Editing Appendix/Ramblings/51BiPolytropeStability/BetterInterface (section)

==Linearized Equations With Preferred Normalizations==

===Review and Elaborate===
<div align="center">
<table border="1" cellpadding="10">
<tr><td align="center">
<font color="#770000">'''Linearized'''</font><br />
<span id="Continuity"><font color="#770000">'''Equation of Continuity'''</font></span><br />
<math>
r_0 \frac{dx}{dr_0} = - 3 x - d ,
</math><br />

<font color="#770000">'''Linearized'''</font><br />
<span id="PGE:Euler"><font color="#770000">'''Euler + Poisson Equations'''</font></span><br />
<math>
\frac{P_0}{\rho_0} \frac{dp}{dr_0}  = (4x + p)g_0 + \omega^2 r_0 x ,
</math><br />

<font color="#770000">'''Linearized'''</font><br />
<span id="PGE:AdiabaticFirstLaw">Adiabatic Form of the<br />
<font color="#770000">'''First Law of Thermodynamics'''</font></span><br />
<math>
p = \gamma_\mathrm{g} d  \, .
</math>
</td></tr>
</table>
</div>

The LHS of the "linearized Euler + Poisson" equation is rewritten as,
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right"><math>\mathrm{LHS} = \frac{P_0}{\rho_0} \frac{dp}{dr_0}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\biggl[K_c^{-10} G^{9} M_\mathrm{tot}^{6}  \biggr]^{-1}\tilde{P}
\biggl[\biggl( \frac{K_c}{G} \biggr)^{3 / 2} \frac{1}{M_\mathrm{tot}} \biggr]^{-5} \tilde{\rho}^{-1}
\biggl[\biggl( \frac{K_c}{G} \biggr)^{5 / 2} M_\mathrm{tot}^{-2}  \biggr] \frac{dp}{d\tilde{r}}
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{\tilde{P}}{\tilde{\rho}} \frac{dp}{d\tilde{r}}
\biggl[K_c^{10} G^{-9} M_\mathrm{tot}^{-6}  \biggr]
\biggl[ G^{15/2} K_c^{-15/2}  M_\mathrm{tot}^5 \biggr] 
\biggl[K_c^{5/2} G^{-5/2} M_\mathrm{tot}^{-2}  \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{\tilde{P}}{\tilde{\rho}} \frac{dp}{d\tilde{r}}
\biggl[K_c^{5} G^{-4} M_\mathrm{tot}^{-3}  \biggr]
\, ;</math>
  </td>
</tr>
</table>
and,
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right"><math>g_0 = \frac{GM_r}{r_0^2}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
GM_\mathrm{tot} \tilde{M}_r
\biggl[\biggl( \frac{K_c}{G} \biggr)^{5 / 2} M_\mathrm{tot}^{-2}  \biggr]^2 \tilde{r}^{-2}
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{\tilde{M}_r}{\tilde{r}^2}
\biggl[\biggl( \frac{K_c}{G} \biggr)^{5 / 2} M_\mathrm{tot}^{-2}  \biggr]^2 
GM_\mathrm{tot}
</math>
  </td>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{\tilde{M}_r}{\tilde{r}^2}
\biggl[K_c^5 G^{-4}M_\mathrm{tot}^{-3}  \biggr] 
\, .
</math>
  </td>
</tr>
</table>
Therefore, multiplying the full equation through by <math>[K_c^{-5} G^4 M_\mathrm{tot}^3]</math> gives,
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right"><math>\frac{\tilde{P}}{\tilde{\rho}} \frac{dp}{d \tilde{r}} </math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
(4x + p)\frac{\tilde{M}_r}{\tilde{r}^2}
+
\biggl[K_c^{-5} G^4 M_\mathrm{tot}^3 \biggr] \biggl[\biggl( \frac{K_c}{G} \biggr)^{5 / 2} M_\mathrm{tot}^{-2}  \biggr]^{-1} \omega^2 \tilde{r} x
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
(4x + p)\frac{\tilde{M}_r}{\tilde{r}^2}
+
\tau_c^2 \omega^2 \tilde{r} x \, ,
</math>
  </td>
</tr>
</table>
where the square of the characteristic timescale, 

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

<tr>
  <td align="right"><math> \tau_c^2</math></td>
  <td align="center"><math>\equiv</math></td>
  <td align="left">
<math>
\biggl[ K_c^{-15/2} G^{13/2} M_\mathrm{tot}^{5}  \biggr] \, .
</math>
  </td>
</tr>
</table>

