Editing SSC/Stability/Isothermal (section)

==Try to Generalize==
===Derivation of Polytropic Displacement Function===
<table border="0" cellpadding="8" align="right">
<tr><td align="center">iPhone snapshot of "whiteboard" Eureka! moment.</td></tr>
<tr><td align="center">
[[File:EurekaWhiteBoardMarch2017sm2.png|center|420px|Whiteboard Eureka moment! (18 March 2017)]]
</td></tr>
</table>
Now, let's turn our attention to the,
<div align="center">
<font color="maroon"><b>Polytropic LAWE</b></font><br />
{{ Math/EQ_RadialPulsation02 }}
</div>

Based on the analytic expression that {{ Yabushita75}} derived for the isothermal case, plus some "white board" poking around, the following general polytropic displacement function looks promising:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>x</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>1 + \biggl(\frac{n-3}{n-1}\biggr) \frac{\theta^' }{\xi \theta^{n}} \, .</math>
  </td>
</tr>
</table>
</div>

Keeping in mind that, according to the [[SSC/Structure/Lane1870#LaneEmden|polytropic Lane-Emden equation]],
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\theta^{''}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>- \theta^n - \frac{2\theta^'}{\xi} \, ,</math>
  </td>
</tr>
</table>
</div>

the first derivative is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\frac{dx}{d\xi}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\biggl(\frac{n-3}{n-1}\biggr) \biggl\{
\frac{\theta^{''} }{\xi \theta^{n}} - \frac{n(\theta^')^2 }{\xi \theta^{n+1}} - \frac{\theta^' }{\xi^2 \theta^{n}} 
\biggr\}
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl(\frac{3-n}{n-1}\biggr) \biggl\{ \frac{1}{\xi} 
+ \frac{n(\theta^')^2 }{\xi \theta^{n+1}} + \frac{3\theta^' }{\xi^2 \theta^{n}} \biggr\} 
\, ;
</math>
  </td>
</tr>
</table>
</div>
and the second derivative is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\frac{d^2 x}{d\xi^2}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\biggl(\frac{3-n}{n-1}\biggr) \biggl\{ 
-\frac{1}{\xi^2} 
+ \frac{2n(\theta^')\theta^{''} }{\xi \theta^{n+1}} 
- \frac{n(\theta^')^2 }{\xi^2 \theta^{n+1}} 
- (n+1) \frac{n(\theta^')^3 }{\xi \theta^{n+2}} 
+ \frac{3\theta^{''} }{\xi^2 \theta^{n}} 
- \frac{6\theta^' }{\xi^3 \theta^{n}} 
- \frac{3n (\theta^')^2 }{\xi^2 \theta^{n+1}} 
\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\biggl(\frac{n-3}{n-1}\biggr) \biggl\{ 
\frac{1}{\xi^2} 
+ \bigg[ \frac{2n(\theta^')}{\xi \theta^{n+1}} + \frac{3 }{\xi^2 \theta^{n}} \biggr] \biggl[ \theta^n + \frac{2\theta^'}{\xi} \biggr]
+ (n+1) \frac{n(\theta^')^3 }{\xi \theta^{n+2}} 
+ \frac{6\theta^' }{\xi^3 \theta^{n}} 
+ \frac{4n (\theta^')^2 }{\xi^2 \theta^{n+1}} 
\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\biggl(\frac{n-3}{n-1}\biggr) \biggl\{ 
\frac{1}{\xi^2} 
+ \frac{2n(\theta^')}{\xi \theta} + \frac{3 }{\xi^2 } 
+ \frac{4n(\theta^')^2}{\xi^2 \theta^{n+1}} + \frac{6\theta^' }{\xi^3 \theta^{n}} 
+ (n+1) \frac{n(\theta^')^3 }{\xi \theta^{n+2}} 
+ \frac{6\theta^' }{\xi^3 \theta^{n}} 
+ \frac{4n (\theta^')^2 }{\xi^2 \theta^{n+1}} 
\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl(\frac{n-3}{n-1}\biggr) \biggl\{ \frac{4}{\xi^2} + \frac{2n(\theta^')}{\xi \theta} 
+ \frac{12\theta^' }{\xi^3 \theta^{n}}+ \frac{8n(\theta^')^2}{\xi^2 \theta^{n+1}}  
+ (n+1) \frac{n(\theta^')^3 }{\xi \theta^{n+2}} \biggr\} 
\, .
</math>
  </td>
</tr>
</table>
</div>
<span id="ForceSigmaZero">Hence, setting</span> <math>\sigma_c^2 = 0</math> in the polytropic LAWE gives,

