Editing SSCpt1/Virial/FormFactors/Pt2

=Structural Form Factors (Pt 2)=
<table border="1" align="center" width="100%" colspan="8">
<tr>
  <td align="center" bgcolor="lightblue" width="33%"><br />[[SSCpt1/Virial/FormFactors|Part I:&nbsp; Synopsis]]
&nbsp;
  </td>
  <td align="center" bgcolor="lightblue" width="33%"><br />[[SSCpt1/Virial/FormFactors/Pt2|Part II:&nbsp; n = 5 Polytrope]]
&nbsp;
  </td>
  <td align="center" bgcolor="lightblue"><br />[[SSCpt1/Virial/FormFactors/Pt3|Part III:&nbsp; n = 1 Polytrope]]
&nbsp;
  </td>
</tr>
</table>

==First Detailed Example (n = 5)==

Here we complete these integrals to derive detailed expressions for the above subset of structural form factors in the case of spherically symmetric configurations that obey an <math>~n=5</math> polytropic equation of state.  The hope is that this will illustrate, in a clear and helpful manner, how the task of calculating form factors is to be carried out, in practice; and, in particular, to provide one nontrivial example for which analytic expressions are derivable.  This should simplify the task of debugging numerical algorithms that are designed to calculate structural form factors for more general cases that cannot be derived analytically.  The limits of integration will be specified in a general enough fashion that the resulting expressions can be applied, not only to the structures of ''isolated'' polytropes, but to [[SSC/Virial/PolytropesSummary#Further_Evaluation_of_n_.3D_5_Polytropic_Structures|''pressure-truncated'' polytropes]] that are embedded in a hot, tenuous external medium and to the [[SSC/Structure/BiPolytropes/Analytic51#Free_Energy|cores of bipolytropes]].

===Foundation (n = 5)===
We use the following normalizations, as drawn from [[SSCpt1/Virial#Normalizations|our more general introductory discussion]]:

<table border="1" align="center" cellpadding="5" width="70%">
<tr><th align="center" colspan="1">
Adopted Normalizations <math>(n=5; \gamma=6/5)</math>
</th></tr>
<tr><td align="center" colspan="1">

<math>
\begin{align}
R_\mathrm{norm} & \equiv \biggl( \frac{G}{K} \biggr)^{5/2} M_\mathrm{tot}^{2} \\
P_\mathrm{norm} & \equiv \biggl( \frac{K^{10}}{G^{9} M_\mathrm{tot}^{6}} \biggr)  \\
\end{align}
</math>
<hr \>
<math>
\begin{align}
E_\mathrm{norm} & \equiv P_\mathrm{norm} R_\mathrm{norm}^3 && = \biggl( \frac{K^5}{G^3} \biggr)^{1 / 2} \\
\rho_\mathrm{norm} & \equiv \frac{3M_\mathrm{tot}}{4\pi R_\mathrm{norm}^3} && = \frac{3}{4\pi} \biggl( \frac{K}{G} \biggr)^{15/2} M_\mathrm{tot}^{-5} \\
c^2_\mathrm{norm} & \equiv \frac{P_\mathrm{norm}}{\rho_\mathrm{norm}} && = \frac{4\pi}{3} \biggl( \frac{K^5}{G^3} \biggr)^{1/2} M_\mathrm{tot}^{-1} \\
\end{align}
</math>


<!-- BEGIN REPLACED 01
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>R_\mathrm{norm}</math>
  </td>
  <td align="center">
<math>\equiv</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>P_\mathrm{norm}</math>
  </td>
  <td align="center">
<math>\equiv</math>
  </td>
  <td align="left">
<math>\biggl( \frac{K^{10}}{G^{9} M_\mathrm{tot}^{6}} \biggr)  </math>
  </td>
</tr>

<tr>
  <td align="center" colspan="3">

----

  </td>
</tr>

<tr>
  <td align="right">
<math>E_\mathrm{norm}</math>
  </td>
  <td align="center"><math> \equiv </math>  </td>
  <td align="left"><math>P_\mathrm{norm} R_\mathrm{norm}^3 = \biggl( \frac{K^5}{G^3} \biggr)^{1/2} </math>  </td>
</tr>

<tr>
  <td align="right">
<math>\rho_\mathrm{norm}</math>
  </td>
  <td align="center"><math> \equiv </math>  </td>
  <td align="left">
<math>\frac{3M_\mathrm{tot}}{4\pi R_\mathrm{norm}^3}
= \frac{3}{4\pi} \biggl( \frac{K}{G} \biggr)^{15/2} M_\mathrm{tot}^{-5} </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>c^2_\mathrm{norm}</math>
  </td>
  <td align="center">
<math>\equiv</math>
  </td>
  <td align="left">
<math>\frac{P_\mathrm{norm}}{\rho_\mathrm{norm}} = \frac{4\pi}{3} \biggl( \frac{K^5}{G^3} \biggr)^{1/2} M_\mathrm{tot}^{-1} </math>
  </td>
</tr>
</table>
END REPLACED 01 -->

</td>
</tr>

<tr><th align="left" colspan="1">
Note that the following relations also hold:
<div align="center">
<math>E_\mathrm{norm} = P_\mathrm{norm} R_\mathrm{norm}^3 = \frac{G M_\mathrm{tot}^2}{ R_\mathrm{norm}} 
= \biggl( \frac{3}{4\pi} \biggr) M_\mathrm{tot} c_\mathrm{norm}^2</math>
</div>
</th></tr>
</table>


