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=Free-Energy Synopsis= All of the self-gravitating configurations considered below have an associated Gibbs-like free-energy that can be expressed analytically as a power-law function of the dimensionless configuration radius, <math>~x</math>. Specifically, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\mathfrak{G}^*_\mathrm{type}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~-ax^{-1} + b x^{-3/n} + c x^{-3/j} + \mathfrak{G}_0 \, .</math> </td> </tr> </table> </div> ==Equilibrium Radii and Critical Radii== The first and second (partial) derivatives with respect to <math>~x</math> are, respectively, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{\partial\mathfrak{G}^*_\mathrm{type}}{\partial x}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ax^{-2} - \biggl(\frac{ 3b}{n}\biggr) x^{-3/n -1} -\biggl(\frac{3 c}{j}\biggr) x^{-3/j-1} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{1}{x^2} \biggl[ a - \biggl(\frac{ 3b}{n}\biggr) x^{(n-3)/n } -\biggl(\frac{3 c}{j}\biggr) x^{(j-3)/j} \biggr] \, ,</math> </td> </tr> <tr> <td align="right"> <math>~\frac{\partial^2 \mathfrak{G}^*_\mathrm{type}}{\partial x^2}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~-2ax^{-3} + \biggl(\frac{ 3b}{n}\biggr) \biggl( \frac{n+3}{n}\biggr) x^{-3/n -2} + \biggl(\frac{3 c}{j}\biggr)\biggl( \frac{j+3}{j}\biggr) x^{-3/j-2} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{1}{x^3} \biggl\{ \biggl(\frac{ 3b}{n}\biggr) \biggl( \frac{n+3}{n}\biggr) x^{(n-3)/n} + \biggl(\frac{3 c}{j}\biggr)\biggl( \frac{j+3}{j}\biggr) x^{(j-3)/j} -2a\biggr\} \, . </math> </td> </tr> </table> </div> Equilibrium configurations are identified by setting the first derivative to zero. This gives, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~0</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~a - \biggl(\frac{ 3b}{n}\biggr) x^{(n-3)/n }_\mathrm{eq} -\biggl(\frac{3 c}{j}\biggr) x^{(j-3)/j}_\mathrm{eq} </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow ~~~x^{(n-3)/n }_\mathrm{eq}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\biggl(\frac{n}{ 3b}\biggr) \biggl[a -\biggl(\frac{3 c}{j}\biggr) x^{(j-3)/j}_\mathrm{eq} \biggr] \, .</math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow ~~~\frac{ b}{nc}\cdot x^{(n-3)/n }_\mathrm{eq} - \frac{a}{3c} + \frac{1}{j}\cdot x^{(j-3)/j}_\mathrm{eq} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ 0 \, .</math> </td> </tr> </table> </div> We conclude, as well, that ''at'' this equilibrium radius, the second (partial) derivative assumes the value, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\biggl[ \frac{\partial^2 \mathfrak{G}^*_\mathrm{type}}{\partial x^2} \biggr]_\mathrm{eq}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{1}{x^3_\mathrm{eq} } \biggl\{ \biggl(\frac{ 3b}{n}\biggr) \biggl( \frac{n+3}{n}\biggr) x^{(n-3)/n} + \biggl(\frac{3 c}{j}\biggr)\biggl( \frac{j+3}{j}\biggr) x^{(j-3)/j} -2a\biggr\}_\mathrm{eq} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{1}{x^3_\mathrm{eq} } \biggl\{ \biggl( \frac{n+3}{n}\biggr) \biggl[a -\biggl(\frac{3 c}{j}\biggr) x^{(j-3)/j}_\mathrm{eq} \biggr] + \biggl(\frac{3 c}{j}\biggr)\biggl( \frac{j+3}{j}\biggr) x^{(j-3)/j}_\mathrm{eq} -2a\biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{1}{x^3_\mathrm{eq} } \biggl\{ \biggl(\frac{3 c}{j}\biggr) \biggl[ \biggl( \frac{j+3}{j}\biggr) -\biggl( \frac{n+3}{n}\biggr) \biggl] x^{(j-3)/j}_\mathrm{eq} + \biggl( \frac{3-n}{n}\biggr) a\biggr\} \, . </math> </td> </tr> </table> </div> Hence, equilibrium configurations for which the ''second'' (as well as first) derivative of the free energy is zero are found at "critical" radii given by the expression, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~0</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl(\frac{3 c}{j}\biggr) \biggl[ \biggl( \frac{j+3}{j}\biggr) -\biggl( \frac{n+3}{n}\biggr) \biggl] [x_\mathrm{eq}^{(j-3)/j}]_\mathrm{crit} + \biggl( \frac{3-n}{n}\biggr) a </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow ~~~[x_\mathrm{eq}^{(j-3)/j}]_\mathrm{crit} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\biggl[ \frac{j^2 a(n-3)}{3 c}\biggr] [ n(j+3) - j(n+3) ]^{-1} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{a}{3^2c}\biggl[ \frac{j^2(n-3)}{n-j} \biggr] \, . </math> </td> </tr> </table> </div> ==Examples== ===Pressure-Truncated Polytropes=== For pressure-truncated polytropes of index <math>~n</math>, we set, <math>~j = -1</math>, in which case, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{ b}{nc}\cdot x^{(n-3)/n }_\mathrm{eq} - \frac{a}{3c} - x^{4}_\mathrm{eq}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~0 \, ;</math> </td> </tr> <tr> <td align="right"> <math>~\biggl[ \biggl(\frac{3}{4\pi}\biggr) \frac{ M_\mathrm{tot}}{M_\mathrm{SWS}} \biggr]^{(n+1)/n} x^{(n-3)/n }_\mathrm{eq} - \frac{3}{20\pi} \biggl( \frac{n+1}{n}\biggr)\biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}} \biggr)^{2} - x^{4}_\mathrm{eq}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~0 \, ;</math> </td> </tr> <tr> <td align="right"> <math>~x^{(n-3)/n }_\mathrm{eq}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\biggl(\frac{n}{ 3b}\biggr) \biggl[a + 3cx^{4}_\mathrm{eq} \biggr] \, ;</math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> and </td> <td align="left"> </td> </tr> <tr> <td align="right"> <math>~[x_\mathrm{eq}]_\mathrm{crit} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\biggl[ \frac{a(n-3)}{3^2 c (n+1)} \biggr]^{1/4} \, . </math> </td> </tr> </table> </div> ====Case M==== <!-- Next segment supports PowerPoint presentation <div align="center"> <math>( {\tilde{\mathfrak{f}}}_M, {\tilde{\mathfrak{f}}}_W, {\tilde{\mathfrak{f}}}_A)</math> </div> <div align="center"> <math>~\frac{d}{dx_\mathrm{eq}}\biggl(\frac{P_e}{P_\mathrm{norm}}\biggr)= 0</math> </div> --> More specifically, the expression that describes the [[#Case_M_Free-Energy_Surface|"Case M" free-energy surface]] is, <div align="center"> <table border="1" cellpadding="5" align="center"> <tr><td align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\mathfrak{G}_{K,M}^* \equiv \frac{\mathfrak{G}_{K,M}}{E_\mathrm{norm}} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ -3\mathcal{A} \biggl(\frac{R}{R_\mathrm{norm}}\biggr)^{-1} +~ n\mathcal{B} \biggl(\frac{R}{R_\mathrm{norm}}\biggr)^{-3/n} +~ \biggl( \frac{4\pi}{3} \biggr) \frac{P_e}{P_\mathrm{norm}} \biggl(\frac{R}{R_\mathrm{norm}}\biggr)^3 \, . </math> </td> </tr> </table> </td></tr> </table> </div> Hence, we have, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~a</math> </td> <td align="center"> <math>~\equiv</math> </td> <td align="left"> <math>~3\mathcal{A} = \frac{3}{5} \cdot \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2}\, , </math> </td> </tr> <tr> <td align="right"> <math>~b</math> </td> <td align="center"> <math>~\equiv</math> </td> <td align="left"> <math>~n\mathcal{B} = n\biggl(\frac{4\pi}{3} \biggr)^{-1/n} \frac{\tilde{\mathfrak{f}}_A}{\tilde{\mathfrak{f}}_M^{(n+1)/n}} \, , </math> </td> </tr> <tr> <td align="right"> <math>~c</math> </td> <td align="center"> <math>~\equiv</math> </td> <td align="left"> <math>~\frac{4\pi}{3}\biggl( \frac{P_e}{P_\mathrm{norm}} \biggr) \, , </math> </td> </tr> </table> </div> where the structural form factors for pressure-truncated polytropes are precisely defined [[SSCpt1/Virial/FormFactors#PTtable|here]]. Therefore, the statement of virial equilibrium is, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~0 </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{ b}{nc}\cdot x^{(n-3)/n }_\mathrm{eq} - \frac{a}{3c} - x^{4}_\mathrm{eq}</math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow ~~~ \biggl(\frac{3}{4\pi}\biggr)c x_\mathrm{eq}^4 </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\biggl(\frac{3}{4\pi}\biggr) \biggl[ \frac{ b}{n}\cdot x^{(n-3)/n }_\mathrm{eq} - \frac{a}{3} \biggr]</math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow ~~~ \biggl(\frac{P_e}{P_\mathrm{norm}}\biggr) x_\mathrm{eq}^4 </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\biggl(\frac{3}{4\pi}\biggr) \biggl[ \biggl(\frac{3}{4\pi} \biggr)^{1/n} \frac{\tilde{\mathfrak{f}}_A}{\tilde{\mathfrak{f}}_M^{(n+1)/n}} \cdot x^{(n-3)/n }_\mathrm{eq} - \frac{1}{5} \cdot \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2} \biggr]</math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\biggl(\frac{3}{4\pi} \biggr)^{(n+1)/n} \frac{\tilde{\mathfrak{f}}_A}{\tilde{\mathfrak{f}}_M^{(n+1)/n}} \cdot x^{(n-3)/n }_\mathrm{eq} - \frac{3}{20\pi} \cdot \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2} \, .</math> </td> </tr> <!