<table border="1" align="center" cellpadding="8" width="75%"><tr><td align="left">
<font color="red">ASIDE:</font>&nbsp;  Building on [[SSC/Stability/Polytropes#Numerical_Integration_from_the_Center,_Outward|an associated discussion]], the square of the dimensionless frequency also can be represented by the expression,

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

<tr>
  <td align="right"><math>\sigma_c^2 \equiv \frac{3\omega^2}{2\pi G\rho_c}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{3\omega^2}{2\pi \tilde{\rho}_c} \biggl[\biggl( \frac{G}{K_c} \biggr)^{15 / 2} M_\mathrm{tot}^5 \biggr]G^{-1} 
=
\frac{3\tau_c^2 \omega^2}{2\pi \tilde{\rho}_c} \, ,
</math>
  </td>
</tr>
</table>

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

<tr>
  <td align="right"><math>\tilde{\rho}_c</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
m_\mathrm{surf}^5 \biggl(\frac{\mu_e}{\mu_c}\biggr)^{-10} \, .
</math>
  </td>
</tr>
</table>
</td></tr></table>

<span id="FromEarlier">Hence we can write,</span>
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right"><math>\frac{dp}{d \tilde{r}} </math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{\tilde{\rho}}{\tilde{P}} \cdot \frac{\tilde{M}_r}{\tilde{r}^2}
\biggl[(4x + p)
+
\tau_c^2 \omega^2 \biggl(\frac{\tilde{r}^3 }{\tilde{M}_r}\biggr) x \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{\tilde{\rho}}{\tilde{P}} \cdot \frac{\tilde{M}_r}{\tilde{r}^2}
\biggl[ (4x + p)
+
\sigma_c^2 \biggl(\frac{2\pi}{3} \cdot \frac{\tilde{\rho}_c \tilde{r}^3 }{\tilde{M}_r}\biggr) x \biggr] \, .
</math>
  </td>
</tr>
</table>

----

Focusing on the core &hellip;

As demonstrated earlier, the leading term on the RHS of this expression can be rewritten to give,

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

<tr>
  <td align="right"><math>\frac{dp}{d \tilde{r}} </math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
2\biggl[\frac{\xi}{\tilde{r}} \biggr]
\biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-1} \xi 
\biggl[ (4x + p)
+
\sigma_c^2 \biggl(\frac{2\pi}{3} \cdot \frac{\tilde{\rho}_c \tilde{r}^3 }{\tilde{M}_r}\biggr) x \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right"><math>\Rightarrow ~~~ \frac{dp}{d \xi} </math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
2\biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-1} \xi 
\biggl[ (4x + p)
+
\sigma_c^2 \biggl(\frac{2\pi}{3} \cdot \frac{\tilde{\rho}_c \tilde{r}^3 }{\tilde{M}_r}\biggr) x \biggr]
\, .
</math>
  </td>
</tr>
</table>
Noting as well that,

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

<tr>
  <td align="right"><math>\tilde{\rho}_c \biggl( \frac{\tilde{r}^3}{\tilde{M}_r}\biggr)</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
m_\mathrm{surf}^5 \biggl(\frac{\mu_e}{\mu_c}\biggr)^{-10} 
\biggl[
\mathcal{m}_\mathrm{surf}^{-5} \biggl(\frac{\mu_e}{\mu_c}\biggr)^{10} 
\biggl( \frac{3}{4\pi } \biggr) \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{3/2}
\biggr] 
=
\frac{3}{4\pi } \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{3/2} \, ,
</math>
  </td>
</tr>

</table>
we have,

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

<tr>
  <td align="right"><math>\frac{dp}{d \xi} </math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
2\biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-1} \xi 
\biggl[ (4x + p)
+
\frac{\sigma_c^2}{2} \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{3/2} x \biggr] \, .
</math>
  </td>
</tr>
</table>

===At the Center===
====All &sigma;<sup>2</sup>====
According to our [[Appendix/Ramblings/PowerSeriesExpressions#Displacement_Function_for_Polytropic_LAWE|discussion in an appendix chapter]], starting from the center of the equilibrium configuration, the displacement function can be represented by a power-series expression of the form,