As is detailed in our [[SSC/Structure/BiPolytropes/Analytic51#Profile|accompanying discussion of bipolytropes]] &#8212; see also our [[SSC/Structure/Polytropes#.3D_5_Polytrope|discussion of the properties of ''isolated'' polytropes]] &#8212; in terms of the dimensionless Lane-Emden coordinate, <math>\xi \equiv r/a_{5}</math>, where,
<div align="center">
<math>
a_{5} =\biggr[ \frac{3K}{2\pi G} \biggr]^{1/2}  \rho_0^{-2/5}  \, ,
</math>
</div>
the radial profile of various physical variables is as follows:
<div align="center">
<math>
\begin{align}
\frac{r}{[K^{1/2}/(G^{1/2}\rho_0^{2/5})]} & = \biggl( \frac{3}{2\pi} \biggr)^{1/2} \xi \, , \\
\frac{\rho}{\rho_0} & = \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-5/2} \, , \\
\frac{P}{K\rho_0^{6/5}} & = \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-3} \, , \\
\frac{M_r}{[K^{3/2}/(G^{3/2}\rho_0^{1/5})]} & = \biggl( \frac{2\cdot 3}{\pi } \biggr)^{1/2} \biggl[ \xi^3 \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-3/2} \biggr] \, . \\
\end{align}
</math>
</div>

<!-- BEGIN REPLACED 02
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\frac{r}{[K^{1/2}/(G^{1/2}\rho_0^{2/5})]}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\biggl( \frac{3}{2\pi} \biggr)^{1/2} \xi \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\frac{\rho}{\rho_0}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-5/2} \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\frac{P}{K\rho_0^{6/5}}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-3} \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\frac{M_r}{[K^{3/2}/(G^{3/2}\rho_0^{1/5})]}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\biggl( \frac{2\cdot 3}{\pi } \biggr)^{1/2} \biggl[ \xi^3 \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-3/2} \biggr] \, .</math>
  </td>
</tr>
</table>

