Difference between revisions of "User:Tohline/SSC/Stability/n5PolytropeLAWE"

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   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl( 1 + \frac{\xi^2}{3} \biggr)^{-1/2} \, .</math>
<math>~\biggl( 1 + \frac{\xi^2}{3} \biggr)^{-1/2} =3^{1/2} ( 3 + \xi^2 )^{-1/2}\, .</math>
   </td>
   </td>
</tr>
</tr>
Line 110: Line 110:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~- \frac{\xi}{3}\biggl( 1 + \frac{\xi^2}{3} \biggr)^{-3/2}\, .</math>
<math>~- \frac{\xi}{3}\biggl( 1 + \frac{\xi^2}{3} \biggr)^{-3/2} = - 3^{1/2}\xi ( 3 + \xi^2 )^{-3/2} \, .</math>
   </td>
   </td>
</tr>
</tr>
Line 128: Line 128:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{d^2x}{d\xi^2} + \frac{5}{3}\biggl[\frac{12}{5} - \xi^2 \biggl( 1 + \frac{\xi^2}{3} \biggr)^{-1}  \biggr] \frac{1}{\xi}\cdot \frac{dx}{d\xi}
<math>~\frac{d^2x}{d\xi^2} + \biggl[\frac{4}{\xi} - \frac{(n+1)}{\theta} \biggl(- \frac{d\theta}{d\xi}\biggr)\biggr] \frac{dx}{d\xi} +
+ \frac{5}{6} \biggl( 1 + \frac{\xi^2}{3} \biggr)^{1/2} \biggl[\frac{\sigma_c^2}{\gamma_g } -  
\frac{(n+1)}{\theta}\biggl[\frac{\sigma_c^2}{6\gamma_g } -  
2\alpha \biggl( 1 + \frac{\xi^2}{3} \biggr)^{-3/2}\biggr]  x \, .</math>
\frac{\alpha}{\xi} \biggl(- \frac{d\theta}{d\xi}\biggr)\biggr]  x </math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
=Search for Analytic Solutions to the LAWE=
==Setup Using Lagrangian Radial Coordinate==
===Individual Terms===
From our [[User:Tohline/SSC/FreeEnergy/PowerPoint#Case_M_Equilibrium_Conditions|accompanying discussion]], we have, for pressure-truncated, <math>~n=5</math> polytropic spheres
<div align="center">
<table border="0" cellpadding="3">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>
&nbsp;
~\frac{R_\mathrm{eq}}{R_\mathrm{norm}}
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl[ \frac{4\pi}{(n+1)^n}\biggr]^{1/(n-3)}
<math>~\frac{d^2x}{d\xi^2} + \biggl[\frac{4}{\xi} - \frac{6}{\theta} \biggl(- \frac{d\theta}{d\xi}\biggr)\biggr] \frac{dx}{d\xi} +
\tilde\xi ( -\tilde\xi^2 \tilde\theta' )^{(1-n)/(n-3)}
\biggl[\frac{\sigma_c^2}{\gamma_g } \cdot \frac{1}{\theta} -  
</math>
\frac{6\alpha}{\xi} \cdot \frac{1}{\theta} \biggl(- \frac{d\theta}{d\xi}\biggr)\biggr]  x</math>
   </td>
   </td>
</tr>
</tr>
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   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl[ \frac{4\pi}{2^5\cdot 3^5}\biggr]^{1/2}
<math>~\frac{d^2x}{d\xi^2} + \biggl[4 - \frac{6\xi^2}{(3+\xi^2)} \biggr] \frac{1}{\xi} \cdot \frac{dx}{d\xi} +
\tilde\xi ( -\tilde\xi^2 \tilde\theta' )^{-2} \, ,
\biggl[\frac{\sigma_c^2}{3^{1/2} \gamma_g } \cdot (3+\xi^2)^{1/2} -  
</math>
\frac{6\alpha}{(3+\xi^2)}\biggr)\biggr]  x \, .</math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</div>
</div>
which matches the expression derived in an [[User:Tohline/SSC/Structure/Polytropes#Lane-Emden_Equation|ASIDE box found with our introduction of the Lane-Emden equation]], and
 
<div align="center">
Or,
<table border="0" cellpadding="3">
<div align="center" id="n5LAWE">
<table border="1" cellpadding="8" align="center">
<tr><th align="center">LAWE for <math>~n=5</math> Polytropes</th></tr>
<tr><td align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>
<math>~0</math>
~\frac{P_\mathrm{e}}{P_\mathrm{norm}}
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">  
<math>~\biggl[ \frac{(n+1)^3}{4\pi}\biggr]^{(n+1)/(n-3)}
<math>~(3+\xi^2) \frac{d^2x}{d\xi^2} + \biggl[12 - 2\xi^2 \biggr] \frac{1}{\xi} \cdot \frac{dx}{d\xi} +
\tilde\theta_n^{n+1}( -\tilde\xi^2 \tilde\theta' )^{2(n+1)/(n-3)}
\biggl[\frac{\sigma_c^2}{3^{1/2} \gamma_g } \cdot (3+\xi^2)^{3/2} -  
</math>
6\alpha \biggr]  x </math>
   </td>
   </td>
</tr>
</tr>
</table>
</td></tr></table>
</div>
==Numerical Integration of LAWE==
By numerically integrating the above LAWE using the algorithm outlined in a [[User:Tohline/SSC/Stability/Polytropes#Numerical_Integration_from_the_Center.2C_Outward|separate chapter]], we have examined the properties of the displacement function that describes radial modes of oscillation in pressure-truncated, n = 5, polytropic configurations. Our brief description, here, of these modes parallels the more detailed description of radial oscillation modes in truncated isothermal spheres that has been [[User:Tohline/SSC/Stability/Isothermal#Previously_Published_Eigenvalues_and_Eigenfunctions|presented in a separate chapter]].
The animation sequence that appears in the right panel of Composite Display 1 shows how our numerically derived displacement function, <math>~x(\xi)</math>, varies with radius &#8212; from the center of the n=5 polytropic sphere, out to <math>~\xi = 10</math> &#8212; for sixteen different values of the square of the eigenfrequency, <math>~\sigma_c^2</math>, as denoted at the top of each animation frame. The segment of the <math>~x(\xi)</math> curve that has been drawn in blue identifies the ''eigenfunction'' that corresponds to the specified value of the eigenfrequency.  In each frame, the radial location at which the blue segment terminates simultaneously identifies: (a) the radius at which the logarithmic derivative of the displacement function, <math>~d\ln x/d\ln\xi </math>, is negative three; and (b) the radius, <math>~\tilde\xi</math>, at which the n = 5 polytropic configuration has been truncated.  As displayed here, in every frame, the <math>~x(\xi)</math> function has been normalized such that the displacement amplitude is unity at the truncated configuration's surface. 
The left panel of Composite Display 1 is also animated and has been provided in support of the animation on the right.  Specifically, the number written at the top of each left-panel frame quantitatively identifies the radial location, <math>~\tilde\xi</math>, of the surface of the relevant truncated polytropic configuration; and, on each frame, "&times;" marks the location of that truncated configuration on the mass-radius equilibrium sequence.


<div align="center" id="n5TruncatedMovie">
<table border="1" align="center" cellpadding="8">
<tr>
<tr>
   <td align="right">
   <th align="center" colspan="2">Composite Display 1: &nbsp; Numerically Generated Fundamental-Mode Eigenvectors</th>
&nbsp;
</tr>
<tr>
  <td align="center">
[[File:N5Truncated2.gif|600px|n5 Truncated movie]]
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
Excel File:<br />
  </td>
[[File:DataFileButton02.png|right|60px|file = Dropbox/WorkFolder/Wiki edits/LinearPerturbation/n5Eigenvectors/n5TruncatedSphere.xlsx --- worksheet = OursPt1]]
  <td align="left">
Movie File:<br />
<math>~\biggl[ \frac{2^3\cdot 3^3}{4\pi}\biggr]^{3}
[[File:DataFileButton02.png|right|60px|file = Dropbox/WorkFolder/Wiki edits/EmbeddedPolytropes/n5movie/ --- worksheet = n5Truncated2.gif]]
\tilde\theta^{6}( -\tilde\xi^2 \tilde\theta' )^{6} \, ,
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</div>
</div>
where,
 
<div align="center">
Each full loop through the left-panel animation sequence can be viewed as evolution along the equilibrium sequence from <math>~\tilde\xi = 0.75</math> to  <math>~\tilde\xi = 5</math>, then back again.  During this evolution, the "&times;" marker moves through both turning points along the sequence:  the maximum radius configuration &#8212; at <math>~\tilde\xi= \sqrt{3}</math> &#8212; and the maximum mass configuration  &#8212; at <math>~\tilde\xi= 3</math>.  Notice that <math>~\sigma_c^2</math> is positive for all models having <math>~\tilde\xi < 3</math> while it is negative for all models having  <math>~\tilde\xi > 3</math>.  Hence, models having <math>~\tilde\xi > 3</math> are dynamically unstable and, as best we have been able to determine via these numerical integrations, the transition from stable to unstable models &#8212; that is, the ''marginally'' unstable model &#8212; occurs at <math>~\tilde\xi = 3</math>.  (Via an ''analytic'' analysis, we prove, below, that this association is precise.)  For emphasis, the "&times;" marker (left panel) and the numerical value recorded for <math>~\sigma_c^2</math> (right panel) have been colored red for models that are not stable.
<table border="0" cellpadding="3">
 
=Search for Analytic Solutions to the LAWE=
 
==Eureka Moment==
 
<font color="red">Note from J. E. Tohline on 3/6/2017:</font>&nbsp;  Yesterday evening, after I finished putting together the [[#n5TruncatedMovie|above animation sequence]] using an Excel workbook, I noticed that the eigenfunction of the fundamental mode for the marginally unstable model <math>~(\sigma_c^2 = 0)</math> resembles a parabola.  In an effort to see how well a parabola fits at least the central portion of this eigenfunction, I returned to my Excel spreedsheet and, in a brute-force manner, began to search for the pair of coefficients that would provide a best fit.  What I discovered was that a parabola with the following formula fits perfectly!
<div align="center" i="AnalyticSoln">
<table border="1" cellpadding="10" align="center">
<tr>
  <th align="center" colspan="1">Fundamental Mode Eigenfunction <br />
when <math>~\sigma_c^2 = 0</math> and <math>~\gamma = 6/5 ~\Rightarrow~\alpha=- 1/3</math></th>
</tr>


<tr>
<tr>
  <td align="right">
<math>~R_\mathrm{norm}</math>
  </td>
   <td align="center">
   <td align="center">
<math>~\equiv</math>
<math>~x = x_0 \biggl[ 1 - \frac{\xi^2}{15} \biggr]</math>
  </td>
  <td align="left">
<math>~\biggl[ \biggl( \frac{G}{K} \biggr)^n M_\mathrm{tot}^{n-1} \biggr]^{1/(n-3)} = \biggl( \frac{G}{K} \biggr)^{5/2} M_\mathrm{tot}^{2} \, ,</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
For the ''specific'' normalization used in the above animation sequence, <math>~x_0 = \tfrac{5}{2}</math>.  Let's demonstrate that this eigenvector provides a solution to the LAWE for <math>~n=5</math> polytropes; for simplicity, we will set <math>~x_0 = 1</math>:
<div align="center" id="Proof">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~P_\mathrm{norm}</math>
<math>~\frac{dx}{d\xi} = -\frac{2\xi}{15} \, ;</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\equiv</math>
&nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp;
   </td>
   </td>
   <td align="left">
   <td align="left">  
<math>~\biggl[ \frac{K^{4n}}{G^{3(n+1)} M_\mathrm{tot}^{2(n+1)}} \biggr]^{1/(n-3)}  = \frac{K^{10}}{G^{9} M_\mathrm{tot}^{6} \, ,</math>
<math>~\frac{d^2x}{d\xi^2} = -\frac{2}{15} \, .</math>
   </td>
   </td>
</tr>
</tr>
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</div>
</div>


and, from [[User:Tohline/SSC/Structure/PolytropesEmbedded#Tabular_Summary_.28n.3D5.29|our more detailed analysis]],
<div align="center" id="Proof">
<table border="0" cellpadding="5" align="center">


<table border="0" cellpadding="3" align="center">
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>
<math>~\Rightarrow ~~~
~{\tilde\theta}_5 = 3^{1 / 2} \biggl( 3 + {\tilde\xi}^2\biggr)^{-1/2}
(3+\xi^2) \frac{d^2x}{d\xi^2} + \biggl[12 - 2\xi^2 \biggr] \frac{1}{\xi} \cdot \frac{dx}{d\xi} +
\biggl[ \cancelto{0}{\frac{\sigma_c^2}{3^{1/2} \gamma_g }} \cdot (3+\xi^2)^{3/2} + 2\biggr]  x
</math>
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp; &nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp; &nbsp;
<math>~=</math>
   </td>
   </td>
 
   <td align="left">  
   <td align="right">
<math>~
<math>
-\frac{2}{15}(3+\xi^2)  -\frac{2}{15} \biggl[12 - 2\xi^2 \biggr]  + 2\biggl[1 - \frac{\xi^2}{15}\biggr]
~\biggl(- {\tilde\xi}^2 {\tilde\theta}^'_5\biggr= 3^{1 / 2} {\tilde\xi}^3 \biggl( 3 + {\tilde\xi}^2\biggr)^{-3/2} \, .
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
Hence,
<div align="center">
<table border="0" cellpadding="3">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>
&nbsp;
~\frac{R_\mathrm{eq}}{R_\mathrm{norm}}
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">  
<math>~\biggl[ \frac{4\pi}{2^5\cdot 3^5}\biggr]^{1/2}
<math>~
\tilde\xi \biggl[ 3^{1 / 2} {\tilde\xi}^3 \biggl( 3 + {\tilde\xi}^2\biggr)^{-3/2} \biggr]^{-2}
\biggl( - \frac{6}{15} - \frac{24}{15} + 2\biggr)
+\xi^2 \biggl( -\frac{2}{15} + \frac{4}{15} - \frac{2}{15} \biggr)
</math>
</math>
   </td>
   </td>
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   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">  
<math>~\biggl[ \frac{4\pi}{2^5\cdot 3^5}\biggr]^{1/2}
<math>~
\tilde\xi \biggl[ 3^{-1} {\tilde\xi}^{-6} \biggl( 3 + {\tilde\xi}^2\biggr)^{3}  \biggr]
0 \, .
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>


<tr>
Q. E. D.  &nbsp;I don't think that anyone has previously appreciated that the LAWE in this case admits to an analytic eigenvector solution.
  <td align="right">
 
&nbsp;
Now, let's see how the boundary condition comes into play.  We see that the logarithmic derivative of the parabolic eigenfunction is,
  </td>
<div align="center">
  <td align="center">
<table border="0" cellpadding="5" align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{4\pi}{2^5\cdot 3^7}\biggr]^{1/2}
{\tilde\xi}^{-5} \biggl( 3 + {\tilde\xi}^2\biggr)^{3}  \, ,
</math>
  </td>
</tr>


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>
<math>~\frac{d\ln x}{d\ln \xi}</math>
~\frac{P_\mathrm{e}}{P_\mathrm{norm}}  
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl[ \frac{2^3\cdot 3^3}{4\pi}\biggr]^{3}
<math>~\frac{\xi}{x} \cdot \frac{dx}{d\xi}</math>
\biggl[  3^{1 / 2} \biggl( 3 + {\tilde\xi}^2\biggr)^{-1/2} \biggr]^{6} \biggl[ 3^{1 / 2} {\tilde\xi}^3 \biggl( 3 + {\tilde\xi}^2\biggr)^{-3/2}  \biggr]^{6}
</math>
   </td>
   </td>
</tr>
</tr>
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   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl[ \frac{2^3\cdot 3^3}{4\pi}\biggr]^{3}
<math>~- \frac{2\xi^2}{15} \biggl[ 1 - \frac{\xi^2}{15}\biggr]^{-1}</math>
\biggl[ 3^{3} \biggl( 3 + {\tilde\xi}^2\biggr)^{-3} \biggr] \biggl[ 3^{3} {\tilde\xi}^{18} \biggl( 3 + {\tilde\xi}^2\biggr)^{-9} \biggr]
</math>
   </td>
   </td>
</tr>
</tr>
Line 348: Line 340:
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl[ \frac{2^3\cdot 3^5}{4\pi}\biggr]^{3}
<math>~- \frac{2\xi^2}{(15-\xi^2)} \, .</math>
{\tilde\xi}^{18} \biggl( 3 + {\tilde\xi}^2\biggr)^{-12} \, .
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</div>
</div>
 
We desire a surface boundary condition that gives, <math>~d\ln x/d\ln\xi = -3</math>.  This will only happen when,
Now, given that the [[User:Tohline/SSC/Virial/FormFactors#Summary_.28n.3D5.29|structural form-factors for <math>~n=5</math> configurations]] are,
<div align="center">
 
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\mathfrak{f}_M</math>
<math>~- \frac{2\xi^2}{(15-\xi^2)}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 371: Line 360:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~-3</math>
( 1 + \ell^2 )^{-3/2}  = 3^{3 / 2} (3 + {\tilde\xi}^2)^{-3 / 2}
</math>
   </td>
   </td>
</tr>
</tr>
Line 379: Line 366:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\mathfrak{f}_W</math>
<math>~\Rightarrow ~~~2\xi^2</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 385: Line 372:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~3(15 - \xi^2)</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>
   </td>
</tr>
</tr>
Line 394: Line 378:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\mathfrak{f}_A</math>
<math>~\Rightarrow ~~~\xi</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 400: Line 384:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~3 \, .</math>
\frac{3}{2^3} \ell^{-3} [ \tan^{-1}(\ell ) + \ell (\ell^2-1) (1+\ell^2)^{-2} ]  \, ,
  </td>
</math>
</tr>
</table>
</div>
Hence, although the parabolic eigenfunction provides an accurate solution to the <math>~n=5</math> LAWE throughout the entire configuration &#8212; that is, for all <math>~\xi</math> &#8212; the desired surface boundary condition will only be satisfied if the polytrope is truncated at <math>~\xi_\mathrm{surf} = 3</math>.  The parabolic eigenfunction is therefore only physically relevant to the model that sits at the point along the equilibrium sequence that is associated with the <math>~P_\mathrm{max}</math> turning point.
 
<!-TRY MORE GENERIC CASE ...
Finally, let's see how this plays out for arbitrary values of <math>~\alpha</math>.  Guess a more general eigenfunction of the form,
<div align="center" i="AnalyticSoln">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="center">
<math>~x = 1 - \frac{\xi^2}{A} </math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
we understand that the central density is,
</div>
In this more general case, we have,


<div align="center" id="Proof">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\rho_c = \frac{\bar\rho}{ {\tilde\mathfrak{f}}_M }</math>
<math>~\frac{dx}{d\xi} = -\frac{2\xi}{A} \, ;</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
&nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp;
   </td>
   </td>
   <td align="left">
   <td align="left">  
<math>~
<math>~\frac{d^2x}{d\xi^2} = -\frac{2}{A} \, .</math>
\biggl[3^{3 / 2} (3 + {\tilde\xi}^2)^{-3 / 2} \biggr]^{-1} \biggl[ \frac{3 M_\mathrm{tot}}{4 \pi R_\mathrm{eq}^3} \biggr]
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
The LAWE then gives,
<div align="center" id="Proof">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~
(3+\xi^2) \frac{d^2x}{d\xi^2} + \biggl[12 - 2\xi^2 \biggr] \frac{1}{\xi} \cdot \frac{dx}{d\xi} +
\biggl[ \cancelto{0}{\frac{\sigma_c^2}{3^{1/2} \gamma_g }} \cdot (3+\xi^2)^{3/2} -6\alpha \biggr]  x
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">  
<math>~ \biggl( \frac{3}{4\pi}\biggr)
<math>~
\biggl[ \frac{2^5\cdot 3^6}{4\pi}\biggr]^{ 3 / 2} (3 + {\tilde\xi}^2)^{3 / 2} M_\mathrm{tot} \biggl[ R_\mathrm{norm}
-\frac{2}{A}(3+\xi^2) -\frac{2}{A} \biggl[12 - 2\xi^2 \biggr] -6\alpha \biggl[1 - \frac{\xi^2}{A}\biggr]
  {\tilde\xi}^{-5} \biggl( 3 + {\tilde\xi}^2\biggr)^{3} \biggr]^{-3}
</math>
</math>
   </td>
   </td>
Line 446: Line 449:
<math>~=</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">  
<math>~
<math>~-\frac{1}{A}\biggl\{  
\biggl[ \frac{2^{5}\cdot 3^{20}}{\pi^5}\biggr]^{ 1 / 2} {\tilde\xi}^{15} (3 + {\tilde\xi}^2)^{-15 / 2} M_\mathrm{tot} R^{-3}_\mathrm{norm}  
2(3+\xi^2) + 2 \biggl[12 - 2\xi^2 \biggr]  + 6\alpha \biggl[A - \xi^2\biggr]
\biggr\}
</math>
</math>
   </td>
   </td>
Line 460: Line 464:
<math>~=</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">  
<math>~
<math>~-
\biggl[ \frac{2^{5}\cdot 3^{20}}{\pi^5}\biggr]^{ 1 / 2} \biggl[{\tilde\xi}  (3 + {\tilde\xi}^2)^{-1 / 2} \biggr]^{15} M_\mathrm{tot}^{-5} \biggl( \frac{G}{K} \biggr)^{-15/2}
\frac{1}{A}\biggl[ \biggl(6 + 24 + 6\alpha A \biggr) + \biggl(2 -4-6\alpha \biggr)\xi^2\biggr]
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
Hence, a parabolic eigenfunction does not work for arbitrary <math>~\alpha</math>.
END MORE GENERIC CASE -->
Let's express the parabolic displacement function, <math>~x</math>, as a function of the Lagrangian mass coordinate, instead of as a function of <math>~\xi</math>.  Drawing upon our [[User:Tohline/Appendix/Ramblings/Nonlinar_Oscillation#Exploration|accompanying discussion]] where we have used <math>~\tilde\xi</math> to denote the truncation edge, we know that,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~r_\xi(\xi)</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 475: Line 488:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~\xi \biggl\{
\biggl[ \frac{2\cdot 3^{4}}{\pi}\biggr]^{ 5 / 2} \biggl[{\tilde\xi} (3 + {\tilde\xi}^2)^{-1 / 2} \biggr]^{15} \biggl( \frac{K^3}{G^3M_\mathrm{tot}^2} \biggr)^{5/2} \, .
\biggl[ \frac{4\pi}{2^5\cdot 3}\biggr]^{1/2} \tilde\xi^{-6}
\biggl( 1+\frac{\tilde\xi^2}{3} \biggr)^{3}\biggr\} \, .
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
 
</div>
<span id="r0">Now let's derive the prescription for the Lagrangian radial coordinate in the context of pressure-truncated,</span> <math>~n=5</math> polytropes. 
and that,
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 488: Line 502:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~r_0 \equiv a_5 \xi</math>
<math>~
r_\xi (m_\xi)
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 494: Line 510:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl[\frac{3K}{2\pi G} \biggr]^{1 / 2} \rho_c^{-2/5} \xi</math>
<math>~\tilde{r}_\mathrm{edge}
\biggl[\frac{3^2m_\xi^{2/3}}{\tilde{C} - 3 m_\xi^{2/3}}\biggr]^{1/2} \, ,
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
<span id="DefineTildeC">where,</span>
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\tilde{C}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~\equiv</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl[\frac{3K}{2\pi G} \biggr]^{1 / 2} \xi \biggl\{
<math>~
\biggl[ \frac{2\cdot 3^{4}}{\pi}\biggr]^{ 5 / 2} \biggl[{\tilde\xi}  (3 + {\tilde\xi}^2)^{-1 / 2} \biggr]^{15} \biggl( \frac{K^3}{G^3M_\mathrm{tot}^2} \biggr)^{5/2}
\frac{3^2}{\tilde\xi^2}\biggl( 1 + \frac{\tilde\xi^2}{3} \biggr) \, .
\biggr\}^{-2/5}
</math>
</math>
   </td>
   </td>
</tr>
</tr>
Line 515: Line 537:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\tilde{r}_\mathrm{edge}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~\equiv</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl[\frac{3K}{2\pi G} \biggr]^{1 / 2} \biggl[ \frac{\pi}{2\cdot 3^{4}}\biggr] \biggl( \frac{G^3M_\mathrm{tot}^2}{K^3} \biggr)
<math>~\biggl[ \frac{\pi}{2^3\cdot 3}\biggr]^{1/2} {\tilde\xi}^{-6} \biggl(1+\frac{\tilde\xi^2}{3}\biggr)^3
\biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3} \xi
\, .
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
R_\mathrm{norm} \biggl[ \frac{\pi}{2^3\cdot 3^{7}}\biggr]^{1 / 2}  \biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3} \xi
</math>
</math>
   </td>
   </td>
Line 543: Line 551:
</div>
</div>


<span id="m0">Also,</span>
By equating <math>~r_\xi(\xi)</math> with <math>~r_\xi(m_\xi)</math>, we find,


<div align="center">
<div align="center">
Line 550: Line 558:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~m_0 \equiv M(r_0)</math>
<math>~\xi \biggl\{
\biggl[ \frac{4\pi}{2^5\cdot 3}\biggr]^{1/2} \tilde\xi^{-6}
\biggl( 1+\frac{\tilde\xi^2}{3} \biggr)^{3}\biggr\}
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 556: Line 567:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl[ 4\pi a_n^3 \rho_c \biggl(-\xi^2 \frac{d\theta}{d\xi}\biggr) \biggr]
<math>~\biggl[ \frac{\pi}{2^3\cdot 3}\biggr]^{1/2} {\tilde\xi}^{-6} \biggl(1+\frac{\tilde\xi^2}{3}\biggr)^3
\, ,</math>
\biggl[\frac{3^2m_\xi^{2/3}}{\tilde{C} - 3 m_\xi^{2/3}}\biggr]^{1/2}
</math>
   </td>
   </td>
</tr>
</tr>
Line 563: Line 575:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\Rightarrow ~~~ \xi
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 569: Line 582:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~2^2\pi \biggl\{ R_\mathrm{norm} \biggl[ \frac{\pi}{2^3\cdot 3^{7}}\biggr]^{1 / 2}  \tilde\xi^{-6} (3 + {\tilde\xi}^2)^{3} \biggr\}^3
<math>~  
\biggl\{ \biggl[ \frac{2\cdot 3^{4}}{\pi}\biggr]^{ 5 / 2} \biggl[{\tilde\xi}  (3 + {\tilde\xi}^2)^{-1 / 2} \biggr]^{15} \biggl( \frac{K^3}{G^3M_\mathrm{tot}^2} \biggr)^{5/2} \biggr\}
\biggl[\frac{3^2m_\xi^{2/3}}{\tilde{C} - 3 m_\xi^{2/3}}\biggr]^{1/2} \, .
\biggl\{  3^{1 / 2} \xi^3 \biggl( 3 + \xi^2\biggr)^{-3/2} \biggr\}
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
This means that,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~x</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 584: Line 601:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ 3^{1 / 2} \biggl[ 2^4 \pi^2\biggr]^{1 / 2} \biggl[ \frac{\pi^3}{2^9\cdot 3^{21}}\biggr]^{1 / 2} \biggl[ \frac{2^5\cdot 3^{20}}{\pi^5}\biggr]^{ 1 / 2} 
<math>~
\biggl\{  \tilde\xi^{-6} (3 + {\tilde\xi}^2)^{3} \biggr\}^3
x_0 \biggl\{1 -
\biggl[{\tilde\xi} (3 + {\tilde\xi}^2)^{-1 / 2} \biggr]^{15} \biggl( \frac{K^3}{G^3M_\mathrm{tot}^2} \biggr)^{5/2} R_\mathrm{norm}^3
\frac{1}{15} \biggl[\frac{3^2m_\xi^{2/3}}{\tilde{C} - 3 m_\xi^{2/3}}\biggr]
\biggl\{ \xi^3 ( 3 + \xi^2 )^{-3/2} \biggr\}
\biggr\} \, ;
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
and, specifically for the critical case of <math>~\tilde\xi = 3</math>, in which case, <math>~\tilde{C} = 4</math>,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~x</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 600: Line 622:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
\biggl\{ \tilde\xi^{-3} (3 + {\tilde\xi}^2)^{3 / 2} \biggr\} M_\mathrm{tot}  
x_0 \biggl\{1 -  
\biggl\{ \xi^3 ( 3 + \xi^2 )^{-3/2} \biggr\} \, .
\frac{1}{15} \biggl[\frac{3^2m_\xi^{2/3}}{4 - 3 m_\xi^{2/3}}\biggr]
\biggr\} \, .
</math>
</math>
   </td>
   </td>
Line 609: Line 632:
</div>
</div>


Hence,
 
{{ LSU_WorkInProgress }}
 
==Setup Using Lagrangian Radial Coordinate==
 
===Individual Terms===
From our [[User:Tohline/SSC/FreeEnergy/PowerPoint#Case_M_Equilibrium_Conditions|accompanying discussion]], we have, for pressure-truncated, <math>~n=5</math> polytropic spheres


<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="3">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~g_0 = \frac{Gm_0}{r_0^2}</math>
<math>
~\frac{R_\mathrm{eq}}{R_\mathrm{norm}}  
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~=~</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ \frac{GM_\mathrm{tot}}{R_\mathrm{norm}^2}
<math>~\biggl[ \frac{4\pi}{(n+1)^n}\biggr]^{1/(n-3)}
\biggl\{ \tilde\xi^{-3} (3 + {\tilde\xi}^2)^{3 / 2} \biggr\} 
\tilde\xi ( -\tilde\xi^2 \tilde\theta' )^{(1-n)/(n-3)}  
\biggl\{ \xi^3 ( 3 + \xi^2 )^{-3/2} \biggr\}
\biggl\{ \biggl[ \frac{\pi}{2^3\cdot 3^{7}}\biggr]^{1 / 2}  \biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3} \xi \biggr\}^{-2}
</math>
</math>
   </td>
   </td>
Line 635: Line 664:
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~=~</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~\biggl[ \frac{4\pi}{2^5\cdot 3^5}\biggr]^{1/2}
\frac{GM_\mathrm{tot}}{R_\mathrm{norm}^2}\biggl[ \frac{2^3\cdot 3^{7}}{\pi}\biggr]  
\tilde\xi ( -\tilde\xi^2 \tilde\theta' )^{-2} \, ,
\biggl[ \tilde\xi (3 + {\tilde\xi}^2)^{-1 / 2} \biggr]^{9} 
\xi ( 3 + \xi^2 )^{-3/2} \, ;
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
which matches the expression derived in an [[User:Tohline/SSC/Structure/Polytropes#Lane-Emden_Equation|ASIDE box found with our introduction of the Lane-Emden equation]], and
<div align="center">
<table border="0" cellpadding="3">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\frac{g_0 }{r_0} </math>
<math>
~\frac{P_\mathrm{e}}{P_\mathrm{norm}}  
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~=~</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ \frac{GM_\mathrm{tot}}{R_\mathrm{norm}^3}\biggl[ \frac{2^3\cdot 3^{7}}{\pi}\biggr]  
<math>~\biggl[ \frac{(n+1)^3}{4\pi}\biggr]^{(n+1)/(n-3)}
\biggl\{  \tilde\xi^{9} (3 + {\tilde\xi}^2)^{-9 / 2} \biggr\}  \biggl\{ \biggl[ \frac{\pi}{2^3\cdot 3^{7}}\biggr]^{1 / 2}   \biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3} \xi \biggr\}^{-1}
\tilde\theta_n^{n+1}( -\tilde\xi^2 \tilde\theta' )^{2(n+1)/(n-3)}  
\xi ( 3 + \xi^2 )^{-3/2}  
</math>
</math>
   </td>
   </td>
Line 666: Line 699:
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~=~</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~\biggl[ \frac{2^3\cdot 3^3}{4\pi}\biggr]^{3}
\frac{GM_\mathrm{tot}}{R_\mathrm{norm}^3}\biggl[ \frac{2^3\cdot 3^{7}}{\pi}\biggr]^{3/2}  
\tilde\theta^{6}( -\tilde\xi^2 \tilde\theta' )^{6} \, ,
\biggl[  \tilde\xi (3 + {\tilde\xi}^2)^{-1 / 2} \biggr]^{15} 
( 3 + \xi^2 )^{-3/2}  
\, ;
</math>
</math>
   </td>
   </td>
Line 679: Line 709:
</table>
</table>
</div>
</div>
 
where,
 
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="3">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\frac{\rho_0}{P_0} = \frac{\rho_0}{K\rho_0^{1+1/n}} </math>
<math>~R_\mathrm{norm}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~\equiv</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~\biggl[ \biggl( \frac{G}{K} \biggr)^n M_\mathrm{tot}^{n-1} \biggr]^{1/(n-3)} = \biggl( \frac{G}{K} \biggr)^{5/2} M_\mathrm{tot}^{2}  \, ,</math>
\biggl[K^5 \rho_c \theta^5 \biggr]^{-1/5}
</math>
   </td>
   </td>
</tr>
</tr>
Line 700: Line 727:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~P_\mathrm{norm}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~\equiv</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ \theta^{-1}
<math>~\biggl[ \frac{K^{4n}}{G^{3(n+1)} M_\mathrm{tot}^{2(n+1)}} \biggr]^{1/(n-3)} = \frac{K^{10}}{G^{9} M_\mathrm{tot}^{6} } \, ,</math>
\biggl\{ K^5 \biggl[ \frac{2\cdot 3^{4}}{\pi}\biggr]^{ 5 / 2} \biggl[{\tilde\xi} (3 + {\tilde\xi}^2)^{-1 / 2} \biggr]^{15} \biggl( \frac{K^3}{G^3M_\mathrm{tot}^2} \biggr)^{5/2}\biggr\}^{-1/5}
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
and, from [[User:Tohline/SSC/Structure/PolytropesEmbedded#Tabular_Summary_.28n.3D5.29|our more detailed analysis]],


<table border="0" cellpadding="3" align="center">
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>
~{\tilde\theta}_5 = 3^{1 / 2} \biggl( 3 + {\tilde\xi}^2\biggr)^{-1/2}
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
&nbsp; &nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp; &nbsp;
   </td>
   </td>
   <td align="left">
 
<math>~ \biggl[ 3^{-1} ( 3 + \xi^2 ) \biggr]^{1/2}
   <td align="right">
\biggl\{ \biggl[ \frac{\pi}{2\cdot 3^{4}}\biggr]^{1 / 2} \cancelto{\mathrm{mistake}}{\biggl[{\tilde\xi}^{-3(3 + {\tilde\xi}^2)^{3 / 2} \biggr]^{-3} } \biggl( \frac{G^3M_\mathrm{tot}^2}{K^5} \biggr)^{1/2}\biggr\}
<math>
~\biggl(- {\tilde\xi}^2 {\tilde\theta}^'_5\biggr)  = 3^{1 / 2} {\tilde\xi}^3 \biggl( 3 + {\tilde\xi}^2\biggr)^{-3/2} \, .
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>