-- NEXT STEP IS FOR POWERPOINT PRESENTATION ... <tr> <td align="right"> <math>~\biggl(\frac{P_e}{P_\mathrm{norm}}\biggr) x_\mathrm{eq}^4 </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{3}{20\pi} \biggl[ 5\biggl(\frac{3}{4\pi} \biggr)^{1/n} \frac{\tilde{\mathfrak{f}}_A}{\tilde{\mathfrak{f}}_M^{(n+1)/n}} \cdot x^{(n-3)/n }_\mathrm{eq} - \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2} \biggr] </math> </td> </tr> --> </table> </div> And we conclude that, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~3c[x_\mathrm{eq}]^4_\mathrm{crit} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{(n-3)}{5(n+1)} \cdot \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2} </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow ~~~ \biggl( \frac{P_e}{P_\mathrm{norm}} \biggr) [x_\mathrm{eq}]^4_\mathrm{crit}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{1}{20\pi} \biggl( \frac{n-3}{n+1} \biggr) \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2} </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow ~~~ \biggl(\frac{3}{4\pi} \biggr)^{(n+1)/n} \frac{\tilde{\mathfrak{f}}_A}{\tilde{\mathfrak{f}}_M^{(n+1)/n}} \cdot [ x_\mathrm{eq} ]^{(n-3)/n }_\mathrm{crit} - \frac{3}{20\pi} \cdot \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{1}{20\pi} \biggl( \frac{n-3}{n+1} \biggr) \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2} </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow ~~~ \biggl(\frac{3}{4\pi} \biggr)^{(n+1)/n} \frac{\tilde{\mathfrak{f}}_A}{\tilde{\mathfrak{f}}_M^{(n+1)/n}} \cdot [ x_\mathrm{eq} ]^{(n-3)/n }_\mathrm{crit} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{1}{20\pi} \biggl( \frac{n-3}{n+1} \biggr) \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2} + \frac{3}{20\pi} \cdot \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2}</math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{1}{20\pi} \biggl( \frac{4n}{n+1} \biggr) \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2} </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow ~~~ [ x_\mathrm{eq} ]^{(n-3)/n }_\mathrm{crit} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{1}{20\pi} \biggl(\frac{4\pi}{3} \biggr)^{(n+1)/n} \biggl( \frac{4n}{n+1} \biggr) \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_A \tilde{\mathfrak{f}}_M^{(n-1)/n} } </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow ~~~ [ x_\mathrm{eq} ]_\mathrm{crit} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl[\frac{4n}{15(n+1)} \biggl(\frac{4\pi}{3} \biggr)^{1/n} \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_A \tilde{\mathfrak{f}}_M^{(n-1)/n} } \biggr]^{n/(n-3)} \, . </math> </td> </tr> </table> </div> <div align="center"> <table border="1" cellpadding="8" width="90%"><tr><td align="left"> <font color="red"><b>ASIDE:</b></font> Let's see what this requires for the case of <math>~n=5</math>, where everything is specifiable analytically. We have gathered together: * Form factors from [[SSCpt1/Virial/FormFactors#Summary_.28n.3D5.29|here]]. * Hoerdt's equilibrium expressions from [[SSC/Structure/PolytropesEmbedded#Tabular_Summary_.28n.3D5.29|here]]. * Conversion from Horedt's units to ours as specified [[SSCpt1/Virial#Choices_Made_by_Other_Researchers|here]]. <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~{\tilde\mathfrak{f}}_M</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ ( 1 + \ell^2 )^{-3/2} </math> </td> </tr> <tr> <td align="right"> <math>~{\tilde\mathfrak{f}}_W</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{5}{2^4} \cdot \ell^{-5} \biggl[ \ell \biggl( \ell^4 - \frac{8}{3}\ell^2 - 1 \biggr)(1 + \ell^2)^{-3} + \tan^{-1}(\ell ) \biggr] </math> </td> </tr> <tr> <td align="right"> <math>~{\tilde\mathfrak{f}}_A</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{3}{2^3} \ell^{-3} [ \tan^{-1}(\ell ) + \ell (\ell^2-1) (1+\ell^2)^{-2} ] </math> </td> </tr> <tr> <td align="right"> <math>~\frac{R_\mathrm{eq}}{R_\mathrm{norm}} = \frac{R_\mathrm{eq}}{R_\mathrm{Horedt}} \biggl[ \frac{4\pi}{(n+1)^n} \biggr]^{1/(n-3)}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\biggl\{ 3 \biggl[ \frac{(\xi_e^2/3)^5}{(1+\xi_e^2/3)^{6}} \biggr] \biggr\}^{-1/2}\biggl[ \frac{4\pi}{(n+1)^n} \biggr]^{1/(n-3)}</math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\biggl[ \frac{(1+\ell^2)^{3}}{\ell^{5}} \biggr] \biggl[ \frac{\pi}{2^3\cdot 3^6} \biggr]^{1/2}</math> </td> </tr> <tr> <td align="right"> <math>~\frac{P_e}{P_\mathrm{norm}} = \frac{P_e}{P_\mathrm{Horedt}} \biggl[ \frac{(n+1)^3}{4\pi} \biggr]^{(n+1)/(n-3)}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~3^3 \biggl[ \frac{(\xi_e^2/3)^3}{(1+\xi_e^2/3)^{4}} \biggr]^3 \biggl[ \frac{(n+1)^3}{4\pi} \biggr]^{(n+1)/(n-3)}</math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\biggl[ \frac{\ell^{18}}{(1+\ell^2)^{12}} \biggr] \biggl[ \frac{2 \cdot 3^4}{\pi} \biggr]^{3}</math> </td> </tr> </table> So, the radius of the critical equilibrium state should be, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\biggl[\frac{R_\mathrm{eq}}{R_\mathrm{norm}} \biggr]^4_\mathrm{crit} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{(n-3)}{3\cdot 5(n+1)} \biggl(\frac{3}{2^2\pi}\biggr) \biggl(\frac{P_e}{P_\mathrm{norm}}\biggr)^{-1}\cdot \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{1}{2^2\cdot 3\cdot 5 \pi} \biggl\{\frac{(1+\ell^2)^{12}}{\ell^{18}} \biggl[ \frac{\pi}{2 \cdot 3^4} \biggr]^{3}\biggr\} (1+\ell^2)^3 \cdot \biggl\{ \frac{5}{2^4} \cdot \ell^{-5} \biggl[ \ell \biggl( \ell^4 - \frac{8}{3}\ell^2 - 1 \biggr)(1 + \ell^2)^{-3} + \tan^{-1}(\ell ) \biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{\pi^2}{2^9\cdot 3^{13}} \biggl\{\frac{(1+\ell^2)^{12}}{\ell^{23}} \biggr\} \cdot \biggl\{ \biggl[ \ell \biggl( \ell^4 - \frac{8}{3}\ell^2 - 1 \biggr) + (1 + \ell^2)^{3}\tan^{-1}(\ell ) \biggr] \biggr\} \, ; </math> </td> </tr> </table> whereas, each equilibrium configuration has, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\biggl[\frac{R_\mathrm{eq}}{R_\mathrm{norm}}\biggr]^4 </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{\pi^2}{2^6\cdot 3^{12}} \biggl[ \frac{(1+\ell^2)^{12}}{\ell^{20}} \biggr] \, .</math> </td> </tr> </table> So the equilibrium state that marks the critical configuration must have a value of <math>~\ell</math> that satisfies the relation, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{\pi^2}{2^6\cdot 3^{12}} \biggl[ \frac{(1+\ell^2)^{12}}{\ell^{20}} \biggr] </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{\pi^2}{2^9\cdot 3^{13}} \biggl\{\frac{(1+\ell^2)^{12}}{\ell^{23}} \biggr\} \cdot \biggl\{ \biggl[ \ell \biggl( \ell^4 - \frac{8}{3}\ell^2 - 1 \biggr) + (1 + \ell^2)^{3}\tan^{-1}(\ell ) \biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow ~~~2^3\cdot 3 \ell^3</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \ell ( \ell^4 - \frac{8}{3}\ell^2 - 1 ) + (1 + \ell^2)^{3}\tan^{-1}(\ell ) </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow ~~~\biggl[ \frac{(1 + \ell^2)^{3}}{\ell} \biggr] \tan^{-1}(\ell ) </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~1 + \frac{80}{3}\cdot \ell^2 -\ell^4 \, . </math> </td> </tr> </table> The solution is: <math>~\ell_\mathrm{crit} \approx 2.223175 \, .</math> </td></tr></table> </div> In addition, we know from [[SSC/Virial/PolytropesEmbedded/SecondEffortAgain#Virial_Equilibrium_of_Adiabatic_Spheres_.28Summary.29|our dissection of Hoerdt's work on detailed force-balance models]] that, in the equilibrium state, <div align="center"> <table border="0" cellpadding="5"> <tr> <td align="right"> <math>~\biggl(\frac{P_e}{P_\mathrm{norm}}\biggr) \biggl(\frac{R_\mathrm{eq}}{R_\mathrm{norm}} \biggr)^4</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math> \biggl[ \frac{\tilde\theta^{n+1} }{(4\pi)(n+1)( -\tilde\theta' )^{2}} \biggr] </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow ~~~ 3c x_\mathrm{eq}^4</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math> \biggl[ \frac{\tilde\theta^{n+1} }{(n+1)( -\tilde\theta' )^{2}} \biggr] \, . </math> </td> </tr> </table> </div> This means that, for any chosen polytropic index, the critical equilibrium state is the equilibrium configuration for which (needs to be checked), <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~2(9-2n){\tilde\theta}^{n+1}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ 3(n-3)\biggl[ (- {\tilde\theta}^')^2 - \frac{\tilde\theta(-{\tilde\theta}^')}{\tilde\xi}\biggr] \, . </math> </td> </tr> </table> </div> We note, as well, that by combining the Horedt expression for <math>~x_\mathrm{eq}</math> with our virial equilibrium expression, we find (needs to be checked), <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~x_\mathrm{eq}^{n-3}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{4\pi}{3}\biggl[ \frac{3}{(n+1)\tilde\xi^2} + \frac{{\tilde\mathfrak{f}}_{W} - {\tilde\mathfrak{f}}_{M}}{5\tilde\mathfrak{f}_A} \biggr]^{n} {\tilde\mathfrak{f}}_{M}^{1-n} \, .