END REPLACED 02-->

Notice that, in these expressions, the central density, <math>\rho_0</math>, has been used instead of <math>M_\mathrm{tot}</math> to normalize the relevant physical variables. We can switch from one normalization to the other by realizing that &#8212; see, again, our [[SSC/Structure/Polytropes#.3D_5_Polytrope|accompanying discussion]] &#8212; in ''isolated'' <math>n=5</math> polytropes, the total mass is given by the expression,
<div align="center">
<math>M_\mathrm{tot} = \biggr[ \frac{2\cdot 3^4 K^3}{\pi G^3} \biggr]^{1/2}  \rho_0^{-1/5}  
~~~~\Rightarrow ~~~~
\rho_0^{1/5} = \biggr[ \frac{2\cdot 3^4 K^3}{\pi G^3} \biggr]^{1/2}  M_\mathrm{tot}^{-1} \, .</math>
</div>
<span id="NormalizedProfiles">Employing this mapping</span> to switch to our "preferred" adopted normalizations, as defined in the above boxed-in table, the four radial profiles become,

<div align="center">
<math>
\begin{align}
r^\dagger & \equiv \frac{r}{R_\mathrm{norm}} && = \biggl( \frac{\pi}{2\cdot 3^4} \biggr) \biggl( \frac{3}{2\pi} \biggr)^{1/2} \xi = \biggl( \frac{\pi}{2^3\cdot 3^7} \biggr)^{1 / 2} \xi \\
\rho^\dagger & \equiv \frac{\rho}{\rho_\mathrm{norm}} && = \biggl( \frac{2^3\cdot 3^6}{\pi} \biggr)^{3/2} \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-5/2} \, , \\
P^\dagger &\equiv \frac{P}{P_\mathrm{norm}} && = \biggl( \frac{2\cdot 3^4}{\pi} \biggr)^3 \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-3} \, , \\
\frac{M_r}{M_\mathrm{tot}}& && = \biggl( \frac{\pi}{2\cdot 3^4} \biggr)^{1/2} 
\biggl( \frac{2\cdot 3}{\pi } \biggr)^{1/2} \biggl[ \xi^3 \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-3/2} \biggr] 
= \biggl[\frac{\xi^2}{3}\biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-1}\biggr]^{3/2} 
\, .\\
\end{align}
</math>
</div>

<!-- BEGIN REPLACED 03
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>r^\dagger \equiv \frac{r}{R_\mathrm{norm}}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl( \frac{\pi}{2\cdot 3^4} \biggr) \biggl( \frac{3}{2\pi} \biggr)^{1/2} \xi 
= \biggl( \frac{\pi}{2^3\cdot 3^7} \biggr)^{1/2} \xi
\, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\rho^\dagger \equiv \frac{\rho}{\rho_\mathrm{norm}}</math>
  </td>
  <td align="center">
<math> = </math>
  </td>
  <td align="left"><math>\biggl( \frac{2^3\cdot 3^6}{\pi} \biggr)^{3/2} \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-5/2} \, ,</math>  </td>
</tr>

<tr>
  <td align="right">
<math>P^\dagger \equiv \frac{P}{P_\mathrm{norm}}</math>
  </td>
  <td align="center">
<math> = </math>
  </td>
  <td align="left">
<math>\biggl( \frac{2\cdot 3^4}{\pi} \biggr)^3 \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-3} \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\frac{M_r}{M_\mathrm{tot}}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl( \frac{\pi}{2\cdot 3^4} \biggr)^{1/2} 
\biggl( \frac{2\cdot 3}{\pi } \biggr)^{1/2} \biggl[ \xi^3 \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-3/2} \biggr] 
= \biggl[\frac{\xi^2}{3}\biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-1}\biggr]^{3/2} 
\, .</math>
  </td>
</tr>
</table>
END REPLACED 03 -->

===Mass1 (n = 5)===
While we already know the expression for the <math>M_r</math> profile, having copied it from our [[SSC/Structure/Polytropes#.3D_5_Polytrope|discussion of detailed force-balanced models of ''isolated'' polytropes]], let's show how that profile can be derived by integrating over the density profile.  After employing the ''norm''-subscripted quantities, as defined above, to normalize the radial coordinate and the mass density in our [[SSCpt1/Virial#Normalize|introductory discussion of the virial theorem]], we obtained the following integral defining the, 

<font color="red">Normalized Mass:</font>
<div align="center">

<math>
\begin{align}
M_r(r^\dagger) & = M_\mathrm{tot} \int_0^{r^\dagger}  3(r^\dagger)^2 \rho^\dagger dr^\dagger  \, . \\
\end{align}
</math>

<!-- BEGIN REPLACED 04
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>M_r(r^\dagger)  </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
M_\mathrm{tot} \int_0^{r^\dagger}  3(r^\dagger)^2 \rho^\dagger dr^\dagger  \, .
</math>
  </td>
</tr>
</table>

END REPLACED 04 -->
</div>
Plugging in the profiles for <math>r^\dagger</math> and <math>\rho^\dagger</math>, and recognizing that,
<div align="center">
<math>dr^\dagger = \biggl( \frac{\pi}{2^3\cdot 3^7} \biggr)^{1/2} d\xi \, ,</math>
</div>
gives, with the help of [http://integrals.wolfram.com/index.jsp Mathematica's Online Integrator], [[Image:OnlineIntegral01.png|250px|right|Mathematica Integral]]
<div align="center">
<math>
\begin{align}
\frac{M_r(\xi)}{M_\mathrm{tot} } & = 3 \biggl( \frac{\pi}{2^3\cdot 3^7} \biggr)^{3/2} \biggl( \frac{2^3\cdot 3^6}{\pi} \biggr)^{3/2}  \int_0^{\xi}  \xi^2 \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-5/2} d\xi \\
& = 3 \biggl( \frac{1}{3} \biggr)^{3/2} \biggl[ \frac{\xi^3}{3}\biggl(1+\frac{\xi^2}{3}\biggr)^{-3/2} \biggr]_0^{\xi} \\
& = \biggl[ \frac{\xi^2}{3}\biggl(1+\frac{\xi^2}{3}\biggr)^{-1}\biggr]^{3/2} \, . \\
\end{align}
</math>

<!-- BEGIN REPLACED 05
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\frac{M_r(\xi)}{M_\mathrm{tot} } </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
3 \biggl( \frac{\pi}{2^3\cdot 3^7} \biggr)^{3/2} \biggl( \frac{2^3\cdot 3^6}{\pi} \biggr)^{3/2}  
