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

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Line 158: Line 158:
<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} +  
<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} -  
\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>
\frac{6\alpha}{(3+\xi^2)}\biggr)\biggr]  x \, .</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
Or,
<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">
&nbsp;
<math>~0</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 170: Line 179:
   </td>
   </td>
   <td align="left">  
   <td align="left">  
<math>~\frac{1}{(3+\xi^2)} \biggl\{(3+\xi^2) \frac{d^2x}{d\xi^2} + \biggl[12 - 2\xi^2 \biggr] \frac{1}{\xi} \cdot \frac{dx}{d\xi} +  
<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} -  
\biggl[\frac{\sigma_c^2}{3^{1/2} \gamma_g } \cdot (3+\xi^2)^{3/2} -  
6\alpha \biggr]  x \biggr\} \, .</math>
6\alpha \biggr]  x </math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</td></tr></table>
</div>
</div>


=Search for Analytic Solutions to the LAWE=
==Numerical Integration of LAWE==
==Setup Using Lagrangian Radial Coordinate==
 
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]].


===Individual Terms===
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. 
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">
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.
<table border="0" cellpadding="3">


<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>
<math>
~\frac{R_\mathrm{eq}}{R_\mathrm{norm}}
</math>
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{4\pi}{(n+1)^n}\biggr]^{1/(n-3)}
\tilde\xi ( -\tilde\xi^2 \tilde\theta' )^{(1-n)/(n-3)}
</math>
  </td>
</tr>
</tr>
<tr>
<tr>
   <td align="right">
   <td align="center">
&nbsp;
[[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{4\pi}{2^5\cdot 3^5}\biggr]^{1/2}
[[File:DataFileButton02.png|right|60px|file = Dropbox/WorkFolder/Wiki edits/EmbeddedPolytropes/n5movie/ --- worksheet = n5Truncated2.gif]]
\tilde\xi ( -\tilde\xi^2 \tilde\theta' )^{-2} \, ,
</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">
<table border="0" cellpadding="3">


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.
=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>
<tr>
   <td align="right">
   <th align="center" colspan="1">Fundamental Mode Eigenfunction <br />
<math>
when <math>~\sigma_c^2 = 0</math> and <math>~\gamma = 6/5 ~\Rightarrow~\alpha=- 1/3</math></th>
~\frac{P_\mathrm{e}}{P_\mathrm{norm}}
</math>
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<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>
  </td>
</tr>
</tr>


<tr>
<tr>
  <td align="right">
&nbsp;
  </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~x = x_0 \biggl[ 1 - \frac{\xi^2}{15} \biggr]</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{2^3\cdot 3^3}{4\pi}\biggr]^{3}
\tilde\theta^{6}( -\tilde\xi^2 \tilde\theta' )^{6} \, ,
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</div>
</div>
where,
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">
<div align="center" id="Proof">
<table border="0" cellpadding="3">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~R_\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[ \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>~\frac{d^2x}{d\xi^2} = -\frac{2}{15} \, .</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
<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>~\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>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\equiv</math>
<math>~=</math>
   </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{2}{15}(3+\xi^2) -\frac{2}{15} \biggl[12 - 2\xi^2 \biggr] + 2\biggl[1 - \frac{\xi^2}{15}\biggr]
</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">
<math>
&nbsp;
~{\tilde\theta}_5 = 3^{1 / 2} \biggl( 3 + {\tilde\xi}^2\biggr)^{-1/2}
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
   </td>
   </td>
</tr>


<tr>
  <td align="right">
&nbsp;
  </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>
0 \, .
~\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>
</table>
</div>


Hence,
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.


Now, let's see how the boundary condition comes into play.  We see that the logarithmic derivative of the parabolic eigenfunction is,
<div align="center">
<div align="center">
<table border="0" cellpadding="3">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>
<math>~\frac{d\ln x}{d\ln \xi}</math>
~\frac{R_\mathrm{eq}}{R_\mathrm{norm}}  
   </td>
</math>
   </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{\xi}{x} \cdot \frac{dx}{d\xi}</math>
\tilde\xi \biggl[ 3^{1 / 2} {\tilde\xi}^3 \biggl( 3 + {\tilde\xi}^2\biggr)^{-3/2}  \biggr]^{-2}
</math>
   </td>
   </td>
</tr>
</tr>
Line 331: Line 328:
   </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{2\xi^2}{15} \biggl[ 1 - \frac{\xi^2}{15}\biggr]^{-1}</math>
\tilde\xi \biggl[ 3^{-1} {\tilde\xi}^{-6} \biggl( 3 + {\tilde\xi}^2\biggr)^{3} \biggr]
</math>
   </td>
   </td>
</tr>
</tr>
Line 345: Line 340:
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~- \frac{2\xi^2}{(15-\xi^2)} \, .</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>
   </td>
</tr>
</tr>
</table>
</div>
We desire a surface boundary condition that gives, <math>~d\ln x/d\ln\xi = -3</math>.  This will only happen when,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>
<math>~- \frac{2\xi^2}{(15-\xi^2)}</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>~-3</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>
Line 373: Line 366:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\Rightarrow ~~~2\xi^2</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>~3(15 - \xi^2)</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 387: Line 378:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\Rightarrow ~~~\xi</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 3^5}{4\pi}\biggr]^{3}
<math>~3 \, .</math>
{\tilde\xi}^{18} \biggl( 3 + {\tilde\xi}^2\biggr)^{-12} \, .
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</div>
</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.


Now, given that the [[User:Tohline/SSC/Virial/FormFactors#Summary_.28n.3D5.29|structural form-factors for <math>~n=5</math> configurations]] are,
<!--  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>
</tr>
</table>
</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>~\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>
( 1 + \ell^2 )^{-3/2}  = 3^{3 / 2} (3 + {\tilde\xi}^2)^{-3 / 2}
</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">
<math>~\mathfrak{f}_W</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[ \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>~
<math>~
\frac{5}{2^4} \cdot \ell^{-5}
-\frac{2}{A}(3+\xi^2)  -\frac{2}{A} \biggl[12 - 2\xi^2 \biggr]  -6\alpha \biggl[1 - \frac{\xi^2}{A}\biggr]
\biggl[ \ell \biggl( \ell^4 - \frac{8}{3}\ell^2 - 1 \biggr)(1 + \ell^2)^{-3} + \tan^{-1}(\ell ) \biggr]
</math>
</math>
   </td>
   </td>
Line 436: Line 444:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\mathfrak{f}_A</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">  
<math>~
<math>~-\frac{1}{A}\biggl\{  
\frac{3}{2^3} \ell^{-3} [ \tan^{-1}(\ell ) + \ell (\ell^2-1) (1+\ell^2)^{-2} ] \, ,
2(3+\xi^2) + 2 \biggl[12 - 2\xi^2 \biggr]  + 6\alpha \biggl[A - \xi^2\biggr]
\biggr\}
</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">
<math>~\rho_c = \frac{\bar\rho}{ {\tilde\mathfrak{f}}_M }</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">  
<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]
\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 474: Line 488:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ \biggl( \frac{3}{4\pi}\biggr)
<math>~\xi \biggl\{  
\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{4\pi}{2^5\cdot 3}\biggr]^{1/2} \tilde\xi^{-6}
{\tilde\xi}^{-5} \biggl( 3 + {\tilde\xi}^2\biggr)^{3} \biggr]^{-3}
\biggl( 1+\frac{\tilde\xi^2}{3} \biggr)^{3}\biggr\} \, .
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
and that,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~
r_\xi (m_\xi)
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 489: Line 510:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~\tilde{r}_\mathrm{edge}
\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}
\biggl[\frac{3^2m_\xi^{2/3}}{\tilde{C} - 3 m_\xi^{2/3}}\biggr]^{1/2} \, ,
</math>
</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>~
<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{3^2}{\tilde\xi^2}\biggl( 1 + \frac{\tilde\xi^2}{3} \biggr) \, .
</math>
</math>
   </td>
   </td>
Line 511: 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>~
<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{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>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</div>
By equating <math>~r_\xi(\xi)</math> with <math>~r_\xi(m_\xi)</math>, we find,


<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">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 530: Line 558:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~r_0 \equiv a_5 \xi</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 536: Line 567:
   </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>~\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>
   </td>
   </td>
</tr>
</tr>
Line 542: Line 575:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\Rightarrow ~~~ \xi
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 548: Line 582:
   </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}
\biggl[\frac{3^2m_\xi^{2/3}}{\tilde{C} - 3 m_\xi^{2/3}}\biggr]^{1/2} \, .
\biggr\}^{-2/5}
</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 563: Line 601:
   </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{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3} \xi
x_0 \biggl\{1 -
\frac{1}{15} \biggl[\frac{3^2m_\xi^{2/3}}{\tilde{C} - 3 m_\xi^{2/3}}\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 578: Line 623:
   <td align="left">
   <td align="left">
<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
x_0 \biggl\{1 -
\frac{1}{15} \biggl[\frac{3^2m_\xi^{2/3}}{4 - 3 m_\xi^{2/3}}\biggr]  
\biggr\} \, .
</math>
</math>
   </td>
   </td>
Line 585: Line 632:
</div>
</div>


