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===Behavior of First Logarithmic Derivative=== <span id="LogarithmicDerivative">For the record,</span> let's develop an expression for the logarithmic derivative of this displacement function. <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>\frac{d\ln x}{d\ln \xi} = \frac{\xi}{x} \cdot \frac{dx}{d\xi}</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \biggl(\frac{3-n}{n-1}\biggr) \biggl[ 1 + \frac{n(\theta^')^2 }{\theta^{n+1}} + \frac{3\theta^' }{\xi \theta^{n}} \biggr] \biggl[ 1 + \biggl(\frac{n-3}{n-1}\biggr) \frac{\theta^' }{\xi \theta^{n}} \biggr]^{-1} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> - (n-3) \biggl[ 1 + \frac{n(\theta^')^2 }{\theta^{n+1}} + \frac{3\theta^' }{\xi \theta^{n}} \biggr] \biggl[ (n-1) + (n-3) \frac{\theta^' }{\xi \theta^{n}} \biggr]^{-1} </math> </td> </tr> <tr> <td align="right"> <math>\Rightarrow ~~~ \biggl[ (n-1) + (n-3) \frac{\theta^' }{\xi \theta^{n}} \biggr] \frac{d\ln x}{d\ln \xi} </math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> - (n-3) \biggl[ 1 + \frac{n(\theta^')^2 }{\theta^{n+1}} + \frac{3\theta^' }{\xi \theta^{n}} \biggr] </math> </td> </tr> <tr> <td align="right"> <math>\Rightarrow ~~~ \biggl[ (n-1) \biggr] \frac{d\ln x}{d\ln \xi} </math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> - (n-3) \biggl\{ \biggl[ 1 + \frac{n(\theta^')^2 }{\theta^{n+1}} + \frac{3\theta^' }{\xi \theta^{n}} \biggr] + \biggl[ \frac{\theta^' }{\xi \theta^{n}} \biggr] \frac{d\ln x}{d\ln \xi} \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> - (n-3) \biggl\{ \biggl[ 1 + \frac{n(\theta^')^2 }{\theta^{n+1}} \biggr] + \biggl[ \frac{\theta^' }{\xi \theta^{n}} \biggr] \biggl[3 + \frac{d\ln x}{d\ln \xi}\biggr] \biggr\} </math> </td> </tr> </table> </div> Now, as we have [[#From_Yabushita.27s_.281992.29_Analysis|discussed above]], {{ Yabushita92 }} has argued that for pressure-truncated configurations, the appropriate surface boundary condition should be, <div align="center"> <math>\frac{d\ln x}{d\ln\xi} = -3 \, .</math> </div> If we adopt this surface boundary condition, then this just-derived expression becomes, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>3 (n-1) </math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> (n-3) \biggl[ 1 + \frac{n(\theta^')^2 }{\theta^{n+1}} \biggr] </math> </td> </tr> <tr> <td align="right"> <math>\Rightarrow~~~ (n-3) \biggl[ \frac{n(\theta^')^2 }{\theta^{n+1}} \biggr] </math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> 2n </math> </td> </tr> <tr> <td align="right"> <math>\Rightarrow~~~ \frac{(\theta^')^2 }{\theta^{n+1}} </math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \frac{2 }{(n-3) } \, . </math> </td> </tr> </table> </div> According to {{ Horedt70 }} — see our [[SSC/Structure/PolytropesEmbedded#Horedt.27s_Derivation|accompanying discussion]] — this is precisely the functional condition that pinpoints the <math>~P_e</math>-max turning point(s) along the equilibrium sequence in a <math>~P-V</math> diagram, for all pressure-truncated, polytropic spheres with <math>~n \ge 3</math>. And, according to {{ Kimura81b }} — again, see our [[SSC/Structure/PolytropesEmbedded#Other_Limits|accompanying discussion]] — this is precisely the functional condition that pinpoints the maximum mass turning point(s) along equlibrium sequences as displayed in a mass-radius diagram. <!-- CHRISTY OR COX BOUNDARY CONDITIONS ... On the other hand, as we have [[SSC/Perturbations#ChristyCox|discussed separately]] — see also, [[#From_Yabushita.27s_.281992.29_Analysis|above]] — in order to align with the separate derivations presented by [http://adsabs.harvard.edu/abs/1966ARA%26A...4..353C Christy (1965)] and [http://adsabs.harvard.edu/abs/1967IAUS...28....3C Cox (1967)] — the corresponding boundary condition at the surface is, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{dh}{dx}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~- \frac{h}{x} \biggr[ 3 - \frac{4}{\gamma} + \cancelto{0}{\frac{x s^2}{\gamma q}} \biggr]</math> </td> <td align="left" colspan="2"> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{n-3}{n+1} \biggl(\frac{h}{x} \biggr) \, .</math> </td> <td align="left"> at </td> <td align="left"> <math>~x = x_0 \, .</math> </td> </tr> </table> </div> -->
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