\int_0^{\xi}  \xi^2 \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-5/2} d\xi 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
3 \biggl( \frac{1}{3} \biggr)^{3/2} 
\biggl[ \frac{\xi^3}{3}\biggl(1+\frac{\xi^2}{3}\biggr)^{-3/2} \biggr]_0^{\xi}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[ \frac{\xi^2}{3}\biggl(1+\frac{\xi^2}{3}\biggr)^{-1}\biggr]^{3/2} \, .
</math>
  </td>
</tr>
</table>
END REPLACED 05 -->

</div>
As it should, this expression exactly matches the normalized <math>M_r</math> profile shown above.  Notice that if we decide to truncate an <math>n = 5</math> polytrope at some radius, <math>\tilde\xi  <  \xi_1</math> &#8212; as in the discussion that follows &#8212; the mass of this truncated configuration will be, simply,
<div align="center">
<math>
\frac{M_\mathrm{limit}}{M_\mathrm{tot} }  = \frac{M_r({\tilde\xi})}{M_\mathrm{tot} } = \biggl[ \frac{\tilde\xi^2}{3}\biggl(1+\frac{\tilde\xi^2}{3}\biggr)^{-1}\biggr]^{3/2} \, .
</math>

<!-- BEGIN REPLACED 06
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\frac{M_\mathrm{limit}}{M_\mathrm{tot} }  = \frac{M_r({\tilde\xi})}{M_\mathrm{tot} } </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[ \frac{\tilde\xi^2}{3}\biggl(1+\frac{\tilde\xi^2}{3}\biggr)^{-1}\biggr]^{3/2} \, .
</math>
  </td>
</tr>
</table>
END REPLACED 06 -->

</div>

===Mass2 (n = 5)===

Alternatively, as has been laid out in our [[SSCpt1/Virial#Summary_of_Normalized_Expressions|accompanying summary of normalized expressions that are relevant to free-energy calculations]],
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\frac{M_r(x)}{M_\mathrm{tot}}  </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\biggl( \frac{\rho_c}{\bar\rho} \biggr)_\mathrm{eq} \biggl( \frac{M_\mathrm{limit}}{M_\mathrm{tot}} \biggr)
\int_0^{x}  3x^2 \biggl[ \frac{\rho(x)}{\rho_0} \biggr]  dx \, ,</math>
  </td>
</tr>
</table>
</div>
where, <math>M_\mathrm{limit}</math> is the "total" mass of the polytropic configuration that is truncated at <math>R_\mathrm{limit}</math>; keep in mind that, here,
<div align="center">
<math>M_\mathrm{tot} = \biggr[ \frac{2\cdot 3^4 K^3}{\pi G^3} \biggr]^{1/2}  \rho_0^{-1/5}  \, ,</math>
</div>
is the total mass of the ''isolated'' <math>n=5</math> polytrope, that is, a polytrope whose ''Lane-Emden'' radius extends all the way to <math>\xi_1</math>.  In our discussions of truncated polytropes, we often will use <math>\tilde\xi \le \xi_1</math> to specify the truncated radius in terms of the familiar, dimensionless Lane-Emden radial coordinate, so here we will set,
<div align="center">
<math>R_\mathrm{limit} = a_5 \tilde\xi ~~~~\Rightarrow ~~~~ x = \frac{r}{R_\mathrm{limit}} = \frac{a_5 \xi}{a_5 \tilde\xi} = \frac{\xi}{\tilde\xi} \, .</math>
</div>
Hence, in terms of the desired integration coordinate, <math>x</math>, the density profile provided above becomes,

<div align="center" id="rhoofx">
<table border="1" cellpadding="10" align="center">
<tr><td align="center">

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

<tr>
  <td align="right">
<math>\frac{\rho(x)}{\rho_0}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\biggl[ 1 + \biggl(\frac{\tilde\xi^2}{3}\biggr)x^2 \biggr]^{-5/2} \, ,</math>
  </td>
</tr>
</table>

</td></tr>
</table>
</div>
and the integral defining <math>M_r(x)</math> becomes,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\frac{M_r(x)}{M_\mathrm{tot}}  </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\biggl( \frac{\rho_c}{\bar\rho} \biggr)_\mathrm{eq} \biggl( \frac{M_\mathrm{limit}}{M_\mathrm{tot}} \biggr)
\int_0^{x}  3x^2 \biggl[ 1 + \biggl(\frac{\tilde\xi^2}{3}\biggr)x^2 \biggr]^{-5/2}  dx </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\biggl( \frac{\rho_c}{\bar\rho} \biggr)_\mathrm{eq} \biggl( \frac{M_\mathrm{limit}}{M_\mathrm{tot}} \biggr)
\biggl\{ x^3 \biggl[ 1 + \biggl(\frac{\tilde\xi^2}{3}\biggr)x^2 \biggr]^{-3/2}  \biggr\} \, .</math>
  </td>
</tr>
</table>
</div>
In this case, integrating "all the way out to the surface" means setting <math>r = R_\mathrm{limit}</math> and, hence, <math>x = 1</math>; by definition, it also means <math>M_r(x) = M_\mathrm{limit}</math>.  Therefore we have,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\frac{M_\mathrm{limit}}{M_\mathrm{tot}}  </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\biggl( \frac{\rho_c}{\bar\rho} \biggr)_\mathrm{eq} \biggl( \frac{M_\mathrm{limit}}{M_\mathrm{tot}} \biggr)
\biggl[ 1 + \frac{\tilde\xi^2}{3} \biggr]^{-3/2}  </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ \biggl( \frac{\bar\rho}{\rho_c} \biggr)_\mathrm{eq} </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\biggl[ 1 + \frac{\tilde\xi^2}{3} \biggr]^{-3/2}  \, .</math>
  </td>
</tr>
</table>
</div>

Using this expression for the mean-to-central density ratio along with the expression for the ratio, <math>M_\mathrm{limit}/M_\mathrm{tot}</math>, derived in the preceding subsection, we also can state that for truncated <math>n=5</math> polytropes,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\frac{M_r(x)}{M_\mathrm{tot}}  </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\biggl[ 1 + \frac{\tilde\xi^2}{3} \biggr]^{3/2} \biggl[ \frac{\tilde\xi^2}{3}\biggl(1+\frac{\tilde\xi^2}{3}\biggr)^{-1}\biggr]^{3/2}
\biggl\{ x^3 \biggl[ 1 + \biggl(\frac{\tilde\xi^2}{3}\biggr)x^2 \biggr]^{-3/2}  \biggr\} </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\biggl[ \frac{\tilde\xi^2}{3}\biggr]^{3/2}