<tr>
Hence,
   <td align="right">
 
&nbsp;
<div align="center">
<table border="0" cellpadding="3">
 
<tr>
   <td align="right">
<math>
~\frac{R_\mathrm{eq}}{R_\mathrm{norm}}
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~=~</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~\biggl[ \frac{4\pi}{2^5\cdot 3^5}\biggr]^{1/2}
\biggl( \frac{G^3M_\mathrm{tot}^2}{K^5} \biggr)^{1 / 2}
\tilde\xi \biggl[ 3^{1 / 2} {\tilde\xi}^3 \biggl( 3 + {\tilde\xi}^2\biggr)^{-3/2} \biggr]^{-2}  
\biggl[ \frac{\pi}{2\cdot 3^{5}}\biggr]^{1 / 2} \cancelto{\mathrm{mistake}}{\biggl[ {\tilde\xi} (3 + {\tilde\xi}^2)^{-1 / 2} \biggr]^{9} }
( 3 + \xi^2 )^{1 / 2}  
\, ;
</math>
</math>
   </td>
   </td>
Line 748: Line 787:
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~=~</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ \biggl[ 3^{-1} ( 3 + \xi^2 ) \biggr]^{1/2}
<math>~\biggl[ \frac{4\pi}{2^5\cdot 3^5}\biggr]^{1/2}
\biggl\{ \biggl[ \frac{\pi}{2\cdot 3^{4}}\biggr]^{1 / 2} \biggl[{\tilde\xi}^{-3} (3 + {\tilde\xi}^2)^{3 / 2} \biggr] \biggl( \frac{G^3M_\mathrm{tot}^2}{K^5} \biggr)^{1/2}\biggr\}
\tilde\xi \biggl[ 3^{-1} {\tilde\xi}^{-6} \biggl( 3 + {\tilde\xi}^2\biggr)^{3} \biggr]
</math>
</math>
   </td>
   </td>
Line 762: Line 801:
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~=~</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~  
<math>~
\biggl( \frac{G^3M_\mathrm{tot}^2}{K^5} \biggr)^{1/2}\biggl[ \frac{\pi}{2\cdot 3^{5}}\biggr]^{1 / 2}  
\biggl[ \frac{4\pi}{2^5\cdot 3^7}\biggr]^{1/2}
\biggl[ \frac{(3 + {\tilde\xi}^2)}{{\tilde\xi}^2 } \biggr]^{3 / 2( 3 + \xi^2 )^{1/2}
{\tilde\xi}^{-5} \biggl( 3 + {\tilde\xi}^2\biggr)^{3}  \, ,
</math>
</math>
   </td>
   </td>
Line 774: Line 813:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\frac{g_0\rho_0}{P_0} </math>
<math>
~\frac{P_\mathrm{e}}{P_\mathrm{norm}}  
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~=~</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~\biggl[ \frac{2^3\cdot 3^3}{4\pi}\biggr]^{3}
\biggl( \frac{G^3M_\mathrm{tot}^2}{K^5} \biggr)^{1/2}\biggl[ \frac{\pi}{2\cdot 3^{5}}\biggr]^{1 / 2}  
\biggl[  3^{1 / 2} \biggl( 3 + {\tilde\xi}^2\biggr)^{-1/2} \biggr]^{6} \biggl[ 3^{1 / 2} {\tilde\xi}^3 \biggl( 3 + {\tilde\xi}^2\biggr)^{-3/2}  \biggr]^{6}  
\biggl[ \frac{(3 + {\tilde\xi}^2)}{{\tilde\xi}^2 } \biggr]^{3 / 2}  ( 3 + \xi^2 )^{1/2}
</math>
</math>
   </td>
   </td>
Line 792: Line 832:
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~=~</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ \times ~
<math>~\biggl[ \frac{2^3\cdot 3^3}{4\pi}\biggr]^{3}
\biggl( \frac{G^2M_\mathrm{tot}^2}{R_\mathrm{norm}^4} \biggr)^{1 / 2}\biggl[ \frac{2^6\cdot 3^{14}}{\pi^2}\biggr]^{1 / 2}  
\biggl[ 3^{3} \biggl( 3 + {\tilde\xi}^2\biggr)^{-3} \biggr] \biggl[ 3^{3} {\tilde\xi}^{18} \biggl( 3 + {\tilde\xi}^2\biggr)^{-9} \biggr]
\biggl[ \frac{(3 + {\tilde\xi}^2)}{{\tilde\xi}^2 } \biggr]^{-9 / 2}
\xi ( 3 + \xi^2 )^{-3/2}  
</math>
</math>
   </td>
   </td>
Line 808: Line 846:
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~=~</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ \biggl( \frac{G^5 M_\mathrm{tot}^4}{K^5} \biggr)^{1 / 2}  R_\mathrm{norm}^{-2}
<math>~\biggl[ \frac{2^3\cdot 3^5}{4\pi}\biggr]^{3}
\biggl[ \frac{{\tilde\xi}^2 }{(3 + {\tilde\xi}^2)} \biggr]^{3} \biggl[ \frac{2^5\cdot 3^{9}}{\pi}\biggr]^{1 / 2} \xi ( 3 + \xi^2 )^{-1}  
{\tilde\xi}^{18} \biggl( 3 + {\tilde\xi}^2\biggr)^{-12} \, .
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
Now, given that the [[User:Tohline/SSC/Virial/FormFactors#Summary_.28n.3D5.29|structural form-factors for <math>~n=5</math> configurations]] are,
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\mathfrak{f}_M</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 826: Line 870:
   <td align="left">
   <td align="left">
<math>~   
<math>~   
\biggl( \frac{K^5}{G^5 M_\mathrm{tot}^4} \biggr)^{1 / 2}   
( 1 + \ell^2 )^{-3/2}  = 3^{3 / 2} (3 + {\tilde\xi}^2)^{-3 / 2}
\biggl[ \frac{{\tilde\xi}^2 }{(3 + {\tilde\xi}^2)} \biggr]^{3} \biggl[ \frac{2^5\cdot 3^{9}}{\pi}\biggr]^{1 / 2}  
\xi ( 3 + \xi^2 )^{-1}
\, .
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>


===The Wave Equation===
<tr>
 
  <td align="right">
====Starting from our Key Adiabatic Wave Equation====
<math>~\mathfrak{f}_W</math>
 
  </td>
The [[#Adiabatic_.28Polytropic.29_Wave_Equation|adiabatic wave equation]] therefore becomes,
  <td align="center">
 
<math>~=</math>
<div align="center">
  </td>
<table border="0" cellpadding="5" align="center">
  <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>
<tr>
   <td align="right">
   <td align="right">
<math>~0</math>
<math>~\mathfrak{f}_A</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 854: Line 899:
   <td align="left">
   <td align="left">
<math>~
<math>~
\frac{d^2x}{dr_0^2} + \biggl[\frac{4}{r_0} - \biggl(\frac{g_0 \rho_0}{P_0}\biggr) \biggr] \frac{dx}{dr_0}
\frac{3}{2^3} \ell^{-3}  [ \tan^{-1}(\ell ) + \ell (\ell^2-1) (1+\ell^2)^{-2} ]  \, ,
+ \biggl(\frac{\rho_0}{\gamma_\mathrm{g} P_0} \biggr)\biggl[\omega^2 + (4 - 3\gamma_\mathrm{g})\frac{g_0}{r_0} \biggrx
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
we understand that the central density is,
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\rho_c = \frac{\bar\rho}{ {\tilde\mathfrak{f}}_M }</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 868: Line 916:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
\frac{d^2x}{dr_0^2} + \frac{1}{R_\mathrm{norm}} \biggl\{
\biggl[3^{3 / 2} (3 + {\tilde\xi}^2)^{-3 / 2} \biggr]^{-1} \biggl[ \frac{3 M_\mathrm{tot}}{4 \pi R_\mathrm{eq}^3} \biggr]
\biggl[ \frac{2^7\cdot 3^{7}}{\pi} \biggr]^{1 / 2}   \biggl[ \frac{ {\tilde\xi}^2}{(3 + {\tilde\xi}^2)} \biggr]^{3} \frac{1}{\xi}
-   \biggl[ \frac{{\tilde\xi}^2 }{(3 + {\tilde\xi}^2)} \biggr]^{3} \biggl[ \frac{2^5\cdot 3^{9}}{\pi}\biggr]^{1 / 2} \xi ( 3 + \xi^2 )^{-1}
\biggr\} \frac{dx}{dr_0}
</math>
</math>
   </td>
   </td>
Line 882: Line 927:
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~ \biggl( \frac{3}{4\pi}\biggr)
+ \frac{(4 - 3\gamma_\mathrm{g})}{\gamma_g R_\mathrm{norm}^2}
\biggl[ \frac{2^5\cdot 3^6}{4\pi}\biggr]^{ 3 / 2} (3 + {\tilde\xi}^2)^{3 / 2} M_\mathrm{tot} \biggl[ R_\mathrm{norm}  
\biggl[ \frac{\pi}{2\cdot 3^{5}}\biggr]^{1 / 2} \biggl[ \frac{{\tilde\xi}^2}{(3 + {\tilde\xi}^2)} \biggr]^{-3/2}  
{\tilde\xi}^{-5} \biggl( 3 + {\tilde\xi}^2\biggr)^{3} \biggr]^{-3}
( 3 + \xi^2 )^{1 / 2} 
\biggl\{ \frac{R_\mathrm{norm}^3}{GM_\mathrm{tot}} \biggl[\frac{\omega^2}{(4 - 3\gamma_\mathrm{g})} \biggr]
+ \biggl[ \frac{2^3\cdot 3^{7}}{\pi}\biggr]^{3/2}  
\biggl[ \frac{{\tilde\xi}^2}{(3 + {\tilde\xi}^2)} \biggr]^{15/2}
( 3 + \xi^2 )^{-3/2}
\biggr\}  x
</math>
</math>
   </td>
   </td>
Line 906: Line 945:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
\frac{d^2x}{dr_0^2} + \frac{1}{R_\mathrm{norm}} \biggl[ \frac{2^3\cdot 3^{7}}{\pi} \biggr]^{1 / 2} \biggl[ \frac{ {\tilde\xi}^2}{(3 + {\tilde\xi}^2)} \biggr]^{3} \biggl[
\biggl[ \frac{2^{5}\cdot 3^{20}}{\pi^5}\biggr]^{ 1 / 2} {\tilde\xi}^{15} (3 + {\tilde\xi}^2)^{-15 / 2} M_\mathrm{tot} R^{-3}_\mathrm{norm}  
\frac{4}{\xi} -   \frac{6 \xi}{ ( 3 + \xi^2 )}
\biggr] \frac{dx}{dr_0}  
</math>
</math>
   </td>
   </td>
Line 919: Line 956:
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
+ \frac{6(4 - 3\gamma_\mathrm{g})}{\gamma_g R_\mathrm{norm}^2}
\biggl[ \frac{2^{5}\cdot 3^{20}}{\pi^5}\biggr]^{ 1 / 2} \biggl[{\tilde\xi} (3 + {\tilde\xi}^2)^{-1 / 2} \biggr]^{15} M_\mathrm{tot}^{-5} \biggl( \frac{G}{K} \biggr)^{-15/2}  
( 3 + \xi^2 )^{1 / 2}  \biggl[ \frac{2^3\cdot 3^{7}}{\pi}\biggr] \biggl[ \frac{{\tilde\xi}^2}{(3 + {\tilde\xi}^2)} \biggr]^{6}
\biggl\{ \frac{R_\mathrm{norm}^3}{GM_\mathrm{tot}} \biggl[\frac{\omega^2}{(4 - 3\gamma_\mathrm{g})} \biggr] \biggl[ \frac{2^3\cdot 3^{7}}{\pi}\biggr]^{-3/2} \biggl[ \frac{{\tilde\xi}^2}{(3 + {\tilde\xi}^2)} \biggr]^{-15/2}
( 3 + \xi^2 )^{-3/2} 
\biggr\}  x
</math>
</math>
   </td>
   </td>
Line 941: Line 973:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
\frac{d^2x}{dr_0^2} + \frac{1}{R_*} \biggl[ \frac{ {\tilde\xi}^2}{(3 + {\tilde\xi}^2)} \biggr]^{3} \biggl[
\biggl[ \frac{2\cdot 3^{4}}{\pi}\biggr]^{ 5 / 2} \biggl[{\tilde\xi} (3 + {\tilde\xi}^2)^{-1 / 2} \biggr]^{15} \biggl( \frac{K^3}{G^3M_\mathrm{tot}^2} \biggr)^{5/2} \, .
\frac{4}{\xi} -  \frac{6 \xi}{ ( 3 + \xi^2 )}
\biggr] \frac{dx}{dr_0}
+ \frac{6}{\gamma_g R_*^2}   \biggl[ \frac{{\tilde\xi}^2}{(3 + {\tilde\xi}^2)} \biggr]^{6}
\biggl\{ \frac{\omega^2 R_*^3}{GM_\mathrm{tot}}  \biggl[ \frac{{\tilde\xi}^2}{(3 + {\tilde\xi}^2)} \biggr]^{-15/2}( 3 + \xi^2 )^{1 / 2}
+  \frac{(4 - 3\gamma_\mathrm{g})}{( 3 + \xi^2 ) }
\biggr\}  x
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</div>
 
where,
<span id="r0">Now let's derive the prescription for the Lagrangian radial coordinate in the context of pressure-truncated,</span> <math>~n=5</math> polytropes. 
<div align="center">
<div align="center">
<math>R_* \equiv R_\mathrm{norm} \biggl[ \frac{\pi}{2^3 \cdot 3^7} \biggr]^{1/2} \, .</math>
<table border="0" cellpadding="5" align="center">
</div>
 
Recognizing that,
 
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~r_0</math>
<math>~r_0 \equiv a_5 \xi</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 972: Line 992:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~\biggl[\frac{3K}{2\pi G} \biggr]^{1 / 2} \rho_c^{-2/5} \xi</math>
R_*  \biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3} \xi \, ,
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
we can write,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~0</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 992: Line 1,004:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{1}{R_*^2} \biggl[ \frac{ {\tilde\xi}^2}{(3 + {\tilde\xi}^2)} \biggr]^{6} \biggl\{
<math>~\biggl[\frac{3K}{2\pi G} \biggr]^{1 / 2} \xi \biggl\{
\frac{d^2x}{d\xi^2} + \biggl[
\biggl[ \frac{2\cdot 3^{4}}{\pi}\biggr]^{ 5 / 2} \biggl[{\tilde\xi} (3 + {\tilde\xi}^2)^{-1 / 2} \biggr]^{15} \biggl( \frac{K^3}{G^3M_\mathrm{tot}^2} \biggr)^{5/2}
\frac{4}{\xi} -  \frac{6 \xi}{ ( 3 + \xi^2 )}
\biggr\}^{-2/5}
\biggr] \frac{dx}{d\xi}
+ \frac{6}{\gamma_g }
\biggl[\sigma^2 ( 3 + \xi^2 )^{1 / 2}
\frac{(4 - 3\gamma_\mathrm{g})}{( 3 + \xi^2 ) }
\biggr]  x \biggr\} \, ,
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
where,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\sigma^2</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\equiv</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ \frac{\omega^2 R_*^3}{GM_\mathrm{tot}}  \biggl( \frac{3 + {\tilde\xi}^2}{{\tilde\xi}^2} \biggr)^{15/2} \, .</math>
<math>~\biggl[\frac{3K}{2\pi G} \biggr]^{1 / 2} \biggl[ \frac{\pi}{2\cdot 3^{4}}\biggr]  \biggl( \frac{G^3M_\mathrm{tot}^2}{K^3} \biggr)
  </td>
\biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3} \xi
</tr>
</table>
</div>
 
Finally, if &#8212; because we are specifically considering the case of <math>~n=5</math> &#8212; we set <math>~\gamma_\mathrm{g} = 1 + 1/n = 6/5</math>, 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{d^2x}{d\xi^2} + \biggl[ \frac{4}{\xi} -  \frac{6 \xi}{ ( 3 + \xi^2 )} \biggr] \frac{dx}{d\xi}
+ \biggl[5\sigma^2 ( 3 + \xi^2 )^{1 / 2} +  \frac{2}{( 3 + \xi^2 ) }\biggr] x
</math>
</math>
   </td>
   </td>
Line 1,052: Line 1,033:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{1}{( 3 + \xi^2 ) } \biggl\{ ( 3 + \xi^2 )\frac{d^2x}{d\xi^2}  
<math>~
+ \biggl[ \frac{2(6 - \xi^2) }{ \xi} \biggr] \frac{dx}{d\xi}  
R_\mathrm{norm} \biggl[ \frac{\pi}{2^3\cdot 3^{7}}\biggr]^{1 / 2}   \biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3} \xi
+ \biggl[5\sigma^2 ( 3 + \xi^2 )^{3 / 2} +  2 \biggr]  x \biggr\} \, .
</math>
</math>
   </td>
   </td>
Line 1,061: Line 1,041:
</div>
</div>


====Starting from the HRW66 Radial Pulsation Equation====
<span id="m0">Also,</span>
More directly, if we begin with the [[#HRW66excerpt| HRW66 radial pulsation equation]] that is already tuned to polytropic configurations, the wave equation appropriate to <math>~n=5</math> polytropes is,
 
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 1,068: Line 1,048:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~0</math>
<math>~m_0 \equiv M(r_0)</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,074: Line 1,054:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~\biggl[ 4\pi a_n^3 \rho_c \biggl(-\xi^2 \frac{d\theta}{d\xi}\biggr) \biggr]
\frac{d^2 X}{d\xi^2} + \biggl[ \frac{4}{\xi} - \frac{6 (-\theta^'_5)}{\theta_5} \biggr]\frac{d X}{d\xi}
\, ,</math>
+ \frac{5(-\theta_5^') }{6\theta_5 \xi} \bigg[ \frac{\xi (s^')^2}{\theta^'_5} + \frac{12}{5} \biggr] X
</math>
   </td>
   </td>
</tr>
</tr>
Line 1,089: Line 1,067:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~2^2\pi \biggl\{ R_\mathrm{norm} \biggl[ \frac{\pi}{2^3\cdot 3^{7}}\biggr]^{1 / 2}   \tilde\xi^{-6} (3 + {\tilde\xi}^2)^{3} \biggr\}^3
\frac{d^2 X}{d\xi^2} + \biggl[ \frac{4}{\xi} - \frac{6 \xi}{(3 + \xi^2)} \biggr]\frac{d X}{d\xi}
\biggl\{ \biggl[ \frac{2\cdot 3^{4}}{\pi}\biggr]^{ 5 / 2} \biggl[{\tilde\xi} (3 + {\tilde\xi}^2)^{-1 / 2} \biggr]^{15} \biggl( \frac{K^3}{G^3M_\mathrm{tot}^2} \biggr)^{5/2}  \biggr\}  
+ \frac{1}{(3 + \xi^2)} \bigg[ -\frac{5(s^')^2(3 + \xi^2)^{3 / 2}}{2 \cdot 3^{3 / 2}} + 2 \biggr] X
\biggl\{  3^{1 / 2} \xi^3 \biggl( 3 + \xi^2\biggr)^{-3/2} \biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 3^{1 / 2} \biggl[ 2^4 \pi^2\biggr]^{1 / 2} \biggl[ \frac{\pi^3}{2^9\cdot 3^{21}}\biggr]^{1 / 2} \biggl[ \frac{2^5\cdot 3^{20}}{\pi^5}\biggr]^{ 1 / 2} 
\biggl\{  \tilde\xi^{-6} (3 + {\tilde\xi}^2)^{3} \biggr\}^3
\biggl[{\tilde\xi}  (3 + {\tilde\xi}^2)^{-1 / 2} \biggr]^{15} \biggl( \frac{K^3}{G^3M_\mathrm{tot}^2} \biggr)^{5/2} R_\mathrm{norm}^3
\biggl\{ \xi^3 ( 3 + \xi^2 )^{-3/2} \biggr\}
</math>
</math>
   </td>
   </td>
Line 1,104: Line 1,098:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{1}{(3+\xi^2)} \biggl\{
<math>~
(3+\xi^2)\frac{d^2 X}{d\xi^2} + \biggl[ \frac{2(6-\xi^2)}{\xi}\biggr]\frac{d X}{d\xi}
\biggl\{ \tilde\xi^{-3} (3 + {\tilde\xi}^2)^{3 / 2} \biggr\} M_\mathrm{tot}  
+ \bigg[ -\frac{5(s^')^2}{2 \cdot 3^{3 / 2}} \cdot (3 + \xi^2)^{3 / 2} + 2 \biggr] X
\biggl\{ \xi^3 ( 3 + \xi^2 )^{-3/2} \biggr\} \, .
\biggr\} \, ,
</math>
</math>
   </td>
   </td>
Line 1,113: Line 1,106:
</table>
</table>
</div>
</div>
which is identical to the brute-force derivation just presented, allowing for the mapping,
<div align="center">
<math>\sigma^2 ~~ \Leftrightarrow ~~ -\frac{(s^')^2}{2 \cdot 3^{3 / 2}} \, .</math>
</div>


====New Independent Variable====
Hence,
Guided by our [[User:Tohline/Appendix/Ramblings/Nonlinar_Oscillation#Conjectures|conjecture]] regarding the proper shape of the radial eigenfunction, let's switch the dependent variable to,
 
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 1,125: Line 1,114:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~u  \equiv 1 + \frac{3}{\xi^2}</math>
<math>~g_0 = \frac{Gm_0}{r_0^2}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp; &nbsp; &nbsp; &nbsp; <math>~\Rightarrow</math> &nbsp; &nbsp; &nbsp; &nbsp;
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~3 + \xi^2 = \frac{3u}{(u-1)} \, ,</math>
<math>~ \frac{GM_\mathrm{tot}}{R_\mathrm{norm}^2}
\biggl\{  \tilde\xi^{-3} (3 + {\tilde\xi}^2)^{3 / 2} \biggr\} 
\biggl\{ \xi^3 ( 3 + \xi^2 )^{-3/2} \biggr\}
\biggl\{ \biggl[ \frac{\pi}{2^3\cdot 3^{7}}\biggr]^{1 / 2}  \biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3} \xi \biggr\}^{-2}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp; &nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp; &nbsp;
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\xi = 3^{1 / 2} (u-1)^{-1 / 2} \, .</math>
<math>~
\frac{GM_\mathrm{tot}}{R_\mathrm{norm}^2}\biggl[ \frac{2^3\cdot 3^{7}}{\pi}\biggr]
\biggl[ \tilde\xi (3 + {\tilde\xi}^2)^{-1 / 2} \biggr]^{9} 
\xi ( 3 + \xi^2 )^{-3/2} \, ;
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
This implies that,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\frac{d}{d\xi}</math>
<math>~\frac{g_0 }{r_0} </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~~~\rightarrow ~~~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~-\frac{2}{\sqrt{3}}(u-1)^{3 / 2} \frac{d}{du} \, ,</math>
<math>~ \frac{GM_\mathrm{tot}}{R_\mathrm{norm}^3}\biggl[ \frac{2^3\cdot 3^{7}}{\pi}\biggr]
\biggl\{  \tilde\xi^{9} (3 + {\tilde\xi}^2)^{-9 / 2} \biggr\}  \biggl\{ \biggl[ \frac{\pi}{2^3\cdot 3^{7}}\biggr]^{1 / 2}   \biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3} \xi \biggr\}^{-1}
\xi ( 3 + \xi^2 )^{-3/2}
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
and,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\frac{d^2}{d\xi^2}</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~~~\rightarrow ~~~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{4}{3}(u-1)^3 \frac{d^2}{du^2} + 2(u-1)^{2} \frac{d}{du} \, .</math>
<math>~
\frac{GM_\mathrm{tot}}{R_\mathrm{norm}^3}\biggl[ \frac{2^3\cdot 3^{7}}{\pi}\biggr]^{3/2}  
\biggl[  \tilde\xi (3 + {\tilde\xi}^2)^{-1 / 2} \biggr]^{15}
( 3 + \xi^2 )^{-3/2}  
\, ;
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</div>
</div>
Hence, the governing wave equation becomes,
 


<div align="center">
<div align="center">
Line 1,183: Line 1,184:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~0</math>
<math>~\frac{\rho_0}{P_0} = \frac{\rho_0}{K\rho_0^{1+1/n}} </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,189: Line 1,190:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~( 3 + \xi^2 )\frac{d^2x}{d\xi^2}
<math>~
+ \biggl[ \frac{2(6 - \xi^2) }{ \xi} \biggr] \frac{dx}{d\xi}
\biggl[K^5 \rho_c \theta^5 \biggr]^{-1/5}
+ \biggl[5\sigma^2 ( 3 + \xi^2 )^{3 / 2} +  2 \biggr]  x
</math>
</math>
   </td>
   </td>
Line 1,204: Line 1,204:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{3u}{(u-1)} \biggl[\frac{4}{3}(u-1)^3 \frac{d^2x}{du^2} + 2(u-1)^{2} \frac{dx}{du}\biggr]
<math>~ \theta^{-1}
+ 4(2u-3)(u-1)\frac{dx}{du}
\biggl\{ K^5 \biggl[ \frac{2\cdot 3^{4}}{\pi}\biggr]^{ 5 / 2} \biggl[{\tilde\xi} (3 + {\tilde\xi}^2)^{-1 / 2} \biggr]^{15} \biggl( \frac{K^3}{G^3M_\mathrm{tot}^2} \biggr)^{5/2}\biggr\}^{-1/5}
+ \biggl\{ 5\sigma^2 \biggl[ \frac{3u}{(u-1)} \biggr]^{3 / 2} +  2 \biggr\} x
</math>
</math>
   </td>
   </td>
Line 1,219: Line 1,218:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~4u(u-1)^2 \frac{d^2x}{du^2}  
<math>~ \biggl[ 3^{-1} ( 3 + \xi^2 ) \biggr]^{1/2}
+ (14u-12)(u-1)\frac{dx}{du}  
\biggl\{ \biggl[ \frac{\pi}{2\cdot 3^{4}}\biggr]^{1 / 2} \cancelto{\mathrm{mistake}}{\biggl[{\tilde\xi}^{-3} (3 + {\tilde\xi}^2)^{3 / 2} \biggr]^{-3} } \biggl( \frac{G^3M_\mathrm{tot}^2}{K^5} \biggr)^{1/2}\biggr\}
+ \biggl\{ 5\sigma^2 \biggl[ \frac{3u}{(u-1)} \biggr]^{3 / 2} +  2 \biggr\} x \, .
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
If we ''assume'' that <math>~\sigma^2 = 0</math>, then the governing relation is,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~0</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,240: Line 1,232:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~4u(u-1)^2 \frac{d^2x}{du^2}  
<math>~
+ (14u-12)(u-1)\frac{dx}{du}  
\biggl( \frac{G^3M_\mathrm{tot}^2}{K^5} \biggr)^{1 / 2}
+ 2 x \, .
\biggl[ \frac{\pi}{2\cdot 3^{5}}\biggr]^{1 / 2} \cancelto{\mathrm{mistake}}{\biggl[ {\tilde\xi}  (3 + {\tilde\xi}^2)^{-1 / 2} \biggr]^{9} }
( 3 + \xi^2 )^{1 / 2}
\, ;
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
Now, again, guided by our  [[User:Tohline/Appendix/Ramblings/Nonlinar_Oscillation#Conjectures|conjecture]], let's guess an eigenfunction of the form:
=====First Guess (n5)=====
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~x</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,262: Line 1,249:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~ \biggl[ 3^{-1} ( 3 + \xi^2 ) \biggr]^{1/2}
A^3 (u - 1)^{1 / 2} (A u - 1 )^{-1 / 2}   \, ,
\biggl\{ \biggl[ \frac{\pi}{2\cdot 3^{4}}\biggr]^{1 / 2} \biggl[{\tilde\xi}^{-3(3 + {\tilde\xi}^2)^{3 / 2} \biggr] \biggl( \frac{G^3M_\mathrm{tot}^2}{K^5} \biggr)^{1/2}\biggr\}
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
in which case,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\frac{dx}{du}</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,281: Line 1,263:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~  
\frac{A^3}{2} \biggl[ (u - 1)^{-1 / 2} (A u - 1 )^{-1 / 2} - A(u - 1)^{1 / 2} (A u - 1 )^{-3 / 2} \biggr]
\biggl( \frac{G^3M_\mathrm{tot}^2}{K^5} \biggr)^{1/2}\biggl[ \frac{\pi}{2\cdot 3^{5}}\biggr]^{1 / 2}  
\biggl[ \frac{(3 + {\tilde\xi}^2)}{{\tilde\xi}^2 } \biggr]^{3 / 2} ( 3 + \xi^2 )^{1/2}
</math>
</math>
   </td>
   </td>
Line 1,289: Line 1,272:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\frac{g_0\rho_0}{P_0} </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,295: Line 1,278:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
\biggl[ \frac{A^3(A-1)}{2} \biggr] (u-1)^{-1 / 2} (Au-1)^{-3 / 2} \, ;
\biggl( \frac{G^3M_\mathrm{tot}^2}{K^5} \biggr)^{1/2}\biggl[ \frac{\pi}{2\cdot 3^{5}}\biggr]^{1 / 2}  
\biggl[ \frac{(3 + {\tilde\xi}^2)}{{\tilde\xi}^2 } \biggr]^{3 / 2} ( 3 + \xi^2 )^{1/2}
</math>
</math>
   </td>
   </td>
Line 1,303: Line 1,287:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\frac{d^2x}{du^2}</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
&nbsp;
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~  \times ~
\biggl[ \frac{A^3(A-1)}{2} \biggr] \biggl\{
\biggl( \frac{G^2M_\mathrm{tot}^2}{R_\mathrm{norm}^4} \biggr)^{1 / 2}\biggl[ \frac{2^6\cdot 3^{14}}{\pi^2}\biggr]^{1 / 2}  
-\frac{1}{2}(u-1)^{-3 / 2} (Au-1)^{-3 / 2} -\frac{3A}{2} (u-1)^{-1 / 2} (Au-1)^{-5 / 2}
\biggl[ \frac{(3 + {\tilde\xi}^2)}{{\tilde\xi}^2 } \biggr]^{-9 / 2}
\biggr\}
\xi ( 3 + \xi^2 )^{-3/2}  
</math>
</math>
   </td>
   </td>
Line 1,325: Line 1,309:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~ \biggl( \frac{G^5 M_\mathrm{tot}^4}{K^5} \biggr)^{1 / 2}  R_\mathrm{norm}^{-2}
-\frac{1}{2} \biggl[ \frac{A^3(A-1)}{2} \biggr] (u-1)^{-3 / 2} (Au-1)^{-5 / 2}\biggl[
\biggl[ \frac{{\tilde\xi}^2 }{(3 + {\tilde\xi}^2)} \biggr]^{3} \biggl[ \frac{2^5\cdot 3^{9}}{\pi}\biggr]^{1 / 2} \xi ( 3 + \xi^2 )^{-1}
(Au-1) +3A (u-1)\biggr]
</math>
</math>
   </td>
   </td>
Line 1,340: Line 1,323:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
\biggl[ \frac{A^3(A-1)}{4} \biggr] (u-1)^{-3 / 2} (Au-1)^{-5 / 2}\biggl[(3A+1) - 4Au \biggr] \, .
\biggl( \frac{K^5}{G^5 M_\mathrm{tot}^4} \biggr)^{1 / 2}
\biggl[ \frac{{\tilde\xi}^2 }{(3 + {\tilde\xi}^2)} \biggr]^{3} \biggl[ \frac{2^5\cdot 3^{9}}{\pi}\biggr]^{1 / 2}  
\xi ( 3 + \xi^2 )^{-1}
\, .
</math>
</math>
   </td>
   </td>
Line 1,348: Line 1,334:
</div>
</div>


<!-- EXTRA 2nd DERIVATIVE
===The Wave Equation===
<tr>
 
  <td align="right">
====Starting from our Key Adiabatic Wave Equation====
<math>~\frac{d^2x}{du^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{A^3}{2^2} \biggl[ -(u - 1)^{-3 / 2} (A u - 1 )^{-1 / 2} - A(u - 1)^{-1 / 2} (A u - 1 )^{-3 / 2}
- A(u - 1)^{-1 / 2} (A u - 1 )^{-3 / 2} + 3A^2(u - 1)^{1 / 2} (A u - 1 )^{-5 / 2} \biggr]
</math>
  </td>
</tr>
END EXTRA 2nd DERIVATIVE -->


So the governing relation becomes:
The [[#Adiabatic_.28Polytropic.29_Wave_Equation|adiabatic wave equation]] therefore becomes,


<div align="center">
<div align="center">
Line 1,376: Line 1,349:
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~=</math>
  </td>
  <td align="left">
<math>~4u(u-1)^2 \biggl\{ \biggl[ \frac{A^3(A-1)}{4} \biggr] (u-1)^{-3 / 2} (Au-1)^{-5 / 2}\biggl[(3A+1) - 4Au \biggr]  \biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
+ (14u-12)(u-1) \biggl\{ \biggl[ \frac{A^3(A-1)}{2} \biggr] (u-1)^{-1 / 2} (Au-1)^{-3 / 2} \biggr\}
\frac{d^2x}{dr_0^2} + \biggl[\frac{4}{r_0} - \biggl(\frac{g_0 \rho_0}{P_0}\biggr) \biggr] \frac{dx}{dr_0}
+ 2 A^3 (u - 1)^{1 / 2} (A u - 1 )^{-1 / 2}  
+ \biggl(\frac{\rho_0}{\gamma_\mathrm{g} P_0} \biggr)\biggl[\omega^2 + (4 - 3\gamma_\mathrm{g})\frac{g_0}{r_0} \biggr]  x
</math>
</math>
   </td>
   </td>
Line 1,406: Line 1,366:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~u(u-1)^{1 / 2} A^3(A-1) (Au-1)^{-5 / 2}\biggl[(3A+1) - 4Au \biggr
<math>~
\frac{d^2x}{dr_0^2} + \frac{1}{R_\mathrm{norm}} \biggl\{
\biggl[ \frac{2^7\cdot 3^{7}}{\pi} \biggr]^{1 / 2}   \biggl[ \frac{ {\tilde\xi}^2}{(3 + {\tilde\xi}^2)} \biggr]^{3} \frac{1}{\xi}
-   \biggl[ \frac{{\tilde\xi}^2 }{(3 + {\tilde\xi}^2)} \biggr]^{3} \biggl[ \frac{2^5\cdot 3^{9}}{\pi}\biggr]^{1 / 2} \xi ( 3 + \xi^2 )^{-1}
\biggr\} \frac{dx}{dr_0}
</math>
</math>
   </td>
   </td>
Line 1,420: Line 1,384:
   <td align="left">
   <td align="left">
<math>~
<math>~
+ (7u-6)(u-1)^{1 / 2}  A^3(A-1)  (Au-1)^{-3 / 2}  
+ \frac{(4 - 3\gamma_\mathrm{g})}{\gamma_g R_\mathrm{norm}^2}
+ 2 A^3 (u - 1)^{1 / 2} (A u - 1 )^{-1 / 2}  
\biggl[ \frac{\pi}{2\cdot 3^{5}}\biggr]^{1 / 2} \biggl[ \frac{{\tilde\xi}^2}{(3 + {\tilde\xi}^2)} \biggr]^{-3/2}
( 3 + \xi^2 )^{1 / 2}   
\biggl\{ \frac{R_\mathrm{norm}^3}{GM_\mathrm{tot}} \biggl[\frac{\omega^2}{(4 - 3\gamma_\mathrm{g})} \biggr]
+ \biggl[ \frac{2^3\cdot 3^{7}}{\pi}\biggr]^{3/2}  
\biggl[ \frac{{\tilde\xi}^2}{(3 + {\tilde\xi}^2)} \biggr]^{15/2}
( 3 + \xi^2 )^{-3/2}
\biggr\}  x
</math>
</math>
   </td>
   </td>
Line 1,434: Line 1,404:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~(u-1)^{1 / 2} \biggl\{ uA^3(A-1) (Au-1)^{-5 / 2}\biggl[(3A+1) - 4Au \biggr]
<math>~
+ (7u-6) A^3(A-1)  (Au-1)^{-3 / 2}
\frac{d^2x}{dr_0^2} + \frac{1}{R_\mathrm{norm}} \biggl[ \frac{2^3\cdot 3^{7}}{\pi} \biggr]^{1 / 2} \biggl[ \frac{ {\tilde\xi}^2}{(3 + {\tilde\xi}^2)} \biggr]^{3} \biggl[
+ 2 A^3 (A u - 1 )^{-1 / 2} \biggr\}
\frac{4}{\xi} -   \frac{6 \xi}{ ( 3 + \xi^2 )}
\biggr] \frac{dx}{dr_0}  
</math>
</math>
   </td>
   </td>
Line 1,446: Line 1,417:
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
&nbsp;
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~A^3(u-1)^{1 / 2} (Au-1)^{-5 / 2} \biggl\{ u(A-1) \biggl[(3A+1) - 4Au \biggr]
<math>~
+ (7u-6) (A-1) (Au-1)
+ \frac{6(4 - 3\gamma_\mathrm{g})}{\gamma_g R_\mathrm{norm}^2}
+ 2 (A u - 1 )^{2} \biggr\}
( 3 + \xi^2 )^{1 / 2}   \biggl[ \frac{2^3\cdot 3^{7}}{\pi}\biggr]  \biggl[ \frac{{\tilde\xi}^2}{(3 + {\tilde\xi}^2)} \biggr]^{6}
\biggl\{ \frac{R_\mathrm{norm}^3}{GM_\mathrm{tot}} \biggl[\frac{\omega^2}{(4 - 3\gamma_\mathrm{g})} \biggr] \biggl[ \frac{2^3\cdot 3^{7}}{\pi}\biggr]^{-3/2} \biggl[ \frac{{\tilde\xi}^2}{(3 + {\tilde\xi}^2)} \biggr]^{-15/2}
+   
( 3 + \xi^2 )^{-3/2}
\biggr\} x
</math>
</math>
   </td>
   </td>
Line 1,464: Line 1,439:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~A^3(u-1)^{1 / 2} (Au-1)^{-5 / 2} \biggl\{ - 4u^2 A(A-1) + u(A-1) (3A+1)
<math>~
+ (7u-6) [A(A-1)u +1 - A]
\frac{d^2x}{dr_0^2} + \frac{1}{R_*} \biggl[ \frac{ {\tilde\xi}^2}{(3 + {\tilde\xi}^2)} \biggr]^{3} \biggl[
+ 2  (A^2u^2 - 2Au +1) \biggr\}
\frac{4}{\xi} -  \frac{6 \xi}{ ( 3 + \xi^2 )}
\biggr] \frac{dx}{dr_0}
+ \frac{6}{\gamma_g R_*^2}  \biggl[ \frac{{\tilde\xi}^2}{(3 + {\tilde\xi}^2)} \biggr]^{6}
\biggl\{ \frac{\omega^2 R_*^3}{GM_\mathrm{tot}}  \biggl[ \frac{{\tilde\xi}^2}{(3 + {\tilde\xi}^2)} \biggr]^{-15/2}( 3 + \xi^2 )^{1 / 2} 
+ \frac{(4 - 3\gamma_\mathrm{g})}{( 3 + \xi^2 ) }
\biggr\} x
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
where,
<div align="center">
<math>R_* \equiv R_\mathrm{norm} \biggl[ \frac{\pi}{2^3 \cdot 3^7} \biggr]^{1/2} \, .</math>
</div>
Recognizing that,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~r_0</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,479: Line 1,470:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~A^3(u-1)^{1 / 2} (Au-1)^{-5 / 2} \biggl\{ u^2 \biggl[ - 4A(A-1) +7A(A-1) +2A^2 \biggr]  
<math>~
+ u\biggl[ (A-1) (3A+1) - 7(A-1) -6A(A-1) - 4A \biggr]  + 2(3A-2)  \biggr\}
R_*  \biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3} \xi \, ,
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
we can write,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~0</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,493: Line 1,490:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~A^3(u-1)^{1 / 2} (Au-1)^{-5 / 2} \biggl\{ Au^2 \biggl[ 5A-3 \biggr]  
<math>~\frac{1}{R_*^2} \biggl[ \frac{ {\tilde\xi}^2}{(3 + {\tilde\xi}^2)} \biggr]^{6} \biggl\{
+ u\biggl[ 3A^2-2A-1-7A+7 -6A^2+6A -4A \biggr]  + 2(3A-2)  \biggr\} \, .
\frac{d^2x}{d\xi^2} + \biggl[
\frac{4}{\xi} -   \frac{6 \xi}{ ( 3 + \xi^2 )}
\biggr] \frac{dx}{d\xi}
+ \frac{6}{\gamma_g } 
\biggl[\sigma^2 ( 3 + \xi^2 )^{1 / 2
+ \frac{(4 - 3\gamma_\mathrm{g})}{( 3 + \xi^2 ) }
\biggr] x \biggr\} \, ,
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
where,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\sigma^2</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~\equiv</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~A^3(u-1)^{1 / 2} (Au-1)^{-5 / 2} \biggl\{ Au^2 \biggl[ 5A-3 \biggr]   
<math>~ \frac{\omega^2 R_*^3}{GM_\mathrm{tot}}  \biggl( \frac{3 + {\tilde\xi}^2}{{\tilde\xi}^2} \biggr)^{15/2} = \frac{\sigma_c^2}{2\cdot 3^{3/2}}\, .</math>
+ u\biggl[ -3A^2 -7A  +6\biggr]  + 2(3A-2)  \biggr\} \, .
</math>
   </td>
   </td>
</tr>
</tr>
Line 1,515: Line 1,522:
</div>
</div>


