Editing SSC/FreeEnergy/EquilibriumSequenceInstabilities (section)

==Close Binary Stars==

<table border="1" cellpadding="5" align="center">
<tr>
  <td align="center">
Extracted from p. 162 of [http://rsta.royalsocietypublishing.org/content/206/402-412/161 G. H. Darwin (1906)]
  </td>
</tr>
<tr>
  <td align="center">
[[File:DarwinPreface.png|700px|Darwin (1906) Preface]]
  </td>
</tr>
</table>



===Simplistic Illustrative Model===
====Using Our Adopted Notation====
Consider the simple model of two spherical stars in circular orbit about one another, as depicted here on the right.  In addition to the physical parameters explicitly labeled in this diagram, we adopt the following variable notation:  
<div align="center">
<table border="0" cellpadding="5" align="center" width="100%">
<tr><td align="left">

* The total system mass is,
<div align="center">
<math>~M_\mathrm{tot} \equiv M + M^' \, ;</math>
</div>
* The ratio of the primary to secondary mass is,
<div align="center">
<math>~\lambda \equiv \frac{M}{M^'} \, ;</math>
</div>
* And the separation between the two centers is,
<div align="center">
<math>~d \equiv r_\mathrm{cm} + r^'_\mathrm{cm} \, .</math>
</div>
</td>
<td align="center">
[[File:BinarySimpleModel02.png|350px|Simple Binary Model]]
</td>
</tr>
</table>
</div>

For a circular orbit, the angular velocity is related to the the system mass and separation via the Kepler relation,
<div align="center">
<math>~\omega^2 d^3 = GM_\mathrm{tot} \, ,</math>
</div>
and the distances, <math>~r_\mathrm{cm}</math> and <math>~r^'_\mathrm{cm}</math>, between the center of each star and the center of mass (cm) of the system must be related to one another via the expression,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{r^'_\mathrm{cm}}{r_\mathrm{cm}} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{M}{M^'} = \lambda \, .</math>
  </td>
</tr>
</table>
</div>

Note that the following relations also hold:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~M = M_\mathrm{tot} \biggl( \frac{\lambda}{1+\lambda}\biggr)</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp; 
  </td>
  <td align="left">
<math>~M^' = M_\mathrm{tot} \biggl( \frac{1}{1+\lambda}\biggr)</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~r_\mathrm{cm} = d \biggl( \frac{1}{1+\lambda}\biggr) \, ;</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp; 
  </td>
  <td align="left">
<math>~r^'_\mathrm{cm} = d \biggl( \frac{\lambda}{1+\lambda}\biggr) \, .</math>
  </td>
</tr>
</table>
</div>

Hence, the orbital angular momentum is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~L_\mathrm{orb}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ [M r^2_\mathrm{cm}  + M^' (r^'_\mathrm{cm})^2]\omega </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ M_\mathrm{tot} d^2 \biggl[\biggl( \frac{\lambda}{1+\lambda}\biggr) \biggl( \frac{1}{1+\lambda}\biggr)^2  + \biggl( \frac{1}{1+\lambda}\biggr) \biggl( \frac{\lambda}{1+\lambda}\biggr)^2 \biggr] \biggl[\frac{GM_\mathrm{tot}}{d^3}\biggr]^{1 / 2} </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ (G M_\mathrm{tot}^3 d)^{1 / 2} \biggl[\frac{\lambda}{(1+\lambda)^2}  \biggr] \, .</math>
  </td>
</tr>
</table>
</div>

Assuming that both stars are rotating synchronously with the orbit, their respective spin angular momenta are,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~L_M = I_M \omega</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2}{5}MR^2 \omega</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2}{5}M_\mathrm{tot} \biggl(\frac{\lambda}{1+\lambda}\biggr) R^2 \biggl[ \frac{G M_\mathrm{tot}}{d^3} \biggr]^{1 / 2}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2}{5}(GM_\mathrm{tot}^3 d)^{1 / 2} \biggl(\frac{\lambda}{1+\lambda}\biggr) \biggl( \frac{R}{d}\biggr)^2 \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~L_{M^'} = I_{M^'} \omega</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2}{5}{M^'}(R^')^2 \omega</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2}{5}M_\mathrm{tot} \biggl(\frac{1}{1+\lambda}\biggr) (R^')^2 \biggl[ \frac{G M_\mathrm{tot}}{d^3} \biggr]^{1 / 2}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2}{5}(GM_\mathrm{tot}^3 d)^{1 / 2} \biggl(\frac{1}{1+\lambda}\biggr) \biggl( \frac{R^'}{d}\biggr)^2 \, .</math>
  </td>
</tr>
</table>
</div>

<span id="SphericalLtot">Hence, the total angular momentum of the system is,</span>
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~L_\mathrm{tot} = L_\mathrm{orb} + L_M + L_{M^'}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(G M_\mathrm{tot}^3 d)^{1 / 2} \biggl[\frac{\lambda}{(1+\lambda)^2}  \biggr] 
+ \frac{2}{5}(GM_\mathrm{tot}^3 d)^{1 / 2} \biggl(\frac{\lambda}{1+\lambda}\biggr) \biggl( \frac{R}{d}\biggr)^2
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \frac{2}{5}(GM_\mathrm{tot}^3 d)^{1 / 2} \biggl(\frac{1}{1+\lambda}\biggr) \biggl( \frac{R^'}{d}\biggr)^2
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(G M_\mathrm{tot}^3 R)^{1 / 2} \biggl\{ \biggl[\frac{\lambda}{(1+\lambda)^2}  \biggr] \biggl( \frac{d}{R}\biggr)^{1 / 2}
+ \frac{2}{5} \biggl(\frac{1}{1+\lambda}\biggr)\biggl[ \lambda
+  \biggl( \frac{R^'}{R}\biggr)^{2} \biggr] \biggl( \frac{d}{R}\biggr)^{-3/2}
\biggr\}
</math>
  </td>
</tr>
</table>
</div>

If we assume that the two stars have the same (uniform) densities, <math>~\rho</math>, then, following Darwin (1906; see immediately below), the two stellar radii can be related to the mass ratio, <math>~\lambda</math> via the expressions,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~R = a\biggl( \frac{\lambda}{1+\lambda}\biggr) \, ,</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp; 
  </td>
  <td align="left">
<math>~R^' = a\biggl( \frac{1}{1+\lambda}\biggr) \, ,</math>
  </td>
</tr>
</table>
</div>
where, the characteristic length scale is,
<div align="center">
<math>a \equiv \biggl(\frac{3M_\mathrm{tot}}{4\pi \rho}\biggr)^{1 / 3} \, .</math>
</div>
Replacing <math>~R</math> and <math>~R^'</math> by these expressions in our equation for <math>~L_\mathrm{tot}</math> results in the simplistic/illustrative expression for <math>~L_1</math> derived by Darwin (1906) and presented in the [[#DarwinL1L2|boxed-in image, below]].  Darwin's expression for <math>~L_2</math> is obtained by using the same expression for <math>~R</math> but treating the secondary as a point mass, that is, setting <math>~R^' = 0</math>.

====Darwin's (1906) Equivalent Illustration====
From Pt. I, &sect;1 (p. 164) of  [http://rsta.royalsocietypublishing.org/content/206/402-412/161 G. H. Darwin (1906)] &#8212; ''verbatum'' text in green:  <font color="green">
It will be useful to make a rough preliminary investigation of the regions in which we shall have to look for cases of limiting stability in the two problems.  For this purpose I consider</font> [1] <font color="green">the case of two spheres as the analogue of</font> [Darwin's] <font color="green">problem of the figure of equilibrium, and </font> [2] <font color="green">the case of a sphere and a particle as the analogue of Roche's problem.  


&hellip; let the mass of the whole system be <math>~M_\mathrm{tot} = \tfrac{4}{3}\pi \rho a^3</math>; let the masses of the two spheres be
</font>
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~M = M_\mathrm{tot} \biggl[\frac{\lambda}{1+\lambda}\biggr] </math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp; 
  </td>
  <td align="left">
<math>~M^' = M_\mathrm{tot} \biggl[\frac{1}{1+\lambda}\biggr] </math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; <math>~\Rightarrow</math> &nbsp; &nbsp; &nbsp; 
  </td>
  <td align="left">
<math>~\lambda = \frac{M}{M^'}</math>
  </td>
</tr>
</table>
</div>

<font color="green">or for Roche's problem let the latter <math>~(M^')</math> be the mass of the particle.</font>  Assuming that both spheres have the same characteristic density, <math>~\rho</math>, that has been used to specify the total mass, we furthermore know that the radius of the first sphere is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~a_M</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl(\frac{M}{\tfrac{4}{3}\pi \rho}\biggr)^{1 / 3} = a \biggl(\frac{\lambda}{1+\lambda}\biggr)^{1 / 3} \, ,</math>
  </td>
</tr>
</table>
</div>
and (for the Darwin problem) the radius of the second sphere is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~a_{M^'}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl(\frac{M^'}{\tfrac{4}{3}\pi \rho}\biggr)^{1 / 3} = a\biggl(\frac{1 }{1+\lambda}\biggr)^{1 / 3} \, .</math>
  </td>
</tr>
</table>
</div>