</math> </td> </tr> </table> </div> ====Case P==== =====First Pass===== Alternatively, let's examine the [[#Case_P_Free-Energy_Surface|"Case P" free-energy surface]]. Drawing on [[SSC/Structure/PolytropesEmbedded#Stahler.27s_Presentation|Stahler's presentation]], we adopt the following radius and mass normalizations: <div align="center"> <math>M_\mathrm{SWS} = \biggl( \frac{n+1}{n} \biggr)^{3/2} G^{-3/2} K_n^{2n/(n+1)} P_\mathrm{e}^{(3-n)/[2(n+1)]} \, ,</math> </div> <div align="center"> <math> R_\mathrm{SWS} = \biggl( \frac{n+1}{n} \biggr)^{1/2} G^{-1/2} K_n^{n/(n+1)} P_\mathrm{e}^{(1-n)/[2(n+1)]} \, . </math> </div> In terms of these new normalizations, we have, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~R_\mathrm{norm} \equiv \biggl[\biggl( \frac{G}{K} \biggr)^n M_\mathrm{tot}^{(n-1)} \biggr]^{1/(n-3)}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl( \frac{G}{K} \biggr)^{n/(n-3)} M_\mathrm{tot}^{(n-1)/(n-3)} R_\mathrm{SWS} \biggl( \frac{n+1}{n} \biggr)^{-1/2} G^{1/2} K_n^{-n/(n+1)} P_\mathrm{e}^{-(1-n)/[2(n+1)]} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~+ M_\mathrm{SWS}^{-(n-1)/(n-3)} \biggl[ \biggl( \frac{n+1}{n} \biggr)^{3/2} G^{-3/2} K_n^{2n/(n+1)} P_\mathrm{e}^{(3-n)/[2(n+1)]} \biggr]^{(n-1)/(n-3)} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~R_\mathrm{SWS} \biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{(n-1)/(n-3)} \biggl( \frac{n+1}{n} \biggr)^{[3(n-1)-(n-3)]/[2(n-3)]} G^{[2n+(n-3)-3(n-1)]/[2(n-3)]} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~+ K_n^{n[2(n-1) - (n+1) - (n-3)]/[(n+1)(n-3)]} P_\mathrm{e}^{-(n-1)(3-n)/[2(n+1)(n-3)]} P_\mathrm{e}^{(n-1)(3-n)/[2(n+1)(n-3)]} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~R_\mathrm{SWS} \biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{(n-1)/(n-3)} \biggl( \frac{n+1}{n} \biggr)^{n/(n-3)} \, . </math> </td> </tr> </table> </div> and, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~P_\mathrm{norm} \equiv \biggl[ \frac{K^{4n}}{G^{3(n+1)} M_\mathrm{tot}^{2(n+1)} } \biggr]^{1/(n-3)}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl[ \frac{K^{4n}}{G^{3(n+1)} } \biggr]^{1/(n-3)} \biggl( \frac{M_\mathrm{tot} }{M_\mathrm{SWS}} \biggr)^{-2(n+1)/(n-3)} \biggl\{ \biggl( \frac{n+1}{n} \biggr)^{3/2} G^{-3/2} K_n^{2n/(n+1)} P_\mathrm{e}^{(3-n)/[2(n+1)]}\biggr\}^{-2(n+1)/(n-3)} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl( \frac{M_\mathrm{tot} }{M_\mathrm{SWS}} \biggr)^{-2(n+1)/(n-3)} \biggl( \frac{n+1}{n} \biggr)^{-3(n+1)/(n-3)} K^{4n/(n-3)} G^{-3(n+1)/(n-3)} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~\times~ G^{3(n+1)/(n-3)} K_n^{-4n/(n-3)} \biggl\{ P_\mathrm{e}^{-(n-3)/[2(n+1)]}\biggr\}^{-2(n+1)/(n-3)} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~P_e \biggl( \frac{M_\mathrm{tot} }{M_\mathrm{SWS}} \biggr)^{-2(n+1)/(n-3)} \biggl( \frac{n+1}{n} \biggr)^{-3(n+1)/(n-3)} \, . </math> </td> </tr> </table> </div> <span id="FirstPassFreeEnergy">Rewriting the expression for the free energy gives,</span> <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\mathfrak{G}_{K,M}^* \equiv \frac{\mathfrak{G}_{K,M}}{E_\mathrm{norm}} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ -3\mathcal{A} \biggl(\frac{R}{R_\mathrm{SWS}}\biggr)^{-1} \biggl(\frac{R_\mathrm{norm}}{R_\mathrm{SWS}} \biggr) +~ n\mathcal{B} \biggl(\frac{R}{R_\mathrm{SWS}}\biggr)^{-3/n} \biggl(\frac{R_\mathrm{norm}}{R_\mathrm{SWS}} \biggr)^{3/n} +~ \biggl( \frac{4\pi}{3} \biggr) \frac{P_e}{P_\mathrm{norm}} \biggl(\frac{R}{R_\mathrm{SWS}}\biggr)^3 \biggl(\frac{R_\mathrm{norm}}{R_\mathrm{SWS}}\biggr)^{-3} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ -3\mathcal{A} \biggl(\frac{R}{R_\mathrm{SWS}}\biggr)^{-1} \biggl[ \biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{(n-1)/(n-3)} \biggl( \frac{n+1}{n} \biggr)^{n/(n-3)}\biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ +~ n\mathcal{B} \biggl(\frac{R}{R_\mathrm{SWS}}\biggr)^{-3/n} \biggl[ \biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{(n-1)/(n-3)} \biggl( \frac{n+1}{n} \biggr)^{n/(n-3)}\biggr]^{3/n} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ +~ \biggl( \frac{4\pi}{3} \biggr) \biggl( \frac{M_\mathrm{tot} }{M_\mathrm{SWS}} \biggr)^{2(n+1)/(n-3)} \biggl( \frac{n+1}{n} \biggr)^{3(n+1)/(n-3)} \biggl(\frac{R}{R_\mathrm{SWS}}\biggr)^3 \biggl[ \biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{(n-1)/(n-3)} \biggl( \frac{n+1}{n} \biggr)^{n/(n-3)}\biggr]^{-3} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ -3\mathcal{A} \biggl( \frac{n+1}{n} \biggr)^{n/(n-3)} \biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{(n-1)/(n-3)} \biggl(\frac{R}{R_\mathrm{SWS}}\biggr)^{-1} +~ n\mathcal{B} \biggl( \frac{n+1}{n} \biggr)^{3/(n-3)} \biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{3(n-1)/[n(n-3)]} \biggl(\frac{R}{R_\mathrm{SWS}}\biggr)^{-3/n} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ +~ \biggl( \frac{4\pi}{3} \biggr) \biggl( \frac{n+1}{n} \biggr)^{3/(n-3)} \biggl( \frac{M_\mathrm{tot} }{M_\mathrm{SWS}} \biggr)^{(5-n)/(n-3)} \biggl(\frac{R}{R_\mathrm{SWS}}\biggr)^3 \, . </math> </td> </tr> </table> </div> Therefore, in this case, we have, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~a</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{3}{5} \cdot \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2} \biggl( \frac{n+1}{n} \biggr)^{n/(n-3)} \biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{(n-1)/(n-3)} \, , </math> </td> </tr> <tr> <td align="right"> <math>~b</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ n\biggl(\frac{4\pi}{3} \biggr)^{-1/n} \frac{\tilde{\mathfrak{f}}_A}{\tilde{\mathfrak{f}}_M^{(n+1)/n}} \biggl( \frac{n+1}{n} \biggr)^{3/(n-3)} \biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{3(n-1)/[n(n-3)]} \, , </math> </td> </tr> <tr> <td align="right"> <math>~c</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{4\pi}{3} \biggl( \frac{n+1}{n} \biggr)^{3/(n-3)} \biggl( \frac{M_\mathrm{tot} }{M_\mathrm{SWS}} \biggr)^{(5-n)/(n-3)} \, , </math> </td> </tr> </table> </div> where the structural form factors for pressure-truncated polytropes are precisely defined [[SSCpt1/Virial/FormFactors#PTtable|here]]. The statement of virial equilibrium is, therefore, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~x^{4}_\mathrm{eq} + \alpha </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \beta x^{(n-3)/n }_\mathrm{eq} \, , </math> </td> </tr> </table> </div> where, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\alpha \equiv \frac{a}{3c}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{1}{5} \cdot \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2} \biggl( \frac{n+1}{n} \biggr)^{n/(n-3)} \biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{(n-1)/(n-3)} \biggl\{ \frac{3}{4\pi} \biggl( \frac{n+1}{n} \biggr)^{-3/(n-3)} \biggl( \frac{M_\mathrm{tot} }{M_\mathrm{SWS}} \biggr)^{(n-5)/(n-3)} \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{3}{20\pi} \cdot \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2} \biggl( \frac{n+1}{n} \biggr) \biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{2} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl( \frac{4\pi}{3\cdot 5}\biggr) \tilde{\mathfrak{f}}_W \biggl( \frac{n+1}{n} \biggr) \mathfrak{m}^{2} \, , </math> </td> </tr> <tr> <td align="right"> <math>~\beta \equiv \frac{b}{nc}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl(\frac{4\pi}{3} \biggr)^{-1/n} \frac{\tilde{\mathfrak{f}}_A}{\tilde{\mathfrak{f}}_M^{(n+1)/n}} \biggl( \frac{n+1}{n} \biggr)^{3/(n-3)} \biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{3(n-1)/[n(n-3)]} \biggl\{ \frac{3}{4\pi} \biggl( \frac{n+1}{n} \biggr)^{-3/(n-3)} \biggl( \frac{M_\mathrm{tot} }{M_\mathrm{SWS}} \biggr)^{(n-5)/(n-3)} \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \tilde{\mathfrak{f}}_A \mathfrak{m}^{(n+1)/n} \, , </math> </td> </tr> <tr> <td align="right"> <math>~\mathfrak{m}</math> </td> <td align="center"> <math>~\equiv</math> </td> <td align="left"> <math>~ \biggl(\frac{3}{4\pi}\biggr) \frac{1}{\tilde{\mathfrak{f}}_M} \biggl( \frac{M_\mathrm{tot} }{M_\mathrm{SWS}} \biggr) \, . </math> </td> </tr> </table> </div> From a previous derivation, we have, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~0 </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{ b}{nc}\cdot x^{(n-3)/n }_\mathrm{eq} - \frac{a}{3c} - x^{4}_\mathrm{eq} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{3}{4\pi} \biggl( \frac{n+1}{n} \biggr)^{-3/(n-3)} \biggl( \frac{M_\mathrm{tot} }{M_\mathrm{SWS}} \biggr)^{(n-5)/(n-3)} \biggl\{ \biggl(\frac{3}{4\pi} \biggr)^{1/n} \frac{\tilde{\mathfrak{f}}_A}{\tilde{\mathfrak{f}}_M^{(n+1)/n}} \biggl( \frac{n+1}{n} \biggr)^{3/(n-3)} \biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{3(n-1)/[n(n-3)]} \biggr\} \cdot x^{(n-3)/n }_\mathrm{eq} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ - \frac{3}{4\pi} \biggl( \frac{n+1}{n} \biggr)^{-3/(n-3)} \biggl( \frac{M_\mathrm{tot} }{M_\mathrm{SWS}} \biggr)^{(n-5)/(n-3)} \biggl\{ \frac{1}{5} \cdot \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2} \biggl( \frac{n+1}{n} \biggr)^{n/(n-3)} \biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{(n-1)/(n-3)} \biggr\} - x^{4}_\mathrm{eq} </math> </td> </tr> <tr> <td align="right"> <math>~0</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl(\frac{3}{4\pi} \biggr)^{(n+1)/n} \frac{\tilde{\mathfrak{f}}_A}{\tilde{\mathfrak{f}}_M^{(n+1)/n}} \biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{(n+1)/n} x^{(n-3)/n }_\mathrm{eq} - \frac{3}{20\pi} \cdot \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2} \biggl( \frac{n+1}{n} \biggr)\biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{2} - x^{4}_\mathrm{eq} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \tilde{\mathfrak{f}}_A \biggl[ \biggl(\frac{3}{4\pi} \biggr) \frac{1}{\tilde{\mathfrak{f}}_M} \biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)\biggr]^{(n+1)/n} x^{(n-3)/n }_\mathrm{eq} - \frac{1}{5} \biggl(\frac{4\pi}{3}\biggr) \cdot \tilde{\mathfrak{f}}_W \biggl( \frac{n+1}{n} \biggr)\biggl[ \biggl(\frac{3}{4\pi} \biggr) \frac{1}{\tilde{\mathfrak{f}}_M} \biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)\biggr]^{2} - x^{4}_\mathrm{eq} </math> </td> </tr> </table> </div> which, thankfully, matches! We conclude as well that the transition from stable to dynamically unstable configurations occurs at, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~[x_\mathrm{eq}]_\mathrm{crit}^4</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\biggl[ \frac{(n-3)}{3 (n+1)} \biggr] \alpha \, . </math> </td> </tr> <!