Editing SSC/Stability/n1PolytropeLAWE/Pt4 (section)

=Stability Analysis=
Here, as well, we draw heavily from our accompanying [[SSC/SynopsisStyleSheet#Stability|"style sheet" synopsis of spherically symmetric configurations]].

This time, we pull the
<div align="center">
<font color="#770000">'''LAWE: &nbsp; Linear Adiabatic Wave''' (or ''Radial Pulsation'') '''Equation'''</font><br />
<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>
\frac{d}{dr}\biggl[ r^4 \gamma P ~\frac{dx}{dr} \biggr] 
+\biggl[ \omega^2 \rho r^4 + (3\gamma - 4) r^3 \frac{dP}{dr} \biggr] x 
\, ,
</math>
  </td>
</tr>
</table>
</div>

from subsection <b><font color="maroon" size="+1">&#x2463;</font></b> of the synopsis; then, guided by subsection <b><font color="maroon" size="+1">&#x2464;</font></b>, we multiply both sides through by <math>4\pi x dr</math> to obtain,

<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>- \biggl[ 4\pi x^2 (\omega^2 \rho r^4 ) \biggr]dr </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
4\pi x\biggl\{ \frac{d}{dr}\biggl[ r^4 \gamma P \biggl( \frac{dx}{dr}\biggr) \biggr] \biggr\} dr
+ \biggl\{ \biggl[ (3\gamma - 4) 4\pi x^2 r^3 \biggl( \frac{dP}{dr}\biggr) \biggr] \biggr\} dr
\, .
</math>
  </td>
</tr>

</table>
Now, given that,

<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>
\frac{d}{dr}\biggl[4\pi x \gamma r^4 P \biggl(\frac{dx}{dr}\biggr) \biggr] 
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
4\pi x \frac{d}{dr}\biggl[\gamma r^4 P \biggl(\frac{dx}{dr}\biggr) \biggr] 
+
\biggl[\gamma r^4 P \biggl(\frac{dx}{dr}\biggr)\biggr]\frac{d}{dr}\biggl[4\pi x \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>
\Rightarrow ~~~ 
4\pi x \frac{d}{dr}\biggl[\gamma r^4 P \biggl(\frac{dx}{dr}\biggr) \biggr] 

</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{d}{dr}\biggl[4\pi x \gamma r^4 P \biggl(\frac{dx}{dr}\biggr) \biggr] 
-
\biggl[4\pi \gamma r^4 P \biggl(\frac{dx}{dr}\biggr)^2\biggr] \, ,
</math>
  </td>
</tr>
</table>
we can rewrite this last expression in the form,

<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>- \biggl[ 4\pi x^2 (\omega^2 \rho r^4 ) \biggr]dr </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{d}{dr}\biggl[4\pi x \gamma r^4 P \biggl(\frac{dx}{dr}\biggr) \biggr]dr 
-
\biggl[4\pi \gamma r^4 P \biggl(\frac{dx}{dr}\biggr)^2\biggr] dr
+ \biggl\{ \biggl[ (3\gamma - 4) 4\pi x^2 r^3 \biggl( \frac{dP}{dr}\biggr) \biggr] \biggr\} dr
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ \biggl[ 4\pi x^2 (\omega^2 \rho r^4 ) \biggr]dr </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
-d\biggl[4\pi x \gamma r^4 P \biggl(\frac{dx}{dr}\biggr) \biggr] 
+
\biggl[4\pi \gamma r^4 P \biggl(\frac{dx}{dr}\biggr)^2\biggr] dr
- \biggl[ (3\gamma - 4) 4\pi x^2 r^3 \biggl( - \frac{GM_r \rho}{r^2}\biggr) \biggr] dr
</math>
  </td>
</tr>

</table>
Note that, in order to obtain the last term on the RHS of this expression, we used the hydrostatic balance relation to replace the pressure gradient in terms of the gravitational potential.  Finally, integrating over the volume of the configuration gives,

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

<tr>
  <td align="right">
<math>\int_0^R \biggl[ 4\pi x^2 (\omega^2 \rho r^4 ) \biggr]dr </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\int_0^R\biggl[4\pi \gamma r^4 P \biggl(\frac{dx}{dr}\biggr)^2\biggr] dr
- \int_0^R \biggl[ (3\gamma - 4) x^2 \biggl( - \frac{GM_r }{r}\biggr)4\pi \rho r^2 \biggr] dr
- \biggl[4\pi x \gamma r^4 P \biggl(\frac{dx}{dr}\biggr) \biggr]_0^R \, ,
</math>
  </td>
</tr>

</table>
or,

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

<tr>
  <td align="right">
<math>\int_0^R \biggl[ 4\pi x^2 (\omega^2 \rho r^4 ) \biggr]dr </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\int_0^R \overbrace{\biggl[x^2 \biggl(\frac{d\ln x}{d\ln r}\biggr)^2 4\pi \gamma r^2 P \biggr] dr}^{\mathrm{TERM1}} 
- \int_0^R \underbrace{\biggl[ (3\gamma - 4) x^2 \biggl( - \frac{GM_r }{r}\biggr)4\pi \rho r^2 \biggr] dr}_{\mathrm{TERM2}}
+ \overbrace{\biggl[\gamma 4\pi x^2 r^2 P \biggl(-\frac{d\ln x}{d\ln r}\biggr) \biggr]_0^R}^{\mathrm{TERM3}} \, .
</math>
  </td>
</tr>

