Editing SSC/Virial/PolytropesSummary (section)

==Take Care Comparing Gravitational Potential Energies==
Does this derived relation for the coefficient, <math>~A</math>, make sense?  Well, we've derived the relation by comparing two separate expressions for the gravitational potential energy that were normalized in slightly different ways, so the leading numerical coefficient may not be correct.  We need to repeat the derivation, checking the relative normalizations carefully.  But before doing this, let's determine what we ''expected'' the relation to be, based on the expressions for the structural form factors that we have been using.

From the [[User:Tohline/SSC/Virial/PolytropesSummary#Serious_Concern|lead-in paragraphs of this subsection]], we have previously assumed that,
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<table border="0" cellpadding="5">
<tr>
  <td align="right">
<math>~\mathcal{A}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>\frac{1}{5} \cdot \biggl( \frac{M_\mathrm{limit}}{M_\mathrm{tot}} \biggr)^2 \frac{\tilde\mathfrak{f}_W}{\tilde\mathfrak{f}_M^2} 
= \frac{1}{5} \cdot \biggl( \frac{M_\mathrm{limit}}{M_\mathrm{tot}} \biggr)^2 \biggl\{ \frac{3^2\cdot 5}{5-n} \biggl[ \frac{\tilde\theta^'}{\tilde\xi} \biggr]^2 \biggr\}
\biggl\{ \biggl[ - \frac{3\tilde\theta^'}{\tilde\xi} \biggr] \biggr\}^{-2} = \frac{1}{(5-n)} \biggl( \frac{M_\mathrm{limit}}{M_\mathrm{tot}} \biggr)^2 \, .
</math>
  </td>
</tr>

</table>
</div>
According to the line of reasoning presented above, the coefficient, <math>~A</math>, is related to the gravitational potential energy via the expression,
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<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~W_\mathrm{grav}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-3A\chi^{-1} E_\mathrm{norm} </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-3A \biggl( \frac{K^5}{G^3} \biggr)^{1/2} \biggl( \frac{G^5 M_\mathrm{tot}^4}{K^5} \biggr)^{1/2} \frac{1}{R}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-3A \biggl(\frac{GM_\mathrm{tot}^2}{R} \biggr) \, .</math>
  </td>
</tr>
</table>
</div>
From our [[User:Tohline/SphericallySymmetricConfigurations/Virial#Expressions_for_Various_Energy_Terms|introductory layout of the free-energy function for polytropes]] &#8212; see, also, p. 64, Equation (12) of [[User:Tohline/Appendix/References|Chandrasekhar [C67]]] &#8212; the gravitational potential energy is,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~W_\mathrm{grav}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ - \int_0^{R_\mathrm{limit}} \biggl( \frac{GM_r}{r} \biggr) 4\pi r^2 \rho dr \, ,</math>
  </td>
</tr>
</table>
</div>
where,
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<tr>
  <td align="right">
<math>~M_r</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\int_0^r 4\pi r^2 \rho dr \, .</math>
  </td>
</tr>
</table>
</div>
Now, independent of the chosen normalization, if we use <math>~M_\mathrm{tot}</math> to represent the total mass of an ''isolated'' <math>~n=5</math> polytrope, then from [[User:Tohline/SSC/Structure/PolytropesEmbedded#Review_2|an earlier review]], we have,
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<tr>
  <td align="right">
<math>~M_\mathrm{tot}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggr[ \frac{2\cdot 3^4 K^3}{\pi G^3} \biggr]^{1/2}  \rho_c^{-1/5} \, ,</math>
  </td>
</tr>
</table>
</div>
and we can write, in terms of the Lane-Emden dimensionless radius, <math>~\xi</math>,
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<math>
\frac{M_r}{M_\mathrm{tot}} = \xi^3 (3 + \xi^2)^{-3/2}  \, .
</math> 
</div>
===Virial Chapter===
Now, in our discussion of the virial equilibrium of embedded polytropes, we used the normalizations specified above and wrote,
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<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~W_\mathrm{grav}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
- 4\pi GM_\mathrm{tot} R_\mathrm{norm}^2 \rho_\mathrm{norm} \int_0^{R_\mathrm{limit}/R_\mathrm{norm}} \biggl[\frac{M_r}{M_\mathrm{tot}} \biggr]  r^* \rho^* dr^* 
</math>
  </td>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
- E_\mathrm{norm} \int_0^{R_\mathrm{limit}/R_\mathrm{norm}} 3\biggl[\frac{M_r}{M_\mathrm{tot}} \biggr]  r^* \rho^* dr^* \, .
</math>
  </td>
</tr>
</table>
</div>
We can replace <math>~r^* \equiv r/R_\mathrm{norm}</math> with <math>~\xi \equiv r/a_5</math> by recognizing that, 
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<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>~
\biggl(  \frac{2\cdot 3^4 K^3}{\pi G^3} \biggr)^{1/2} \rho_c^{-1/5} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \rho_c^{-2/5} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl(  \frac{\pi G^3}{2\cdot 3^4 K^3} \biggr)  M_\mathrm{tot}^2 \, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~a_5</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl(  \frac{3K}{2\pi G} \biggr)^{1/2} \rho_c^{-2/5} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl(  \frac{3K}{2\pi G} \biggr)^{1/2} \biggl(  \frac{\pi G^3}{2\cdot 3^4 K^3} \biggr)  M_\mathrm{tot}^2
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl(  \frac{\pi}{2^3\cdot 3^7} \biggr)^{1/2}  \biggl(  \frac{G}{K} \biggr)^{5/2} M_\mathrm{tot}^2 \, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~R_\mathrm{norm}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl(  \frac{G}{K} \biggr)^{5/2} M_\mathrm{tot}^{2} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~ \frac{a_5}{R_\mathrm{norm}}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl(  \frac{\pi}{2^3\cdot 3^7} \biggr)^{1/2}  \, .
</math>
  </td>
</tr>
</table>
</div>
Hence,
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<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~r^*</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl(  \frac{\pi}{2^3\cdot 3^7} \biggr)^{1/2} \xi \, .
</math>
  </td>
</tr>
</table>
</div>

