Editing SSC/BipolytropeGeneralization/Pt4 (section)

===Summary===
====Understanding Free-Energy Behavior====
'''<font color="#997700">Step 1:</font>''' Pick values for the separate coefficients, <math>\mathcal{A}, \mathcal{B},</math> and <math>\mathcal{C},</math> of the three terms in the normalized free-energy expression,
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<math>~\mathfrak{G}^*</math>
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<math>~=</math>
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<math>~ -~ 3\mathcal{A} \chi^{-1} - \frac{\mathcal{B}}{(1-\gamma_c)} ~\chi^{3-3\gamma_c} - \frac{\mathcal{C}}{(1-\gamma_e)} ~\chi^{3-3\gamma_e} </math>
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then plot the function, <math>\mathfrak{G}^*(\chi)</math>, and identify the value(s) of <math>~\chi_\mathrm{eq}</math> at which the function has an extremum (or multiple extrema).

'''<font color="#997700">Step 2:</font>''' Note that,
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<math>~\mathcal{A}</math>
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<math>~\equiv</math>
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<math>~\frac{\nu^2}{5q} \biggl\{  
1 + \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) \biggl(\frac{1}{q^2} - 1 \biggr)
+ \biggl( \frac{\rho_e}{\rho_c} \biggr)^2 \biggl[ \biggl(\frac{1}{q^5} - 1 \biggr) - \frac{5}{2}\biggl(\frac{1}{q^2} - 1 \biggr) \biggr] 
\biggr\}</math>
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&nbsp;
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<math>~=</math>
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<math>~\frac{1}{5} \biggl( \frac{\nu}{q^3} \biggr)^2 \biggl[  
q^5 + \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) (1 - q^2 )q^3
+ \biggl( \frac{\rho_e}{\rho_c} \biggr)^2 \biggl( 1 - \frac{5}{2}q^3 + \frac{3}{2}q^5\biggr) 
\biggr]</math>
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<math>~\mathcal{B}</math>
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<math>~\equiv</math>
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<math>~\nu \biggl[ \biggl( \frac{3}{4\pi} \biggr)\frac{\nu}{q^3} \biggr]^{\gamma_c-1}  \biggl[  
1+\Lambda_\mathrm{eq}
\biggr]</math>
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<math>~\mathcal{C}</math>
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<math>~\equiv</math>
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<math>~(1-\nu)\biggl( \frac{K_e}{K_c} \biggr)^*
\biggl[\biggl( \frac{3}{4\pi} \biggr) \frac{(1-\nu)}{(1-q^3)} \biggr]^{\gamma_e-1}  \biggl\{  
1 + \frac{\Lambda_\mathrm{eq}}{(1-q^3)}\biggl[ \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c}\biggr) (-2 + 3q - q^3) +
\frac{3}{2q^2} \biggl( \frac{\rho_e}{\rho_c}\biggr)^2 (-1 +5q^2 - 5q^3 + q^5) \biggr] 
\biggr\}</math>
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&nbsp;
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<math>~\equiv</math>
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<math>~ \nu \biggl[ \biggl( \frac{3}{4\pi} \biggr)\frac{\nu}{q^3} \biggr]^{\gamma_c-1} 
\biggl\{  
\frac{(1-q^3)}{q^3} + \frac{\Lambda_\mathrm{eq}}{q^3} \biggl[ \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c}\biggr) (-2 + 3q - q^3) +
\frac{3}{2q^2} \biggl( \frac{\rho_e}{\rho_c}\biggr)^2 (-1 +5q^2 - 5q^3 + q^5) \biggr] 
\biggr\} \chi_\mathrm{eq}^{3(\gamma_e - \gamma_c)}</math>
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where (see, for example, [[SSC/Structure/BiPolytropes/Analytic00#Expression_for_Free_Energy|in the context of its original definition]]),
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<math>~\Lambda_\mathrm{eq} \equiv
 \frac{3}{2^2\pi \cdot 5} \biggl( \frac{GM_\mathrm{tot}^2}{R_\mathrm{eq}^4 P_i} \biggr) \frac{\nu^2}{q^4} 
</math> 
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<math>~=~</math>
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<math>
 \frac{1}{5} \biggl( \frac{\nu}{q} \biggr) \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{1-\gamma_c} 
\chi_\mathrm{eq}^{3\gamma_c - 4} 
</math>
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and, where,
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<math>\biggl( \frac{K_e}{K_c} \biggr)^* \equiv \frac{K_e}{K_c} \biggl[ \frac{K_c^3}{G^3 M_\mathrm{tot}^2} \biggr]^{(\gamma_c - \gamma_e)/(3\gamma_c -4)}  
</math>
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<math>~=</math>
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<math>
\biggl[\biggl( \frac{3}{4\pi} \biggr) \frac{1-\nu}{1-q^3} \biggr]^{-\gamma_e}
\biggl[ \biggl( \frac{3}{4\pi} \biggr)\frac{\nu}{q^3} \biggr]^{\gamma_c} 
\chi_\mathrm{eq}^{3(\gamma_e - \gamma_c)} \, .