<font color="green">Let <math>~r</math> be the distance from the centre of one sphere to that of the other, or to the particle, as the case may be; and <math>~\omega</math> the orbital angular velocity,</font> where
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\omega^2 r^3</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~GM_\mathrm{tot} \, .</math>
  </td>
</tr>
</table>
</div>
<font color="green">The centre of inertia of the two masses is distant <math>~r/(1+\lambda)</math> and <math>~\lambda r/(1+\lambda)</math> from their respective centres, and we easily find the orbital momentum to be
</font>
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~L_\mathrm{orb} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~M \biggl(\frac{r}{1+\lambda}\biggr)^2\omega + M^' \biggl(\frac{\lambda r}{1+\lambda}\biggr)^2\omega</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~M_\mathrm{tot} \biggl[ \biggl(\frac{\lambda}{1+\lambda}\biggr)\biggl(\frac{1}{1+\lambda}\biggr)^2\omega 
+ \biggl(\frac{1}{1+\lambda}\biggr) \biggl(\frac{\lambda }{1+\lambda}\biggr)^2 \biggr] r^2 \omega</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~M_\mathrm{tot}  \biggl[ \frac{\lambda}{(1+\lambda)^2} \biggr] r^2\omega \, .</math>
  </td>
</tr>
</table>
</div>
<font color="green">In both problems the rotational momentum of the first sphere is</font>
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~L_\mathrm{M} = \frac{2}{5} Ma_M^2 \omega</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{2}{5} M_\mathrm{tot}\biggl( \frac{\lambda}{1+\lambda} \biggr) \biggl(\frac{\lambda}{1+\lambda}\biggr)^{2 / 3}a^2 \omega
=\frac{2}{5} M_\mathrm{tot} \biggl(\frac{\lambda}{1+\lambda}\biggr)^{5 / 3}a^2 \omega \, .
</math>
  </td>
</tr>
</table>
</div>
<font color="green">In the</font> [Darwin] <font color="green">problem the rotational momentum of the second sphere is
</font>
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~L_\mathrm{M^'} = \frac{2}{5} M^' a_{M^'}^2 \omega</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{2}{5} M_\mathrm{tot}\biggl( \frac{1}{1+\lambda} \biggr) \biggl(\frac{1}{1+\lambda}\biggr)^{2 / 3}a^2 \omega
=\frac{2}{5} M_\mathrm{tot} \biggl(\frac{1}{1+\lambda}\biggr)^{5 / 3}a^2 \omega \, ,
</math>
  </td>
</tr>
</table>
</div>
<font color="green">and in the</font> [Roche] <font color="green">problem it is nil.

If, then, we write <math>~L_1</math> for the total angular momentum of the two spheres, and <math>~L_2</math> for that of the sphere and particle, we have
</font>

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

<tr>
  <td align="right">
<math>~L_1 </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
L_\mathrm{orb} + L_M + L_{M^'}  
</math>
</td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
M_\mathrm{tot}  \biggl[ \frac{\lambda}{(1+\lambda)^2} \biggr] r^2\omega
+ \frac{2}{5} M_\mathrm{tot} \biggl(\frac{\lambda}{1+\lambda}\biggr)^{5 / 3}a^2 \omega
+ \frac{2}{5} M_\mathrm{tot} \biggl(\frac{1}{1+\lambda}\biggr)^{5 / 3}a^2 \omega</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
M_\mathrm{tot} a^2 \omega \biggl[ \frac{\lambda r^2}{(1+\lambda)^2a^2} 
+ \frac{2}{5} \frac{1+ \lambda^{5/3}}{(1+\lambda)^{5 / 3}} \biggr] \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~L_2 </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
L_\mathrm{orb} + L_M  
</math>
</td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
M_\mathrm{tot} a^2 \omega \biggl[ \frac{\lambda r^2}{(1+\lambda)^2a^2} 
+ \frac{2}{5} \frac{\lambda^{5/3}}{(1+\lambda)^{5 / 3}} \biggr] \, .
</math>
  </td>
</tr>
</table>
</div>

<div align="center">
<table border="1" cellpadding="5" align="center" width="85%">
<tr>
  <td align="left">
For comparison, in the context of the LRS93b discussion of ''Compressible Roche Ellipsoids'', the total angular momentum is,

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

<tr>
  <td align="right">
<math>~J</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[\frac{Mr^2}{(1+\lambda)} + I \biggr] \Omega \, ,</math>
  </td>
</tr>
</table>

where (see their equation 4.8),

<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>~\frac{1}{5} \kappa_n (a_1^2 + a_2^2)M \, ,</math>
  </td>
</tr>
</table>
and, for incompressible configurations, <math>~\kappa_{n=0} = 1</math>.  This gives,

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

<tr>
  <td align="right">
<math>~J</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~M_\mathrm{tot} \biggl( \frac{\lambda}{1+\lambda}\biggr) a_1^2 \Omega \biggl[\frac{r^2}{(1+\lambda)a_1^2} + \frac{1}{5}\biggl(1 + \frac{a_2^2}{a_1^2}\biggr)  \biggr]  \, .</math>
  </td>
</tr>
</table>

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

<font color="green">On substituting for <math>~\omega</math> its value in terms of <math>~r</math>, these expressions become</font>

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

<tr>
  <td align="right">
<math>~L_1 </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(GM_\mathrm{tot}^3 a)^{1 / 2} \biggl( \frac{a}{r} \biggr)^{3/2}  \biggl[ 
\frac{2}{5} \frac{1+ \lambda^{5/3}}{(1+\lambda)^{5 / 3}} + \frac{\lambda r^2}{(1+\lambda)^2a^2} \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(GM_\mathrm{tot}^3 a)^{1 / 2} \frac{1}{(1+\lambda)^2} \biggl[ 
\frac{2}{5} (1+ \lambda^{5/3}) (1+\lambda)^{1 / 3}\biggl( \frac{a}{r} \biggr)^{3/2} + \lambda \biggl( \frac{r}{a} \biggr)^{1 / 2}\biggr] \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~L_2 </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(GM_\mathrm{tot}^3 a)^{1 / 2} \biggl( \frac{a}{r} \biggr)^{3/2}  \biggl[ 
\frac{2}{5} \frac{\lambda^{5/3}}{(1+\lambda)^{5 / 3}} + \frac{\lambda r^2}{(1+\lambda)^2a^2} \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(GM_\mathrm{tot}^3 a)^{1 / 2} \frac{1}{(1+\lambda)^2} \biggl[ 
\frac{2}{5} \lambda^{5/3} (1+\lambda)^{1 / 3}\biggl( \frac{a}{r} \biggr)^{3/2} + \lambda \biggl( \frac{r}{a} \biggr)^{1 / 2} \biggr] \, .
</math>
  </td>
</tr>
</table>
</div>

<span id="DarwinL1L2">As is shown in the following boxed-in image,</span> (after setting <math>~G = 1</math>) this matches the pair of equations that appears immediately following equation (1) in [http://rsta.royalsocietypublishing.org/content/206/402-412/161 G. H. Darwin (1906)].

<table border="1" cellpadding="5" align="center">
<tr>
  <td align="center">
Equations extracted without modification from [http://rsta.royalsocietypublishing.org/content/206/402-412/161 G. H. Darwin (1906)]
  </td>
</tr>
<tr>
  <td align="center">
[[File:Darwin1906Eq1b.png|Darwin (1906) Eq. 1b]]
  </td>
</tr>
</table>

In the following figure we have plotted, for both problems, how the total angular momentum varies  with orbital separation.  In order to facilitate a direct comparison with Figure 1 from LRS, in place of the dimensionless separation <math>~r/a</math> we plot along the abscissa the quantity, <math>~r_\mathrm{LRS}/R</math>, where, <math>~r_\mathrm{LRS}</math> is the radius of the circular orbit and <math>~R</math> is the radius of the primary star; that is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~r_\mathrm{LRS} = \tfrac{1}{2}r</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp; 
  </td>
  <td align="left">
<math>~R = a_M</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; <math>~\Rightarrow</math> &nbsp; &nbsp; &nbsp; 
  </td>
  <td align="left">
<math>~\frac{r_\mathrm{LRS}}{R} = \frac{1}{2} \biggl(\frac{1+\lambda}{\lambda}\biggr)^{1 / 3} \frac{r}{a} \, .</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~r_\mathrm{LRS} = \tfrac{1}{2}r</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp; 
  </td>
  <td align="left">
<math>~R = a_M</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; <math>~\Rightarrow</math> &nbsp; &nbsp; &nbsp; 
  </td>
  <td align="left">
<math>~\frac{r}{a} = 2 \biggl(\frac{\lambda}{1+\lambda}\biggr)^{1 / 3} \frac{r_\mathrm{LRS}}{R}  \, .</math>
  </td>
</tr>
</table>
</div>

Rewriting Darwin's pair of angular momentum expressions in terms of this preferred dimensionless separation, we have,

