Editing SSC/Structure/OtherAnalyticModels (section)

==Parabolic Density Distribution==

===Equilibrium Structure===
In an article titled, "Radial Oscillations of a Stellar Model," {{ Prasad49full }} investigated the properties of an equilibrium configuration with a prescribed density distribution given by the expression,
<div align="center">
<math>\rho(r) = \rho_c\biggl[ 1 - \biggl(\frac{r}{R} \biggr)^2 \biggr] \, ,</math>
</div>
where, <math>\rho_c</math> is the central density and, <math>R</math> is the radius of the star.  Both the mass distribution and the pressure distribution can be obtained analytically from this specified density distribution.  Specifically, following our [[SSCpt2/SolutionStrategies#Solution_Strategies|general solution strategy]] for determining the equilibrium structure of spherically symmetric, self-gravitating configurations,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~M_r(r)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\int_0^r 4\pi r^2 \rho(r) dr</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{4\pi\rho_c r^3}{3} \biggl[1 - \frac{3}{5} \biggl( \frac{r}{R} \biggr)^2 \biggr] \, ,</math>
  </td>
</tr>
</table>
</div>
in which case we can write,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~g_0(r) \equiv \frac{G M_r(r) }{r^2} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{4\pi G \rho_c r}{3}  
\biggl[1 - \frac{3}{5} \biggl( \frac{r}{R} \biggr)^2\biggr] \, .</math>
  </td>
</tr>
</table>
</div>
<span id="TotalMass">Note that the total mass</span> is obtained by setting <math>r = R</math> in the expression for <math>M_r(r )</math>, namely,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>M_\mathrm{tot}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\frac{4\pi\rho_c R^3}{3} \biggl[\frac{2}{5}\biggr] 
=
\frac{8\pi\rho_c R^3}{15} 
</math>
&nbsp; &nbsp; &nbsp; <math>\Rightarrow</math> &nbsp; &nbsp; &nbsp; 
<math>
2\pi \rho_c = \frac{15 M_\mathrm{tot}}{4R^3}
\, .
</math>
  </td>
</tr>
</table>


----


Following [[SSCpt2/SolutionStrategies#Technique_1|"Technique 1"]], we appreciate that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\frac{d\Phi}{dr}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>g_0</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ \Phi </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\frac{4\pi G \rho_c }{3}  \int 
\biggl[1 - \frac{3}{5} \biggl( \frac{r}{R} \biggr)^2\biggr] r~dr
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\frac{4\pi G \rho_c }{3}  \biggl\{ \int r~dr
- \biggl(\frac{3}{5R^2}\biggr)\int r^3~dr \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\frac{4\pi G \rho_c }{3}  \biggl\{ \frac{1}{2} r^2
- \biggl(\frac{3}{5R^2}\biggr)\frac{1}{4} r^4 \biggr\}
+ C_\Phi
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\frac{2\pi G \rho_c R^2}{15}  \biggl\{ 5 \biggl(\frac{r}{R}\biggr)^2
- \frac{3}{2}\biggl(\frac{r}{R}\biggr)^4\biggr\}
+ C_\Phi
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{G M_\mathrm{tot}}{4R} \biggl\{ 5 \biggl(\frac{r}{R}\biggr)^2
- \frac{3}{2}\biggl(\frac{r}{R}\biggr)^4\biggr\}
+ C_\Phi 
\, .
</math>
  </td>
</tr>
</table>
Let's choose a constant, <math>C_\Phi</math>, such that the potential is <math>-GM_\mathrm{tot}/R</math> at the surface <math>(r/R = 1)</math>, that is,

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

<tr>
  <td align="right">
<math>- \frac{GM_\mathrm{tot}}{R}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\frac{GM_\mathrm{tot}}{4R}  \biggl\{ 5 - \frac{3}{2}\biggr\}
+ C_\Phi 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ C_\Phi</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>-\frac{7GM_\mathrm{tot}}{8R}  - \frac{GM_\mathrm{tot}}{R} = -\frac{15GM_\mathrm{tot}}{8R} \, .
</math>
  </td>
</tr>
</table>
Hence, the normalized gravitational potential is given by the expression,