-- FOR POWERPOINT SYNOPSIS <tr> <td align="right"> <math>~[x_\mathrm{eq}]_\mathrm{crit}^4</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\biggl[ \frac{(n-3)}{3 (n+1)} \biggr] \frac{a}{3c} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{(n-3)}{20\pi n} \cdot \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2} \biggl( \frac{n+1}{n} \biggr) \biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{2} </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow ~~~ [x_\mathrm{eq}]_\mathrm{crit}^{2(n+1)}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl( \frac{n}{n-3} \biggr)^{(1-n)} \biggl( \frac{3^2\cdot 5}{4\pi \tilde{\mathfrak{f}}_W} \biggr)^{(n+1)} \biggl(\frac{\tilde{\mathfrak{f}}_A}{4} \biggr)^{2n} \, . </math> </td> </tr> --> </table> </div> When combined with the statement of virial equilibrium ''at'' this critical point, we find, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~ \biggl\{ \biggl[ \frac{(n-3)}{3 (n+1)} \biggr] + 1\biggr\}\frac{ \alpha }{\beta} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ [x_\mathrm{eq}]^{(n-3)/n }_\mathrm{crit} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl\{ \biggl[ \frac{(n-3)}{3 (n+1)} \biggr] \alpha \biggr\}^{(n-3)/(4n) } </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow~~~ \biggl[ \frac{4n}{3 (n+1)} \biggr]^{4n} \biggl( \frac{ \alpha }{\beta} \biggr)^{4n} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl[ \frac{(n-3)}{3 (n+1)} \biggr]^{(n-3)} \alpha^{(n-3)} </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow~~~ \biggl[ \frac{3 n}{(n-3)} \biggl( \frac{n+1}{n}\biggr) \biggr]^{(3-n)} \biggl[ \frac{3 }{4} \biggl( \frac{n+1}{n}\biggr) \biggr]^{4n} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \alpha^{3(n+1)} \beta^{-4n} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl\{ \biggl( \frac{4\pi}{3\cdot 5}\biggr) \tilde{\mathfrak{f}}_W \biggl( \frac{n+1}{n} \biggr) \mathfrak{m}^{2} \biggr\}^{3(n+1)} \biggl\{ \tilde{\mathfrak{f}}_A \mathfrak{m}^{(n+1)/n} \biggr\}^{-4n} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\tilde{\mathfrak{f}}_A^{-4n} \biggl[ \biggl( \frac{4\pi}{3\cdot 5}\biggr) \tilde{\mathfrak{f}}_W \biggl( \frac{n+1}{n} \biggr)\biggr]^{3(n+1)} \mathfrak{m}^{2(n+1)} </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow ~~~ \mathfrak{m}^{2(n+1)} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl[ \biggl( \frac{4\pi}{3\cdot 5}\biggr) \tilde{\mathfrak{f}}_W \biggl( \frac{n+1}{n} \biggr)\biggr]^{-3(n+1)} \biggl[ \frac{3 n}{(n-3)} \biggl( \frac{n+1}{n}\biggr) \biggr]^{(3-n)} \biggl[ \frac{3\tilde{\mathfrak{f}}_A }{4} \biggl( \frac{n+1}{n}\biggr) \biggr]^{4n} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl[ \biggl( \frac{3\cdot 5}{4\pi}\biggr) \frac{1}{\tilde{\mathfrak{f}}_W} \biggr]^{3(n+1)} \biggl[ \frac{3 n}{(n-3)} \biggr]^{(3-n)} \biggl[ \frac{3\tilde{\mathfrak{f}}_A }{4} \biggr]^{4n} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl[ \frac{3^2 \cdot 5 n}{ 4\pi(n-3)} \cdot \frac{1}{\tilde{\mathfrak{f}}_W} \biggr]^{(3-n)} \biggl[ \biggl( \frac{3^2\cdot 5}{2^4\pi}\biggr) \frac{\tilde{\mathfrak{f}}_A}{\tilde{\mathfrak{f}}_W} \biggr]^{4n} \, . </math> </td> </tr> </table> </div> This also means that the critical radius is, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~[x_\mathrm{eq}]_\mathrm{crit}^4</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\biggl[ \frac{(n-3)}{3 (n+1)} \biggr] \biggl( \frac{4\pi}{3\cdot 5}\biggr) \tilde{\mathfrak{f}}_W \biggl( \frac{n+1}{n} \biggr) \mathfrak{m}^{2} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl[ \frac{3^2\cdot 5 n}{4\pi(n-3)} \cdot \frac{1}{\tilde{\mathfrak{f}}_W}\biggr]^{-1} \mathfrak{m}^{2} </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow ~~~ [x_\mathrm{eq}]_\mathrm{crit}^{4(n+1)}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\biggl[ \frac{3^2\cdot 5 n}{4\pi(n-3)} \cdot \frac{1}{\tilde{\mathfrak{f}}_W} \biggr]^{-(n+1)} \biggl[ \frac{3^2 \cdot 5 n}{ 4\pi(n-3)} \cdot \frac{1}{\tilde{\mathfrak{f}}_W} \biggr]^{(3-n)} \biggl[ \biggl( \frac{3^2\cdot 5}{2^4\pi}\biggr) \frac{\tilde{\mathfrak{f}}_A}{\tilde{\mathfrak{f}}_W} \biggr]^{4n} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\biggl[ \frac{4\pi(n-3)}{3^2\cdot 5 n} \cdot \tilde{\mathfrak{f}}_W \biggr]^{2(n-1)} \biggl[ \biggl( \frac{3^2\cdot 5}{2^4\pi}\biggr) \frac{\tilde{\mathfrak{f}}_A}{\tilde{\mathfrak{f}}_W} \biggr]^{4n} </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow ~~~ [x_\mathrm{eq}]_\mathrm{crit}^{2(n+1)}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl[ \frac{n}{(n-3)} \biggl( \frac{3^2\cdot 5}{4\pi \tilde{\mathfrak{f}}_W} \biggr) \biggr]^{(1-n)} \biggl[ \biggl( \frac{3^2\cdot 5}{4\pi \tilde{\mathfrak{f}}_W}\biggr) \frac{\tilde{\mathfrak{f}}_A}{4} \biggr]^{2n} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl( \frac{n}{n-3} \biggr)^{(1-n)} \biggl( \frac{3^2\cdot 5}{4\pi \tilde{\mathfrak{f}}_W} \biggr)^{(n+1)} \biggl(\frac{\tilde{\mathfrak{f}}_A}{4} \biggr)^{2n} \, . </math> </td> </tr> </table> </div> <!-- THERE IS A MISTAKE IN THIS OMITTED SUBSECTION From an earlier derivation, we obtained, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\biggl[\frac{R_\mathrm{eq}}{R_\mathrm{SWS}} \biggr]^4_\mathrm{crit} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\biggl[ \frac{(n-3)}{3^2 (n+1)} \biggr] \frac{a}{c} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\biggl[ \frac{(n-3)}{3^2 (n+1)} \biggr] \biggl[ \frac{3}{5} \cdot \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2} \biggl( \frac{n+1}{n} \biggr)^{n/(n-3)} \biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{(n-1)/(n-3)} \biggr] \biggl[ \frac{3}{4\pi} \biggl( \frac{n+1}{n} \biggr)^{-3/(n-3)} \biggl( \frac{M_\mathrm{tot} }{M_\mathrm{SWS}} \biggr)^{(n-5)/(n-3)} \biggr] </math> </td> </tr> <tr> <td align="right"> <math>~[x_\mathrm{eq}]_\mathrm{crit}^4</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{(n-3)}{20\pi n} \biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{2} \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2} = \tilde{\mathfrak{f}}_W \cdot \frac{(n-3)}{15 n} \biggl(\frac{4\pi}{3}\biggr) \biggl[\biggl(\frac{3}{4\pi}\biggr)\frac{1}{\tilde{\mathfrak{f}}_M} \biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)\biggr]^{2} </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow ~~~\biggl[\biggl(\frac{3}{4\pi}\biggr)\frac{1}{\tilde{\mathfrak{f}}_M} \biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)\biggr] </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl[ \frac{1}{\tilde{\mathfrak{f}}_W} \cdot \frac{15 n}{(n-3)} \biggl(\frac{3}{4\pi}\biggr)\biggr]^{1/2} [x_\mathrm{eq}]_\mathrm{crit}^2</math> </td> </tr> </table> </div> which, when combined with the statement of virial equilibrium ''at'' this critical point, implies, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\tilde{\mathfrak{f}}_A \biggl\{ \biggl[ \frac{1}{\tilde{\mathfrak{f}}_W} \cdot \frac{15 n}{(n-3)} \biggl(\frac{3}{4\pi}\biggr)\biggr]^{1/2} [x_\mathrm{eq}]_\mathrm{crit}^2\biggr\}^{(n+1)/n} [x_\mathrm{eq}]_\mathrm{crit}^{(n-3)/n }</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ [x_\mathrm{eq}]_\mathrm{crit}^4 + \frac{1}{5} \biggl(\frac{4\pi}{3}\biggr) \cdot \tilde{\mathfrak{f}}_W \biggl( \frac{n+1}{n} \biggr) \biggl\{ \biggl[ \frac{1}{\tilde{\mathfrak{f}}_W} \cdot \frac{15 n}{(n-3)} \biggl(\frac{3}{4\pi}\biggr)\biggr]^{1/2} [x_\mathrm{eq}]_\mathrm{crit}^2\biggr\}^{2} </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow~~~ \tilde{\mathfrak{f}}_A \biggl[ \frac{1}{\tilde{\mathfrak{f}}_W} \cdot \frac{15 n}{(n-3)} \biggl(\frac{3}{4\pi}\biggr)\biggr]^{(n+1)/(2n)} [x_\mathrm{eq}]_\mathrm{crit}^{(3n-1)/n }</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ [x_\mathrm{eq}]_\mathrm{crit}^4 \biggl\{ 1 + \frac{1}{5} \biggl(\frac{4\pi}{3}\biggr) \cdot \tilde{\mathfrak{f}}_W \biggl( \frac{n+1}{n} \biggr) \biggl[ \frac{1}{\tilde{\mathfrak{f}}_W} \cdot \frac{15 n}{(n-3)} \biggl(\frac{3}{4\pi}\biggr)\biggr]^{2} \biggr\} </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow~~~ [x_\mathrm{eq}]_\mathrm{crit}^{(n+1)/n} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \tilde{\mathfrak{f}}_A \biggl[ \frac{1}{\tilde{\mathfrak{f}}_W} \cdot \frac{15 n}{(n-3)} \biggl(\frac{3}{4\pi}\biggr)\biggr]^{(n+1)/(2n)} \biggl\{ 1+ \frac{1}{5} \biggl(\frac{4\pi}{3}\biggr) \cdot \tilde{\mathfrak{f}}_W \biggl( \frac{n+1}{n} \biggr) \biggl[ \frac{1}{\tilde{\mathfrak{f}}_W} \cdot \frac{15 n}{(n-3)} \biggl(\frac{3}{4\pi}\biggr)\biggr]^{2} \biggr\}^{-1} </math> </td> </tr> </table> </div> END OMITTED SUBSECTION --> The following parallel derivation was done independently. [<font color="red">Note that a factor of 2n/(n-1) appears to correct a mistake made during the original derivation.