Editing Appendix/Ramblings/NonlinarOscillation (section)

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From our [[#Run_of_Mass|discussion, below]], for any value of the truncation radius, <math>~\tilde\xi</math>, the fractional mass <math>~(0 \le m_\xi \le 1)</math> that lies interior to <math>~\xi</math> is given by the expression,

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<math>~m_\xi \equiv \frac{M(\xi)}{M_\mathrm{tot}}</math>
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<math>~=</math>
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<math>~
\biggl(\frac{\xi}{\tilde\xi}\biggr)^3 \biggl(1 + \frac{\xi^2}{3}\biggr)^{-3/2} 
\biggl(1 + \frac{\tilde\xi^2}{3}\biggr)^{3/2} </math>
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&nbsp;
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<math>~=</math>
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<math>~\biggl(\frac{\tilde{C}}{3}\biggr)^{3 / 2}
\xi^3 \biggl(3 + \xi^2\biggr)^{-3/2} \, ,
</math>
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where,
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<math>~\tilde{C} \equiv \frac{3^2}{\tilde\xi^2}\biggl( 1 + \frac{\tilde\xi^2}{3} \biggr)</math>
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&nbsp; &nbsp; &nbsp; <math>~\Rightarrow</math> &nbsp; &nbsp; &nbsp; 
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<math>~
{\tilde\xi}^2 = \frac{9}{\tilde{C} - 3} \, .
</math>
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And, when normalized to <math>~R_\mathrm{SWS}</math>, the corresponding radius is,

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<math>~r_\mathrm{SWS}(\xi)</math>
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<math>~=</math>
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<math>~\biggl(\frac{\xi}{\tilde\xi} \biggr) \frac{R_\mathrm{eq}}{R_\mathrm{SWS}} </math>
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&nbsp;
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<math>~=</math>
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<math>~\biggl(\frac{\xi}{\tilde\xi} \biggr) \biggl\{ \frac{3\cdot 5}{2^2 \pi} \biggl[ \frac{ \tilde\xi^2/3}{(1+ \tilde\xi^2/3)^{2}} \biggr] \biggr\}^{1/2}</math>
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&nbsp;
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<math>~=</math>
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<math>~\xi\biggl[ \frac{3^2 \cdot 5}{2^2 \pi} \biggr]^{1 / 2} (3+{\tilde\xi}^2)^{-1}  </math>
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&nbsp;
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<math>~=</math>
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<math>~\xi\biggl[ \frac{3^2 \cdot 5}{2^2 \pi} \biggr]^{1 / 2} \biggl[ \frac{\tilde{C} - 3}{3\tilde{C}} \biggr]  \, .</math>
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Now, this works fine in the sense that, for any choice of <math>~\tilde\xi</math>, and therefore <math>~\tilde{C}</math>, this pair of parametric relations can be used to generate a plot of <math>~r_\mathrm{SWS}</math> versus <math>~m_\xi</math> that correctly displays how the mass enclosed within a given radius varies with radial location throughout the spherical configuration.  But, in order to ''compare'' one of these configurations to another, we really need to identify how this function varies across a Lagrangian mass grid that is the same for both configurations.  The easiest way to accomplish this is to derive an expression for <math>~r_\mathrm{SWS}</math> that is directly a function of <math>~m_\xi</math>.  Fortunately, this can be done analytically. First, we invert the mass expression to find,

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<math>~m_\xi^{2/3}</math>
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<math>~=</math>
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<math>~\biggl(\frac{\tilde{C}}{3}\biggr)
\xi^2 (3 + \xi^2)^{-1} 
</math>
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<math>~\Rightarrow ~~~ 3 + \xi^2</math>
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<math>~=</math>
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<math>~
m_\xi^{-2/3}\biggl(\frac{\tilde{C}}{3}\biggr)\xi^2 
</math>
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<math>~\Rightarrow ~~~ \xi^2\biggl[ 1 - m_\xi^{-2/3}\biggl(\frac{\tilde{C}}{3}\biggr) \biggr]</math>
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<math>~=</math>
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<math>~
-3
</math>
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<math>~\Rightarrow ~~~ \xi^2</math>
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<math>~=</math>
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<math>~
3^2 [ \tilde{C}~m_\xi^{-2/3} -3 ]^{-1} \, .
</math>
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<span id="ExactProfile">Inserting this into the radial equation, then, gives,</span>

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<math>~r_\mathrm{SWS}(m_\xi)</math>
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<math>~=</math>
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<math>~\biggl[ \frac{3^2 \cdot 5}{2^2 \pi} \biggr]^{1 / 2} \biggl[ \frac{\tilde{C} - 3}{\tilde{C}} \biggr] \biggl[ \tilde{C}~m_\xi^{-2/3} -3 \biggr]^{-1 / 2} \, .</math>
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