\biggl\{ x^3 \biggl[ 1 + \biggl(\frac{\tilde\xi^2}{3}\biggr)x^2 \biggr]^{-3/2}  \biggr\} \, .</math>
  </td>
</tr>
</table>
</div>
By making the substitution, <math>x \rightarrow \xi/\tilde\xi</math>, this expression becomes identical to the <math>M_r/M_\mathrm{tot}</math> [[SSCpt1/Virial/FormFactors#NormalizedProfiles|profile presented just before the "Mass1" subsection]], above.  In summary, then, we have the following two equally valid expressions for the <math>M_r</math> profile &#8212; one expressed as a function of <math>\xi</math> and the other expressed as a function of <math>x</math>:

<div align="center" id="2MassProfiles">
<table border="1" cellpadding="10" align="center">
<tr><td align="center">

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

<tr>
  <td align="right">
<math>\frac{M_r(\xi)}{M_\mathrm{tot}}</math> 
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\biggl[\frac{\xi^2}{3}\biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-1}\biggr]^{3/2} \, ;</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\frac{M_r(x)}{M_\mathrm{tot}}</math> 
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\biggl[ \frac{\tilde\xi^2}{3}\biggr]^{3/2}
\biggl\{ x^3 \biggl[ 1 + \biggl(\frac{\tilde\xi^2}{3}\biggr)x^2 \biggr]^{-3/2}  \biggr\} \, .</math>
  </td>
</tr>
</table>

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

===Mean-to-Central Density (n = 5)===

From the above line of reasoning we appreciate that, for any spherically symmetric configuration, the ratio of the configuration's mean density to its central density can be obtained by setting the upper limit of our just-completed "Mass2" integration to <math>x=1</math>.  That is to say, quite generally,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\frac{M_\mathrm{limit}}{M_\mathrm{tot}}  </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\biggl( \frac{\rho_c}{\bar\rho} \biggr)_\mathrm{eq} \biggl( \frac{M_\mathrm{limit}}{M_\mathrm{tot}} \biggr)
\int_0^{1}  3x^2 \biggl[ \frac{\rho(x)}{\rho_0} \biggr]  dx </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ \biggl( \frac{\bar\rho}{\rho_c} \biggr)_\mathrm{eq} </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math> 
\int_0^{1}  3x^2 \biggl[ \frac{\rho(x)}{\rho_0} \biggr]  dx </math>
  </td>
</tr>
</table>
</div>
But the integral expression on the righthand side of this relation is also the definition of the structural form factor, <math>~\mathfrak{f}_M</math>, given at the [[SSCpt1/Virial/FormFactors#Structural_Form_Factors|top of this page]].  Hence, we can say, quite generally, that,
<div align="center">
<math>\mathfrak{f}_M = \frac{\bar\rho}{\rho_c} \, .</math>
</div>
And, given that we have just completed this integral for the case of truncated <math>n=5</math> polytropic structures, we can state, specifically, that,
<div align="center">
<math>\mathfrak{f}_M\biggr|_{n=5} = \biggl[ 1 + \frac{\tilde\xi^2}{3} \biggr]^{-3/2} \, .</math>
</div>

===Gravitational Potential Energy (n = 5)===
As presented at the [[SSCpt1/Virial/FormFactors#Structural_Form_Factors|top of this page]], the structural form factor associated with determination of the gravitational potential energy is,

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

<tr>
  <td align="right">
<math>\mathfrak{f}_W</math>
  </td>
  <td align="center">
<math>\equiv</math>
  </td>
  <td align="left">
<math>3\cdot 5 \int_0^1 \biggl\{ \int_0^x  \biggl[ \frac{\rho(x)}{\rho_0}\biggr] x^2 dx \biggr\}  \biggl[ \frac{\rho(x)}{\rho_0}\biggr] x dx\, .</math>
  </td>
</tr>
</table>
</div>
[[File:OnlineIntegral02.png|225px|right|Mathematica Integral]]Given that an expression for the normalized density profile, <math>\rho(x)/\rho_0</math>, has already [[SSCpt1/Virial/FormFactors#rhoofx|been determined, above]], we can carry out the nested pair of integrals immediately.  Indeed, the integral contained inside of the curly braces has already been completed [[SSCpt1/Virial/FormFactors#Mass2|in the "Mass2" subsection, above]], in order to determine the radial mass profile.  Specifically, we have already determined that,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\biggl\{ \int_0^x  \biggl[ \frac{\rho(x)}{\rho_0}\biggr] x^2 dx \biggr\} </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\frac{1}{3} \biggl\{ \int_0^{x}  3x^2 \biggl[ 1 + \biggl(\frac{\tilde\xi^2}{3}\biggr)x^2 \biggr]^{-5/2}  dx\biggr\}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\frac{1}{3} \biggl\{ x^3 \biggl[ 1 + \biggl(\frac{\tilde\xi^2}{3}\biggr)x^2 \biggr]^{-3/2}  \biggr\} \, .</math>
  </td>
</tr>
</table>
</div>
Hence, with the help of [http://integrals.wolfram.com/index.jsp Mathematica's Online Integrator], we have,

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

<tr>
  <td align="right">
<math>\mathfrak{f}_W</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math> 
5 \int_0^1 \biggl\{ x^3 \biggl[ 1 + \biggl(\frac{\tilde\xi^2}{3}\biggr)x^2 \biggr]^{-3/2}  \biggr\}  
\biggl[ 1 + \biggl(\frac{\tilde\xi^2}{3}\biggr)x^2 \biggr]^{-5/2} x dx
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math> 
5 \int_0^1 \biggl[ 1 + \biggl(\frac{\tilde\xi^2}{3}\biggr)x^2 \biggr]^{-4} x^4 dx
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>  
\frac{5}{2^4\cdot 3} \biggl( \frac{\tilde\xi^2}{3}\biggr)^{-5/2} \biggl(1 + \frac{\tilde\xi^2}{3}\biggr)^{-3} 
\biggl\{
