Editing SSC/Stability/UniformDensity (section)

===Setup as Presented by Sterne (1937)===

In &sect;2 of his paper, {{ Sterne37 }} details the structural properties of an equilibrium, uniform-density sphere as follows.  (Text taken verbatim from {{ Sterne37hereafter }} are presented here in green.)  Given that <font color="green">the undisturbed density is constant and equal to the mean density</font>, <math>~\bar\rho</math>, the <font color="green">mass within any radius is</font>,
<div align="center">
<math>M_r = \biggl( \frac{4\pi}{3} \biggr) \bar\rho \xi_0^3 \, ;</math>
</div>
<font color="green">the undisturbed values of gravity and the pressure are</font>, respectively,
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<math>g_0 \equiv \frac{GM_r}{\xi_0^2} = \biggl( \frac{4\pi}{3} \biggr) G\bar\rho R x \, </math>
</div>
and 
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<math>P_0 = \biggl( \frac{2\pi}{3} \biggr) G R^2 \bar\rho^2(1 - x^2) \, ;</math>
</div>
and the quantity,
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<math>\mu \equiv \frac{g_0 \bar\rho \xi_0}{P_0} = \frac{2x^2}{(1-x^2)} \, .</math>
</div>

Hence, for this particular equilibrium model, the wave equation derived by {{ Sterne37hereafter }} &#8212; his equation (1.91), as displayed above &#8212; becomes,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\xi_1^{ ' ' } + \biggl[\frac{4-\mu}{x} \biggr]\xi_1^' + \frac{R\bar\rho}{P_0} \biggl( \frac{n^2 R}{\gamma} - \frac{\alpha g_0}{x} \biggr) \xi_1</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\xi_1^{ ' ' } + \frac{1}{x}\biggl[4 -\frac{2x^2}{(1-x^2)} \biggr]\xi_1^' + 
\frac{3}{2\pi G R \bar\rho (1 - x^2)}\biggl[ \frac{n^2 R}{\gamma} - \biggl( \frac{4\pi}{3} \biggr) \alpha G\bar\rho R \biggr] \xi_1</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(1-x^2) \xi_1^{ ' ' } + \frac{1}{x}\biggl[4(1-x^2) - 2x^2 \biggr]\xi_1^' + \biggl[ \frac{3n^2 }{2\pi \gamma  G \bar\rho} - 2 \alpha \biggr] \xi_1</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(1-x^2) \xi_1^{ ' ' } + \frac{1}{x}\biggl[4 - 6x^2 \biggr]\xi_1^' + \mathfrak{F} \xi_1 \, ,</math>
  </td>
</tr>
</table>
</div>
where,
<div align="center">
<math>~\mathfrak{F}  \equiv \frac{3n^2 }{2\pi \gamma  G \bar\rho} - 2 \alpha \, .</math>
</div>

Note that, once the value of the parameter, <math>~\mathfrak{F}</math>, has been determined for a given eigenvector, the square of the eigenfrequency will also be known via the inversion of this last expression.  Specifically, in terms of <math>~\mathfrak{F}</math>,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~n^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2\pi \gamma  G \bar\rho}{3} \biggl[ \mathfrak{F} + 2 \biggl(3-\frac{4}{\gamma}\biggr)\biggr]</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~~~ \frac{n^2}{4\pi G \bar\rho}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl(1 + \frac{\mathfrak{F}}{6} \biggr)\gamma -\frac{4}{3}  \, .</math>
  </td>
</tr>
</table>
</div>

As a reminder, in these terms the inner boundary condition is
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<math>~\xi_1^'\biggr|_{x = 0} = 0 \, .</math>
</div>
And the outer boundary condition becomes,
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<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\xi_1^'</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\xi_1 \biggl( \frac{n^2 R}{\gamma g_0} - \alpha \biggr)</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\xi_1 \biggl[ \frac{3}{x}\biggl( \frac{n^2 }{4\pi \gamma G\bar\rho } \biggr) - 3 + \frac{4}{\gamma} \biggr]</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~3\xi_1 \biggl\{ \frac{1}{x}\biggl[ \biggl(1 + \frac{\mathfrak{F}}{6} \biggr) -\frac{4}{3\gamma} \biggr] - 1 + \frac{4}{3\gamma} \biggr\}</math>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; at &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 
<math>~x = 1 \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~~~\biggl[ \xi_1^' - \frac{\mathfrak{F} \xi_1}{2} \biggr]_{x=1}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~0 \, .</math>
  </td>
</tr>
</table>
</div>


<!--  OMIT NEXT TEXT
==Analytic Solution==

===First few lowest-order modes===
* <font color="purple">Mode 0</font>:
: <math>x_0 = \mathrm{constant}</math>, in which case,
<div align="center">
<math>
\omega_0^2 = - 2(4 - 3\gamma_\mathrm{g})\biggl[ \frac{2\pi G\rho_c}{3} \biggr] = 4\pi G \rho_c \biggl[ \gamma_\mathrm{g}- \frac{4}{3} \biggr]
</math>
</div>

* <font color="purple">Mode 1</font>:
: <math>x_1 = a + b\chi_0^2</math>, in which case,
<div align="center">
<math>
\frac{dx}{d\chi_0} = 2b\chi_0; ~~~~ \frac{d^2 x}{d\chi_0^2} =  2b;
</math>
</div>

<div align="center">
<math>
\frac{1}{(1 - \chi_0^2)}  \biggl\{ 2b (1 - \chi_0^2) + 8b \biggl[1 -  \frac{3}{2}\chi_0^2 \biggr]  +  A_1 \biggl(1 + \frac{b}{a}\chi_0^2 \biggr) \biggr\} = 0 ,
</math><br />
</div>
where,

<div align="center">
<math>
A_1 \equiv \frac{a}{\gamma_\mathrm{g}}\biggl[ \biggl( \frac{3}{2\pi G\rho_c} \biggr) \omega_1^2+ 2(4 - 3\gamma_\mathrm{g}) \biggr] .
</math>
</div>
Therefore,

<div align="center">
<math>
(A_1 + 10b) + \biggl[ \biggl(\frac{b}{a}\biggr) A_1 - 14b \biggr] \chi_0^2  = 0 ,
</math>
<br />
<br />
<math>
\Rightarrow ~~~~~ A_1  = - 10b ~~~~~\mathrm{and} ~~~~~ A_1 = 14a
</math>
<br />
<br />
<math>
\Rightarrow ~~~~~ \frac{b}{a} = -\frac{7}{5}  ~~~~~\mathrm{and} ~~~~~ \frac{A_1}{a} = 14 = \frac{1}{\gamma_\mathrm{g}}\biggl[ \biggl( \frac{3}{2\pi G\rho_c} \biggr) \omega_1^2+ 2(4 - 3\gamma_\mathrm{g}) \biggr] .

</math>
</div>
Hence,
<div align="center">
<math>
\biggl( \frac{3}{2\pi G\rho_c} \biggr) \omega_1^2 = 20\gamma_\mathrm{g}  -8
</math>
<br />
<br />
<math>
\Rightarrow ~~~~~ \omega_1^2 = \frac{2}{3}\biggl( 4\pi G\rho_c \biggr) (5\gamma_\mathrm{g}  -2)
</math>
</div>
and, to within an arbitrary normalization factor,
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<math>
x_1 = 1 - \frac{7}{5}\chi_0^2 .
</math>
</div>
<br />

END OMISSION -->