Also,
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<tr>
  <td align="right">
<math>~\rho_\mathrm{norm}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{3}{4\pi} \biggl[ \frac{K}{G} \biggr]^{15/2} M_\mathrm{tot}^{-5}
</math>
  </td>
</tr>

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

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl(\frac{3^2}{2^4\pi^2} \biggr)^{1/2} \biggl(  \frac{\pi^5 }{2^5 \cdot 3^{20}} \biggr)^{1/2} \rho_c
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl(  \frac{\pi^3 }{2^9 \cdot 3^{18}} \biggr)^{1/2} \rho_c \, .
</math>
  </td>
</tr>
</table>
</div>
Hence,
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<tr>
  <td align="right">
<math>~\rho^* \equiv \frac{\rho}{\rho_\mathrm{norm}}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl(  \frac{2^9 \cdot 3^{18}}{\pi^3 } \biggr)^{1/2} \frac{\rho}{\rho_c} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl(  \frac{2^9 \cdot 3^{18}}{\pi^3 } \biggr)^{1/2} \biggl[ 1 + \frac{\xi^2}{3} \biggr]^{-5/2} \, .
</math>
  </td>
</tr>
</table>
</div>
So the energy integral becomes,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{W_\mathrm{grav}}{E_\mathrm{norm} }</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
- 3 \biggl(  \frac{\pi}{2^3\cdot 3^7} \biggr) \biggl(  \frac{2^9 \cdot 3^{18}}{\pi^3 } \biggr)^{1/2} 
\int_0^{\tilde\xi} \biggl[\xi^3 (3 + \xi^2)^{-3/2}  \biggr]  \xi \biggl[ 1 + \frac{\xi^2}{3} \biggr]^{-5/2} d\xi 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
- \biggl(  \frac{2^3 \cdot 3^{6}}{\pi } \biggr)^{1/2} 
\int_0^{\tilde\xi} \xi^4 \biggl[ 1 + \frac{\xi^2}{3} \biggr]^{-4} d\xi \, .
</math>
  </td>
</tr>
</table>
</div>
This needs to be compared with the <math>~W_\mathrm{grav}^*</math> integral that we previously have handled in [[User:Tohline/SSC/Structure/BiPolytropes/Analytic5_1#Expression_for_Free_Energy|the chapter discussing bipolytrope models]].