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==Solution Strategy== For a given set of free-energy coefficients, <math>~\mathcal{A}, \mathcal{B},</math> and <math>~\mathcal{C}</math>, along with a choice of the two adiabatic exponents <math>~(\gamma_c, \gamma_e)</math>, here's how to determine all of the physical parameters that are detailed in the above example table. * '''<font color="darkgreen">Step 1:</font>''' Guess a value of <math>~0 < q < 1</math>. * '''<font color="darkgreen">Step 2:</font>''' Given the pair of parameter values, <math>~(\mathcal{A}, q)</math>, determine the interface-density ratio, <math>~\rho_e/\rho_c</math>, by finding the appropriate root of the expression that defines the function, <math>~\mathcal{A}(q, \rho_e/\rho_c)</math>. This can be straightforwardly accomplished because, [[#Explain_Logic|as demonstrated below]], the relevant expression can be written as a quadratic function of <math>~(\rho_e/\rho_c)</math>. * '''<font color="darkgreen">Step 3:</font>''' Given the pair of parameter values, <math>~(q, \rho_e/\rho_c)</math>, determine the value of the core-to-total mass ratio, <math>~\nu</math>, from the expression that was obtained from an integration over the mass, namely, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{1}{\nu}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~1 + \biggl( \frac{\rho_e}{\rho_c} \biggr) \biggl( \frac{1}{q^3} - 1 \biggr) \, .</math> </td> </tr> </table> </div> * '''<font color="darkgreen">Step 4:</font>''' Given the value of <math>~\mathcal{B}</math> along with the pair of parameter values, <math>~(q, \nu)</math>, the above expression that defines <math>~\mathcal{B}</math> can be solved to give the relevant value of the dimensionless parameter, <math>~\Lambda_\mathrm{eq}.</math> * '''<font color="darkgreen">Step 5:</font>''' The value of <math>~\mathcal{C}^'</math> — the coefficient that appears on the right-hand-side of the above expression that defines <math>~\mathcal{C}</math> — can be determined, given the values of parameter triplet, <math>~(q, \nu, \Lambda_\mathrm{eq})</math>. * '''<font color="darkgreen">Step 6:</font>''' Given the value of <math>~\mathcal{C}</math> and the just-determined value of the coefficient <math>~\mathcal{C}^'</math>, the normalized equilibrium radius, <math>~\chi_\mathrm{eq},</math> that corresponds to the value of <math>~q</math> that was ''guessed'' in Step #1 can be determined from the above definition of <math>~\mathcal{C}</math>, specifically, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\chi_\mathrm{eq}\biggr|_\mathrm{guess} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\biggl( \frac{\mathcal{C}}{\mathcal{C}^'} \biggr)^{1/(3\gamma_e - 3\gamma_c)} \, .</math> </td> </tr> </table> </div> * '''<font color="darkgreen">Step 7:</font>''' But, independent of this ''guessed'' value of <math>~\chi_\mathrm{eq},</math> the condition for virial equilibrium — which identifies extrema in the free-energy function — gives the following expression for the normalized equilibrium radius: <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\chi_\mathrm{eq}\biggr|_\mathrm{virial} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\biggl[ \frac{\mathcal{A}}{\mathcal{B} + \mathcal{C}^'} \biggr]^{1/(4 - 3\gamma_c)} \, .</math> </td> </tr> </table> </div> * '''<font color="darkgreen">Step 8:</font>''' If <math>~\chi_\mathrm{eq}|_\mathrm{guess} \ne \chi_\mathrm{eq}|_\mathrm{virial}</math>, return to Step #1 and guess a different value of <math>~q</math>. Repeat Steps #1 through #7 until the two independently derived values of the normalized radius match, to a desired level of precision. * '''<font color="darkgreen">Keep in mind:</font>''' (A) A graphical representation of the free-energy function, <math>~\mathfrak{G}(\chi)</math>, can also be used to identify the "correct" value of <math>~\chi_\mathrm{eq}</math> and, ultimately, the above-described iteration loop should converge on this value. (B) The free-energy function may exhibit more than one (or, actually, no) extrema, in which case more than one (or no) value of <math>~q</math> should lead to convergence of the above-described iteration loop.
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