Editing ThreeDimensionalConfigurations/RiemannStype (section)

===Basic Relations===

====Expressions Supplied by EFE====

In Chapter 7, &sect;48(b) (pp. 137 - 138) of [[Appendix/References#EFE|[<font color="red">EFE</font>] ]], Chandrasekhar shows that <font color="orange">"&hellip; a ''stable'' [[Apps/MaclaurinSpheroidSequence|Maclaurin spheroid]]</font> can be considered as <font color="orange">a limiting <math>(a_2/a_1 \rightarrow 1)</math> form of a [S-type] Riemann ellipsoid."</font>  First, we [[Apps/MaclaurinSpheroidSequence#Equilibrium_Angular_Velocity|recall]] that, as viewed from the inertial frame, each Maclaurin spheroid of eccentricity, <math>e = (1 - a_3^2/a_1^2)^{1 / 2}</math>, rotates uniformly with angular velocity,
<table border="0" align="center" cellpadding="8">
<tr>
  <td align="right"><math>\Omega_\mathrm{Mc}^2 \equiv \frac{\omega_0^2}{\pi G \rho} = 2e^2 B_{13} </math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
2(3-2e^2)(1 - e^2)^{1 / 2} \cdot \frac{\sin^{-1} e}{e^3} - \frac{6(1-e^2)}{e^2} \, .</math>
  </td>
</tr>
<tr>
  <td align="center" colspan="3">
[<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>], &sect;32, pp. 77-78, Eqs. (4) &amp; (6)
</td>
</tr>
</table>

Given the specified value of the semi-axis ratio, <math>a_3/a_1</math>, the properties of the limiting S-type Riemann ellipsoid are given by the expressions,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>2\Omega_3</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\Omega_\mathrm{Mc} \pm \biggl[4 B_{11} - \Omega^2_\mathrm{Mc}\biggr]^{1 / 2} \, ,</math>
  </td>
</tr>
<tr><td align="center" colspan="3">[<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>], <font color="#00CC00">Chapter 7, &sect;48(b), p. 137, Eq. (60)</font> </td></tr>
</table>

<span id="zeta3">and,</span>
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\zeta_3^2</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>4(\Omega_\mathrm{Mc} - \Omega_3)^2 \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ \zeta_3</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\Omega_\mathrm{Mc} \mp \biggl[ 4B_{11} - \Omega^2_\mathrm{Mc} \biggr]^{1 / 2}
\, .</math>
  </td>
</tr>

<tr><td align="center" colspan="3">[<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>], <font color="#00CC00">Chapter 7, &sect;48(b), p. 137, Eqs. (58) &amp; (61)</font> </td></tr>
</table>
<span id="B11">where, </span>
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>B_{11}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>A_1 - a_1^2 A_{11} \, .</math>
  </td>
</tr>
<tr><td align="center" colspan="3">[<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>], <font color="#00CC00">Chapter 3, &sect;21, Eq. (105)</font></td></tr>
</table>

====Evaluating A<sub>11</sub>====

As has been stated in [[ThreeDimensionalConfigurations/HomogeneousEllipsoids#Evaluating_Aℓℓ|a separate derivation]], quite generally,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\frac{A_{\ell\ell}}{a_\ell a_m a_s}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\int_0^\infty \biggl[ (a_\ell^2 + u)^5(a_m^2 + u)(a_s^2 + u) \biggr]^{-1 / 2} du \, .
</math>
  </td>
</tr>
</table>
Now, in the case of Maclaurin spheroids, <math>a_m = a_\ell</math>, so this integral expression becomes,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\frac{A_{\ell\ell}}{a_\ell^2 a_s}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\int_0^\infty \biggl[ (a_\ell^2 + u)^6 (a_s^2 + u) \biggr]^{-1 / 2} du
=
\int_0^\infty \frac{du}{(a_\ell^2 + u)^3 (a_s^2 + u)^{1 / 2} } 
=
\int_0^\infty \frac{du}{v^m \sqrt{w}} \, ,
</math>
  </td>
</tr>
</table>
where, <math>m = 3, v = (a_\ell^2 + u), w = (a_s^2 + u), b = d = 1, a = a_s^2, c = a_\ell^2, k = (a_s^2 - a_\ell^2).</math>  Before applying the limits, carrying out the integration gives,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\int \frac{du}{v^m \sqrt{w}}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
-~\frac{1}{(m-1)(a_s^2 - a_\ell^2)} \biggl[
\frac{\sqrt{w}}{v^{m-1}} + \biggl(m - \frac{3}{2}\biggr) \int \frac{du}{v^{m-1} \sqrt{w}}
\biggr]
</math>
  </td>
</tr>
<tr>
  <td align="center" colspan="3">
[<b>[[Appendix/References#CRC|<font color="red">CRC</font>]]</b>] chapter on INTEGRALS, p. 408, Eq. (154)<br />
[https://www.academia.edu/36550954/I_S_Gradshteyn_and_I_M_Ryzhik_Table_of_integrals_series_and_products_Academic_Press_2007_ GR7<sup>th</sup>], p. 91, Eq. (2.248.8)</td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{1}{2(a_\ell^2 - a_s^2)} \biggl[
\frac{\sqrt{w}}{v^{2}} + \biggl(\frac{3}{2}\biggr) \int \frac{du}{v^{2} \sqrt{w}}
\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{1}{2(a_\ell^2 - a_s^2)} \biggl[
\frac{\sqrt{w}}{v^{2}}\biggr] 
+
\frac{3}{4(a_\ell^2 - a_s^2)} 
\biggl\{
\frac{1}{(a_\ell^2 - a_s^2)} \biggl[
\frac{\sqrt{w}}{v} + \biggl(\frac{1}{2}\biggr) \int \frac{du}{v \sqrt{w}}
\biggr]\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{1}{2(a_\ell^2 - a_s^2)} \biggl[\frac{\sqrt{w}}{v^{2}}\biggr] 
+
\frac{3}{4(a_\ell^2 - a_s^2)^2} \biggl[\frac{\sqrt{w}}{v}\biggr] 
+
\frac{3}{8(a_\ell^2 - a_s^2)^2} 
\biggl\{
\int \frac{du}{v \sqrt{w}}
\biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="center" colspan="3">
[<b>[[Appendix/References#CRC|<font color="red">CRC</font>]]</b>] chapter on INTEGRALS, p. 407, Eq. (148)<br />
[https://www.academia.edu/36550954/I_S_Gradshteyn_and_I_M_Ryzhik_Table_of_integrals_series_and_products_Academic_Press_2007_ GR7<sup>th</sup>], p. 90, Eq. (2.246)
</td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{1}{2(a_\ell^2 - a_s^2)} \biggl[\frac{\sqrt{w}}{v^{2}}\biggr] 
+
\frac{3}{4(a_\ell^2 - a_s^2)^2} \biggl[\frac{\sqrt{w}}{v}\biggr] 
+
\frac{3}{8(a_\ell^2 - a_s^2)^2} 
\biggl\{
\frac{2}{(a_\ell^2 - a_s^2)^{1 / 2}} \tan^{-1}\biggl[ \frac{\sqrt{w}}{(a_\ell^2 - a_s^2)^{1 / 2}} \biggr]
\biggr\}\, .
</math>
  </td>
</tr>
</table>
Applying the limits then gives,