</table>

==TERM1==
Given that (from above),
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right"><math>dS^* </math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{3}{2} \biggl[ \frac{4\pi r^2 P dr}{[K_c^5/G^3]^{1 / 2}} \biggr] 
=
6\pi \biggl( \frac{3}{2\pi} \biggr)^{3 / 2} ~ 
\xi^2 \biggl(1+\frac{\xi^2}{3}\biggr)^{-3}  d\xi \, ,
</math>
  </td>
</tr>
</table>
we can write,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right"><math>\frac{\mathrm{TERM1}}{[K_c^5 / G^3]^{1 / 2}}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\gamma x_P^2 \biggl(\frac{d\ln x_P}{d\ln r} \biggr)^2 \biggl[ \frac{4\pi r^2 P dr}{[K_c^5/G^3]^{1 / 2}} \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\gamma x_P^2 \biggl(\frac{d\ln x_P}{d\ln r} \biggr)^2 \biggl\{
\biggl( \frac{2}{3} \biggr) 6\pi \biggl( \frac{3}{2\pi} \biggr)^{3 / 2} ~ 
\xi^2 \biggl(1+\frac{\xi^2}{3}\biggr)^{-3}  d\xi\biggr\} \, ,
</math>
  </td>
</tr>
</table>
Furthermore, [[SSC/Stability/InstabilityOnsetOverview#Analyses_of_Radial_Oscillations|given that]] for a truncated <math>n=5</math> configuration,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right"><math>x_p</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
1 - \frac{\xi^2}{15}
</math>
  </td>
</tr>

<tr>
  <td align="right"><math>\Rightarrow ~~~ \frac{dx_p}{d\xi}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
- \frac{2}{15}~\xi 
</math>
  </td>
</tr>

<tr>
  <td align="right"><math>\Rightarrow ~~~ \frac{d\ln x_p}{d\ln\xi}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
- \frac{\xi}{x_P}\biggl[\frac{2}{15}~\xi \biggr]
=
- \biggl[\frac{2}{15}~\xi^2 \biggr]\biggl[ \frac{15}{15-\xi^2}\biggr]
=
- \biggl[ \frac{2\xi^2}{15-\xi^2}\biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right"><math>\Rightarrow ~~~ x_P^2 \biggl(\frac{d\ln x_p}{d\ln\xi}\biggr)^2</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\biggl[\frac{15 - \xi^2}{15}\biggr]^2
\biggl[ \frac{2\xi^2}{15-\xi^2}\biggr]^2 
=
\biggl[ \frac{2\xi^2}{15} \biggr]^2
\, ,
</math>
  </td>
</tr>
</table>
we have,

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

<tr>
  <td align="right"><math>\frac{\mathrm{TERM1}}{[K_c^5 / G^3]^{1 / 2}}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\gamma \biggl[ \frac{2\xi^2}{15} \biggr]^2 \biggl\{
\biggl( \frac{2}{3} \biggr) 6\pi \biggl( \frac{3}{2\pi} \biggr)^{3 / 2} ~ 
\xi^2 \biggl(1+\frac{\xi^2}{3}\biggr)^{-3}  d\xi\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\gamma \biggl[ \frac{2^2}{3^2 5^2} \biggr]
\biggl( \frac{2}{3} \biggr) 6\pi \biggl( \frac{3}{2\pi} \biggr)^{3 / 2} 
\biggl\{
\xi^6 \biggl(1+\frac{\xi^2}{3}\biggr)^{-3}  d\xi\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\gamma \biggl[ \frac{2^4 \pi}{3^2 5^2} \biggr]
\biggl[ \frac{3^3}{2^3\pi^3} \biggr]^{1 / 2} 
\biggl\{
\xi^6 \biggl(1+\frac{\xi^2}{3}\biggr)^{-3}  d\xi\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\gamma \biggl[ \frac{2^5 }{3 \cdot 5^4 \pi} \biggr]^{1 /2}
\biggl\{
\xi^6 \biggl(1+\frac{\xi^2}{3}\biggr)^{-3}  d\xi\biggr\} \, .
</math>
  </td>
</tr>
</table>

