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====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]].
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