black curve &#8212; that represents a nonrotating <math>(L_* = 0)</math> sphere has a normalized energy, <math>(E_\mathrm{tot}/E_0) = -1.</math>  We conclude, therefore, that the normalization adopted by {{ MPT77hereafter }} is,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>E_0</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math> 
\tfrac{3}{5}E_\mathrm{T78} \, .
</math>
  </td>
</tr>
</table>
Our Figure 2 (immediately above) attempts to quantitatively replicate the behavior of the Maclaurin spheroid sequence that is shown in Figure 4 (p. 213) of {{ MPT77hereafter }}; the ordinate depicts, on a base-10 logarithmic scale, how the total energy varies with the spheroid's angular momentum over the range, <math>0 \le L_* \le 1.50</math>.  More specifically, for eccentricities over the range, <math>0 \le e \le 0.99998967881</math>, the corresponding value of the spheroid's normalized angular momentum is obtained from the [[#L*|above expression for]] <math>L_*</math>, and the normalized energy is given by the relation,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\frac{E_\mathrm{tot}}{E_0}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math> 
\frac{5}{3}\biggl[ \frac{T_\mathrm{rot}}{E_\mathrm{T78}} + \frac{W_\mathrm{grav}}{E_\mathrm{T78}}\biggr]
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math> 
\frac{(1-e^2)^{1 / 6}}{2e^2}\biggl\{
\biggl[ (3 - 2e^2)\frac{\sin^{-1}e}{e} - 3(1-e^2)^{1 / 2} \biggr] 
- 
\frac{2e^2\sin^{-1}e }{e} 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math> 
\frac{(1-e^2)^{1 / 6}}{2e^2}
\biggl[ (3 - 4e^2)\frac{\sin^{-1}e}{e} 
- 3(1-e^2)^{1 / 2} \biggr] 
\, .
</math>
  </td>
</tr>
</table>