=====Second Guess (n5)=====
Finally, if &#8212; because we are specifically considering the case of <math>~n=5</math> &#8212; we set <math>~\gamma_\mathrm{g} = 1 + 1/n = 6/5</math>, we have,
 
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 1,521: Line 1,529:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~x</math>
<math>~0</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,528: Line 1,536:
   <td align="left">
   <td align="left">
<math>~
<math>~
(u - 1)^{b / 2} (A u - 1 )^{-a / 2}   \, ,
\frac{d^2x}{d\xi^2} + \biggl[ \frac{4}{\xi} -  \frac{6 \xi}{ ( 3 + \xi^2 )} \biggr] \frac{dx}{d\xi}
+ \biggl[\frac{5\sigma_c^2}{2\cdot 3^{3/2}} ( 3 + \xi^2 )^{1 / 2} +  \frac{2}{( 3 + \xi^2 ) }\biggr]  x
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
 
</div>
in which case,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\frac{dx}{du}</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,546: Line 1,550:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~\frac{1}{( 3 + \xi^2 ) } \biggl\{ ( 3 + \xi^2 )\frac{d^2x}{d\xi^2}  
\frac{b}{2}(u-1)^{b/2-1} (A u - 1 )^{-a / 2}
+ \biggl[ \frac{2(6 - \xi^2) }{ \xi} \biggr] \frac{dx}{d\xi}  
- \frac{aA}{2}(u - 1)^{b / 2} (A u - 1 )^{-a / 2-1}
+ \biggl[\frac{5\sigma_c^2}{2\cdot 3^{3/2}( 3 + \xi^2 )^{3 / 2} +  2 \biggr]  x \biggr\} \, ,
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
which matches exactly the [[#n5LAWE|form of the LAWE derived above]], if in ''that'' expression, <math>~\gamma_g</math> is also forced to align with our specification of the polytropic index, that is, if <math>~\gamma_g = (n+1)/n = 6/5</math> and, in turn, <math>~\alpha = (3-4/\gamma) = -1/3</math>.
====Starting from the HRW66 Radial Pulsation Equation====
More directly, if we begin with the [[User:Tohline/SSC/Stability/Polytropes#HRW66excerpt|HRW66 radial pulsation equation]] that is already tuned to polytropic configurations, the wave equation appropriate to <math>~n=5</math> polytropes is,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~0</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,561: Line 1,574:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~x \biggl[  
<math>~
\frac{b}{2}(u-1)^{-1}  
\frac{d^2 X}{d\xi^2} + \biggl[ \frac{4}{\xi} - \frac{6 (-\theta^'_5)}{\theta_5} \biggr]\frac{d X}{d\xi}
- \frac{aA}{2} (A u - 1 )^{-1} \biggr]
+ \frac{5(-\theta_5^') }{6\theta_5 \xi} \bigg[ \frac{\xi (s^')^2}{\theta^'_5} + \frac{12}{5} \biggr] X
</math>
</math>
   </td>
   </td>
Line 1,570: Line 1,583:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\Rightarrow ~~~ \frac{(u-1)}{x} \frac{dx}{du}</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,576: Line 1,589:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ (A u - 1 )^{-1} \biggl[  
<math>~
\frac{b}{2} (A u - 1 )
\frac{d^2 X}{d\xi^2} + \biggl[ \frac{4}{\xi} - \frac{6 \xi}{(3 + \xi^2)} \biggr]\frac{d X}{d\xi}
- \frac{aA}{2} (u-1) \biggr]
+ \frac{1}{(3 + \xi^2)} \bigg[ -\frac{5(s^')^2(3 + \xi^2)^{3 / 2}}{2 \cdot 3^{3 / 2}} + 2 \biggr] X
</math>
</math>
   </td>
   </td>
Line 1,591: Line 1,604:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{1 }{2(A u - 1 )} \biggl[  
<math>~\frac{1}{(3+\xi^2)} \biggl\{
b (A u - 1 ) - aA (u-1) \biggr]
(3+\xi^2)\frac{d^2 X}{d\xi^2} + \biggl[ \frac{2(6-\xi^2)}{\xi}\biggr]\frac{d X}{d\xi}
+ \bigg[ -\frac{5(s^')^2}{2 \cdot 3^{3 / 2}} \cdot (3 + \xi^2)^{3 / 2} + 2 \biggr] X
\biggr\} \, ,
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
which is identical to the brute-force derivation just presented, allowing for the mapping,
<div align="center">
<math>\sigma^2 ~~ \Leftrightarrow ~~ -\frac{(s^')^2}{2 \cdot 3^{3 / 2}} \, .</math>
</div>
Finally, remembering that the [[User:Tohline/SSC/Stability/Polytropes#HRW66frequency|HRW66 dimensionless frequency definition]] is,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~(s^')^2</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,605: Line 1,630:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~-\omega^2 \biggl[\frac{n+1 }{4\pi G \rho_c} \biggr] \, ,</math>
\frac{1 }{2(A u - 1 )} \biggl[
(aA - b) + A(b    - a)u \biggr] \, ;
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</div>
</div>
and,
we recognize that, specifically for the case of <math>~n=5</math>, we can make the substitution, <math>~(s^')^2 \rightarrow -\sigma_c^2</math>, in which case the LAWE becomes,
 
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 1,619: Line 1,642:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\frac{d^2x}{du^2}</math>
<math>~0</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,625: Line 1,648:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~\frac{1}{(3+\xi^2)} \biggl\{
\biggl[ \frac{b}{2}(u-1)^{-1} - \frac{aA}{2(A u - 1 )^{-1} \biggr]\frac{dx}{du}  
(3+\xi^2)\frac{d^2 X}{d\xi^2} + \biggl[ \frac{2(6-\xi^2)}{\xi}\biggr]\frac{d X}{d\xi}
+ x \frac{d}{du}\biggl[ \frac{b}{2}(u-1)^{-1} - \frac{aA}{2}  (A u - 1 )^{-1} \biggr]
+ \bigg[ \frac{5\sigma_c^2}{2 \cdot 3^{3 / 2}} \cdot (3 + \xi^2)^{3 / 2} + 2 \biggr] X
\biggr\} \, ,
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
which matches exactly the [[#n5LAWE|form of the LAWE derived above]], if in ''that'' expression, <math>~\gamma_g</math> is also forced to align with our specification of the polytropic index, that is, if <math>~\gamma_g = (n+1)/n = 6/5</math> and, in turn, <math>~\alpha = (3-4/\gamma) = -1/3</math>.
====New Independent Variable====
Guided by our [[User:Tohline/Appendix/Ramblings/Nonlinar_Oscillation#Conjectures|conjecture]] regarding the proper shape of the radial eigenfunction, let's switch the dependent variable to,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~u  \equiv 1 + \frac{3}{\xi^2}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
&nbsp; &nbsp; &nbsp; &nbsp; <math>~\Rightarrow</math> &nbsp; &nbsp; &nbsp; &nbsp;
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~3 + \xi^2 = \frac{3u}{(u-1)}  \, ,</math>
x\biggl[ \frac{b}{2}(u-1)^{-1} - \frac{aA}{2(A u - 1 )^{-1} \biggr]^2
  </td>
+ x  \biggl[ -\frac{b}{2}(u-1)^{-2} + \frac{aA^2}{2}  (A u - 1 )^{-2} \biggr]
  <td align="center">
</math>
&nbsp; &nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp; &nbsp;
  </td>
  <td align="left">
<math>~\xi = 3^{1 / 2} (u-1)^{-1 / 2} \, .</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
This implies that,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\frac{d}{d\xi}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~~~\rightarrow ~~~</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ \frac{x}{4(u-1)^2 (Au-1)^2} \biggl\{
<math>~-\frac{2}{\sqrt{3}}(u-1)^{3 / 2} \frac{d}{du} \, ,</math>
\biggl[ b(Au-1) - aA  (u - 1 ) \biggr]^2
\biggl[  2aA^2 (u-1)^{2} -2b (A u - 1 )^{2} \biggr]
\biggr\}
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
and,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\Rightarrow ~~~ \frac{(1-u)^2}{x}\frac{d^2x}{du^2}</math>
<math>~\frac{d^2}{d\xi^2}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~~~\rightarrow ~~~</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~  
<math>~\frac{4}{3}(u-1)^3 \frac{d^2}{du^2} + 2(u-1)^{2} \frac{d}{du} \, .</math>
\frac{1}{4 (Au-1)^2} \biggl\{
\biggl[ b(Au-1) - aA  (u - 1 ) \biggr]^2
+ \biggl[  2aA^2 (u-1)^{2} -2b (A u - 1 )^{2} \biggr]
\biggr\}
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</div>
</div>
Hence, the governing wave equation becomes,
Hence, the governing wave equation becomes,


Line 1,695: Line 1,730:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~2u \biggl\{ \frac{(u-1)^2}{x} \frac{d^2x}{du^2} \biggr\}
<math>~( 3 + \xi^2 )\frac{d^2x}{d\xi^2}  
+ (7u-6)\biggl\{ \frac{(u-1)}{x} \frac{dx}{du} \biggl\} + 1
+ \biggl[ \frac{2(6 - \xi^2) }{ \xi} \biggr] \frac{dx}{d\xi}  
+ \biggl[5\sigma^2 ( 3 + \xi^2 )^{3 / 2} + 2 \biggr]  x
</math>
</math>
   </td>
   </td>
Line 1,709: Line 1,745:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~\frac{3u}{(u-1)} \biggl[\frac{4}{3}(u-1)^3 \frac{d^2x}{du^2} + 2(u-1)^{2} \frac{dx}{du}\biggr]
\frac{2u}{4 (Au-1)^2} \biggl\{
+ 4(2u-3)(u-1)\frac{dx}{du}
\biggl[ (aA - b) + A(b    - a)u \biggr]^2
+ \biggl\{ 5\sigma^2 \biggl[ \frac{3u}{(u-1)} \biggr]^{3 / 2} +  2 \biggr\} x
+ \biggl[  2aA^2 (u-1)^{2} -2b (A u - 1 )^{2} \biggr]
\biggr\}
</math>
</math>
   </td>
   </td>
Line 1,723: Line 1,757:
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~4u(u-1)^2 \frac{d^2x}{du^2}
+ \frac{(7u-6) }{2(A u - 1 )} \biggl[  
+ (14u-12)(u-1)\frac{dx}{du}  
(aA - b) + A(b    - a)u \biggr]  
+ \biggl\{ 5\sigma^2 \biggl[ \frac{3u}{(u-1)} \biggr]^{3 / 2} + 2 \biggr\}  x \, .
  + 1
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
{{ LSU_WorkInProgress }}
If we ''assume'' that <math>~\sigma^2 = 0</math>, then the governing relation is,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~0</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,742: Line 1,785:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~4u(u-1)^2 \frac{d^2x}{du^2}
\frac{1}{4 (Au-1)^2} \biggl\{
+ (14u-12)(u-1)\frac{dx}{du}
2u\biggl[ (aA - b)^2 + 2(aA - b)A(b    - a)u + A^2(b    - a)^2u^2 \biggr]
+ 2 x \, .
+ 2u\biggl[  2aA^2 (u^2 - 2u + 1)  -2b (A^2 u^2 - 2Au + 1 ) \biggr]
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
Now, again, guided by our  [[User:Tohline/Appendix/Ramblings/Nonlinar_Oscillation#Conjectures|conjecture]], let's guess an eigenfunction of the form:
=====First Guess (n5)=====
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~x</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
+ 2(A u - 1 )(7u-6) \biggl[
A^3 (u - 1)^{1 / 2} (A u - 1 )^{-1 / 2\, ,
(aA - b) + A(b    - a)u \biggr]
+ 4 (Au-1)^2 \biggr\}
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
in which case,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\frac{dx}{du}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,775: Line 1,827:
   <td align="left">
   <td align="left">
<math>~
<math>~
\frac{1}{4 (Au-1)^2} \biggl\{
\frac{A^3}{2} \biggl[ (u - 1)^{-1 / 2} (A u - 1 )^{-1 / 2} - A(u - 1)^{1 / 2} (A u - 1 )^{-3 / 2} \biggr]
2u\biggl[ (aA - b)^2 + 2(aA - b)A(b    - a)u + A^2(b    - a)^2u^2 \biggr]
+  2u\biggl[ 2A^2(a-b)u^2 + 4A(b - aA) u + 2(aA^2  -b)  \biggr]
</math>
</math>
   </td>
   </td>
Line 1,787: Line 1,837:
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
+ 2\biggl[7Au^2 - (6A+7)u  +6 \biggr]\biggl[
\biggl[ \frac{A^3(A-1)}{2} \biggr] (u-1)^{-1 / 2} (Au-1)^{-3 / 2} \, ;
(aA - b) + A(b    - a)u \biggr]
+  (4A^2u^2-8Au + 4) \biggr\}
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
If <math>~b=a</math>,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~0</math>
<math>~\frac{d^2x}{du^2}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,814: Line 1,855:
   <td align="left">
   <td align="left">
<math>~
<math>~
2u\biggl[ (aA - b)^2   \biggr]
\biggl[ \frac{A^3(A-1)}{2} \biggr] \biggl\{
+  2u\biggl[ 4A(b - aA) u + 2(aA^2   -b) \biggr]
-\frac{1}{2}(u-1)^{-3 / 2} (Au-1)^{-3 / 2} -\frac{3A}{2} (u-1)^{-1 / 2} (Au-1)^{-5 / 2}
\biggr\}
</math>
</math>
   </td>
   </td>
Line 1,825: Line 1,867:
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
+ 2\biggl[7Au^2 - (6A+7)u  +6 \biggr]\biggl[  
-\frac{1}{2} \biggl[ \frac{A^3(A-1)}{2} \biggr] (u-1)^{-3 / 2} (Au-1)^{-5 / 2}\biggl[
(aA - b) \biggr]  + (4A^2u^2-8Au + 4)  
(Au-1) +3A (u-1)\biggr]
</math>
</math>
   </td>
   </td>
Line 1,844: Line 1,886:
   <td align="left">
   <td align="left">
<math>~
<math>~
2a^2u (A - 1)^2
\biggl[ \frac{A^3(A-1)}{4} \biggr] (u-1)^{-3 / 2} (Au-1)^{-5 / 2}\biggl[(3A+1) - 4Au \biggr] \, .
+  2au [ 4A(1 - A) u + 2(A^2  -1) ]
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>


<!-- EXTRA 2nd DERIVATIVE
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\frac{d^2x}{du^2}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
+ 2a(A - 1) \biggl[7Au^2 - (6A+7)u +6 \biggr] +  (4A^2u^2-8Au + 4)  
\frac{A^3}{2^2} \biggl[ -(u - 1)^{-3 / 2} (A u - 1 )^{-1 / 2} - A(u - 1)^{-1 / 2} (A u - 1 )^{-3 / 2}
- A(u - 1)^{-1 / 2} (A u - 1 )^{-3 / 2} + 3A^2(u - 1)^{1 / 2} (A u - 1 )^{-5 / 2} \biggr]
</math>
</math>
   </td>
   </td>
</tr>
</tr>
END EXTRA 2nd DERIVATIVE -->
 
So the governing relation becomes:
 
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~0</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,872: Line 1,923:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~4u(u-1)^2 \biggl\{ \biggl[ \frac{A^3(A-1)}{4} \biggr] (u-1)^{-3 / 2} (Au-1)^{-5 / 2}\biggl[(3A+1) - 4Au \biggr] \biggr\}
2Au^2 [4a (1 - A) +  7a(A - 1) + 2A] + 2u [ a^2 (A - 1)^2 +  2a(A^2  -1) - a(A - 1) (6A+7) - 4A] + 4[ 3a(A-1) + 1]
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
 
</div>
<tr>
This should then match the [[#First_Guess_.28n5.29|"first guess"]] algebraic condition if we set <math>~a=1</math>.  Let's see.
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
   <td align="right">
   <td align="right">
<math>~0</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
&nbsp;
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
2Au^2 [4 (1 - A) +  7(A - 1) + 2A] + 2u [ (A - 1)^2 +  2(A^2  -1) - (A - 1) (6A+7) - 4A] + 4[ 3(A-1) + 1]
+ (14u-12)(u-1) \biggl\{ \biggl[ \frac{A^3(A-1)}{2} \biggr] (u-1)^{-1 / 2} (Au-1)^{-3 / 2} \biggr\}
+ 2 A^3 (u - 1)^{1 / 2} (A u - 1 )^{-1 / 2}
</math>
</math>
   </td>
   </td>
Line 1,906: Line 1,951:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~u(u-1)^{1 / 2} A^3(A-1) (Au-1)^{-5 / 2}\biggl[(3A+1) - 4Au \biggr]
2Au^2 [4  - 4A +  7A - 7 + 2A] + 2u [ (A^2 - 2A + 1) +  2A^2   -2 + (1-A ) (6A+7) -4A] + 4[ 3A-2]
</math>
</math>
   </td>
   </td>
Line 1,917: Line 1,961:
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
&nbsp;
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
2Au^2 [5A - 3] + 2u [  - 3A^2 - 7A + 6 ] + 4[ 3A-2] \, .
+ (7u-6)(u-1)^{1 / 2}  A^3(A-1)  (Au-1)^{-3 / 2}
+ 2 A^3 (u - 1)^{1 / 2} (A u - 1 )^{-1 / 2}
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
And we see that this expression ''does'' match the one derived earlier.
Going back a bit, before setting <math>~a=1</math>, we have the expression:
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~0</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,943: Line 1,979:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~(u-1)^{1 / 2} \biggl\{ uA^3(A-1) (Au-1)^{-5 / 2}\biggl[(3A+1) - 4Au \biggr]  
2Au^2 [4a (1 - A) + 7a(A - 1) + 2A] + 2u [ a^2 (A - 1)^2 +  2a(A^-1) - a(A - 1) (6A+7) - 4A] + 4[ 3a(A-1) + 1]
+ (7u-6) A^3(A-1) (Au-1)^{-3 / 2}
+ 2 A^3 (A u - 1 )^{-1 / 2} \biggr\}
</math>
</math>
   </td>
   </td>
Line 1,957: Line 1,994:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~A^3(u-1)^{1 / 2} (Au-1)^{-5 / 2} \biggl\{ u(A-1) \biggl[(3A+1) - 4Au \biggr]
2Au^2 [  3aA  -3a + 2A] + 2u [ a^2 (A - 1)^2 +  2a(A^2  -1) - a(6A^2+A-7) - 4A] + 4[ 3a(A-1) + 1]
+ (7u-6) (A-1) (Au-1)
+ 2 (A u - 1 )^{2} \biggr\}
</math>
</math>
   </td>
   </td>
Line 1,971: Line 2,009:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~A^3(u-1)^{1 / 2} (Au-1)^{-5 / 2} \biggl\{ - 4u^2 A(A-1)  + u(A-1) (3A+1)  
2Au^2 [  3a(- 1) + 2A] + 2u [ a^2 (A - 1)^2 +  a(  -4A^2-A+5) - 4A] + 4[ 3a(A-1) + 1] \, .
+ (7u-6)  [A(A-1)u +1 - A]
+ 2  (A^2u^2 - 2Au +1) \biggr\}
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
Now, in order for all three expressions inside the square-bracket pairs to be zero, we need, first,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~3a(A  - 1) + 2A</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,990: Line 2,024:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~0</math>
<math>~A^3(u-1)^{1 / 2} (Au-1)^{-5 / 2} \biggl\{ u^2 \biggl[ - 4A(A-1) +7A(A-1) +2A^2 \biggr]   
+ u\biggl[ (A-1) (3A+1) - 7(A-1) -6A(A-1) - 4A \biggr]  + 2(3A-2)  \biggr\}
</math>
   </td>
   </td>
</tr>
</tr>
Line 1,996: Line 2,032:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\Rightarrow ~~~ a</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,002: Line 2,038:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{2A}{3(1-A)} \, ;</math>
<math>~A^3(u-1)^{1 / 2} (Au-1)^{-5 / 2} \biggl\{ Au^2 \biggl[ 5A-3 \biggr]   
+ u\biggl[ 3A^2-2A-1-7A+7 -6A^2+6A -4A \biggr]  + 2(3A-2) \biggr\} \, .
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
and, third, by simple visual comparison with the first expression,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~3a(A-1) + 1</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,019: Line 2,052:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~3a(A-1) + 2A</math>
<math>~A^3(u-1)^{1 / 2} (Au-1)^{-5 / 2} \biggl\{ Au^2 \biggl[ 5A-3 \biggr]   
+ u\biggl[ -3A^2 -7A  +6\biggr]  + 2(3A-2)  \biggr\} \, .
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>


<tr>
=====Second Guess (n5)=====
  <td align="right">
<div align="center">
<math>~\Rightarrow A</math>
<table border="0" cellpadding="5" align="center">
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{2} </math>
  </td>
</tr>


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\Rightarrow ~~~ a</math>
<math>~x</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,043: Line 2,072:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{2}{3} \, ;</math>
<math>~
(u - 1)^{b / 2} (A u - 1 )^{-a / 2}   \, ,
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</div>
</div>
which forces the second expression to the value,
in which case,
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 2,054: Line 2,085:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~a^2 (A - 1)^2 +  a(  -4A^2-A+5) - 4A</math>
<math>~\frac{dx}{du}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,060: Line 2,091:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl(\frac{2}{3}\biggr)^2 \biggl(-\frac{1}{2} \biggr)^2 +  \frac{2}{3}\biggl[  -1-\frac{1}{2} +5 \biggr] - 2</math>
<math>~
\frac{b}{2}(u-1)^{b/2-1} (A u - 1 )^{-a / 2}
- \frac{aA}{2}(u - 1)^{b / 2} (A u - 1 )^{-a / 2-1}
</math>
   </td>
   </td>
</tr>
</tr>
Line 2,072: Line 2,106:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{1}{9} \frac{7}{3}  - 2</math>
<math>~x \biggl[
\frac{b}{2}(u-1)^{-1}  
- \frac{aA}{2(A u - 1 )^{-1} \biggr]
</math>
   </td>
   </td>
</tr>
</tr>
Line 2,078: Line 2,115:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\Rightarrow ~~~ \frac{(u-1)}{x} \frac{dx}{du}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,084: Line 2,121:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{4}{9} \, ,</math>
<math>~ (A u - 1 )^{-1} \biggl[
\frac{b}{2} (A u - 1 )
- \frac{aA}{2} (u-1) \biggr]
</math>
   </td>
   </td>
</tr>
</tr>
</table>
 
</div>
which is ''not'' zero.  Hence our pair of unknown parameters &#8212; <math>~a </math> and <math>~A</math> &#8212; do not simultaneously satisfy all three conditions.  (Not really a surprise.)
 
==Setup Using Lagrangian Mass Coordinate==
 
===Alternative Terms===
Let's change the independent coordinate from <math>~r_0</math> to <math>~m_0</math>.  In particular, the derivative operation will change as follows:
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\frac{d}{dr_0}</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~~\rightarrow~~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl( \frac{dm_0}{dr_0} \biggr)\frac{d}{dm_0}  
<math>~\frac{1 }{2(A u - 1 )} \biggl[
= \biggl( \frac{dm_0}{d\xi} \cdot \frac{d\xi}{dr_0}  \biggr)\frac{d}{dm_0}
b (A u - 1 ) - aA (u-1) \biggr]
\, ,</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
so what is the expression for the leading coefficient?  From [[#r0|above]], we have,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~r_0</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,127: Line 2,151:
   <td align="left">
   <td align="left">
<math>~
<math>~
R_*  \biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3} \xi
\frac{1 }{2(A u - 1 )} \biggl[
(aA - b) + A(b    - a)u \biggr] \, ;
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
and,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\Rightarrow ~~~ \xi</math>
<math>~\frac{d^2x}{du^2}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,141: Line 2,171:
   <td align="left">
   <td align="left">
<math>~
<math>~
\frac{1}{R_*}  \biggl[ \frac{ {\tilde\xi}^2}{(3 + {\tilde\xi}^2)}\biggr]^{3} r_0 \, .
\biggl[ \frac{b}{2}(u-1)^{-1} - \frac{aA}{2(A u - 1 )^{-1} \biggr]\frac{dx}{du}
+ x \frac{d}{du}\biggl[ \frac{b}{2}(u-1)^{-1} - \frac{aA}{2}  (A u - 1 )^{-1} \biggr]
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
Also, from [[#m0|above]], we know that,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~m_0</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,161: Line 2,185:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
M_\mathrm{tot} \biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3 / 2}
x\biggl[ \frac{b}{2}(u-1)^{-1} - \frac{aA}{2} (A u - 1 )^{-1} \biggr]^2
\biggl\{ \xi^3 ( 3 + \xi^2 )^{-3/2} \biggr\}
+ x  \biggl[ -\frac{b}{2}(u-1)^{-2} + \frac{aA^2}{2}  (A u - 1 )^{-2} \biggr]
</math>
</math>
   </td>
   </td>
Line 2,170: Line 2,194:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\Rightarrow ~~~ \frac{dm_0}{d\xi}</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,176: Line 2,200:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~ \frac{x}{4(u-1)^2 (Au-1)^2} \biggl\{
M_\mathrm{tot} \biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3 / 2}
\biggl[ b(Au-1) - aA  (u - 1 ) \biggr]^2
\biggl\{ 3\xi^2 ( 3 + \xi^2 )^{-3/2}  
+ \biggl[  2aA^2 (u-1)^{2} -2b (A u - 1 )^{2} \biggr]
- 3 \xi^4 ( 3 + \xi^2 )^{-5/2}\biggr\}  
\biggr\}
</math>
</math>
   </td>
   </td>
Line 2,186: Line 2,210:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\Rightarrow ~~~ \frac{(1-u)^2}{x}\frac{d^2x}{du^2}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,192: Line 2,216:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~  
M_\mathrm{tot} \biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3 / 2} 3\xi^2 (3 + \xi^2)^{-5/2}  
\frac{1}{4 (Au-1)^2} \biggl\{
\biggl\{ ( 3 + \xi^2 )
\biggl[ b(Au-1) - aA  (u - 1 ) \biggr]^2
- \xi^2  \biggr\}  
\biggl[  2aA^2 (u-1)^{2}  -2b (A u - 1 )^{2} \biggr]
\biggr\}
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
Hence, the governing wave equation becomes,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~0</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,208: Line 2,240:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~2u \biggl\{ \frac{(u-1)^2}{x} \frac{d^2x}{du^2} \biggr\}
M_\mathrm{tot} \biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3 / 2} 3^2\xi^2 (3 + \xi^2)^{-5/2}  
+ (7u-6)\biggl\{ \frac{(u-1)}{x} \frac{dx}{du} \biggl\} + 1
</math>
</math>
   </td>
   </td>
Line 2,216: Line 2,248:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\Rightarrow ~~~ \frac{dm_0}{dr_0}</math>
&nbsp;
</td>
  </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
M_\mathrm{tot} \biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3 / 2} 3^2\xi^2 (3 + \xi^2)^{-5/2}
\frac{2u}{4 (Au-1)^2} \biggl\{
\frac{1}{R_*} \biggl[ \frac{ {\tilde\xi}^2}{(3 + {\tilde\xi}^2)}\biggr]^{3}
\biggl[ (aA - b) + A(b    - a)u \biggr]^2
+ \biggl[ 2aA^2 (u-1)^{2} -2b (A u - 1 )^{2} \biggr]
\biggr\}
</math>
</math>
   </td>
   </td>
Line 2,232: Line 2,266:
   <td align="right">
   <td align="right">
&nbsp;
&nbsp;
</td>
  </td>
   <td align="center">
   <td align="center">
<math>~=</math>
&nbsp;
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ \frac{M_\mathrm{tot} }{R_*}  
<math>~
\biggl[ \frac{ {\tilde\xi}^2}{(3 + {\tilde\xi}^2)}\biggr]^{3 / 2} 3^2\xi^2 (3 + \xi^2)^{-5/2} \, .
+ \frac{(7u-6) }{2(A u - 1 )} \biggl[  
(aA - b) + A(b    - a)u \biggr]  
+ 1
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>


To simplify expressions, let's [[User:Tohline/Appendix/Ramblings/Nonlinar_Oscillation#DefineTildeC|borrow from an accompanying derivation]] and define,
<tr>
<div align="center">
   <td align="right">
<math>\tilde{C} \equiv \frac{3^2}{{\tilde\xi}^2} \biggl( 1 + \frac{ {\tilde\xi}^2}{3} \biggr) = 3 \biggl[ \frac{( 3 + {\tilde\xi}^2 )}{ {\tilde\xi}^2} \biggr] \, .</math>
&nbsp;
</div>
Then we have,
 
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
   <td align="right">
<math>~\frac{m_0}{M_\mathrm{tot}}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,262: Line 2,287:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
\biggl[ \frac{\tilde{C}}{ 3}\biggr]^{3 / 2}
\frac{1}{4 (Au-1)^2} \biggl\{
\biggl[ \frac{\xi^2}{ ( 3 + \xi^2 )} \biggr]^{3/2}
2u\biggl[ (aA - b)^2 + 2(aA - b)A(b    - a)u + A^2(b    - a)^2u^2 \biggr]
+  2u\biggl[  2aA^2 (u^2 - 2u + 1)  -2b (A^2 u^2 - 2Au + 1 ) \biggr]
</math>
</math>
   </td>
   </td>
Line 2,271: Line 2,297:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\Rightarrow ~~~\biggl[ \frac{ 3}{\tilde{C}}\biggr] \biggl[\frac{m_0}{M_\mathrm{tot}}\biggr]^{2 / 3}</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
&nbsp;
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
\frac{\xi^2}{ ( 3 + \xi^2 )}  
+ 2(A u - 1 )(7u-6) \biggl[
(aA - b) + A(b    - a)u \biggr]
+ 4 (Au-1)^2 \biggr\}
</math>
</math>
   </td>
   </td>
Line 2,285: Line 2,313:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\Rightarrow ~~~( 3 + \xi^2 )\biggl[ \frac{ 3}{\tilde{C}}\biggr] \biggl[\frac{m_0}{M_\mathrm{tot}}\biggr]^{2 / 3}</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,291: Line 2,319:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~   
<math>~
\xi^2  
\frac{1}{4 (Au-1)^2} \biggl\{
2u\biggl[ (aA - b)^2 + 2(aA - b)A(b    - a)u + A^2(b    - a)^2u^2 \biggr]
+ 2u\biggl[ 2A^2(a-b)u^2 + 4A(b - aA) u + 2(aA^2  -b)  \biggr]
</math>
</math>
   </td>
   </td>
Line 2,299: Line 2,329:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\Rightarrow ~~~3 m_*</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
&nbsp;
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~  \xi^2 (1-m_*)
<math>~
+ 2\biggl[7Au^2 - (6A+7)u +6 \biggr]\biggl[
(aA - b) + A(b    - a)u \biggr]
+  (4A^2u^2-8Au + 4) \biggr\}
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
If <math>~b=a</math>,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\Rightarrow ~~~\xi^2 </math>
<math>~0</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,318: Line 2,358:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~  \frac{3m_*}{(1-m_*)} \, ,
<math>~
2u\biggl[ (aA - b)^2   \biggr]
+  2u\biggl[ 4A(b - aA) u + 2(aA^2  -b) \biggr]
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
where,
<div align="center">
<math>~m_* \equiv \biggl[ \frac{ 3}{\tilde{C}}\biggr] \biggl[\frac{m_0}{M_\mathrm{tot}}\biggr]^{2 / 3} \, .</math>
</div>
In summary:
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~ 
&nbsp;
\frac{\xi^2}{ ( 3 + \xi^2 )} = m_* \, ;
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp; &nbsp; &nbsp; while, &nbsp; &nbsp; &nbsp;  
&nbsp;
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
\frac{ {\tilde\xi}^2}{ ( 3 + {\tilde\xi}^2 )} = \frac{3}{\tilde{C}} \, ;
+ 2\biggl[7Au^2 - (6A+7)u  +6 \biggr]\biggl[
(aA - b) \biggr]  +  (4A^2u^2-8Au + 4)  
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~r_0</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,363: Line 2,389:
   <td align="left">
   <td align="left">
<math>~
<math>~
R_*  \biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3} \xi
2a^2u (A - 1)^2
= R_* \biggl( \frac{ \tilde{C} }{ 3}\biggr)^{3} \biggr[ \frac{3m_*}{ (1-m_*) }\biggr]^{1 / 2} \, ;
+ 2au [ 4A(1 - A) u + 2(A^2  -1) ]
</math>
</math>
   </td>
   </td>
Line 2,371: Line 2,397:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\frac{g_0\rho_0}{P_0} </math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
&nbsp;
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
\frac{6}{R_*} \biggl[ \frac{ {\tilde\xi}^2 }{ (3 + {\tilde\xi}^2) }\biggr]^{9} \frac{\xi}{ ( 3 + \xi^2 )}
+ 2a(A - 1) \biggl[7Au^2 - (6A+7)u +6 \biggr] + (4A^2u^2-8Au + 4)  
=
\frac{6}{R_*} \biggl[  \frac{ 3 }{ \tilde{C} }\biggr]^{9}  \frac{m_*}{ \xi }
=
\frac{6}{R_*} \biggl[  \frac{ 3 }{ \tilde{C} }\biggr]^{9} m_* \biggl[  \frac{(1-m_*)}{3m_*}  \biggr]^{1 / 2} \, ;
</math>
</math>
   </td>
   </td>
</tr>
</tr>
 