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

<tr>
  <td align="right">
<math>~L_1 </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(GM_\mathrm{tot}^3 a)^{1 / 2} \frac{1}{(1+\lambda)^2} \biggl\{ 
\frac{2}{5} (1+ \lambda^{5/3}) (1+\lambda)^{1 / 3}\biggl[ 2 \biggl(\frac{\lambda}{1+\lambda}\biggr)^{1 / 3} \frac{r_\mathrm{LRS}}{R}  \biggr]^{-3/2} 
+ \lambda \biggl[ 2 \biggl(\frac{\lambda}{1+\lambda}\biggr)^{1 / 3} \frac{r_\mathrm{LRS}}{R} \biggr]^{1 / 2}
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(GM_\mathrm{tot}^3 a)^{1 / 2} \frac{1}{(1+\lambda)^2} \biggl\{ 
\biggl( \frac{1}{2\cdot 5^2} \biggr)^{1 / 2} (1+ \lambda^{5/3}) (1+\lambda)^{1 / 3} \biggl(\frac{1+\lambda}{\lambda}\biggr)^{1 / 2}  \biggl(  \frac{r_\mathrm{LRS}}{R}  \biggr)^{-3/2}
+ 2^{1 / 2}\lambda \biggl(\frac{\lambda}{1+\lambda}\biggr)^{1 / 6} \biggl( \frac{r_\mathrm{LRS}}{R} \biggr)^{1 / 2}
\biggr\} \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~L_2 </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(GM_\mathrm{tot}^3 a)^{1 / 2} \frac{1}{(1+\lambda)^2} \biggl\{ 
\frac{2}{5} \lambda^{5/3} (1+\lambda)^{1 / 3}\biggl[ 2 \biggl(\frac{\lambda}{1+\lambda}\biggr)^{1 / 3} \frac{r_\mathrm{LRS}}{R}  \biggr]^{-3/2} 
+ \lambda \biggl[ 2 \biggl(\frac{\lambda}{1+\lambda}\biggr)^{1 / 3} \frac{r_\mathrm{LRS}}{R}  \biggr]^{1 / 2} 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(GM_\mathrm{tot}^3 a)^{1 / 2} \frac{1}{(1+\lambda)^2} \biggl\{ 
\biggl( \frac{1}{2\cdot 5^2} \biggr)^{1 / 2} \lambda^{5/3} (1+\lambda)^{1 / 3} \biggl(\frac{1+\lambda}{\lambda}\biggr)^{1 / 2}  \biggl(  \frac{r_\mathrm{LRS}}{R}  \biggr)^{-3/2}
+ 2^{1 / 2}\lambda \biggl(\frac{\lambda}{1+\lambda}\biggr)^{1 / 6} \biggl( \frac{r_\mathrm{LRS}}{R} \biggr)^{1 / 2}
\biggr\} \, .
</math>
  </td>
</tr>
</table>
</div>

Note that the two binary components come into contact when, for the Darwin problem,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~a_M + a_{M^'} = r</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; <math>~\Rightarrow</math> &nbsp; &nbsp; &nbsp; 
  </td>
  <td align="left">
<math>~\frac{r}{a} =
\frac{1 + \lambda^{1 / 3}}{(1+\lambda)^{1 / 3}}  \, ;
</math>
  </td>
</tr>
</table>
</div>
and, for the Roche problem,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~a_M = r</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; <math>~\Rightarrow</math> &nbsp; &nbsp; &nbsp; 
  </td>
  <td align="left">
<math>~\frac{r}{a} =
\frac{\lambda^{1 / 3}}{(1+\lambda)^{1 / 3}}  \, .
</math>
  </td>
</tr>
</table>
</div>

===Setup===
====Jeans (1919)====
From &sect; 50 (p. 46) of  [http://adsabs.harvard.edu/abs/1919pcsd.book.....J J. H. Jeans (1919)] &#8212; ''verbatum'' text in green:  <font color="green">
Let the two bodies be spoken of as primary and secondary, and let their masses be <math>~M</math>, <math>~M^'</math> respectively; let the distance apart of their centres of gravity be <math>~R</math>, and let the angular velocity of rotation of the line joining them be <math>~\omega</math>.  It will be sufficient to fix our attention on the conditions of equilibrium of one of the two masses, say the primary.  Let its centre of gravity be taken as origin, let the line joining it to the centre of the secondary be axis of <math>~x</math>, and let the plane in which the rotation takes place be that of <math>~xy</math>.  Then the equation of the axis of rotation is
</font>

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

<tr>
  <td align="right">
<math>~x = \frac{M^'}{M + M^'} ~ R</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp; 
  </td>
  <td align="left">
<math>~y = 0 \, .</math>
  </td>
</tr>
</table>
</div>
<font color="green">
The problem may be reduced to a statical one (cf. &sect; 31) by supposing the masses acted on by a field of force of</font> [the centrifugal] <font color="green">potential
</font>
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{1}{2}\omega^2\biggl[ \biggl( x - \frac{M^' }{M + M^'} ~R \biggr)^2 + y^2\biggr] \, .</math>
  </td>
</tr>
</table>
</div>

====Chandrasekhar (1969)====
From pp. 189-190 of [<b>[[User:Tohline/Appendix/References#EFE|<font color="red">EFE</font>]]</b>] &#8212; ''verbatum'' text in green:  <font color="green">
Let the masses of the primary and the secondary be <math>~M</math> and <math>~M^'</math>, respectively; let the distance between their centers of mass be <math>~R</math>; and let the constant angular velocity of rotation about their common center of mass be <math>~\Omega</math>.  Choose a coordinate system in which the origin is at the center of mass of the primary, the <math>~x_1-</math>axis points to the center of mass of the secondary, and the <math>~x_3-</math>axis is parallel to the direction of <math>~\vec\Omega</math>.  In this coordinate system, the equation of motion governing fluid elements of <math>~M</math></font> includes (see EFE's equation 1) a gradient of the centrifugal potential,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{1}{2}\Omega^2\biggl[ \biggl( x_1 - \frac{M^' R}{M + M^'} \biggr)^2 + x_2^2\biggr] \, .</math>
  </td>
</tr>
</table>
</div>

====Tassoul (1978)====
From p. 449 of [<b>[[User:Tohline/Appendix/References#T78|<font color="red">T78</font>]]</b>] &#8212; ''verbatum'' text in green:  <font color="green">
Let the masses of the primary and the secondary be <math>~M</math> and <math>~M^'</math>, respectively; let the distance between their centers of mass be <math>~d</math>; and let the angular velocity of rotation about their common center of mass be <math>~\Omega</math>.  Next choose a system of reference in which the origin is at the center of mass of the primary; for convenience, the <math>~x_1-</math>axis points toward the center of mass of the secondary, and the <math>~x_3-</math>axis is parallel to the direction of <math>~\vec\Omega</math>.  Then, the equation of the rotation axis, which of course passes through the center of mass of the two bodies, is
</font>

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

<tr>
  <td align="right">
<math>~x_1 = \frac{M^' }{M + M^'} ~d</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp; 
  </td>
  <td align="left">
<math>~x_2 = 0 \, .</math>
  </td>
</tr>
</table>
</div>

<font color="green">
Accordingly, the centrifugal force acting on the mass <math>~M</math> may be derived from the potential
</font>
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~-\frac{1}{2}\Omega^2\biggl[ \biggl( x_1 - \frac{M^'}{M + M^'} ~ d\biggr)^2 + x_2^2\biggr] \, .</math>
  </td>
</tr>
</table>
</div>

===Roche Ellipsoids===
====Jeans (1919)====
From &sect; 51 (p. 47) of  [http://adsabs.harvard.edu/abs/1919pcsd.book.....J J. H. Jeans (1919)] &#8212; ''verbatum'' text in green:  <font color="green">
The simplest problem occurs when the secondary may be treated as a rigid sphere;  this is the special problem dealt with by Roche.  As in &sect; 47 the tide-generating potential acting on the primary may be supposed to be
</font>
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="left">
<math>~\frac{M^'}{R} + \frac{M^'}{R^2} x + \frac{M^'}{R^3}(x^2 - \tfrac{1}{2}y^2 - \tfrac{1}{2}z^2) + \cdots </math>
  </td>
</tr>
</table>
</div>
<font color="green">We shall for the present be content to omit all terms beyond those written down.  The correction required by the neglect of these terms will be discussed later, and will be found to be so small that the results now to be obtained are hardly affected.</font>

<font color="green">On omitting these terms, and combining the two potentials &hellip; it appears that the primary may be supposed influenced by a statical field of potential
</font>
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="left">
<math>~\frac{M^'}{R} x\biggr(1 - \frac{\omega^2 R^3}{M + M^'}\biggr) + \frac{M^'}{R^3}(x^2 - \tfrac{1}{2}y^2 - \tfrac{1}{2}z^2) 
+ \tfrac{1}{2}\omega^2(x^2 + y^2)       \, .</math>
  </td>
</tr>
</table>
</div>
<font color="green">The terms in <math>~x</math> may immediately be removed by supposing <math>~\omega</math> to have the appropriate value given by
</font>
<div align="center">
<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{M+M^'}{R^3}</math>
  </td>
</tr>
</table>
</div>
<font color="green">and the condition for equilibrium is now seen to be that we shall have, at every point of the surface,
</font>
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~V_b + \mu (x^2 - \tfrac{1}{2}y^2 - \tfrac{1}{2}z^2) + \tfrac{1}{2}\omega^2(x^2 + y^2)     </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
constant
  </td>
</tr>
</table>
</div>
<font color="green">where <math>~\mu</math> … stands for <math>~M^'/R^3</math> .
</font>