</math>
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Also, keep in mind that, if the envelope and core both have uniform (but different) densities, then <math>~\rho_{ic} = \rho_c</math>, <math>~\rho_{ie} = \rho_e</math>, and
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<math>
\frac{\rho_c}{\bar\rho} = \frac{\nu}{q^3} \, ; ~~~~~ \frac{\rho_e}{\bar\rho} = \frac{1-\nu}{1-q^3} \, ; ~~~~~ 
\frac{\rho_e}{\rho_c} = \frac{q^3(1-\nu)}{\nu (1-q^3)} ~~\Rightarrow ~~~ \frac{q^3}{\nu} = \biggl( \frac{\rho_e}{\rho_c} \biggr) (1-q^3) + q^3 \, .
</math>
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'''<font color="#997700">Step 3:</font>'''  An analytic evaluation tells us that the following ''should'' happen.  Using the numerically derived value for <math>~\chi_\mathrm{eq}</math>, define,
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<math>~\mathcal{C}^' \equiv \mathcal{C} \chi_\mathrm{eq}^{3(\gamma_c - \gamma_e)} \, .</math>
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We should then discover that,
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<math>\frac{\mathcal{A}}{\mathcal{B} + \mathcal{C}^'} = \chi_\mathrm{eq}^{4-3\gamma_c} = 
\frac{1}{\Lambda_\mathrm{eq}} \cdot \frac{1}{5} \biggl( \frac{\nu}{q} \biggr) \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{1-\gamma_c} 
\, .</math>
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====Check It====
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<math>~\mathcal{B} + \mathcal{C}^'</math>
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<math>~=</math>
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<math>~
\nu \biggl[ \biggl( \frac{3}{4\pi} \biggr)\frac{\nu}{q^3} \biggr]^{\gamma_c-1} \biggl\{ \biggl[  1+\Lambda_\mathrm{eq} \biggr]
+ \frac{(1-q^3)}{q^3} + \frac{\Lambda_\mathrm{eq}}{q^3} \biggl[ \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c}\biggr) (-2 + 3q - q^3) +
\frac{3}{2q^2} \biggl( \frac{\rho_e}{\rho_c}\biggr)^2 (-1 +5q^2 - 5q^3 + q^5) \biggr]  \biggr\} 
</math>
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<math>\Rightarrow~~~~\mathcal{A} \biggl[ \Lambda_\mathrm{eq} \cdot 5\biggl( \frac{q}{\nu^2} \biggr)  \biggr]</math>
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<math>~=</math>
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<math>~
1+\Lambda_\mathrm{eq} 
+ \frac{(1-q^3)}{q^3} + \frac{\Lambda_\mathrm{eq}}{q^3} \biggl[ \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c}\biggr) (-2 + 3q - q^3) +
\frac{3}{2q^2} \biggl( \frac{\rho_e}{\rho_c}\biggr)^2 (-1 +5q^2 - 5q^3 + q^5) \biggr]  
</math>
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<math>
\Rightarrow~~~~\Lambda_\mathrm{eq}  \biggl\{  1 + \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) \biggl(\frac{1}{q^2} - 1 \biggr)
+ \biggl( \frac{\rho_e}{\rho_c} \biggr)^2 \biggl[ \biggl(\frac{1}{q^5} - 1 \biggr) - \frac{5}{2}\biggl(\frac{1}{q^2} - 1 \biggr) \biggr]   \biggr\}
</math>
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<math>~=</math>
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<math>~
1+\Lambda_\mathrm{eq} 
+ \frac{(1-q^3)}{q^3} + \frac{\Lambda_\mathrm{eq}}{q^3} \biggl[ \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c}\biggr) (-2 + 3q - q^3) +
\frac{3}{2q^2} \biggl( \frac{\rho_e}{\rho_c}\biggr)^2 (-1 +5q^2 - 5q^3 + q^5) \biggr]  
</math>
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</table>
</div>

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<math>
\Rightarrow~~~~\Lambda_\mathrm{eq}  \biggl\{ \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) \biggl(\frac{1}{q^2} - 1 \biggr)