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\frac{g_0 }{r_0} </math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,395: Line 2,417:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
\frac{GM_\mathrm{tot}}{R_*^3}
2Au^2 [4a (1 - A) + 7a(A - 1) + 2A] + 2u [ a^2 (A - 1)^2 + 2a(A^2   -1) - a(A - 1) (6A+7) - 4A] + 4[ 3a(A-1) + 1]
\biggl\frac{ {\tilde\xi}^2 }{ (3 + {\tilde\xi}^2)}\biggr]^{15/2} \frac{1}{\xi^3}
\biggl[ \frac{ \xi^2 }{ ( 3 + \xi^2 ) }\biggr]^{3/2}
=
\frac{GM_\mathrm{tot}}{R_*^3} \biggl[ \frac{3 }{ \tilde{C} }\biggr]^{15/2}  (1-m_*)^{3 / 2} \, ;
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
This should then match the [[#First_Guess_.28n5.29|"first guess"]] algebraic condition if we set <math>~a=1</math>.  Let's see.
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\frac{\rho_0}{\gamma_g P_0} </math>
<math>~0</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,413: Line 2,437:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~   
<math>~
\frac{6R_* }{\gamma_g GM_\mathrm{tot} }\biggl( \frac{ 3}{ \tilde{C} } \biggr)^{9 / 2} \biggl[ \frac{3}{(1-m_*)}\biggr]^{1 / 2} \, .
2Au^2 [4 (1 - A) + 7(A - 1) + 2A] + 2u [ (A - 1)^2 +  2(A^2  -1) - (A - 1) (6A+7) - 4A] + 4[ 3(A-1) + 1]
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
So, the wave equation may be written as,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~0</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,434: Line 2,452:
   <td align="left">
   <td align="left">
<math>~
<math>~
\frac{d^2x}{dr_0^2} + \biggl[\frac{4}{r_0} - \biggl(\frac{g_0 \rho_0}{P_0}\biggr) \biggr] \frac{dx}{dr_0}
2Au^2 [4 - 4A +  7A - 7 + 2A] + 2u [ (A^2 - 2A + 1) +  2A^2  -2 + (1-A ) (6A+7) -4A] + 4[ 3A-2]
+ \biggl(\frac{\rho_0}{\gamma_\mathrm{g} P_0} \biggr)\biggl[\omega^2 + (4 - 3\gamma_\mathrm{g})\frac{g_0}{r_0} \biggr] x
</math>
</math>
   </td>
   </td>
Line 2,449: Line 2,466:
   <td align="left">
   <td align="left">
<math>~
<math>~
\frac{d^2x}{dr_0^2}
2Au^2 [5A - 3] + 2u [ - 3A^2 - 7A + 6 ] + 4[ 3A-2] \, .
+ \biggl\{ \frac{4}{R_*} \biggl( \frac{ 3}{ \tilde{C} }\biggr)^{3} \biggr[ \frac{ (1-m_*) }{3m_*}\biggr]^{1 / 2
- \frac{6}{R_*} \biggl[  \frac{ 3 }{ \tilde{C} }\biggr]^{9}  m_* \biggl[ \frac{(1-m_*)}{3m_*}  \biggr]^{1 / 2}  \biggr\} \frac{dx}{dr_0}
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
And we see that this expression ''does'' match the one derived earlier.
Going back a bit, before setting <math>~a=1</math>, we have the expression:
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~0</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
+ \frac{6R_* }{\gamma_g GM_\mathrm{tot} }\biggl( \frac{ 3}{ \tilde{C} } \biggr)^{9 / 2} \biggl[ \frac{3}{(1-m_*)}\biggr]^{1 / 2}
2Au^2 [4a (1 - A) +  7a(A - 1) + 2A] + 2u [ a^2 (A - 1)^2 + 2a(A^2   -1) - a(A - 1) (6A+7) - 4A] + 4[ 3a(A-1) + 1]
\biggl\{ \omega^2 + (4 - 3\gamma_\mathrm{g})\frac{GM_\mathrm{tot}}{R_*^3} \biggl[ \frac{3 }{ \tilde{C} }\biggr]^{15/2(1-m_*)^{3 / 2} \biggr\}  x
</math>
</math>
   </td>
   </td>
Line 2,480: Line 2,503:
   <td align="left">
   <td align="left">
<math>~
<math>~
\frac{d^2x}{dr_0^2}
2Au^2 [  3aA  -3a + 2A] + 2u [ a^2 (A - 1)^2 + 2a(A^2  -1) - a(6A^2+A-7) - 4A] + 4[ 3a(A-1) + 1]
+ \frac{1}{R_*} \biggl( \frac{ 3}{ \tilde{C} }\biggr)^{3} \biggl\{ 4 
- 6\biggl[  \frac{ 3 }{ \tilde{C} }\biggr]^{6}  m_* \biggr\} \biggr[ \frac{ (1-m_*) }{3m_*}\biggr]^{1 / 2}\frac{dx}{dr_0}
</math>
</math>
   </td>
   </td>
Line 2,492: Line 2,513:
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
+ \frac{6(4 - 3\gamma_\mathrm{g}) }{\gamma_g } \cdot \frac{1 }{R_*^2} \biggl\frac{3 }{ \tilde{C} }\biggr)^{3} \biggl[ \frac{3}{(1-m_*)}\biggr]^{1 / 2}
2Au^2 [  3a(A - 1) + 2A] + 2u [ a^2 (A - 1)^2 +  a(  -4A^2-A+5) - 4A] + 4[ 3a(A-1) + 1] \, .
\biggl\{\sigma^2 + (1-m_*)^{3 / 2} \biggr\}  x
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
Now, in order for all three expressions inside the square-bracket pairs to be zero, we need, first,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~3a(A  - 1) + 2A</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,510: Line 2,535:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{1 }{R_*^2} \biggl(  \frac{3 }{ \tilde{C} }\biggr)^{3} \biggl\{
<math>~0</math>
R_*^2 \biggl(  \frac{ \tilde{C} }{3 }\biggr)^{3}  \frac{d^2x}{dr_0^2}
+ R_* \biggl[ 4  - 6\biggl(  \frac{ 3 }{ \tilde{C} }\biggr)^{6}  m_* \biggr] \biggr[ \frac{ (1-m_*) }{3m_*}\biggr]^{1 / 2}\frac{dx}{dr_0}
</math>
   </td>
   </td>
</tr>
</tr>
Line 2,519: Line 2,541:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\Rightarrow ~~~ a</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~\frac{2A}{3(1-A)} \, ;</math>
+ \frac{6(4 - 3\gamma_\mathrm{g}) }{\gamma_g } \cdot \biggl[ \frac{3}{(1-m_*)}\biggr]^{1 / 2}
\biggl[ \sigma^2  +  (1-m_*)^{3 / 2} \biggr]  x  \biggr\}
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
and, third, by simple visual comparison with the first expression,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~3a(A-1) + 1</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,540: Line 2,564:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{1 }{R_*^2} \biggl( \frac{3 }{ \tilde{C} }\biggr)^{3} \biggl[ \frac{1}{3m_*(1-m_*)}\biggr]^{1 / 2} \biggl\{ [ 3m_*(1-m_*) ]^{1 / 2}
<math>~3a(A-1) + 2A</math>
R_*^2 \biggl(  \frac{ \tilde{C} }{3 }\biggr)^{3}  \frac{d^2x}{dr_0^2}
+ R_* \biggl[ 4  - 6\biggl(  \frac{ 3 }{ \tilde{C} }\biggr)^{6}  m_* \biggr] (1-m_*) \frac{dx}{dr_0}
</math>
   </td>
   </td>
</tr>
</tr>
Line 2,549: Line 2,570:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\Rightarrow A</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{2} </math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ a</math>
  </td>
  <td align="center">
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~\frac{2}{3} \, ;</math>
+ \frac{18(4 - 3\gamma_\mathrm{g}) }{\gamma_g } \cdot m_*^{1 / 2}
\biggl[ \sigma^2  +  (1-m_*)^{3 / 2} \biggr]  x  \biggr\} \, ,
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</div>
</div>
 
which forces the second expression to the value,
where,
<div align="center">
<math>~\sigma^2 \equiv (4 - 3\gamma_\mathrm{g})^{-1} \frac{R_*^3}{GM_\mathrm{tot}} \biggl[  \frac{ \tilde{C} }{3 } \biggr]^{15/2} \omega^2 \, .</math>
</div>
 
Now, let's look at the differential operators, after defining.
<div align="center">
<math>~c_0 \equiv  3^{1 / 2} R_*  \biggl( \frac{ \tilde{C} }{ 3}\biggr)^{3} ~~~~\Rightarrow ~~~~R_*  = c_0 3^{-1 / 2}  \biggl( \frac{ \tilde{C} }{ 3}\biggr)^{-3} \, .</math>
</div>
 
We find,
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 2,580: Line 2,599:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~dr_0</math>
<math>~a^2 (A - 1)^2 +  a(  -4A^2-A+5) - 4A</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,586: Line 2,605:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~\biggl(\frac{2}{3}\biggr)^2 \biggl(-\frac{1}{2} \biggr)^2 +  \frac{2}{3}\biggl[  -1-\frac{1}{2} +5 \biggr] - 2</math>
c_0 ~d[ m_*^{1 / 2} (1-m_*)^{-1 / 2} ]
</math>
   </td>
   </td>
</tr>
</tr>
Line 2,600: Line 2,617:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~\frac{1}{9} + \frac{7}{3} - 2</math>
c_0 ~\biggl[\frac{1}{2} ~m_*^{-1 / 2}( 1 - m_*)^{-1 / 2} + \frac{1}{2} ~m_*^{1 / 2} (1 - m_*)^{-3 / 2}
\biggr] dm_*
</math>
   </td>
   </td>
</tr>
</tr>
Line 2,615: Line 2,629:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~\frac{4}{9} \, ,</math>
\frac{c_0}{2} ~m_*^{-1 / 2}( 1 - m_*)^{-3 / 2}~ dm_*
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
which is ''not'' zero.  Hence our pair of unknown parameters &#8212; <math>~a </math> and <math>~A</math> &#8212; do not simultaneously satisfy all three conditions.  (Not really a surprise.)
==Setup Using Lagrangian Mass Coordinate==
===Alternative Terms===
Let's change the independent coordinate from <math>~r_0</math> to <math>~m_0</math>.  In particular, the derivative operation will change as follows:
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
Line 2,626: Line 2,648:
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~~\rightarrow~~</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~\biggl( \frac{dm_0}{dr_0} \biggr)\frac{d}{dm_0}
\frac{2}{c_0} ~m_*^{1 / 2}( 1 - m_*)^{3 / 2}~ \frac{d}{dm_*}
= \biggl( \frac{dm_0}{d\xi} \cdot \frac{d\xi}{dr_0} \biggr)\frac{d}{dm_0}
</math>
\, ,</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
so what is the expression for the leading coefficient?  From [[#r0|above]], we have,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\Rightarrow ~~~ R_*\frac{dx}{dr_0}</math>
<math>~r_0</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,644: Line 2,672:
   <td align="left">
   <td align="left">
<math>~
<math>~
\frac{2}{3^{1 / 2}}\biggl( \frac{ \tilde{C} }{ 3}\biggr)^{-3} ~m_*^{1 / 2}( 1 - m_*)^{3 / 2}~ \frac{dx}{dm_*} \, .
R_*  \biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3} \xi
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>


Also,
<tr>
<div align="center">
   <td align="right">
<table border="0" cellpadding="5" align="center">
<math>~\Rightarrow ~~~ \xi</math>
 
<tr>
   <td align="right">
<math>~\frac{d^2}{dr_0^2}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,664: Line 2,686:
   <td align="left">
   <td align="left">
<math>~
<math>~
\biggl( \frac{2}{c_0} \biggr)^{2}~m_*^{1 / 2}( 1 - m_*)^{3 / 2}~ \frac{d}{dm_*} \biggl[ m_*^{1 / 2}( 1 - m_*)^{3 / 2}~ \frac{d}{dm_*}  \biggr]
\frac{1}{R_*} \biggl[ \frac{ {\tilde\xi}^2}{(3 + {\tilde\xi}^2)}\biggr]^{3} r_0 \, .
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
Also, from [[#m0|above]], we know that,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~m_0</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,677: Line 2,706:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
\biggl( \frac{2}{c_0} \biggr)^{2}~m_* ( 1 - m_*)^{3 }~ \frac{d^2}{dm_*^2} 
M_\mathrm{tot} \biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3 / 2}
+\biggl( \frac{2}{c_0} \biggr)^{2}~m_*^{1 / 2}( 1 - m_*)^{3 / 2} \biggl[ \frac{1}{2} m_*^{-1 / 2}( 1 - m_*)^{3 / 2} - \frac{3}{2} m_*^{1 / 2}( 1 - m_*)^{1 / 2}~
\biggl\{ \xi^3 ( 3 + \xi^2 )^{-3/2} \biggr\}  
\biggr] ~ \frac{d}{dm_*}
</math>
</math>
   </td>
   </td>
Line 2,687: Line 2,715:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\Rightarrow ~~~ \frac{dm_0}{d\xi}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,693: Line 2,721:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
\biggl( \frac{2}{c_0} \biggr)^{2}~m_* ( 1 - m_*)^{3 }~ \frac{d^2}{dm_*^2} +\frac{1}{2} \biggl( \frac{2}{c_0} \biggr)^{2}~ ( 1 - m_*)^{2} ( 1 - 4m_*) \frac{d}{dm_*}  
M_\mathrm{tot} \biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3 / 2}
\biggl\{ 3\xi^2 ( 3 + \xi^2 )^{-3/2}  
- 3 \xi^4 ( 3 + \xi^2 )^{-5/2}\biggr\}  
</math>
</math>
   </td>
   </td>
Line 2,701: Line 2,731:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\Rightarrow ~~~ R_*^2 \biggl(  \frac{ \tilde{C} }{3 }\biggr)^{3}  \frac{d^2x}{dr_0^2}</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,707: Line 2,737:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
\biggl[\frac{2^2}{3} \biggl(\frac{ \tilde{C} }{3} \biggr)^{-3} \biggr]  
M_\mathrm{tot} \biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3 / 2} 3\xi^2 (3 + \xi^2)^{-5/2}  
\biggl[ ~m_* ( 1 - m_*)^{3 }~ \frac{d^2x}{dm_*^2+\frac{1}{2} ~ ( 1 - m_*)^{2} ( 1 - 4m_*) \frac{dx}{dm_*} \biggr]
\biggl\{ ( 3 + \xi^2 )
- \xi^2  \biggr\}
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
So, the wave equation becomes,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~0</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,729: Line 2,753:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{1 }{R_*^2} \biggl\frac{3 }{ \tilde{C} }\biggr)^{3} \biggl[ \frac{1}{3m_*(1-m_*)}\biggr]^{1 / 2} \biggl\{ [ 3m_*(1-m_*) ]^{1 / 2}  
<math>~
\biggl[\frac{2^2}{3} \biggl(\frac{ \tilde{C} }{3} \biggr)^{-3} \biggr]  
M_\mathrm{tot} \biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3 / 2} 3^2\xi^2 (3 + \xi^2)^{-5/2}  
\biggl[ ~m_* ( 1 - m_*)^{3 }~ \frac{d^2x}{dm_*^2+\frac{1}{2} ~ ( 1 - m_*)^{2} ( 1 - 4m_*) \frac{dx}{dm_*} \biggr]
</math>
</math>
   </td>
   </td>
Line 2,738: Line 2,761:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\Rightarrow ~~~ \frac{dm_0}{dr_0}</math>
  </td>
</td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
+ \biggl[ 4  - 6\biggl(  \frac{ 3 }{ \tilde{C} }\biggr)^{6} m_* \biggr] (1-m_*) \biggl[ \frac{2}{3^{1 / 2}}\biggl( \frac{ \tilde{C} }{ 3}\biggr)^{-3} ~m_*^{1 / 2}( 1 - m_*)^{3 / 2}~ \frac{dx}{dm_*}  \biggr]
M_\mathrm{tot} \biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3 / 2} 3^2\xi^2 (3 + \xi^2)^{-5/2}  
+ \frac{18(4 - 3\gamma_\mathrm{g}) }{\gamma_g } \cdot m_*^{1 / 2}
\frac{1}{R_*}  \biggl[ \frac{ {\tilde\xi}^2}{(3 + {\tilde\xi}^2)}\biggr]^{3}
\biggl[ \sigma^2 +  (1-m_*)^{3 / 2} \biggr] x  \biggr\}  
</math>
</math>
   </td>
   </td>
Line 2,755: Line 2,777:
   <td align="right">
   <td align="right">
&nbsp;
&nbsp;
  </td>
</td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{1 }{R_*^2} \biggl\frac{3 }{ \tilde{C} }\biggr)^{6} \biggl[ \frac{1}{3m_*(1-m_*)}\biggr]^{1 / 2} \biggl\{ [ 3m_*(1-m_*) ]^{1 / 2}
<math>~ \frac{M_\mathrm{tot} }{R_*}  
\biggl[\frac{2^2}{3}  \biggr] 
\biggl[ \frac{ {\tilde\xi}^2}{(3 + {\tilde\xi}^2)}\biggr]^{3 / 2} 3^2\xi^2 (3 + \xi^2)^{-5/2} \, .
\biggl[ ~m_* ( 1 - m_*)^{3 }~ \frac{d^2x}{dm_*^2}  +\frac{1}{2} ~ ( 1 - m_*)^{2} ( 1 - 4m_*) \frac{dx}{dm_*} \biggr]
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>


<tr>
To simplify expressions, let's [[User:Tohline/Appendix/Ramblings/Nonlinar_Oscillation#DefineTildeC|borrow from an accompanying derivation]] and define,
<div align="center">
<math>\tilde{C} \equiv \frac{3^2}{{\tilde\xi}^2} \biggl( 1 + \frac{ {\tilde\xi}^2}{3} \biggr) = 3 \biggl[ \frac{( 3 + {\tilde\xi}^2 )}{ {\tilde\xi}^2} \biggr] \, .</math>
</div>
Then we have,
 
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\frac{m_0}{M_\mathrm{tot}}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
+ \biggl[ 4  - 6\biggl(  \frac{ 3 }{ \tilde{C} }\biggr)^{6} m_* \biggr] (1-m_*) \biggl[ \frac{2}{3^{1 / 2}} ~m_*^{1 / 2}( 1 - m_*)^{3 / 2}~ \frac{dx}{dm_*}  \biggr]
\biggl[ \frac{\tilde{C}}{ 3}\biggr]^{3 / 2}
+ \frac{18(4 - 3\gamma_\mathrm{g}) }{\gamma_g } \biggl( \frac{ \tilde{C} }{ 3}\biggr)^{3} m_*^{1 / 2} \biggl[ \sigma^2  +  (1-m_*)^{3 / 2} \biggr]  x  \biggr\}  
\biggl[ \frac{\xi^2}{ ( 3 + \xi^2 )} \biggr]^{3/2}  
</math>
</math>
   </td>
   </td>
Line 2,784: Line 2,816:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\Rightarrow ~~~\biggl[ \frac{ 3}{\tilde{C}}\biggr] \biggl[\frac{m_0}{M_\mathrm{tot}}\biggr]^{2 / 3}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,790: Line 2,822:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{2 }{3R_*^2} \biggl( \frac{3 }{ \tilde{C} }\biggr)^{6}
<math>~   
\biggl\{   2m_* ( 1 - m_*)^{3 }~ \frac{d^2x}{dm_*^2}  +  ( 1 - m_*)^{2} ( 1 - 4m_*) \frac{dx}{dm_*}  
\frac{\xi^2}{ ( 3 + \xi^2 )}  
</math>
</math>
   </td>
   </td>
Line 2,798: Line 2,830:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\Rightarrow ~~~( 3 + \xi^2 )\biggl[ \frac{ 3}{\tilde{C}}\biggr] \biggl[\frac{m_0}{M_\mathrm{tot}}\biggr]^{2 / 3}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~   
+ \biggl[ 4  - 6\biggl(  \frac{ 3 }{ \tilde{C} }\biggr)^{6}  m_* \biggr] (1-m_*)^2 \frac{dx}{dm_*}  
\xi^2  
+ \frac{9(4 - 3\gamma_\mathrm{g}) }{\gamma_g } \biggl( \frac{ \tilde{C} }{ 3}\biggr)^{3} \biggl[ \frac{3}{(1-m_*)}\biggr]^{1 / 2}\biggl[ \sigma^2  +  (1-m_*)^{3 / 2} \biggr]  x  \biggr\}
</math>
</math>
   </td>
   </td>
Line 2,813: Line 2,844:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\Rightarrow ~~~3 m_*</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,819: Line 2,850:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{2 }{3R_*^2} \biggl(  \frac{3 }{ \tilde{C} }\biggr)^{6} 
<math>~ \xi^2 (1-m_*)
\biggl\{  2m_* ( 1 - m_*)^{3 }~ \frac{d^2x}{dm_*^2} 
</math>
</math>
   </td>
   </td>
Line 2,827: Line 2,857:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\Rightarrow ~~~\xi^2 </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~   \frac{3m_*}{(1-m_*)} \, ,
+ \biggl[ 5 - 4m_*  - 6\biggl(  \frac{ 3 }{ \tilde{C} }\biggr)^{6}  m_* \biggr] (1-m_*)^2 \frac{dx}{dm_*} 
+ \frac{9(4 - 3\gamma_\mathrm{g}) }{\gamma_g } \biggl( \frac{ \tilde{C} }{ 3}\biggr)^{3} \biggl[ \frac{3}{(1-m_*)}\biggr]^{1 / 2}\biggl[ \sigma^2  +  (1-m_*)^{3 / 2} \biggr]  x  \biggr\}
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
where,
<div align="center">
<math>~m_* \equiv \biggl[ \frac{ 3}{\tilde{C}}\biggr] \biggl[\frac{m_0}{M_\mathrm{tot}}\biggr]^{2 / 3} \, .</math>
</div>
In summary:
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~ 
\frac{\xi^2}{ ( 3 + \xi^2 )} = m_* \, ;
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
&nbsp; &nbsp; &nbsp; while, &nbsp; &nbsp; &nbsp;
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{2 }{3R_*^2} \biggl( \frac{3 }{ \tilde{C} }\biggr)^{6} \biggl\{ 2m_* ( 1 - m_*)^{3 }~ \frac{d^2x}{dm_*^2}
<math>~   
+ (5 - \mathcal{A}  m_*) (1-m_*)^2 \frac{dx}{dm_*} 
\frac{ {\tilde\xi}^2}{ ( 3 + {\tilde\xi}^2 )} = \frac{3}{\tilde{C}} \, ;
+ \mathcal{B} \biggl[ \frac{\sigma^2}{(1-m_*)^{1 / 2}}  +  (1-m_*) \biggr]  x  \biggr\} \, ,
</math>
</math>
   </td>
   </td>
Line 2,856: Line 2,895:
</table>
</table>
</div>
</div>
where,
 
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 2,862: Line 2,901:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\mathcal{A}</math>
<math>~r_0</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\equiv</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~4 + 6\biggl( \frac{ 3 }{ \tilde{C} }\biggr)^{6} \, ,</math>
<math>~
R_*  \biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3} \xi
= R_*  \biggl( \frac{ \tilde{C} }{ 3}\biggr)^{3} \biggr[ \frac{3m_*}{ (1-m_*) }\biggr]^{1 / 2} \, ;
</math>
   </td>
   </td>
</tr>
</tr>
Line 2,874: Line 2,916:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\mathcal{B}</math>
<math>~\frac{g_0\rho_0}{P_0} </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\equiv</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{3^{5/2}(4 - 3\gamma_\mathrm{g}) }{\gamma_g } \biggl( \frac{ \tilde{C} }{ 3}\biggr)^{3} \, .</math>
<math>~
\frac{6}{R_*} \biggl[  \frac{ {\tilde\xi}^2 }{ (3 + {\tilde\xi}^2) }\biggr]^{9} \frac{\xi}{ ( 3 + \xi^2 )}
=
\frac{6}{R_*} \biggl\frac{ 3 }{ \tilde{C} }\biggr]^{9}  \frac{m_*}{ \xi }
=
\frac{6}{R_*} \biggl[  \frac{ 3 }{ \tilde{C} }\biggr]^{9}  m_* \biggl[  \frac{(1-m_*)}{3m_*}  \biggr]^{1 / 2} \, ;
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
==Try Again==
This time, let's adopt the notation used in a [[User:Tohline/Appendix/Ramblings/Nonlinar_Oscillation#n_.3D_5_Mass-Radius_Relation|related chapter in our ''Ramblings'' appendix]].  Specifically, the parametric relationship between <math>~m_\xi</math> and <math>~r_\xi</math> in pressure-truncated, <math>~n=5</math>  polytropes is,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~m_\xi \equiv \frac{m_0}{ M_\mathrm{tot} } = \frac{M_r(\xi)}{M_\mathrm{tot}}</math>
<math>~\frac{g_0 }{r_0} </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,901: Line 2,940:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
\biggl(\frac{\xi}{\tilde\xi}\biggr)^3 \biggl(3 + \xi^2 \biggr)^{-3/2}  
\frac{GM_\mathrm{tot}}{R_*^3}
\biggl(3 {\tilde\xi}^2 \biggr)^{3/2} </math>
\biggl\frac{ {\tilde\xi}^2 }{ (3 + {\tilde\xi}^2)}\biggr]^{15/2}  \frac{1}{\xi^3}
\biggl[ \frac{ \xi^2 }{ ( 3 + \xi^2 ) }\biggr]^{3/2}  
=
\frac{GM_\mathrm{tot}}{R_*^3} \biggl[  \frac{3 }{ \tilde{C} }\biggr]^{15/2}  (1-m_*)^{3 / 2} \, ;
</math>
   </td>
   </td>
</tr>
</tr>
Line 2,909: Line 2,952:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\frac{\rho_0}{\gamma_g P_0} </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,915: Line 2,958:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
\biggl[ \frac{( 3+\tilde\xi^2)}{ {\tilde\xi}^2} \biggr]^{3 / 2}\biggl[ \frac{( 3+\xi^2)}{ {\xi}^2} \biggr]^{- 3 / 2}
\frac{6R_* }{\gamma_g GM_\mathrm{tot} }\biggl( \frac{ 3}{ \tilde{C} } \biggr)^{9 / 2} \biggl[ \frac{3}{(1-m_*)}\biggr]^{1 / 2} \, .
\, ,</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
So, the wave equation may be written as,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~r_\xi \equiv \frac{r_0}{R_\mathrm{norm}} = \biggl(\frac{\xi}{\tilde\xi} \biggr) \frac{R_\mathrm{eq}}{R_\mathrm{norm}}</math>
<math>~0</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,929: Line 2,978:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\xi \biggl\{  
<math>~
\biggl[ \frac{4\pi}{2^5\cdot 3}\biggr]^{1/2} \tilde\xi^{-6}
\frac{d^2x}{dr_0^2} + \biggl[\frac{4}{r_0} - \biggl(\frac{g_0 \rho_0}{P_0}\biggr) \biggr] \frac{dx}{dr_0}  
\biggl( 1+\frac{\tilde\xi^2}{3} \biggr)^{3}\biggr\}
+ \biggl(\frac{\rho_0}{\gamma_\mathrm{g} P_0} \biggr)\biggl[\omega^2 + (4 - 3\gamma_\mathrm{g})\frac{g_0}{r_0} \biggr]  x
</math>
</math>
   </td>
   </td>
Line 2,945: Line 2,994:
   <td align="left">
   <td align="left">
<math>~
<math>~
\biggl[ \frac{\pi}{2^3\cdot 3^7}\biggr]^{1/2}  
\frac{d^2x}{dr_0^2}
\biggl[ \frac{( 3+\tilde\xi^2)}{ {\tilde\xi}^2} \biggr]^{3} \xi \, .
+ \biggl\{ \frac{4}{R_*} \biggl( \frac{ 3}{ \tilde{C} }\biggr)^{3} \biggr[ \frac{ (1-m_*) }{3m_*}\biggr]^{1 / 2}
- \frac{6}{R_*} \biggl[ \frac{ 3 }{ \tilde{C} }\biggr]^{9} m_* \biggl[  \frac{(1-m_*)}{3m_*} \biggr]^{1 / 2} \biggr\} \frac{dx}{dr_0}
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
And we are in the fortunate situation of being able to eliminate <math>~\xi</math> to obtain the direct relation,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~
&nbsp;
r_\xi (m_\xi)
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
&nbsp;
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\tilde{r}_\mathrm{edge}
<math>~
\biggl[\frac{3^2m_\xi^{2/3}}{\tilde{C} - 3 m_\xi^{2/3}}\biggr]^{1/2} \, ,
+ \frac{6R_* }{\gamma_g GM_\mathrm{tot} }\biggl( \frac{ 3}{ \tilde{C} } \biggr)^{9 / 2} \biggl[ \frac{3}{(1-m_*)}\biggr]^{1 / 2}
\biggl\{ \omega^2 + (4 - 3\gamma_\mathrm{g})\frac{GM_\mathrm{tot}}{R_*^3} \biggl[  \frac{3 }{ \tilde{C} }\biggr]^{15/2}  (1-m_*)^{3 / 2} \biggr\}  x
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{d^2x}{dr_0^2}
+ \frac{1}{R_*} \biggl( \frac{ 3}{ \tilde{C} }\biggr)^{3} \biggl\{ 4 
- 6\biggl[  \frac{ 3 }{ \tilde{C} }\biggr]^{6}  m_* \biggr\} \biggr[ \frac{ (1-m_*) }{3m_*}\biggr]^{1 / 2}\frac{dx}{dr_0}
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
where,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\tilde{C}</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\equiv</math>
&nbsp;
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
\frac{3^2}{\tilde\xi^2}\biggl( 1 + \frac{\tilde\xi^2}{3} \biggr)  
+ \frac{6(4 - 3\gamma_\mathrm{g}) }{\gamma_g } \cdot \frac{1 }{R_*^2} \biggl( \frac{3 }{ \tilde{C} }\biggr)^{3} \biggl[ \frac{3}{(1-m_*)}\biggr]^{1 / 2}  
= 3 \biggl[ \frac{( 3+\tilde\xi^2)}{ {\tilde\xi}^2} \biggr]
\biggl\{\sigma^2  +  (1-m_*)^{3 / 2} \biggr\}  x
\, ,
</math>
</math>
   </td>
   </td>
Line 2,996: Line 3,049:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\tilde{r}_\mathrm{edge}</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\equiv</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl[ \frac{\pi}{2^3\cdot 3}\biggr]^{1/2} {\tilde\xi}^{-6} \biggl(1+\frac{\tilde\xi^2}{3}\biggr)^3
<math>~\frac{1 }{R_*^2} \biggl\frac{3 }{ \tilde{C} }\biggr)^{3} \biggl\{
= \biggl[ \frac{\pi}{2^3\cdot 3^7}\biggr]^{1 / 2} \biggl[ \frac{\tilde{C}}{ 3} \biggr]^{3}  
R_*^2 \biggl( \frac{ \tilde{C} }{3 }\biggr)^{3}  \frac{d^2x}{dr_0^2}
\, .
+ R_* \biggl[ 4  - 6\biggl(  \frac{ 3 }{ \tilde{C} }\biggr)^{6} m_* \biggr] \biggr[ \frac{ (1-m_*) }{3m_*}\biggr]^{1 / 2}\frac{dx}{dr_0}
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>


If we furthermore define,
<tr>
<div align="center">
<math>m_* \equiv \frac{3}{\tilde{C}} \cdot m_\xi^{2 / 3} \, ,</math>
</div>
then,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
   <td align="right">
   <td align="right">
<math>~
&nbsp;
r_\xi (m_*)
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
&nbsp;
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
3^{1 / 2} \tilde{r}_\mathrm{edge} \biggl[\frac{m_*}{1-m_*}\biggr]^{1/2}  
+ \frac{6(4 - 3\gamma_\mathrm{g}) }{\gamma_g } \cdot \biggl[ \frac{3}{(1-m_*)}\biggr]^{1 / 2}  
\, .
\biggl[ \sigma^2  +  (1-m_*)^{3 / 2} \biggr]  x  \biggr\}
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
Hence,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~
&nbsp;
\frac{dr_0}{R_\mathrm{norm}} = dr_\xi
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 3,052: Line 3,085:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~3^{1 / 2} \tilde{r}_\mathrm{edge} \biggl\{
<math>~\frac{1 }{R_*^2} \biggl(  \frac{3 }{ \tilde{C} }\biggr)^{3} \biggl[ \frac{1}{3m_*(1-m_*)}\biggr]^{1 / 2} \biggl\{ [ 3m_*(1-m_*) ]^{1 / 2}  
\frac{1}{2} (1-m_*)^{- 1 / 2} m_*^{-1 / 2} + \frac{1}{2}m_*^{1 / 2}(1-m_*)^{-3 / 2}
R_*^2 \biggl(  \frac{ \tilde{C} }{3 }\biggr)^{3}  \frac{d^2x}{dr_0^2}  
\biggr\} dm_*
+ R_* \biggl[ 4  - 6\biggl(  \frac{ 3 }{ \tilde{C} }\biggr)^{6} m_* \biggr] (1-m_*) \frac{dx}{dr_0}  
</math>
</math>
   </td>
   </td>
Line 3,064: Line 3,097:
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
&nbsp;
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ \biggl( \frac{3^{1 / 2}}{2} \biggr) \tilde{r}_\mathrm{edge} m_*^{-1 / 2} (1-m_*)^{-3 / 2} dm_*
<math>~
+ \frac{18(4 - 3\gamma_\mathrm{g}) }{\gamma_g } \cdot m_*^{1 / 2}
\biggl[ \sigma^2  +  (1-m_*)^{3 / 2} \biggr]  x  \biggr\} \, ,
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
where,
<div align="center">
<math>~\sigma^2 \equiv (4 - 3\gamma_\mathrm{g})^{-1} \frac{R_*^3}{GM_\mathrm{tot}} \biggl[  \frac{ \tilde{C} }{3 } \biggr]^{15/2} \omega^2 \, .</math>
</div>


<tr>
Now, let's look at the differential operators, after defining.
  <td align="right">
<div align="center">
<math>\Rightarrow ~~~ R_\mathrm{norm} \cdot \frac{d}{dr_0} </math>
<math>~c_0 \equiv  3^{1 / 2} R_*  \biggl( \frac{ \tilde{C} }{ 3}\biggr)^{3} ~~~~\Rightarrow ~~~~R_*  = c_0 3^{-1 / 2} \biggl( \frac{ \tilde{C} }{ 3}\biggr)^{-3} \, .</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~  \frac{1}{ \tilde{r}_\mathrm{edge}} \biggl( \frac{2}{3^{1 / 2}} \biggr) m_*^{1 / 2} (1-m_*)^{3 / 2} \frac{d}{dm_*} \, .
</math>
  </td>
</tr>
</table>
</div>
</div>
We therefore also have,


We find,
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 3,093: Line 3,125:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~ R^2_\mathrm{norm} \cdot \frac{d^2}{dr_0^2} </math>
<math>~dr_0</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 3,099: Line 3,131:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ \frac{1}{ {\tilde{r}}^2_\mathrm{edge}} \biggl( \frac{2^2}{3} \biggr) m_*^{1 / 2} (1-m_*)^{3 / 2} \frac{d}{dm_*}\biggl[ m_*^{1 / 2} (1-m_*)^{3 / 2} \frac{d}{dm_*}\biggr]  
<math>~
c_0 ~d[ m_*^{1 / 2} (1-m_*)^{-1 / 2} ]
</math>
</math>
   </td>
   </td>
Line 3,112: Line 3,145:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ \frac{1}{ {\tilde{r}}^2_\mathrm{edge}} \biggl( \frac{2^2}{3} \biggr) m_*^{1 / 2} (1-m_*)^{3 / 2}
<math>~
\biggl\{
c_0 ~\biggl[\frac{1}{2} ~m_*^{-1 / 2}( 1 - m_*)^{-1 / 2} + \frac{1}{2} ~m_*^{1 / 2} (1 - m_*)^{-3 / 2}
\biggl[ m_*^{1 / 2} (1-m_*)^{3 / 2} \frac{d^2}{dm_*^2}\biggr]
\biggr] dm_*
+ \biggl[ \frac{1}{2} m_*^{-1 / 2} (1-m_*)^{3 / 2} + \frac{3}{2}m_*^{1 / 2} (1-m_*)^{1 / 2}\biggr] \frac{d}{dm_*}
\biggr\}
</math>
</math>
   </td>
   </td>
Line 3,129: Line 3,160:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ \frac{1}{ {\tilde{r}}^2_\mathrm{edge}} \biggl( \frac{2}{3} \biggr) \biggl\{ 
<math>~
\biggl[ 2m_* (1-m_*)^{3} \frac{d^2}{dm_*^2}\biggr]
\frac{c_0}{2} ~m_*^{-1 / 2}( 1 - m_*)^{-3 / 2}~ dm_*
+ \biggl[  (1-m_*)^{3 } + 3m_* (1-m_*)^{2}\biggr]  \frac{d}{dm_*}
</math>
\biggr\}
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\frac{d}{dr_0}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{2}{c_0} ~m_*^{1 / 2}( 1 - m_*)^{3 / 2}~ \frac{d}{dm_*}
</math>
</math>
   </td>
   </td>
Line 3,139: Line 3,182:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\Rightarrow ~~~ R_*\frac{dx}{dr_0}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 3,145: Line 3,188:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
\frac{1}{ {\tilde{r}}^2_\mathrm{edge}} \biggl( \frac{2}{3} \biggr) \biggl\{
\frac{2}{3^{1 / 2}}\biggl( \frac{ \tilde{C} }{ 3}\biggr)^{-3} ~m_*^{1 / 2}( 1 - m_*)^{3 / 2}~ \frac{dx}{dm_*} \, .
2m_* (1-m_*)^{3} \frac{d^2}{dm_*^2}
+ (1-m_*)^{2} ( 1 + 2m_* )  \frac{d}{dm_*} \biggr\}
\, .
</math>
</math>
   </td>
   </td>
Line 3,156: Line 3,196:
</div>
</div>