<span id="m0">Also,</span>
 
{{ 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>~m_0 \equiv M(r_0)</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>~\biggl[ 4\pi a_n^3 \rho_c \biggl(-\xi^2 \frac{d\theta}{d\xi}\biggr) \biggr]
<math>~\biggl[ \frac{4\pi}{(n+1)^n}\biggr]^{1/(n-3)}
\, ,</math>
\tilde\xi ( -\tilde\xi^2 \tilde\theta' )^{(1-n)/(n-3)}  
</math>
   </td>
   </td>
</tr>
</tr>
Line 608: Line 664:
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~=~</math>
   </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[ \frac{4\pi}{2^5\cdot 3^5}\biggr]^{1/2}
\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}  \biggr\}
\tilde\xi ( -\tilde\xi^2 \tilde\theta' )^{-2} \, ,
\biggl\{  3^{1 / 2} \xi^3 \biggl( 3 + \xi^2\biggr)^{-3/2} \biggr\}
</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">
&nbsp;
<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>~ 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[ \frac{(n+1)^3}{4\pi}\biggr]^{(n+1)/(n-3)}
\biggl\{  \tilde\xi^{-6} (3 + {\tilde\xi}^2)^{3} \biggr\}^3
\tilde\theta_n^{n+1}( -\tilde\xi^2 \tilde\theta' )^{2(n+1)/(n-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 639: 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}
\biggl\{ \tilde\xi^{-3} (3 + {\tilde\xi}^2)^{3 / 2} \biggr\} M_\mathrm{tot}  
\tilde\theta^{6}( -\tilde\xi^2 \tilde\theta' )^{6} \, ,
\biggl\{ \xi^3 ( 3 + \xi^2 )^{-3/2} \biggr\} \, .
</math>
</math>
   </td>
   </td>
Line 650: Line 709:
</table>
</table>
</div>
</div>
 
where,
Hence,
 
<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>~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>~ \frac{GM_\mathrm{tot}}{R_\mathrm{norm}^2}
<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\{ \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>
   </td>
</tr>
</tr>
Line 674: 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>~
<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>
\frac{GM_\mathrm{tot}}{R_\mathrm{norm}^2}\biggl[ \frac{2^3\cdot 3^{7}}{\pi}\biggr]  
  </td>
\biggl[ \tilde\xi (3 + {\tilde\xi}^2)^{-1 / 2} \biggr]^{9}
</tr>
\xi ( 3 + \xi^2 )^{-3/2} \, ;
</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>
  <td align="right">
<math>
~{\tilde\theta}_5 = 3^{1 / 2} \biggl( 3 + {\tilde\xi}^2\biggr)^{-1/2}
</math>
  </td>
 
  <td align="center">
&nbsp; &nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp; &nbsp;
  </td>
 
  <td align="right">
<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>
Hence,
<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{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}^3}\biggl[ \frac{2^3\cdot 3^{7}}{\pi}\biggr]  
<math>~\biggl[ \frac{4\pi}{2^5\cdot 3^5}\biggr]^{1/2}
\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\xi \biggl[ 3^{1 / 2} {\tilde\xi}^3 \biggl( 3 + {\tilde\xi}^2\biggr)^{-3/2} \biggr]^{-2}  
\xi ( 3 + \xi^2 )^{-3/2}  
</math>
</math>
   </td>
   </td>
Line 708: Line 787:
   </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}^3}\biggl[ \frac{2^3\cdot 3^{7}}{\pi}\biggr]^{3/2}  
\tilde\xi \biggl[ 3^{-1} {\tilde\xi}^{-6} \biggl( 3 + {\tilde\xi}^2\biggr)^{3} \biggr]
\biggl[  \tilde\xi (3 + {\tilde\xi}^2)^{-1 / 2} \biggr]^{15} 
( 3 + \xi^2 )^{-3/2}  
\, ;
</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>~\frac{\rho_0}{P_0} = \frac{\rho_0}{K\rho_0^{1+1/n}} </math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~=~</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
\biggl[K^5 \rho_c \theta^5 \biggr]^{-1/5}
\biggl[ \frac{4\pi}{2^5\cdot 3^7}\biggr]^{1/2}
{\tilde\xi}^{-5} \biggl( 3 + {\tilde\xi}^2\biggr)^{3}  \, ,
</math>
</math>
   </td>
   </td>
Line 742: Line 813:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<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>~ \theta^{-1}
<math>~\biggl[ \frac{2^3\cdot 3^3}{4\pi}\biggr]^{3}
\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[ 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>
   </td>
   </td>
Line 759: Line 832:
   </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{2^3\cdot 3^3}{4\pi}\biggr]^{3}
\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[ 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>
   </td>
   </td>
Line 773: Line 846:
   </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 3^5}{4\pi}\biggr]^{3}
\biggl( \frac{G^3M_\mathrm{tot}^2}{K^5} \biggr)^{1 / 2}
{\tilde\xi}^{18} \biggl( 3 + {\tilde\xi}^2\biggr)^{-12} \, .
\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, 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 793: Line 869:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~  \biggl[ 3^{-1} ( 3 + \xi^2 ) \biggr]^{1/2}
<math>~   
\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\}
( 1 + \ell^2 )^{-3/2} = 3^{3 / 2} (3 + {\tilde\xi}^2)^{-3 / 2}
</math>
</math>
   </td>
   </td>
Line 801: Line 877:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\mathfrak{f}_W</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 807: Line 883:
   </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}
\frac{5}{2^4} \cdot \ell^{-5}
\biggl[ \frac{(3 + {\tilde\xi}^2)}{{\tilde\xi}^2 } \biggr]^{3 / 2} ( 3 + \xi^2 )^{1/2}
\biggl[ \ell \biggl( \ell^4 - \frac{8}{3}\ell^2 - 1 \biggr)(1 + \ell^2)^{-3} + \tan^{-1}(\ell ) \biggr]
</math>
</math>
   </td>
   </td>
Line 816: Line 892:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\frac{g_0\rho_0}{P_0} </math>
<math>~\mathfrak{f}_A</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 822: Line 898:
   </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}  
\frac{3}{2^3} \ell^{-3} [ \tan^{-1}(\ell ) + \ell (\ell^2-1) (1+\ell^2)^{-2} ]  \, ,
\biggl[ \frac{(3 + {\tilde\xi}^2)}{{\tilde\xi}^2 } \biggr]^{3 / 2}  ( 3 + \xi^2 )^{1/2}
</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">
&nbsp;
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~  \times ~
<math>~   
\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 / 2} (3 + {\tilde\xi}^2)^{-3 / 2} \biggr]^{-1} \biggl[ \frac{3 M_\mathrm{tot}}{4 \pi R_\mathrm{eq}^3} \biggr]
\biggl[ \frac{(3 + {\tilde\xi}^2)}{{\tilde\xi}^2 } \biggr]^{-9 / 2} 
\xi ( 3 + \xi^2 )^{-3/2}
</math>
</math>
   </td>
   </td>
Line 853: Line 930:
   </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{3}{4\pi}\biggr)
\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}  
\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>
   </td>
   </td>
Line 868: Line 946:
   <td align="left">
   <td align="left">
<math>~   
<math>~   
\biggl( \frac{K^5}{G^5 M_\mathrm{tot}^4} \biggr)^{1 / 2}
\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}  
\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===
====Starting from our Key Adiabatic Wave Equation====
The [[#Adiabatic_.28Polytropic.29_Wave_Equation|adiabatic wave equation]] therefore 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 895: Line 959:
   </td>
   </td>
   <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}  
\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}  
+ \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 910: Line 973:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
\frac{d^2x}{dr_0^2} + \frac{1}{R_\mathrm{norm}} \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{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>
</tr>
</tr>
</table>
<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">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~r_0 \equiv a_5 \xi</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~=</math>
   </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>
+ \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>
   </td>
   </td>
</tr>
</tr>
Line 948: Line 1,004:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~\biggl[\frac{3K}{2\pi G} \biggr]^{1 / 2} \xi \biggl\{
\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\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}{dr_0}  
</math>
</math>
   </td>
   </td>
Line 961: Line 1,016:
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<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)
+ \frac{6(4 - 3\gamma_\mathrm{g})}{\gamma_g R_\mathrm{norm}^2}  
\biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3} \xi
( 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}
</math>
\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}
   </td>
( 3 + \xi^2 )^{-3/2} 
\biggr\}  x
</math>
   </td>
</tr>
</tr>


Line 984: Line 1,034:
   <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[
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
\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,
<div align="center">
<math>R_* \equiv R_\mathrm{norm} \biggl[ \frac{\pi}{2^3 \cdot 3^7} \biggr]^{1/2} \, .</math>
</div>
</div>


Recognizing that,
<span id="m0">Also,</span>


<div align="center">
<div align="center">
Line 1,008: Line 1,048:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~r_0</math>
<math>~m_0 \equiv M(r_0)</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,014: 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]
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">
<math>~0</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,034: Line 1,067:
   </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>~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^2x}{d\xi^2} + \biggl[
\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{4}{\xi} -  \frac{6 \xi}{ ( 3 + \xi^2 )}
\biggl\3^{1 / 2} \xi^3 \biggl( 3 + \xi^2\biggr)^{-3/2} \biggr\}
\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,
<tr>
<div align="center">
  <td align="right">
<table border="0" cellpadding="5" align="center">
&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>
  </td>
</tr>


