Editing SSCpt1/Virial/FormFactors/Pt2 (section)

===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>


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<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>

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<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>
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