====Chandrasekhar (1969)====
From p. 190 of [<b>[[User:Tohline/Appendix/References#EFE|<font color="red">EFE</font>]]</b>] &#8212; ''verbatum'' text in green:  <font color="green">
In Roche's particular problem, the secondary is treated as a rigid sphere.  Then, over the primary, the tide-generating potential, <math>~\mathfrak{B}^'</math> can be expanded in the form
</font>
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\mathfrak{B}^'</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{GM^'}{R} \biggl( 1 + \frac{x_1}{R} + \frac{x_1^2 - \tfrac{1}{2}x_2^2 - \tfrac{1}{2}x_3^2}{R^2} + \cdots \biggr) \, ;</math>
  </td>
</tr>
</table>
</div>
<font color="green">and the approximation which underlies this theory is to retain, in this expansion for <math>~\mathfrak{B}^'</math>, only the terms which have been explicitly written down and ignore all the terms which are of higher order.  On this assumption, the equation of motion becomes
</font>
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{du_i}{dt} + \frac{1}{\rho} \frac{\partial p}{\partial x_i}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{\partial}{\partial x_i} \biggl[ \mathfrak{B} + \tfrac{1}{2}\Omega^2(x_1^2 + x_2^2) + \mu(x_1^2 - \tfrac{1}{2}x_2^2 - \tfrac{1}{2}x_3^2)
+ \biggl( \frac{GM^'}{R^2} - \frac{M^' R}{M+M^'} ~\Omega^2 \biggr)x_1
\biggr] + 2\Omega \epsilon_{i\ell 3} u_\ell
</math>
  </td>
</tr>
</table>
</div>
<font color="green">where we have introduced the abbreviation
</font>
<div align="center">
<math>~\mu = \frac{GM^'}{R^3} \, .</math>
</div>

<font color="green">So far, we have left <math>~\Omega^2</math> unspecified.  If we now let <math>~\Omega^2</math> have the "Keplerian value"
</font>
<div align="center">
<math>~\Omega^2 = \frac{G(M+ M^')}{R^3} = \mu \biggl(1 + \frac{M}{M^'} \biggr) \, ,</math>
</div>

<font color="green">the "unwanted" term in <math>~x_1</math>, on the right-hand side of</font> [this equation,] <font color="green">vanishes and we are left with
</font>
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{du_i}{dt} + \frac{1}{\rho} \frac{\partial p}{\partial x_i}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{\partial}{\partial x_i} \biggl[ \mathfrak{B} + \tfrac{1}{2}\Omega^2(x_1^2 + x_2^2) + \mu(x_1^2 - \tfrac{1}{2}x_2^2 - \tfrac{1}{2}x_3^2)
\biggr] + 2\Omega \epsilon_{i\ell 3} u_\ell \, .
</math>
  </td>
</tr>
</table>
</div>
<font color="green">This is the basic equation of this theory;  and Roche's problem is concerned with the equilibrium and the stability of homogeneous masses governed by</font> [this relation].


====Tassoul (1978)====
From pp. 449-450 of [<b>[[User:Tohline/Appendix/References#T78|<font color="red">T78</font>]]</b>] &#8212; ''verbatum'' text in green:  <font color="green">
In Roche's particular problem, the secondary is treated as a rigid sphere; hence, over the primary, the tide-generating potential can be expanded in the form
</font>
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="left">
<math>~ -\frac{GM^'}{d} \biggl( 1 + \frac{x_1}{d} + \frac{x_1^2 - \tfrac{1}{2}x_2^2 - \tfrac{1}{2}x_3^2}{d^2} + \cdots \biggr) \, .</math>
  </td>
</tr>
</table>
</div>
<font color="green">The approximation that underlies the theory is to omit all terms beyond those written down.  On this assumption, we find that, apart from its own gravitation, the primary may be supposed to be acted upon by a total field of force derived from the potential
</font>
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~-\tfrac{1}{2}\Omega^2(x_1^2 + x_x^2) - \mu(x_1^2 - \tfrac{1}{2}x_2^2 - \tfrac{1}{2}x_3^2) - \biggl(\mu - \frac{M^'}{M+M^'} \Omega^2\biggr) dx_1 \, ,</math>
  </td>
</tr>
</table>
</div>
<font color="green">where</font>
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\mu</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{GM^'}{d^3} \, .</math>
  </td>
</tr>
</table>
</div>

<font color="green">Further letting <math>~\Omega^2</math> have its "Keplerian value"</font>
<div align="center">
<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{G(M+M^')}{d^3} \, ,</math>
  </td>
</tr>
</table>
</div>
<font color="green">we can thus write the conditions of relative equilibrium for the primary in the form</font>
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{1}{\rho} \nabla p</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-\nabla [ V
-\tfrac{1}{2}\Omega^2(x_1^2 + x_x^2) - \mu(x_1^2 - \tfrac{1}{2}x_2^2 - \tfrac{1}{2}x_3^2) ] \, ,
</math>
  </td>
</tr>
</table>
</div>
<font color="green">where <math>~V</math> is the self-gravitating potential of the primary.</font>

===Incompressible Roche Ellipsoids (&lambda; &#8800; 0)===

Let's see if we can understand the relationship between tabulated data presented by [http://adsabs.harvard.edu/abs/1993ApJS...88..205L Lai, Rasio, &amp; Shapiro (1993b, ApJS, 88, 205)] &#8212; hereafter, LRS93S &#8212; for the case of incompressible Roche ellipsoids, when <math>~\lambda = p = 1</math>.  After setting <math>~\kappa_{n=0} = 1</math> in their equation (4.8), we have,
<div align="center">
<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>~\frac{1}{5}a_1^2 M \biggl( 1 + \frac{a_2^2}{a_1^2}\biggr) \, .</math>
  </td>
</tr>
</table>
</div>
And, from their equation (7.12),
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~J_\mathrm{tot}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{Mr^2}{(1+p)} + I \biggr]\Omega \, .</math>
  </td>
</tr>
</table>
</div>

<span id="Kepler">If we adopt the Keplerian orbital frequency, the expression for the total angular momentum is,</span>
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~J_\mathrm{Kep}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{Mr^2}{(1+p)} + \frac{1}{5}a_1^2 M \biggl( 1 + \frac{a_2^2}{a_1^2}\biggr) \biggr]\biggl[ \frac{GM_\mathrm{tot}}{r^3} \biggr]^{1/2} </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(GM^3)^{1/2} a_1^2\biggl[ \frac{1}{(1+p)}\biggl(\frac{r}{a_1}\biggr)^2 + \frac{1}{5}\biggl( 1 + \frac{a_2^2}{a_1^2}\biggr) \biggr]\biggl[ \frac{1}{r^3} \biggl( \frac{1+p}{p} \biggr)\biggr]^{1/2} </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(GM^3 R)^{1/2} \biggl( \frac{a_1^4}{r^{3}R} \biggr)^{1/2} \biggl[ \frac{1}{(1+p)}\biggl(\frac{r}{a_1}\biggr)^2 + \frac{1}{5}\biggl( 1 + \frac{a_2^2}{a_1^2}\biggr) \biggr]
\biggl[ \biggl( \frac{1+p}{p} \biggr)\biggr]^{1/2} \, . </math>
  </td>
</tr>
</table>
</div>

For the specific case of an equal-mass binary sequence &#8212; that is, <math>~\lambda = p = 1</math> &#8212; as considered in the following table, we have,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\bar{J}_\mathrm{Kep}\biggr|_{p=1} \equiv (GM^3 R)^{-1/2} J_\mathrm{Kep}\biggr|_{p=1} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2^{-1 / 2}  \biggl( \frac{a_1}{r} \biggr)^{2}\biggl( \frac{r^3}{a_1^3} \cdot \frac{a_1^3}{R^3}\biggr)^{1/6} \biggl[ \biggl(\frac{r}{a_1}\biggr)^2 + \frac{2}{5}\biggl( 1 + \frac{a_2^2}{a_1^2}\biggr) \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2^{-1 / 2}  \biggl( \frac{a_1}{r} \biggr)^{3/2} \biggl( \frac{a_2}{a_1}\cdot \frac{a_3}{a_1}\biggr)^{-1/6} \biggl[ \biggl(\frac{r}{a_1}\biggr)^2 + \frac{2}{5}\biggl( 1 + \frac{a_2^2}{a_1^2}\biggr) \biggr] \, ,
</math>
  </td>
</tr>
</table>
</div>

where we also have involved the expression for an equivalent spherical radius given just before their equation (7.21), namely,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~R^3</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~a_1 a_2 a_3 = a_1^3\biggl( \frac{a_2}{a_1}\cdot \frac{a_3}{a_1}\biggr) \, .</math>
  </td>
</tr>
</table>
</div>


<table border="1" align="center" cellpadding="5">

<tr>
  <th align="center" colspan="9"><font size="+1"><b>Table 1:</b></font>&nbsp; Incompressible <math>~(n=0)</math> Roche Ellipsoids with <math>~\lambda = p = 1</math></th>
</tr>