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

<tr>
  <td align="right">
LAWE
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{d^2x}{d\xi^2} + \biggl[ 4 + (n+1) \biggl(\frac{\xi \theta^'}{\theta} \biggr) \biggr] \frac{1}{\xi} \cdot \frac{dx}{d\xi}  
+ \alpha (n+1) \biggl(\frac{\xi \theta^'}{\theta} \biggr)  \frac{x}{\xi^2}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl(\frac{n-3}{n-1}\biggr) \biggl\{ \frac{4}{\xi^2} + \frac{2n(\theta^')}{\xi \theta} 
+ \frac{12\theta^' }{\xi^3 \theta^{n}}+ \frac{8n(\theta^')^2}{\xi^2 \theta^{n+1}}  
+ (n+1) \frac{n(\theta^')^3 }{\xi \theta^{n+2}} \biggr\}  
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\alpha (n+1) \biggl(\frac{\theta^'}{\xi \theta} \biggr) + 
\biggl(\frac{n-3}{n-1}\biggr) \biggl\{ \frac{4}{\xi^2} + \frac{2n(\theta^')}{\xi \theta} 
+ \frac{12\theta^' }{\xi^3 \theta^{n}}+ \frac{8n(\theta^')^2}{\xi^2 \theta^{n+1}}  
+ (n+1) \frac{n(\theta^')^3 }{\xi \theta^{n+2}}  
- \frac{4}{\xi^2} 
- \frac{4n(\theta^')^2 }{\xi^2 \theta^{n+1}} - \frac{12\theta^' }{\xi^3 \theta^{n}} 
</math>
  </td>
</tr>

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

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

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

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>- [ \alpha (n+1) +  (n-3) ] \biggl(\frac{x}{\xi^2} \biggr) Q \, .
</math>
  </td>
</tr>
</table>
</div>

Hence, if <math>\gamma = (n+1)/n</math>, then, <math>\alpha = (3-4/\gamma) = (3-n)/(n+1)</math>, and the polytropic LAWE is satisfied.  <font color="red">'''HOOOORAAAAY!''' (18 March 2017)</font>

===Behavior of First Logarithmic Derivative===

<span id="LogarithmicDerivative">For the record,</span> let's develop an expression for the logarithmic derivative of this displacement function.
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\frac{d\ln x}{d\ln \xi} = \frac{\xi}{x} \cdot \frac{dx}{d\xi}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl(\frac{3-n}{n-1}\biggr) \biggl[ 1 
+ \frac{n(\theta^')^2 }{\theta^{n+1}} + \frac{3\theta^' }{\xi \theta^{n}} \biggr] 
\biggl[ 1 + \biggl(\frac{n-3}{n-1}\biggr) \frac{\theta^' }{\xi \theta^{n}}  \biggr]^{-1} 
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
<math>\Rightarrow ~~~ \biggl[ (n-1) + (n-3) \frac{\theta^' }{\xi \theta^{n}}  \biggr] \frac{d\ln x}{d\ln \xi} </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
- (n-3) \biggl[ 1 
+ \frac{n(\theta^')^2 }{\theta^{n+1}} + \frac{3\theta^' }{\xi \theta^{n}} \biggr]  
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ \biggl[ (n-1)  \biggr] \frac{d\ln x}{d\ln \xi} </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
- (n-3) \biggl\{ \biggl[ 1 
+ \frac{n(\theta^')^2 }{\theta^{n+1}} + \frac{3\theta^' }{\xi \theta^{n}} \biggr]  
+ \biggl[ \frac{\theta^' }{\xi \theta^{n}}  \biggr] \frac{d\ln x}{d\ln \xi} \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
- (n-3) \biggl\{ \biggl[ 1 
+ \frac{n(\theta^')^2 }{\theta^{n+1}}  \biggr]  
+ \biggl[ \frac{\theta^' }{\xi \theta^{n}}  \biggr] \biggl[3 + \frac{d\ln x}{d\ln \xi}\biggr] \biggr\}
</math>
  </td>
</tr>
</table>
</div>