\biggl( \frac{\tilde\xi^2}{3}\biggr)^{1/2} \biggl[ 3\biggl( \frac{\tilde\xi^2}{3}\biggr)^2 - 8\biggl( \frac{\tilde\xi^2}{3}\biggr) - 3 \biggr]
+ 3\biggl( 1 + \frac{\tilde\xi^2}{3} \biggr)^3\tan^{-1}\biggl[ \biggl( \frac{\tilde\xi^2}{3}\biggr)^{1/2} \biggr]
\biggr\} \, .
</math>
  </td>
</tr>
</table>
</div>

<table border="1" width="90%" align="center" cellpadding="10">
<tr><td align="left">

<font color="maroon">'''ASIDE:'''</font> Now that we have expressions for, both, <math>\mathfrak{f}_M</math> and <math>\mathfrak{f}_W</math>, we can determine an analytic expression for the normalized gravitational potential energy for truncated, <math>n=5</math> polytropes.  As is shown in [[SSCpt1/Virial#Structural_Form_Factors|a companion discussion]],
<div align="center">
<table border="0" cellpadding="8" 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>
- \frac{3}{5} \chi^{-1} \biggl( \frac{M_\mathrm{limit}}{M_\mathrm{tot}} \biggr)^2 \cdot \frac{\mathfrak{f}_W}{\mathfrak{f}^2_M} \, ,
</math>
  </td>
</tr>
</table>
</div>
where,
<div align="center">
<table border="0" cellpadding="8" align="center">

<tr>
  <td align="right">
<math>\chi</math>
  </td>
  <td align="center">
<math>\equiv</math>
  </td>
 <td align="left">
<math>
\frac{R_\mathrm{limit}}{R_\mathrm{norm}} = \biggl(\frac{\pi}{2^3\cdot 3^7}\biggr)^{1/2} \tilde\xi \, .
</math>
  </td>
</tr>
</table>
</div>
In order to simplify typing, we will switch to the variable, 
<div align="center">
<math>\ell \equiv \frac{\tilde\xi}{\sqrt{3}} ~~~\Rightarrow~~~~ \frac{\tilde\xi^2}{3} = \ell^2 \, ,</math>
</div>
in which case a summary of derived expressions, from above, gives,
<div align="center">
<table border="0" cellpadding="8" align="center">

<tr>
  <td align="right">
<math>\chi</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
 <td align="left">
<math>\biggl(\frac{\pi}{2^3\cdot 3^6}\biggr)^{1/2} \ell \, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\mathfrak{f}_M</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
 <td align="left">
<math>( 1 + \ell^2 )^{-3/2}  \, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\mathfrak{f}_W</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
 <td align="left">
<math>  
\frac{5}{2^4\cdot 3} \cdot \ell^{-5} (1 + \ell^2)^{-3} 
\biggl\{ \ell [ 3\ell^4 - 8\ell^2 - 3 ] + 3( 1 + \ell^2 )^3\tan^{-1}(\ell ) \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
 <td align="left">
<math>  
\frac{5}{2^4} \cdot \ell^{-5}  
\biggl[ \ell \biggl( \ell^4 - \frac{8}{3}\ell^2 - 1 \biggr)(1 + \ell^2)^{-3} + \tan^{-1}(\ell ) \biggr] \, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\frac{M_\mathrm{limit}}{M_\mathrm{tot} } </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\ell^3 (1+\ell^2)^{-3/2} \, .
</math>
  </td>
</tr>
</table>
</div>

Hence,
<div align="center">
<table border="0" cellpadding="8" align="center">

<tr>
  <td align="right">
<math>\biggl( \frac{M_\mathrm{limit}}{M_\mathrm{tot}} \biggr)^{-2} \frac{W_\mathrm{grav}}{E_\mathrm{norm}}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
 <td align="left">
<math>- \frac{3}{5} \biggl(\frac{2^3\cdot 3^6}{\pi}\biggr)^{1/2} \frac{1}{\ell} \cdot (1 + \ell^2)^3
\mathfrak{f}_W 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
 <td align="left">
<math>- \biggl(\frac{3^8}{2^5 \pi}\biggr)^{1/2} \cdot \ell^{-6}  (1 + \ell^2)^3
\biggl[ \ell \biggl( \ell^4 - \frac{8}{3}\ell^2 - 1 \biggr)(1 + \ell^2)^{-3} + \tan^{-1}(\ell ) \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ \frac{W_\mathrm{grav}}{E_\mathrm{norm}}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
 <td align="left">
<math>- \biggl(\frac{3^8}{2^5\pi}\biggr)^{1/2} \cdot  
\biggl[\ell \biggl( \ell^4 - \frac{8}{3}\ell^2 - 1 \biggr)(1+\ell^2)^{-3} + \tan^{-1}(\ell ) \biggr] \, .
</math>
  </td>
</tr>
</table>
</div>
This exactly matches the normalized gravitational potential energy derived independently in the context of our [[SSC/Structure/BiPolytropes/Analytic51#Expression_for_Free_Energy|exploration of <math>(n_c, n_e) = (5,1)</math> bipolytropes]], referred to in that discussion as <math>W_\mathrm{core}^*</math>.

Hence, also, as defined in the [[SSCpt1/Virial#Gathering_it_All_Together|accompanying introductory discussion]], the constant, <math>\mathcal{A}</math>, that appears in our general free-energy equation is (for <math>n=5</math> polytropic configurations),
<div align="center">
<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[ \biggl( \frac{M_\mathrm{limit}}{M_\mathrm{tot}} \biggr) \frac{1}{\mathfrak{f}_M} \biggr]^2 \cdot \mathfrak{f}_W </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{\ell}{2^4} \biggl[ \ell \biggl( \ell^4 - \frac{8}{3}\ell^2 - 1 \biggr)(1 + \ell^2)^{-3} + \tan^{-1}(\ell ) \biggr] \, .
</math>
  </td>
</tr>
</table>
</div>

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

===Thermal Energy (n = 5)===
As presented at the [[#Structural_Form_Factors|top of this page]], the structural form factor associated with determination of the configuration's thermal energy is,

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

<tr>
  <td align="right">
<math>\mathfrak{f}_A</math>
  </td>
  <td align="center">
<math>\equiv</math>
  </td>
  <td align="left">
<math>\int_0^1 3\biggl[ \frac{P(x)}{P_0}\biggr]  x^2 dx \, ,</math>
  </td>
</tr>
</table>
</div>