Hence, after making the replacement <math>\chi \equiv \xi/\sqrt{3} ~\Rightarrow ~ \xi = 3^{1 / 2}\chi</math>, we find that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right"><math>\int_0^{\chi_\mathrm{surf}}\frac{\mathrm{TERM1}}{[K_c^5 / G^3]^{1 / 2}}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\gamma \biggl[ \frac{2^5 }{3 \cdot 5^4 \pi} \biggr]^{1 /2}
\cdot 3^{7/2}\int_0^{\chi_\mathrm{surf}}\biggl\{
\chi^6 \biggl(1+\chi^2\biggr)^{-3} d\chi
\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\gamma \biggl[ \frac{2^5 }{3 \cdot 5^4 \pi} \biggr]^{1 /2}
\cdot 3^{7/2}
\cdot \frac{1}{8}\biggl\{
\frac{\chi_\mathrm{surf}(8\chi^4_\mathrm{surf} + 25 \chi^2_\mathrm{surf} 
+ 15)}{(1+\chi^2_\mathrm{surf})^2} - 15\tan^{-1}(\chi_\mathrm{surf})
\biggr\} 
</math>
  </td>
</tr>
</table>
where, we have completed the integral with the assistance of the [https://wolframalpha.com WolframAlpha] online integrator:
[[File:Mathematica05.png|500px|center|Mathematica Integral]]

==TERM2==
Given that (from above),

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

<tr>
  <td align="right"><math>dW^* </math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
[K_c^{-5/2} G^{3/2}] \biggl(-\frac{GM_r}{r}\biggr) 4\pi r^2 \rho dr 
=
- \biggl[ \frac{2^3 \cdot 3^3}{\pi} \biggr]^{1 / 2}
\biggl[\xi^4 \biggl(1 + \frac{\xi^2}{3}\biggr)^{-4}  \biggr] d\xi 
\, ,
</math>
  </td>
</tr>
</table>
we can write,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right"><math>\frac{\mathrm{TERM2}}{[K_c^5 / G^3]^{1 / 2}}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
[K_c^{-5 / 2} G^{3 / 2}]\biggl[ (3\gamma - 4) x^2 \biggl( - \frac{GM_r }{r}\biggr)4\pi \rho r^2 \biggr] dr
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
- (3\gamma - 4) x_P^2 \biggl\{
\biggl[ \frac{2^3 \cdot 3^3}{\pi} \biggr]^{1 / 2}
\biggl[\xi^4 \biggl(1 + \frac{\xi^2}{3}\biggr)^{-4}  \biggr] d\xi\biggr\}
\, .
</math>
  </td>
</tr>
</table>
Furthermore, [[SSC/Stability/InstabilityOnsetOverview#Analyses_of_Radial_Oscillations|given that]] (as above) for a truncated <math>n=5</math> configuration,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right"><math>x_p</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{15 - \xi^2}{15} \, ,
</math>
  </td>
</tr>
</table>
we have,

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

<tr>
  <td align="right"><math>\frac{\mathrm{TERM2}}{[K_c^5 / G^3]^{1 / 2}}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
- (3\gamma - 4) \biggl[ \frac{2^3 \cdot 3^3}{\pi} \cdot \frac{1}{3^4\cdot 5^4}\biggr]^{1 / 2}
\biggl\{
\biggl[(15 - \xi^2)^2 \xi^4 \biggl(1 + \frac{\xi^2}{3}\biggr)^{-4}  \biggr] d\xi
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
- (3\gamma - 4) \biggl[ \frac{2^3 }{3\cdot 5^4\pi} \biggr]^{1 / 2}
\biggl\{
\biggl[(15 - \xi^2)^2 \xi^4 \biggl(1 + \frac{\xi^2}{3}\biggr)^{-4}  \biggr] d\xi
\biggr\}
\, .
</math>
  </td>
</tr>
</table>

Hence, after making the replacement <math>\chi \equiv \xi/\sqrt{3} ~\Rightarrow ~ \xi = 3^{1 / 2}\chi</math>, we find that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right"><math>\int_0^{\chi_\mathrm{surf}}\frac{\mathrm{TERM2}}{[K_c^5 / G^3]^{1 / 2}}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
- (3\gamma - 4) \biggl[ \frac{2^3 }{3\cdot 5^4\pi} \biggr]^{1 / 2}
\int_0^{\chi_\mathrm{surf}}
\biggl\{
\biggl[(15 - \xi^2)^2 \xi^4 \biggl(1 + \frac{\xi^2}{3}\biggr)^{-4}  \biggr] d\xi
\biggr\}

</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
- (3\gamma - 4) \biggl[ \frac{2^3 }{3\cdot 5^4\pi} \biggr]^{1 / 2} \cdot 3^{9 / 2}
\int_0^{\chi_\mathrm{surf}}
\biggl\{
\biggl[(5 - \chi^2)^2 \chi^4 \biggl(1 + \chi^2\biggr)^{-4}  \biggr] d\chi
\biggr\}

</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
- (3\gamma - 4) \biggl[ \frac{2^3 \cdot 3^8 }{5^4\pi} \biggr]^{1 / 2} 
\cdot \frac{1}{4}\biggl\{
\frac{\chi_\mathrm{surf}(4\chi^6_\mathrm{surf} + 53\chi^4_\mathrm{surf} + 40 \chi^2_\mathrm{surf} 
+ 15)}{(1+\chi^2_\mathrm{surf})^3} - 15\tan^{-1}(\chi_\mathrm{surf})
\biggr\} 
</math>
  </td>
</tr>
</table>
where, we have completed the integral with the assistance of the [https://wolframalpha.com WolframAlpha] online integrator:
[[File:Mathematica06.png|500px|center|Mathematica Integral]]