<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} = \frac{\sigma_c^2}{2\cdot 3^{3/2}}\, .</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>
   </td>
   </td>
</tr>
</tr>
Line 1,066: Line 1,107:
</div>
</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,
Hence,


<div align="center">
<div align="center">
Line 1,073: Line 1,114:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~0</math>
<math>~g_0 = \frac{Gm_0}{r_0^2}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,079: Line 1,120:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~ \frac{GM_\mathrm{tot}}{R_\mathrm{norm}^2}
\frac{d^2x}{d\xi^2} + \biggl[ \frac{4}{\xi} \frac{6 \xi}{ ( 3 + \xi^2 )} \biggr] \frac{dx}{d\xi}
\biggl\{ \tilde\xi^{-3} (3 + {\tilde\xi}^2)^{3 / 2} \biggr\} 
+ \biggl[\frac{5\sigma_c^2}{2\cdot 3^{3/2}} ( 3 + \xi^2 )^{1 / 2} \frac{2}{( 3 + \xi^2 ) }\biggr] x
\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 1,094: Line 1,136:
   </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}
\frac{GM_\mathrm{tot}}{R_\mathrm{norm}^2}\biggl[ \frac{2^3\cdot 3^{7}}{\pi}\biggr]  
+ \biggl[\frac{5\sigma_c^2}{2\cdot 3^{3/2}}  ( 3 + \xi^2 )^{3 / 2} +  2 \biggr]  x \biggr\} \, ,
\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 exactly the [[#Specifically_for_n.3D5_Configurations|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 [[#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">
<math>~0</math>
<math>~\frac{g_0 }{r_0} </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,118: Line 1,152:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~ \frac{GM_\mathrm{tot}}{R_\mathrm{norm}^3}\biggl[ \frac{2^3\cdot 3^{7}}{\pi}\biggr]
\frac{d^2 X}{d\xi^2} + \biggl[ \frac{4}{\xi} - \frac{6 (-\theta^'_5)}{\theta_5} \biggr]\frac{d X}{d\xi}
\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}
+ \frac{5(-\theta_5^') }{6\theta_5 \xi} \bigg[ \frac{\xi (s^')^2}{\theta^'_5} + \frac{12}{5} \biggr] X
\xi ( 3 + \xi^2 )^{-3/2}  
</math>
</math>
   </td>
   </td>
Line 1,133: Line 1,167:
   </td>
   </td>
   <td align="left">
   <td align="left">
<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{GM_\mathrm{tot}}{R_\mathrm{norm}^3}\biggl[ \frac{2^3\cdot 3^{7}}{\pi}\biggr]^{3/2}  
+ \frac{1}{(3 + \xi^2)} \bigg[ -\frac{5(s^')^2(3 + \xi^2)^{3 / 2}}{2 \cdot 3^{3 / 2}} + 2 \biggr] X
\biggl[  \tilde\xi (3 + {\tilde\xi}^2)^{-1 / 2} \biggr]^{15} 
( 3 + \xi^2 )^{-3/2}  
\, ;
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>


<tr>
 
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<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 1,148: Line 1,190:
   </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[K^5 \rho_c \theta^5 \biggr]^{-1/5}
+ \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">
<math>~(s^')^2</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,174: Line 1,204:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~-\omega^2 \biggl[\frac{n+1 }{4\pi G \rho_c} \biggr] \, ,</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>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
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">
<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 1,192: Line 1,218:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{1}{(3+\xi^2)} \biggl\{
<math>~ \biggl[ 3^{-1} ( 3 + \xi^2 ) \biggr]^{1/2}
(3+\xi^2)\frac{d^2 X}{d\xi^2} + \biggl[ \frac{2(6-\xi^2)}{\xi}\biggr]\frac{d X}{d\xi}
\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\}
+ \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 [[#Specifically_for_n.3D5_Configurations|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">
<math>~u  \equiv 1 + \frac{3}{\xi^2}</math>
&nbsp;
   </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>~
\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>
  </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>~ \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>
   </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>
&nbsp;
   </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>~  
\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>
   </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>
<math>~\frac{g_0\rho_0}{P_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{4}{3}(u-1)^3 \frac{d^2}{du^2} + 2(u-1)^{2} \frac{d}{du} \, .</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>
   </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">
<math>~0</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
&nbsp;
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~( 3 + \xi^2 )\frac{d^2x}{d\xi^2}  
<math>~ \times ~
+ \biggl[ \frac{2(6 - \xi^2) }{ \xi} \biggr] \frac{dx}{d\xi}
\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[5\sigma^2 ( 3 + \xi^2 )^{3 / 2} +  2 \biggr]  x
\biggl[ \frac{(3 + {\tilde\xi}^2)}{{\tilde\xi}^2 } \biggr]^{-9 / 2}
\xi ( 3 + \xi^2 )^{-3/2}  
</math>
</math>
   </td>
   </td>
Line 1,289: Line 1,309:
   </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>~ \biggl( \frac{G^5 M_\mathrm{tot}^4}{K^5} \biggr)^{1 / 2} R_\mathrm{norm}^{-2}
+ 4(2u-3)(u-1)\frac{dx}{du}
\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}  
+ \biggl\{ 5\sigma^2 \biggl[ \frac{3u}{(u-1)} \biggr]^{3 / 2} + 2 \biggr\} x
</math>
</math>
   </td>
   </td>
Line 1,304: Line 1,323:
   </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{K^5}{G^5 M_\mathrm{tot}^4} \biggr)^{1 / 2}
+ \biggl\{ 5\sigma^2 \biggl[ \frac{3u}{(u-1)} \biggr]^{3 / 2} + 2 \biggr\} x \, .
\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 1,313: Line 1,334:
</div>
</div>


{{ LSU_WorkInProgress }}
===The Wave Equation===


====Starting from our Key Adiabatic Wave Equation====


If we ''assume'' that <math>~\sigma^2 = 0</math>, then the governing relation is,
The [[#Adiabatic_.28Polytropic.29_Wave_Equation|adiabatic wave equation]] therefore becomes,


<div align="center">
<div align="center">
Line 1,329: Line 1,351:
   </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}  
\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 x \, .
+ \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>
</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 1,352: Line 1,367:
   <td align="left">
   <td align="left">
<math>~
<math>~
A^3 (u - 1)^{1 / 2} (A u - 1 )^{-1 / 2}   \, ,
\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>
</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">
<math>~=</math>
&nbsp;
   </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]
+ \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>
   </td>
   </td>
Line 1,385: Line 1,405:
   <td align="left">
   <td align="left">
<math>~
<math>~
\biggl[ \frac{A^3(A-1)}{2} \biggr] (u-1)^{-1 / 2} (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[
\frac{4}{\xi} -   \frac{6 \xi}{ ( 3 + \xi^2 )}
\biggr] \frac{dx}{dr_0}
</math>
</math>
   </td>
   </td>
Line 1,392: Line 1,414:
<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>~
\biggl[ \frac{A^3(A-1)}{2} \biggr] \biggl\{
+ \frac{6(4 - 3\gamma_\mathrm{g})}{\gamma_g R_\mathrm{norm}^2}
-\frac{1}{2}(u-1)^{-3 / 2} (Au-1)^{-3 / 2} -\frac{3A}{2} (u-1)^{-1 / 2} (Au-1)^{-5 / 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}
\biggr\}
\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 1,415: Line 1,440:
   <td align="left">
   <td align="left">
<math>~
<math>~
-\frac{1}{2} \biggl[ \frac{A^3(A-1)}{2} \biggr] (u-1)^{-3 / 2} (Au-1)^{-5 / 2}\biggl[
\frac{d^2x}{dr_0^2} + \frac{1}{R_*} \biggl[ \frac{ {\tilde\xi}^2}{(3 + {\tilde\xi}^2)} \biggr]^{3} \biggl[
(Au-1) +3A (u-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 1,430: Line 1,471:
   <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] \, .
R_*  \biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3} \xi \, ,
</math>
</math>
   </td>
   </td>
Line 1,436: Line 1,477:
</table>
</table>
</div>
</div>
we can write,
<div align="center">
<table border="0" cellpadding="5" align="center">


<!-- EXTRA 2nd DERIVATIVE
<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 1,446: Line 1,490:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~\frac{1}{R_*^2} \biggl[ \frac{ {\tilde\xi}^2}{(3 + {\tilde\xi}^2)} \biggr]^{6} \biggl\{
\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}  
\frac{d^2x}{d\xi^2} + \biggl[
- A(u - 1)^{-1 / 2} (A u - 1 )^{-3 / 2} + 3A^2(u - 1)^{1 / 2} (A u - 1 )^{-5 / 2} \biggr]
\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>
END EXTRA 2nd DERIVATIVE -->
</table>
 
