Editing Appendix/Ramblings/BordeauxSequences (section)

===Model Sequences===

====Fixed &Omega;====
<span id="jSquared">The Maclaurin sequence (MLS) is defined by the relations:</span>
<table border="0" cellpadding="5" align="center">

<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>~
\frac{1}{2}(3 - 2e^2)(1 - e^2)^{1 / 2} \cdot \frac{\sin^{-1} e}{e^3} - \frac{3(1-e^2)}{2e^2} \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~j^2 \equiv \biggl( \frac{3}{2^8 \pi^4} \biggr)^{1 / 3} \frac{L_c^2}{G M_c^3 } \cdot R_\mathrm{eq}^{-1} (1-e^2)^{-1 / 6}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl( \frac{3}{2^8 \pi^4} \biggr)^{1 / 3} \frac{\Omega^2}{G R_\mathrm{eq}(1-e^2)^{1 / 6} }  
\biggl[ \biggl( \frac{2^8 \pi^3}{3^2 \cdot 5^2}\biggr) (1 - e^2) G R_\mathrm{eq}^{10} \rho^3   \biggr]
\biggl[ \frac{4\pi}{3} \rho R_\mathrm{eq}^3 (1 - e^2)^{1 / 2} \biggr]^{-3}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl( \frac{2^{24} \pi^9}{3^6 \cdot 5^6}\biggr)^{1 / 3} \biggl( \frac{3^9}{2^{18}\pi^9}\biggr)^{1 / 3}
\biggl( \frac{3}{2^8 \pi^4} \biggr)^{1 / 3} \frac{\Omega^2}{(1-e^2)^{2 / 3} }  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl( \frac{3^4}{2^2 \cdot 5^6\pi^4}\biggr)^{1 / 3}
\frac{\Omega^2}{(1-e^2)^{2 / 3} }  \, .
</math>
  </td>
</tr>
</table>

=====Central Object=====
Assume that the central object is exactly a Maclaurin spheroid.  Then from [[Apps/MaclaurinSpheroidSequence#Equilibrium_Angular_Velocity|Figure 1 (and Table 1) of our review of equilibrium models along the Maclaurin spheroid sequence]], we appreciate that all we have to do is specify the eccentricity, <math>~0 \le e \le 1</math>, and <math>~\Omega^2 \equiv \omega_0^2/(4\pi G\rho)</math> is known.  For example, if we choose <math>~e = 0.60</math>, then from that Table 1, <math>~\omega_0^2/(2\pi G\rho) = 0.1007~~\Rightarrow~~\Omega^2 \approx 0.05</math>.  Other properties of this "central" spheroid &#8212; such as its mass, moment of inertia, and angular momentum &#8212; are given by the following expressions:
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~M_c</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{4\pi}{3} \rho R_\mathrm{eq}^2 Z = \frac{4\pi}{3} \rho R_\mathrm{eq}^3 (1 - e^2)^{1 / 2} \, ;</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~I_c</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2}{5} M_c R_\mathrm{eq}^2 \, ;</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~L_c</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~I_c \omega_0 = \frac{2}{5} M_c R_\mathrm{eq}^2 \omega_0 </math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2^3\pi}{3\cdot 5} (1 - e^2)^{1 / 2} R_\mathrm{eq}^5 \rho \omega_0 </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~L^2_c</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~4\pi G\biggl( \frac{2^3\pi}{3\cdot 5}\biggr)^2 (1 - e^2) R_\mathrm{eq}^{10} \rho^3 \Omega^2 </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl( \frac{2^8 \pi^3}{3^2 \cdot 5^2}\biggr) (1 - e^2) G R_\mathrm{eq}^{10} \rho^3 \Omega^2 \, . </math>
  </td>
</tr>
</table>

We note as well that the (square of the) Keplerian frequency for a massless particle orbiting in the equatorial plane at a distance, <math>~r</math>, from the center of this central object will be,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\omega_K^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{GM_c}{r^3}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{G}{r^3} \biggl[ \frac{4\pi}{3} \rho R_\mathrm{eq}^3 (1 - e^2)^{1 / 2} \biggr] \, .</math>
  </td>
</tr>
</table>

