Editing SSC/Structure/OtherAnalyticModels (section)

===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>