<tr>
  <th align="center" colspan="4">
Extracted from Table 1 of [http://adsabs.harvard.edu/abs/1963ApJ…138.1182C Chandrasekhar (1963)]<br />
same as [<b>[[User:Tohline/Appendix/References#EFE|<font color="red">EFE</font>]]</b>] Table XVI</th>
  <th align="center" colspan="5">EFE Check</th>
</tr>
<tr>
  <td align="center"> (1) </td>
  <td align="center"> (2) </td>
  <td align="center"> (3) </td>
  <td align="center"> (4) </td>
  <td align="center"> (5) </td>
  <td align="center"> (6) </td>
  <td align="center"> (7) </td>
  <td align="center"> (8) </td>
  <td align="center"> (9) </td>
</tr>
<tr>
<td align="center"><math>~\cos^{-1}(a_3/a_1)</math>
<td align="center"><math>~a_2/a_1</math>
<td align="center"><math>~a_3/a_1</math>
<td align="center"><math>~\Omega^2</math>
<td align="center"><math>~r/a_1</math>
<td align="center"><math>~\bar{J}_\mathrm{Kep}=L_\mathrm{tot}/(GM^3 R)^{1/2}</math>
<td align="center"><math>~r/R</math>
<td align="center"><math>~\mathfrak{J}</math>
<td align="center"><math>~L_\mathrm{tot}/(GM_\mathrm{tot}^3 R)^{1/2}</math>
</tr>
<tr>
  <td align="center"> 12&deg; </td>
  <td align="center"> 0.98660 </td>
  <td align="center"> 0.97815 </td>
  <td align="center"> 0.009293 </td>
  <td align="center"> 6.5181</td>
  <td align="center"> 1.8498 </td>
  <td align="center"> 6.5959 </td>
  <td align="center"> 1.0104 </td>
  <td align="center"> 0.6540 </td>
</tr>
<tr>
  <td align="center"> 24&deg; </td>
  <td align="center"> 0.94376 </td>
  <td align="center"> 0.91355 </td>
  <td align="center"> 0.036152 </td>
  <td align="center"> 3.9916 </td>
  <td align="center"> 1.5168 </td>
  <td align="center"> 4.1938 </td>
  <td align="center"> 1.0436 </td>
  <td align="center"> 0.5363 </td>
</tr>
<tr>
  <td align="center"> 36&deg; </td>
  <td align="center"> 0.86345 </td>
  <td align="center"> 0.80902 </td>
  <td align="center"> 0.076342 </td>
  <td align="center"> 2.9005 </td>
  <td align="center"> 1.3846 </td>
  <td align="center"> 3.2689 </td>
  <td align="center"> 1.1086 </td>
  <td align="center"> 0.4895 </td>
</tr>
<tr>
  <td align="center"> 48&deg; </td>
  <td align="center"> 0.73454 </td>
  <td align="center"> 0.66913 </td>
  <td align="center"> 0.118726 </td>
  <td align="center"> 2.2266 </td>
  <td align="center"> 1.3353 </td>
  <td align="center"> 2.8215 </td>
  <td align="center"> 1.2360 </td>
  <td align="center"> 0.4721 </td>
</tr>
<tr>
  <td align="center"> 54&deg; </td>
  <td align="center"> 0.64956 </td>
  <td align="center"> 0.58779 </td>
  <td align="center"> 0.134284 </td>
  <td align="center"> 1.9645 </td>
  <td align="center"> 1.3351 </td>
  <td align="center"> 2.7080 </td>
  <td align="center"> 1.3510 </td>
  <td align="center"> 0.4720 </td>
</tr>
<tr>
  <td align="center"> 59&deg; </td>
  <td align="center"> 0.56892 </td>
  <td align="center"> 0.51504 </td>
  <td align="center"> 0.140854 </td>
  <td align="center"> 1.7702 </td>
  <td align="center"> 1.3494 </td>
  <td align="center"> 2.6652 </td>
  <td align="center"> 1.5003 </td>
  <td align="center"> 0.4771 </td>
</tr>
<tr>
  <td align="center"> 60&deg; </td>
  <td align="center"> 0.55186 </td>
  <td align="center"> 0.50000 </td>
  <td align="center"> 0.141250 </td>
  <td align="center"> 1.7335 </td>
  <td align="center"> 1.3542 </td>
  <td align="center"> 2.6627 </td>
  <td align="center"> 1.5390 </td>
  <td align="center"> 0.4788 </td>
</tr>
<tr>
  <td align="center"> 61&deg; </td>
  <td align="center"> 0.53451 </td>
  <td align="center"> 0.48481 </td>
  <td align="center"> 0.141298 </td>
  <td align="center"> 1.6974 </td>
  <td align="center"> 1.3597 </td>
  <td align="center"> 2.6624 </td>
  <td align="center"> 1.5816 </td>
  <td align="center"> 0.4807 </td>
</tr>
<tr>
  <td align="center"> 66&deg; </td>
  <td align="center"> 0.44429 </td>
  <td align="center"> 0.40674 </td>
  <td align="center"> 0.135785 </td>
  <td align="center"> 1.5253 </td>
  <td align="center"> 1.4006 </td>
  <td align="center"> 2.6980 </td>
  <td align="center"> 1.8732 </td>
  <td align="center"> 0.4952 </td>
</tr>
<tr>
  <td align="center"> 69&deg; </td>
  <td align="center"> 0.38813 </td>
  <td align="center"> 0.35837 </td>
  <td align="center"> 0.127424 </td>
  <td align="center"> 1.4388 </td>
  <td align="center"> 1.4278 </td>
  <td align="center"> 2.7557 </td>
  <td align="center"> 2.1105 </td>
  <td align="center"> 0.5073 </td>
</tr>
<tr>
  <td align="center"> 71&deg; </td>
  <td align="center"> 0.35022 </td>
  <td align="center"> 0.32557 </td>
  <td align="center"> 0.119625 </td>
  <td align="center"> 1.3647 </td>
  <td align="center"> 1.4723 </td>
  <td align="center"> 2.8144 </td>
  <td align="center"> 2.3873 </td>
  <td align="center"> 0.5205 </td>
</tr>
<tr>
  <td align="center"> 72&deg; </td>
  <td align="center"> 0.33119 </td>
  <td align="center"> 0.30902 </td>
  <td align="center"> 0.115054 </td>
  <td align="center"> 1.3337 </td>
  <td align="center"> 1.4919 </td>
  <td align="center"> 2.8512 </td>
  <td align="center"> 2.5357 </td>
  <td align="center"> 0.5275 </td>
</tr>
<tr>
  <td align="center"> 73&deg; </td>
  <td align="center"> 0.31213 </td>
  <td align="center"> 0.29237 </td>
  <td align="center"> 0.110044 </td>
  <td align="center"> 1.3028 </td>
  <td align="center"> 1.5140 </td>
  <td align="center"> 2.8938 </td>
  <td align="center"> 2.7072 </td>
  <td align="center"> 0.5353 </td>
</tr>
<tr>
  <td align="center"> 75&deg; </td>
  <td align="center"> 0.27405 </td>
  <td align="center"> 0.25882 </td>
  <td align="center"> 0.098753 </td>
  <td align="center"> 1.2419 </td>
  <td align="center"> 1.5663 </td>
  <td align="center"> 3.0001 </td>
  <td align="center"> 3.1371 </td>
  <td align="center"> 0.5538 </td>
</tr>
<tr>
  <td align="center"> 78&deg; </td>
  <td align="center"> 0.21726 </td>
  <td align="center"> 0.20791 </td>
  <td align="center"> 0.078934 </td>
  <td align="center"> 1.1513 </td>
  <td align="center"> 1.6731 </td>
  <td align="center"> 3.2327 </td>
  <td align="center"> 4.1282 </td>
  <td align="center"> 0.5915 </td>
</tr>
<tr>
  <td align="center"> 81&deg; </td>
  <td align="center"> 0.16126 </td>
  <td align="center"> 0.15643 </td>
  <td align="center"> 0.056499 </td>
  <td align="center"> 1.0599 </td>
  <td align="center"> 1.8353 </td>
  <td align="center"> 3.6139 </td>
  <td align="center"> 5.9641 </td>
  <td align="center"> 0.6489 </td>
</tr>
</table>


<table border="1" align="center" cellpadding="5">

<tr>
  <th align="center" colspan="10"><font size="+1"><b>Table 2:</b></font>&nbsp; Incompressible <math>~(n=0)</math> Roche Ellipsoids with <math>~\lambda = p = 1</math></th>
</tr>