</div>
So the governing relation becomes:


where,
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 1,461: Line 1,510:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~0</math>
<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>~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>~ \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>
</math>
   </td>
   </td>
</tr>
</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>
<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>~
+ (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}{d\xi^2} + \biggl[ \frac{4}{\xi} -  \frac{6 \xi}{ ( 3 + \xi^2 )} \biggr] \frac{dx}{d\xi}
+ 2 A^3 (u - 1)^{1 / 2} (A u - 1 )^{-1 / 2}  
+ \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>
Line 1,495: Line 1,550:
   </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{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>
</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">
&nbsp;
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
+ (7u-6)(u-1)^{1 / 2} A^3(A-1)  (Au-1)^{-3 / 2}  
\frac{d^2 X}{d\xi^2} + \biggl[ \frac{4}{\xi} - \frac{6 (-\theta^'_5)}{\theta_5} \biggr]\frac{d X}{d\xi}
+ 2 A^3 (u - 1)^{1 / 2} (A u - 1 )^{-1 / 2}  
+ \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 1,523: Line 1,589:
   </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^2 X}{d\xi^2} + \biggl[ \frac{4}{\xi} - \frac{6 \xi}{(3 + \xi^2)} \biggr]\frac{d X}{d\xi}
+ 2 A^3 (A u - 1 )^{-1 / 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
</math>
</math>
   </td>
   </td>
Line 1,538: Line 1,604:
   </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>~\frac{1}{(3+\xi^2)} \biggl\{
+ (7u-6) (A-1) (Au-1)
(3+\xi^2)\frac{d^2 X}{d\xi^2} + \biggl[ \frac{2(6-\xi^2)}{\xi}\biggr]\frac{d X}{d\xi}
+ 2 (A u - 1 )^{2} \biggr\}
+ \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 1,553: Line 1,630:
   </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>~-\omega^2 \biggl[\frac{n+1 }{4\pi G \rho_c} \biggr] \, ,</math>
+ (7u-6)  [A(A-1)u +1 - A]
+ 2  (A^2u^2 - 2Au +1) \biggr\}
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
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">
<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 1,568: Line 1,648:
   </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>~\frac{1}{(3+\xi^2)} \biggl\{
+ u\biggl[ (A-1) (3A+1) - 7(A-1) -6A(A-1) - 4A \biggr] + 2(3A-2)  \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\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 align="center">
&nbsp; &nbsp; &nbsp; &nbsp; <math>~\Rightarrow</math> &nbsp; &nbsp; &nbsp; &nbsp;
  </td>
  <td align="left">
<math>~3 + \xi^2 = \frac{3u}{(u-1)}  \, ,</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">
   <td align="left">
<math>~A^3(u-1)^{1 / 2} (Au-1)^{-5 / 2} \biggl\{ Au^2 \biggl[ 5A-3 \biggr]   
<math>~\xi = 3^{1 / 2} (u-1)^{-1 / 2} \, .</math>
+ u\biggl[ 3A^2-2A-1-7A+7 -6A^2+6A -4A \biggr]  + 2(3A-2)  \biggr\} \, .
</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>~A^3(u-1)^{1 / 2} (Au-1)^{-5 / 2} \biggl\{ Au^2 \biggl[ 5A-3 \biggr]   
<math>~-\frac{2}{\sqrt{3}}(u-1)^{3 / 2} \frac{d}{du} \, ,</math>
+ u\biggl[ -3A^2 -7A  +6\biggr]  + 2(3A-2)  \biggr\} \, .
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</div>
</div>
 
and,
=====Second Guess (n5)=====
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 1,610: Line 1,706:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~x</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>
(u - 1)^{b / 2} (A u - 1 )^{-a / 2}   \, ,
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</div>
</div>
in which case,
Hence, the governing wave equation becomes,
 
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 1,629: Line 1,724:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\frac{dx}{du}</math>
<math>~0</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,635: Line 1,730:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~( 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[5\sigma^2 ( 3 + \xi^2 )^{3 / 2} +  2 \biggr]  x
</math>
</math>
   </td>
   </td>
Line 1,650: Line 1,745:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~x \biggl[  
<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{b}{2}(u-1)^{-1}  
+ 4(2u-3)(u-1)\frac{dx}{du}
- \frac{aA}{2} (A u - 1 )^{-1} \biggr]
+ \biggl\{ 5\sigma^2 \biggl[ \frac{3u}{(u-1)} \biggr]^{3 / 2} +  2 \biggr\}  x
</math>
</math>
   </td>
   </td>
Line 1,659: Line 1,754:
<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 1,665: Line 1,760:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ (A u - 1 )^{-1} \biggl[
<math>~4u(u-1)^2 \frac{d^2x}{du^2}  
\frac{b}{2} (A u - 1 )
+ (14u-12)(u-1)\frac{dx}{du}
- \frac{aA}{2} (u-1) \biggr]
+ \biggl\{ 5\sigma^2 \biggl[ \frac{3u}{(u-1)} \biggr]^{3 / 2}  +  2 \biggr\}  x \, .
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
{{ LSU_WorkInProgress }}


<tr>
 
   <td align="right">
If we ''assume'' that <math>~\sigma^2 = 0</math>, then the governing relation is,
&nbsp;
 
<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">
Line 1,680: Line 1,785:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{1 }{2(A u - 1 )} \biggl[
<math>~4u(u-1)^2 \frac{d^2x}{du^2}  
b (A u - 1 ) - aA (u-1) \biggr]
+ (14u-12)(u-1)\frac{dx}{du}
+ 2 x \, .
</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">
Line 1,695: Line 1,808:
   <td align="left">
   <td align="left">
<math>~
<math>~
\frac{1 }{2(A u - 1 )} \biggl[
A^3 (u - 1)^{1 / 2} (A u - 1 )^{-1 / 2}   \, ,
(aA - b) + A(b    - a)u \biggr] \, ;
</math>
</math>
   </td>
   </td>
Line 1,702: Line 1,814:
</table>
</table>
</div>
</div>
and,
in which case,
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 1,708: Line 1,820:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\frac{d^2x}{du^2}</math>
<math>~\frac{dx}{du}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,715: Line 1,827:
   <td align="left">
   <td align="left">
<math>~
<math>~
\biggl[ \frac{b}{2}(u-1)^{-1} - \frac{aA}{2} (A u - 1 )^{-1} \biggr]\frac{dx}{du}
\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]
+ x \frac{d}{du}\biggl[ \frac{b}{2}(u-1)^{-1} - \frac{aA}{2} (A u - 1 )^{-1} \biggr]
</math>
</math>
   </td>
   </td>
Line 1,730: Line 1,841:
   <td align="left">
   <td align="left">
<math>~
<math>~
x\biggl[ \frac{b}{2}(u-1)^{-1} - \frac{aA}{2}  (A u - 1 )^{-1} \biggr]^2
\biggl[ \frac{A^3(A-1)}{2} \biggr] (u-1)^{-1 / 2} (Au-1)^{-3 / 2} \, ;
+ x  \biggl[ -\frac{b}{2}(u-1)^{-2} + \frac{aA^2}{2}  (A u - 1 )^{-2} \biggr]
</math>
</math>
   </td>
   </td>
Line 1,738: Line 1,848:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\frac{d^2x}{du^2}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,744: Line 1,854:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ \frac{x}{4(u-1)^2 (Au-1)^2} \biggl\{
<math>~
\biggl[ b(Au-1) - aA  (u - 1 ) \biggr]^2
\biggl[ \frac{A^3(A-1)}{2} \biggr] \biggl\{
\biggl[  2aA^2 (u-1)^{2} -2b (A u - 1 )^{2} \biggr]
-\frac{1}{2}(u-1)^{-3 / 2} (Au-1)^{-3 / 2} -\frac{3A}{2} (u-1)^{-1 / 2} (Au-1)^{-5 / 2}
\biggr\}
\biggr\}
</math>
</math>
Line 1,754: Line 1,864:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\Rightarrow ~~~ \frac{(1-u)^2}{x}\frac{d^2x}{du^2}</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,760: Line 1,870:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~  
<math>~
\frac{1}{4 (Au-1)^2} \biggl\{
-\frac{1}{2} \biggl[ \frac{A^3(A-1)}{2} \biggr] (u-1)^{-3 / 2} (Au-1)^{-5 / 2}\biggl[
\biggl[ b(Au-1) - aA  (u - 1 ) \biggr]^2
(Au-1) +3A (u-1)\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">
<math>~0</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,784: Line 1,885:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~2u \biggl\{ \frac{(u-1)^2}{x} \frac{d^2x}{du^2} \biggr\}
<math>~
+ (7u-6)\biggl\{ \frac{(u-1)}{x} \frac{dx}{du} \biggl\} + 1  
\biggl[ \frac{A^3(A-1)}{4} \biggr] (u-1)^{-3 / 2} (Au-1)^{-5 / 2}\biggl[(3A+1) - 4Au \biggr] \, .
</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">
Line 1,799: Line 1,903:
   <td align="left">
   <td align="left">
<math>~
<math>~
\frac{2u}{4 (Au-1)^2} \biggl\{
\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}
\biggl[ (aA - b) + A(b    - a)u \biggr]^2
- A(u - 1)^{-1 / 2} (A u - 1 )^{-3 / 2} + 3A^2(u - 1)^{1 / 2} (A u - 1 )^{-5 / 2} \biggr]
\biggl[ 2aA^2 (u-1)^{2} -2b (A u - 1 )^{2} \biggr]
</math>
\biggr\}
  </td>
</tr>
END EXTRA 2nd DERIVATIVE -->
 
