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====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!
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