<tr>
  <th align="center" colspan="7">LRS93 Supplements</th>
  <th align="center" colspan="3">LRS93 Check</th>
</tr>
<tr>
  <td align="center"> (1) </td>
  <td align="center"> (2) </td>
  <td align="center"> (3) </td>
  <td align="center"> (4) </td>
  <td align="center"> (5)</td>
  <td align="center"> (6) </td>
  <td align="center"> (7) </td>
  <td align="center">(8)</td>
  <td align="center">(9)</td>
  <td align="center">(10)</td>
</tr>
<tr>
<td align="center"><math>~r/a_1</math>
<td align="center"><math>~r/R</math>
<td align="center"><math>~a_2/a_1</math>
<td align="center"><math>~a_3/a_1</math>
<td align="center"><math>~\bar\Omega</math>
<td align="center"><math>~\bar{J}</math>
<td align="center"><math>~\bar{E}</math>
<td align="center"><math>~\biggl(\frac{rp^{1/3}}{a_1}\biggr)^3\biggl(\frac{R}{rp^{1/3}}\biggr)^3\biggl( \frac{a_2}{a_1}\cdot \frac{a_3}{a_1}\biggr)^{-1}</math>
<td align="center"><math>~\bar\Omega_\mathrm{Kep}</math>
<td align="center"><math>~\bar{J}_\mathrm{Kep}</math>
</tr>
<tr>
  <td align="center"> 5.0 </td>
  <td align="center"> 5.131 </td>
  <td align="center"> 0.9707 </td>
  <td align="center"> 0.9533 </td>
  <td align="center"> 0.1406 </td>
  <td align="center"> 1.653 </td>
  <td align="center"> -0.6943 </td>
  <td align="center">1.0000</td>
  <td align="center">0.1405</td>
  <td align="center">1.6515</td>
</tr>
<tr>
  <td align="center"> 4.0 </td>
  <td align="center"> 4.202 </td>
  <td align="center"> 0.9441 </td>
  <td align="center"> 0.9139 </td>
  <td align="center"> 0.1901 </td>
  <td align="center"> 1.522 </td>
  <td align="center"> -0.7128 </td>
  <td align="center">0.9998</td>
  <td align="center">0.1896</td>
  <td align="center">1.5180</td>
</tr>
<tr>
  <td align="center"> 3.0 </td>
  <td align="center"> 3.348 </td>
  <td align="center"> 0.8750 </td>
  <td align="center"> 0.8222 </td>
  <td align="center"> 0.2690 </td>
  <td align="center"> 1.408 </td>
  <td align="center"> -0.7349 </td>
  <td align="center">1.0001</td>
  <td align="center">0.2666</td>
  <td align="center">1.3954</td>
</tr>
<tr>
  <td align="center"> 2.7 </td>
  <td align="center"> 3.124 </td>
  <td align="center"> 0.8345 </td>
  <td align="center"> 0.7738 </td>
  <td align="center"> 0.3000 </td>
  <td align="center"> 1.386 </td>
  <td align="center"> -0.7404 </td>
  <td align="center">0.9998</td>
  <td align="center">0.2958</td>
  <td align="center">1.3661</td>
</tr>
<tr>
  <td align="center"> 2.5 </td>
  <td align="center"> 2.989 </td>
  <td align="center"> 0.7981 </td>
  <td align="center"> 0.7330 </td>
  <td align="center"> 0.3222 </td>
  <td align="center"> 1.377 </td>
  <td align="center"> -0.7427 </td>
  <td align="center">1.0002</td>
  <td align="center">0.3160</td>
  <td align="center">1.3506</td>
</tr>
<tr>
  <td align="center"> 2.380 </td>
  <td align="center"> 2.916 </td>
  <td align="center"> 0.7715 </td>
  <td align="center"> 0.7044 </td>
  <td align="center"> 0.3358 </td>
  <td align="center"> 1.375 </td>
  <td align="center"> -0.7432 </td>
  <td align="center">1.0005</td>
  <td align="center">0.3279</td>
  <td align="center">1.3436</td>
</tr>
<tr>
  <td align="center"> 2.2 </td>
  <td align="center"> 2.821 </td>
  <td align="center"> 0.7236 </td>
  <td align="center"> 0.6553 </td>
  <td align="center"> 0.3556 </td>
  <td align="center"> 1.380 </td>
  <td align="center"> -0.7418 </td>
  <td align="center">1.0003</td>
  <td align="center">0.3446</td>
  <td align="center">1.3373</td>
</tr>
<tr>
  <td align="center"> 2.112 </td>
  <td align="center"> 2.783 </td>
  <td align="center"> 0.6960 </td>
  <td align="center"> 0.6281 </td>
  <td align="center"> 0.3648 </td>
  <td align="center"> 1.386 </td>
  <td align="center"> -0.7399 </td>
  <td align="center">0.9998</td>
  <td align="center">0.3518</td>
  <td align="center">1.3366</td>
</tr>
<tr>
  <td align="center"> 2.0 </td>
  <td align="center"> 2.744 </td>
  <td align="center"> 0.6561 </td>
  <td align="center"> 0.5901 </td>
  <td align="center"> 0.3753 </td>
  <td align="center"> 1.399 </td>
  <td align="center"> -0.7358 </td>
  <td align="center">1.0001</td>
  <td align="center">0.3592</td>
  <td align="center">1.3389</td>
</tr>
<tr>
  <td align="center"> 1.801 </td>
  <td align="center"> 2.713 </td>
  <td align="center"> 0.5708 </td>
  <td align="center"> 0.5123 </td>
  <td align="center"> 0.3886 </td>
  <td align="center"> 1.441 </td>
  <td align="center"> -0.7218 </td>
  <td align="center">1.0004</td>
  <td align="center">0.3654</td>
  <td align="center">1.3552</td>
</tr>
<tr>
  <td align="center"> 1.697 </td>
  <td align="center"> 2.724 </td>
  <td align="center"> 0.5184 </td>
  <td align="center"> 0.4664 </td>
  <td align="center"> 0.3908 </td>
  <td align="center"> 1.477 </td>
  <td align="center"> -0.7097 </td>
  <td align="center">1.0000</td>
  <td align="center">0.3632</td>
  <td align="center">1.3727</td>
</tr>
<tr>
  <td align="center"> 1.6 </td>
  <td align="center"> 2.759 </td>
  <td align="center"> 0.4644 </td>
  <td align="center"> 0.4198 </td>
  <td align="center"> 0.3884 </td>
  <td align="center"> 1.524 </td>
  <td align="center"> -0.6939 </td>
  <td align="center">1.0004</td>
  <td align="center">0.3563</td>
  <td align="center">1.3977</td>
</tr>
<tr>
  <td align="center"> 1.5 </td>
  <td align="center"> 2.831 </td>
  <td align="center"> 0.4040 </td>
  <td align="center"> 0.3682 </td>
  <td align="center"> 0.3798 </td>
  <td align="center"> 1.590 </td>
  <td align="center"> -0.6717 </td>
  <td align="center">1.0000</td>
  <td align="center">0.3428</td>
  <td align="center">1.4358</td>
</tr>
<tr>
  <td align="center"> 1.0 </td>
  <td align="center"> 5.312 </td>
  <td align="center"> 0.0823 </td>
  <td align="center"> 0.0811 </td>
  <td align="center"> 0.1685 </td>
  <td align="center"> 2.888 </td>
  <td align="center"> -0.3772 </td>
  <td align="center">0.9996</td>
  <td align="center">0.1334</td>
  <td align="center">2.2859</td>
</tr>
</table>


====Digesting LRS93S Results====

In column (12) of the above table we have combined the data from columns (5), (6), (7), &amp; (8) to demonstrate that LRS93S indeed used the definition of <math>~R</math>, given above for their incompressible, Roche ellipsoid configurations; these terms should zombie to give unity, and they appear to do so, within the accuracy presented by the data from LRS93S.  In column 14 of the table, we have listed the value of <math>~\bar{J}_\mathrm{Kep}</math>, as given by the above expression.  Column 13 lists <math>~\bar{\Omega}_\mathrm{Kep}</math>, as follows:

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

<tr>
  <td align="right">
<math>~\Omega_K</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{GM_\mathrm{tot}}{r^3} \biggr]^{1/2}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{GM}{R^3} \biggl( \frac{1+p}{p}\biggr) \biggl( \frac{R}{r}\biggr)^3\biggr]^{1/2}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[\frac{4\pi G \rho}{3}\biggr]^{1/ 2} \biggl( \frac{1+p}{p}\biggr)^{1 / 2} \biggl[\biggl( \frac{a_1}{r}\biggr)^3 \frac{a_2}{a_1} \cdot \frac{a_3}{a_1}\biggr]^{1/2}</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~ \bar\Omega_\mathrm{Kep}\biggr|_{p=1} \equiv \frac{\Omega_\mathrm{Kep}}{(\pi G \rho)^{1/2}}\biggr|_{p=1}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{8}{3}\biggl( \frac{a_1}{r}\biggr)^3 \frac{a_2}{a_1} \cdot \frac{a_3}{a_1}\biggr]^{1/2} \, .</math>
  </td>
</tr>
</table>
</div>
The value of the LRS93S correction factor, <math>~(1+\delta)</math>, can be obtained either from the ratio, <math>~\bar{J}/\bar{J}_\mathrm{Kep}</math>, or from the ratio, <math>~\bar{\Omega}/\bar{\Omega}_\mathrm{Kep}</math>.

====Digesting the EFE Results====

The EFE table lists values of <math>~\bar\Omega_\mathrm{Kep}</math>, but not values of <math>~r/a_1</math>.  Inverting the expression just provided gives,

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

<tr>
  <td align="right">
<math>~\biggl( \frac{a_1}{r}\biggr)^3 \frac{a_2}{a_1} \cdot \frac{a_3}{a_1} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{3}{8}\bar\Omega^2_\mathrm{Kep}</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{r}{a_1}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{8}{3\bar\Omega^2_\mathrm{Kep} }\biggl( \frac{a_2}{a_1} \cdot \frac{a_3}{a_1} \biggr) \biggr]^{1 / 3} \, .</math>
  </td>
</tr>
</table>
</div>

In the above "EFE Check" column, we've listed these inferred values of <math>~r/a_1</math>.  Or, we might prefer the ratio, <math>~r/R</math>, which is obtained from the Keplerian frequency via the expression,

<div align="center">
<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{4\pi G \rho}{3} \biggl( \frac{1+p}{p}\biggr) \biggl[\frac{R}{r} \biggr]^{3} </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~  \biggl[\frac{R}{r} \biggr]^{3} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>\biggl[\frac{3}{4}\biggr] \biggl( \frac{p}{1+p}\biggr) ~\biggl( \frac{\Omega_K^2}{\pi G\rho}\biggr)</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~  \biggl(\frac{r}{R} \biggr)_{p=1}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>\biggl[ \frac{3}{8} \biggl( \frac{\Omega_K^2}{\pi G\rho}\biggr)_{p=1} \biggr]^{-1/3}  \, .</math>
  </td>
</tr>
</table>
</div>


====Comparison====
In the context of our [[#SphericalLtot|simplistic spherical model, above]], we derived the following expression for the total angular momentum:

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

<tr>
  <td align="right">
<math>~L_\mathrm{tot} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(G M_\mathrm{tot}^3 R)^{1 / 2} \biggl\{ \frac{1}{(1+\lambda)}  \biggl( \frac{d}{R}\biggr)^{1 / 2}
+ \frac{2}{5}\biggl[ 1
+  \cancelto{0}{\frac{1}{\lambda}\biggl( \frac{R^'}{R}\biggr)^{2}} \biggr] \biggl( \frac{d}{R}\biggr)^{-3/2}
\biggr\}\biggl(\frac{\lambda}{1+\lambda} \biggr) \, .
</math>
  </td>
</tr>
</table>
</div>