[[File:OnlineIntegral03.png|225px|right|Mathematica Integral]]Given that an expression for the normalized pressure profile, <math>P/P_0</math>, has already [[SSCpt1/Virial/FormFactors#rhoofx|been provided, above]], we can carry out the integral immediately.  Specifically, we have,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\frac{P(\xi)}{P_0} </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\biggl( 1 + \frac{\xi^2}{3} \biggr)^{-3}</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~~ \frac{P(x)}{P_0} </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\biggl[ 1 + \biggl(\frac{\tilde\xi^2}{3}\biggr)x \biggr]^{-3} \, .</math>
  </td>
</tr>
</table>
</div>
Hence, with the aid of [http://integrals.wolfram.com/index.jsp Mathematica's Online Integrator], the relevant integral gives,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\mathfrak{f}_A</math>
  </td>
  <td align="center">
<math>\equiv</math>
  </td>
  <td align="left">
<math>3\int_0^1 \biggl[ 1 + \biggl(\frac{\tilde\xi^2}{3}\biggr)x \biggr]^{-3}  x^2 dx </math>
  </td>
</tr>

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

<table border="1" width="90%" align="center" cellpadding="10">
<tr><td align="left">

<font color="maroon">'''ASIDE:'''</font> Having this expression for <math>\mathfrak{f}_A</math> allows us to determine an analytic expression for the coefficient, <math>\mathcal{B}</math>, that appears in our general expression for the free energy, and that can be straightforwardly used to obtain an expression for the thermal energy content of <math>n=5 (\gamma=6/5)</math> polytropic configurations.  From our [[SSCpt1/Virial#Gathering_it_all_Together|accompanying introductory discussion]], we have,
<div align="center">
<table border="0" cellpadding="8" align="center">

<tr>
  <td align="right">
<math>\mathcal{B}</math>
  </td>
  <td align="center">
<math>\equiv</math>
  </td>
  <td align="left">
<math>
\biggl(\frac{3}{2^2 \pi} \biggr)^{1/5}
\biggl[ \biggl( \frac{M_\mathrm{limit}}{M_\mathrm{tot}}\biggr)  \frac{1}{\mathfrak{f}_M} \biggr]_\mathrm{eq}^{6/5}
\cdot \mathfrak{f}_A \, .
</math>
  </td>
</tr>
</table>
</div>

If, as above, we adopt the simplifying variable notation,
<div align="center">
<math>\ell \equiv \frac{\tilde\xi}{\sqrt{3}} ~~~\Rightarrow~~~~ \frac{\tilde\xi^2}{3} = \ell^2 \, ,</math>
</div>
the various factors in the definition of <math>\mathcal{B}</math> and <math>S_\mathrm{therm}</math> are (see above),
<div align="center">
<table border="0" cellpadding="8" align="center">

<tr>
  <td align="right">
<math>\chi</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
 <td align="left">
<math>\biggl(\frac{\pi}{2^3\cdot 3^6}\biggr)^{1/2} \ell \, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\biggl(\frac{M_\mathrm{limit}}{M_\mathrm{tot} }\biggr)\frac{1}{\mathfrak{f}_M} </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\ell^3 \, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\mathfrak{f}_A</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
 <td align="left">
<math> 
\frac{3}{2^3}  [ \ell^{-3} \tan^{-1}(\ell ) + \ell^{-2}(1+\ell^2)^{-1} - 2\ell^{-2}(1+\ell^2)^{-2} ] \, .
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
 <td align="left">
<math>  
\frac{3}{2^3} \ell^{-3}  [ \tan^{-1}(\ell ) + \ell (\ell^2-1) (1+\ell^2)^{-2} ] \, .
</math>
  </td>
</tr>
</table>
</div>

Hence,
<div align="center">
<table border="0" cellpadding="8" align="center">

<tr>
  <td align="right">
<math>\mathcal{B}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl(\frac{3^6}{2^{17} \pi} \biggr)^{1/5}~\ell^{3/5} [ \tan^{-1}(\ell ) + \ell (\ell^2-1) (1+\ell^2)^{-2} ]  \, ;
</math>
  </td>
</tr>
</table>
</div>
and (see [[VE#Adiabatic_Systems|here]] and [[SSCpt1/Virial#Structural_Form_Factors|here]]),
<div align="center">
<table border="0" cellpadding="8" align="center">

<tr>
  <td align="right">
<math>\frac{S_\mathrm{therm}}{E_\mathrm{norm}}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{3}{2}(\gamma - 1)\biggl[ \frac{\mathfrak{S}_\mathrm{therm}}{E_\mathrm{norm}}\biggr] 
= \frac{3}{2} \cdot \chi^{3(1-\gamma)} \mathcal{B}
= \frac{3}{2} \cdot \chi^{-3/5} \mathcal{B}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{3}{2} \cdot \biggl[ \biggl(\frac{\pi}{2^3\cdot 3^6}\biggr)^{1/2} \ell \biggr]^{-3/5} 
\biggl(\frac{3^6}{2^{17} \pi} \biggr)^{1/5} \ell^{3/5} [ \tan^{-1}(\ell ) + \ell (\ell^2-1) (1+\ell^2)^{-2} ] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[ \frac{3^{10}}{2^{10}} \biggl(\frac{2^9\cdot 3^{18}}{\pi^3}\biggr)  
\biggl(\frac{3^{12}}{2^{34} \pi^2} \biggr) \biggr]^{1/10}  [ \tan^{-1}(\ell ) + \ell (\ell^2-1) (1+\ell^2)^{-2} ] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl(\frac{3^{8}}{2^{7}\pi}\biggr)^{1/2} [ \tan^{-1}(\ell ) + \ell (\ell^2-1) (1+\ell^2)^{-2} ] \, .
</math>
  </td>
</tr>
</table>
</div>
This exactly matches the normalized thermal energy derived independently in the context of our [[SSC/Structure/BiPolytropes/Analytic51#Expression_for_Free_Energy|exploration of <math>(n_c, n_e) = (5,1)</math> bipolytropes]], referred to in that discussion as <math>S_\mathrm{core}^*</math>.  Its similarity to the expression for the gravitational potential energy &#8212; which is relevant to the virial theorem &#8212; is more apparent if it is rewritten in the following form:
<div align="center">
<table border="0" cellpadding="8" align="center">

<tr>
  <td align="right">
<math>\frac{S_\mathrm{therm}}{E_\mathrm{norm}}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\frac{1}{2}