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

<tr>
  <td align="right">
<math>\Phi_\mathrm{grav}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{G M_\mathrm{tot}}{4R} \biggl\{ 5 \biggl(\frac{r}{R}\biggr)^2
- \frac{3}{2}\biggl(\frac{r}{R}\biggr)^4\biggr\}
- \frac{15 GM_\mathrm{tot}}{8R}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{G M_\mathrm{tot}}{8R} \biggl\{- 15 + 10 \biggl(\frac{r}{R}\biggr)^2- 3\biggl(\frac{r}{R}\biggr)^4\biggr\}
\, .
</math>
  </td>
</tr>
</table>
<span id="ParabolicPotential">Note that in terms of Cartesian coordinates,</span> this expression becomes,

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

<tr>
  <td align="right">
<math>\Phi_\mathrm{grav}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
-~\frac{\pi G\rho R^2}{15}  
\biggl\{15 - \frac{10}{R^2} \biggl(x^2 + y^2 + z^2\biggr)
+ \frac{3}{R^4}\biggl(x^2 + y^2 + z^2\biggr)\biggl(x^2 + y^2 + z^2\biggr)\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
-~\frac{\pi G\rho R^2}{15}  
\biggl\{15 - \frac{10}{R^2} \biggl(x^2 + y^2 + z^2\biggr)
+ \frac{3}{R^4}
\biggl( x^4 + x^2y^2 + x^2z^2 + y^2x^2 + y^4 + y^2z^2 + x^2z^2 + y^2z^2 + z^4\biggr)
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
-~\frac{\pi G\rho R^2}{15}  
\biggl\{15 - \frac{10}{R^2} \biggl(x^2 + y^2 + z^2\biggr)
+ 
\frac{6}{R^4}\biggl(x^2y^2 + x^2z^2  + y^2z^2\biggr) + \frac{3}{R^4}\biggl( x^4 + y^4 + z^4 \biggr)  
\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
-~\pi G\rho R^2  
\biggl\{1 - \frac{2}{3R^2} \biggl(x^2 + y^2 + z^2\biggr)
+ 
\frac{2}{5R^4}\biggl(x^2y^2 + x^2z^2  + y^2z^2\biggr) + \frac{1}{5R^4}\biggl( x^4 + y^4 + z^4 \biggr)  
\biggr\} 
\, .
</math>
  </td>
</tr>
</table>

<font color="red">This matches the gravitational potential</font> [[ParabolicDensity/GravPot#ParabolicPotential|derived in a separate chapter]] using some of the <math>A_i</math> and <math>A_{ij}</math> coefficients adopted by [<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>]; see also the [[ThreeDimensionalConfigurations/HomogeneousEllipsoids#For_Spheres_(aℓ_=_am_=_as)|raw derivation of the relevant coefficients]].


----


<span id="Pressure">Hence</span>, proceeding via what we have labeled as [[SSCpt2/SolutionStrategies#Technique_1|"Technique 1"]], and enforcing the surface boundary condition, <math>~P(R) = 0</math>, {{ Prasad49 }} determines that,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~P(r)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- \int_0^r g_0(r) \rho(r) dr</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- \frac{4\pi G \rho_c^2 R^2}{15} \int_0^r \biggl[ 1 - \biggl(\frac{r}{R} \biggr)^2 \biggr]\biggl[5 - 3\biggl( \frac{r}{R} \biggr)^2\biggr]  \biggl( \frac{r}{R} \biggr) \frac{dr}{R}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- \frac{4\pi G \rho_c^2 R^2}{15} \int_0^r \biggl[ 5\biggl(\frac{r}{R} \biggr) - 8\biggl(\frac{r}{R} \biggr)^3 + 3\biggl(\frac{r}{R} \biggr)^5\biggr]  \frac{dr}{R}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2\pi G\rho_c^2 R^2}{15} \biggl[2 - 5 \biggl( \frac{r}{R} \biggr)^2
+ 4 \biggl( \frac{r}{R} \biggr)^4 - \biggl( \frac{r}{R} \biggr)^6 \biggr] </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{4\pi G\rho_c^2 R^2}{15} 
\biggl[1-\biggl(\frac{r}{R}\biggr)^2\biggr]^2 
\biggl[1-\frac{1}{2}\biggl(\frac{r}{R}\biggr)^2\biggr] 
\, ,</math>
  </td>
</tr>
</table>
</div>
where, it can readily be deduced, as well, that the central pressure is,
<div align="center">
<math>~P_c = \frac{4\pi}{15} G\rho_c^2 R^2 \, .</math>
</div>