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

<tr>
  <td align="right">
<math>\int_0^\infty \frac{du}{v^m \sqrt{w}}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{3\pi}{8(a_\ell^2 - a_s^2)^{5 / 2}} 
-~\frac{a_s}{2(a_\ell^2 - a_s^2)a_\ell^4}  
-
\frac{3a_s}{4(a_\ell^2 - a_s^2)^2 a_\ell^2}  
-
\frac{3}{4(a_\ell^2 - a_s^2)^{5 / 2}} 
\biggl\{
\tan^{-1}\biggl[ \frac{a_s}{(a_\ell^2 - a_s^2)^{1 / 2}} \biggr]
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{3\pi}{8 a_\ell^5 e^5} 
-~\frac{(1-e^2)^{1 / 2}}{2a_\ell^5 e^2}  
-
\frac{3(1-e^2)^{1 / 2}}{4a_\ell^5 e^4 }  
-
\frac{3}{4a_\ell^5 e^5} 
\biggl\{
\sin^{-1}(1-e^2)^{1 / 2}
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
1.49957 - 0.16703 - 0.27593 - 0.29421 = 0.76239
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ \frac{A_{\ell\ell} }{a_\ell^3 (1 - e^2)^{1 / 2} }</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{1}{4a_\ell^5 e^5}\biggl\{
\frac{3\pi}{2 } 
-3 \biggl[\sin^{-1}(1-e^2)^{1 / 2}\biggr]
-~2e^3(1-e^2)^{1 / 2}  
- 3e(1-e^2)^{1 / 2}
\biggr\}  
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ a_\ell^2 A_{\ell\ell} </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{3(1-e^2)^{1 / 2}}{4 e^5}
\underbrace{\biggl[ \frac{\pi}{2 } - \sin^{-1}(1-e^2)^{1 / 2}\biggr]}_{\sin^{-1}e}
-  
\frac{1}{4 e^4}( 2e^2   + 3)(1-e^2)  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
0.36562 - 0.13436 = 0.23126
</math>
  </td>
</tr>
</table>

====Derivation Check====

At the top of p. 141 (&sect;48) of [<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>] &#8212; see also Table VI on p. 142 &#8212; Chandrasekhar emphasizes that the point at which the <math>x = +1</math> self-adjoint (S-type) Riemann ellipsoid intersects the Maclaurin spheroid has <math>f = +2, a_3 = 0.30333, e = 0.95289, \Omega_3^2 = 0.11005</math>. As Table I (p. 78) records, the Maclaurin spheroid having this eccentricity has <math>\Omega_\mathrm{Mc}^2 = 0.44022.</math>  At this point, therefore,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>B_{11}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\tfrac{1}{4}\Omega^2_\mathrm{Mc} = 0.110055 \, .</math>
  </td>
</tr>
<tr><td align="center" colspan="3">[<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>], <font color="#00CC00">Chapter 7, &sect;48(c), Eq. (141)</font></td></tr>
</table>

Note as well that, from the [[#zeta3|above expression for (&zeta;<sub>3</sub>)<sup>2</sup>]], we have,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\zeta_3 </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>2(\Omega_\mathrm{Mc}-\Omega_3) = 0.66034 \, ;</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ f </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\frac{\zeta_3}{\Omega_3} = 2.0 \, ,</math>
  </td>
</tr>
</table>
as it should be.

For this same value of the eccentricity, we recognize as well that, <math>A_{1} = 0.34132</math>.  Therefore, drawing from the [[#B11|above expression for B<sub>11</sub>]], we also deduce that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>a_1^2 A_{11}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>A_1 - B_{11} = 0.23127 \, .</math>
  </td>
</tr>
<tr><td align="center" colspan="3">[<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>], <font color="#00CC00">Chapter 3, &sect;21, Eq. (105)</font></td></tr>
</table>
<font color="red"><b>Hooray!!</b></font> &nbsp; This matches the numerical value of <math>a_\ell^2 A_{\ell\ell} = 0.23126</math> determined above for <math>e = 0.95289</math>.  So it seems reasonable to conclude that the general expression, itself, is correct.