Rewriting our [[#Kepler|just-derived "Keplerian" expression]] to emphasize the ratio <math>~r/R</math> instead of <math>~r/a_1</math>, and to highlight the system's total mass in the leading ''dimensional'' coefficient, allows us to more readily recognize the overlap with this simpler expression.
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~J_\mathrm{Kep}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(GM^3 R)^{1/2} 
\biggl[ \frac{1}{(1+p)}\biggl( \frac{r}{R} \biggr)^{1/2}  + \frac{1}{5}\biggl( 1 + \frac{a_2^2}{a_1^2}\biggr) \biggl( \frac{a_1}{R} \cdot  \frac{R}{r} \biggr)^{2} \biggl( \frac{r}{R} \biggr)^{1/2} \biggr]
\biggl( \frac{1+p}{p} \biggr)^{1/2} </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(GM_\mathrm{tot}^3 R)^{1/2} \biggl( \frac{p}{1+p} \biggr)^{3/2}
\biggl[ \frac{1}{(1+p)}\biggl( \frac{r}{R} \biggr)^{1/2}  + \frac{1}{5}\biggl( 1 + \frac{a_2^2}{a_1^2}\biggr) \biggl( \frac{a_1}{R}  \biggr)^{2} \biggl( \frac{r}{R} \biggr)^{-3/2} \biggr]
\biggl( \frac{1+p}{p} \biggr)^{1/2} </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(GM_\mathrm{tot}^3 R)^{1/2} 
\biggl[ \frac{1}{(1+p)}\biggl( \frac{r}{R} \biggr)^{1/2}  + \frac{1}{5}\biggl( 1 + \frac{a_2^2}{a_1^2}\biggr) \biggl( \frac{a_1}{R}  \biggr)^{2} \biggl( \frac{r}{R} \biggr)^{-3/2} \biggr]
\biggl( \frac{p}{1+p} \biggr) </math>
  </td>
</tr>
</table>
</div>

It makes sense, then, to write the total angular momentum as,

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

<tr>
  <td align="right">
<math>~L_\mathrm{tot} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(G M_\mathrm{tot}^3 R)^{1 / 2} \biggl\{ \frac{1}{(1+\lambda)}  \biggl( \frac{d}{R}\biggr)^{1 / 2}
+ \frac{2}{5} \cdot \mathfrak{J}\biggl( \frac{d}{R}\biggr)^{-3/2}
\biggr\}\biggl(\frac{\lambda}{1+\lambda} \biggr) \, ,
</math>
  </td>
</tr>
</table>
</div>
where, <math>~\mathfrak{J} =1 </math> when one assumes that the primary star is spherical, but when tidal distortions are taken into account,
<div align="center">
<math>~\mathfrak{J} =  \frac{1}{2} \biggl( 1 + \frac{a_2^2}{a_1^2}\biggr) \biggl( \frac{a_1}{R}  \biggr)^{2} \, .</math>
</div>

<table border="1" cellpadding="5" align="center" width="80%">
<tr>
  <td align="center" colspan="2">
<font size="+1"><b>Figure 1:</b></font> "Roche" Binary Sequences with Point-Mass Secondary and <math>~M/M^' = 1</math>
  </td>
</tr>
<tr>
  <td align="center">
Our Constructed Diagram
  </td>
  <td align="center">
Extracted from Fig. 10 of  [http://adsabs.harvard.edu/abs/1993ApJS...88..205L LRS93Supplement] 
  </td>
</tr>
<tr>
  <td align="center">
[[File:DarwinP1compare.png|600px|Compare to LRS93S Fig10]]
  </td>
  <td align="center">
[[File:LRS93SFig10.png|400px|LRS93S Fig10]]
  </td>
</tr>
<tr>
  <td align="left" colspan="2">
<b>Left:</b> &nbsp; Curves showing how the total system angular momentum varies with binary separation when <math>~n=0</math> and the secondary star <math>~(M^')</math> is treated as a point mass.  (Blue dashed curve) Primary star assumed to be a sphere and, hence, <math>~\mathfrak{J} = 1</math>; (Green filled circular markers) Primary star is an (EFE) ellipsoidal configuration with axis ratios specified by columns 2 and 3 of our Table 1, normalized angular momentum specified by column 6 of our Table 1, and binary separation specified by column 7 of our Table 1; (Solid red curve connecting red filled circular markers) Primary star is an (LRS93S) ellipsoidal configuration with axis ratios specified by columns 3 and 4 of our Table 2, normalized angular momentum specified by column 6 of our Table 2, and binary separation specified by column 2 of our Table 2.  The green filled circular markers define the same (EFE) sequence that is presented as a dot-dashed curve in the right-hand panel; the red filled circular markers and associated smoothed curve define the same (LRS93S) sequence that is presented as a solid curve in the right-hand panel.  The purple filled circlular marker identifies the turning point along the (LRS93S) sequence associated with the minimum system angular momentum; the yellow filled circular marker identifies the turning point along the same sequence that is associated with the minimum separation &#8212; the so-called "Roche" limit.

<b>Right:</b> &nbsp; (The following text is largely taken from the Fig. 10 caption of LRS93S)  Equilibrium curves generated by LRS93S showing total angular momentum as a function of binary separation along two incompressible, and three compressible Roche sequences with <math>~M/M^' = 1</math>.  The various curves display results from polytropic configurations having <math>~n=0</math> (''solid line''), <math>~n=1</math> (''dotted line''), <math>~n=1.5</math> (''short-dashed line''), and <math>~n=2.5</math> (''long-dashed line'').  For comparison, the sequence obtained by EFE for <math>~n=0</math> is also drawn (''dotted-dashed line'').
  </td>
</tr>
</table>

===Incompressible Roche Ellipsoids (&lambda; = 0)===

<table border="1" cellpadding="5" align="center">
<tr>
  <td align="center">
Extracted from p. 229 of [http://rsta.royalsocietypublishing.org/content/206/402-412/161 G. H. Darwin (1906)]
  </td>
</tr>
<tr>
  <td align="center">
[[File:DarwinText01.png|700px|Roche limit]]
  </td>
</tr>
<tr>
  <td align="center">
Extracted from p. 242 of [http://rsta.royalsocietypublishing.org/content/206/402-412/161 G. H. Darwin (1906)]
  </td>
</tr>
<tr>
  <td align="center">
[[File:DarwinText02.png|700px|Roche limit]]
  </td>
</tr>
</table>

Here we examine the results presented by Roche, by Darwin, and by EFE for the case of a point-mass secondary (<math>~(M^')</math> and a primary whose mass <math>~(M)</math> is formally zero.  In this case, we must use a different scheme for normalizing physical quantities.  Because the secondary is not spinning and it has no orbital motion, only the primary contributes to the system's "angular momentum"; but because the primary has no mass, we need to examine its (and, hence, the system's) ''specific'' angular momentum.  Specifically,

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

<tr>
  <td align="right">
<math>~\frac{I}{M}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{5}a_1^2  \biggl( 1 + \frac{a_2^2}{a_1^2}\biggr) \, ,</math>
  </td>
</tr>
</table>
</div>
and,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~j \equiv \frac{J_\mathrm{tot}}{M}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{r^2}{(1+\cancelto{0}{p})} + \frac{I}{M} \biggr]\Omega_\mathrm{Kep} </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~R^2\biggl[ \biggl(\frac{r}{R}\biggr)^2 + \frac{2}{5} \cdot \mathfrak{J} \biggr]\biggl(\frac{GM^'}{r^3}\biggr)^{1 / 2} </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(GM^' R)^{1 / 2} \biggl[ \biggl(\frac{r}{R}\biggr)^{1 / 2} + \frac{2}{5} \cdot \mathfrak{J} \biggl(\frac{r}{R}\biggr)^{-3 / 2} \biggr] \, , </math>
  </td>
</tr>
</table>
</div>
where, in order to ensure that the density of the primary remains constant along an equilibrium sequece, the adopted normalizing length scale is customarily,
<div align="center">
<math>~R^3 \equiv a_1 a_2 a_3 ~~~\Rightarrow ~~~ \frac{R}{a_1} = \biggl( \frac{a_2}{a_1}\cdot \frac{a_3}{a_1} \biggr)^{1 / 3} \, ,</math>
</div>

in which case, 
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\mathfrak{J} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{2} \biggl( 1 + \frac{a_2^2}{a_1^2}\biggr) \biggl( \frac{a_1}{R}  \biggr)^{2} </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{2} \biggl( 1 + \frac{a_2^2}{a_1^2}\biggr) \biggl( \frac{a_2}{a_1}\cdot \frac{a_3}{a_1} \biggr)^{-2 / 3} </math>
  </td>
</tr>
</table>
</div>


<table border="1" align="center" cellpadding="5">

<tr>
  <th align="center" colspan="8"><font size="+1"><b>Table 3:</b></font>&nbsp; Incompressible <math>~(n=0)</math> Roche Ellipsoids with <math>~\lambda = p = 0</math></th>
</tr>