</font>] Beginning with the virial expression, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\beta x^{(n-3)/n }_\mathrm{eq} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \alpha + x^{4}_\mathrm{eq} </math> </td> </tr> </table> </div> <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~ \biggl(\frac{3}{4\pi} \biggr)^{(n+1)/n} \frac{\tilde{\mathfrak{f}}_A}{\tilde{\mathfrak{f}}_M^{(n+1)/n}} \biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{(n+1)/n} [x_\mathrm{eq} ]^{(n-3)/n }_\mathrm{crit} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{3}{20\pi} \cdot \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2} \biggl( \frac{n+1}{n} \biggr)\biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{2} + \frac{(n-3)}{20\pi n} \biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{2} \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{(n-1)}{10\pi n} \biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{2} \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2} \biggl[ \frac{2n}{(n-1)}\biggr] </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow~~~ [x_\mathrm{eq} ]^{(n-3)/n }_\mathrm{crit} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{2(n-1)}{15 n} \biggl(\frac{4\pi}{3} \biggr)^{1/n} \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_A\tilde{\mathfrak{f}}_M^{(n-1)/n}} \cdot \biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{(n-1)/n} \biggl[ \frac{2n}{(n-1)}\biggr] </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow~~~ [x_\mathrm{eq} ]^{(n-3) }_\mathrm{crit} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl[\frac{2(n-1)}{15 n} \biggr]^n \biggl(\frac{4\pi}{3} \biggr) \frac{\tilde{\mathfrak{f}}_W^n}{\tilde{\mathfrak{f}}_A^n \tilde{\mathfrak{f}}_M^{(n-1)}} \biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{(n-1)} \biggl[ \frac{2n}{(n-1)}\biggr]^n </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl[\frac{2(n-1)}{15 n} \biggr]^n \biggl(\frac{4\pi}{3} \biggr) \frac{\tilde{\mathfrak{f}}_W^n}{\tilde{\mathfrak{f}}_A^n \tilde{\mathfrak{f}}_M^{(n-1)}} \biggl\{ \biggl[ \frac{20\pi n}{(n-3)} \biggr]^{(n-1)/2} \biggl( \frac{\tilde{\mathfrak{f}}_M^2}{\tilde{\mathfrak{f}}_W} \biggr)^{(n-1)/2} [x_\mathrm{eq} ]^{2(n-1) }_\mathrm{crit} \biggr\}\biggl[ \frac{2n}{(n-1)}\biggr]^n </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl[\frac{2(n-1)}{15 n} \biggr]^n \biggl(\frac{4\pi}{3} \biggr) \frac{\tilde{\mathfrak{f}}_W^{(n+1)/2}}{\tilde{\mathfrak{f}}_A^n } \biggl\{ \biggl[ \frac{20\pi n}{(n-3)} \biggr]^{(n-1)/2} [x_\mathrm{eq} ]^{2(n-1) }_\mathrm{crit} \biggr\}\biggl[ \frac{2n}{(n-1)}\biggr]^n </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow~~~ [x_\mathrm{eq} ]^{(n+1) }_\mathrm{crit} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl(\frac{3}{4\pi} \biggr) \biggl[\frac{15 n}{2(n-1)} \biggr]^n \biggl[ \frac{(n-3)}{20\pi n} \biggr]^{(n-1)/2} \frac{\tilde{\mathfrak{f}}_A^n }{\tilde{\mathfrak{f}}_W^{(n+1)/2}} \biggl[ \frac{(n-1)}{2n} \biggr]^n </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow~~~ [x_\mathrm{eq} ]^{(n+1) }_\mathrm{crit} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl(\frac{3}{4\pi} \biggr) \biggl[\frac{15 }{2^2} \biggr]^n \biggl[ \frac{(n-3)}{20\pi n} \biggr]^{(n-1)/2} \frac{\tilde{\mathfrak{f}}_A^n }{\tilde{\mathfrak{f}}_W^{(n+1)/2}} </math> </td> </tr> </table> </div> Also from [[SSC/Structure/PolytropesEmbedded#Stahler.27s_Presentation|Stahler's work]] we know that the equilibrium mass and radius are, <div align="center"> <table border="0" cellpadding="3"> <tr> <td align="right"> <math> ~\frac{M_\mathrm{tot}}{M_\mathrm{SWS}} </math> </td> <td align="center"> <math>~=~</math> </td> <td align="left"> <math> \biggl( \frac{n^3}{4\pi} \biggr)^{1/2} \biggl[ {\tilde\theta}_n^{(n-3)/2} {\tilde\xi}^2 (-{\tilde\theta}^') \biggr] \, , </math> </td> </tr> <tr> <td align="right"> <math> ~\frac{R_\mathrm{eq}}{R_\mathrm{SWS}} </math> </td> <td align="center"> <math>~=~</math> </td> <td align="left"> <math> \biggl( \frac{n}{4\pi} \biggr)^{1/2} \biggl[ \tilde\xi {\tilde\theta}_n^{(n-1)/2} \biggr] \, . </math> </td> </tr> </table> </div> Additional details in support of an associated PowerPoint presentation can be found [[SSC/FreeEnergy/PowerPoint|here]]. ====Reconcile==== <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\biggl[\frac{R_\mathrm{eq}}{R_\mathrm{SWS}} \biggr]^4_\mathrm{crit} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\biggl[ \frac{(n-3)}{20\pi (n+1)} \biggr] \biggl(\frac{n+1}{n}\biggr) \biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{2} \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2} </math> </td> </tr> </table> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\biggl( \frac{R_\mathrm{eq}}{R_\mathrm{norm}} \biggr)^4_\mathrm{crit}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{1}{20\pi} \biggl( \frac{n-3}{n+1} \biggr) \biggl( \frac{P_e}{P_\mathrm{norm}} \biggr)^{-1} \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2} </math> </td> </tr> </table> </div> Taking the ratio, the RHS is, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{2} \biggl( \frac{P_e}{P_\mathrm{norm}} \biggr)</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~P_e M_\mathrm{tot}^2 \biggl[ \frac{G^{3(n+1)} M_\mathrm{tot}^{2(n+1)} }{K^{4n}} \biggr]^{1/(n-3)} \biggl[ \biggl( \frac{n+1}{n} \biggr)^{3/2} G^{-3/2} K_n^{2n/(n+1)} P_\mathrm{e}^{(3-n)/[2(n+1)]} \biggr]^{-2} \biggl( \frac{n+1}{n}\biggr)</math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\biggl( \frac{n+1}{n} \biggr)^{-2}P_e M_\mathrm{tot}^2 \biggl[ G^{3} M_\mathrm{tot}^{2} \biggr]^{(n+1)/(n-3)} K_n^{-4n/(n-3)} \biggl[ G^{3} K_n^{-4n/(n+1)} P_\mathrm{e}^{(n-3)/(n+1)} \biggr]</math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\biggl( \frac{n+1}{n} \biggr)^{-2} \biggl[ G^{3} M_\mathrm{tot}^{2} \biggr]^{[(n-3)+(n+1)]/(n-3)} \biggl[ K_n^{[(n+1)+(n-3)]/[(n+1)(n-3)] } \biggr]^{-4n} P_\mathrm{e}^{[(n+1)+ (n-3)]/(n+1)} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\biggl( \frac{n+1}{n} \biggr)^{-2} M_\mathrm{tot}^{4(n-1)/(n-3)} G^{[6(n-1)]/(n-3)} K_n^{-8(n-1)/[(n+1)(n-3)] } P_\mathrm{e}^{2(n-1)/(n+1)} \, ;</math> </td> </tr> </table> </div> while the LHS is, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\biggl( \frac{R_\mathrm{norm}}{R_\mathrm{SWS}} \biggr)^{4}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl[\biggl( \frac{G}{K} \biggr)^n M_\mathrm{tot}^{(n-1)} \biggr]^{4/(n-3)} \biggl\{\biggl( \frac{n+1}{n} \biggr)^{1/2} G^{-1/2} K_n^{n/(n+1)} P_\mathrm{e}^{(1-n)/[2(n+1)]}\biggr\}^{-4} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\biggl( \frac{n+1}{n} \biggr)^{-2} M_\mathrm{tot}^{4(n-1)/(n-3)} G^{[6(n-1)]/(n-3)} K^{-8n(n-1)/[(n-3)(n+1)] } P_\mathrm{e}^{2(n-1)/(n+1)} \, . </math> </td> </tr> </table> </div> Q.E.D. Now, given that, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~M_\mathrm{SWS}^{-4(n-1)/(n-3)}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\biggl[\biggl( \frac{n+1}{n} \biggr)^{3/2} G^{-3/2} K_n^{2n/(n+1)} P_\mathrm{e}^{(3-n)/[2(n+1)]}\biggr]^{-4(n-1)/(n-3)} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\biggl( \frac{n+1}{n} \biggr)^{-6(n-1)/(n-3)} G^{6(n-1)/(n-3)} K_n^{-8n(n-1)/[(n+1)(n-3)]} P_\mathrm{e}^{2(n-1)/(n+1)} </math> </td> </tr> </table> </div> we have, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\biggl( \frac{R_\mathrm{norm}}{R_\mathrm{SWS}} \biggr)^{4}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\biggl( \frac{n+1}{n} \biggr)^{-2} \biggl(\frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{4(n-1)/(n-3)} \biggl( \frac{n+1}{n} \biggr)^{6(n-1)/(n-3)} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl(\frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{4(n-1)/(n-3)} \biggl( \frac{n+1}{n} \biggr)^{4n/(n-3)} </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow ~~~\biggl( \frac{R_\mathrm{norm}}{R_\mathrm{SWS}} \biggr)^{n-3}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl(\frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{n-1} \biggl( \frac{n+1}{n} \biggr)^{n} </math> </td> </tr> </table> </div> By inspection, in the specific case of <math>~n=5</math> (see above), this critical configuration appears to coincide with one of the [[SSC/Structure/PolytropesEmbedded#Other_Limits|"turning points" identified by Kimura]]. Specifically, it appears to coincide with the "extremal in r<sub>1</sub>" along an M<sub>1</sub> sequence, which satisfies the condition, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\biggl[ \frac{n-3}{n-1} \biggr]_{n=5}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\biggl[ \frac{\tilde\xi \tilde\theta^{n}}{(-\tilde\theta^')}\biggr]_{n=5}</math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow ~~~\frac{1}{2} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~3^{1/2}\ell \biggl[ (1 + \ell^2)^{-1/2} \biggr]^5 \biggl[ \frac{\ell}{3^{1/2}} (1+\ell^2 )^{-3/2} \biggr]^{-1}</math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~3(1 + \ell^2)^{-1} </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow~~~ \ell </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~5^{1/2} \, .