So, if we force this orbital frequency to also equal the spin-frequency of the Maclaurin spheroid, the radius of the orbit must be,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\Omega^2 (4\pi G\rho)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{G}{r^3} \biggl[ \frac{4\pi}{3} \rho R_\mathrm{eq}^3 (1 - e^2)^{1 / 2} \biggr] </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~\Omega^2 </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{r^3} \biggl[ \frac{1}{3} R_\mathrm{eq}^3 (1 - e^2)^{1 / 2} \biggr] </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~\frac{r}{R_\mathrm{eq}} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[\frac{1}{3\Omega^2} (1 - e^2)^{1 / 2} \biggr]^{1 / 3} \, .</math>
  </td>
</tr>
</table>
For example, when <math>~(e, \Omega^2) = (0.6, 0.05)</math>, we have, <math>~r/R_\mathrm{eq} = (5.3333)^{1 / 3} = 1.747</math>.

=====Surrounding Torus=====
We'll assume that the surrounding ''2<sup>nd</sup>'' object is a thin torus (1) with the same density as the central object, (2) with a major axis, <math>~a</math>, which ensures that the torus is spinning with the Keplerian frequency prescribed by the mass of the central object, (3) and with a minor cross-sectional radius, <math>~b</math>.  The second of these constraints means that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\alpha_t \equiv \frac{a}{R_\mathrm{eq}} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[\frac{1}{3\Omega^2} (1 - e^2)^{1 / 2} \biggr]^{1 / 3} \, .</math>
  </td>
</tr>
</table>

The mass of the torus is given by the expression,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~M_t</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2\pi a (\pi b^2)\rho 
=
2\pi^2 R_\mathrm{eq}^3 \rho \biggl[ \alpha_t \beta_t^2 \biggr]
\, ,</math>
  </td>
</tr>
</table>
where,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\beta_t</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\frac{b}{R_\mathrm{eq}} \, .</math>
  </td>
</tr>
</table>

Given that <math>~\alpha_t</math> is known once the eccentricity of the central Maclaurin spheroid has been selected and, given that the density of the torus must match the density of the central object, the mass of the torus will only depend on the choice of <math>~0 < \beta_t \le \beta_\mathrm{max}</math>.  The maximum allowed value, <math>~\beta_\mathrm{max}</math>, is set by ensuring that equatorial-plane location of the inner edge of the torus is no smaller than the equatorial radius of the central spheroid, <math>~R_\mathrm{eq}</math>.  This means,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\beta_\mathrm{max}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\alpha_t - 1 \, .</math>
  </td>
</tr>
</table>
So, the maximum torus mass is,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~M_\mathrm{max}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2\pi^2 R_\mathrm{eq}^3 \rho \biggl[ \alpha_t \beta_\mathrm{max}^2 \biggr]
=
2\pi^2 R_\mathrm{eq}^3 \rho ~\alpha_t (\alpha_t - 1)^2 \, .
</math>
  </td>
</tr>
</table>

The moment of inertia of the torus is,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~I_t</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~M_t R_\mathrm{eq}^2\biggl[ \biggl(\frac{a}{R_\mathrm{eq}} \biggr)^2 + \frac{3}{4} \biggl( \frac{b}{R_\mathrm{eq}} \biggr)^2 \biggr]</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2\pi^2 R_\mathrm{eq}^5 \rho \alpha_t \beta_t^2 \biggl( \alpha_t^2 + \frac{3}{4} \beta_t^2 \biggr) \, .</math>
  </td>
</tr>
</table>
Hence, the (square of the) angular momentum of the torus is,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~L_t^2 = I_t^2 \omega_0^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ 2\pi^2 R_\mathrm{eq}^5 \rho \alpha_t \beta_t^2 \biggl( \alpha_t^2 + \frac{3}{4} \beta_t^2 \biggr) \biggr]^2 \Omega^2 (4\pi G\rho)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2^4 \pi^5 \alpha_t^2 \beta_t^4 \biggl( \alpha_t^2 + \frac{3}{4} \beta_t^2 \biggr)^2 G R_\mathrm{eq}^{10} \rho^3  \Omega^2 \, .
</math>
  </td>
</tr>
</table>