Now, as we have [[#From_Yabushita.27s_.281992.29_Analysis|discussed above]], {{ Yabushita92 }} has argued that for pressure-truncated configurations, the appropriate surface boundary condition should be,
<div align="center">
<math>\frac{d\ln x}{d\ln\xi} = -3 \, .</math>
</div>
If we adopt this surface boundary condition, then this just-derived expression becomes,

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

<tr>
  <td align="right">
<math>3 (n-1)   </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
(n-3) \biggl[ 1 
+ \frac{n(\theta^')^2 }{\theta^{n+1}}  \biggr]  
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow~~~
(n-3) \biggl[ \frac{n(\theta^')^2 }{\theta^{n+1}}  \biggr]
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
2n 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow~~~
\frac{(\theta^')^2 }{\theta^{n+1}} 
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{2 }{(n-3) } \, .
</math>
  </td>
</tr>
</table>
</div>
According to {{ Horedt70 }} &#8212; see our [[SSC/Structure/PolytropesEmbedded#Horedt.27s_Derivation|accompanying discussion]] &#8212; this is precisely the functional condition that pinpoints the <math>~P_e</math>-max turning point(s) along the equilibrium sequence in a <math>~P-V</math> diagram, for all pressure-truncated, polytropic spheres with <math>~n \ge 3</math>.  And, according to {{ Kimura81b }} &#8212; again, see our [[SSC/Structure/PolytropesEmbedded#Other_Limits|accompanying discussion]] &#8212; this is precisely the functional condition that pinpoints the maximum mass turning point(s) along equlibrium sequences as displayed in a mass-radius diagram.

<!-- CHRISTY OR COX BOUNDARY CONDITIONS ...
On the other hand, as we have [[SSC/Perturbations#ChristyCox|discussed separately]] &#8212; see also, [[#From_Yabushita.27s_.281992.29_Analysis|above]] &#8212; in order to align with the separate derivations presented by [http://adsabs.harvard.edu/abs/1966ARA%26A...4..353C Christy (1965)] and [http://adsabs.harvard.edu/abs/1967IAUS...28....3C Cox (1967)]  &#8212; the corresponding boundary condition at the surface is, 
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{dh}{dx}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- \frac{h}{x} \biggr[  3 - \frac{4}{\gamma}  + \cancelto{0}{\frac{x s^2}{\gamma q}} \biggr]</math>
  </td>
  <td align="left" colspan="2">
&nbsp;
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{n-3}{n+1} \biggl(\frac{h}{x} \biggr) \, .</math>
  </td>
  <td align="left">
&nbsp; &nbsp; &nbsp; &nbsp; at &nbsp; &nbsp; &nbsp; &nbsp; 
  </td>
  <td align="left">
<math>~x = x_0 \, .</math>
  </td>
</tr>
</table>
</div>
-->

===Behavior Approaching Center===
<span id="CentralValue">It would also be good to know</span> what the limiting value of this displacement function is at the center of the polytropic configuration.  Drawing on some of our already-developed [[Appendix/Ramblings/PowerSeriesExpressions#Approximate_Power-Series_Expressions|power-series expressions]], we have,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~x </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~1 + \biggl(\frac{n-3}{n-1}\biggr) \frac{\theta^' }{\xi \theta^{n}} \, .</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~1 + \biggl(\frac{n-3}{n-1}\biggr)\frac{1}{\xi} 
\biggl[1- \frac{\xi^2}{6} + \frac{n\xi^4}{120} + \cdots \biggr]^{-n}
\frac{d}{d\xi}\biggl[1- \frac{\xi^2}{6} + \frac{n\xi^4}{120} + \cdots \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~1 + \biggl(\frac{n-3}{n-1}\biggr) 
\biggl[1 + \frac{n\xi^2}{6} \biggr]
\biggl[- \frac{1}{3} + \frac{n\xi^2}{30} \biggr]
</math>
  </td>
</tr>

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

So, let's rescale the polytropic displacement function such that the central value is unity for all values of the polytropic index.  We have,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~x_P</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~A_0 \biggl[ 1 + \biggl(\frac{n-3}{n-1}\biggr) \frac{\theta^' }{\xi \theta^{n}} \biggr] \, ,</math>
  </td>
</tr>
</table>
</div>
where,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~A_0</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\biggl[ 1 - \frac{1}{3}\cdot \biggl(\frac{n-3}{n-1}\biggr) \biggr]^{-1}
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{3(n-1)}{2n} \, .
</math>
  </td>
</tr>
</table>
</div>
Note that this even provides the correct normalization factor for the isothermal case <math>~(n=\infty)</math>.