</math> </td> </tr> </table> </div> Hence, according to Kimura, the turning point associated with the configuration with the largest equilibrium radius, corresponds to the equilibrium configuration having, <div align="center"> <math>~\ell |_\mathrm{R_{max}} = \sqrt{5} \approx 2.2360680 \, .</math> </div> This is, indeed, very close to — but decidedly different from — the value of <math>~\ell_\mathrm{crit}</math> determined, above! ====Streamlined==== Let's copy the expression for the [[#FirstPassFreeEnergy|"Case P" free energy derived above]], then factor out a common term: <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{\mathfrak{G}_{K,M}}{E_\mathrm{norm}} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ -3\mathcal{A} \biggl( \frac{n+1}{n} \biggr)^{n/(n-3)} \biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{(n-1)/(n-3)} \biggl(\frac{R}{R_\mathrm{SWS}}\biggr)^{-1} +~ n\mathcal{B} \biggl( \frac{n+1}{n} \biggr)^{3/(n-3)} \biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{3(n-1)/[n(n-3)]} \biggl(\frac{R}{R_\mathrm{SWS}}\biggr)^{-3/n} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ +~ \biggl( \frac{4\pi}{3} \biggr) \biggl( \frac{n+1}{n} \biggr)^{3/(n-3)} \biggl( \frac{M_\mathrm{tot} }{M_\mathrm{SWS}} \biggr)^{(5-n)/(n-3)} \biggl(\frac{R}{R_\mathrm{SWS}}\biggr)^3 \, . </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{(5-n)/(n-3)} \biggl(\frac{n+1}{n}\biggr)^{3/(n-3)} \biggl\{ -3\mathcal{A} \biggl( \frac{n+1}{n} \biggr) \biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{2} \biggl(\frac{R}{R_\mathrm{SWS}}\biggr)^{-1} +~ n\mathcal{B} \biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{(n+1)/n} \biggl(\frac{R}{R_\mathrm{SWS}}\biggr)^{-3/n} +\frac{4\pi}{3} \biggl(\frac{R}{R_\mathrm{SWS}}\biggr)^3 \biggr\} </math> </td> </tr> </table> </div> Defining a new normalization energy, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~E_\mathrm{SWS}</math> </td> <td align="center"> <math>~\equiv</math> </td> <td align="left"> <math>~E_\mathrm{norm} \biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{(5-n)/(n-3)} \biggl(\frac{n+1}{n}\biggr)^{3/(n-3)} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl(\frac{n+1}{n}\biggr)^{3/2} K^{3n/(n+1)} G^{-3/2} P_e^{(5-n)/[2(n+1)]} \, , </math> </td> </tr> </table> </div> we can write, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\mathfrak{G}_{K,M}^* \equiv \frac{\mathfrak{G}_{K,M}}{E_\mathrm{SWS}} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ -3\mathcal{A} \biggl( \frac{n+1}{n} \biggr) \biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{2} \biggl(\frac{R}{R_\mathrm{SWS}}\biggr)^{-1} +~ n\mathcal{B} \biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{(n+1)/n} \biggl(\frac{R}{R_\mathrm{SWS}}\biggr)^{-3/n} +\frac{4\pi}{3} \biggl(\frac{R}{R_\mathrm{SWS}}\biggr)^3 \, , </math> </td> </tr> </table> </div> in which case the coefficients of the generic free-energy expression are, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~a</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{3}{5} \cdot \frac{ \tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2} \biggl(\frac{n+1}{n}\biggr)\biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^2 = \frac{3}{5} \cdot \biggl( \frac{4\pi }{3}\biggr)^2 \biggl(\frac{n+1}{n}\biggr)\tilde{\mathfrak{f}}_W \mathfrak{m}^{2}</math> </td> </tr> <tr> <td align="right"> <math>~b</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ n\biggl(\frac{3}{4\pi}\biggr)^{1/n} \frac{\tilde{\mathfrak{f}}_A}{\tilde{\mathfrak{f}}_M^{(n+1)/n}} \biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{(n+1)/n} = \biggl( \frac{4\pi n}{3}\biggr) \tilde{\mathfrak{f}}_A \mathfrak{m}^{(n+1)/n} </math> </td> </tr> <tr> <td align="right"> <math>~c</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{4\pi}{3} \, , </math> </td> </tr> </table> </div> where, as above, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\mathfrak{m}</math> </td> <td align="center"> <math>~\equiv</math> </td> <td align="left"> <math>~\biggl( \frac{3}{4\pi} \biggr) \frac{ 1}{\tilde{\mathfrak{f}}_M} \biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr) \, .</math> </td> </tr> </table> </div> Now, if we define the pair of parameters, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\alpha</math> </td> <td align="center"> <math>~\equiv</math> </td> <td align="left"> <math>~\frac{a}{3c}</math> </td> </tr> <tr> <td align="right"> <math>~\beta</math> </td> <td align="center"> <math>~\equiv</math> </td> <td align="left"> <math>~\frac{b}{nc} \, ,</math> </td> </tr> </table> </div> then the statement of virial equilibrium is, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~x_\mathrm{eq}^4 + \alpha</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\beta x_\mathrm{eq}^{(n-3)/n} \, ,</math> </td> </tr> </table> </div> and the boundary between dynamical stability and instability occurs at, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~[x_\mathrm{eq}]_\mathrm{crit}^4</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\biggl[ \frac{n-3}{3(n+1)} \biggr]\alpha \, .</math> </td> </tr> </table> </div> Combining these last two expressions means that the boundary between dynamical stability and instability is associated with the parameter condition, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~[x_\mathrm{eq}]^{(n-3)/n}_\mathrm{crit} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\biggl[ \frac{n-3}{3(n+1)} + 1\biggr] \frac{\alpha}{\beta} </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow ~~~ \biggl\{ \biggl[ \frac{n-3}{3(n+1)} \biggr]\alpha \biggr\}^{(n-3)/(4n)} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl[ \frac{ 4n }{3(n+1)}\biggr] \frac{\alpha}{\beta} </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow ~~~ \beta \alpha^{-3(n+1)/(4n)} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl[ \frac{ 4n }{3(n+1)}\biggr] \biggl[ \frac{n-3}{n} \cdot \frac{n}{3(n+1)} \biggr]^{(3-n)/(4n)} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ 4 \biggl[ \frac{ n }{3(n+1)}\biggr]^{3(n+1)/(4n)} \biggl[ \frac{n-3}{n} \biggr]^{(3-n)/(4n)} </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow ~~~ \biggl( \frac{\beta}{4}\biggr)^{4n} \alpha^{-3(n+1)} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl[ \frac{ n }{3(n+1)}\biggr]^{3(n+1)} \biggl[ \frac{n-3}{n} \biggr]^{(3-n)} </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow ~~~ \biggl( \frac{\beta}{4}\biggr)^{4n} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl[ \frac{ n\alpha }{3(n+1)}\biggr]^{3(n+1)} \biggl[ \frac{n}{n-3} \biggr]^{n-3} \, . </math> </td> </tr> </table> </div> ======Case M====== <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~a</math> </td> <td align="center"> <math>~\equiv</math> </td> <td align="left"> <math>~3\mathcal{A} = \frac{3}{5} \cdot \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2}\, , </math> </td> </tr> <tr> <td align="right"> <math>~b</math> </td> <td align="center"> <math>~\equiv</math> </td> <td align="left"> <math>~n\mathcal{B} = n\biggl(\frac{4\pi}{3} \biggr)^{-1/n} \frac{\tilde{\mathfrak{f}}_A}{\tilde{\mathfrak{f}}_M^{(n+1)/n}} \, , </math> </td> </tr> <tr> <td align="right"> <math>~c</math> </td> <td align="center"> <math>~\equiv</math> </td> <td align="left"> <math>~\frac{4\pi}{3}\biggl( \frac{P_e}{P_\mathrm{norm}} \biggr) \, . </math> </td> </tr> </table> </div> Hence, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\alpha</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\biggl(\frac{4\pi }{15} \biggr) \tilde{\mathfrak{f}}_W \biggl(\frac{3}{4\pi \tilde{\mathfrak{f}}_M}\biggr)^2 \biggl( \frac{P_e}{P_\mathrm{norm}} \biggr)^{-1}</math> </td> </tr> <tr> <td align="right"> <math>~\beta</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\tilde{\mathfrak{f}}_A \biggl(\frac{3}{4\pi \tilde{\mathfrak{f}}_M}\biggr)^{(n+1)/n} \biggl( \frac{P_e}{P_\mathrm{norm}} \biggr)^{-1} \, .