So the wave equation may be written,
Also,
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 3,162: Line 3,202:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~0</math>
<math>~\frac{d^2}{dr_0^2}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 3,169: Line 3,209:
   <td align="left">
   <td align="left">
<math>~
<math>~
R_\mathrm{norm}^2 \cdot \frac{d^2x}{dr_0^2} + \biggl[\frac{4R_\mathrm{norm}}{r_0} - \biggl(\frac{g_0 \rho_0 R_\mathrm{norm}}{P_0}\biggr) \biggr] R_\mathrm{norm} \cdot \frac{dx}{dr_0}  
\biggl( \frac{2}{c_0} \biggr)^{2}~m_*^{1 / 2}( 1 - m_*)^{3 / 2}~ \frac{d}{dm_*} \biggl[ m_*^{1 / 2}( 1 - m_*)^{3 / 2}~ \frac{d}{dm_*} \biggr]
+ \biggl(\frac{\rho_0 R_\mathrm{norm}}{\gamma_\mathrm{g} P_0} \biggr)\biggl[R_\mathrm{norm} \omega^2 + (4 - 3\gamma_\mathrm{g})\frac{g_0 R_\mathrm{norm}}{r_0} \biggr] x
</math>
</math>
   </td>
   </td>
Line 3,184: Line 3,223:
   <td align="left">
   <td align="left">
<math>~
<math>~
\frac{1}{ {\tilde{r}}^2_\mathrm{edge}} \biggl( \frac{2}{3} \biggr) \biggl\{
\biggl( \frac{2}{c_0} \biggr)^{2}~m_* ( 1 - m_*)^{3 }~ \frac{d^2}{dm_*^2}
2m_* (1-m_*)^{3} \frac{d^2x}{dm_*^2}
+\biggl( \frac{2}{c_0} \biggr)^{2}~m_*^{1 / 2}( 1 - m_*)^{3 / 2} \biggl[ \frac{1}{2} m_*^{-1 / 2}( 1 - m_*)^{3 / 2} - \frac{3}{2} m_*^{1 / 2}( 1 - m_*)^{1 / 2}~
+ (1-m_*)^{2} ( 1 + 2m_* ) \frac{dx}{dm_*} \biggr\}  
\biggr] ~ \frac{d}{dm_*}
</math>
</math>
   </td>
   </td>
Line 3,196: Line 3,235:
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
+\frac{1}{ \tilde{r}_\mathrm{edge}} \biggl( \frac{2}{3^{1 / 2}} \biggr) \biggl\{ \frac{4}{r_\xi}  
\biggl( \frac{2}{c_0} \biggr)^{2}~m_* ( 1 - m_*)^{3 }~ \frac{d^2}{dm_*^2} +\frac{1}{2} \biggl( \frac{2}{c_0} \biggr)^{2}~ ( 1 - m_*)^{2} ( 1 - 4m_*) \frac{d}{dm_*}  
- \biggl[\frac{6R_\mathrm{norm}}{R_*} \biggl( \frac{ 3 }{ \tilde{C} }\biggr)^{9} m_* \biggl[  \frac{(1-m_*)}{3m_*}  \biggr]^{1 / 2} \biggr] \biggr\}
m_*^{1 / 2} (1-m_*)^{3 / 2} \frac{dx}{dm_*}  
</math>
</math>
   </td>
   </td>
Line 3,209: Line 3,246:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\Rightarrow ~~~ R_*^2 \biggl(  \frac{ \tilde{C} }{3 }\biggr)^{3}  \frac{d^2x}{dr_0^2}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
+ \frac{6R_* R_\mathrm{norm}}{\gamma_g GM_\mathrm{tot} }\biggl( \frac{ 3}{ \tilde{C} } \biggr)^{9 / 2} \biggl[ \frac{3}{(1-m_*)}\biggr]^{1 / 2}
\biggl[\frac{2^2}{3} \biggl(\frac{ \tilde{C} }{3} \biggr)^{-3} \biggr] 
\biggl\{
\biggl[ ~m_* ( 1 - m_*)^{3 }~ \frac{d^2x}{dm_*^2+\frac{1}{2} ~ ( 1 - m_*)^{2} ( 1 - 4m_*) \frac{dx}{dm_*} \biggr]
R_\mathrm{norm} \omega^2 + (4 - 3\gamma_\mathrm{g}) \frac{GM_\mathrm{tot} R_\mathrm{norm}}{R_*^3} \biggl[ \frac{3 }{ \tilde{C} }\biggr]^{15/2} (1-m_*)^{3 / 2}
\biggr\} x \, .
</math>
</math>
   </td>
   </td>
Line 3,225: Line 3,260:
</table>
</table>
</div>
</div>
Keeping in mind that,
<div align="center">
<math>~\frac{R_*}{R_\mathrm{norm}} = \biggl[ \frac{\pi}{2^3 \cdot 3^7} \biggr]^{1 / 2} = {\tilde{r}}_\mathrm{edge} \biggl( \frac{3}{\tilde{C}} \biggr)^3 \, ,</math>
</div>


we therefore have,
So, the wave equation becomes,
 
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 3,242: Line 3,274:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~\frac{1 }{R_*^2} \biggl(  \frac{3 }{ \tilde{C} }\biggr)^{3} \biggl[ \frac{1}{3m_*(1-m_*)}\biggr]^{1 / 2} \biggl\{ [ 3m_*(1-m_*) ]^{1 / 2}
\frac{1}{ {\tilde{r}}^2_\mathrm{edge}} \biggl( \frac{2}{3} \biggr) \biggl\{ 
\biggl[\frac{2^2}{3} \biggl(\frac{ \tilde{C} }{3} \biggr)^{-3} \biggr] 
2m_* (1-m_*)^{3} \frac{d^2x}{dm_*^2}
\biggl[ ~m_* ( 1 - m_*)^{3 }~ \frac{d^2x}{dm_*^2} +\frac{1}{2} ~ ( 1 - m_*)^{2} ( 1 - 4m_*) \frac{dx}{dm_*} \biggr]
+ (1-m_*)^{2} ( 1 + 2m_* ) \frac{dx}{dm_*} \biggr\}
</math>
</math>
   </td>
   </td>
Line 3,259: Line 3,290:
   <td align="left">
   <td align="left">
<math>~
<math>~
+\frac{1}{ \tilde{r}_\mathrm{edge}} \biggl( \frac{2}{3^{1 / 2}} \biggr) \biggl\{ 4 \biggl[3^{1 / 2} \tilde{r}_\mathrm{edge} \biggl[\frac{m_*}{1-m_*}\biggr]^{1/2} \biggr]^{-1}
+ \biggl[ 4  - 6\biggl( \frac{ 3 }{ \tilde{C} }\biggr)^{6} m_* \biggr] (1-m_*) \biggl[ \frac{2}{3^{1 / 2}}\biggl( \frac{ \tilde{C} }{ 3}\biggr)^{-3} ~m_*^{1 / 2}( 1 - m_*)^{3 / 2}~ \frac{dx}{dm_*} \biggr]
- 6 \biggl( \frac{ 3 }{ \tilde{C} }\biggr)^{9} \biggl[{\tilde{r}}_\mathrm{edge} \biggl( \frac{3}{\tilde{C}} \biggr)^3 \biggr]^{-1} m_* \biggl[  \frac{(1-m_*)}{3m_*} \biggr]^{1 / 2} \biggr\}
+ \frac{18(4 - 3\gamma_\mathrm{g}) }{\gamma_g } \cdot m_*^{1 / 2}
m_*^{1 / 2} (1-m_*)^{3 / 2} \frac{dx}{dm_*}  
\biggl[ \sigma^2 (1-m_*)^{3 / 2} \biggr]  x  \biggr\}  
</math>
</math>
   </td>
   </td>
Line 3,271: Line 3,302:
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~\frac{1 }{R_*^2} \biggl( \frac{3 }{ \tilde{C} }\biggr)^{6} \biggl[ \frac{1}{3m_*(1-m_*)}\biggr]^{1 / 2} \biggl\{ [ 3m_*(1-m_*) ]^{1 / 2}
+ 6 \biggl( \frac{ 3}{ \tilde{C} } \biggr)^{9 / 2} \biggl[{\tilde{r}}_\mathrm{edge} \biggl( \frac{3}{\tilde{C}} \biggr)^3 \biggr]^{-2} \biggl[ \frac{3}{(1-m_*)}\biggr]^{1 / 2}
\biggl[\frac{2^2}{3} \biggr
\biggl\{
\biggl[ ~m_* ( 1 - m_*)^{3 }~ \frac{d^2x}{dm_*^2} +\frac{1}{2} ~ ( 1 - m_*)^{2} ( 1 - 4m_*) \frac{dx}{dm_*} \biggr]  
\biggl[ \frac{R_*^3}{\gamma_g GM_\mathrm{tot} } \biggr] \omega^2
</math>
+ \frac{(4 - 3\gamma_\mathrm{g})}{\gamma_g} \biggl\frac{3 }{ \tilde{C} }\biggr]^{15/2}  (1-m_*)^{3 / 2}   
  </td>
\biggr\}  x
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \biggl[ 4  - 6\biggl\frac{ 3 }{ \tilde{C} }\biggr)^{6}  m_* \biggr] (1-m_*) \biggl[ \frac{2}{3^{1 / 2}} ~m_*^{1 / 2}( 1 - m_*)^{3 / 2}~ \frac{dx}{dm_*} \biggr]  
+ \frac{18(4 - 3\gamma_\mathrm{g}) }{\gamma_g } \biggl( \frac{ \tilde{C} }{ 3}\biggr)^{3} m_*^{1 / 2} \biggl[ \sigma^2  + (1-m_*)^{3 / 2} \biggr] x  \biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2 }{3R_*^2} \biggl(  \frac{3 }{ \tilde{C} }\biggr)^{6} 
\biggl\{    2m_* ( 1 - m_*)^{3 }~ \frac{d^2x}{dm_*^2+  ( 1 - m_*)^{2} ( 1 - 4m_*) \frac{dx}{dm_*}
</math>
</math>
   </td>
   </td>
Line 3,287: Line 3,344:
   <td align="right">
   <td align="right">
&nbsp;
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \biggl[ 4  - 6\biggl(  \frac{ 3 }{ \tilde{C} }\biggr)^{6}  m_* \biggr] (1-m_*)^2 \frac{dx}{dm_*} 
+ \frac{9(4 - 3\gamma_\mathrm{g}) }{\gamma_g } \biggl( \frac{ \tilde{C} }{ 3}\biggr)^{3} \biggl[ \frac{3}{(1-m_*)}\biggr]^{1 / 2}\biggl[ \sigma^2  +  (1-m_*)^{3 / 2} \biggr]  x  \biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2 }{3R_*^2} \biggl(  \frac{3 }{ \tilde{C} }\biggr)^{6} 
\biggl\{  2m_* ( 1 - m_*)^{3 }~ \frac{d^2x}{dm_*^2} 
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \biggl[ 5 - 4m_*  - 6\biggl(  \frac{ 3 }{ \tilde{C} }\biggr)^{6}  m_* \biggr] (1-m_*)^2 \frac{dx}{dm_*} 
+ \frac{9(4 - 3\gamma_\mathrm{g}) }{\gamma_g } \biggl( \frac{ \tilde{C} }{ 3}\biggr)^{3} \biggl[ \frac{3}{(1-m_*)}\biggr]^{1 / 2}\biggl[ \sigma^2  +  (1-m_*)^{3 / 2} \biggr]  x  \biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2 }{3R_*^2} \biggl(  \frac{3 }{ \tilde{C} }\biggr)^{6}  \biggl\{  2m_* ( 1 - m_*)^{3 }~ \frac{d^2x}{dm_*^2} 
+ (5 - \mathcal{A}  m_*) (1-m_*)^2 \frac{dx}{dm_*} 
+ \mathcal{B} \biggl[ \frac{\sigma^2}{(1-m_*)^{1 / 2}}  +  (1-m_*) \biggr]  x  \biggr\} \, ,
</math>
  </td>
</tr>
</table>
</div>
where,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\mathcal{A}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~4 + 6\biggl(  \frac{ 3 }{ \tilde{C} }\biggr)^{6} \, ,</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\mathcal{B}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\frac{3^{5/2}(4 - 3\gamma_\mathrm{g}) }{\gamma_g } \biggl( \frac{ \tilde{C} }{ 3}\biggr)^{3}  \, .</math>
  </td>
</tr>
</table>
</div>
==Try Again==
This time, let's adopt the notation used in a [[User:Tohline/Appendix/Ramblings/Nonlinar_Oscillation#n_.3D_5_Mass-Radius_Relation|related chapter in our ''Ramblings'' appendix]].  Specifically, the parametric relationship between <math>~m_\xi</math> and <math>~r_\xi</math> in pressure-truncated, <math>~n=5</math>  polytropes is,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~m_\xi \equiv \frac{m_0}{ M_\mathrm{tot} } = \frac{M_r(\xi)}{M_\mathrm{tot}}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl(\frac{\xi}{\tilde\xi}\biggr)^3 \biggl(3 + \xi^2 \biggr)^{-3/2}
\biggl(3 +  {\tilde\xi}^2 \biggr)^{3/2} </math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{( 3+\tilde\xi^2)}{ {\tilde\xi}^2} \biggr]^{3 / 2}\biggl[ \frac{( 3+\xi^2)}{ {\xi}^2} \biggr]^{- 3 / 2}
\, ,</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~r_\xi \equiv \frac{r_0}{R_\mathrm{norm}} = \biggl(\frac{\xi}{\tilde\xi} \biggr) \frac{R_\mathrm{eq}}{R_\mathrm{norm}}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\xi \biggl\{
\biggl[ \frac{4\pi}{2^5\cdot 3}\biggr]^{1/2} \tilde\xi^{-6}
\biggl( 1+\frac{\tilde\xi^2}{3} \biggr)^{3}\biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{\pi}{2^3\cdot 3^7}\biggr]^{1/2}
\biggl[ \frac{( 3+\tilde\xi^2)}{ {\tilde\xi}^2} \biggr]^{3} \xi \, .
</math>
  </td>
</tr>
</table>
</div>
And we are in the fortunate situation of being able to eliminate <math>~\xi</math> to obtain the direct relation,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~
r_\xi (m_\xi)
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\tilde{r}_\mathrm{edge}
\biggl[\frac{3^2m_\xi^{2/3}}{\tilde{C} - 3 m_\xi^{2/3}}\biggr]^{1/2} \, ,
</math>
  </td>
</tr>
</table>
</div>
where,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\tilde{C}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
\frac{3^2}{\tilde\xi^2}\biggl( 1 + \frac{\tilde\xi^2}{3} \biggr)
= 3 \biggl[ \frac{( 3+\tilde\xi^2)}{ {\tilde\xi}^2} \biggr]
\, ,
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\tilde{r}_\mathrm{edge}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{\pi}{2^3\cdot 3}\biggr]^{1/2} {\tilde\xi}^{-6} \biggl(1+\frac{\tilde\xi^2}{3}\biggr)^3
= \biggl[ \frac{\pi}{2^3\cdot 3^7}\biggr]^{1 / 2} \biggl[ \frac{\tilde{C}}{ 3} \biggr]^{3}
\, .
</math>
  </td>
</tr>
</table>
</div>
If we furthermore define,
<div align="center">
<math>m_* \equiv \frac{3}{\tilde{C}} \cdot m_\xi^{2 / 3} \, ,</math>
</div>
then,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~
r_\xi (m_*)
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
3^{1 / 2} \tilde{r}_\mathrm{edge} \biggl[\frac{m_*}{1-m_*}\biggr]^{1/2}
\, .
</math>
  </td>
</tr>
</table>
</div>
Hence,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~
\frac{dr_0}{R_\mathrm{norm}} = dr_\xi
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~3^{1 / 2} \tilde{r}_\mathrm{edge} \biggl\{
\frac{1}{2} (1-m_*)^{- 1 / 2} m_*^{-1 / 2} + \frac{1}{2}m_*^{1 / 2}(1-m_*)^{-3 / 2}
\biggr\} dm_*
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \biggl( \frac{3^{1 / 2}}{2} \biggr) \tilde{r}_\mathrm{edge} m_*^{-1 / 2} (1-m_*)^{-3 / 2} dm_*
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>\Rightarrow ~~~ R_\mathrm{norm} \cdot \frac{d}{dr_0} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~  \frac{1}{ \tilde{r}_\mathrm{edge}} \biggl( \frac{2}{3^{1 / 2}} \biggr) m_*^{1 / 2} (1-m_*)^{3 / 2} \frac{d}{dm_*} \, .
</math>
  </td>
</tr>
</table>
</div>
We therefore also have,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~ R^2_\mathrm{norm} \cdot \frac{d^2}{dr_0^2} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~  \frac{1}{ {\tilde{r}}^2_\mathrm{edge}} \biggl( \frac{2^2}{3} \biggr) m_*^{1 / 2} (1-m_*)^{3 / 2} \frac{d}{dm_*}\biggl[ m_*^{1 / 2} (1-m_*)^{3 / 2} \frac{d}{dm_*}\biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~  \frac{1}{ {\tilde{r}}^2_\mathrm{edge}} \biggl( \frac{2^2}{3} \biggr) m_*^{1 / 2} (1-m_*)^{3 / 2}
\biggl\{
\biggl[ m_*^{1 / 2} (1-m_*)^{3 / 2} \frac{d^2}{dm_*^2}\biggr]
+ \biggl[ \frac{1}{2} m_*^{-1 / 2} (1-m_*)^{3 / 2} + \frac{3}{2}m_*^{1 / 2} (1-m_*)^{1 / 2}\biggr]  \frac{d}{dm_*}
\biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~  \frac{1}{ {\tilde{r}}^2_\mathrm{edge}} \biggl( \frac{2}{3} \biggr) \biggl\{ 
\biggl[ 2m_* (1-m_*)^{3} \frac{d^2}{dm_*^2}\biggr]
+ \biggl[  (1-m_*)^{3 } + 3m_* (1-m_*)^{2}\biggr]  \frac{d}{dm_*}
\biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\frac{1}{ {\tilde{r}}^2_\mathrm{edge}} \biggl( \frac{2}{3} \biggr) \biggl\{ 
2m_* (1-m_*)^{3} \frac{d^2}{dm_*^2}
+ (1-m_*)^{2} ( 1 + 2m_* )  \frac{d}{dm_*} \biggr\}
\, .
</math>
  </td>
</tr>
</table>
</div>
So the wave equation may be written,
<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>~
R_\mathrm{norm}^2 \cdot \frac{d^2x}{dr_0^2} + \biggl[\frac{4R_\mathrm{norm}}{r_0} - \biggl(\frac{g_0 \rho_0 R_\mathrm{norm}}{P_0}\biggr) \biggr] R_\mathrm{norm} \cdot \frac{dx}{dr_0}
+ \biggl(\frac{\rho_0 R_\mathrm{norm}}{\gamma_\mathrm{g} P_0} \biggr)\biggl[R_\mathrm{norm} \omega^2 + (4 - 3\gamma_\mathrm{g})\frac{g_0 R_\mathrm{norm}}{r_0} \biggr]  x
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{ {\tilde{r}}^2_\mathrm{edge}} \biggl( \frac{2}{3} \biggr) \biggl\{ 
2m_* (1-m_*)^{3} \frac{d^2x}{dm_*^2}
+ (1-m_*)^{2} ( 1 + 2m_* )  \frac{dx}{dm_*} \biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+\frac{1}{ \tilde{r}_\mathrm{edge}} \biggl( \frac{2}{3^{1 / 2}} \biggr)  \biggl\{ \frac{4}{r_\xi}
- \biggl[\frac{6R_\mathrm{norm}}{R_*} \biggl(  \frac{ 3 }{ \tilde{C} }\biggr)^{9}  m_* \biggl[  \frac{(1-m_*)}{3m_*}  \biggr]^{1 / 2} \biggr] \biggr\}
m_*^{1 / 2} (1-m_*)^{3 / 2} \frac{dx}{dm_*}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \frac{6R_* R_\mathrm{norm}}{\gamma_g GM_\mathrm{tot} }\biggl( \frac{ 3}{ \tilde{C} } \biggr)^{9 / 2} \biggl[ \frac{3}{(1-m_*)}\biggr]^{1 / 2}
\biggl\{
R_\mathrm{norm} \omega^2 + (4 - 3\gamma_\mathrm{g}) \frac{GM_\mathrm{tot} R_\mathrm{norm}}{R_*^3} \biggl[  \frac{3 }{ \tilde{C} }\biggr]^{15/2}  (1-m_*)^{3 / 2} 
\biggr\}  x \, .
</math>
  </td>
</tr>
</table>
</div>
Keeping in mind that,
<div align="center">
<math>~\frac{R_*}{R_\mathrm{norm}} = \biggl[ \frac{\pi}{2^3 \cdot 3^7} \biggr]^{1 / 2} = {\tilde{r}}_\mathrm{edge} \biggl( \frac{3}{\tilde{C}} \biggr)^3 \, ,</math>
</div>
we therefore 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{1}{ {\tilde{r}}^2_\mathrm{edge}} \biggl( \frac{2}{3} \biggr) \biggl\{ 
2m_* (1-m_*)^{3} \frac{d^2x}{dm_*^2}
+ (1-m_*)^{2} ( 1 + 2m_* )  \frac{dx}{dm_*} \biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+\frac{1}{ \tilde{r}_\mathrm{edge}} \biggl( \frac{2}{3^{1 / 2}} \biggr)  \biggl\{ 4 \biggl[3^{1 / 2} \tilde{r}_\mathrm{edge} \biggl[\frac{m_*}{1-m_*}\biggr]^{1/2}  \biggr]^{-1}
- 6 \biggl(  \frac{ 3 }{ \tilde{C} }\biggr)^{9} \biggl[{\tilde{r}}_\mathrm{edge} \biggl( \frac{3}{\tilde{C}} \biggr)^3 \biggr]^{-1} m_* \biggl[  \frac{(1-m_*)}{3m_*}  \biggr]^{1 / 2}  \biggr\}
m_*^{1 / 2} (1-m_*)^{3 / 2} \frac{dx}{dm_*}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ 6 \biggl( \frac{ 3}{ \tilde{C} } \biggr)^{9 / 2} \biggl[{\tilde{r}}_\mathrm{edge} \biggl( \frac{3}{\tilde{C}} \biggr)^3 \biggr]^{-2} \biggl[ \frac{3}{(1-m_*)}\biggr]^{1 / 2}
\biggl\{
\biggl[ \frac{R_*^3}{\gamma_g GM_\mathrm{tot} } \biggr] \omega^2
+ \frac{(4 - 3\gamma_\mathrm{g})}{\gamma_g}  \biggl[  \frac{3 }{ \tilde{C} }\biggr]^{15/2}  (1-m_*)^{3 / 2} 
\biggr\}  x
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{ {\tilde{r}}^2_\mathrm{edge}} \biggl( \frac{2}{3} \biggr) \biggl\{ 
2m_* (1-m_*)^{3} \frac{d^2x}{dm_*^2}
+ (1-m_*)^{2} ( 1 + 2m_* )  \frac{dx}{dm_*} \biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+\frac{1}{ \tilde{r}_\mathrm{edge}^2} \biggl( \frac{2^3}{3} \biggr) 
\biggl[ 1  - \frac{3}{2} \biggl(  \frac{ 3 }{ \tilde{C} }\biggr)^{6}  m_*  \biggr]
(1-m_*)^{2} \frac{dx}{dm_*}
+ \frac{6}{ {\tilde{r}}_\mathrm{edge}^2 } 
\biggl(  \frac{3 }{ \tilde{C} }\biggr)^{6}  \biggl[ \frac{3}{(1-m_*)}\biggr]^{1 / 2}
\frac{(4 - 3\gamma_\mathrm{g})}{\gamma_g}
\biggl[ \sigma^2  +  (1-m_*)^{3 / 2}  \biggr]  x
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{ {\tilde{r}}^2_\mathrm{edge}} \biggl( \frac{2}{3} \biggr) \biggl\{ 
2m_* (1-m_*)^{3} \frac{d^2x}{dm_*^2}
+ \biggl[  5  - 6 \biggl(  \frac{ 3 }{ \tilde{C} }\biggr)^{6}  m_* + 2m_* \biggr]
(1-m_*)^{2} \frac{dx}{dm_*}
+ 3^{5 / 2} \biggl(  \frac{3 }{ \tilde{C} }\biggr)^{6} 
\frac{(4 - 3\gamma_\mathrm{g})}{\gamma_g}
\biggl[ \frac{\sigma^2 }{(1-m_*)^{1 / 2}} +  (1-m_*)  \biggr]  x \biggr\}
\, ,
</math>
  </td>
</tr>
</table>
</div>
where, as before,
<div align="center">
<math>\sigma^2 \equiv \biggl(  \frac{ \tilde{C} }{3 } \biggr)^{15/2} \biggl[ \frac{R_*^3}{(4 - 3\gamma_g) GM_\mathrm{tot} } \biggr] \omega^2 \, .</math>
</div>
==Take Another Approach Using Logarithmic Derivatives==
===Change Independent Variable===
Returning to the [[#n5LAWE|LAWE for n = 3 polytropes, as given, above]], and repeated here,
<table border="1" cellpadding="8" align="center">
<tr><th align="center">LAWE for <math>~n=5</math> Polytropes</th></tr>
<tr><td 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>~(3+\xi^2) \frac{d^2x}{d\xi^2} + \biggl[12 - 2\xi^2 \biggr] \frac{1}{\xi} \cdot \frac{dx}{d\xi} +
\biggl[\frac{\sigma_c^2}{3^{1/2} \gamma_g } \cdot (3+\xi^2)^{3/2} -
6\alpha \biggr]  x </math>
  </td>
</tr>
</table>
</td></tr></table>
let's make the substitution,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~u \equiv (3 + \xi^2)^{1/2}</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; <math>~\Rightarrow</math>&nbsp; &nbsp; &nbsp;
  </td>
  <td align="left">
<math>~\xi^2 = u^2-3 \, .</math>
  </td>
</tr>
</table>
</div>
We must therefore also make the operator substitution,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\frac{d}{d\xi}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{du}{d\xi} \cdot \frac{d}{du}</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \xi (3+\xi^2)^{-1/2}  \biggr]  \frac{d}{du} = \biggl[ 1 - \frac{3}{u^2} \biggr]^{1/2}  \frac{d}{du}</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow~~~ \frac{1}{\xi} \cdot \frac{dx}{d\xi}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{u}\cdot  \frac{dx}{du} \, ;</math>
  </td>
</tr>
</table>
</div>
and,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\frac{d^2}{d\xi^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ 1 - \frac{3}{u^2} \biggr]^{1/2}  \frac{d}{du} \biggl\{  \biggl[ 1 - \frac{3}{u^2} \biggr]^{1/2}  \frac{d}{du} \biggr\}</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ 1 - \frac{3}{u^2} \biggr]^{1/2}  \biggl\{  \frac{3}{u^3} \biggl[ 1 - \frac{3}{u^2} \biggr]^{-1/2} \frac{d}{du} + \biggl[ 1 - \frac{3}{u^2} \biggr]^{1/2}  \frac{d^2}{du^2}\biggr\}</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{3}{u^3} \frac{d}{du} + \biggl[ 1 - \frac{3}{u^2} \biggr]  \frac{d^2}{du^2}</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{d^2x}{d\xi^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{3}{u^3} \frac{dx}{du} + \biggl[ 1 - \frac{3}{u^2} \biggr]  \frac{d^2x}{du^2} \, .</math>
  </td>
</tr>
</table>
</div>
The rewritten LAWE is therefore,
<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>~u^2 \biggl\{ \frac{3}{u^3} \frac{dx}{du} + \biggl[ 1 - \frac{3}{u^2} \biggr]  \frac{d^2x}{du^2}  \biggr\}
+ 2\biggl[9 - u^2 \biggr] \frac{1}{u} \cdot \frac{dx}{du} +
\biggl[\Omega^2 u^3 -
6\alpha \biggr]  x </math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(u^2-3)  \frac{d^2x}{du^2} 
+ (21 - 2u^2 ) \frac{1}{u} \cdot \frac{dx}{du} +
(\Omega^2  u^3 - 6\alpha )  x \, ,</math>
  </td>
</tr>
</table>
</div>
where we have adopted the shorthand notation,
<div align="center">
<math>~\Omega^2 \equiv \frac{\sigma_c^2}{3^{1/2} \gamma_g } \, .</math>
</div>
===Look at Logarithmic Derivative===
Multiplying through by <math>~(u^2/x)</math> 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>~(u^2-3) \frac{u^2}{x} \cdot \frac{d^2x}{du^2} 
+ (21 - 2u^2 ) \frac{d\ln x}{d\ln u} +
(\Omega^2  u^5 - 6\alpha u^2 )  \, .</math>
  </td>
</tr>
</table>
</div>
Now, in the context of a [[User:Tohline/SSC/Stability/BiPolytrope0_0Details#Idea_Involving_Logarithmic_Derivatives|separate derivation]], we showed that, quite generally we can make the substitution,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\frac{u^2}{x} \cdot \frac{d^2x}{du^2}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{d}{d\ln u} \biggl[ \frac{d\ln x}{d\ln u} \biggr]
+ \biggl[  \frac{d\ln x}{d\ln u}-1 \biggr]\cdot \frac{d\ln x}{d\ln u} \, .
</math>
</td>
</tr>
</table>
</div>
Hence, if we assume that the displacement function can be expressed as a power-law in <math>~u</math>, such that,
<div align="center">
<math>\frac{d\ln x}{d\ln u} = c_0 \, ,</math>
</div>
then the LAWE for <math>~n=5</math> polytropes simplifies as follows,
<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>~(u^2-3) c_0(c_0-1) + (21 - 2u^2 ) c_0 + (\Omega^2  u^5 - 6\alpha u^2 )  \, .</math>
  </td>
</tr>
</table>
</div>
This polynomial equation will be satisfied if, simultaneously, we set:
<ul>
<li><math>\Omega^2 = 0 \, ;</math></li>
<li><math>c_0^2 -3c_0 -6\alpha = 0 </math>&nbsp; &nbsp; &nbsp; <math>~\Rightarrow</math>&nbsp; &nbsp; &nbsp; <math>c_0 = \frac{3}{2}\biggl[1 \pm \biggl(1+\frac{8\alpha}{3} \biggr)^{1/2} \biggl]\, ;</math></li>
<li><math>~\alpha = 20/3 \, .</math>
</ul>
This gives us some hope that a more general solution of the following form will work:
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~x</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~u^{c_0} \biggl[ a + bu + cu^2 + du^3 + \cdots\biggr] \, .</math>
  </td>
</tr>
</table>
</div>
This means that, for example,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\frac{dx}{du}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
c_0 u^{c_0-1} \biggl[ a + bu + cu^2 + du^3 \biggr]
+ u^{c_0} \biggl[ b + 2cu + 3du^2 \biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow ~~~\frac{d\ln x}{d\ln u}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{c_0(a + bu + cu^2 + du^3) + bu + 2cu^2 + 3du^3}{a + bu + cu^2 + du^3}
</math>
  </td>
</tr>
</table>
</div>
and,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\frac{d^2x}{du^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
c_0(c_0-1) u^{c_0-2} \biggl[ a + bu + cu^2 + du^3 \biggr]
+ 2c_0 u^{c_0-1} \biggl[ b + 2cu + 3du^2 \biggr]
+ u^{c_0} \biggl[ 2c + 6du \biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow~~~ \frac{u^2}{x} \cdot \frac{d^2x}{du^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{c_0(c_0-1) ( a + bu + cu^2 + du^3 ) + 2c_0 ( bu + 2cu^2 + 3du^3 ) + ( 2cu^2 + 6du^3 ) }{ a + bu + cu^2 + du^3}
</math>
  </td>
</tr>
</table>
</div>
So the LAWE becomes,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~- (\Omega^2  u^5 - 6\alpha u^2 ) (a + bu + cu^2 + du^3)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(u^2-3)  [c_0(c_0-1) ( a + bu + cu^2 + du^3 ) + 2c_0 ( bu + 2cu^2 + 3du^3 ) + ( 2cu^2 + 6du^3 )] 
+ (21 - 2u^2 ) [c_0(a + bu + cu^2 + du^3) + bu + 2cu^2 + 3du^3] \,.
</math>
  </td>
</tr>
</table>
</div>
This is cute, but I don't see any way that this approach will provide an avenue to cancel the <math>~\Omega^2 u^5</math> term.
==Yet Another Guess==
Let's try,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~x</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~e^{a + b\ln\xi + c(\ln\xi)^2} \, ,</math>
  </td>
</tr>
</table>
</div>
and examine the specific case of <math>~\sigma_c^2 = 0</math>, and, <math>~\gamma = (n+1)/n = 6/5 ~~\Rightarrow~~ \alpha = (3-20/6) = -1/3</math>.  Under these conditions, the LAWE for <math>~n=5</math> polytropes becomes,
<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>~(3+\xi^2) \frac{d^2x}{d\xi^2} + \biggl[12 - 2\xi^2 \biggr] \frac{1}{\xi} \cdot \frac{dx}{d\xi} + 2x
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(3+\xi^2) \frac{\xi^2}{x} \cdot \frac{d^2x}{d\xi^2} + \biggl[12 - 2\xi^2 \biggr] \frac{\xi}{x} \cdot \frac{dx}{d\xi} + 2\xi^2 \, .
</math>
  </td>
</tr>
</table>
And the derivatives give,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\frac{dx}{d\xi}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~x \frac{d}{d\xi}\biggl[ a + b\ln\xi + c(\ln\xi)^2 \biggr]</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~x \biggl[ \frac{b}{\xi}+ \frac{2c\ln\xi}{\xi} \biggr]</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{\xi}{x} \cdot \frac{dx}{d\xi}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~b + 2c\ln\xi \, ;</math>
  </td>
</tr>
</table>
</div>
and,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\frac{d^2x}{d\xi^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \biggl[ \frac{b}{\xi}+ \frac{2c\ln\xi}{\xi} \biggr] \frac{dx}{d\xi} + x \frac{d}{d\xi}\biggl[ \frac{b}{\xi}+ \frac{2c\ln\xi}{\xi} \biggr]</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{\xi^2}{x} \cdot \frac{d^2x}{d\xi^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \xi \biggl[ \frac{b}{\xi}+ \frac{2c\ln\xi}{\xi} \biggr] \frac{\xi}{x} \cdot \frac{dx}{d\xi} + \xi^2 \frac{d}{d\xi}\biggl[ \frac{b+ 2c\ln\xi}{\xi} \biggr]</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \biggl[ b + 2c\ln\xi \biggr]^2 + \xi \frac{d}{d\xi}\biggl[ b+ 2c\ln\xi \biggr]
+ (b+ 2c\ln\xi) \xi^2 \biggl[ - \frac{1}{\xi^2} \biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ (b + 2c\ln\xi )^2 + 2c- (b+ 2c\ln\xi)
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~[ b^2 + 2c - b] + [4bc  - 2c] \ln\xi+4c^2 (\ln\xi)^2 \, .
</math>
  </td>
</tr>
</table>
</div>
Hence the "fundamental mode" LAWE becomes,
<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>~(3+\xi^2) \biggl[ ( b^2 + 2c - b ) + (4bc  - 2c) \ln\xi+4c^2 (\ln\xi)^2 \biggr]
+ (12 - 2\xi^2 ) \biggl[ b + 2c\ln\xi  \biggr] \, .
+ 2\xi^2
</math>
  </td>
</tr>
</table>
</div>
Now, this expression cannot be satisfied for arbitrary <math>~\xi</math>.  But, here we seek a solution only at the surface for the ''specific'' model, <math>~\xi = 3</math>.  Plugging this value into the expression 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>~12 \biggl[ ( b^2 + 2c - b ) + (4bc  - 2c) \ln 3+4c^2 (\ln 3)^2 \biggr]
+ (12 - 18 ) \biggl[ b + 2c\ln 3 \biggr] + 18
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2 \biggl[ ( b^2 + 2c - b ) + (4bc  - 2c) \ln 3+4c^2 (\ln 3)^2 \biggr]
-\biggl[ b + 2c\ln 3 \biggr] + 3 \, .
</math>
  </td>
</tr>
</table>
</div>
It appears as though one perfectly satisfactory solution is, <math>~c = 0</math>, in which case, we need,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~0</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">  
<math>~
<math>~2 b^2   - 3b + 3  
\frac{1}{ {\tilde{r}}^2_\mathrm{edge}} \biggl( \frac{2}{3} \biggr) \biggl\{ 
2m_* (1-m_*)^{3} \frac{d^2x}{dm_*^2}
+ (1-m_*)^{2} ( 1 + 2m_* )  \frac{dx}{dm_*} \biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+\frac{1}{ \tilde{r}_\mathrm{edge}^2} \biggl( \frac{2^3}{3} \biggr) 
\biggl[ 1  - \frac{3}{2} \biggl(  \frac{ 3 }{ \tilde{C} }\biggr)^{6}  m_*  \biggr]
(1-m_*)^{2} \frac{dx}{dm_*}
+ \frac{6}{ {\tilde{r}}_\mathrm{edge}^2 } 
\biggl(  \frac{3 }{ \tilde{C} }\biggr)^{6}  \biggl[ \frac{3}{(1-m_*)}\biggr]^{1 / 2}
\frac{(4 - 3\gamma_\mathrm{g})}{\gamma_g}
\biggl[ \sigma^2  +  (1-m_*)^{3 / 2}  \biggr]  x
</math>
</math>
   </td>
   </td>
Line 3,322: Line 4,482:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\Rightarrow~~~b</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">  
<math>~
<math>~
\frac{1}{ {\tilde{r}}^2_\mathrm{edge}} \biggl( \frac{2}{3} \biggr) \biggl\{
\frac{3}{4}\biggl[1 \pm \sqrt{1-\frac{8}{3} }  \biggr] \, .
2m_* (1-m_*)^{3} \frac{d^2x}{dm_*^2}
+ \biggl[ 5  - 6 \biggl(  \frac{ 3 }{ \tilde{C} }\biggr)^{6}  m_* + 2m_* \biggr]
(1-m_*)^{2} \frac{dx}{dm_*}
+ 3^{5 / 2} \biggl(  \frac{3 }{ \tilde{C} }\biggr)^{6} 
\frac{(4 - 3\gamma_\mathrm{g})}{\gamma_g}  
\biggl[ \frac{\sigma^2 }{(1-m_*)^{1 / 2}} +  (1-m_*) \biggr] x \biggr\}
\, ,
</math>
</math>
   </td>
   </td>
Line 3,342: Line 4,495:
</table>
</table>
</div>
</div>
where, as before,
Thus, <math>~b</math> is an complex number.
<div align="center">
<math>\sigma^2 \equiv \biggl(  \frac{ \tilde{C} }{3 } \biggr)^{15/2} \biggl[ \frac{R_*^3}{(4 - 3\gamma_g) GM_\mathrm{tot} } \biggr] \omega^2 \, .</math>
</div>
 