{{ SGFworkInProgress }}

===What if Eigenfrequency not Zero===
Here we should keep in mind that the [[SSC/Perturbations#Ensure_Finite_Amplitude_Fluctuations|''conventional'' surface boundary condition]] is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{d\ln x}{d\ln\xi}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{\gamma_\mathrm{g}} \biggl( 4 - 3\gamma_\mathrm{g} + \frac{\omega^2 R^3}{GM_\mathrm{tot}} \biggr)</math>
  </td>
</tr>
</table>
</div>

Recalling that the ratio of the mean to central density is given by the [[SSCpt1/Virial/FormFactors#PTtable|structural form factor]],
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{\bar\rho}{\rho_c} = f_\mathrm{M}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl( - \frac{3\theta^'}{\xi} \biggr) \, ,</math>
  </td>
</tr>
</table>
</div>
we can write,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{\omega^2 R^3}{GM_\mathrm{tot}}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{\omega^2 R^3}{G}\biggl(\frac{3}{4\pi \bar\rho R^3}  \biggr)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{3\omega^2 }{4\pi G}\biggl[\rho_c\biggl( - \frac{3\theta^'}{\xi} \biggr)  \biggr]^{-1}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\sigma_c^2 \biggl( - \frac{\xi}{6\theta^'} \biggr) \, ,
</math>
  </td>
</tr>
</table>
</div>
where,
<div align="center">
<math>~\sigma_c^2 \equiv \frac{3\omega^2}{2\pi G\rho_c} \, .</math>
</div>
Hence, the conventional surface boundary can be written as,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~-~\frac{d\ln x}{d\ln\xi}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\alpha + \frac{\sigma_c^2}{6 \gamma_\mathrm{g}} \biggl( \frac{\xi}{\theta^'} \biggr) </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[
\biggl( \frac{\sigma_c^2}{6 \gamma_\mathrm{g}}\biggr) \frac{\xi^2}{\theta} -\alpha Q \biggr] \frac{\theta}{\xi \theta^'}
</math>
  </td>
</tr>
</table>
</div>



<div align="center">
{{ Math/EQ_RadialPulsation02 }}
</div>

Now, if we return to the [[#ForceSigmaZero|above derivation]] but, this time, do ''not'' set <math>~\sigma_c^2 = 0</math>, the polytropic LAWE gives,