As has been explained in the context of [[SR#Barotropic_Structure|our discussion of supplemental relations]], the enthalpy distribution <math>(H)</math> can be obtained from the relation,
<table border="0" align="center" cellpadding="5">
<tr>
  <td align="right"><math>dH</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>\frac{dP}{\rho} \, .</math></td>
</tr>

</table>

That is,
<table border="0" align="center" cellpadding="5">
<tr>
  <td align="right"><math>\frac{dH}{dr}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{4\pi G\rho_c R^2}{15}\biggl[1 - \biggl(\frac{r}{R}\biggr)^2\biggr]^{-1} \frac{d}{dr}\biggl\{
\biggl[1-\biggl(\frac{r}{R}\biggr)^2\biggr]^2 
\biggl[1-\frac{1}{2}\biggl(\frac{r}{R}\biggr)^2\biggr] 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{4\pi G\rho_c R^2}{15}\biggl[1 - \biggl(\frac{r}{R}\biggr)^2\biggr]^{-1} \biggl\{
\biggl[1-\biggl(\frac{r}{R}\biggr)^2\biggr]^2 \biggl[-\biggl(\frac{r}{R^2}\biggr)\biggr] 
+
2\biggl[1-\biggl(\frac{r}{R}\biggr)^2\biggr]\biggl[-2\biggl(\frac{r}{R^2}\biggr)\biggr] \biggl[1-\frac{1}{2}\biggl(\frac{r}{R}\biggr)^2\biggr] 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{4\pi G\rho_c R^2}{15} \biggl\{
\biggl[1-\biggl(\frac{r}{R}\biggr)^2\biggr] \biggl[-\biggl(\frac{r}{R^2}\biggr)\biggr] 
+
2\biggl[-2\biggl(\frac{r}{R^2}\biggr)\biggr] \biggl[1-\frac{1}{2}\biggl(\frac{r}{R}\biggr)^2\biggr] 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
-\frac{4\pi G\rho_c R^2}{15} \biggl\{
\biggl[1-\biggl(\frac{r}{R}\biggr)^2\biggr] \biggl(\frac{r}{R^2}\biggr) 
+
4\biggl(\frac{r}{R^2}\biggr) \biggl[1-\frac{1}{2}\biggl(\frac{r}{R}\biggr)^2\biggr] 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
-\frac{4\pi G\rho_c R^2}{15} \biggl\{
\biggl[1-\biggl(\frac{r}{R}\biggr)^2\biggr]  
+
\biggl[4 - 2\biggl(\frac{r}{R}\biggr)^2\biggr] 
\biggr\}\biggl(\frac{r}{R^2}\biggr)
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
-\frac{4\pi G\rho_c R^2}{15} 
\biggl[5 - 3\biggl(\frac{r}{R}\biggr)^2\biggr] 
\biggl(\frac{r}{R^2}\biggr)
\, .
</math>
  </td>
</tr>
</table>

