Editing SSC/BipolytropeGeneralization/Pt4 (section)

==Playing With One Example==
By setting,
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<math>~\gamma_c = 6/5; ~~~ \gamma_e = 2</math>
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<math>~\mathcal{A}</math>
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<math>~\mathcal{B}</math>
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<math>~\mathcal{C}</math>
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2.5
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1.0
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2.0
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a plot of <math>~\mathfrak{G}^*</math> versus <math>~\chi</math> exhibits the following, two extrema:
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extremum
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<math>~\chi_\mathrm{eq}</math>
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<math>~\mathfrak{G}^*</math>
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&nbsp;
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<math>~\chi_\mathrm{eq}^{3(\gamma_c - \gamma_e)}</math>
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<math>~\mathcal{C}^'</math>
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<math>~\chi_\mathrm{eq}^{4-3\gamma_c}</math>
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<math>~\frac{\mathcal{A}}{\mathcal{B} +\mathcal{C}^'}</math>
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MIN
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<math>1.1824</math>
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<math>-0.611367</math>
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&nbsp;<math>~~~~\Rightarrow~~~~</math>&nbsp;
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<math>0.66891</math>
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<math>1.3378</math>
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<math>1.0693</math>
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<math>1.0694</math>
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MAX
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<math>9.6722</math>
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<math>+0.508104</math>
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<math>~\Rightarrow</math>
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<math>0.004313</math>
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<math>0.008625</math>
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<math>2.4786</math>
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<math>2.4786</math>
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The last two columns of this table confirm the internal consistency of the relationships presented in '''<font color="#997700">Step 3</font>''', above.  But what does this mean in terms of the values of <math>~\nu</math>, <math>~q</math>, and the related ratio of densities at the interface, <math>~\rho_e/\rho_c</math>?

Let's assume that what we're trying to display and examine is the behavior of the free-energy ''surface'' for a fixed value of the ratio of densities at the interface.  Once the value of <math>~\rho_e/\rho_c</math> has been specified, it is clear that the value of <math>~q</math> (and, hence, also <math>~\nu</math>) is set because <math>~\mathcal{A}</math> has also been specified.  But our specification of <math>~\mathcal{B}</math> along with <math>~\rho_e/\rho_c</math> also forces a particular value of <math>~q</math>.  It is unlikely that these two values of <math>~q</math> will be the same.  In reality, once <math>~\mathcal{A}</math> and <math>~\mathcal{B}</math> have both been specified, they force a particular <math>~(\nu, q)</math> pair.  How do we (easily) figure out what this pair is?