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

<tr>
  <td align="right">
LAWE
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{d^2x}{d\xi^2} + \biggl[ 4 + (n+1) \biggl(\frac{\xi \theta^'}{\theta} \biggr) \biggr] \frac{1}{\xi} \cdot \frac{dx}{d\xi}  
+ \alpha (n+1) \biggl(\frac{\xi \theta^'}{\theta} \biggr)  \frac{x}{\xi^2} + \biggl( \frac{\sigma_c^2}{6}\biggr) \frac{n x}{\theta}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl(\frac{n-3}{n-1}\biggr) \biggl\{ \frac{4}{\xi^2} + \frac{2n(\theta^')}{\xi \theta} 
+ \frac{12\theta^' }{\xi^3 \theta^{n}}+ \frac{8n(\theta^')^2}{\xi^2 \theta^{n+1}}  
+ (n+1) \frac{n(\theta^')^3 }{\xi \theta^{n+2}} \biggr\}  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
- \biggl(\frac{n-3}{n-1}\biggr) \biggl[ 4 + (n+1) \biggl(\frac{\xi \theta^'}{\theta} \biggr) \biggr] \biggl[ \frac{1}{\xi^2} 
+ \frac{n(\theta^')^2 }{\xi^2 \theta^{n+1}} + \frac{3\theta^' }{\xi^3 \theta^{n}} \biggr]
+ \alpha (n+1) \biggl(\frac{\theta^'}{\xi \theta} \biggr) \biggl[ 1 + \biggl(\frac{n-3}{n-1}\biggr) \frac{\theta^' }{\xi \theta^{n}} \biggr] 
+ \biggl( \frac{\sigma_c^2}{6}\biggr) \frac{n}{\theta}\biggl[ 1 + \biggl(\frac{n-3}{n-1}\biggr) \frac{\theta^' }{\xi \theta^{n}} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\alpha (n+1) \biggl(\frac{\theta^'}{\xi \theta} \biggr) + 
\biggl(\frac{n-3}{n-1}\biggr) \biggl\{ \frac{4}{\xi^2} + \frac{2n(\theta^')}{\xi \theta} 
+ \frac{12\theta^' }{\xi^3 \theta^{n}}+ \frac{8n(\theta^')^2}{\xi^2 \theta^{n+1}}  
+ (n+1) \frac{n(\theta^')^3 }{\xi \theta^{n+2}}  
- \frac{4}{\xi^2} 
- \frac{4n(\theta^')^2 }{\xi^2 \theta^{n+1}} - \frac{12\theta^' }{\xi^3 \theta^{n}} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
-  (n+1) \biggl(\frac{ \theta^'}{\xi\theta} \biggr) 
-  (n+1)  \biggl[  \frac{n(\theta^')^3 }{\xi \theta^{n+2}}  \biggr]
- (n+1) \biggl[  \frac{3 (\theta^')^2 }{\xi^2 \theta^{n+1}} \biggr]
+ \alpha (n+1) \biggl[ \frac{(\theta^')^2 }{\xi^2 \theta^{n+1}} \biggr] \biggr\}
+ \biggl( \frac{\sigma_c^2}{6}\biggr) \frac{n}{\theta}\biggl[ 1 + \biggl(\frac{n-3}{n-1}\biggr) \frac{\theta^' }{\xi \theta^{n}} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\alpha (n+1) \biggl(\frac{\theta^'}{\xi \theta} \biggr) + 
\biggl(\frac{n-3}{n-1}\biggr) \biggl\{ (n-1) \biggl(\frac{ \theta^'}{\xi\theta} \biggr) 
+ \frac{(\theta^')^2}{\xi^2 \theta^{n+1}}  \biggl[4n - 3(n+1)  + \alpha (n+1) \biggr]\biggr\}
+ \biggl( \frac{\sigma_c^2}{6}\biggr) \frac{n}{\theta}\biggl[ 1 + \biggl(\frac{n-3}{n-1}\biggr) \frac{\theta^' }{\xi \theta^{n}} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \alpha (n+1) +  (n-3) \biggr] \biggl[ \biggl(\frac{ \theta^'}{\xi\theta} \biggr) 
+ \biggl( \frac{n-3}{n-1}\biggr) \frac{(\theta^')^2}{\xi^2 \theta^{n+1}}  \biggr] 
+ \biggl( \frac{\sigma_c^2}{6}\biggr) \frac{n}{\theta}\biggl[ 1 + \biggl(\frac{n-3}{n-1}\biggr) \frac{\theta^' }{\xi \theta^{n}} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \alpha (n+1) +  (n-3) \biggr]  \biggl(\frac{ \theta^'}{\xi\theta} \biggr)\biggl[ 1
+ \biggl( \frac{n-3}{n-1}\biggr) \frac{\theta^'}{\xi \theta^{n}}  \biggr] 
+ \biggl( \frac{\sigma_c^2}{6}\biggr) \frac{n}{\theta}\biggl[ 1 + \biggl(\frac{n-3}{n-1}\biggr) \frac{\theta^' }{\xi \theta^{n}} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ 1 + \biggl( \frac{n-3}{n-1}\biggr) \frac{\theta^'}{\xi \theta^{n}}  \biggr] \biggl\{[ \alpha (n+1) +  (n-3) ]  \biggl(\frac{ \theta^'}{\xi\theta} \biggr)
+ \biggl( \frac{\sigma_c^2}{6}\biggr) \frac{n}{\theta}\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{x}{\xi^2} \biggl\{[ \alpha (n+1) +  (n-3) ]  \biggl(\frac{ \xi \theta^'}{\theta} \biggr)
+ \biggl( \frac{\sigma_c^2}{6}\biggr) \frac{n\xi^2 }{\theta}\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{x}{\xi^2} \biggl\{ \biggl( \frac{\sigma_c^2}{6}\biggr) \frac{n\xi^2 }{\theta} -  [ \alpha (n+1) +  (n-3) ]  Q
\biggr\}
</math>
  </td>
</tr>
</table>
</div>