Hence,
<table border="0" align="center" cellpadding="5">
<tr>
  <td align="right"><math>H</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
-\frac{4\pi G\rho_c }{15} \int
\biggl[5r - \frac{3}{R^2}\biggl(r^3\biggr)\biggr] dr
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
-\frac{4\pi G\rho_c R^2}{15} 
\biggl[\frac{5}{2}\biggl(\frac{r}{R}\biggr)^2 - \frac{3}{4}\biggl(\frac{r}{R}\biggr)^4\biggr] 
+ C
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
- \frac{1}{15} (2\pi G\rho_c R^2) 
\biggl[5\biggl(\frac{r}{R}\biggr)^2 - \frac{3}{2}\biggl(\frac{r}{R}\biggr)^4\biggr] 
+ C
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
- \biggl( \frac{GM_\mathrm{tot}}{4 R} \biggr) 
\biggl[5\biggl(\frac{r}{R}\biggr)^2 - \frac{3}{2}\biggl(\frac{r}{R}\biggr)^4\biggr] 
+ C \, .
</math>
  </td>
</tr>
</table>
Let's normalize by setting <math>H = 0</math> when <math>r=R</math>; that is,

<table border="0" align="center" cellpadding="5">
<tr>
  <td align="right"><math>C</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{7GM_\mathrm{tot}}{8 R} \, .  
</math>
  </td>
</tr>
</table>
<span id="SphericalEnthalpyProfile">So, the enthalpy is given by the expression,</span>

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

<tr>
  <td align="right"><math>H</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
-\biggl( \frac{GM_\mathrm{tot}}{4 R} \biggr) 
\biggl[5\biggl(\frac{r}{R}\biggr)^2 - \frac{3}{2}\biggl(\frac{r}{R}\biggr)^4\biggr] 
+ \frac{7GM}{8 R} 
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{GM_\mathrm{tot}}{8 R} \biggl[7 - 10\biggl(\frac{r}{R}\biggr)^2 + 3\biggl(\frac{r}{R}\biggr)^4\biggr] 
\, .
</math>
  </td>
</tr>
</table>

<table border="1" align="center" cellpadding="8"><tr><td align="left">
Note that the quantity, <math>(\Phi_\mathrm{grav}+H)</math>, is constant throughout the configuration.  Specifically,

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

<tr>
  <td align="right"><math>\Phi_\mathrm{grav} + H</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{G M_\mathrm{tot}}{8R} \biggl\{- 15 + 10 \biggl(\frac{r}{R}\biggr)^2- 3\biggl(\frac{r}{R}\biggr)^4\biggr\}
+
\frac{GM_\mathrm{tot}}{8 R} \biggl[7 - 10\biggl(\frac{r}{R}\biggr)^2 + 3\biggl(\frac{r}{R}\biggr)^4\biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
-~\frac{GM_\mathrm{tot}}{R}  \, . 
</math>
  </td>
</tr>
</table>

</td></tr></table>

===Specific Entropy Distribution===

From an [[PGE/FirstLawOfThermodynamics#EntropyLL75|accompanying discussion]] of specific entropy distributions, <math>s</math>, we realize that to within an additive constant <math>(s_0)</math>,
<div align="center" id="LL75">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>s - s_0</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>c_P \ln \biggl(\frac{P^{1/\gamma_g}}{\rho} \biggr)\, .</math>
  </td>
</tr>
</table>
[<b>[[Appendix/References#LL75|<font color="red">LL75</font>]]</b>], &sect;80, Eq. (80.12)
</div>
Or (see a [[Appendix/Ramblings/PatrickMotl#Tying_Expressions_into_H_Book_Context|related discussion]]), given that 
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~c_P </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\gamma_g}{(\gamma_g-1)} \biggl( \frac{\Re}{\bar\mu} \biggr) \, ,</math>
  </td>
</tr>
</table>
we can also write,