So the governing relation 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>~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>
   </td>
   </td>
Line 1,816: Line 1,937:
   <td align="left">
   <td align="left">
<math>~
<math>~
+ \frac{(7u-6) }{2(A u - 1 )} \biggl[
+ (14u-12)(u-1) \biggl\{ \biggl[ \frac{A^3(A-1)}{2} \biggr] (u-1)^{-1 / 2} (Au-1)^{-3 / 2} \biggr\}
(aA - b) + A(b    - a)u \biggr]
+ 2 A^3 (u - 1)^{1 / 2} (A u - 1 )^{-1 / 2}
+ 1  
</math>
</math>
   </td>
   </td>
Line 1,831: 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]
\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>
   </td>
   </td>
Line 1,848: Line 1,965:
   <td align="left">
   <td align="left">
<math>~
<math>~
+ 2(A u - 1 )(7u-6) \biggl[
+ (7u-6)(u-1)^{1 / 2}  A^3(A-1) (Au-1)^{-3 / 2}
(aA - b) + A(b    - a)u \biggr]
+ 2 A^3 (u - 1)^{1 / 2} (A u - 1 )^{-1 / 2}  
+ 4 (Au-1)^2 \biggr\}
</math>
</math>
   </td>
   </td>
Line 1,863: 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] 
\frac{1}{4 (Au-1)^2} \biggl\{
+ (7u-6) A^3(A-1) (Au-1)^{-3 / 2}
2u\biggl[ (aA - b)^2 + 2(aA - b)A(b    - a)u + A^2(b    - a)^2u^2 \biggr]
+ 2 A^3 (A u - 1 )^{-1 / 2} \biggr\}
+ 2u\biggl[ 2A^2(a-b)u^2 + 4A(b - aA) u + 2(aA^2  -b)  \biggr]
</math>
</math>
   </td>
   </td>
Line 1,876: Line 1,991:
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~=</math>
   </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]
+ 2\biggl[7Au^2 - (6A+7)u  +6 \biggr]\biggl[
+ (7u-6) (A-1)  (Au-1)  
(aA - b) + A(b    - a)u \biggr]
+ 2 (A u - 1 )^{2} \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>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,902: 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)
2u\biggl[ (aA - b)^2   \biggr]
+ (7u-6) [A(A-1)u +1 - A]
2u\biggl[ 4A(b - aA) u + 2(aA^2   -b) \biggr]
+ 2 (A^2u^2 - 2Au +1) \biggr\}
</math>
</math>
   </td>
   </td>
Line 1,914: Line 2,021:
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<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]  
+ 2\biggl[7Au^2 - (6A+7)+6 \biggr]\biggl[  
+ u\biggl[ (A-1) (3A+1) - 7(A-1) -6A(A-1) - 4A \biggr]  + 2(3A-2) \biggr\}
(aA - b) \biggr]  + (4A^2u^2-8Au + 4)  
</math>
</math>
   </td>
   </td>
Line 1,932: Line 2,038:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~A^3(u-1)^{1 / 2} (Au-1)^{-5 / 2} \biggl\{ Au^2 \biggl[ 5A-3 \biggr]   
2a^2u (A - 1)^2
+ u\biggl[ 3A^2-2A-1-7A+7 -6A^2+6A -4A \biggr]  + 2(3A-2) \biggr\} \, .
+ 2au [ 4A(1 - A) u + 2(A^2   -1) ]
</math>
</math>
   </td>
   </td>
Line 1,944: Line 2,049:
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~A^3(u-1)^{1 / 2} (Au-1)^{-5 / 2} \biggl\{ Au^2 \biggl[ 5A-3 \biggr]   
+ 2a(A - 1) \biggl[7Au^2 - (6A+7)u  +6 \biggr] + (4A^2u^2-8Au + 4)  
+ u\biggl[ -3A^2 -7A +6\biggr] + 2(3A-2) \biggr\} \, .
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
 
=====Second 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">
Line 1,962: Line 2,073:
   <td align="left">
   <td align="left">
<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]
(u - 1)^{b / 2} (A u - 1 )^{-a / 2}   \, ,
</math>
</math>
   </td>
   </td>
Line 1,968: Line 2,079:
</table>
</table>
</div>
</div>
This should then match the [[#First_Guess_.28n5.29|"first guess"]] algebraic condition if we set <math>~a=1</math>.  Let's see.
in which case,
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 1,975: Line 2,085:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~0</math>
<math>~\frac{dx}{du}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,982: Line 2,092:
   <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]
\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>
   </td>
   </td>
Line 1,995: Line 2,106:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~x \biggl[
2Au^2 [4  - 4A +  7A - 7 + 2A] + 2u [ (A^2 - 2A + 1) +  2A^-2 + (1-A ) (6A+7) -4A] + 4[ 3A-2]
\frac{b}{2}(u-1)^{-1}
- \frac{aA}{2(A u - 1 )^{-1} \biggr]
</math>
</math>
   </td>
   </td>
Line 2,003: 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 2,009: Line 2,121:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~ (A u - 1 )^{-1} \biggl[  
2Au^2 [5A - 3] + 2u [ - 3A^2 - 7A + 6 ] + 4[ 3A-2] \, .
\frac{b}{2} (A u - 1 )
- \frac{aA}{2} (u-1) \biggr]
</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:
<tr>
   <td align="right">
 
&nbsp;
<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">
Line 2,032: Line 2,136:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~\frac{1 }{2(A u - 1 )} \biggl[  
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]
b (A u - 1 ) - aA (u-1) \biggr]
</math>
</math>
   </td>
   </td>
Line 2,047: Line 2,151:
   <td align="left">
   <td align="left">
<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]
\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">
&nbsp;
<math>~\frac{d^2x}{du^2}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,061: Line 2,171:
   <td align="left">
   <td align="left">
<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] \, .
\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>
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 2,079: Line 2,185:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~0</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>
   </td>
   </td>
</tr>
</tr>
Line 2,085: Line 2,194:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\Rightarrow ~~~ a</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,091: Line 2,200:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{2A}{3(1-A)} \, ;</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>
   </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>
<math>~\Rightarrow ~~~ \frac{(1-u)^2}{x}\frac{d^2x}{du^2}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,108: Line 2,216:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~3a(A-1) + 2A</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>
   </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">
<math>~\Rightarrow A</math>
<math>~0</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,120: Line 2,240:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{1}{2} </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>
   </td>
   </td>
</tr>
</tr>
Line 2,126: Line 2,248:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\Rightarrow ~~~ a</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,132: Line 2,254:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{2}{3} \, ;</math>
<math>~
   </td>
\frac{2u}{4 (Au-1)^2} \biggl\{
</tr>
\biggl[ (aA - b) + A(b    - a)u \biggr]^2
</table>
+  \biggl[  2aA^2 (u-1)^{2} -2b (A u - 1 )^{2} \biggr]
</div>
\biggr\}
which forces the second expression to the value,
</math>
<div align="center">
   </td>
<table border="0" cellpadding="5" align="center">
</tr>


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~a^2 (A - 1)^2 +  a(  -4A^2-A+5) - 4A</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
&nbsp;
   </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{(7u-6) }{2(A u - 1 )} \biggl[  
(aA - b) + A(b    - a)u \biggr]  
+ 1
</math>
   </td>
   </td>
</tr>
</tr>
Line 2,161: Line 2,287:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{1}{9} +  \frac{7}{3} - 2</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>
   </td>
   </td>
</tr>
</tr>
Line 2,170: Line 2,300:
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
&nbsp;
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{4}{9}  \, ,</math>
<math>~
+ 2(A u - 1 )(7u-6) \biggl[
(aA - b) + A(b    - a)u \biggr]
+ 4 (Au-1)^2 \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==
<tr>
 
  <td align="right">
===Alternative Terms===
&nbsp;
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:
  </td>
<div align="center">
  <td align="center">
<table border="0" cellpadding="5" align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
  </td>
</tr>


<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>
&nbsp;
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl( \frac{dm_0}{dr_0} \biggr)\frac{d}{dm_0}
<math>~
= \biggl( \frac{dm_0}{d\xi} \cdot \frac{d\xi}{dr_0} \biggr)\frac{d}{dm_0}
+ 2\biggl[7Au^2 - (6A+7)u  +6 \biggr]\biggl[
\, ,</math>
(aA - b) + A(b    - a)u \biggr]
+ (4A^2u^2-8Au + 4) \biggr\}
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</div>
</div>
so what is the expression for the leading coefficient?  From [[#r0|above]], we have,
 
If <math>~b=a</math>,


<div align="center">
<div align="center">
Line 2,209: Line 2,352:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~r_0</math>
<math>~0</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,216: Line 2,359:
   <td align="left">
   <td align="left">
<math>~
<math>~
R_*  \biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3} \xi
2u\biggl[ (aA - b)^2   \biggr]
+  2u\biggl[ 4A(b - aA) u + 2(aA^2   -b)  \biggr]
</math>
</math>
   </td>
   </td>
Line 2,223: Line 2,367:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\Rightarrow ~~~ \xi</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
&nbsp;
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
\frac{1}{R_*}  \biggl[ \frac{ {\tilde\xi}^2}{(3 + {\tilde\xi}^2)}\biggr]^{3} r_0 \, .
+ 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>
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 2,250: Line 2,388:
   </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}
2a^2u (A - 1)^2
\biggl\{ \xi^3 ( 3 + \xi^2 )^{-3/2} \biggr\}
+  2au [ 4A(1 - A) u + 2(A^2   -1) ]
</math>
</math>
   </td>
   </td>
Line 2,259: Line 2,397:
<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">
<math>~=</math>
&nbsp;
   </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}
+ 2a(A - 1) \biggl[7Au^2 - (6A+7)u  +6 \biggr] (4A^2u^2-8Au + 4)  
\biggl\{ 3\xi^2 ( 3 + \xi^2 )^{-3/2}
- 3 \xi^4 ( 3 + \xi^2 )^{-5/2}\biggr\}
</math>
</math>
   </td>
   </td>
</tr>
</tr>
 