=Related Discussions=
=Related Discussions=

Latest revision as of 03:50, 18 April 2017

Radial Oscillations of n = 5 Polytropic Spheres

Whitworth's (1981) Isothermal Free-Energy Surface
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Background

General Form of the LAWE for Spherical Polytropes

In an accompanying discussion, we derived the so-called,

Adiabatic Wave (or Radial Pulsation) Equation

LSU Key.png

<math>~ \frac{d^2x}{dr_0^2} + \biggl[\frac{4}{r_0} - \biggl(\frac{g_0 \rho_0}{P_0}\biggr) \biggr] \frac{dx}{dr_0} + \biggl(\frac{\rho_0}{\gamma_\mathrm{g} P_0} \biggr)\biggl[\omega^2 + (4 - 3\gamma_\mathrm{g})\frac{g_0}{r_0} \biggr] x = 0 </math>

whose solution gives eigenfunctions that describe various radial modes of oscillation in spherically symmetric, self-gravitating fluid configurations. Because this widely used form of the radial pulsation equation is not dimensionless but, rather, has units of inverse length-squared, we have found it useful to also recast it in the following dimensionless form:

<math> \frac{d^2x}{d\chi_0^2} + \biggl[\frac{4}{\chi_0} - \biggl(\frac{\rho_0}{\rho_c}\biggr) \biggl(\frac{P_0}{P_c}\biggr)^{-1} \biggl(\frac{g_0}{g_\mathrm{SSC}}\biggr) \biggr] \frac{dx}{d\chi_0} + \biggl(\frac{\rho_0}{\rho_c}\biggr) \biggl(\frac{P_0}{P_c}\biggr)^{-1} \biggl(\frac{1}{\gamma_\mathrm{g}} \biggr)\biggl[\tau_\mathrm{SSC}^2 \omega^2 + (4 - 3\gamma_\mathrm{g})\biggl(\frac{g_0}{g_\mathrm{SSC}}\biggr) \frac{1}{\chi_0} \biggr] x = 0 , </math>

where,

<math>~g_\mathrm{SSC} \equiv \frac{P_c}{R\rho_c} \, ,</math>       and       <math>~\tau_\mathrm{SSC} \equiv \biggl[\frac{R^2 \rho_c}{P_c}\biggr]^{1/2} \, .</math>

In a separate discussion, we showed that specifically for isolated, polytropic configurations, this linear adiabatic wave equation (LAWE) can be rewritten as,

<math>~0 </math>

<math>~=</math>

<math>~\frac{d^2x}{d\xi^2} + \biggl[\frac{4 - (n+1)V(\xi)}{\xi} \biggr] \frac{dx}{d\xi} + \biggl[\frac{\omega^2}{\gamma_g \theta} \biggl(\frac{n+1 }{4\pi G \rho_c} \biggr) - \biggl(3-\frac{4}{\gamma_g}\biggr) \cdot \frac{(n+1)V(x)}{\xi^2} \biggr] x </math>

 

<math>~=</math>

<math>~\frac{d^2x}{d\xi^2} + \biggl[\frac{4}{\xi} - \frac{(n+1)}{\theta} \biggl(- \frac{d\theta}{d\xi}\biggr)\biggr] \frac{dx}{d\xi} + \frac{(n+1)}{\theta}\biggl[\frac{\sigma_c^2}{6\gamma_g } - \frac{\alpha}{\xi} \biggl(- \frac{d\theta}{d\xi}\biggr)\biggr] x \, ,</math>

where we have adopted the dimensionless frequency notation,

<math>~\sigma_c^2</math>

<math>~\equiv</math>

<math>~\frac{3\omega^2}{2\pi G \rho_c} \, .</math>

Specifically for n=5 Configurations

Here we focus on an analysis of the specific case of isolated, <math>~n=5</math> polytropic configurations, whose unperturbed equilibrium structure can be prescribed in terms of analytic functions. Our hope — as yet unfulfilled — is that we can discover an analytically prescribed eigenvector solution to the governing LAWE.

From our discussion of the equilibrium structure of isolated, <math>~n=5</math> polytropes, we know that,

<math>~\theta</math>

<math>~=</math>

<math>~\biggl( 1 + \frac{\xi^2}{3} \biggr)^{-1/2} =3^{1/2} ( 3 + \xi^2 )^{-1/2}\, .</math>

Hence, we know as well that,

<math>~\frac{d\theta}{d\xi}</math>

<math>~=</math>

<math>~- \frac{\xi}{3}\biggl( 1 + \frac{\xi^2}{3} \biggr)^{-3/2} = - 3^{1/2}\xi ( 3 + \xi^2 )^{-3/2} \, .</math>

The LAWE therefore becomes,

<math>~0 </math>

<math>~=</math>

<math>~\frac{d^2x}{d\xi^2} + \biggl[\frac{4}{\xi} - \frac{(n+1)}{\theta} \biggl(- \frac{d\theta}{d\xi}\biggr)\biggr] \frac{dx}{d\xi} + \frac{(n+1)}{\theta}\biggl[\frac{\sigma_c^2}{6\gamma_g } - \frac{\alpha}{\xi} \biggl(- \frac{d\theta}{d\xi}\biggr)\biggr] x </math>

 

<math>~=</math>

<math>~\frac{d^2x}{d\xi^2} + \biggl[\frac{4}{\xi} - \frac{6}{\theta} \biggl(- \frac{d\theta}{d\xi}\biggr)\biggr] \frac{dx}{d\xi} + \biggl[\frac{\sigma_c^2}{\gamma_g } \cdot \frac{1}{\theta} - \frac{6\alpha}{\xi} \cdot \frac{1}{\theta} \biggl(- \frac{d\theta}{d\xi}\biggr)\biggr] x</math>

 

<math>~=</math>

<math>~\frac{d^2x}{d\xi^2} + \biggl[4 - \frac{6\xi^2}{(3+\xi^2)} \biggr] \frac{1}{\xi} \cdot \frac{dx}{d\xi} + \biggl[\frac{\sigma_c^2}{3^{1/2} \gamma_g } \cdot (3+\xi^2)^{1/2} - \frac{6\alpha}{(3+\xi^2)}\biggr)\biggr] x \, .</math>

Or,

LAWE for <math>~n=5</math> Polytropes

<math>~0</math>

<math>~=</math>

<math>~(3+\xi^2) \frac{d^2x}{d\xi^2} + \biggl[12 - 2\xi^2 \biggr] \frac{1}{\xi} \cdot \frac{dx}{d\xi} + \biggl[\frac{\sigma_c^2}{3^{1/2} \gamma_g } \cdot (3+\xi^2)^{3/2} - 6\alpha \biggr] x </math>

Numerical Integration of LAWE

By numerically integrating the above LAWE using the algorithm outlined in a separate chapter, we have examined the properties of the displacement function that describes radial modes of oscillation in pressure-truncated, n = 5, polytropic configurations. Our brief description, here, of these modes parallels the more detailed description of radial oscillation modes in truncated isothermal spheres that has been presented in a separate chapter.

The animation sequence that appears in the right panel of Composite Display 1 shows how our numerically derived displacement function, <math>~x(\xi)</math>, varies with radius — from the center of the n=5 polytropic sphere, out to <math>~\xi = 10</math> — for sixteen different values of the square of the eigenfrequency, <math>~\sigma_c^2</math>, as denoted at the top of each animation frame. The segment of the <math>~x(\xi)</math> curve that has been drawn in blue identifies the eigenfunction that corresponds to the specified value of the eigenfrequency. In each frame, the radial location at which the blue segment terminates simultaneously identifies: (a) the radius at which the logarithmic derivative of the displacement function, <math>~d\ln x/d\ln\xi </math>, is negative three; and (b) the radius, <math>~\tilde\xi</math>, at which the n = 5 polytropic configuration has been truncated. As displayed here, in every frame, the <math>~x(\xi)</math> function has been normalized such that the displacement amplitude is unity at the truncated configuration's surface.

The left panel of Composite Display 1 is also animated and has been provided in support of the animation on the right. Specifically, the number written at the top of each left-panel frame quantitatively identifies the radial location, <math>~\tilde\xi</math>, of the surface of the relevant truncated polytropic configuration; and, on each frame, "×" marks the location of that truncated configuration on the mass-radius equilibrium sequence.

Composite Display 1:   Numerically Generated Fundamental-Mode Eigenvectors

n5 Truncated movie

Excel File:

file = Dropbox/WorkFolder/Wiki edits/LinearPerturbation/n5Eigenvectors/n5TruncatedSphere.xlsx --- worksheet = OursPt1

Movie File:

file = Dropbox/WorkFolder/Wiki edits/EmbeddedPolytropes/n5movie/ --- worksheet = n5Truncated2.gif

Each full loop through the left-panel animation sequence can be viewed as evolution along the equilibrium sequence from <math>~\tilde\xi = 0.75</math> to <math>~\tilde\xi = 5</math>, then back again. During this evolution, the "×" marker moves through both turning points along the sequence: the maximum radius configuration — at <math>~\tilde\xi= \sqrt{3}</math> — and the maximum mass configuration — at <math>~\tilde\xi= 3</math>. Notice that <math>~\sigma_c^2</math> is positive for all models having <math>~\tilde\xi < 3</math> while it is negative for all models having <math>~\tilde\xi > 3</math>. Hence, models having <math>~\tilde\xi > 3</math> are dynamically unstable and, as best we have been able to determine via these numerical integrations, the transition from stable to unstable models — that is, the marginally unstable model — occurs at <math>~\tilde\xi = 3</math>. (Via an analytic analysis, we prove, below, that this association is precise.) For emphasis, the "×" marker (left panel) and the numerical value recorded for <math>~\sigma_c^2</math> (right panel) have been colored red for models that are not stable.

Search for Analytic Solutions to the LAWE

Eureka Moment

Note from J. E. Tohline on 3/6/2017:  Yesterday evening, after I finished putting together the above animation sequence using an Excel workbook, I noticed that the eigenfunction of the fundamental mode for the marginally unstable model <math>~(\sigma_c^2 = 0)</math> resembles a parabola. In an effort to see how well a parabola fits at least the central portion of this eigenfunction, I returned to my Excel spreedsheet and, in a brute-force manner, began to search for the pair of coefficients that would provide a best fit. What I discovered was that a parabola with the following formula fits perfectly!

Fundamental Mode Eigenfunction
when <math>~\sigma_c^2 = 0</math> and <math>~\gamma = 6/5 ~\Rightarrow~\alpha=- 1/3</math>

<math>~x = x_0 \biggl[ 1 - \frac{\xi^2}{15} \biggr]</math>

For the specific normalization used in the above animation sequence, <math>~x_0 = \tfrac{5}{2}</math>. Let's demonstrate that this eigenvector provides a solution to the LAWE for <math>~n=5</math> polytropes; for simplicity, we will set <math>~x_0 = 1</math>:

<math>~\frac{dx}{d\xi} = -\frac{2\xi}{15} \, ;</math>

      and      

<math>~\frac{d^2x}{d\xi^2} = -\frac{2}{15} \, .</math>

<math>~\Rightarrow ~~~ (3+\xi^2) \frac{d^2x}{d\xi^2} + \biggl[12 - 2\xi^2 \biggr] \frac{1}{\xi} \cdot \frac{dx}{d\xi} + \biggl[ \cancelto{0}{\frac{\sigma_c^2}{3^{1/2} \gamma_g }} \cdot (3+\xi^2)^{3/2} + 2\biggr] x </math>

<math>~=</math>

<math>~ -\frac{2}{15}(3+\xi^2) -\frac{2}{15} \biggl[12 - 2\xi^2 \biggr] + 2\biggl[1 - \frac{\xi^2}{15}\biggr] </math>

 

<math>~=</math>

<math>~ \biggl( - \frac{6}{15} - \frac{24}{15} + 2\biggr) +\xi^2 \biggl( -\frac{2}{15} + \frac{4}{15} - \frac{2}{15} \biggr) </math>

 

<math>~=</math>

<math>~ 0 \, . </math>

Q. E. D.  I don't think that anyone has previously appreciated that the LAWE in this case admits to an analytic eigenvector solution.

Now, let's see how the boundary condition comes into play. We see that the logarithmic derivative of the parabolic eigenfunction is,

<math>~\frac{d\ln x}{d\ln \xi}</math>

<math>~=</math>

<math>~\frac{\xi}{x} \cdot \frac{dx}{d\xi}</math>

 

<math>~=</math>

<math>~- \frac{2\xi^2}{15} \biggl[ 1 - \frac{\xi^2}{15}\biggr]^{-1}</math>

 

<math>~=</math>

<math>~- \frac{2\xi^2}{(15-\xi^2)} \, .</math>

We desire a surface boundary condition that gives, <math>~d\ln x/d\ln\xi = -3</math>. This will only happen when,

<math>~- \frac{2\xi^2}{(15-\xi^2)}</math>

<math>~=</math>

<math>~-3</math>

<math>~\Rightarrow ~~~2\xi^2</math>

<math>~=</math>

<math>~3(15 - \xi^2)</math>

<math>~\Rightarrow ~~~\xi</math>

<math>~=</math>

<math>~3 \, .</math>

Hence, although the parabolic eigenfunction provides an accurate solution to the <math>~n=5</math> LAWE throughout the entire configuration — that is, for all <math>~\xi</math> — the desired surface boundary condition will only be satisfied if the polytrope is truncated at <math>~\xi_\mathrm{surf} = 3</math>. The parabolic eigenfunction is therefore only physically relevant to the model that sits at the point along the equilibrium sequence that is associated with the <math>~P_\mathrm{max}</math> turning point.


Let's express the parabolic displacement function, <math>~x</math>, as a function of the Lagrangian mass coordinate, instead of as a function of <math>~\xi</math>. Drawing upon our accompanying discussion where we have used <math>~\tilde\xi</math> to denote the truncation edge, we know that,

<math>~r_\xi(\xi)</math>

<math>~=</math>

<math>~\xi \biggl\{ \biggl[ \frac{4\pi}{2^5\cdot 3}\biggr]^{1/2} \tilde\xi^{-6} \biggl( 1+\frac{\tilde\xi^2}{3} \biggr)^{3}\biggr\} \, . </math>

and that,

<math>~ r_\xi (m_\xi) </math>

<math>~=</math>

<math>~\tilde{r}_\mathrm{edge} \biggl[\frac{3^2m_\xi^{2/3}}{\tilde{C} - 3 m_\xi^{2/3}}\biggr]^{1/2} \, , </math>

where,

<math>~\tilde{C}</math>

<math>~\equiv</math>

<math>~ \frac{3^2}{\tilde\xi^2}\biggl( 1 + \frac{\tilde\xi^2}{3} \biggr) \, . </math>

<math>~\tilde{r}_\mathrm{edge}</math>

<math>~\equiv</math>

<math>~\biggl[ \frac{\pi}{2^3\cdot 3}\biggr]^{1/2} {\tilde\xi}^{-6} \biggl(1+\frac{\tilde\xi^2}{3}\biggr)^3 \, . </math>

By equating <math>~r_\xi(\xi)</math> with <math>~r_\xi(m_\xi)</math>, we find,

<math>~\xi \biggl\{ \biggl[ \frac{4\pi}{2^5\cdot 3}\biggr]^{1/2} \tilde\xi^{-6} \biggl( 1+\frac{\tilde\xi^2}{3} \biggr)^{3}\biggr\} </math>

<math>~=</math>

<math>~\biggl[ \frac{\pi}{2^3\cdot 3}\biggr]^{1/2} {\tilde\xi}^{-6} \biggl(1+\frac{\tilde\xi^2}{3}\biggr)^3 \biggl[\frac{3^2m_\xi^{2/3}}{\tilde{C} - 3 m_\xi^{2/3}}\biggr]^{1/2} </math>

<math>~\Rightarrow ~~~ \xi </math>

<math>~=</math>

<math>~ \biggl[\frac{3^2m_\xi^{2/3}}{\tilde{C} - 3 m_\xi^{2/3}}\biggr]^{1/2} \, . </math>

This means that,

<math>~x</math>

<math>~=</math>

<math>~ x_0 \biggl\{1 - \frac{1}{15} \biggl[\frac{3^2m_\xi^{2/3}}{\tilde{C} - 3 m_\xi^{2/3}}\biggr] \biggr\} \, ; </math>

and, specifically for the critical case of <math>~\tilde\xi = 3</math>, in which case, <math>~\tilde{C} = 4</math>,

<math>~x</math>

<math>~=</math>

<math>~ x_0 \biggl\{1 - \frac{1}{15} \biggl[\frac{3^2m_\xi^{2/3}}{4 - 3 m_\xi^{2/3}}\biggr] \biggr\} \, . </math>



Work-in-progress.png

Material that appears after this point in our presentation is under development and therefore
may contain incorrect mathematical equations and/or physical misinterpretations.
|   Go Home   |


Setup Using Lagrangian Radial Coordinate

Individual Terms

From our accompanying discussion, we have, for pressure-truncated, <math>~n=5</math> polytropic spheres

<math> ~\frac{R_\mathrm{eq}}{R_\mathrm{norm}} </math>

<math>~=~</math>

<math>~\biggl[ \frac{4\pi}{(n+1)^n}\biggr]^{1/(n-3)} \tilde\xi ( -\tilde\xi^2 \tilde\theta' )^{(1-n)/(n-3)} </math>

 

<math>~=~</math>

<math>~\biggl[ \frac{4\pi}{2^5\cdot 3^5}\biggr]^{1/2} \tilde\xi ( -\tilde\xi^2 \tilde\theta' )^{-2} \, , </math>

which matches the expression derived in an ASIDE box found with our introduction of the Lane-Emden equation, and

<math> ~\frac{P_\mathrm{e}}{P_\mathrm{norm}} </math>

<math>~=~</math>

<math>~\biggl[ \frac{(n+1)^3}{4\pi}\biggr]^{(n+1)/(n-3)} \tilde\theta_n^{n+1}( -\tilde\xi^2 \tilde\theta' )^{2(n+1)/(n-3)} </math>

 

<math>~=~</math>

<math>~\biggl[ \frac{2^3\cdot 3^3}{4\pi}\biggr]^{3} \tilde\theta^{6}( -\tilde\xi^2 \tilde\theta' )^{6} \, , </math>

where,

<math>~R_\mathrm{norm}</math>

<math>~\equiv</math>

<math>~\biggl[ \biggl( \frac{G}{K} \biggr)^n M_\mathrm{tot}^{n-1} \biggr]^{1/(n-3)} = \biggl( \frac{G}{K} \biggr)^{5/2} M_\mathrm{tot}^{2} \, ,</math>

<math>~P_\mathrm{norm}</math>

<math>~\equiv</math>

<math>~\biggl[ \frac{K^{4n}}{G^{3(n+1)} M_\mathrm{tot}^{2(n+1)}} \biggr]^{1/(n-3)} = \frac{K^{10}}{G^{9} M_\mathrm{tot}^{6} } \, ,</math>

and, from our more detailed analysis,

<math> ~{\tilde\theta}_5 = 3^{1 / 2} \biggl( 3 + {\tilde\xi}^2\biggr)^{-1/2} </math>

        and        

<math> ~\biggl(- {\tilde\xi}^2 {\tilde\theta}^'_5\biggr) = 3^{1 / 2} {\tilde\xi}^3 \biggl( 3 + {\tilde\xi}^2\biggr)^{-3/2} \, . </math>

Hence,

<math> ~\frac{R_\mathrm{eq}}{R_\mathrm{norm}} </math>

<math>~=~</math>

<math>~\biggl[ \frac{4\pi}{2^5\cdot 3^5}\biggr]^{1/2} \tilde\xi \biggl[ 3^{1 / 2} {\tilde\xi}^3 \biggl( 3 + {\tilde\xi}^2\biggr)^{-3/2} \biggr]^{-2} </math>

 

<math>~=~</math>

<math>~\biggl[ \frac{4\pi}{2^5\cdot 3^5}\biggr]^{1/2} \tilde\xi \biggl[ 3^{-1} {\tilde\xi}^{-6} \biggl( 3 + {\tilde\xi}^2\biggr)^{3} \biggr] </math>

 

<math>~=~</math>

<math>~ \biggl[ \frac{4\pi}{2^5\cdot 3^7}\biggr]^{1/2}

{\tilde\xi}^{-5} \biggl( 3 + {\tilde\xi}^2\biggr)^{3}  \, ,

</math>

<math> ~\frac{P_\mathrm{e}}{P_\mathrm{norm}} </math>

<math>~=~</math>

<math>~\biggl[ \frac{2^3\cdot 3^3}{4\pi}\biggr]^{3} \biggl[ 3^{1 / 2} \biggl( 3 + {\tilde\xi}^2\biggr)^{-1/2} \biggr]^{6} \biggl[ 3^{1 / 2} {\tilde\xi}^3 \biggl( 3 + {\tilde\xi}^2\biggr)^{-3/2} \biggr]^{6} </math>

 

<math>~=~</math>

<math>~\biggl[ \frac{2^3\cdot 3^3}{4\pi}\biggr]^{3} \biggl[ 3^{3} \biggl( 3 + {\tilde\xi}^2\biggr)^{-3} \biggr] \biggl[ 3^{3} {\tilde\xi}^{18} \biggl( 3 + {\tilde\xi}^2\biggr)^{-9} \biggr] </math>

 

<math>~=~</math>

<math>~\biggl[ \frac{2^3\cdot 3^5}{4\pi}\biggr]^{3} {\tilde\xi}^{18} \biggl( 3 + {\tilde\xi}^2\biggr)^{-12} \, . </math>

Now, given that the structural form-factors for <math>~n=5</math> configurations are,

<math>~\mathfrak{f}_M</math>

<math>~=</math>

<math>~ ( 1 + \ell^2 )^{-3/2} = 3^{3 / 2} (3 + {\tilde\xi}^2)^{-3 / 2} </math>

<math>~\mathfrak{f}_W</math>

<math>~=</math>

<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>

<math>~\mathfrak{f}_A</math>

<math>~=</math>

<math>~ \frac{3}{2^3} \ell^{-3} [ \tan^{-1}(\ell ) + \ell (\ell^2-1) (1+\ell^2)^{-2} ] \, , </math>

we understand that the central density is,

<math>~\rho_c = \frac{\bar\rho}{ {\tilde\mathfrak{f}}_M }</math>

<math>~=</math>

<math>~ \biggl[3^{3 / 2} (3 + {\tilde\xi}^2)^{-3 / 2} \biggr]^{-1} \biggl[ \frac{3 M_\mathrm{tot}}{4 \pi R_\mathrm{eq}^3} \biggr] </math>

 

<math>~=</math>

<math>~ \biggl( \frac{3}{4\pi}\biggr) \biggl[ \frac{2^5\cdot 3^6}{4\pi}\biggr]^{ 3 / 2} (3 + {\tilde\xi}^2)^{3 / 2} M_\mathrm{tot} \biggl[ R_\mathrm{norm}

{\tilde\xi}^{-5} \biggl( 3 + {\tilde\xi}^2\biggr)^{3} \biggr]^{-3}

</math>

 

<math>~=</math>

<math>~ \biggl[ \frac{2^{5}\cdot 3^{20}}{\pi^5}\biggr]^{ 1 / 2} {\tilde\xi}^{15} (3 + {\tilde\xi}^2)^{-15 / 2} M_\mathrm{tot} R^{-3}_\mathrm{norm} </math>

 

<math>~=</math>

<math>~ \biggl[ \frac{2^{5}\cdot 3^{20}}{\pi^5}\biggr]^{ 1 / 2} \biggl[{\tilde\xi} (3 + {\tilde\xi}^2)^{-1 / 2} \biggr]^{15} M_\mathrm{tot}^{-5} \biggl( \frac{G}{K} \biggr)^{-15/2} </math>

 

<math>~=</math>

<math>~ \biggl[ \frac{2\cdot 3^{4}}{\pi}\biggr]^{ 5 / 2} \biggl[{\tilde\xi} (3 + {\tilde\xi}^2)^{-1 / 2} \biggr]^{15} \biggl( \frac{K^3}{G^3M_\mathrm{tot}^2} \biggr)^{5/2} \, . </math>

Now let's derive the prescription for the Lagrangian radial coordinate in the context of pressure-truncated, <math>~n=5</math> polytropes.

<math>~r_0 \equiv a_5 \xi</math>

<math>~=</math>

<math>~\biggl[\frac{3K}{2\pi G} \biggr]^{1 / 2} \rho_c^{-2/5} \xi</math>

 

<math>~=</math>

<math>~\biggl[\frac{3K}{2\pi G} \biggr]^{1 / 2} \xi \biggl\{ \biggl[ \frac{2\cdot 3^{4}}{\pi}\biggr]^{ 5 / 2} \biggl[{\tilde\xi} (3 + {\tilde\xi}^2)^{-1 / 2} \biggr]^{15} \biggl( \frac{K^3}{G^3M_\mathrm{tot}^2} \biggr)^{5/2} \biggr\}^{-2/5} </math>

 

<math>~=</math>

<math>~\biggl[\frac{3K}{2\pi G} \biggr]^{1 / 2} \biggl[ \frac{\pi}{2\cdot 3^{4}}\biggr] \biggl( \frac{G^3M_\mathrm{tot}^2}{K^3} \biggr) \biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3} \xi </math>

 

<math>~=</math>

<math>~ R_\mathrm{norm} \biggl[ \frac{\pi}{2^3\cdot 3^{7}}\biggr]^{1 / 2} \biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3} \xi </math>

Also,

<math>~m_0 \equiv M(r_0)</math>

<math>~=</math>

<math>~\biggl[ 4\pi a_n^3 \rho_c \biggl(-\xi^2 \frac{d\theta}{d\xi}\biggr) \biggr] \, ,</math>

 

<math>~=</math>

<math>~2^2\pi \biggl\{ R_\mathrm{norm} \biggl[ \frac{\pi}{2^3\cdot 3^{7}}\biggr]^{1 / 2} \tilde\xi^{-6} (3 + {\tilde\xi}^2)^{3} \biggr\}^3 \biggl\{ \biggl[ \frac{2\cdot 3^{4}}{\pi}\biggr]^{ 5 / 2} \biggl[{\tilde\xi} (3 + {\tilde\xi}^2)^{-1 / 2} \biggr]^{15} \biggl( \frac{K^3}{G^3M_\mathrm{tot}^2} \biggr)^{5/2} \biggr\} \biggl\{ 3^{1 / 2} \xi^3 \biggl( 3 + \xi^2\biggr)^{-3/2} \biggr\} </math>

 

<math>~=</math>

<math>~ 3^{1 / 2} \biggl[ 2^4 \pi^2\biggr]^{1 / 2} \biggl[ \frac{\pi^3}{2^9\cdot 3^{21}}\biggr]^{1 / 2} \biggl[ \frac{2^5\cdot 3^{20}}{\pi^5}\biggr]^{ 1 / 2} \biggl\{ \tilde\xi^{-6} (3 + {\tilde\xi}^2)^{3} \biggr\}^3 \biggl[{\tilde\xi} (3 + {\tilde\xi}^2)^{-1 / 2} \biggr]^{15} \biggl( \frac{K^3}{G^3M_\mathrm{tot}^2} \biggr)^{5/2} R_\mathrm{norm}^3 \biggl\{ \xi^3 ( 3 + \xi^2 )^{-3/2} \biggr\} </math>

 

<math>~=</math>

<math>~ \biggl\{ \tilde\xi^{-3} (3 + {\tilde\xi}^2)^{3 / 2} \biggr\} M_\mathrm{tot} \biggl\{ \xi^3 ( 3 + \xi^2 )^{-3/2} \biggr\} \, . </math>

Hence,

<math>~g_0 = \frac{Gm_0}{r_0^2}</math>

<math>~=</math>

<math>~ \frac{GM_\mathrm{tot}}{R_\mathrm{norm}^2} \biggl\{ \tilde\xi^{-3} (3 + {\tilde\xi}^2)^{3 / 2} \biggr\} \biggl\{ \xi^3 ( 3 + \xi^2 )^{-3/2} \biggr\} \biggl\{ \biggl[ \frac{\pi}{2^3\cdot 3^{7}}\biggr]^{1 / 2} \biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3} \xi \biggr\}^{-2} </math>

 

<math>~=</math>

<math>~ \frac{GM_\mathrm{tot}}{R_\mathrm{norm}^2}\biggl[ \frac{2^3\cdot 3^{7}}{\pi}\biggr] \biggl[ \tilde\xi (3 + {\tilde\xi}^2)^{-1 / 2} \biggr]^{9} \xi ( 3 + \xi^2 )^{-3/2} \, ; </math>

<math>~\frac{g_0 }{r_0} </math>

<math>~=</math>

<math>~ \frac{GM_\mathrm{tot}}{R_\mathrm{norm}^3}\biggl[ \frac{2^3\cdot 3^{7}}{\pi}\biggr] \biggl\{ \tilde\xi^{9} (3 + {\tilde\xi}^2)^{-9 / 2} \biggr\} \biggl\{ \biggl[ \frac{\pi}{2^3\cdot 3^{7}}\biggr]^{1 / 2} \biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3} \xi \biggr\}^{-1} \xi ( 3 + \xi^2 )^{-3/2} </math>

 

<math>~=</math>

<math>~ \frac{GM_\mathrm{tot}}{R_\mathrm{norm}^3}\biggl[ \frac{2^3\cdot 3^{7}}{\pi}\biggr]^{3/2} \biggl[ \tilde\xi (3 + {\tilde\xi}^2)^{-1 / 2} \biggr]^{15} ( 3 + \xi^2 )^{-3/2} \, ; </math>


<math>~\frac{\rho_0}{P_0} = \frac{\rho_0}{K\rho_0^{1+1/n}} </math>

<math>~=</math>

<math>~ \biggl[K^5 \rho_c \theta^5 \biggr]^{-1/5} </math>

 

<math>~=</math>

<math>~ \theta^{-1} \biggl\{ K^5 \biggl[ \frac{2\cdot 3^{4}}{\pi}\biggr]^{ 5 / 2} \biggl[{\tilde\xi} (3 + {\tilde\xi}^2)^{-1 / 2} \biggr]^{15} \biggl( \frac{K^3}{G^3M_\mathrm{tot}^2} \biggr)^{5/2}\biggr\}^{-1/5} </math>

 

<math>~=</math>

<math>~ \biggl[ 3^{-1} ( 3 + \xi^2 ) \biggr]^{1/2} \biggl\{ \biggl[ \frac{\pi}{2\cdot 3^{4}}\biggr]^{1 / 2} \cancelto{\mathrm{mistake}}{\biggl[{\tilde\xi}^{-3} (3 + {\tilde\xi}^2)^{3 / 2} \biggr]^{-3} } \biggl( \frac{G^3M_\mathrm{tot}^2}{K^5} \biggr)^{1/2}\biggr\} </math>

 

<math>~=</math>

<math>~ \biggl( \frac{G^3M_\mathrm{tot}^2}{K^5} \biggr)^{1 / 2} \biggl[ \frac{\pi}{2\cdot 3^{5}}\biggr]^{1 / 2} \cancelto{\mathrm{mistake}}{\biggl[ {\tilde\xi} (3 + {\tilde\xi}^2)^{-1 / 2} \biggr]^{9} } ( 3 + \xi^2 )^{1 / 2} \, ; </math>

 

<math>~=</math>

<math>~ \biggl[ 3^{-1} ( 3 + \xi^2 ) \biggr]^{1/2} \biggl\{ \biggl[ \frac{\pi}{2\cdot 3^{4}}\biggr]^{1 / 2} \biggl[{\tilde\xi}^{-3} (3 + {\tilde\xi}^2)^{3 / 2} \biggr] \biggl( \frac{G^3M_\mathrm{tot}^2}{K^5} \biggr)^{1/2}\biggr\} </math>

 

<math>~=</math>

<math>~ \biggl( \frac{G^3M_\mathrm{tot}^2}{K^5} \biggr)^{1/2}\biggl[ \frac{\pi}{2\cdot 3^{5}}\biggr]^{1 / 2} \biggl[ \frac{(3 + {\tilde\xi}^2)}{{\tilde\xi}^2 } \biggr]^{3 / 2} ( 3 + \xi^2 )^{1/2} </math>

<math>~\frac{g_0\rho_0}{P_0} </math>

<math>~=</math>

<math>~ \biggl( \frac{G^3M_\mathrm{tot}^2}{K^5} \biggr)^{1/2}\biggl[ \frac{\pi}{2\cdot 3^{5}}\biggr]^{1 / 2} \biggl[ \frac{(3 + {\tilde\xi}^2)}{{\tilde\xi}^2 } \biggr]^{3 / 2} ( 3 + \xi^2 )^{1/2} </math>

 

 

<math>~ \times ~ \biggl( \frac{G^2M_\mathrm{tot}^2}{R_\mathrm{norm}^4} \biggr)^{1 / 2}\biggl[ \frac{2^6\cdot 3^{14}}{\pi^2}\biggr]^{1 / 2} \biggl[ \frac{(3 + {\tilde\xi}^2)}{{\tilde\xi}^2 } \biggr]^{-9 / 2} \xi ( 3 + \xi^2 )^{-3/2} </math>

 