+ \biggl( \frac{\rho_e}{\rho_c} \biggr)^2 \biggl[ \biggl(\frac{1}{q^5} - 1 \biggr) - \frac{5}{2}\biggl(\frac{1}{q^2} - 1 \biggr) \biggr]   \biggr\}
</math>
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<math>~=</math>
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<math>~
\frac{1}{q^3} + \frac{\Lambda_\mathrm{eq}}{q^3} \biggl[ \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c}\biggr) (-2 + 3q - q^3) +
\frac{3}{2q^2} \biggl( \frac{\rho_e}{\rho_c}\biggr)^2 (-1 +5q^2 - 5q^3 + q^5) \biggr]  
</math>
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</table>
</div>

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<math>\Rightarrow~~~~
\frac{1}{\Lambda_\mathrm{eq}} 
</math>
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<math>~=</math>
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<math>
q^3 \biggl\{ \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) \biggl(\frac{1}{q^2} - 1 \biggr)
+ \biggl( \frac{\rho_e}{\rho_c} \biggr)^2 \biggl[ \biggl(\frac{1}{q^5} - 1 \biggr) - \frac{5}{2}\biggl(\frac{1}{q^2} - 1 \biggr) \biggr]   \biggr\}
- \biggl[ \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c}\biggr) (-2 + 3q - q^3) +
\frac{3}{2q^2} \biggl( \frac{\rho_e}{\rho_c}\biggr)^2 (-1 +5q^2 - 5q^3 + q^5) \biggr]  
</math>
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<math>\Rightarrow~~~~
\frac{2}{\Lambda_\mathrm{eq}} 
</math>
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<math>~=</math>
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<math>
5\biggl( \frac{\rho_e}{\rho_c} \biggr) ( q - q^3 )
+ \frac{2}{q^2}\biggl( \frac{\rho_e}{\rho_c} \biggr)^2 \biggl[ (1 - q^5 ) - \frac{5}{2}( q^3 - q^5 ) \biggr]  
- 5 \biggl( \frac{\rho_e}{\rho_c}\biggr) (-2 + 3q - q^3) 
- \frac{3}{q^2} \biggl( \frac{\rho_e}{\rho_c}\biggr)^2 (-1 +5q^2 - 5q^3 + q^5) 
</math>
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&nbsp;
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<math>~=</math>
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<math>
5\biggl( \frac{\rho_e}{\rho_c} \biggr) \biggl[ ( q - q^3 ) - (-2 + 3q - q^3) \biggr]
+ \frac{1}{q^2}\biggl( \frac{\rho_e}{\rho_c} \biggr)^2 
\biggl[ 2(1 - q^5 ) - 5( q^3 - q^5 ) - 3(-1 +5q^2 - 5q^3 + q^5) \biggr]
</math>
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&nbsp;
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<math>~=</math>
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<math>
10\biggl( \frac{\rho_e}{\rho_c} \biggr) \biggl[ 1-q \biggr]
+ \frac{5}{q^2}\biggl( \frac{\rho_e}{\rho_c} \biggr)^2 
\biggl[ 1  - 3q^2 + 2q^3  \biggr]
</math>
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<math>\Rightarrow~~~~
\frac{1}{\Lambda_\mathrm{eq}} \biggl[ \frac{2q^2}{5} \biggl( \frac{\rho_e}{\rho_c} \biggr)^{-1} \biggr]
</math>
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<math>~=</math>
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<math>
2q^2 (1-q )+ \biggl( \frac{\rho_e}{\rho_c} \biggr) ( 1  - 3q^2 + 2q^3 )
</math>
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Fortunately, this precisely matches our [[SSC/Structure/BiPolytropes/Analytic00#LambdaDeff|earlier derivation]], which states that,
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<math>~\frac{1}{\Lambda}</math>
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&nbsp; <math>~=</math>&nbsp; 
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<math>\frac{5}{2}(g^2-1) =
\frac{5}{2}\biggl(\frac{\rho_e}{\rho_0}\biggr)  \biggl[ 2 \biggl(1 - \frac{\rho_e}{\rho_0} \biggr) \biggl( 1-q \biggr) + 
\frac{\rho_e}{\rho_0} \biggl(\frac{1}{q^2} - 1\biggr) \biggr] \, .</math>
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</table>
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