</math> </td> </tr> </table> </div> So the dynamical stability conditions are: <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\biggl( \frac{P_e}{P_\mathrm{norm}} \biggr)\biggl( \frac{n}{n-3} \biggr) [x_\mathrm{eq}]_\mathrm{crit}^4</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\biggl[ \biggl( \frac{n}{n+1} \biggr) \frac{4\pi \tilde{\mathfrak{f}}_W}{3^2\cdot 5} \biggr] \biggl(\frac{3}{4\pi \tilde{\mathfrak{f}}_M}\biggr)^2 \, ;</math> </td> </tr> </table> </div> and, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~ \biggl(\frac{\tilde{\mathfrak{f}}_A}{4}\biggr)^{4n} \biggl(\frac{3}{4\pi \tilde{\mathfrak{f}}_M}\biggr)^{4(n+1)} \biggl( \frac{P_e}{P_\mathrm{norm}} \biggr)^{-4n} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl[ \frac{ n}{3(n+1)}\biggr]^{3(n+1)} \biggl[ \frac{n}{n-3} \biggr]^{n-3} \biggl(\frac{4\pi \tilde{\mathfrak{f}}_W}{15} \biggr)^{3(n+1)} \biggl(\frac{3}{4\pi \tilde{\mathfrak{f}}_M}\biggr)^{6(n+1)} \biggl( \frac{P_e}{P_\mathrm{norm}} \biggr)^{-3(n+1)} </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow~~~ \biggl(\frac{\tilde{\mathfrak{f}}_A}{4}\biggr)^{4n} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl[\biggl(\frac{ n}{n+1}\biggr) \frac{4\pi \tilde{\mathfrak{f}}_W}{3^2\cdot 5} \biggr]^{3(n+1)} \biggl(\frac{3}{4\pi \tilde{\mathfrak{f}}_M}\biggr)^{2(n+1)} \biggl[ \frac{n}{n-3} \biggl( \frac{P_e}{P_\mathrm{norm}} \biggr)\biggr]^{n-3} </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow~~~\biggl[ \frac{n}{n-3} \biggl( \frac{P_e}{P_\mathrm{norm}} \biggr)\biggr]^{n-3} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl(\frac{\tilde{\mathfrak{f}}_A}{4}\biggr)^{4n} \biggl[\biggl(\frac{ n}{n+1}\biggr) \frac{4\pi \tilde{\mathfrak{f}}_W}{3^2\cdot 5} \biggr]^{-3(n+1)} \biggl(\frac{3}{4\pi \tilde{\mathfrak{f}}_M}\biggr)^{-2(n+1)} \, . </math> </td> </tr> </table> </div> Together, then, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~[x_\mathrm{eq}]_\mathrm{crit}^{4(n-3)}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\biggl[ \biggl( \frac{n}{n+1} \biggr) \frac{4\pi \tilde{\mathfrak{f}}_W}{3^2\cdot 5} \biggr]^{n-3} \biggl(\frac{3}{4\pi \tilde{\mathfrak{f}}_M}\biggr)^{2(n-3)} \biggl[ \biggl( \frac{P_e}{P_\mathrm{norm}} \biggr) \frac{n}{n-3} \biggr]^{-(n-3)} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\biggl[ \biggl( \frac{n}{n+1} \biggr) \frac{4\pi \tilde{\mathfrak{f}}_W}{3^2\cdot 5} \biggr]^{n-3} \biggl(\frac{3}{4\pi \tilde{\mathfrak{f}}_M}\biggr)^{2(n-3)} \biggl(\frac{\tilde{\mathfrak{f}}_A}{4}\biggr)^{-4n} \biggl[\biggl(\frac{ n}{n+1}\biggr) \frac{4\pi \tilde{\mathfrak{f}}_W}{3^2\cdot 5} \biggr]^{3(n+1)} \biggl(\frac{3}{4\pi \tilde{\mathfrak{f}}_M}\biggr)^{2(n+1)} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\biggl[ \biggl( \frac{n}{n+1} \biggr) \frac{4\pi \tilde{\mathfrak{f}}_W}{3^2\cdot 5} \biggr]^{4n} \biggl(\frac{3}{4\pi \tilde{\mathfrak{f}}_M}\biggr)^{4(n-1)} \biggl(\frac{\tilde{\mathfrak{f}}_A}{4}\biggr)^{-4n} </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow ~~~[x_\mathrm{eq}]_\mathrm{crit}^{(n-3)}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\biggl[ \frac{4}{\tilde{\mathfrak{f}}_A}\biggl( \frac{n}{n+1} \biggr) \frac{4\pi \tilde{\mathfrak{f}}_W}{3^2\cdot 5} \biggr]^n \biggl(\frac{3}{4\pi \tilde{\mathfrak{f}}_M}\biggr)^{(n-1)} \, . </math> </td> </tr> </table> </div> ======Case P====== <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~a</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{3}{5} \cdot \biggl( \frac{4\pi }{3}\biggr)^2 \biggl(\frac{n+1}{n}\biggr)\tilde{\mathfrak{f}}_W \mathfrak{m}^{2}</math> </td> </tr> <tr> <td align="right"> <math>~b</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\biggl( \frac{4\pi n}{3}\biggr) \tilde{\mathfrak{f}}_A \mathfrak{m}^{(n+1)/n} </math> </td> </tr> <tr> <td align="right"> <math>~c</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{4\pi}{3} \, , </math> </td> </tr> </table> </div> where, as above, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\mathfrak{m}</math> </td> <td align="center"> <math>~\equiv</math> </td> <td align="left"> <math>~\biggl( \frac{3}{4\pi} \biggr) \frac{ 1}{\tilde{\mathfrak{f}}_M} \biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr) \, .</math> </td> </tr> </table> </div> Hence, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\alpha</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{1}{5} \cdot \biggl( \frac{4\pi }{3}\biggr) \biggl(\frac{n+1}{n}\biggr)\tilde{\mathfrak{f}}_W \mathfrak{m}^{2} </math> </td> </tr> <tr> <td align="right"> <math>~\beta</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \tilde{\mathfrak{f}}_A \mathfrak{m}^{(n+1)/n} \, . </math> </td> </tr> </table> </div> So the dynamical stability conditions are: <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~[x_\mathrm{eq}]_\mathrm{crit}^4</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\biggl[ \frac{n}{3(n+1)} \biggr]\biggl[ \frac{n-3}{n} \biggr]\frac{1}{5} \cdot \biggl( \frac{4\pi }{3}\biggr) \biggl(\frac{n+1}{n}\biggr)\tilde{\mathfrak{f}}_W \mathfrak{m}^{2} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\biggl[ \frac{n-3}{n} \biggr]\biggl( \frac{4\pi \tilde{\mathfrak{f}}_W}{3^2\cdot 5}\biggr) \mathfrak{m}^{2} </math> </td> </tr> </table> </div> and, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~ \biggl( \frac{\tilde{\mathfrak{f}}_A }{4}\biggr)^{4n} \mathfrak{m}^{4(n+1)} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl[ \biggl( \frac{4\pi }{3^2\cdot 5}\biggr) \tilde{\mathfrak{f}}_W \mathfrak{m}^{2}\biggr]^{3(n+1)} \biggl[ \frac{n}{n-3} \biggr]^{n-3} </math> </td> </tr> <tr> <td align="right"> <math>~ \Rightarrow~~~ \mathfrak{m}^{2(n+1)} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl[ \biggl( \frac{4\pi }{3^2\cdot 5}\biggr) \tilde{\mathfrak{f}}_W \biggr]^{-3(n+1)} \biggl[ \frac{n}{n-3} \biggr]^{-(n-3)} \biggl( \frac{\tilde{\mathfrak{f}}_A }{4}\biggr)^{4n} \, . </math> </td> </tr> </table> </div> Together, then, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~[x_\mathrm{eq}]_\mathrm{crit}^{4(n+1)}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl[ \frac{n-3}{n} \biggr]^{(n+1)} \biggl( \frac{4\pi \tilde{\mathfrak{f}}_W}{3^2\cdot 5}\biggr)^{(n+1)} \biggl( \frac{4\pi \tilde{\mathfrak{f}}_W }{3^2\cdot 5}\biggr)^{-3(n+1)} \biggl[ \frac{n-3}{n} \biggr]^{(n-3)} \biggl( \frac{\tilde{\mathfrak{f}}_A }{4}\biggr)^{4n} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl[ \frac{n-3}{n} \biggr]^{2(n-1)} \biggl( \frac{4\pi \tilde{\mathfrak{f}}_W }{3^2\cdot 5}\biggr)^{-2(n+1)} \biggl( \frac{\tilde{\mathfrak{f}}_A }{4}\biggr)^{4n} \, . </math> </td> </tr> </table> </div> ======Compare====== Let's see if the two cases, in fact, provide the same answer. <!-- <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\biggl( \frac{R_\mathrm{norm}}{R_\mathrm{SWS}}\biggr)^{n-3} = \biggl[ \frac{x_\mathrm{P}}{x_\mathrm{M}} \biggr]_\mathrm{crit}^{n-3}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl\{ \biggl[ \frac{n-3}{n} \biggr]^{2(n-1)} \biggl( \frac{4\pi \tilde{\mathfrak{f}}_W }{3^2\cdot 5}\biggr)^{-2(n+1)} \biggl( \frac{\tilde{\mathfrak{f}}_A }{4}\biggr)^{4n} \biggr\}^{(n-3)/[4(n+1)]} \biggl\{ \biggl[ \frac{4}{\tilde{\mathfrak{f}}_A}\biggl( \frac{n}{n+1} \biggr) \frac{4\pi \tilde{\mathfrak{f}}_W}{3^2\cdot 5} \biggr]^n \biggl(\frac{3}{4\pi \tilde{\mathfrak{f}}_M}\biggr)^{(n-1)} \biggr\}^{-1} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\biggl(\frac{n-3}{n}\biggr)^{2(n-1)(n-3)/[4(n+1)]} \biggl( \frac{4\pi \tilde{\mathfrak{f}}_W }{3^2\cdot 5}\biggr)^{-n -(n-3)/2} \biggl( \frac{\tilde{\mathfrak{f}}_A }{4}\biggr)^{n+n(n-3)/(n+1)} \biggl( \frac{n+1}{n} \biggr)^{n} \biggl(\frac{3}{4\pi \tilde{\mathfrak{f}}_M}\biggr)^{1-n} </math> </td> </tr> </table> </div> --> <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\biggl( \frac{R_\mathrm{norm}}{R_\mathrm{SWS}}\biggr)^{n-3} = \biggl[ \frac{x_\mathrm{P}}{x_\mathrm{M}} \biggr]_\mathrm{crit}^{n-3}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl\{ \biggl[ \frac{n-3}{n} \biggr]\biggl( \frac{4\pi \tilde{\mathfrak{f}}_W}{3^2\cdot 5}\biggr) \mathfrak{m}^{2} \biggr\}^{(n-3)/4} \biggl\{ \biggl[ \frac{4}{\tilde{\mathfrak{f}}_A}\biggl( \frac{n}{n+1} \biggr) \frac{4\pi \tilde{\mathfrak{f}}_W}{3^2\cdot 5} \biggr]^n \biggl(\frac{3}{4\pi \tilde{\mathfrak{f}}_M}\biggr)^{(n-1)} \biggr\}^{-1} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl\{ \biggl[ \frac{n-3}{n} \biggr]\biggl( \frac{4\pi \tilde{\mathfrak{f}}_W}{3^2\cdot 5}\biggr) \mathfrak{m}^{2} \biggr\}^{(n-3)/4} \biggl\{ \biggl[ \frac{4}{\tilde{\mathfrak{f}}_A}\biggl( \frac{n}{n+1} \biggr) \frac{4\pi \tilde{\mathfrak{f}}_W}{3^2\cdot 5} \biggr]^n \biggl(\frac{3}{4\pi \tilde{\mathfrak{f}}_M}\biggr)^{(n-1)} \biggr\}^{-1} </math> </td> </tr> </table> </div> ===Five-One Bipolytropes=== For analytically prescribed, "five-one" bipolytropes, <math>~n = 5</math> and <math>~j =1</math>, in which case, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>x^{2/5 }_\mathrm{eq}</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math>\biggl(\frac{5}{ 3b}\biggr) \biggl[a -3 c x^{-2}_\mathrm{eq} \biggr] \, ;</math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> and </td> <td align="left"> </td> </tr> <tr> <td align="right"> <math>[x_\mathrm{eq}]_\mathrm{crit} </math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math>\biggl[ \frac{18 