<tr>
<tr>
   <td align="right">
   <td align="right">
Line 2,281: Line 2,417:
   </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}
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\{ ( 3 + \xi^2 )  
- \xi^2  \biggr\}
</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">
&nbsp;
<math>~0</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,297: Line 2,437:
   </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}
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>
Line 2,305: Line 2,445:
<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}
2Au^2 [4  - 4A + 7A - 7 + 2A] + 2u [ (A^2 - 2A + 1) +  2A^2   -2 + (1-A ) (6A+7) -4A] + 4[ 3A-2]
\frac{1}{R_*}  \biggl[ \frac{ {\tilde\xi}^2}{(3 + {\tilde\xi}^2)}\biggr]^{3}
</math>
</math>
   </td>
   </td>
Line 2,321: Line 2,460:
   <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{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} \, .
2Au^2 [5A - 3] + 2u [  - 3A^2 - 7A + 6 ] + 4[ 3A-2] \, .
</math>
</math>
   </td>
   </td>
Line 2,333: Line 2,472:
</table>
</table>
</div>
</div>
And we see that this expression ''does'' match the one derived earlier.


To simplify expressions, let's [[User:Tohline/Appendix/Ramblings/Nonlinar_Oscillation#DefineTildeC|borrow from an accompanying derivation]] and define,
Going back a bit, before setting <math>~a=1</math>, we have the expression:
<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">
<div align="center">
Line 2,345: Line 2,482:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\frac{m_0}{M_\mathrm{tot}}</math>
<math>~0</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,351: Line 2,488:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
\biggl[ \frac{\tilde{C}}{ 3}\biggr]^{3 / 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[ \frac{\xi^2}{ ( 3 + \xi^2 )} \biggr]^{3/2}
</math>
</math>
   </td>
   </td>
Line 2,360: Line 2,496:
<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">
Line 2,366: Line 2,502:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~   
<math>~
\frac{\xi^2}{ ( 3 + \xi^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]
</math>
</math>
   </td>
   </td>
Line 2,374: Line 2,510:
<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 2,380: Line 2,516:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~   
<math>~
\xi^2  
2Au^2 [ 3a(A  - 1) + 2A] + 2u [ a^2 (A - 1)^2 +  a(  -4A^2-A+5) - 4A] + 4[ 3a(A-1) + 1] \, .
</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>~\Rightarrow ~~~3 m_*</math>
<math>~3a(A  - 1) + 2A</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,394: Line 2,535:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ \xi^2 (1-m_*)
<math>~0</math>
</math>
   </td>
   </td>
</tr>
</tr>
Line 2,401: Line 2,541:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\Rightarrow ~~~\xi^2 </math>
<math>~\Rightarrow ~~~ a</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,407: Line 2,547:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~   \frac{3m_*}{(1-m_*)} \, ,
<math>~\frac{2A}{3(1-A)} \, ;</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</div>
</div>
where,
and, third, by simple visual comparison with the first expression,
<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">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 2,424: Line 2,558:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~
<math>~3a(A-1) + 1</math>
\frac{\xi^2}{ ( 3 + \xi^2 )} = m_* \, ;
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp; &nbsp; &nbsp; while, &nbsp; &nbsp; &nbsp;
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~3a(A-1) + 2A</math>
\frac{ {\tilde\xi}^2}{ ( 3 + {\tilde\xi}^2 )} = \frac{3}{\tilde{C}} \, ;
</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>
<math>~\Rightarrow A</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,451: Line 2,576:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~\frac{1}{2} </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 2,460: Line 2,582:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\frac{g_0\rho_0}{P_0} </math>
<math>~\Rightarrow ~~~ a</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,466: Line 2,588:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~\frac{2}{3} \, ;</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>
which forces the second expression to the value,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\frac{g_0 }{r_0} </math>
<math>~a^2 (A - 1)^2 +  a(  -4A^2-A+5) - 4A</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,484: 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>
\frac{GM_\mathrm{tot}}{R_*^3}
   </td>
\biggl\frac{ {\tilde\xi}^2 }{ (3 + {\tilde\xi}^2)}\biggr]^{15/2}  \frac{1}{\xi^3}
</tr>
\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>
</tr>


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\frac{\rho_0}{\gamma_g P_0} </math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,502: Line 2,617:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~\frac{1}{9} \frac{7}{3} - 2</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>
   </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 2,522: Line 2,629:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~\frac{4}{9} \, ,</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>
   </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">
&nbsp;
<math>~\frac{d}{dr_0}</math>
   </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{d^2x}{dr_0^2}
= \biggl( \frac{dm_0}{d\xi} \cdot \frac{d\xi}{dr_0}  \biggr)\frac{d}{dm_0}
+ \biggl\{ \frac{4}{R_*} \biggl( \frac{ 3}{ \tilde{C} }\biggr)^{3} \biggr[ \frac{ (1-m_*) }{3m_*}\biggr]^{1 / 2}
\, ,</math>
- \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>
   </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">
&nbsp;
<math>~r_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}
R_* \biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3} \xi
\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 2,562: Line 2,679:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\Rightarrow ~~~ \xi</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,569: Line 2,686:
   <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} r_0 \, .
+ \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>
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">
&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}  
M_\mathrm{tot} \biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3 / 2}
\biggl\{\sigma^2 +  (1-m_*)^{3 / 2} \biggr\} x
\biggl\{ \xi^3 ( 3 + \xi^2 )^{-3/2} \biggr\}  
</math>
</math>
   </td>
   </td>
Line 2,593: 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 2,599: Line 2,721:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{1 }{R_*^2} \biggl\frac{3 }{ \tilde{C} }\biggr)^{3} \biggl\{
<math>~
R_*^2 \biggl( \frac{ \tilde{C} }{3 }\biggr)^{3}  \frac{d^2x}{dr_0^2}  
M_\mathrm{tot} \biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3 / 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}  
\biggl\{ 3\xi^2 ( 3 + \xi^2 )^{-3/2}  
- 3 \xi^4 ( 3 + \xi^2 )^{-5/2}\biggr\}  
</math>
</math>
   </td>
   </td>
Line 2,611: Line 2,734:
   </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 \biggl[ \frac{3}{(1-m_*)}\biggr]^{1 / 2}  
M_\mathrm{tot} \biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3 / 2} 3\xi^2 (3 + \xi^2)^{-5/2}  
\biggl[ \sigma^2 +  (1-m_*)^{3 / 2} \biggr]  x \biggr\}  
\biggl\{ ( 3 + \xi^2 )
- \xi^2 \biggr\}  
</math>
</math>
   </td>
   </td>
Line 2,629: 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>~
R_*^2 \biggl( \frac{ \tilde{C} }{3 }\biggr)^{3}  \frac{d^2x}{dr_0^2}
M_\mathrm{tot} \biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3 / 2} 3^2\xi^2 (3 + \xi^2)^{-5/2}  
+ R_* \biggl[ 4  - 6\biggl(  \frac{ 3 }{ \tilde{C} }\biggr)^{6}  m_* \biggr] (1-m_*) \frac{dx}{dr_0}  
</math>
</math>
   </td>
   </td>
Line 2,638: 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>~
+ \frac{18(4 - 3\gamma_\mathrm{g}) }{\gamma_g } \cdot m_*^{1 / 2}
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[ \sigma^2  + (1-m_*)^{3 / 2} \biggr] \biggr\} \, ,
\frac{1}{R_*}  \biggl[ \frac{ {\tilde\xi}^2}{(3 + {\tilde\xi}^2)}\biggr]^{3}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
</math>
   </td>
   </td>
Line 2,653: Line 2,790:
</div>
</div>


where,
To simplify expressions, let's [[User:Tohline/Appendix/Ramblings/Nonlinar_Oscillation#DefineTildeC|borrow from an accompanying derivation]] and define,
<div align="center">
<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>
<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>
 