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

<tr>
  <td align="right">
<math>x</math>
  </td>
  <td align="center">
<math>\approx</math>
  </td>
  <td align="left">
<math> 1 - \biggl[\frac{(n+1)\mathfrak{F}}{60}\biggr]\xi^2 
\, ,</math>
  </td>
</tr>
</table>
where, <math>\mathfrak{F} \equiv (\sigma_c^2/\gamma - 2\alpha)</math>, and (see, for example, [[SSC/Structure/BiPolytropes/Analytic51Renormalize#Core|here]]),

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

<tr>
  <td align="right">
<math>\xi</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math> 
m_\mathrm{surf}^2 \biggl(\frac{\mu_e}{\mu_c}\biggr)^{-4} \biggl(\frac{2\pi}{3}\biggr)^{1 / 2} \tilde{r}
\, .</math>
  </td>
</tr>
</table>
Note that, at the center of our <math>(n_c, n_e) = (5, 1)</math> bipolytrope, <math>\gamma_g = 6/5</math>, so <math>\alpha = -1/3</math>.  Hence, for this particular investigation, the central boundary condition is,

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

<tr>
  <td align="right">
<math>x</math>
  </td>
  <td align="center">
<math>\approx</math>
  </td>
  <td align="left">
<math> 1 -  \biggl[ \frac{\sigma_c^2}{12} + \frac{1}{15} \biggr]\xi^2
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>\approx</math>
  </td>
  <td align="left">
<math> 1 -  \biggl[ \frac{\sigma_c^2}{12} + \frac{1}{15} \biggr]
\biggl[ m_\mathrm{surf}^4 \biggl(\frac{\mu_e}{\mu_c}\biggr)^{-8} \biggl(\frac{2\pi}{3}\biggr) \biggr] \tilde{r}^2
\, . 
</math>
  </td>
</tr>
</table>
Also, the derivative of this displacement function is,

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

<tr>
  <td align="right">
<math>\frac{dx}{d\xi}</math>
  </td>
  <td align="center">
<math>\approx</math>
  </td>
  <td align="left">
<math> -  2\biggl[ \frac{\sigma_c^2}{12} + \frac{1}{15} \biggr]\xi \, .
</math>
  </td>
</tr>
</table>
Hence,

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

<tr>
  <td align="right">
<math>p = -\gamma_c\biggl[3x + \xi \cdot \frac{dx}{d\xi}\biggr]</math>
  </td>
  <td align="center">
<math>\approx</math>
  </td>
  <td align="left">
<math>-\frac{6}{5} \biggl\{ 3 \biggl[ 1 -  \biggl( \frac{\sigma_c^2}{12} + \frac{1}{15} \biggr)\xi^2 \biggr] 
-  2\biggl( \frac{\sigma_c^2}{12} + \frac{1}{15} \biggr)\xi^2 \biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>-\frac{18}{5} \biggl[ 1 -  \frac{5}{3}\biggl( \frac{\sigma_c^2}{12} + \frac{1}{15} \biggr)\xi^2 \biggr] \, .
</math>
  </td>
</tr>
</table>
Furthermore, we find,

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

<tr>
  <td align="right">
<math>(4x + p)</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
4\biggl[ 1 - \biggl( \frac{\sigma_c^2}{12} + \frac{1}{15} \biggr)\xi^2  \biggr]
-\frac{18}{5} \biggl[ 1 -  \frac{5}{3}\biggl( \frac{\sigma_c^2}{12} + \frac{1}{15} \biggr)\xi^2 \biggr] \, .
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[ \frac{20}{5} - 4\biggl( \frac{\sigma_c^2}{12} + \frac{1}{15} \biggr)\xi^2  \biggr]
+ \biggl[ - \frac{18}{5} + 6\biggl( \frac{\sigma_c^2}{12} + \frac{1}{15} \biggr)\xi^2 \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{2}{5} + 2\biggl( \frac{\sigma_c^2}{12} + \frac{1}{15} \biggr)\xi^2 \, .
</math>
  </td>
</tr>
</table>
Hence we have,

