Editing SSCpt1/Virial/FormFactors/Pt3 (section)

===Mass1 (n = 1)===
While we already know the expression for the <math>M_r</math> profile, having copied it from our [[SSC/Structure/Polytropes#.3D_1_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>
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<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>
</div>
Plugging in the profiles for <math>r^\dagger</math> and <math>\rho^\dagger</math> gives, with the help of [http://integrals.wolfram.com/index.jsp Mathematica's Online Integrator], 
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  <td align="right">
<math>\frac{M_r(\xi)}{M_\mathrm{tot} } </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
3 \int_0^{\xi}  \frac{\xi^2}{2\pi} \biggl( \frac{2^3\pi}{3^2} \biggr)^{1/2} \frac{\sin\xi}{\xi} \cdot \frac{d\xi }{(2\pi)^{1/2}}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
3( 2\pi)^{-3/2}\biggl( \frac{2^3\pi}{3^2} \biggr)^{1/2}   \int_0^{\xi}  \xi \sin\xi d\xi 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
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<math>
\frac{1}{\pi} (\sin\xi - \xi\cos\xi) \, .
</math>
  </td>
</tr>
</table>
</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=1</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,
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<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>
\frac{1}{\pi} (\sin\tilde\xi - \tilde\xi \cos\tilde\xi) \, .
</math>
  </td>
</tr>
</table>
</div>