Editing Apps/MaclaurinSpheroids (section)

===Apply Boundary Conditions===
The enthalpy should go to zero everywhere on the surface of the spheroid.  By pinning the surface down at two points and setting <math>~H=0</math> at both of these locations, we can determine the two unknown constants in the above expression.  We choose to pin down the edge of the configuration in the equatorial plane &#8212; ''i.e.'', at <math>(\varpi,z) = (a_1,0)</math> &#8212;  and along the symmetry axis at the pole &#8212; ''i.e.'', at <math>(\varpi,z) = (0,a_3)</math>. From the boundary condition at the pole, we derive the Bernoulli constant, specifically,
<div align="center">
<math>
C_\mathrm{B} = - \pi G \rho \biggl[ I_\mathrm{BT} a_1^2 - A_3 a_3^2  \biggr] = 
- \pi G \rho a_1^2 \biggl[ I_\mathrm{BT} - A_3 (1-e^2)  \biggr] ;
</math>
</div> 
and from the boundary condition in the equatorial plane we derive the rotational angular velocity, specifically,
<div align="center" id="EquilibriumFrequency">
<table align="center" border="0" cellpadding="5">
<tr>
  <td align="right">
<math>
\frac{1}{2}a_1^2 \omega_0^2 
</math>
  </td>
  <td align="center">
<math>
~=
</math>
  </td>
  <td align="left">
<math>
- C_\mathrm{B} - \pi G \rho \biggl[ I_\mathrm{BT} a_1^2 - A_1 a_1^2 \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>
\Rightarrow ~~~~~ \omega_0^2
</math>
  </td>
  <td align="center">
<math>
~=
</math>
  </td>
  <td align="left">
<math>
2\pi G \rho \biggl[ A_1 - A_3 (1-e^2) \biggr] \, .
</math>
  </td>
</tr>
<tr>
  <td align="center" colspan="3">
[<b>[[Appendix/References#T78|<font color="red">T78</font>]]</b>], &sect;4.5, p. 86, Eq. (52)<br />
[<b>[[Appendix/References#ST83|<font color="red">ST83</font>]]</b>], &sect;7.3, p. 172, Eq. (7.3.18)
  </td>
</tr>
</table> 
</div>

Plugging these constants into the expression for the enthalpy results in the desired solution,
<table align="center" border="0" cellpadding="5">
<tr>
  <td align="right">
<math>
H(\varpi,z)
</math>
  </td>
  <td align="center">
<math>
~=
</math>
  </td>
  <td align="left">
<math>
\pi G \rho a_1^2 A_3 (1-e^2)\biggl[1 - \biggl( \frac{\varpi}{a_1} \biggr)^2 - \biggl( \frac{z}{a_3} \biggr)^2  \biggr] .
</math>
  </td>
</tr>
</table>

[[File:CommentButton02.png|right|100px|Comment by J. E. Tohline:  In Tassoul (1978), the leading coefficient in the expression for the pressure &#8212; and, hence, the central pressure &#8212; is too large by a factor of 2.]]We know from our [[SR#Barotropic_Structure|separate discussion of supplemental, barotropic equations of state]] that, for a uniform-density, <math>~n = 0</math> polytropic configuration, the pressure is related to the enthalpy via the expression, <math>~P = H\rho</math>.  Hence, we conclude that,
<table align="center" border="0" cellpadding="5">
<tr>
  <td align="right">
<math>
P(\varpi,z)
</math>
  </td>
  <td align="center">
<math>
~=
</math>
  </td>
  <td align="left">
<math>
\pi G \rho^2 a_1^2 A_3 (1-e^2)\biggl[1 - \biggl( \frac{\varpi}{a_1} \biggr)^2 - \biggl( \frac{z}{a_3} \biggr)^2  
\biggr] 
</math>
  </td>
</tr>
<tr>
  <td align="center" colspan="3">
[<b>[[Appendix/References#T78|<font color="red">T78</font>]]</b>], &sect;4.5, p. 86, Eq. (51)<br />
[<b>[[Appendix/References#ST83|<font color="red">ST83</font>]]</b>], &sect;7.3, p. 172, Eqs. (7.3.16) &amp; (7.3.17)
  </td>
</tr>

<tr>
  <td align="right">
<math>
\Rightarrow ~~~~ \frac{P}{P_0}
</math>
  </td>
  <td align="center">
<math>
~=
</math>
  </td>
  <td align="left">
<math>
\frac{3}{2} A_3 (1-e^2)^{2/3}\biggl[1 - \biggl( \frac{\varpi}{a_1} \biggr)^2 - \biggl( \frac{z}{a_3} \biggr)^2  
\biggr] ,
</math>
  </td>
</tr>
</table>
where,
<div align="center">
<math>~
P_0 \equiv \frac{2}{3}\pi G \rho^2 a_\mathrm{mean}^2 = \frac{2}{3}\pi G \rho^2 a_1^2 (1-e^2)^{1/3} .
</math>
</div>