=====Combined Configuration=====

Given that, for the chosen Maclaurin spheroid,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\bar{a}^3</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~R_\mathrm{eq}^2 Z = R_\mathrm{eq}^3(1-e^2)^{1 / 2} \, ,</math>
  </td>
</tr>
</table>

and that the total mass of the system is,

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

<tr>
  <td align="right">
<math>~M_\mathrm{tot}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~M_c + M_t</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{2^2\pi}{3} \rho R_\mathrm{eq}^3 (1 - e^2)^{1 / 2} 
+ 
2\pi^2 R_\mathrm{eq}^3 \rho \alpha_t \beta_t^2  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2\pi R_\mathrm{eq}^3 \rho \biggl[ \frac{2}{3}  (1 - e^2)^{1 / 2} + \pi \alpha_t \beta_t^2 \biggr] 
\, ,
</math>
  </td>
</tr>
</table>
the (square of the) dimensionless total angular momentum of the combined system is,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~j_\mathrm{tot}^2</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
\biggl(\frac{3}{2^8 \pi^4}\biggr)^{1 / 3} \frac{(L_c + L_t)^2}{GM_\mathrm{tot}^3 \bar{a}}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl(\frac{3}{2^8 \pi^4}\biggr)^{1 / 3} 
\biggl\{ L_c + L_t \biggr\}^2 \biggl\{ M_\mathrm{tot} \biggr\}^{-3}G^{-1} R_\mathrm{eq}^{-1} (1 - e^2)^{-1 / 6}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl(\frac{3}{2^8 \pi^4}\biggr)^{1 / 3} 
\biggl\{ \biggl( \frac{2^8 \pi^3}{3^2 \cdot 5^2}\biggr)^{1 / 2} (1 - e^2)^{1 / 2} G^{1 / 2} R_\mathrm{eq}^{5} \rho^{3 / 2} \Omega 
+ 2^2 \pi^{5 / 2} \alpha_t \beta_t^2 \biggl( \alpha_t^2 + \frac{3}{4} \beta_t^2 \biggr) G^{1 / 2} R_\mathrm{eq}^{5} \rho^{3 / 2}  \Omega \biggr\}^2 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~\times
\biggl\{ 2\pi R_\mathrm{eq}^3 \rho \biggl[ \frac{2}{3}  (1 - e^2)^{1 / 2} + \pi \alpha_t \beta_t^2 \biggr] \biggr\}^{-3}G^{-1} R_\mathrm{eq}^{-1} (1 - e^2)^{-1 / 6}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{2^3 \pi^3}
\biggl(\frac{3}{2^8 \pi^4}\biggr)^{1 / 3} 
\biggl\{ \biggl( \frac{2^8 \pi^3}{3^2 \cdot 5^2}\biggr)^{1 / 2} (1 - e^2)^{1 / 2} \Omega 
+ 2^2 \pi^{5 / 2} \alpha_t \beta_t^2 \biggl( \alpha_t^2 + \frac{3}{4} \beta_t^2 \biggr)   \Omega \biggr\}^2 
\biggl[ \frac{2}{3}  (1 - e^2)^{1 / 2} + \pi \alpha_t \beta_t^2 \biggr]^{-3}  (1 - e^2)^{-1 / 6}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2\Omega^2
\biggl(\frac{3}{2^8 \pi^4}\biggr)^{1 / 3} 
\biggl\{ \biggl( \frac{2^2 }{3 \cdot 5}\biggr) (1 - e^2)^{1 / 2} 
+ \pi \alpha_t \beta_t^2 \biggl( \alpha_t^2 + \frac{3}{4} \beta_t^2 \biggr) \biggr\}^2 
\biggl[ \frac{2}{3}  (1 - e^2)^{1 / 2} + \pi \alpha_t \beta_t^2 \biggr]^{-3}  (1 - e^2)^{-1 / 6} \, .
</math>
  </td>
</tr>
</table>
Notice that when the toroidal component is omitted, this expression reduces to,