<math>~=</math>

<math>~ \biggl( \frac{G^5 M_\mathrm{tot}^4}{K^5} \biggr)^{1 / 2} R_\mathrm{norm}^{-2} \biggl[ \frac{{\tilde\xi}^2 }{(3 + {\tilde\xi}^2)} \biggr]^{3} \biggl[ \frac{2^5\cdot 3^{9}}{\pi}\biggr]^{1 / 2} \xi ( 3 + \xi^2 )^{-1} </math>

 

<math>~=</math>

<math>~ \biggl( \frac{K^5}{G^5 M_\mathrm{tot}^4} \biggr)^{1 / 2} \biggl[ \frac{{\tilde\xi}^2 }{(3 + {\tilde\xi}^2)} \biggr]^{3} \biggl[ \frac{2^5\cdot 3^{9}}{\pi}\biggr]^{1 / 2} \xi ( 3 + \xi^2 )^{-1} \, . </math>

The Wave Equation

Starting from our Key Adiabatic Wave Equation

The adiabatic wave equation therefore becomes,

<math>~0</math>

<math>~=</math>

<math>~ \frac{d^2x}{dr_0^2} + \biggl[\frac{4}{r_0} - \biggl(\frac{g_0 \rho_0}{P_0}\biggr) \biggr] \frac{dx}{dr_0} + \biggl(\frac{\rho_0}{\gamma_\mathrm{g} P_0} \biggr)\biggl[\omega^2 + (4 - 3\gamma_\mathrm{g})\frac{g_0}{r_0} \biggr] x </math>

 

<math>~=</math>

<math>~ \frac{d^2x}{dr_0^2} + \frac{1}{R_\mathrm{norm}} \biggl\{ \biggl[ \frac{2^7\cdot 3^{7}}{\pi} \biggr]^{1 / 2} \biggl[ \frac{ {\tilde\xi}^2}{(3 + {\tilde\xi}^2)} \biggr]^{3} \frac{1}{\xi} - \biggl[ \frac{{\tilde\xi}^2 }{(3 + {\tilde\xi}^2)} \biggr]^{3} \biggl[ \frac{2^5\cdot 3^{9}}{\pi}\biggr]^{1 / 2} \xi ( 3 + \xi^2 )^{-1} \biggr\} \frac{dx}{dr_0} </math>

 

 

<math>~ + \frac{(4 - 3\gamma_\mathrm{g})}{\gamma_g R_\mathrm{norm}^2} \biggl[ \frac{\pi}{2\cdot 3^{5}}\biggr]^{1 / 2} \biggl[ \frac{{\tilde\xi}^2}{(3 + {\tilde\xi}^2)} \biggr]^{-3/2} ( 3 + \xi^2 )^{1 / 2} \biggl\{ \frac{R_\mathrm{norm}^3}{GM_\mathrm{tot}} \biggl[\frac{\omega^2}{(4 - 3\gamma_\mathrm{g})} \biggr] + \biggl[ \frac{2^3\cdot 3^{7}}{\pi}\biggr]^{3/2} \biggl[ \frac{{\tilde\xi}^2}{(3 + {\tilde\xi}^2)} \biggr]^{15/2} ( 3 + \xi^2 )^{-3/2} \biggr\} x </math>

 

<math>~=</math>

<math>~ \frac{d^2x}{dr_0^2} + \frac{1}{R_\mathrm{norm}} \biggl[ \frac{2^3\cdot 3^{7}}{\pi} \biggr]^{1 / 2} \biggl[ \frac{ {\tilde\xi}^2}{(3 + {\tilde\xi}^2)} \biggr]^{3} \biggl[ \frac{4}{\xi} - \frac{6 \xi}{ ( 3 + \xi^2 )} \biggr] \frac{dx}{dr_0} </math>

 

 

<math>~ + \frac{6(4 - 3\gamma_\mathrm{g})}{\gamma_g R_\mathrm{norm}^2} ( 3 + \xi^2 )^{1 / 2} \biggl[ \frac{2^3\cdot 3^{7}}{\pi}\biggr] \biggl[ \frac{{\tilde\xi}^2}{(3 + {\tilde\xi}^2)} \biggr]^{6} \biggl\{ \frac{R_\mathrm{norm}^3}{GM_\mathrm{tot}} \biggl[\frac{\omega^2}{(4 - 3\gamma_\mathrm{g})} \biggr] \biggl[ \frac{2^3\cdot 3^{7}}{\pi}\biggr]^{-3/2} \biggl[ \frac{{\tilde\xi}^2}{(3 + {\tilde\xi}^2)} \biggr]^{-15/2} + ( 3 + \xi^2 )^{-3/2} \biggr\} x </math>

 

<math>~=</math>

<math>~ \frac{d^2x}{dr_0^2} + \frac{1}{R_*} \biggl[ \frac{ {\tilde\xi}^2}{(3 + {\tilde\xi}^2)} \biggr]^{3} \biggl[ \frac{4}{\xi} - \frac{6 \xi}{ ( 3 + \xi^2 )} \biggr] \frac{dx}{dr_0} + \frac{6}{\gamma_g R_*^2} \biggl[ \frac{{\tilde\xi}^2}{(3 + {\tilde\xi}^2)} \biggr]^{6} \biggl\{ \frac{\omega^2 R_*^3}{GM_\mathrm{tot}} \biggl[ \frac{{\tilde\xi}^2}{(3 + {\tilde\xi}^2)} \biggr]^{-15/2}( 3 + \xi^2 )^{1 / 2} + \frac{(4 - 3\gamma_\mathrm{g})}{( 3 + \xi^2 ) } \biggr\} x </math>

where,

<math>R_* \equiv R_\mathrm{norm} \biggl[ \frac{\pi}{2^3 \cdot 3^7} \biggr]^{1/2} \, .</math>

Recognizing that,

<math>~r_0</math>

<math>~=</math>

<math>~ R_* \biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3} \xi \, , </math>

we can write,

<math>~0</math>

<math>~=</math>

<math>~\frac{1}{R_*^2} \biggl[ \frac{ {\tilde\xi}^2}{(3 + {\tilde\xi}^2)} \biggr]^{6} \biggl\{ \frac{d^2x}{d\xi^2} + \biggl[ \frac{4}{\xi} - \frac{6 \xi}{ ( 3 + \xi^2 )} \biggr] \frac{dx}{d\xi} + \frac{6}{\gamma_g } \biggl[\sigma^2 ( 3 + \xi^2 )^{1 / 2} + \frac{(4 - 3\gamma_\mathrm{g})}{( 3 + \xi^2 ) } \biggr] x \biggr\} \, , </math>

where,

<math>~\sigma^2</math>

<math>~\equiv</math>

<math>~ \frac{\omega^2 R_*^3}{GM_\mathrm{tot}} \biggl( \frac{3 + {\tilde\xi}^2}{{\tilde\xi}^2} \biggr)^{15/2} = \frac{\sigma_c^2}{2\cdot 3^{3/2}}\, .</math>

Finally, if — because we are specifically considering the case of <math>~n=5</math> — we set <math>~\gamma_\mathrm{g} = 1 + 1/n = 6/5</math>, we have,

<math>~0</math>

<math>~=</math>

<math>~ \frac{d^2x}{d\xi^2} + \biggl[ \frac{4}{\xi} - \frac{6 \xi}{ ( 3 + \xi^2 )} \biggr] \frac{dx}{d\xi} + \biggl[\frac{5\sigma_c^2}{2\cdot 3^{3/2}} ( 3 + \xi^2 )^{1 / 2} + \frac{2}{( 3 + \xi^2 ) }\biggr] x </math>

 

<math>~=</math>

<math>~\frac{1}{( 3 + \xi^2 ) } \biggl\{ ( 3 + \xi^2 )\frac{d^2x}{d\xi^2} + \biggl[ \frac{2(6 - \xi^2) }{ \xi} \biggr] \frac{dx}{d\xi} + \biggl[\frac{5\sigma_c^2}{2\cdot 3^{3/2}} ( 3 + \xi^2 )^{3 / 2} + 2 \biggr] x \biggr\} \, , </math>

which matches exactly the form of the LAWE derived above, if in that expression, <math>~\gamma_g</math> is also forced to align with our specification of the polytropic index, that is, if <math>~\gamma_g = (n+1)/n = 6/5</math> and, in turn, <math>~\alpha = (3-4/\gamma) = -1/3</math>.


Starting from the HRW66 Radial Pulsation Equation

More directly, if we begin with the HRW66 radial pulsation equation that is already tuned to polytropic configurations, the wave equation appropriate to <math>~n=5</math> polytropes is,

<math>~0</math>

<math>~=</math>

<math>~ \frac{d^2 X}{d\xi^2} + \biggl[ \frac{4}{\xi} - \frac{6 (-\theta^'_5)}{\theta_5} \biggr]\frac{d X}{d\xi} + \frac{5(-\theta_5^') }{6\theta_5 \xi} \bigg[ \frac{\xi (s^')^2}{\theta^'_5} + \frac{12}{5} \biggr] X </math>

 

<math>~=</math>

<math>~ \frac{d^2 X}{d\xi^2} + \biggl[ \frac{4}{\xi} - \frac{6 \xi}{(3 + \xi^2)} \biggr]\frac{d X}{d\xi} + \frac{1}{(3 + \xi^2)} \bigg[ -\frac{5(s^')^2(3 + \xi^2)^{3 / 2}}{2 \cdot 3^{3 / 2}} + 2 \biggr] X </math>

 

<math>~=</math>

<math>~\frac{1}{(3+\xi^2)} \biggl\{ (3+\xi^2)\frac{d^2 X}{d\xi^2} + \biggl[ \frac{2(6-\xi^2)}{\xi}\biggr]\frac{d X}{d\xi} + \bigg[ -\frac{5(s^')^2}{2 \cdot 3^{3 / 2}} \cdot (3 + \xi^2)^{3 / 2} + 2 \biggr] X \biggr\} \, , </math>

which is identical to the brute-force derivation just presented, allowing for the mapping,

<math>\sigma^2 ~~ \Leftrightarrow ~~ -\frac{(s^')^2}{2 \cdot 3^{3 / 2}} \, .</math>

Finally, remembering that the HRW66 dimensionless frequency definition is,

<math>~(s^')^2</math>

<math>~=</math>

<math>~-\omega^2 \biggl[\frac{n+1 }{4\pi G \rho_c} \biggr] \, ,</math>

we recognize that, specifically for the case of <math>~n=5</math>, we can make the substitution, <math>~(s^')^2 \rightarrow -\sigma_c^2</math>, in which case the LAWE becomes,

<math>~0</math>

<math>~=</math>

<math>~\frac{1}{(3+\xi^2)} \biggl\{ (3+\xi^2)\frac{d^2 X}{d\xi^2} + \biggl[ \frac{2(6-\xi^2)}{\xi}\biggr]\frac{d X}{d\xi} + \bigg[ \frac{5\sigma_c^2}{2 \cdot 3^{3 / 2}} \cdot (3 + \xi^2)^{3 / 2} + 2 \biggr] X \biggr\} \, , </math>

which matches exactly the form of the LAWE derived above, if in that expression, <math>~\gamma_g</math> is also forced to align with our specification of the polytropic index, that is, if <math>~\gamma_g = (n+1)/n = 6/5</math> and, in turn, <math>~\alpha = (3-4/\gamma) = -1/3</math>.

New Independent Variable

Guided by our conjecture regarding the proper shape of the radial eigenfunction, let's switch the dependent variable to,

<math>~u \equiv 1 + \frac{3}{\xi^2}</math>

        <math>~\Rightarrow</math>        

<math>~3 + \xi^2 = \frac{3u}{(u-1)} \, ,</math>

        and        

<math>~\xi = 3^{1 / 2} (u-1)^{-1 / 2} \, .</math>

This implies that,

<math>~\frac{d}{d\xi}</math>

<math>~~~\rightarrow ~~~</math>

<math>~-\frac{2}{\sqrt{3}}(u-1)^{3 / 2} \frac{d}{du} \, ,</math>

and,

<math>~\frac{d^2}{d\xi^2}</math>

<math>~~~\rightarrow ~~~</math>

<math>~\frac{4}{3}(u-1)^3 \frac{d^2}{du^2} + 2(u-1)^{2} \frac{d}{du} \, .</math>

Hence, the governing wave equation becomes,

<math>~0</math>

<math>~=</math>

<math>~( 3 + \xi^2 )\frac{d^2x}{d\xi^2} + \biggl[ \frac{2(6 - \xi^2) }{ \xi} \biggr] \frac{dx}{d\xi} + \biggl[5\sigma^2 ( 3 + \xi^2 )^{3 / 2} + 2 \biggr] x </math>

 

<math>~=</math>

<math>~\frac{3u}{(u-1)} \biggl[\frac{4}{3}(u-1)^3 \frac{d^2x}{du^2} + 2(u-1)^{2} \frac{dx}{du}\biggr] + 4(2u-3)(u-1)\frac{dx}{du} + \biggl\{ 5\sigma^2 \biggl[ \frac{3u}{(u-1)} \biggr]^{3 / 2} + 2 \biggr\} x </math>

 

<math>~=</math>

<math>~4u(u-1)^2 \frac{d^2x}{du^2} + (14u-12)(u-1)\frac{dx}{du} + \biggl\{ 5\sigma^2 \biggl[ \frac{3u}{(u-1)} \biggr]^{3 / 2} + 2 \biggr\} x \, . </math>


Work-in-progress.png

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If we assume that <math>~\sigma^2 = 0</math>, then the governing relation is,

<math>~0</math>

<math>~=</math>

<math>~4u(u-1)^2 \frac{d^2x}{du^2} + (14u-12)(u-1)\frac{dx}{du} + 2 x \, . </math>

Now, again, guided by our conjecture, let's guess an eigenfunction of the form:

First Guess (n5)

<math>~x</math>

<math>~=</math>

<math>~ A^3 (u - 1)^{1 / 2} (A u - 1 )^{-1 / 2} \, , </math>

in which case,

<math>~\frac{dx}{du}</math>

<math>~=</math>

<math>~ \frac{A^3}{2} \biggl[ (u - 1)^{-1 / 2} (A u - 1 )^{-1 / 2} - A(u - 1)^{1 / 2} (A u - 1 )^{-3 / 2} \biggr] </math>

 

<math>~=</math>

<math>~ \biggl[ \frac{A^3(A-1)}{2} \biggr] (u-1)^{-1 / 2} (Au-1)^{-3 / 2} \, ; </math>

<math>~\frac{d^2x}{du^2}</math>

<math>~=</math>

<math>~ \biggl[ \frac{A^3(A-1)}{2} \biggr] \biggl\{ -\frac{1}{2}(u-1)^{-3 / 2} (Au-1)^{-3 / 2} -\frac{3A}{2} (u-1)^{-1 / 2} (Au-1)^{-5 / 2} \biggr\} </math>

 

<math>~=</math>

<math>~ -\frac{1}{2} \biggl[ \frac{A^3(A-1)}{2} \biggr] (u-1)^{-3 / 2} (Au-1)^{-5 / 2}\biggl[ (Au-1) +3A (u-1)\biggr] </math>

 

<math>~=</math>

<math>~ \biggl[ \frac{A^3(A-1)}{4} \biggr] (u-1)^{-3 / 2} (Au-1)^{-5 / 2}\biggl[(3A+1) - 4Au \biggr] \, . </math>


So the governing relation becomes:

<math>~0</math>

<math>~=</math>

<math>~4u(u-1)^2 \biggl\{ \biggl[ \frac{A^3(A-1)}{4} \biggr] (u-1)^{-3 / 2} (Au-1)^{-5 / 2}\biggl[(3A+1) - 4Au \biggr] \biggr\} </math>

 

 

<math>~ + (14u-12)(u-1) \biggl\{ \biggl[ \frac{A^3(A-1)}{2} \biggr] (u-1)^{-1 / 2} (Au-1)^{-3 / 2} \biggr\} + 2 A^3 (u - 1)^{1 / 2} (A u - 1 )^{-1 / 2} </math>

 

<math>~=</math>

<math>~u(u-1)^{1 / 2} A^3(A-1) (Au-1)^{-5 / 2}\biggl[(3A+1) - 4Au \biggr] </math>

 

 

<math>~ + (7u-6)(u-1)^{1 / 2} A^3(A-1) (Au-1)^{-3 / 2} + 2 A^3 (u - 1)^{1 / 2} (A u - 1 )^{-1 / 2} </math>

 

<math>~=</math>

<math>~(u-1)^{1 / 2} \biggl\{ uA^3(A-1) (Au-1)^{-5 / 2}\biggl[(3A+1) - 4Au \biggr] + (7u-6) A^3(A-1) (Au-1)^{-3 / 2} + 2 A^3 (A u - 1 )^{-1 / 2} \biggr\} </math>

 

<math>~=</math>

<math>~A^3(u-1)^{1 / 2} (Au-1)^{-5 / 2} \biggl\{ u(A-1) \biggl[(3A+1) - 4Au \biggr] + (7u-6) (A-1) (Au-1) + 2 (A u - 1 )^{2} \biggr\} </math>

 

<math>~=</math>

<math>~A^3(u-1)^{1 / 2} (Au-1)^{-5 / 2} \biggl\{ - 4u^2 A(A-1) + u(A-1) (3A+1) + (7u-6) [A(A-1)u +1 - A] + 2 (A^2u^2 - 2Au +1) \biggr\} </math>

 

<math>~=</math>

<math>~A^3(u-1)^{1 / 2} (Au-1)^{-5 / 2} \biggl\{ u^2 \biggl[ - 4A(A-1) +7A(A-1) +2A^2 \biggr] + u\biggl[ (A-1) (3A+1) - 7(A-1) -6A(A-1) - 4A \biggr] + 2(3A-2) \biggr\} </math>

 

<math>~=</math>

<math>~A^3(u-1)^{1 / 2} (Au-1)^{-5 / 2} \biggl\{ Au^2 \biggl[ 5A-3 \biggr] + u\biggl[ 3A^2-2A-1-7A+7 -6A^2+6A -4A \biggr] + 2(3A-2) \biggr\} \, . </math>

 

<math>~=</math>

<math>~A^3(u-1)^{1 / 2} (Au-1)^{-5 / 2} \biggl\{ Au^2 \biggl[ 5A-3 \biggr] + u\biggl[ -3A^2 -7A +6\biggr] + 2(3A-2) \biggr\} \, . </math>

Second Guess (n5)

<math>~x</math>

<math>~=</math>

<math>~ (u - 1)^{b / 2} (A u - 1 )^{-a / 2} \, , </math>

in which case,

<math>~\frac{dx}{du}</math>

<math>~=</math>

<math>~ \frac{b}{2}(u-1)^{b/2-1} (A u - 1 )^{-a / 2} - \frac{aA}{2}(u - 1)^{b / 2} (A u - 1 )^{-a / 2-1} </math>

 

<math>~=</math>

<math>~x \biggl[ \frac{b}{2}(u-1)^{-1} - \frac{aA}{2} (A u - 1 )^{-1} \biggr] </math>

<math>~\Rightarrow ~~~ \frac{(u-1)}{x} \frac{dx}{du}</math>

<math>~=</math>

<math>~ (A u - 1 )^{-1} \biggl[ \frac{b}{2} (A u - 1 ) - \frac{aA}{2} (u-1) \biggr] </math>

 

<math>~=</math>

<math>~\frac{1 }{2(A u - 1 )} \biggl[ b (A u - 1 ) - aA (u-1) \biggr] </math>

 

<math>~=</math>

<math>~ \frac{1 }{2(A u - 1 )} \biggl[ (aA - b) + A(b - a)u \biggr] \, ; </math>

and,

<math>~\frac{d^2x}{du^2}</math>

<math>~=</math>

<math>~ \biggl[ \frac{b}{2}(u-1)^{-1} - \frac{aA}{2} (A u - 1 )^{-1} \biggr]\frac{dx}{du} + x \frac{d}{du}\biggl[ \frac{b}{2}(u-1)^{-1} - \frac{aA}{2} (A u - 1 )^{-1} \biggr] </math>

 

<math>~=</math>

<math>~ x\biggl[ \frac{b}{2}(u-1)^{-1} - \frac{aA}{2} (A u - 1 )^{-1} \biggr]^2 + x \biggl[ -\frac{b}{2}(u-1)^{-2} + \frac{aA^2}{2} (A u - 1 )^{-2} \biggr] </math>

 

<math>~=</math>

<math>~ \frac{x}{4(u-1)^2 (Au-1)^2} \biggl\{ \biggl[ b(Au-1) - aA (u - 1 ) \biggr]^2 + \biggl[ 2aA^2 (u-1)^{2} -2b (A u - 1 )^{2} \biggr] \biggr\} </math>

<math>~\Rightarrow ~~~ \frac{(1-u)^2}{x}\frac{d^2x}{du^2}</math>

<math>~=</math>

<math>~ \frac{1}{4 (Au-1)^2} \biggl\{ \biggl[ b(Au-1) - aA (u - 1 ) \biggr]^2 + \biggl[ 2aA^2 (u-1)^{2} -2b (A u - 1 )^{2} \biggr] \biggr\} </math>

Hence, the governing wave equation becomes,

<math>~0</math>

<math>~=</math>

<math>~2u \biggl\{ \frac{(u-1)^2}{x} \frac{d^2x}{du^2} \biggr\} + (7u-6)\biggl\{ \frac{(u-1)}{x} \frac{dx}{du} \biggl\} + 1 </math>

 

<math>~=</math>

<math>~ \frac{2u}{4 (Au-1)^2} \biggl\{ \biggl[ (aA - b) + A(b - a)u \biggr]^2 + \biggl[ 2aA^2 (u-1)^{2} -2b (A u - 1 )^{2} \biggr] \biggr\} </math>

 

 

<math>~ + \frac{(7u-6) }{2(A u - 1 )} \biggl[ (aA - b) + A(b - a)u \biggr]

+ 1 

</math>

 

<math>~=</math>

<math>~ \frac{1}{4 (Au-1)^2} \biggl\{ 2u\biggl[ (aA - b)^2 + 2(aA - b)A(b - a)u + A^2(b - a)^2u^2 \biggr] + 2u\biggl[ 2aA^2 (u^2 - 2u + 1) -2b (A^2 u^2 - 2Au + 1 ) \biggr] </math>

 

 

<math>~ + 2(A u - 1 )(7u-6) \biggl[ (aA - b) + A(b - a)u \biggr]

+ 4 (Au-1)^2 \biggr\}

</math>

 

<math>~=</math>

<math>~ \frac{1}{4 (Au-1)^2} \biggl\{ 2u\biggl[ (aA - b)^2 + 2(aA - b)A(b - a)u + A^2(b - a)^2u^2 \biggr] + 2u\biggl[ 2A^2(a-b)u^2 + 4A(b - aA) u + 2(aA^2 -b) \biggr] </math>

 

 

<math>~ + 2\biggl[7Au^2 - (6A+7)u +6 \biggr]\biggl[ (aA - b) + A(b - a)u \biggr]

+  (4A^2u^2-8Au + 4) \biggr\}

</math>

If <math>~b=a</math>,

<math>~0</math>

<math>~=</math>

<math>~ 2u\biggl[ (aA - b)^2 \biggr] + 2u\biggl[ 4A(b - aA) u + 2(aA^2 -b) \biggr] </math>

 

 

<math>~ + 2\biggl[7Au^2 - (6A+7)u +6 \biggr]\biggl[ (aA - b) \biggr] + (4A^2u^2-8Au + 4) </math>

 

<math>~=</math>

<math>~ 2a^2u (A - 1)^2 + 2au [ 4A(1 - A) u + 2(A^2 -1) ] </math>

 

 

<math>~ + 2a(A - 1) \biggl[7Au^2 - (6A+7)u +6 \biggr] + (4A^2u^2-8Au + 4) </math>

 

<math>~=</math>

<math>~ 2Au^2 [4a (1 - A) + 7a(A - 1) + 2A] + 2u [ a^2 (A - 1)^2 + 2a(A^2 -1) - a(A - 1) (6A+7) - 4A] + 4[ 3a(A-1) + 1] </math>

This should then match the "first guess" algebraic condition if we set <math>~a=1</math>. Let's see.

<math>~0</math>

<math>~=</math>

<math>~ 2Au^2 [4 (1 - A) + 7(A - 1) + 2A] + 2u [ (A - 1)^2 + 2(A^2 -1) - (A - 1) (6A+7) - 4A] + 4[ 3(A-1) + 1] </math>

 

<math>~=</math>

<math>~ 2Au^2 [4 - 4A + 7A - 7 + 2A] + 2u [ (A^2 - 2A + 1) + 2A^2 -2 + (1-A ) (6A+7) -4A] + 4[ 3A-2] </math>

 

<math>~=</math>

<math>~ 2Au^2 [5A - 3] + 2u [ - 3A^2 - 7A + 6 ] + 4[ 3A-2] \, . </math>

And we see that this expression does match the one derived earlier.

Going back a bit, before setting <math>~a=1</math>, we have the expression:


<math>~0</math>

<math>~=</math>

<math>~ 2Au^2 [4a (1 - A) + 7a(A - 1) + 2A] + 2u [ a^2 (A - 1)^2 + 2a(A^2 -1) - a(A - 1) (6A+7) - 4A] + 4[ 3a(A-1) + 1] </math>

 

<math>~=</math>

<math>~ 2Au^2 [ 3aA -3a + 2A] + 2u [ a^2 (A - 1)^2 + 2a(A^2 -1) - a(6A^2+A-7) - 4A] + 4[ 3a(A-1) + 1] </math>

 

<math>~=</math>

<math>~ 2Au^2 [ 3a(A - 1) + 2A] + 2u [ a^2 (A - 1)^2 + a( -4A^2-A+5) - 4A] + 4[ 3a(A-1) + 1] \, . </math>

Now, in order for all three expressions inside the square-bracket pairs to be zero, we need, first,

<math>~3a(A - 1) + 2A</math>

<math>~=</math>

<math>~0</math>

<math>~\Rightarrow ~~~ a</math>

<math>~=</math>

<math>~\frac{2A}{3(1-A)} \, ;</math>

and, third, by simple visual comparison with the first expression,

<math>~3a(A-1) + 1</math>

<math>~=</math>

<math>~3a(A-1) + 2A</math>

<math>~\Rightarrow A</math>

<math>~=</math>

<math>~\frac{1}{2} </math>

<math>~\Rightarrow ~~~ a</math>

<math>~=</math>

<math>~\frac{2}{3} \, ;</math>

which forces the second expression to the value,

<math>~a^2 (A - 1)^2 + a( -4A^2-A+5) - 4A</math>

<math>~=</math>

<math>~\biggl(\frac{2}{3}\biggr)^2 \biggl(-\frac{1}{2} \biggr)^2 + \frac{2}{3}\biggl[ -1-\frac{1}{2} +5 \biggr] - 2</math>

 

<math>~=</math>

<math>~\frac{1}{9} + \frac{7}{3} - 2</math>

 

<math>~=</math>

<math>~\frac{4}{9} \, ,</math>

which is not zero. Hence our pair of unknown parameters — <math>~a </math> and <math>~A</math> — do not simultaneously satisfy all three conditions. (Not really a surprise.)

Setup Using Lagrangian Mass Coordinate

Alternative Terms

Let's change the independent coordinate from <math>~r_0</math> to <math>~m_0</math>. In particular, the derivative operation will change as follows:

<math>~\frac{d}{dr_0}</math>

<math>~~\rightarrow~~</math>

<math>~\biggl( \frac{dm_0}{dr_0} \biggr)\frac{d}{dm_0} = \biggl( \frac{dm_0}{d\xi} \cdot \frac{d\xi}{dr_0} \biggr)\frac{d}{dm_0} \, ,</math>

so what is the expression for the leading coefficient? From above, we have,

<math>~r_0</math>

<math>~=</math>

<math>~ R_* \biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3} \xi </math>

<math>~\Rightarrow ~~~ \xi</math>

<math>~=</math>

<math>~ \frac{1}{R_*} \biggl[ \frac{ {\tilde\xi}^2}{(3 + {\tilde\xi}^2)}\biggr]^{3} r_0 \, . </math>

Also, from above, we know that,

<math>~m_0</math>

<math>~=</math>

<math>~ M_\mathrm{tot} \biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3 / 2} \biggl\{ \xi^3 ( 3 + \xi^2 )^{-3/2} \biggr\} </math>

<math>~\Rightarrow ~~~ \frac{dm_0}{d\xi}</math>

<math>~=</math>

<math>~ M_\mathrm{tot} \biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3 / 2} \biggl\{ 3\xi^2 ( 3 + \xi^2 )^{-3/2} - 3 \xi^4 ( 3 + \xi^2 )^{-5/2}\biggr\} </math>

 

<math>~=</math>

<math>~ M_\mathrm{tot} \biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3 / 2} 3\xi^2 (3 + \xi^2)^{-5/2} \biggl\{ ( 3 + \xi^2 ) - \xi^2 \biggr\} </math>

 

<math>~=</math>

<math>~ M_\mathrm{tot} \biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3 / 2} 3^2\xi^2 (3 + \xi^2)^{-5/2} </math>

<math>~\Rightarrow ~~~ \frac{dm_0}{dr_0}</math>

<math>~=</math>

<math>~ M_\mathrm{tot} \biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3 / 2} 3^2\xi^2 (3 + \xi^2)^{-5/2} \frac{1}{R_*} \biggl[ \frac{ {\tilde\xi}^2}{(3 + {\tilde\xi}^2)}\biggr]^{3} </math>

 

<math>~=</math>

<math>~ \frac{M_\mathrm{tot} }{R_*} \biggl[ \frac{ {\tilde\xi}^2}{(3 + {\tilde\xi}^2)}\biggr]^{3 / 2} 3^2\xi^2 (3 + \xi^2)^{-5/2} \, . </math>

To simplify expressions, let's borrow from an accompanying derivation and define,

<math>\tilde{C} \equiv \frac{3^2}{{\tilde\xi}^2} \biggl( 1 + \frac{ {\tilde\xi}^2}{3} \biggr) = 3 \biggl[ \frac{( 3 + {\tilde\xi}^2 )}{ {\tilde\xi}^2} \biggr] \, .</math>

Then we have,

<math>~\frac{m_0}{M_\mathrm{tot}}</math>

<math>~=</math>

<math>~ \biggl[ \frac{\tilde{C}}{ 3}\biggr]^{3 / 2} \biggl[ \frac{\xi^2}{ ( 3 + \xi^2 )} \biggr]^{3/2} </math>

<math>~\Rightarrow ~~~\biggl[ \frac{ 3}{\tilde{C}}\biggr] \biggl[\frac{m_0}{M_\mathrm{tot}}\biggr]^{2 / 3}</math>

<math>~=</math>

<math>~ \frac{\xi^2}{ ( 3 + \xi^2 )} </math>

<math>~\Rightarrow ~~~( 3 + \xi^2 )\biggl[ \frac{ 3}{\tilde{C}}\biggr] \biggl[\frac{m_0}{M_\mathrm{tot}}\biggr]^{2 / 3}</math>

<math>~=</math>

<math>~ \xi^2 </math>

<math>~\Rightarrow ~~~3 m_*</math>

<math>~=</math>

<math>~ \xi^2 (1-m_*) </math>

<math>~\Rightarrow ~~~\xi^2 </math>

<math>~=</math>

<math>~ \frac{3m_*}{(1-m_*)} \, , </math>

where,

<math>~m_* \equiv \biggl[ \frac{ 3}{\tilde{C}}\biggr] \biggl[\frac{m_0}{M_\mathrm{tot}}\biggr]^{2 / 3} \, .</math>

In summary:

<math>~ \frac{\xi^2}{ ( 3 + \xi^2 )} = m_* \, ; </math>

      while,      

<math>~ \frac{ {\tilde\xi}^2}{ ( 3 + {\tilde\xi}^2 )} = \frac{3}{\tilde{C}} \, ; </math>

<math>~r_0</math>

<math>~=</math>

<math>~ R_* \biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3} \xi = R_* \biggl( \frac{ \tilde{C} }{ 3}\biggr)^{3} \biggr[ \frac{3m_*}{ (1-m_*) }\biggr]^{1 / 2} \, ; </math>

<math>~\frac{g_0\rho_0}{P_0} </math>

<math>~=</math>

<math>~ \frac{6}{R_*} \biggl[ \frac{ {\tilde\xi}^2 }{ (3 + {\tilde\xi}^2) }\biggr]^{9} \frac{\xi}{ ( 3 + \xi^2 )} = \frac{6}{R_*} \biggl[ \frac{ 3 }{ \tilde{C} }\biggr]^{9} \frac{m_*}{ \xi } = \frac{6}{R_*} \biggl[ \frac{ 3 }{ \tilde{C} }\biggr]^{9} m_* \biggl[ \frac{(1-m_*)}{3m_*} \biggr]^{1 / 2} \, ; </math>

<math>~\frac{g_0 }{r_0} </math>

<math>~=</math>

<math>~ \frac{GM_\mathrm{tot}}{R_*^3} \biggl[ \frac{ {\tilde\xi}^2 }{ (3 + {\tilde\xi}^2)}\biggr]^{15/2} \frac{1}{\xi^3} \biggl[ \frac{ \xi^2 }{ ( 3 + \xi^2 ) }\biggr]^{3/2} = \frac{GM_\mathrm{tot}}{R_*^3} \biggl[ \frac{3 }{ \tilde{C} }\biggr]^{15/2} (1-m_*)^{3 / 2} \, ; </math>

<math>~\frac{\rho_0}{\gamma_g P_0} </math>

<math>~=</math>

<math>~ \frac{6R_* }{\gamma_g GM_\mathrm{tot} }\biggl( \frac{ 3}{ \tilde{C} } \biggr)^{9 / 2} \biggl[ \frac{3}{(1-m_*)}\biggr]^{1 / 2} \, . </math>

So, the wave equation may be written as,

<math>~0</math>

<math>~=</math>

<math>~ \frac{d^2x}{dr_0^2} + \biggl[\frac{4}{r_0} - \biggl(\frac{g_0 \rho_0}{P_0}\biggr) \biggr] \frac{dx}{dr_0} + \biggl(\frac{\rho_0}{\gamma_\mathrm{g} P_0} \biggr)\biggl[\omega^2 + (4 - 3\gamma_\mathrm{g})\frac{g_0}{r_0} \biggr] x </math>

 

<math>~=</math>

<math>~ \frac{d^2x}{dr_0^2} + \biggl\{ \frac{4}{R_*} \biggl( \frac{ 3}{ \tilde{C} }\biggr)^{3} \biggr[ \frac{ (1-m_*) }{3m_*}\biggr]^{1 / 2} - \frac{6}{R_*} \biggl[ \frac{ 3 }{ \tilde{C} }\biggr]^{9} m_* \biggl[ \frac{(1-m_*)}{3m_*} \biggr]^{1 / 2} \biggr\} \frac{dx}{dr_0} </math>

 

 

<math>~ + \frac{6R_* }{\gamma_g GM_\mathrm{tot} }\biggl( \frac{ 3}{ \tilde{C} } \biggr)^{9 / 2} \biggl[ \frac{3}{(1-m_*)}\biggr]^{1 / 2} \biggl\{ \omega^2 + (4 - 3\gamma_\mathrm{g})\frac{GM_\mathrm{tot}}{R_*^3} \biggl[ \frac{3 }{ \tilde{C} }\biggr]^{15/2} (1-m_*)^{3 / 2} \biggr\} x </math>

 

<math>~=</math>

<math>~ \frac{d^2x}{dr_0^2} + \frac{1}{R_*} \biggl( \frac{ 3}{ \tilde{C} }\biggr)^{3} \biggl\{ 4 - 6\biggl[ \frac{ 3 }{ \tilde{C} }\biggr]^{6} m_* \biggr\} \biggr[ \frac{ (1-m_*) }{3m_*}\biggr]^{1 / 2}\frac{dx}{dr_0} </math>

 

 

<math>~ + \frac{6(4 - 3\gamma_\mathrm{g}) }{\gamma_g } \cdot \frac{1 }{R_*^2} \biggl( \frac{3 }{ \tilde{C} }\biggr)^{3} \biggl[ \frac{3}{(1-m_*)}\biggr]^{1 / 2} \biggl\{\sigma^2 + (1-m_*)^{3 / 2} \biggr\} x </math>

 

<math>~=</math>

<math>~\frac{1 }{R_*^2} \biggl( \frac{3 }{ \tilde{C} }\biggr)^{3} \biggl\{ R_*^2 \biggl( \frac{ \tilde{C} }{3 }\biggr)^{3} \frac{d^2x}{dr_0^2} + R_* \biggl[ 4 - 6\biggl( \frac{ 3 }{ \tilde{C} }\biggr)^{6} m_* \biggr] \biggr[ \frac{ (1-m_*) }{3m_*}\biggr]^{1 / 2}\frac{dx}{dr_0} </math>

 

 

<math>~ + \frac{6(4 - 3\gamma_\mathrm{g}) }{\gamma_g } \cdot \biggl[ \frac{3}{(1-m_*)}\biggr]^{1 / 2} \biggl[ \sigma^2 + (1-m_*)^{3 / 2} \biggr] x \biggr\} </math>

 

<math>~=</math>

<math>~\frac{1 }{R_*^2} \biggl( \frac{3 }{ \tilde{C} }\biggr)^{3} \biggl[ \frac{1}{3m_*(1-m_*)}\biggr]^{1 / 2} \biggl\{ [ 3m_*(1-m_*) ]^{1 / 2} R_*^2 \biggl( \frac{ \tilde{C} }{3 }\biggr)^{3} \frac{d^2x}{dr_0^2} + R_* \biggl[ 4 - 6\biggl( \frac{ 3 }{ \tilde{C} }\biggr)^{6} m_* \biggr] (1-m_*) \frac{dx}{dr_0} </math>

 

 

<math>~ + \frac{18(4 - 3\gamma_\mathrm{g}) }{\gamma_g } \cdot m_*^{1 / 2} \biggl[ \sigma^2 + (1-m_*)^{3 / 2} \biggr] x \biggr\} \, , </math>

where,

<math>~\sigma^2 \equiv (4 - 3\gamma_\mathrm{g})^{-1} \frac{R_*^3}{GM_\mathrm{tot}} \biggl[ \frac{ \tilde{C} }{3 } \biggr]^{15/2} \omega^2 \, .</math>

Now, let's look at the differential operators, after defining.