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

<tr>
  <td align="right">
<math>\frac{(s - s_0)}{\Re/\bar{\mu}}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\frac{\gamma_g}{(\gamma_g-1)} \biggl\{ 
\frac{1}{\gamma_g}\ln P - \ln\rho
\biggr\}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\frac{1}{(\gamma_g-1)} \biggl\{ 
\ln P - \gamma_g\ln\rho
\biggr\}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\frac{1}{(\gamma_g-1)} \biggl\{ 
\ln P - \gamma_g\ln\rho - \ln(\gamma_g-1)
\biggr\} + \frac{\ln(\gamma_g-1)}{(\gamma_g - 1)}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{1}{(\gamma_g-1)}\ln \biggl[ \frac{P}{(\gamma_g-1)\rho^{\gamma_g}} \biggr]
+ \frac{\ln(\gamma_g-1)}{(\gamma_g - 1)} \, .</math>
  </td>
</tr>
</table>
Notice that if we set the constant,

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

<tr>
  <td align="right">
<math>\frac{s_0}{\Re/\bar{\mu}}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
- \frac{\ln(\gamma_g-1)}{(\gamma_g - 1)} \, ,
</math>
  </td>
</tr>
</table>

we obtain the same expression for the entropy distribution as we used in our [[Appendix/Ramblings/PatrickMotl#Tying_Expressions_into_H_Book_Context|discussions with Patrick Motl]], namely,

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

<tr>
  <td align="right">
<math>\frac{s}{\Re/\bar{\mu}}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{1}{(\gamma_g-1)}\ln \biggl[ \frac{P}{(\gamma_g-1)\rho^{\gamma_g}} \biggr] \, .
</math>
  </td>
</tr>
</table>

Shifting this expression for the specific entropy by another constant (not written out explicitly here) leads us to the more "normalized" expression,

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

<tr>
  <td align="right">
<math>\frac{s}{\Re/\bar{\mu}}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{1}{(\gamma_g-1)}\ln \biggl[ \frac{P/P_c}{(\gamma_g-1)(\rho/\rho_c)^{\gamma_g}} \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ (\gamma_g - 1)\exp\biggl[\frac{(\gamma_g - 1)s}{\Re/\bar{\mu}} \biggr]</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[ \frac{P/P_c}{(\rho/\rho_c)^{\gamma_g}} \biggr] 
</math>
  </td>
</tr>
</table>
Plugging in the expressions for the pressure and density distributions that are relevant to the {{ Prasad49 }} model having a "parabolic" density distribution, then gives,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>(\gamma_g-1)\exp\biggl[\frac{(\gamma_g - 1)s}{\Re/\bar{\mu}} \biggr]</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl\{ 
\biggl[1-\biggl(\frac{r}{R}\biggr)^2\biggr]^2 
\biggl[1-\frac{1}{2}\biggl(\frac{r}{R}\biggr)^2\biggr] 
\biggr\}
\biggl[ 1 - \biggl(\frac{r}{R} \biggr)^2 \biggr]^{-\gamma_g} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[1-\biggl(\frac{r}{R}\biggr)^2\biggr]^{(2 - \gamma_g)} 
\biggl[1-\frac{1}{2}\biggl(\frac{r}{R}\biggr)^2\biggr] \, .
</math>
  </td>
</tr>
</table>

===Stabililty===

As has been derived in [[SSC/Perturbations#Eigen_Value_Problem|an accompanying discussion]], the second-order ODE that defines the relevant Eigenvalue problem is,