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>
</div>
Then we have,


We find,
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 2,669: Line 2,801:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~dr_0</math>
<math>~\frac{m_0}{M_\mathrm{tot}}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,675: Line 2,807:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
c_0 ~d[ m_*^{1 / 2} (1-m_*)^{-1 / 2} ]
\biggl[ \frac{\tilde{C}}{ 3}\biggr]^{3 / 2}
\biggl[ \frac{\xi^2}{ ( 3 + \xi^2 )} \biggr]^{3/2}  
</math>
</math>
   </td>
   </td>
Line 2,683: 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 2,689: Line 2,822:
   </td>
   </td>
   <td align="left">
   <td align="left">
<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}
\frac{\xi^2}{ ( 3 + \xi^2 )}  
\biggr] dm_*
</math>
</math>
   </td>
   </td>
Line 2,698: 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">
Line 2,704: Line 2,836:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
\frac{c_0}{2} ~m_*^{-1 / 2}( 1 - m_*)^{-3 / 2}~ dm_*
\xi^2  
</math>
</math>
   </td>
   </td>
Line 2,712: Line 2,844:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\frac{d}{dr_0}</math>
<math>~\Rightarrow ~~~3 m_*</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,718: Line 2,850:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~ \xi^2 (1-m_*)
\frac{2}{c_0} ~m_*^{1 / 2}( 1 - m_*)^{3 / 2}~ \frac{d}{dm_*}
</math>
</math>
   </td>
   </td>
Line 2,726: Line 2,857:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\Rightarrow ~~~ R_*\frac{dx}{dr_0}</math>
<math>~\Rightarrow ~~~\xi^2 </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,732: Line 2,863:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~   \frac{3m_*}{(1-m_*)} \, ,
\frac{2}{3^{1 / 2}}\biggl( \frac{ \tilde{C} }{ 3}\biggr)^{-3} ~m_*^{1 / 2}( 1 - m_*)^{3 / 2}~ \frac{dx}{dm_*} \, .
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</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>
</div>


Also,
In summary:
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 2,746: Line 2,880:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\frac{d^2}{dr_0^2}</math>
<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>~
<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{ {\tilde\xi}^2}{ ( 3 + {\tilde\xi}^2 )} = \frac{3}{\tilde{C}} \, ;
</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">
&nbsp;
<math>~r_0</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,767: Line 2,908:
   <td align="left">
   <td align="left">
<math>~
<math>~
\biggl( \frac{2}{c_0} \biggr)^{2}~m_* ( 1 - m_*)^{3 }~ \frac{d^2}{dm_*^2}   
R_*  \biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3} \xi
+\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}~
= R_*  \biggl( \frac{ \tilde{C} }{ 3}\biggr)^{3} \biggr[ \frac{3m_*}{ (1-m_*) }\biggr]^{1 / 2} \, ;
\biggr] ~ \frac{d}{dm_*}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\frac{g_0\rho_0}{P_0} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
   </td>
   </td>
Line 2,776: Line 2,934:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\frac{g_0 }{r_0} </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,782: Line 2,940:
   </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_*}
\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>
   </td>
   </td>
Line 2,790: Line 2,952:
<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>
<math>~\frac{\rho_0}{\gamma_g P_0} </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,796: Line 2,958:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
\biggl[\frac{2^2}{3} \biggl(\frac{ \tilde{C} }{3} \biggr)^{-3} \biggr] 
\frac{6R_* }{\gamma_g GM_\mathrm{tot} }\biggl( \frac{ 3}{ \tilde{C} } \biggr)^{9 / 2} \biggl[ \frac{3}{(1-m_*)}\biggr]^{1 / 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 2,804: Line 2,965:
</table>
</table>
</div>
</div>
 
So, the wave equation may be written as,
So, the wave equation becomes,


<div align="center">
<div align="center">
Line 2,818: Line 2,978:
   </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
\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[ ~m_* ( 1 - m_*)^{3 }~ \frac{d^2x}{dm_*^2+\frac{1}{2} ~ ( 1 - m_*)^{2} ( 1 - 4m_*) \frac{dx}{dm_*} \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 2,830: Line 2,990:
   </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]  
\frac{d^2x}{dr_0^2}
+ \frac{18(4 - 3\gamma_\mathrm{g}) }{\gamma_g } \cdot m_*^{1 / 2}
+ \biggl\{ \frac{4}{R_*} \biggl( \frac{ 3}{ \tilde{C} }\biggr)^{3} \biggr[ \frac{ (1-m_*) }{3m_*}\biggr]^{1 / 2}
\biggl[ \sigma^2 +  (1-m_*)^{3 / 2} \biggr]  x  \biggr\}  
- \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>
Line 2,846: Line 3,006:
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
&nbsp;
   </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>~
\biggl[\frac{2^2}{3} \biggr] 
+ \frac{6R_* }{\gamma_g GM_\mathrm{tot} }\biggl( \frac{ 3}{ \tilde{C} } \biggr)^{9 / 2} \biggl[ \frac{3}{(1-m_*)}\biggr]^{1 / 2}  
\biggl[ ~m_* ( 1 - m_*)^{3 }~ \frac{d^2x}{dm_*^2} +\frac{1}{2} ~ ( 1 - m_*)^{2} ( 1 - 4m_*) \frac{dx}{dm_*} \biggr]
\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 2,861: Line 3,021:
   </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)^{6m_* \biggr] (1-m_*) \biggl[ \frac{2}{3^{1 / 2}} ~m_*^{1 / 2}( 1 - m_*)^{3 / 2}~ \frac{dx}{dm_*} \biggr]
\frac{d^2x}{dr_0^2}
+ \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\}  
+ \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 2,876: Line 3,037:
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
&nbsp;
   </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{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>
   </td>
   </td>
Line 2,890: Line 3,052:
   </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)^{3} \biggl\{
+ \biggl[ 4  - 6\biggl(  \frac{ 3 }{ \tilde{C} }\biggr)^{6} m_* \biggr] (1-m_*)^2 \frac{dx}{dm_*}
R_*^2 \biggl(  \frac{ \tilde{C} }{3 }\biggr)^{3\frac{d^2x}{dr_0^2}  
+ \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\}  
+ 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>
Line 2,905: Line 3,067:
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
&nbsp;
   </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} 
+ \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>
Line 2,919: Line 3,082:
   </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)^{3} \biggl[ \frac{1}{3m_*(1-m_*)}\biggr]^{1 / 2} \biggl\{ [ 3m_*(1-m_*) ]^{1 / 2}  
+ \biggl[ 5 - 4m_* - 6\biggl(  \frac{ 3 }{ \tilde{C} }\biggr)^{6} m_* \biggr] (1-m_*)^2 \frac{dx}{dm_*
R_*^2 \biggl( \frac{ \tilde{C} }{3 }\biggr)^{3} \frac{d^2x}{dr_0^2}
+ \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\}  
+ R_* \biggl[ 4  - 6\biggl(  \frac{ 3 }{ \tilde{C} }\biggr)^{6} m_* \biggr] (1-m_*) \frac{dx}{dr_0}  
</math>
</math>
   </td>
   </td>
Line 2,934: Line 3,097:
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
&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{18(4 - 3\gamma_\mathrm{g}) }{\gamma_g } \cdot m_*^{1 / 2}
+ \mathcal{B} \biggl[ \frac{\sigma^2}{(1-m_*)^{1 / 2}}  +  (1-m_*) \biggr]  x  \biggr\} \, ,
\biggl[ \sigma^2 (1-m_*)^{3 / 2} \biggr]  x  \biggr\} \, ,
</math>
</math>
   </td>
   </td>
Line 2,945: Line 3,108:
</table>
</table>
</div>
</div>
where,
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 2,951: Line 3,125:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\mathcal{A}</math>
<math>~dr_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>~
c_0 ~d[ m_*^{1 / 2} (1-m_*)^{-1 / 2} ]
</math>
   </td>
   </td>
</tr>
</tr>
Line 2,963: Line 3,139:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\mathcal{B}</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{3^{5/2}(4 - 3\gamma_\mathrm{g}) }{\gamma_g } \biggl( \frac{ \tilde{C} }{ 3}\biggr)^{3} \, .</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>
   </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>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,991: Line 3,161:
   <td align="left">
   <td align="left">
<math>~
<math>~
\biggl(\frac{\xi}{\tilde\xi}\biggr)^3 \biggl(3 + \xi^2 \biggr)^{-3/2}  
\frac{c_0}{2} ~m_*^{-1 / 2}( 1 - m_*)^{-3 / 2}~ dm_*
\biggl(3 +  {\tilde\xi}^2 \biggr)^{3/2} </math>
</math>
   </td>
   </td>
</tr>
</tr>
Line 2,998: Line 3,168:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\frac{d}{dr_0}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 3,005: Line 3,175:
   <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{2}{c_0} ~m_*^{1 / 2}( 1 - m_*)^{3 / 2}~ \frac{d}{dm_*}
\, ,</math>
</math>
   </td>
   </td>
</tr>
</tr>
Line 3,012: Line 3,182:
<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>~\Rightarrow ~~~ R_*\frac{dx}{dr_0}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 3,018: Line 3,188:
   </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{2}{3^{1 / 2}}\biggl( \frac{ \tilde{C} }{ 3}\biggr)^{-3} ~m_*^{1 / 2}( 1 - m_*)^{3 / 2}~ \frac{dx}{dm_*} \, .
\biggl( 1+\frac{\tilde\xi^2}{3} \biggr)^{3}\biggr\}
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
Also,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\frac{d^2}{dr_0^2}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 3,034: Line 3,209:
   <td align="left">
   <td align="left">
<math>~
<math>~
\biggl[ \frac{\pi}{2^3\cdot 3^7}\biggr]^{1/2}  
\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{( 3+\tilde\xi^2)}{ {\tilde\xi}^2} \biggr]^{3} \xi \, .
</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">
Line 3,056: Line 3,222:
   </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} \, ,
\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>
   </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>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
\frac{3^2}{\tilde\xi^2}\biggl( 1 + \frac{\tilde\xi^2}{3} \biggr)
\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_*}
= 3 \biggl[ \frac{( 3+\tilde\xi^2)}{ {\tilde\xi}^2} \biggr]
\, ,
</math>
</math>
   </td>
   </td>
Line 3,085: Line 3,246:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\tilde{r}_\mathrm{edge}</math>
<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">
<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>~
= \biggl[ \frac{\pi}{2^3\cdot 3^7}\biggr]^{1 / 2} \biggl[ \frac{\tilde{C}}{ 3} \biggr]^{3}
\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>
   </td>
   </td>
Line 3,100: Line 3,261:
</div>
</div>


