Editing SSC/Structure/BiPolytropes/Analytic00 (section)

=Plotting Free-Energy Surfaces=
Thus far, we have been able to determine what the equilibrium radius of the bipolytrope is for any specified set of parameters, <math>~(M_\mathrm{tot}, \nu, q)</math>.  Now let's examine how the free energy varies across the two-dimensional plane defined by the parameters, <math>~(q, \nu)</math>.  We begin by copying an informative segment from the [[#Expression_for_Free_Energy|above subsection where we developed an expression for the free energy]].

The free energy is, 
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<math>~\mathfrak{G}</math>
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<math>~=~</math>
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<math>~U_\mathrm{tot} + W</math>
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&nbsp;
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<math>~=~</math>
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<math>\biggl[ \frac{2}{3(\gamma_c -1)} \biggr] S_\mathrm{core} + \biggl[ \frac{2}{3(\gamma_e -1)} \biggr] S_\mathrm{env} + W</math>
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&nbsp;
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<math>~=~</math>
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<math>B_\mathrm{core}R^3 P_{ic} + B_\mathrm{env} R^3 P_{ie}  - A_\mathrm{grav} R^{-1}  \, ,
</math>
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</tr>

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&nbsp;
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<math>~=~</math>
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<math>\frac{GM_\mathrm{tot}^2}{R_\mathrm{eq}} \biggl[ (B_\mathrm{core}+B_\mathrm{env} ) \frac{R^4 P_i}{GM_\mathrm{tot}^2}  - A^*_\mathrm{grav} \biggr] \, ,
</math>
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where, for a given choice of the three parameters <math>~(M_\mathrm{tot}, \nu, q)</math>, the constant coefficients in this expression are,
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<math>~A^*_\mathrm{grav}</math>
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<math>~\equiv~</math>
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<math>
\frac{3}{2\cdot 5} \biggl(\frac{\nu^2}{q^6} \biggr) \biggl\{
2q^5 + 5q^3\biggl( \frac{\rho_e}{\rho_c} \biggr) (1-q^2)
+ \biggl( \frac{\rho_e}{\rho_c} \biggr)^2 \biggl[ 2(1-q^5) 
- 5q^3(1-q^2) \biggr] \biggr\} \, ,
</math>
  </td>
</tr>

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<math>~B_\mathrm{core}</math>
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<math>~\equiv~</math>
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<math>
 \biggl[ \frac{4\pi n_c q^3}{3} \biggr]\biggl(1 + \Lambda\biggr)  \, ,
</math>
  </td>
</tr>

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<math>~B_\mathrm{env}</math>
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<math>~\equiv~</math>
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<math>
 \biggl[ \frac{2\pi n_e}{3} \biggr] \biggl\{ 2 (1-q^3)  
+ 5 \Lambda \biggl( \frac{\rho_e}{\rho_0} \biggr) \biggl[-2 + 3q - q^3  \biggr]
  + \frac{3 \Lambda}{q^2} \biggl( \frac{\rho_e}{\rho_0} \biggr)^2 \biggl[ -1 + 5q^2 -5q^3 +  q^5   \biggr] \biggr\}  \, .
</math>
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Now, the condition derived above for the equilibrium radius is,
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<math>
\biggl[ \frac{R^4}{GM_\mathrm{tot}^2} \biggr]  P_i = \biggl( \frac{3}{2^3\pi} \biggr) \frac{\nu^2}{q^4} (g^2 -1) \, .
</math>
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Hence, the free energy expression becomes,
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<math>~\mathfrak{G}</math>
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<math>~=~</math>
  </td>
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<math>\frac{GM_\mathrm{tot}^2}{R_\mathrm{eq}} \biggl[ (B_\mathrm{core}+B_\mathrm{env} ) 
\biggl( \frac{3}{2^3\pi} \biggr) \frac{\nu^2}{q^4} (g^2 -1)  - A^*_\mathrm{grav} \biggr] \, .
</math>
  </td>
</tr>
</table>
</div>
Combining the same condition for the equilibrium radius with the expression for the pressure at the interface in the equilibrium configuration, namely,
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<math>~P_i</math>
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<math>~=~</math>
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<math>K_c \biggl[ \frac{3M_\mathrm{tot} \nu}{4\pi q^3} \biggr]^{\gamma_c} R^{-3\gamma_c} \, ,</math>
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</table>
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allows us to write the equilibrium radius in terms of <math>~M_\mathrm{tot}</math> and <math>~K_c</math>.  Specifically, we obtain,
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<math>R^{4-3\gamma_c} </math>
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<math>~~=~~</math>
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<math>\biggl[ \frac{GM_\mathrm{tot}^{2-\gamma_c}}{K_c } \biggr] \biggl( \frac{3\nu}{4\pi q^3} \biggr)^{1-\gamma_c} 
\biggl(\frac{\nu}{2q}\biggr) (g^2 -1) </math>
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</tr>

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<math>\Rightarrow ~~~~~ \frac{GM_\mathrm{tot}^2}{R}</math>
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<math>~~=~~</math>
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<math>
\biggl[ \frac{G^{3\gamma_c - 3} M^{5\gamma_c -6}_\mathrm{tot}}{K_c} \biggr]^{1/(3\gamma_c - 4)} \biggl[ \biggl( \frac{4\pi q^3}{3\nu}\biggr)^{\gamma_c-1} 
\biggl(\frac{\nu}{2q}\biggr) (g^2 -1) \biggr]^{1/(3\gamma_c - 4)} \, .</math>
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</table>
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Finally, then, recognizing that <math>~\gamma_c = 1 + 1/n_c</math> and, hence, <math>~1/(3\gamma_c - 4) = n_c/(3-n_c)</math>, we obtain,
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<math>~\mathfrak{G}</math>
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<math>~=~</math>
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<math>
\biggl[ \frac{K_c^{n_c}}{G^{3} M^{5-n_c}_\mathrm{tot}} \biggr]^{1/(n_c-3)} 
\biggl[ \biggl( \frac{4\pi q^3}{3\nu}\biggr)^{1/n_c} \biggl(\frac{\nu}{2q}\biggr) (g^2 -1) \biggr]^{-n_c/(n_c-3)}  
\biggl[ (B_\mathrm{core}+B_\mathrm{env} ) 
\biggl( \frac{3}{2^3\pi} \biggr) \frac{\nu^2}{q^4} (g^2 -1)  - A^*_\mathrm{grav} \biggr] \, .
</math>
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</table>
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