<div align="center">
<math>
\biggl(\frac{P_0}{P_c}\biggr)\frac{d^2x}{d\chi_0^2} 
+ \biggl[\biggl(\frac{P_0}{P_c}\biggr)\frac{4}{\chi_0} - \biggl(\frac{\rho_0}{\rho_c}\biggr) \biggl(\frac{g_0}{g_\mathrm{SSC}}\biggr) \biggr] \frac{dx}{d\chi_0} 
+ \biggl(\frac{\rho_0}{\rho_c}\biggr) \biggl(\frac{1}{\gamma_\mathrm{g}} \biggr)\biggl[\tau_\mathrm{SSC}^2 \omega^2 
+ (4 - 3\gamma_\mathrm{g})\biggl(\frac{g_0}{g_\mathrm{SSC}}\biggr) \frac{1}{\chi_0} \biggr]  x = 0 \, ,
</math><br />
</div>
where the dimensionless radius,
<div align="center">
<math> 
\chi_0 \equiv \frac{r_0}{R} \, ,
</math><br />
</div>
<div align="center">
<math> 
g_\mathrm{SSC} \equiv \frac{P_c}{R\rho_c}</math> &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 
<math>\tau_\mathrm{SSC} \equiv \biggl( \frac{R^2\rho_c}{P_c}\biggr)^{1/2} \, .
</math>
</div>

For Prasad's configuration with a parabolic density distribution, 
<div align="center">
<math> 
g_\mathrm{SSC} = \frac{4\pi G\rho_c R}{15}</math> &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 
<math>\tau_\mathrm{SSC} \equiv \biggl( \frac{15}{4\pi G \rho_c }\biggr)^{1/2} = \biggl( \frac{2R^3}{GM_\mathrm{tot} }\biggr)^{1/2} 
= \biggl( \frac{3}{2\pi G\bar\rho}\biggr)^{1/2}\, .
</math>
</div>
Hence,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{g_0}{g_\mathrm{SSC}} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(5 - 3 \chi_0^2)\chi_0  \, ,</math>
  </td>
</tr>
</table>
</div>
and the governing adiabatic wave equation takes the form,

<div align="center">
<math>
(1-\chi_0^2) \biggl( 1 - \frac{1}{2}\chi_0^2 \biggr)\frac{d^2x}{d\chi_0^2} 
+ \frac{1}{\chi_0}\biggl[4 (1-\chi_0^2) \biggl( 1 - \frac{1}{2}\chi_0^2 \biggr) - (5 - 3 \chi_0^2)\chi_0^2\biggr] \frac{dx}{d\chi_0} 
+ \biggl[\frac{\tau_\mathrm{SSC}^2 \omega^2}{\gamma_\mathrm{g}} 
-\alpha (5 - 3 \chi_0^2)\biggr]  x = 0 \, ,
</math><br />
</div>
where,
<div align="center">
<math>~\alpha \equiv 3 - \frac{4}{\gamma_\mathrm{g}} \, .</math>
</div>
In keeping with Prasad's presentation &#8212; see, specifically, his equations (2) &amp; (3) &#8212; this wave equation can also be written as,

<div align="center">
<math>
(1-\chi_0^2) \biggl( 1 - \frac{1}{2}\chi_0^2 \biggr)\frac{d^2x}{d\chi_0^2} 
+ \frac{1}{\chi_0}\biggl[4 - 11\chi_0^2 + 5\chi_0^4\biggr] \frac{dx}{d\chi_0} 
+ \biggl[\mathfrak{J}+3\alpha \chi_0^2 \biggr]  x = 0 \, ,
</math><br />
</div>
where,
<div align="center">
<math>~\mathfrak{J} \equiv \frac{3\omega^2}{2\pi G \gamma_\mathrm{g} \bar\rho} - 5\alpha \, .</math>
</div>
For what it's worth, we have also deduced that this expression can be written as,

<div align="center">
<math>
(1-\chi_0^2) \biggl( 1 - \frac{1}{2}\chi_0^2 \biggr)\chi_0^{-4} \frac{d}{d\chi_0} \biggl[\chi_0^4 \frac{dx}{d\chi_0}  \biggr]
-(5-3\chi_0^2)\chi_0^{1+\alpha} \frac{d}{d\chi_0} \biggl[ \chi_0^{-\alpha} x \biggr]
+ \biggl(\frac{\tau_\mathrm{SSC}^2~ \omega^2}{\gamma_\mathrm{g}}\biggr)  x = 0 \, ,
</math>
</div>