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

<tr>
  <td align="right"><math>\frac{dp}{d \xi} </math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
2\biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-1} \xi 
\biggl\{ 
\frac{2}{5} + 2\biggl( \frac{\sigma_c^2}{12} + \frac{1}{15} \biggr)\xi^2 
+
\frac{\sigma_c^2}{2} \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{3/2} 
\biggl[
1 -  \biggl( \frac{\sigma_c^2}{12} + \frac{1}{15} \biggr)\xi^2
\biggr] 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
2\biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-1} \xi 
\biggl\{ 
24 + 2\biggl( 5\sigma_c^2 + 4 \biggr)\xi^2 
+
\frac{\sigma_c^2}{2} \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{3/2} 
\biggl[
60 -  \biggl( 5\sigma_c^2 + 4 \biggr)\xi^2
\biggr] 
\biggr\}\frac{1}{60}
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{1}{30}\biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-1} \xi 
\biggl\{ 
24 + 8\xi^2 +  10\sigma_c^2 \xi^2 
+
30\sigma_c^2\biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{3/2}  
-
\frac{\sigma_c^2}{2} \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{3/2} \biggl( 4 + 5\sigma_c^2 \biggr)\xi^2 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{1}{30}\biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-1} \xi 
\biggl\{ 
\biggl[ 24 + 8\xi^2 +  10\sigma_c^2 \xi^2 \biggr]
+
\biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{3/2}\biggl[
30\sigma_c^2  
-
2\sigma_c^2  \xi^2 
-
\frac{5}{2} \sigma_c^4 \xi^2 \biggr]
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{1}{15}\biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-1} \xi 
\biggl[ 12 + 4\xi^2 +  5\sigma_c^2 \xi^2 \biggr]

+
\frac{\sigma_c^2 \xi}{30} 
\biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{1/2}\biggl[
30  
-
2  \xi^2 
-
\frac{5}{2} \sigma_c^2 \xi^2 \biggr]
</math>
  </td>
</tr>
</table>

====Just &sigma;<sup>2</sup> = 0====
If we set <math>\sigma_c^2 = 0</math>, we have,

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

<tr>
  <td align="right">
<math>p</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>-\frac{18}{5} +  \frac{2}{5} \xi^2 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ \frac{dp}{d\xi}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\frac{4}{5} \xi \, .
</math>
  </td>
</tr>
</table>

Alternatively, from immediately above,

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

<tr>
  <td align="right"><math>\frac{dp}{d \xi}\biggr|_{\sigma_c^2=0} </math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{1}{15}\biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-1} \xi 
\biggl[ 12 + 4\xi^2 +  \cancelto{0}{5\sigma_c^2 \xi^2} \biggr]

+
\cancelto{0}{\frac{\sigma_c^2 \xi}{30} }
\biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{1/2}\biggl[
30  
-
2  \xi^2 
-
\frac{5}{2} \sigma_c^2 \xi^2 \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{12}{15}\biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-1} \xi 
\biggl[ 1 + \frac{1}{3}\xi^2 \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{4}{5}\xi \, .
</math>
  </td>
</tr>
</table>

Yes!  It  matches!

====Summary====

Moving from the center, outward thorough the core &#8212; that is, interior to the interface &#8212; we can assign values of <math>x(\xi)</math> and <math>p(\xi)</math> using the following approximate (''exact'' if <math>\sigma_c^2 = 0</math>) relations:

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

<tr>
  <td align="right">
<math>x</math>
  </td>
  <td align="center">
<math>\approx</math>
  </td>
  <td align="left">
<math> 1 -  \biggl[ \frac{\sigma_c^2}{12} + \frac{1}{15} \biggr]\xi^2 \, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>p </math>
  </td>
  <td align="center">
<math>\approx</math>
  </td>
  <td align="left">
<math>-\frac{18}{5} \biggl[ 1 -  \frac{5}{3}\biggl( \frac{\sigma_c^2}{12} + \frac{1}{15} \biggr)\xi^2 \biggr] \, .
</math>
  </td>
</tr>
</table>
For all radial shells throughout the entire bipolytropic configuration, the pair of first derivatives can be evaluated using the following relations:

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

<tr>
  <td align="right">
<math>\frac{dx}{d\tilde{r}}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math> -\frac{1}{\tilde{r}}\biggl[3x + \frac{p}{\gamma_g}\biggr] \, ;
</math>
  </td>
</tr>

<tr>
  <td align="right"><math>\frac{dp}{d \tilde{r}} </math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{\tilde{\rho}}{\tilde{P}} \cdot \frac{\tilde{M}_r}{\tilde{r}^2}
\biggl[ (4x + p)
+
\sigma_c^2 \biggl(\frac{2\pi}{3} \cdot \frac{\tilde{\rho}_c \tilde{r}^3 }{\tilde{M}_r}\biggr) x \biggr] \, .
</math>
  </td>
</tr>
</table>