<math>~c_0 \equiv 3^{1 / 2} R_* \biggl( \frac{ \tilde{C} }{ 3}\biggr)^{3} ~~~~\Rightarrow ~~~~R_* = c_0 3^{-1 / 2} \biggl( \frac{ \tilde{C} }{ 3}\biggr)^{-3} \, .</math>

We find,

<math>~dr_0</math>

<math>~=</math>

<math>~ c_0 ~d[ m_*^{1 / 2} (1-m_*)^{-1 / 2} ] </math>

 

<math>~=</math>

<math>~ c_0 ~\biggl[\frac{1}{2} ~m_*^{-1 / 2}( 1 - m_*)^{-1 / 2} + \frac{1}{2} ~m_*^{1 / 2} (1 - m_*)^{-3 / 2} \biggr] dm_* </math>

 

<math>~=</math>

<math>~ \frac{c_0}{2} ~m_*^{-1 / 2}( 1 - m_*)^{-3 / 2}~ dm_* </math>

<math>~\frac{d}{dr_0}</math>

<math>~=</math>

<math>~ \frac{2}{c_0} ~m_*^{1 / 2}( 1 - m_*)^{3 / 2}~ \frac{d}{dm_*} </math>

<math>~\Rightarrow ~~~ R_*\frac{dx}{dr_0}</math>

<math>~=</math>

<math>~ \frac{2}{3^{1 / 2}}\biggl( \frac{ \tilde{C} }{ 3}\biggr)^{-3} ~m_*^{1 / 2}( 1 - m_*)^{3 / 2}~ \frac{dx}{dm_*} \, . </math>

Also,

<math>~\frac{d^2}{dr_0^2}</math>

<math>~=</math>

<math>~ \biggl( \frac{2}{c_0} \biggr)^{2}~m_*^{1 / 2}( 1 - m_*)^{3 / 2}~ \frac{d}{dm_*} \biggl[ m_*^{1 / 2}( 1 - m_*)^{3 / 2}~ \frac{d}{dm_*} \biggr] </math>

 

<math>~=</math>

<math>~ \biggl( \frac{2}{c_0} \biggr)^{2}~m_* ( 1 - m_*)^{3 }~ \frac{d^2}{dm_*^2} +\biggl( \frac{2}{c_0} \biggr)^{2}~m_*^{1 / 2}( 1 - m_*)^{3 / 2} \biggl[ \frac{1}{2} m_*^{-1 / 2}( 1 - m_*)^{3 / 2} - \frac{3}{2} m_*^{1 / 2}( 1 - m_*)^{1 / 2}~ \biggr] ~ \frac{d}{dm_*} </math>

 

<math>~=</math>

<math>~ \biggl( \frac{2}{c_0} \biggr)^{2}~m_* ( 1 - m_*)^{3 }~ \frac{d^2}{dm_*^2} +\frac{1}{2} \biggl( \frac{2}{c_0} \biggr)^{2}~ ( 1 - m_*)^{2} ( 1 - 4m_*) \frac{d}{dm_*} </math>

<math>~\Rightarrow ~~~ R_*^2 \biggl( \frac{ \tilde{C} }{3 }\biggr)^{3} \frac{d^2x}{dr_0^2}</math>

<math>~=</math>

<math>~ \biggl[\frac{2^2}{3} \biggl(\frac{ \tilde{C} }{3} \biggr)^{-3} \biggr] \biggl[ ~m_* ( 1 - m_*)^{3 }~ \frac{d^2x}{dm_*^2} +\frac{1}{2} ~ ( 1 - m_*)^{2} ( 1 - 4m_*) \frac{dx}{dm_*} \biggr] </math>

So, the wave equation becomes,

<math>~0</math>

<math>~=</math>

<math>~\frac{1 }{R_*^2} \biggl( \frac{3 }{ \tilde{C} }\biggr)^{3} \biggl[ \frac{1}{3m_*(1-m_*)}\biggr]^{1 / 2} \biggl\{ [ 3m_*(1-m_*) ]^{1 / 2} \biggl[\frac{2^2}{3} \biggl(\frac{ \tilde{C} }{3} \biggr)^{-3} \biggr] \biggl[ ~m_* ( 1 - m_*)^{3 }~ \frac{d^2x}{dm_*^2} +\frac{1}{2} ~ ( 1 - m_*)^{2} ( 1 - 4m_*) \frac{dx}{dm_*} \biggr] </math>

 

 

<math>~ + \biggl[ 4 - 6\biggl( \frac{ 3 }{ \tilde{C} }\biggr)^{6} m_* \biggr] (1-m_*) \biggl[ \frac{2}{3^{1 / 2}}\biggl( \frac{ \tilde{C} }{ 3}\biggr)^{-3} ~m_*^{1 / 2}( 1 - m_*)^{3 / 2}~ \frac{dx}{dm_*} \biggr] + \frac{18(4 - 3\gamma_\mathrm{g}) }{\gamma_g } \cdot m_*^{1 / 2} \biggl[ \sigma^2 + (1-m_*)^{3 / 2} \biggr] x \biggr\} </math>

 

<math>~=</math>

<math>~\frac{1 }{R_*^2} \biggl( \frac{3 }{ \tilde{C} }\biggr)^{6} \biggl[ \frac{1}{3m_*(1-m_*)}\biggr]^{1 / 2} \biggl\{ [ 3m_*(1-m_*) ]^{1 / 2} \biggl[\frac{2^2}{3} \biggr] \biggl[ ~m_* ( 1 - m_*)^{3 }~ \frac{d^2x}{dm_*^2} +\frac{1}{2} ~ ( 1 - m_*)^{2} ( 1 - 4m_*) \frac{dx}{dm_*} \biggr] </math>

 

 

<math>~ + \biggl[ 4 - 6\biggl( \frac{ 3 }{ \tilde{C} }\biggr)^{6} m_* \biggr] (1-m_*) \biggl[ \frac{2}{3^{1 / 2}} ~m_*^{1 / 2}( 1 - m_*)^{3 / 2}~ \frac{dx}{dm_*} \biggr] + \frac{18(4 - 3\gamma_\mathrm{g}) }{\gamma_g } \biggl( \frac{ \tilde{C} }{ 3}\biggr)^{3} m_*^{1 / 2} \biggl[ \sigma^2 + (1-m_*)^{3 / 2} \biggr] x \biggr\} </math>

 

<math>~=</math>

<math>~\frac{2 }{3R_*^2} \biggl( \frac{3 }{ \tilde{C} }\biggr)^{6} \biggl\{ 2m_* ( 1 - m_*)^{3 }~ \frac{d^2x}{dm_*^2} + ( 1 - m_*)^{2} ( 1 - 4m_*) \frac{dx}{dm_*} </math>

 

 

<math>~ + \biggl[ 4 - 6\biggl( \frac{ 3 }{ \tilde{C} }\biggr)^{6} m_* \biggr] (1-m_*)^2 \frac{dx}{dm_*} + \frac{9(4 - 3\gamma_\mathrm{g}) }{\gamma_g } \biggl( \frac{ \tilde{C} }{ 3}\biggr)^{3} \biggl[ \frac{3}{(1-m_*)}\biggr]^{1 / 2}\biggl[ \sigma^2 + (1-m_*)^{3 / 2} \biggr] x \biggr\} </math>

 

<math>~=</math>

<math>~\frac{2 }{3R_*^2} \biggl( \frac{3 }{ \tilde{C} }\biggr)^{6} \biggl\{ 2m_* ( 1 - m_*)^{3 }~ \frac{d^2x}{dm_*^2} </math>

 

 

<math>~ + \biggl[ 5 - 4m_* - 6\biggl( \frac{ 3 }{ \tilde{C} }\biggr)^{6} m_* \biggr] (1-m_*)^2 \frac{dx}{dm_*} + \frac{9(4 - 3\gamma_\mathrm{g}) }{\gamma_g } \biggl( \frac{ \tilde{C} }{ 3}\biggr)^{3} \biggl[ \frac{3}{(1-m_*)}\biggr]^{1 / 2}\biggl[ \sigma^2 + (1-m_*)^{3 / 2} \biggr] x \biggr\} </math>

 

<math>~=</math>

<math>~\frac{2 }{3R_*^2} \biggl( \frac{3 }{ \tilde{C} }\biggr)^{6} \biggl\{ 2m_* ( 1 - m_*)^{3 }~ \frac{d^2x}{dm_*^2} + (5 - \mathcal{A} m_*) (1-m_*)^2 \frac{dx}{dm_*} + \mathcal{B} \biggl[ \frac{\sigma^2}{(1-m_*)^{1 / 2}} + (1-m_*) \biggr] x \biggr\} \, , </math>

where,

<math>~\mathcal{A}</math>

<math>~\equiv</math>

<math>~4 + 6\biggl( \frac{ 3 }{ \tilde{C} }\biggr)^{6} \, ,</math>

<math>~\mathcal{B}</math>

<math>~\equiv</math>

<math>~\frac{3^{5/2}(4 - 3\gamma_\mathrm{g}) }{\gamma_g } \biggl( \frac{ \tilde{C} }{ 3}\biggr)^{3} \, .</math>

Try Again

This time, let's adopt the notation used in a related chapter in our Ramblings appendix. Specifically, the parametric relationship between <math>~m_\xi</math> and <math>~r_\xi</math> in pressure-truncated, <math>~n=5</math> polytropes is,

<math>~m_\xi \equiv \frac{m_0}{ M_\mathrm{tot} } = \frac{M_r(\xi)}{M_\mathrm{tot}}</math>

<math>~=</math>

<math>~ \biggl(\frac{\xi}{\tilde\xi}\biggr)^3 \biggl(3 + \xi^2 \biggr)^{-3/2} \biggl(3 + {\tilde\xi}^2 \biggr)^{3/2} </math>

 

<math>~=</math>

<math>~ \biggl[ \frac{( 3+\tilde\xi^2)}{ {\tilde\xi}^2} \biggr]^{3 / 2}\biggl[ \frac{( 3+\xi^2)}{ {\xi}^2} \biggr]^{- 3 / 2} \, ,</math>

<math>~r_\xi \equiv \frac{r_0}{R_\mathrm{norm}} = \biggl(\frac{\xi}{\tilde\xi} \biggr) \frac{R_\mathrm{eq}}{R_\mathrm{norm}}</math>

<math>~=</math>

<math>~\xi \biggl\{ \biggl[ \frac{4\pi}{2^5\cdot 3}\biggr]^{1/2} \tilde\xi^{-6} \biggl( 1+\frac{\tilde\xi^2}{3} \biggr)^{3}\biggr\} </math>

 

<math>~=</math>

<math>~ \biggl[ \frac{\pi}{2^3\cdot 3^7}\biggr]^{1/2} \biggl[ \frac{( 3+\tilde\xi^2)}{ {\tilde\xi}^2} \biggr]^{3} \xi \, . </math>

And we are in the fortunate situation of being able to eliminate <math>~\xi</math> to obtain the direct relation,

<math>~ r_\xi (m_\xi) </math>

<math>~=</math>

<math>~\tilde{r}_\mathrm{edge} \biggl[\frac{3^2m_\xi^{2/3}}{\tilde{C} - 3 m_\xi^{2/3}}\biggr]^{1/2} \, , </math>

where,

<math>~\tilde{C}</math>

<math>~\equiv</math>

<math>~ \frac{3^2}{\tilde\xi^2}\biggl( 1 + \frac{\tilde\xi^2}{3} \biggr) = 3 \biggl[ \frac{( 3+\tilde\xi^2)}{ {\tilde\xi}^2} \biggr] \, , </math>

<math>~\tilde{r}_\mathrm{edge}</math>

<math>~\equiv</math>

<math>~\biggl[ \frac{\pi}{2^3\cdot 3}\biggr]^{1/2} {\tilde\xi}^{-6} \biggl(1+\frac{\tilde\xi^2}{3}\biggr)^3 = \biggl[ \frac{\pi}{2^3\cdot 3^7}\biggr]^{1 / 2} \biggl[ \frac{\tilde{C}}{ 3} \biggr]^{3} \, . </math>

If we furthermore define,

<math>m_* \equiv \frac{3}{\tilde{C}} \cdot m_\xi^{2 / 3} \, ,</math>

then,

<math>~ r_\xi (m_*) </math>

<math>~=</math>

<math>~ 3^{1 / 2} \tilde{r}_\mathrm{edge} \biggl[\frac{m_*}{1-m_*}\biggr]^{1/2} \, . </math>

Hence,

<math>~ \frac{dr_0}{R_\mathrm{norm}} = dr_\xi </math>

<math>~=</math>

<math>~3^{1 / 2} \tilde{r}_\mathrm{edge} \biggl\{ \frac{1}{2} (1-m_*)^{- 1 / 2} m_*^{-1 / 2} + \frac{1}{2}m_*^{1 / 2}(1-m_*)^{-3 / 2} \biggr\} dm_* </math>

 

<math>~=</math>

<math>~ \biggl( \frac{3^{1 / 2}}{2} \biggr) \tilde{r}_\mathrm{edge} m_*^{-1 / 2} (1-m_*)^{-3 / 2} dm_* </math>

<math>\Rightarrow ~~~ R_\mathrm{norm} \cdot \frac{d}{dr_0} </math>

<math>~=</math>

<math>~ \frac{1}{ \tilde{r}_\mathrm{edge}} \biggl( \frac{2}{3^{1 / 2}} \biggr) m_*^{1 / 2} (1-m_*)^{3 / 2} \frac{d}{dm_*} \, . </math>

We therefore also have,

<math>~ R^2_\mathrm{norm} \cdot \frac{d^2}{dr_0^2} </math>

<math>~=</math>

<math>~ \frac{1}{ {\tilde{r}}^2_\mathrm{edge}} \biggl( \frac{2^2}{3} \biggr) m_*^{1 / 2} (1-m_*)^{3 / 2} \frac{d}{dm_*}\biggl[ m_*^{1 / 2} (1-m_*)^{3 / 2} \frac{d}{dm_*}\biggr] </math>

 

<math>~=</math>

<math>~ \frac{1}{ {\tilde{r}}^2_\mathrm{edge}} \biggl( \frac{2^2}{3} \biggr) m_*^{1 / 2} (1-m_*)^{3 / 2} \biggl\{ \biggl[ m_*^{1 / 2} (1-m_*)^{3 / 2} \frac{d^2}{dm_*^2}\biggr] + \biggl[ \frac{1}{2} m_*^{-1 / 2} (1-m_*)^{3 / 2} + \frac{3}{2}m_*^{1 / 2} (1-m_*)^{1 / 2}\biggr] \frac{d}{dm_*} \biggr\} </math>

 

<math>~=</math>

<math>~ \frac{1}{ {\tilde{r}}^2_\mathrm{edge}} \biggl( \frac{2}{3} \biggr) \biggl\{ \biggl[ 2m_* (1-m_*)^{3} \frac{d^2}{dm_*^2}\biggr] + \biggl[ (1-m_*)^{3 } + 3m_* (1-m_*)^{2}\biggr] \frac{d}{dm_*} \biggr\} </math>

 

<math>~=</math>

<math>~ \frac{1}{ {\tilde{r}}^2_\mathrm{edge}} \biggl( \frac{2}{3} \biggr) \biggl\{ 2m_* (1-m_*)^{3} \frac{d^2}{dm_*^2} + (1-m_*)^{2} ( 1 + 2m_* ) \frac{d}{dm_*} \biggr\} \, . </math>

So the wave equation may be written,

<math>~0</math>

<math>~=</math>

<math>~ R_\mathrm{norm}^2 \cdot \frac{d^2x}{dr_0^2} + \biggl[\frac{4R_\mathrm{norm}}{r_0} - \biggl(\frac{g_0 \rho_0 R_\mathrm{norm}}{P_0}\biggr) \biggr] R_\mathrm{norm} \cdot \frac{dx}{dr_0} + \biggl(\frac{\rho_0 R_\mathrm{norm}}{\gamma_\mathrm{g} P_0} \biggr)\biggl[R_\mathrm{norm} \omega^2 + (4 - 3\gamma_\mathrm{g})\frac{g_0 R_\mathrm{norm}}{r_0} \biggr] x </math>

 

<math>~=</math>

<math>~ \frac{1}{ {\tilde{r}}^2_\mathrm{edge}} \biggl( \frac{2}{3} \biggr) \biggl\{ 2m_* (1-m_*)^{3} \frac{d^2x}{dm_*^2} + (1-m_*)^{2} ( 1 + 2m_* ) \frac{dx}{dm_*} \biggr\} </math>

 

 

<math>~ +\frac{1}{ \tilde{r}_\mathrm{edge}} \biggl( \frac{2}{3^{1 / 2}} \biggr) \biggl\{ \frac{4}{r_\xi} - \biggl[\frac{6R_\mathrm{norm}}{R_*} \biggl( \frac{ 3 }{ \tilde{C} }\biggr)^{9} m_* \biggl[ \frac{(1-m_*)}{3m_*} \biggr]^{1 / 2} \biggr] \biggr\} m_*^{1 / 2} (1-m_*)^{3 / 2} \frac{dx}{dm_*} </math>

 

 

<math>~ + \frac{6R_* R_\mathrm{norm}}{\gamma_g GM_\mathrm{tot} }\biggl( \frac{ 3}{ \tilde{C} } \biggr)^{9 / 2} \biggl[ \frac{3}{(1-m_*)}\biggr]^{1 / 2} \biggl\{ R_\mathrm{norm} \omega^2 + (4 - 3\gamma_\mathrm{g}) \frac{GM_\mathrm{tot} R_\mathrm{norm}}{R_*^3} \biggl[ \frac{3 }{ \tilde{C} }\biggr]^{15/2} (1-m_*)^{3 / 2} \biggr\} x \, . </math>

Keeping in mind that,

<math>~\frac{R_*}{R_\mathrm{norm}} = \biggl[ \frac{\pi}{2^3 \cdot 3^7} \biggr]^{1 / 2} = {\tilde{r}}_\mathrm{edge} \biggl( \frac{3}{\tilde{C}} \biggr)^3 \, ,</math>

we therefore have,

<math>~0</math>

<math>~=</math>

<math>~ \frac{1}{ {\tilde{r}}^2_\mathrm{edge}} \biggl( \frac{2}{3} \biggr) \biggl\{ 2m_* (1-m_*)^{3} \frac{d^2x}{dm_*^2} + (1-m_*)^{2} ( 1 + 2m_* ) \frac{dx}{dm_*} \biggr\} </math>

 

 

<math>~ +\frac{1}{ \tilde{r}_\mathrm{edge}} \biggl( \frac{2}{3^{1 / 2}} \biggr) \biggl\{ 4 \biggl[3^{1 / 2} \tilde{r}_\mathrm{edge} \biggl[\frac{m_*}{1-m_*}\biggr]^{1/2} \biggr]^{-1} - 6 \biggl( \frac{ 3 }{ \tilde{C} }\biggr)^{9} \biggl[{\tilde{r}}_\mathrm{edge} \biggl( \frac{3}{\tilde{C}} \biggr)^3 \biggr]^{-1} m_* \biggl[ \frac{(1-m_*)}{3m_*} \biggr]^{1 / 2} \biggr\} m_*^{1 / 2} (1-m_*)^{3 / 2} \frac{dx}{dm_*} </math>

 

 

<math>~ + 6 \biggl( \frac{ 3}{ \tilde{C} } \biggr)^{9 / 2} \biggl[{\tilde{r}}_\mathrm{edge} \biggl( \frac{3}{\tilde{C}} \biggr)^3 \biggr]^{-2} \biggl[ \frac{3}{(1-m_*)}\biggr]^{1 / 2} \biggl\{ \biggl[ \frac{R_*^3}{\gamma_g GM_\mathrm{tot} } \biggr] \omega^2 + \frac{(4 - 3\gamma_\mathrm{g})}{\gamma_g} \biggl[ \frac{3 }{ \tilde{C} }\biggr]^{15/2} (1-m_*)^{3 / 2} \biggr\} x </math>

 

<math>~=</math>

<math>~ \frac{1}{ {\tilde{r}}^2_\mathrm{edge}} \biggl( \frac{2}{3} \biggr) \biggl\{ 2m_* (1-m_*)^{3} \frac{d^2x}{dm_*^2} + (1-m_*)^{2} ( 1 + 2m_* ) \frac{dx}{dm_*} \biggr\} </math>

 

 

<math>~ +\frac{1}{ \tilde{r}_\mathrm{edge}^2} \biggl( \frac{2^3}{3} \biggr) \biggl[ 1 - \frac{3}{2} \biggl( \frac{ 3 }{ \tilde{C} }\biggr)^{6} m_* \biggr] (1-m_*)^{2} \frac{dx}{dm_*} + \frac{6}{ {\tilde{r}}_\mathrm{edge}^2 } \biggl( \frac{3 }{ \tilde{C} }\biggr)^{6} \biggl[ \frac{3}{(1-m_*)}\biggr]^{1 / 2} \frac{(4 - 3\gamma_\mathrm{g})}{\gamma_g} \biggl[ \sigma^2 + (1-m_*)^{3 / 2} \biggr] x </math>

 

<math>~=</math>

<math>~ \frac{1}{ {\tilde{r}}^2_\mathrm{edge}} \biggl( \frac{2}{3} \biggr) \biggl\{ 2m_* (1-m_*)^{3} \frac{d^2x}{dm_*^2} + \biggl[ 5 - 6 \biggl( \frac{ 3 }{ \tilde{C} }\biggr)^{6} m_* + 2m_* \biggr] (1-m_*)^{2} \frac{dx}{dm_*} + 3^{5 / 2} \biggl( \frac{3 }{ \tilde{C} }\biggr)^{6} \frac{(4 - 3\gamma_\mathrm{g})}{\gamma_g} \biggl[ \frac{\sigma^2 }{(1-m_*)^{1 / 2}} + (1-m_*) \biggr] x \biggr\} \, , </math>

where, as before,

<math>\sigma^2 \equiv \biggl( \frac{ \tilde{C} }{3 } \biggr)^{15/2} \biggl[ \frac{R_*^3}{(4 - 3\gamma_g) GM_\mathrm{tot} } \biggr] \omega^2 \, .</math>

Take Another Approach Using Logarithmic Derivatives

Change Independent Variable

Returning to the LAWE for n = 3 polytropes, as given, above, and repeated here,

LAWE for <math>~n=5</math> Polytropes

<math>~0</math>

<math>~=</math>

<math>~(3+\xi^2) \frac{d^2x}{d\xi^2} + \biggl[12 - 2\xi^2 \biggr] \frac{1}{\xi} \cdot \frac{dx}{d\xi} + \biggl[\frac{\sigma_c^2}{3^{1/2} \gamma_g } \cdot (3+\xi^2)^{3/2} - 6\alpha \biggr] x </math>

let's make the substitution,

<math>~u \equiv (3 + \xi^2)^{1/2}</math>

      <math>~\Rightarrow</math>     

<math>~\xi^2 = u^2-3 \, .</math>

We must therefore also make the operator substitution,

<math>~\frac{d}{d\xi}</math>

<math>~=</math>

<math>~\frac{du}{d\xi} \cdot \frac{d}{du}</math>

 

<math>~=</math>

<math>~\biggl[ \xi (3+\xi^2)^{-1/2} \biggr] \frac{d}{du} = \biggl[ 1 - \frac{3}{u^2} \biggr]^{1/2} \frac{d}{du}</math>

<math>~\Rightarrow~~~ \frac{1}{\xi} \cdot \frac{dx}{d\xi}</math>

<math>~=</math>

<math>~\frac{1}{u}\cdot \frac{dx}{du} \, ;</math>

and,

<math>~\frac{d^2}{d\xi^2}</math>

<math>~=</math>

<math>~\biggl[ 1 - \frac{3}{u^2} \biggr]^{1/2} \frac{d}{du} \biggl\{ \biggl[ 1 - \frac{3}{u^2} \biggr]^{1/2} \frac{d}{du} \biggr\}</math>

 

<math>~=</math>

<math>~\biggl[ 1 - \frac{3}{u^2} \biggr]^{1/2} \biggl\{ \frac{3}{u^3} \biggl[ 1 - \frac{3}{u^2} \biggr]^{-1/2} \frac{d}{du} + \biggl[ 1 - \frac{3}{u^2} \biggr]^{1/2} \frac{d^2}{du^2}\biggr\}</math>

 

<math>~=</math>

<math>~ \frac{3}{u^3} \frac{d}{du} + \biggl[ 1 - \frac{3}{u^2} \biggr] \frac{d^2}{du^2}</math>

<math>~\Rightarrow ~~~ \frac{d^2x}{d\xi^2}</math>

<math>~=</math>

<math>~ \frac{3}{u^3} \frac{dx}{du} + \biggl[ 1 - \frac{3}{u^2} \biggr] \frac{d^2x}{du^2} \, .</math>

The rewritten LAWE is therefore,

<math>~0</math>

<math>~=</math>

<math>~u^2 \biggl\{ \frac{3}{u^3} \frac{dx}{du} + \biggl[ 1 - \frac{3}{u^2} \biggr] \frac{d^2x}{du^2} \biggr\} + 2\biggl[9 - u^2 \biggr] \frac{1}{u} \cdot \frac{dx}{du} + \biggl[\Omega^2 u^3 - 6\alpha \biggr] x </math>

 

<math>~=</math>

<math>~(u^2-3) \frac{d^2x}{du^2} + (21 - 2u^2 ) \frac{1}{u} \cdot \frac{dx}{du} + (\Omega^2 u^3 - 6\alpha ) x \, ,</math>

where we have adopted the shorthand notation,

<math>~\Omega^2 \equiv \frac{\sigma_c^2}{3^{1/2} \gamma_g } \, .</math>

Look at Logarithmic Derivative

Multiplying through by <math>~(u^2/x)</math> gives,

<math>~0</math>

<math>~=</math>

<math>~(u^2-3) \frac{u^2}{x} \cdot \frac{d^2x}{du^2} + (21 - 2u^2 ) \frac{d\ln x}{d\ln u} + (\Omega^2 u^5 - 6\alpha u^2 ) \, .</math>

Now, in the context of a separate derivation, we showed that, quite generally we can make the substitution,

<math>~\frac{u^2}{x} \cdot \frac{d^2x}{du^2} </math>

<math>~=</math>

<math>~ \frac{d}{d\ln u} \biggl[ \frac{d\ln x}{d\ln u} \biggr] + \biggl[ \frac{d\ln x}{d\ln u}-1 \biggr]\cdot \frac{d\ln x}{d\ln u} \, . </math>

Hence, if we assume that the displacement function can be expressed as a power-law in <math>~u</math>, such that,

<math>\frac{d\ln x}{d\ln u} = c_0 \, ,</math>

then the LAWE for <math>~n=5</math> polytropes simplifies as follows,

<math>~0</math>

<math>~=</math>

<math>~(u^2-3) c_0(c_0-1) + (21 - 2u^2 ) c_0 + (\Omega^2 u^5 - 6\alpha u^2 ) \, .</math>

This polynomial equation will be satisfied if, simultaneously, we set:

  • <math>\Omega^2 = 0 \, ;</math>
  • <math>c_0^2 -3c_0 -6\alpha = 0 </math>      <math>~\Rightarrow</math>      <math>c_0 = \frac{3}{2}\biggl[1 \pm \biggl(1+\frac{8\alpha}{3} \biggr)^{1/2} \biggl]\, ;</math>
  • <math>~\alpha = 20/3 \, .</math>

This gives us some hope that a more general solution of the following form will work:

<math>~x</math>

<math>~=</math>

<math>~u^{c_0} \biggl[ a + bu + cu^2 + du^3 + \cdots\biggr] \, .</math>

This means that, for example,

<math>~\frac{dx}{du}</math>

<math>~=</math>

<math>~ c_0 u^{c_0-1} \biggl[ a + bu + cu^2 + du^3 \biggr] + u^{c_0} \biggl[ b + 2cu + 3du^2 \biggr] </math>

<math>~\Rightarrow ~~~\frac{d\ln x}{d\ln u}</math>

<math>~=</math>

<math>~ \frac{c_0(a + bu + cu^2 + du^3) + bu + 2cu^2 + 3du^3}{a + bu + cu^2 + du^3} </math>

and,

<math>~\frac{d^2x}{du^2}</math>

<math>~=</math>

<math>~ c_0(c_0-1) u^{c_0-2} \biggl[ a + bu + cu^2 + du^3 \biggr] + 2c_0 u^{c_0-1} \biggl[ b + 2cu + 3du^2 \biggr] + u^{c_0} \biggl[ 2c + 6du \biggr] </math>

<math>~\Rightarrow~~~ \frac{u^2}{x} \cdot \frac{d^2x}{du^2}</math>

<math>~=</math>

<math>~ \frac{c_0(c_0-1) ( a + bu + cu^2 + du^3 ) + 2c_0 ( bu + 2cu^2 + 3du^3 ) + ( 2cu^2 + 6du^3 ) }{ a + bu + cu^2 + du^3} </math>

So the LAWE becomes,

<math>~- (\Omega^2 u^5 - 6\alpha u^2 ) (a + bu + cu^2 + du^3)</math>

<math>~=</math>

<math>~(u^2-3) [c_0(c_0-1) ( a + bu + cu^2 + du^3 ) + 2c_0 ( bu + 2cu^2 + 3du^3 ) + ( 2cu^2 + 6du^3 )] + (21 - 2u^2 ) [c_0(a + bu + cu^2 + du^3) + bu + 2cu^2 + 3du^3] \,. </math>

This is cute, but I don't see any way that this approach will provide an avenue to cancel the <math>~\Omega^2 u^5</math> term.


Yet Another Guess

Let's try,

<math>~x</math>

<math>~=</math>

<math>~e^{a + b\ln\xi + c(\ln\xi)^2} \, ,</math>

and examine the specific case of <math>~\sigma_c^2 = 0</math>, and, <math>~\gamma = (n+1)/n = 6/5 ~~\Rightarrow~~ \alpha = (3-20/6) = -1/3</math>. Under these conditions, the LAWE for <math>~n=5</math> polytropes becomes,

<math>~0</math>

<math>~=</math>

<math>~(3+\xi^2) \frac{d^2x}{d\xi^2} + \biggl[12 - 2\xi^2 \biggr] \frac{1}{\xi} \cdot \frac{dx}{d\xi} + 2x </math>

 

<math>~=</math>

<math>~(3+\xi^2) \frac{\xi^2}{x} \cdot \frac{d^2x}{d\xi^2} + \biggl[12 - 2\xi^2 \biggr] \frac{\xi}{x} \cdot \frac{dx}{d\xi} + 2\xi^2 \, . </math>

And the derivatives give,

<math>~\frac{dx}{d\xi}</math>

<math>~=</math>

<math>~x \frac{d}{d\xi}\biggl[ a + b\ln\xi + c(\ln\xi)^2 \biggr]</math>

 

<math>~=</math>

<math>~x \biggl[ \frac{b}{\xi}+ \frac{2c\ln\xi}{\xi} \biggr]</math>

<math>~\Rightarrow ~~~ \frac{\xi}{x} \cdot \frac{dx}{d\xi}</math>

<math>~=</math>

<math>~b + 2c\ln\xi \, ;</math>

and,

<math>~\frac{d^2x}{d\xi^2}</math>

<math>~=</math>

<math>~ \biggl[ \frac{b}{\xi}+ \frac{2c\ln\xi}{\xi} \biggr] \frac{dx}{d\xi} + x \frac{d}{d\xi}\biggl[ \frac{b}{\xi}+ \frac{2c\ln\xi}{\xi} \biggr]</math>

<math>~\Rightarrow ~~~ \frac{\xi^2}{x} \cdot \frac{d^2x}{d\xi^2}</math>

<math>~=</math>

<math>~ \xi \biggl[ \frac{b}{\xi}+ \frac{2c\ln\xi}{\xi} \biggr] \frac{\xi}{x} \cdot \frac{dx}{d\xi} + \xi^2 \frac{d}{d\xi}\biggl[ \frac{b+ 2c\ln\xi}{\xi} \biggr]</math>

 

<math>~=</math>

<math>~ \biggl[ b + 2c\ln\xi \biggr]^2 + \xi \frac{d}{d\xi}\biggl[ b+ 2c\ln\xi \biggr] + (b+ 2c\ln\xi) \xi^2 \biggl[ - \frac{1}{\xi^2} \biggr] </math>

 

<math>~=</math>

<math>~ (b + 2c\ln\xi )^2 + 2c- (b+ 2c\ln\xi) </math>

 

<math>~=</math>

<math>~[ b^2 + 2c - b] + [4bc - 2c] \ln\xi+4c^2 (\ln\xi)^2 \, . </math>

Hence the "fundamental mode" LAWE becomes,

<math>~0</math>

<math>~=</math>

<math>~(3+\xi^2) \biggl[ ( b^2 + 2c - b ) + (4bc - 2c) \ln\xi+4c^2 (\ln\xi)^2 \biggr] + (12 - 2\xi^2 ) \biggl[ b + 2c\ln\xi \biggr] \, . + 2\xi^2 </math>

Now, this expression cannot be satisfied for arbitrary <math>~\xi</math>. But, here we seek a solution only at the surface for the specific model, <math>~\xi = 3</math>. Plugging this value into the expression gives,

<math>~0</math>

<math>~=</math>

<math>~12 \biggl[ ( b^2 + 2c - b ) + (4bc - 2c) \ln 3+4c^2 (\ln 3)^2 \biggr] + (12 - 18 ) \biggl[ b + 2c\ln 3 \biggr] + 18 </math>

 

<math>~=</math>

<math>~2 \biggl[ ( b^2 + 2c - b ) + (4bc - 2c) \ln 3+4c^2 (\ln 3)^2 \biggr] -\biggl[ b + 2c\ln 3 \biggr] + 3 \, . </math>

It appears as though one perfectly satisfactory solution is, <math>~c = 0</math>, in which case, we need,

<math>~0</math>

<math>~=</math>

<math>~2 b^2 - 3b + 3 </math>

<math>~\Rightarrow~~~b</math>

<math>~=</math>

<math>~ \frac{3}{4}\biggl[1 \pm \sqrt{1-\frac{8}{3} } \biggr] \, . </math>

Thus, <math>~b</math> is an complex number.

Related Discussions

  • In an accompanying Chapter within our "Ramblings" Appendix, we have played with the adiabatic wave equation for polytropes, examining its form when the primary perturbation variable is an enthalpy-like quantity, rather than the radial displacement of a spherical mass shell. This was done in an effort to mimic the approach that has been taken in studies of the stability of Papaloizou-Pringle tori.
  • <math>~n=3</math> … M. Schwarzschild (1941, ApJ, 94, 245), Overtone Pulsations of the Standard Model: This work is referenced in §38.3 of [KW94]. It contains an analysis of the radial modes of oscillation of <math>~n=3</math> polytropes, assuming various values of the adiabatic exponent.
  • <math>~n=\tfrac{3}{2}</math> … D. Lucas (1953, Bul. Soc. Roy. Sci. Liege, 25, 585) … Citation obtained from the Prasad & Gurm (1961) article.
  • <math>~n=1</math> … L. D. Chatterji (1951, Proc. Nat. Inst. Sci. [India], 17, 467) … Citation obtained from the Prasad & Gurm (1961) article.


Whitworth's (1981) Isothermal Free-Energy Surface

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