\biggl(\frac{3^{8}}{2^{5}\pi}\biggr)^{1/2} [ \ell (\ell^4-1) (1+\ell^2)^{-3} + \tan^{-1}(\ell )] \, .
</math>
  </td>
</tr>
</table>
</div>

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

===Summary (n = 5)===
In summary, for <math>n=5</math> structures we have,

<div align="center">
<table border="1" align="center" cellpadding="10">
<tr><th align="center">
Structural Form Factors (n = 5)
</th></tr>
<tr><td align="center">

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

<tr>
  <td align="right">
<math>\mathfrak{f}_M</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>  
( 1 + \ell^2 )^{-3/2}  
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\mathfrak{f}_W</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>  
\frac{5}{2^4} \cdot \ell^{-5}  
\biggl[ \ell \biggl( \ell^4 - \frac{8}{3}\ell^2 - 1 \biggr)(1 + \ell^2)^{-3} + \tan^{-1}(\ell ) \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\mathfrak{f}_A</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{3}{2^3} \ell^{-3}  [ \tan^{-1}(\ell ) + \ell (\ell^2-1) (1+\ell^2)^{-2} ]  
</math>
  </td>
</tr>
</table>

</td></tr>
<tr><th align="center">
Free-Energy Coefficients (n = 5)
</th></tr>
<tr><td align="center">

<table border="0" cellpadding="8" align="center">

<tr>
  <td align="right">
<math>\mathcal{A}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
 <td align="left">
<math>
\frac{\ell}{2^4} \biggl[ \ell \biggl( \ell^4 - \frac{8}{3}\ell^2 - 1 \biggr)(1 + \ell^2)^{-3} + \tan^{-1}(\ell ) \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\mathcal{B}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl(\frac{3^6}{2^{17} \pi} \biggr)^{1/5}~\ell^{3/5} [ \tan^{-1}(\ell ) + \ell (\ell^2-1) (1+\ell^2)^{-2} ] 
</math>
  </td>
</tr>
</table>
<tr><th align="center">
Normalized Energies (n = 5)
</th></tr>
<tr><td align="center">
<table border="0" cellpadding="8" align="center">

<tr>
  <td align="right">
<math>\frac{S_\mathrm{therm}}{E_\mathrm{norm}}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{1}{2} \biggl(\frac{3^{8}}{2^{5}\pi}\biggr)^{1/2} [ \ell (\ell^4-1) (1+\ell^2)^{-3} + \tan^{-1}(\ell )]
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\frac{W_\mathrm{grav}}{E_\mathrm{norm}}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
 <td align="left">
<math>
- \biggl(\frac{3^8}{2^5\pi}\biggr)^{1/2} \cdot  
\biggl[\ell \biggl( \ell^4 - \frac{8}{3}\ell^2 - 1 \biggr)(1+\ell^2)^{-3} + \tan^{-1}(\ell ) \biggr]
</math>
  </td>
</tr>
</table>

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

===Reality Check (n = 5)===

<div align="center">
<table border="0" cellpadding="8" align="center">

<tr>
  <td align="right">
<math>2\biggl(\frac{S_\mathrm{therm}}{E_\mathrm{norm}}\biggr)+ \frac{W_\mathrm{grav}}{E_\mathrm{norm}}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
 <td align="left">
<math>\biggl(\frac{3^{8}}{2^{5}\pi}\biggr)^{1/2}\biggl\{ [ \ell (\ell^4-1) (1+\ell^2)^{-3} + \tan^{-1}(\ell )]
-  
\biggl[\ell \biggl( \ell^4 - \frac{8}{3}\ell^2 - 1 \biggr)(1+\ell^2)^{-3} + \tan^{-1}(\ell ) \biggr] \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
 <td align="left">
<math>\biggl(\frac{3^{8}}{2^{5}\pi}\biggr)^{1/2}
\biggl[\frac{8}{3}\ell^3 (1+\ell^2)^{-3}\biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
 <td align="left">
<math>\biggl(\frac{2 \cdot 3^{6}}{\pi}\biggr)^{1/2}
\biggl[\frac{\ell}{ (1+\ell^2)} \biggr]^3 \, .
</math>
  </td>
</tr>
</table>
</div>
For embedded polytropes, this should be compared against the expectation (prediction) [[#Generic_Reality_Check|provided by Stahler's equilibrium models, as detailed above]].  Given that, for <math>n=5</math> polytropes &#8212; see the [[#Mass1|"Mass1" discussion above]] and our accompanying [[SSC/Structure/PolytropesEmbedded#Tabular_Summary_.28n.3D5.29|tabular summary of relevant properties]],
<div align="center">
<table border="0" cellpadding="3">
<tr>
  <td align="right">
<math>
\frac{M_\mathrm{limit}}{M_\mathrm{tot}} = \biggl[ \ell^2(1+\ell^2)^{-1} \biggr]^{3/2}
</math>
  </td>

  <td align="center">
&nbsp; ; &nbsp; &nbsp; &nbsp; &nbsp;
  </td>

  <td align="right">
<math>
\theta_5 = ( 1 + \ell^2 )^{-1/2}
</math>
  </td>

  <td align="center">
&nbsp; &nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp; &nbsp;
  </td>

  <td align="right">
<math>
-\frac{d\theta_5}{d\xi} \biggr|_{\xi_e} = 3^{1/2} \ell ( 1 + \ell^2 )^{-3/2} \, ,
</math>
  </td>
</tr>
</table>

</div>

the expectation is that,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\frac{4\pi P_e R_\mathrm{eq}^3}{E_\mathrm{norm}} </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[ \frac{(n+1)^3}{4\pi}\biggr]^{1/(n-3)} \biggl[ \biggl( \frac{M_\mathrm{limit}}{M_\mathrm{tot}} \biggr)\frac{1}{(-\theta'_n)_{\tilde\xi}}\biggr]^{(n-5)/(n-3)} 
(\theta_n)_{\tilde\xi}^{(n+1)} \tilde\xi^{(n+1)/(n-3)} 
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl( \frac{2\cdot 3^6}{\pi}\biggr)^{1/2}  \biggl[ \frac{\ell}{( 1 + \ell^2 )} \biggr]^{3} \, .
</math>
  </td>
</tr>
</table>
</div>
This precisely matches our sum of the thermal and gravitational potential energies, as just determined using our expressions for the structural form factors.  This gives us confidence that our form-factor expressions are correct, at least in the case of embedded <math>n=5</math> polytropic structures.

=See Also=

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