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

<tr>
  <td align="right">
<math>~j_c^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2\Omega^2
\biggl(\frac{3}{2^8 \pi^4}\biggr)^{1 / 3} 
\biggl\{ \biggl( \frac{2^2}{3 \cdot 5}\biggr) (1 - e^2)^{1 / 2} 
\biggr\}^2 
\biggl[ \frac{2}{3}  (1 - e^2)^{1 / 2} \biggr]^{-3}  (1 - e^2)^{-1 / 6} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2\Omega^2
\biggl(\frac{3}{2^8 \pi^4}\biggr)^{1 / 3} 
\biggl( \frac{2^4}{3^2 \cdot 5^2}\biggr)  
\biggl(\frac{3^3}{2^3}\biggr) 
(1 - e^2)^{-3 / 2}   (1 - e^2)^{-1 / 6} (1 - e^2)
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\Omega^2
\biggl(\frac{3^4}{2^2 \cdot 5^6 \pi^4}\biggr)^{1 / 3} 
(1 - e^2)^{-2 / 3} \, .
</math>
  </td>
</tr>
</table>
This matches the expression for the isolated Maclaurin spheroid [[#jSquared|derived above]].

====Fixed Mass Ratio====

Let's define the mass ratio,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~q</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
\frac{M_t}{M_c} \, ,
</math>
  </td>
</tr>
</table>

and build a sequence along which this ratio is held constant.  For the problem being considered here, the relevant expression for <math>~q</math> is,

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

<tr>
  <td align="right">
<math>~q</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ 2\pi^2 R_\mathrm{eq}^3 \rho \alpha_t \beta_t^2  \biggr]
\biggl[ \frac{4\pi}{3} \rho R_\mathrm{eq}^3 (1 - e^2)^{1 / 2} \biggr]^{-1}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl( \frac{3\pi}{2} \biggr)  \alpha_t \beta_t^2 (1 - e^2)^{- 1 / 2}\, .
</math>
  </td>
</tr>
</table>

The sequence is constructed by choosing <math>~q</math> (held fixed), and varying the value of <math>~0 < \beta_t \le \beta_\mathrm{max}</math>; then, for each chosen parameter pair, recognize that,

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

<tr>
  <td align="right">
<math>~
\alpha_t (1 - e^2)^{- 1 / 2}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl( \frac{2}{3\pi} \biggr)  \frac{q}{ \beta_t^2  } \, ,</math>
  </td>
</tr>
</table>
that is, from the [[#Surrounding_Torus|above expression for <math>~\alpha_t</math>]],
<table border="0" cellpadding="5" align="center">

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

<tr>
  <td align="right">
<math>~\Rightarrow~~~
\Omega^2 (1 - e^2) 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{3}\biggl[ \biggl( \frac{3\pi}{2} \biggr)  \frac{ \beta_t^2  }{q} \biggr]^3 \, .</math>
  </td>
</tr>
</table>

But, for central models along the MLS, we also must satisfy the relation [[#Model_Sequences|given above]], namely,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\Omega^2 </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{2}(3 - 2e^2)(1 - e^2)^{1 / 2} \cdot \frac{\sin^{-1} e}{e^3} - \frac{3(1-e^2)}{2e^2} \, .
</math>
  </td>
</tr>
</table>
Hence, for each <math>~(q, \beta_t)</math> parameter pair, the relevant central-object eccentricity is given by the root of the relation,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~
\frac{1}{2}(3 - 2e^2)(1 - e^2)^{3 / 2} \cdot \frac{\sin^{-1} e}{e^3} - \frac{3(1-e^2)^2 }{2e^2}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{3}\biggl[ \biggl( \frac{3\pi}{2} \biggr)  \frac{ \beta_t^2  }{q} \biggr]^3 \, .</math>
  </td>
</tr>
</table>

Note that, for the limiting value, <math>~\beta_\mathrm{max} = (\alpha_t - 1)</math>, the relevant relation becomes,

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

<tr>
  <td align="right">
<math>~
(\alpha_t - 1)^2\alpha_t (1 - e^2)^{- 1 / 2}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2q}{3\pi}    </math>
  </td>
</tr>
</table>
where [[#Surrounding_Torus|again, as above]]
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\alpha_t \equiv \frac{a}{R_\mathrm{eq}} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[\frac{1}{3\Omega^2} (1 - e^2)^{1 / 2} \biggr]^{1 / 3} \, .</math>
  </td>
</tr>
</table>