SSC/Stability/BiPolytropes/RedGiantToPN/Pt3: Difference between revisions

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</tr>  
</tr>  
<td align="center">
<td align="center">
<math>C_0 = \frac{d^2x}{d\xi^2} + \biggl[ 4 - (n+1) Q \biggr] \frac{1}{\xi} \cdot \frac{dx}{d\xi}   
<math>C_0 = \frac{d^2x}{d\eta^2} + \biggl[ 4 - (n+1) Q \biggr] \frac{1}{\eta} \cdot \frac{dx}{d\eta}   
+ (n+1) \biggl[ \biggl( \frac{\sigma_c^2}{6\gamma_g } \biggr)  \frac{\xi^2}{\theta}  
+ (n+1) \biggl[ \biggl( \frac{\sigma_c^2}{6\gamma_g } \biggr)  \frac{\eta^2}{\theta}  
- \alpha Q\biggr]  \frac{x}{\xi^2} </math>
- \alpha Q\biggr]  \frac{x}{\eta^2} </math>
</td>
</td>
</tr>
</table>
After setting <math>\sigma_c^2 = 0</math>, let's guess an eigenfunction of the form,
<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>
\eta^{-m}~~~\Rightarrow ~~~ \frac{dx}{d\eta} = -m\eta^{-m-1}
</math>
&nbsp; &nbsp; &nbsp; and &nbsp; &nbsp;
<math> \frac{d^2x}{d\eta^2} = -m(-m-1)\eta^{-m-2} \, ,</math>
  </td>
</tr>
</table>
in which case we find that,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>C_0</math>
  </td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
-m(-m-1)\eta^{-m-2}
+ \biggl[ 4 - (n+1) Q \biggr] \frac{1}{\eta} \biggl[ -m\eta^{-m-1} \biggr]
+ (n+1) \biggl[  - \alpha Q\biggr]  \eta^{-m - 2}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\biggl\{~
m(m+1)
- \biggl[ 4 - (n+1) Q \biggr]  \biggl[ m \biggr]
+ (n+1) \biggl[  - \alpha Q\biggr]
~\biggr\} \eta^{-m - 2}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\biggl\{~
m(m+1) -4m
+\biggl[ m(n+1) \biggr]  Q
- \biggl[  \alpha(n+1)  \biggr]Q
~\biggr\} \eta^{-m - 2}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\biggl\{~
m^2 - 3m
+ (n+1)( m - \alpha )  Q
~\biggr\} \eta^{-m - 2}
\, .
</math>
  </td>
</tr>
</table>
The dependence on <math>Q</math> can be eliminated if we set <math>m = \alpha</math>; in which case we obtain,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>C_0</math>
  </td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\biggl\{~
\alpha^2 - 3\alpha
~\biggr\} \eta^{-m - 2}
\, .
</math>
  </td>
</tr>
</tr>
</table>
</table>

Latest revision as of 13:13, 6 January 2026

Main Sequence to Red Giant to Planetary Nebula (Part 3)[edit]


Part I:  Background & Objective

 


Part II: 

 


Yabushita68-Motivated Analysis

 


Part IV: 

 

Yabushita68-Motivated Analysis[edit]

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


Polytropic LAWE

0=d2xdξ2+[4−(n+1)Q]1ξ⋅dxdξ+(n+1)[(σc26γg)ξ2θ−αQ]xξ2

where:    Q(ξ)≡−dln⁡θdln⁡ξ,    σc2≡3ω22πGρc,     and,     α≡(3−4γg)

Motivated by the derivation presented by 📚 S. Yabushita (1968, MNRAS, Vol. 140, pp. 109 - 120), let's now insert an integration constant, C0, into this 2nd-order ODE to obtain what we henceforth will refer to as the,

Yabushita68-Motivated Polytropic LAWE

C0=d2xdη2+[4−(n+1)Q]1η⋅dxdη+(n+1)[(σc26γg)η2θ−αQ]xη2

After setting σc2=0, let's guess an eigenfunction of the form,

x

=

η−m⇒dxdη=−mη−m−1       and     d2xdη2=−m(−m−1)η−m−2,

in which case we find that,

C0

=

−m(−m−1)η−m−2+[4−(n+1)Q]1η[−mη−m−1]+(n+1)[−αQ]η−m−2

 

=

{m(m+1)−[4−(n+1)Q][m]+(n+1)[−αQ]}η−m−2

 

=

{m(m+1)−4m+[m(n+1)]Q−[α(n+1)]Q}η−m−2

 

=

{m2−3m+(n+1)(m−α)Q}η−m−2.

The dependence on Q can be eliminated if we set m=α; in which case we obtain,

C0

=

{α2−3α}η−m−2.

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