If we furthermore define,
So, the wave equation becomes,
<div align="center">
 
<math>m_* \equiv \frac{3}{\tilde{C}} \cdot m_\xi^{2 / 3} \, ,</math>
</div>
then,
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 3,110: Line 3,268:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~
<math>~0</math>
r_\xi (m_*)
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 3,118: 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}
3^{1 / 2} \tilde{r}_\mathrm{edge} \biggl[\frac{m_*}{1-m_*}\biggr]^{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>
   </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">
<math>~=</math>
&nbsp;
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~3^{1 / 2} \tilde{r}_\mathrm{edge} \biggl\{
<math>~
\frac{1}{2} (1-m_*)^{- 1 / 2} m_*^{-1 / 2} + \frac{1}{2}m_*^{1 / 2}(1-m_*)^{-3 / 2}
+ \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]
\biggr\} dm_*
+ \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>
Line 3,156: Line 3,305:
   </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{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>
   </td>
   </td>
Line 3,163: Line 3,314:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>\Rightarrow ~~~ R_\mathrm{norm} \cdot \frac{d}{dr_0} </math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
&nbsp;
   </td>
   </td>
   <td align="left">
   <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>~
+ \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>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
We therefore also have,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~ R^2_\mathrm{norm} \cdot \frac{d^2}{dr_0^2} </math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 3,188: Line 3,335:
   </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>~\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 3,198: Line 3,346:
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
&nbsp;
   </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\{  
+ \biggl[ 4  - 6\biggl( \frac{ 3 }{ \tilde{C} }\biggr)^{6} m_* \biggr] (1-m_*)^2 \frac{dx}{dm_*} 
\biggl[ m_*^{1 / 2} (1-m_*)^{3 / 2} \frac{d^2}{dm_*^2}\biggr]
+ \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]  \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>
   </td>
   </td>
Line 3,218: Line 3,364:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ \frac{1}{ {\tilde{r}}^2_\mathrm{edge}} \biggl( \frac{2}{3} \biggr) \biggl\{   
<math>~\frac{2 }{3R_*^2} \biggl( \frac{3 }{ \tilde{C} }\biggr)^{6} 
\biggl[ 2m_* (1-m_*)^{3} \frac{d^2}{dm_*^2}\biggr]
\biggl\{  2m_* ( 1 - m_*)^{3 }~ \frac{d^2x}{dm_*^2}   
+ \biggl[  (1-m_*)^{3 } + 3m_* (1-m_*)^{2}\biggr] \frac{d}{dm_*}
\biggr\}
</math>
</math>
   </td>
   </td>
Line 3,231: Line 3,375:
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
&nbsp;
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~   
<math>~
\frac{1}{ {\tilde{r}}^2_\mathrm{edge}} \biggl( \frac{2}{3} \biggr) \biggl\{
+ \biggl[ 5 - 4m_*  - 6\biggl( \frac{ 3 }{ \tilde{C} }\biggr)^{6}  m_* \biggr] (1-m_*)^2 \frac{dx}{dm_*} 
2m_* (1-m_*)^{3} \frac{d^2}{dm_*^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\}  
+ (1-m_*)^{2} ( 1 + 2m_* ) \frac{d}{dm_*} \biggr\}  
\, .
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
So the wave equation may be written,
<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 3,257: Line 3,393:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~\frac{2 }{3R_*^2} \biggl\frac{3 }{ \tilde{C} }\biggr)^{6} \biggl\{ 2m_* ( 1 - m_*)^{3 }~ \frac{d^2x}{dm_*^2}
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}  
+ (5 - \mathcal{A}  m_*) (1-m_*)^2 \frac{dx}{dm_*}
+ \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
+ \mathcal{B} \biggl[ \frac{\sigma^2}{(1-m_*)^{1 / 2}} +  (1-m_*) \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>~\mathcal{A}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~\equiv</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~4 + 6\biggl( \frac{ 3 }{ \tilde{C} }\biggr)^{6} \, ,</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>
   </td>
</tr>
</tr>
Line 3,282: Line 3,419:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\mathcal{B}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~\equiv</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~\frac{3^{5/2}(4 - 3\gamma_\mathrm{g}) }{\gamma_g } \biggl( \frac{ \tilde{C} }{ 3}\biggr)^{3} \, .</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>
   </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">
&nbsp;
<math>~m_\xi \equiv \frac{m_0}{ M_\mathrm{tot} } = \frac{M_r(\xi)}{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>~
+ \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{\xi}{\tilde\xi}\biggr)^3 \biggl(3 + \xi^2 \biggr)^{-3/2}  
\biggl\{
\biggl(3 {\tilde\xi}^2 \biggr)^{3/2} </math>
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>
   </td>
</tr>
</tr>
</table>
 
</div>
<tr>
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">
   <td align="right">
<math>~0</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 3,332: Line 3,461:
   <td align="left">
   <td align="left">
<math>~
<math>~
\frac{1}{ {\tilde{r}}^2_\mathrm{edge}} \biggl( \frac{2}{3} \biggr) \biggl\{
\biggl[ \frac{( 3+\tilde\xi^2)}{ {\tilde\xi}^2} \biggr]^{3 / 2}\biggl[ \frac{( 3+\xi^2)}{ {\xi}^2} \biggr]^{- 3 / 2}
2m_* (1-m_*)^{3} \frac{d^2x}{dm_*^2}
\, ,</math>
+ (1-m_*)^{2} ( 1 + 2m_* )  \frac{dx}{dm_*} \biggr\}
</math>
   </td>
   </td>
</tr>
</tr>
Line 3,341: Line 3,468:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<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>
   <td align="center">
   <td align="center">
&nbsp;
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~\xi \biggl\{  
+\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[ \frac{4\pi}{2^5\cdot 3}\biggr]^{1/2} \tilde\xi^{-6}
- 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\}
\biggl( 1+\frac{\tilde\xi^2}{3} \biggr)^{3}\biggr\}
m_*^{1 / 2} (1-m_*)^{3 / 2} \frac{dx}{dm_*}  
</math>
</math>
   </td>
   </td>
Line 3,360: Line 3,486:
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<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[ \frac{\pi}{2^3\cdot 3^7}\biggr]^{1/2}  
\biggl\{
\biggl[ \frac{( 3+\tilde\xi^2)}{ {\tilde\xi}^2} \biggr]^{3} \xi \, .
\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>
   </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">
&nbsp;
<math>~
r_\xi (m_\xi)
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 3,381: Line 3,512:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~\tilde{r}_\mathrm{edge}
\frac{1}{ {\tilde{r}}^2_\mathrm{edge}} \biggl( \frac{2}{3} \biggr) \biggl\
\biggl[\frac{3^2m_\xi^{2/3}}{\tilde{C} - 3 m_\xi^{2/3}}\biggr]^{1/2} \, ,
2m_* (1-m_*)^{3} \frac{d^2x}{dm_*^2}
+ (1-m_*)^{2} ( 1 + 2m_* )  \frac{dx}{dm_*} \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>~\tilde{C}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~\equiv</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
+\frac{1}{ \tilde{r}_\mathrm{edge}^2} \biggl( \frac{2^3}{3} \biggr)
\frac{3^2}{\tilde\xi^2}\biggl( 1 + \frac{\tilde\xi^2}{3} \biggr)  
\biggl[ 1  - \frac{3}{2} \biggl( \frac{ 3 }{ \tilde{C} }\biggr)^{6}  m_*  \biggr]
= 3 \biggl[ \frac{( 3+\tilde\xi^2)}{ {\tilde\xi}^2} \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 3,411: Line 3,541:
<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>~
<math>~\biggl[ \frac{\pi}{2^3\cdot 3}\biggr]^{1/2} {\tilde\xi}^{-6} \biggl(1+\frac{\tilde\xi^2}{3}\biggr)^3
\frac{1}{ {\tilde{r}}^2_\mathrm{edge}} \biggl( \frac{2}{3} \biggr) \biggl\{ 
= \biggl[ \frac{\pi}{2^3\cdot 3^7}\biggr]^{1 / 2} \biggl[ \frac{\tilde{C}}{ 3} \biggr]^{3}  
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 3,431: Line 3,555:
</table>
</table>
</div>
</div>
where, as before,
 
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">
<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>
<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>
</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 align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2 b^2    - 3b + 3
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow~~~b</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{3}{4}\biggl[1 \pm \sqrt{1-\frac{8}{3} }  \biggr] \, .
</math>
  </td>
</tr>
</table>
</div>
Thus, <math>~b</math> is an complex number.


=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>


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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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