<tr>
  <th align="center" colspan="4">
Extracted from Table 1 of [http://adsabs.harvard.edu/abs/1963ApJ…138.1182C Chandrasekhar (1963)]<br />
same as [<b>[[User:Tohline/Appendix/References#EFE|<font color="red">EFE</font>]]</b>] Table XVI</th>
  <th align="center" colspan="5">EFE Check</th>
</tr>
<tr>
  <td align="center"> (1) </td>
  <td align="center"> (2) </td>
  <td align="center"> (3) </td>
  <td align="center"> (4) </td>
  <td align="center"> (5) </td>
  <td align="center"> (6) </td>
  <td align="center"> (7) </td>
  <td align="center"> (8) </td>
</tr>
<tr>
<td align="center"><math>~\cos^{-1}(a_3/a_1)</math></td>
<td align="center"><math>~a_2/a_1</math></td>
<td align="center"><math>~a_3/a_1</math></td>
<td align="center"><math>~\Omega^2</math></td>
<td align="center"><math>~r/R</math></td>
<td align="center"><math>~R/a_1</math></td>
<td align="center"><math>~\mathfrak{J}</math></td>
<td align="center"><math>~j/(GM^' R)^{1/2}</math></td>
</tr>
<tr>
  <td align="center"> 24&deg; </td>
  <td align="center"> 0.93188 </td>
  <td align="center"> 0.91355 </td>
  <td align="center"> 0.022624 </td>
  <td align="center"> 3.8916 </td>
  <td align="center"> 0.9478 </td>
  <td align="center"> 1.0400 </td>
  <td align="center"> 2.0269 </td>
</tr>
<tr>
  <td align="center"> 36&deg; </td>
  <td align="center"> 0.84112 </td>
  <td align="center"> 0.80902 </td>
  <td align="center"> 0.047871 </td>
  <td align="center"> 3.0312 </td>
  <td align="center"> 0.8796 </td>
  <td align="center"> 1.1035 </td>
  <td align="center"> 1.8247 </td>
</tr>
<tr>
  <td align="center"> 48&deg; </td>
  <td align="center"> 0.70687 </td>
  <td align="center"> 0.66913 </td>
  <td align="center"> 0.074799 </td>
  <td align="center"> 2.6122 </td>
  <td align="center"> 0.7791 </td>
  <td align="center"> 1.2352 </td>
  <td align="center"> 1.7333 </td>
</tr>
<tr>
  <td align="center"> 57&deg; </td>
  <td align="center"> 0.57787 </td>
  <td align="center"> 0.54464 </td>
  <td align="center"> 0.088267 </td>
  <td align="center"> 2.4720 </td>
  <td align="center"> 0.68022 </td>
  <td align="center"> 1.4415 </td>
  <td align="center"> 1.7206 </td>
</tr>
<tr>
  <td align="center"> 60&deg; </td>
  <td align="center"> 0.53013 </td>
  <td align="center"> 0.50000 </td>
  <td align="center"> 0.089946 </td>
  <td align="center"> 2.4565 </td>
  <td align="center"> 0.6424 </td>
  <td align="center"> 1.5523 </td>
  <td align="center"> 1.7286 </td>
</tr>
<tr>
  <td align="center"> 61&deg; </td>
  <td align="center"> 0.51373 </td>
  <td align="center"> 0.48481 </td>
  <td align="center"> 0.090068 </td>
  <td align="center"> 2.4554 </td>
  <td align="center"> 0.6292 </td>
  <td align="center"> 1.5964 </td>
  <td align="center"> 1.7329 </td>
</tr>
<tr>
  <td align="center"> 62&deg; </td>
  <td align="center"> 0.49714 </td>
  <td align="center"> 0.46947 </td>
  <td align="center"> 0.089977 </td>
  <td align="center"> 2.4562 </td>
  <td align="center"> 0.6157 </td>
  <td align="center"> 1.6450 </td>
  <td align="center"> 1.7382 </td>
</tr>
<tr>
  <td align="center"> 63&deg; </td>
  <td align="center"> 0.48040 </td>
  <td align="center"> 0.45399 </td>
  <td align="center"> 0.089689 </td>
  <td align="center"> 2.4589 </td>
  <td align="center"> 0.6019 </td>
  <td align="center"> 1.6984 </td>
  <td align="center"> 1.7443 </td>
</tr>
<tr>
  <td align="center"> 66&deg; </td>
  <td align="center"> 0.42898 </td>
  <td align="center"> 0.40674 </td>
  <td align="center"> 0.087201 </td>
  <td align="center"> 2.48202 </td>
  <td align="center"> 0.5588 </td>
  <td align="center"> 1.8959 </td>
  <td align="center"> 1.7694 </td>
</tr>
<tr>
  <td align="center"> 71&deg; </td>
  <td align="center"> 0.34052 </td>
  <td align="center"> 0.32557 </td>
  <td align="center"> 0.077474 </td>
  <td align="center"> 2.5818 </td>
  <td align="center"> 0.4804 </td>
  <td align="center"> 2.4178 </td>
  <td align="center"> 1.8399 </td>
</tr>
<tr>
  <td align="center"> 72&deg; </td>
  <td align="center"> 0.32254 </td>
  <td align="center"> 0.30902 </td>
  <td align="center"> 0.074648 </td>
  <td align="center"> 2.6140 </td>
  <td align="center"> 0.4636 </td>
  <td align="center"> 2.5679 </td>
  <td align="center"> 1.8598 </td>
</tr>
<tr>
  <td align="center"> 75&deg; </td>
  <td align="center"> 0.26827 </td>
  <td align="center"> 0.25882 </td>
  <td align="center"> 0.064426 </td>
  <td align="center"> 2.7455 </td>
  <td align="center"> 0.4110 </td>
  <td align="center"> 3.1728 </td>
  <td align="center"> 1.9359 </td>
</tr>
<tr>
  <td align="center"> 79&deg; </td>
  <td align="center"> 0.19569 </td>
  <td align="center"> 0.19081 </td>
  <td align="center"> 0.047111 </td>
  <td align="center"> 3.0475 </td>
  <td align="center"> 0.3342 </td>
  <td align="center"> 4.6471 </td>
  <td align="center"> 2.0951 </td>
</tr>
<tr>
  <td align="center" colspan="8">
[[File:P0Diagram.png|600px|Roche, Darwin, and Chandrasekhar p=0]]
  </td>
</tr>
</table>

===References===

* [http://rsta.royalsocietypublishing.org/content/206/402-412/161 G. H. Darwin (1906)] ''On the Figure and Stability of a Liquid Satellite'', Philosophical Transactions of the Royal Society A, 206, 161.
* [http://adsabs.harvard.edu/abs/1919pcsd.book.....J J. H. Jeans (1919)] ''Problems of Cosmogony and Stellar Dynamics'' -- see especially beginning in Ch. III (p. 46)
* [http://adsabs.harvard.edu/abs/1959cbs..book.....K Z. Kopal (1959)], ''Close Binary Systems'' -- see especially, Ch. III (p. 125); but not particularly useful.
* Chapter 8 (pp. 189-241) of [<b>[[User:Tohline/Appendix/References#EFE|<font color="red">EFE</font>]]</b>]
* Chapter 16 (pp. 448-469) of [<b>[[User:Tohline/Appendix/References#T78|<font color="red">T78</font>]]</b>]
* [http://adsabs.harvard.edu/abs/1993ApJ...406L..63L Lai, Rasio, &amp; Shapiro (1993a, ApJL, 406, L63)]
* [http://adsabs.harvard.edu/abs/1993ApJS...88..205L Lai, Rasio, &amp; Shapiro (1993b, ApJS, 88, 205)]
* [https://arxiv.org/abs/gr-qc/9909055 Y. Eriguchi &amp; K. Uryu (1999, Prog. Theor. Phys. Suppl., 136, 199-215)] ''Darwin-Riemann Problems in Newtonian Gravity''; see also [http://adsabs.harvard.edu/cgi-bin/bib_query?arXiv:gr-qc/9909055 ADS reference]


<div align="center">
<table border="1" cellpadding="5">
<tr>
  <td align="center" colspan="2">
<b>
Figures from (left) [http://adsabs.harvard.edu/abs/1993ApJ...406L..63L Lai, Rasio, &amp; Shapiro (1993)]
and (right) [http://adsabs.harvard.edu/abs/1997ApJ...490..311N New &amp; Tohline (1997)]
</b>
  </td>
</tr>

<tr>
  <td align="center" colspan="">
[[File:LRS93BinarySequence.png|350px|center|Lai, Rasio &amp; Shapiro (1993a)]]
<!-- [[Image:AAAwaiting01.png|600px|center]] -->
  </td>
  <td align="center" colspan="">
[[File:NewTohline97Fig1c.png|350px|center|New &amp; Tohline (1997)]]
  </td>
</tr>

</table>
</div>


<!--
====Coincidence Between Points of Secular and Dynamical Instability====
From the [[User:Tohline/SSC/Structure/BiPolytropes/Analytic5_1#Model_Sequences|accompanying graphical display of equilibrium sequences]], it seems that a <math>~\nu_\mathrm{max}</math> turning point will only exist in five-one bipolytropes for <math>~\mu_e/\mu_c</math> less than some value &#8212; call it, <math>~(\mu_e/\mu_c)_\mathrm{begin}</math> &#8212; which is less than but approximately equal to <math>~\tfrac{1}{3}</math>.  As we move along any sequence for which <math>~\mu_e/\mu_c < (\mu_e/\mu_c)_\mathrm{begin}</math>, in the direction of increasing <math>~\ell_i</math>, it is fair to ask whether the system becomes dynamically unstable (at <math>~[x_\mathrm{eq}]_\mathrm{crit}</math>) before or after it encounters the point of secular instability marked by <math>~\nu_\mathrm{max}</math>.
-->