Let's begin by rewriting the expressions for <math>~\mathcal{A}</math> and <math>~\mathcal{B}</math> in terms of just <math>~q</math> and the ratio, <math>~\rho_e/\rho_c</math>, keeping in mind that, for the case of a uniform-density core (of density, <math>~\rho_c</math>) and a uniform-density envelope (of density, <math>~\rho_e</math>), 
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<math>~\frac{\rho_e}{\rho_c}</math>
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<math>~=</math>
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<math>~\frac{q^3(1-\nu)}{\nu(1-q^3)} \, ,</math>
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hence,
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<math>~\nu</math>
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<math>~=</math>
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<math>~\biggl[ 1 + \biggl( \frac{\rho_e}{\rho_c} \biggr) \frac{(1-q^3)}{q^3} \biggr]^{-1}</math> 
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  <td align="center">&nbsp;&nbsp; and &nbsp; &nbsp;</td>
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<math>~\frac{q^3}{\nu}</math>
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<math>~=</math>
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<math>~\biggl[ \biggl( \frac{\rho_e}{\rho_c} \biggr) (1-q^3) + q^3 \biggr] \, .</math> 
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Putting the expression for <math>~\mathcal{A}</math> in the desired form is simple because <math>~\nu</math> only appears as a leading factor.  Specifically, we have,
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<math>~\mathcal{A}</math>
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<math>~=</math>
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<math>~\frac{\pi q^5}{5} \biggl[ \biggl( \frac{\rho_e}{\rho_c} \biggr) (1-q^3) + q^3 \biggr]^{-2}
\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{\pi }{5} \biggl[ \biggl( \frac{\rho_e}{\rho_c} \biggr) (1-q^3) + q^3 \biggr]^{-2}
\biggl\{  q^5 + \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) (q^3 - q^5 ) + \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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The expression for <math>~\mathcal{B}</math> can be written in the form,
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<math>~\mathcal{B}</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\{  
1+\frac{\pi}{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} 
\biggr\}</math>
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&nbsp;
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<math>~=</math>
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<math>~\nu \biggl[ \biggl( \frac{4\pi}{3} \biggr)\frac{q^3}{\nu} \biggr]^{1-\gamma_c}   
+\frac{\pi q^5}{5} \biggl( \frac{\nu^2}{q^6} \biggr) \chi_\mathrm{eq}^{3\gamma_c - 4} 
</math>
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&nbsp;
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<math>~=</math>
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<math>~q^3 \biggl( \frac{4\pi}{3} \biggr)^{1-\gamma_c} \biggl[ \frac{q^3}{\nu} \biggr]^{-\gamma_c}   
+\frac{\pi q^5}{5} \biggl( \frac{q^3}{\nu} \biggr)^{-2} \chi_\mathrm{eq}^{3\gamma_c - 4} 
</math>
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&nbsp;
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<math>~=</math>
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<math>~q^3 \biggl( \frac{4\pi}{3} \biggr)^{1-\gamma_c} \biggl[ \biggl( \frac{\rho_e}{\rho_c} \biggr) (1-q^3) + q^3  \biggr]^{-\gamma_c}   
+\frac{\pi q^5}{5} \biggl[ \biggl( \frac{\rho_e}{\rho_c} \biggr) (1-q^3) + q^3 \biggr]^{-2} \chi_\mathrm{eq}^{3\gamma_c - 4} \, .
</math>
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Generally speaking, the equilibrium radius, <math>~\chi_\mathrm{eq}</math>, which appears in the expression for <math>~\mathcal{B}</math>, is not known ahead of time.  Indeed, as is illustrated in our simple example immediately above, the normal path is to pick values for the coefficients, <math>~\mathcal{A}</math>, <math>~\mathcal{B}</math>, and <math>~\mathcal{C}</math>, and determine the equilibrium radius by looking for extrema in the free-energy function.  And because <math>~\chi_\mathrm{eq}</math> is not known ahead of time, it isn't clear how to (easily) figure out what pair of physical parameter values, <math>~(\nu, q)</math>, give self-consistent values for the coefficient pair, <math>~(\mathcal{A}, \mathcal{B})</math>.

Because we are using a uniform density core and uniform density envelope as our base model, however, we ''do'' know the analytic solution for <math>~\chi_\mathrm{eq}</math>.  As stated above, it is,
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<math>~\chi_\mathrm{eq}^{4-3\gamma_c}</math>
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<math>~=</math>
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<math>~\frac{1}{\Lambda_\mathrm{eq}} \cdot \frac{\pi}{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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&nbsp;
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<math>~=</math>
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<math>~\frac{\pi q^2}{2} \biggl( \frac{3}{4\pi} \biggr)^{1-\gamma_c}\biggl( \frac{\nu}{q^3} \biggr) 
\biggl(\frac{\rho_e}{\rho_c}\biggr)  \biggl[ 2 \biggl(1 - \frac{\rho_e}{\rho_c} \biggr) \biggl( 1-q \biggr) + 
\frac{\rho_e}{\rho_c} \biggl(\frac{1}{q^2} - 1\biggr) \biggr] 
\cdot \biggl[ \frac{q^3}{\nu} \biggr]^{\gamma_c-1} </math>
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&nbsp;
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<math>~=</math>
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<math>~\frac{\pi}{2} \biggl( \frac{3}{4\pi} \biggr)^{1-\gamma_c}
\biggl(\frac{\rho_e}{\rho_c}\biggr)  \biggl[ 2 \biggl(1 - \frac{\rho_e}{\rho_c} \biggr) ( q^2-q^3 ) + 
\frac{\rho_e}{\rho_c} (1 - q^2) \biggr] 
\cdot \biggl[ \biggl( \frac{\rho_e}{\rho_c} \biggr) (1-q^3) + q^3\biggr]^{\gamma_c-2} </math>
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Combining this expression with the one for <math>~\mathcal{B}</math> gives us the desired result &#8212; although, strictly speaking, it is cheating!  We can now methodically choose <math>~(\nu, q)</math> pairs and map them into the corresponding values of <math>~\mathcal{A}</math> and <math>~\mathcal{B}</math>.  And, via an analogous "cheat," the choice of <math>~(\nu, q)</math> also gives us the self-consistent value of <math>~\mathcal{C}</math>.  In this manner, we should be able to map out the free-energy surface for any desired set of physical parameters.