c}{a }\biggr]^{1/2} \, . </math> </td> </tr> </table> More specifically, [[#BiPolytrope51|the expression that describes the free-energy surface]] is, <div align="center" id="FreeEnergy51"> <table border="1" cellpadding="5" align="center"> <tr><td align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>\mathfrak{G}^*_{51} \equiv 2^4\biggl( \frac{q}{\nu^2}\biggr) \chi_\mathrm{eq} \biggl[\frac{\mathfrak{G}_{51}}{E_\mathrm{norm}} \biggr]</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math>\frac{1}{\ell_i^2} \biggl[ \Chi^{-3/5} (5 \mathfrak{L}_i) +\Chi^{-3} (4\mathfrak{K}_i) -\Chi^{-1} (3\mathfrak{L}_i +12\mathfrak{K}_i ) \biggr] \, . </math> </td> </tr> </table> </td></tr> </table> </div> Hence, we have, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>a</math> </td> <td align="center"> <math>\equiv</math> </td> <td align="left"> <math>3\chi_\mathrm{eq}(\mathfrak{L}_i + 4\mathfrak{K}_i) \, , </math> </td> </tr> <tr> <td align="right"> <math>b</math> </td> <td align="center"> <math>\equiv</math> </td> <td align="left"> <math>5 \mathfrak{L}_i \chi_\mathrm{eq}^{3/5} \, , </math> </td> </tr> <tr> <td align="right"> <math>c</math> </td> <td align="center"> <math>\equiv</math> </td> <td align="left"> <math>4 \mathfrak{K}_i \chi_\mathrm{eq}^{3} \, , </math> </td> </tr> </table> and conclude that, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>[\chi_\mathrm{eq}]_\mathrm{crit}</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math>\biggl[ \frac{18 (4 \mathfrak{K}_i \chi_\mathrm{eq}^{3} )}{ 3\chi_\mathrm{eq}(\mathfrak{L}_i + 4\mathfrak{K}_i)} \biggr]^{1/2}_\mathrm{crit} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math>[\chi_\mathrm{eq}]_\mathrm{crit}\biggl[ \frac{24 \mathfrak{K}_i }{ (\mathfrak{L}_i + 4\mathfrak{K}_i)} \biggr]^{1/2} </math> </td> </tr> <tr> <td align="right"> <math>\Rightarrow~~~\biggl[ \frac{24 \mathfrak{K}_i }{ (\mathfrak{L}_i + 4\mathfrak{K}_i)} \biggr]_\mathrm{crit}</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math>1 </math> </td> </tr> <tr> <td align="right"> <math>\Rightarrow~~~\biggl[ \frac{\mathfrak{L}_i }{ \mathfrak{K}_i } \biggr]_\mathrm{crit}</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math>20 \, . </math> </td> </tr> </table> </div> <span id="FiveOneRadius">Also, from our [[#Summary51|detailed force balance derivations]], we know that,</span> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>\chi_\mathrm{eq} \equiv \frac{ R_\mathrm{eq}}{R_\mathrm{norm}}</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math>\biggl(\frac{\pi}{2^3}\biggr)^{1/2} \frac{\nu^2}{q} \cdot \frac{(1+\ell_i^2)^3}{3^3\ell_i^5} \, .</math> </td> </tr> </table> ===Zero-Zero Bipolytropes=== ====General Form==== In this case, we retain full generality making the substitutions, <math>~n \rightarrow n_c</math> and <math>~j \rightarrow n_e</math>, to obtain, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~x^{(n_c-3)/n_c }_\mathrm{eq}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{n_c}{ 3b} \biggl[a -\biggl(\frac{3 c}{n_e}\biggr) x^{(n_e-3)/n_e}_\mathrm{eq} \biggr] \, ;</math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> and </td> <td align="left"> </td> </tr> <tr> <td align="right"> <math>~[x_\mathrm{eq}^{(n_e-3)/n_e}]_\mathrm{crit} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\biggl\{\frac{n_e^2(n_c-3)}{3[ n_c (n_e+3) - n_e(n_c+3) ]}\biggr\} \frac{a}{c} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\biggl[\frac{n_e^2(n_c-3)}{3^2(n_c - n_e)}\biggr] \frac{a}{c} \, . </math> </td> </tr> </table> </div> And here, [[#BiPolytrope00|the expression that describes the free-energy surface]] is, <div align="center" id="FreeEnergy00"> <table border="1" cellpadding="5" align="center"> <tr><td align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\mathfrak{G}^*_{00} \equiv 5 \biggl(\frac{q}{\nu^2}\biggr) \chi_\mathrm{eq} \biggl[\frac{\mathfrak{G}_{00}}{E_\mathrm{norm}} \biggr]</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{5}{2q^3} \biggl[ n_c A_2\Chi^{-3/n_c} + n_e B_2\Chi^{-3/n_e} - 3C_2\Chi^{-1} \biggr] \, . </math> </td> </tr> </table> </td></tr> </table> </div> Hence, we have, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~a \equiv 3\chi_\mathrm{eq} \biggl(\frac{5}{2q^3} \biggr) C_2 </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math> 3f \chi_\mathrm{eq} \, , </math> </td> </tr> <tr> <td align="right"> <math>~b \equiv n_c \chi_\mathrm{eq}^{3/n_c} \biggl(\frac{5}{2q^3} \biggr) A_2 </math> </td> <td align="center"> <math>~\equiv</math> </td> <td align="left"> <math> n_c \chi_\mathrm{eq}^{3/n_c} \biggl[1 + q^3 (f - 1-\mathfrak{F} ) \biggr] \, , </math> </td> </tr> <tr> <td align="right"> <math>~c \equiv n_e \chi_\mathrm{eq}^{3/n_e} \biggl(\frac{5}{2q^3} \biggr) B_2 </math> </td> <td align="center"> <math>~\equiv</math> </td> <td align="left"> <math>~ n_e \chi_\mathrm{eq}^{3/n_e} \biggl(\frac{5}{2q^3} \biggr) \biggl[\frac{2}{5} q^3 f - A_2\biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ n_e \chi_\mathrm{eq}^{3/n_e} \biggl\{ f - \biggl[1 + q^3 (f - 1-\mathfrak{F} ) \biggr] \biggr\} \, , </math> </td> </tr> </table> </div> where the definitions of <math>~f</math> and <math>~\mathfrak{F}</math> are [[#BiPolytrope00|given below]]. We immediately deduce that the ''critical'' equilibrium state is identified by, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~[x_\mathrm{eq}^{(n_e-3)/n_e}]_\mathrm{crit} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\biggl\{\frac{fn_e(n_c-3)}{3(n_c - n_e)}\biggr\} [\chi_\mathrm{eq}^{(n_e-3)/n_e}]_\mathrm{crit} \biggl\{ f - \biggl[1 + q^3 (f - 1-\mathfrak{F} ) \biggr] \biggr\}^{-1} </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow ~~~\frac{1}{f}\biggl[1 + q^3 (f - 1-\mathfrak{F} ) \biggr]</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ 1 - \biggl[ \frac{n_e(n_c-3)}{3(n_c-n_e)} \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{n_c(3-n_e)}{3(n_c-n_e)} \, .</math> </td> </tr> </table> </div> From our [[#Equilibrium_Radius_2|associated detailed-force-balance derivation]], we know that the associated equilibrium radius is, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\chi_\mathrm{eq}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl\{ \biggl(\frac{\pi}{3}\biggr) 2^{2-n_c} \nu^{n_c-1} q^{3-n_c} \biggl[1 + \frac{2}{5} q^3(f-1-\mathfrak{F})\biggr]^{n_c} \biggr\}^{1/(n_c-3)} \, . </math> </td> </tr> </table> </div> <!-- Coefficient mistake, I think! We have deduced that the system is unstable if, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{n_e}{3}\biggl[ \frac{3-n_e}{n_c-n_e} \biggr] </math> </td> <td align="center"> <math>~< </math> </td> <td align="left"> <math>~ \frac{A_2}{C_2} = \frac{1}{f} \biggl[1 + q^3 (f - 1-\mathfrak{F} ) \biggr] \, . </math> </td> </tr> </table> </div> --> ====Compare with Five-One==== It is worthwhile to set <math>~n_c = 5</math> and <math>~n_e = 1</math> in this expression and compare the result to the [[#FiveOneRadius|comparable expression shown above for the "Five-One" Bipolytrope]]. Here we have, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\biggl[\chi_\mathrm{eq}\biggr]_{51}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl\{ \biggl(\frac{\pi}{3}\biggr) 2^{-3} \nu^{4} q^{-2} \biggl[1 + \frac{2}{5} q^3(f-1-\mathfrak{F})\biggr]^{5} \biggr\}^{1/2} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl(\frac{\pi}{2^3}\biggr)^{1/2} \frac{\nu^2}{q} \cdot \frac{1}{\sqrt{3}} \biggl[1 + \frac{2}{5} q^3(f-1-\mathfrak{F})\biggr]^{5/2} \, ; </math> </td> </tr> </table> </div> whereas, rewriting the [[#FiveOneRadius|above relation]] gives, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\chi_\mathrm{eq}\biggr|_{51}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\biggl(\frac{\pi}{2^3}\biggr)^{1/2} \frac{\nu^2}{q} \cdot \frac{1}{\sqrt{3}} \biggl[\frac{(1+\ell_i^2)^{6/5}}{3\ell_i^2}\biggr]^{5/2} \, .</math> </td> </tr> </table> </div> And, here, we should conclude that the ''critical'' equilibrium configuration is associated with, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{1}{f}\biggl[1 + q^3 (f - 1-\mathfrak{F} ) \biggr]</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{5}{6} \, .</math> </td> </tr> <!-- <tr> <td align="right"> <math>~\Rightarrow~~~q^3 (f - 1-\mathfrak{F} )</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{5}{6} \cdot f - 1</math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow~~~\biggl[1 + \frac{2}{5} q^3(f-1-\mathfrak{F})\biggr]_\mathrm{crit} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ 1 + \frac{2}{5}\biggl(\frac{5}{6} \cdot f - 1\biggr)</math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{f}{3} + \frac{3}{5}</math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow ~~~ [\chi_\mathrm{eq}]_\mathrm{crit}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl(\frac{\pi}{2^3}\biggr)^{1/2} \frac{\nu^2}{q} \cdot \frac{1}{\sqrt{3}} \biggl[\frac{f}{3} + \frac{3}{5}\biggr]^{5/2} \, . </math> </td> </tr> --> </table> </div>
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