Editing SSC/FreeEnergy/EquilibriumSequenceInstabilities (section)

====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>