Near the center, this pressure-derivative expression can be checked against the relation,

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

<tr>
  <td align="right"><math>\frac{dp}{d \xi} </math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{1}{15}\biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-1} \xi 
\biggl[ 12 + 4\xi^2 +  5\sigma_c^2 \xi^2 \biggr]

+
\frac{\sigma_c^2 \xi}{30} 
\biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{1/2}\biggl[
30  
-
2  \xi^2 
-
\frac{5}{2} \sigma_c^2 \xi^2 \biggr] \, ;
</math>
  </td>
</tr>
</table>
notice that, in order to make this comparison, you need to multiply this last expression through by the ratio,

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

<tr>
  <td align="right"><math>\frac{\xi}{\tilde{r}} </math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\mathcal{m}_\mathrm{surf}^{2} \biggl(\frac{\mu_e}{\mu_c}\biggr)^{-4} 
\biggl(\frac{2\pi}{3}\biggr)^{1 / 2} \, .</math>
  </td>
</tr>
</table>

The comparison should be especially accurate in the case of <math>\sigma_c^2 = 0</math>.

===At the Interface===

See [[#Interface|below]].

===At the Surface===

Drawing from a [[SSC/Stability/Polytropes#Boundary_Conditions|separate discussion]], the surface boundary condition is,

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

<tr>
  <td align="right">
<math>~r_0 \frac{dx}{dr_0}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl( 4 - 3\gamma_g + \frac{\omega^2 R^3}{GM_\mathrm{tot}}\biggr) \frac{x}{\gamma_g}</math>&nbsp; &nbsp; &nbsp; &nbsp; at &nbsp; &nbsp; &nbsp; &nbsp; <math>~r_0 = R \, ,</math>
  </td>
</tr>
</table>
that is,

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

<tr>
  <td align="right">
<math>\frac{d\ln x}{d\ln \tilde{r}}\biggr|_s</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>- \alpha  + \frac{\omega^2 R^3}{\gamma_g GM_\mathrm{tot}} 
\, ,</math>
  </td>
</tr>
</table>
where (see also, [[SSC/Stability/BiPolytropes#Review_of_the_Analysis_by_Murphy_&_Fiedler_(1985b)|here]]),

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

<tr>
  <td align="right">
<math>\alpha</math>
  </td>
  <td align="center">
<math>\equiv</math>
  </td>
  <td align="left">
<math>3 - \frac{4}{\gamma_g} \, .</math>
  </td>
</tr>
</table>
Note that since,

<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>
\tilde{r}_s^3 \biggl[ G^{15/2} K_c^{-15/2} M_\mathrm{tot}^6\biggr]
\, ,</math>
  </td>
</tr>
</table>
in terms of our adopted normalizations, the frequency-squared term should be rewritten as,

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

<tr>
  <td align="right">
<math>\frac{\omega^2 R^3}{\gamma_g G M_\mathrm{tot}}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\tilde{r}_s^3 \biggl[ \frac{\omega^2 \tau_c^2}{\gamma_g}\biggr]
\, .</math>
  </td>
</tr>
</table>
Note as well that, at the surface of our <math>(n_c, n_e) = (5, 1)</math> bipolytrope, <math>\gamma_g = 2</math>, so <math>\alpha = +1</math>.  Hence, for this particular investigation, the surface boundary condition is,

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

<tr>
  <td align="right">
<math>\frac{d\ln x}{d\ln \tilde{r}}\biggr|_s</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math> \frac{1}{2}\biggl[ \tilde{r}_s^3\omega^2 \tau_c^2\biggr] -1
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math> \frac{\tilde{r}_s^3}{6}\biggl[ 2\pi \tilde{\rho}_c\sigma_c^2\biggr] -1
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math> \frac{\tilde{r}_s^3}{6}\biggl[ 2\pi \tilde{\rho}_c\sigma_c^2\biggr]\frac{ 3 }{4\pi \tilde{r}_s^3 \tilde{\rho}_\mathrm{mean}} -1
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math> \frac{1}{4}\biggl( \frac{ \tilde{\rho}_c }{\tilde{\rho}_\mathrm{mean}}\biggr)\sigma_c^2  -1 \, .
</math>
  </td>
</tr>
</table>

This result should be compared with our [[SSC/Stability/BiPolytropes#Eigenfunction_Details|separate discussion of ''eigenfunction details'']].