Editing Appendix/Ramblings/SphericalWaveEquation

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=Playing With Spherical Wave Equation=

The [[SSC/Perturbations#Spherically_Symmetric_Configurations_.28Stability_.E2.80.94_Part_II.29|traditional presentation of the (spherically symmetric) adiabatic wave equation]] focuses on fractional radial displacements, <math>~x \equiv \delta r/r_0</math>, of spherical mass shells.  After studying in depth various [[Apps/PapaloizouPringle84#Nonaxisymmetric_Instability_in_Papaloizou-Pringle_Tori|stability analyses of Papaloizou-Pringle tori]], I have begun to wonder whether the wave equation for spherical polytropes might look simpler if we focus, instead, on fluctuations in the fluid entropy.

==Assembling the Key Relations==
In the traditional approach, the following three linearized equations describe the physical relationship between the three dimensionless perturbation amplitudes <math>~p(r_0) \equiv P_1/P_0</math>, <math>~d(r_0) \equiv \rho_1/\rho_0</math> and <math>~x(r_0) \equiv r_1/r_0</math>, for various characteristic eigenfrequencies, <math>~\omega</math>:

<div align="center">
<table border="1" cellpadding="10">
<tr><td align="center">
<font color="#770000">'''Linearized'''</font><br />
<span id="Continuity"><font color="#770000">'''Equation of Continuity'''</font></span><br />
<math>
r_0 \frac{dx}{dr_0} = - 3 x - d ,
</math><br />

<font color="#770000">'''Linearized'''</font><br />
<span id="PGE:Euler"><font color="#770000">'''Euler + Poisson Equations'''</font></span><br />
<math>
\frac{P_0}{\rho_0} \frac{dp}{dr_0}  = (4x + p)g_0 + \omega^2 r_0 x ,
</math><br />

<font color="#770000">'''Linearized'''</font><br />
<span id="PGE:AdiabaticFirstLaw">Adiabatic Form of the<br />
<font color="#770000">'''First Law of Thermodynamics'''</font></span><br />
<math>
p = \gamma_\mathrm{g} d  \, .
</math>
</td></tr>
</table>
</div>


<!--
Let's use the second expression to define the radial perturbation, <math>~x</math>.  That is,
<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>~\frac{1}{(4g_0 + \omega^2 r_0 ) } \biggl[\frac{P_0}{\rho_0} \frac{dp}{dr_0} - p g_0\biggr] \, .</math>
  </td>
</tr>
</table>
</div>

Differentiating this relation gives,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~ \frac{dx}{dr_0}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{(4g_0 + \omega^2 r_0 ) } \frac{d}{dr_0}\biggl[\frac{P_0}{\rho_0} \frac{dp}{dr_0} - p g_0\biggr] 
+ \biggl[\frac{P_0}{\rho_0} \frac{dp}{dr_0} - p g_0\biggr] \frac{d}{dr_0}\biggl(4g_0 + \omega^2 r_0 \biggr)^{-1}  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{(4g_0 + \omega^2 r_0 ) } \biggl\{ \frac{d}{dr_0}\biggl[\frac{P_0}{\rho_0} \frac{dp}{dr_0}\biggr] - \frac{d}{dr_0}\biggl[p g_0\biggr] \biggr\}
- \biggl(4g_0 + \omega^2 r_0 \biggr)^{-2} \biggl[\frac{P_0}{\rho_0} \frac{dp}{dr_0} - p g_0\biggr] \frac{d}{dr_0}\biggl(4g_0 + \omega^2 r_0 \biggr)  
</math>
  </td>
</tr>
</table>
</div>
-->

===First Effort===
Let's switch from the perturbation variable, <math>~p</math>, to an enthalpy-related variable,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~W</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\frac{P_1}{\rho_0} = \biggl(\frac{P_0}{\rho_0}\biggr) p \, .</math>
  </td>
</tr>
</table>
</div>

The second expression then becomes,

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

<tr>
  <td align="right">
<math>~x(4g_0 + \omega^2 r_0)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{P_0}{\rho_0} \frac{d}{dr_0}\biggl(\frac{W\rho_0}{P_0}\biggr)  - \biggl(\frac{g_0 \rho_0}{P_0}\biggr)W</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{dW}{dr_0}
+ \frac{W}{\rho_0} \frac{d\rho_0}{dr_0} 
- \frac{W }{P_0} \frac{dP_0}{dr_0} 
- \biggl(\frac{g_0 \rho_0}{P_0}\biggr)W</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{dW}{dr_0}+ \frac{W}{\rho_0} \frac{d\rho_0}{dr_0} \, .
</math>
  </td>
</tr>
</table>
</div>
Taking the derivative of this expression with respect to <math>~r_0</math> gives,

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

<tr>
  <td align="right">
<math>~\frac{dx}{dr_0}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{d}{dr_0}\biggl\{
(4g_0 + \omega^2 r_0)^{-1}\biggl[\frac{dW}{dr_0} + \frac{W}{\rho_0} \frac{d\rho_0}{dr_0} \biggr]
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(4g_0 + \omega^2 r_0)^{-1}\frac{d}{dr_0}
\biggl[\frac{dW}{dr_0} + \frac{W}{\rho_0} \frac{d\rho_0}{dr_0} \biggr]
+\biggl[\frac{dW}{dr_0} + \frac{W}{\rho_0} \frac{d\rho_0}{dr_0} \biggr]\frac{d}{dr_0}
(4g_0 + \omega^2 r_0)^{-1}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(4g_0 + \omega^2 r_0)^{-1} \biggl\{
\frac{d^2W}{dr^2_0} + \frac{d}{dr_0}\biggl[\frac{W}{\rho_0} \frac{d\rho_0}{dr_0} \biggr] \biggr\}
-(4g_0 + \omega^2 r_0)^{-2}\biggl[\frac{dW}{dr_0} + \frac{W}{\rho_0} \frac{d\rho_0}{dr_0} \biggr]
\biggl\{ 4\frac{dg_0}{dr_0} + \omega^2
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~~
(4g_0 + \omega^2 r_0)^{2} \biggl[ \frac{dx}{dr_0} \biggr]
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(4g_0 + \omega^2 r_0)\biggl\{
\frac{d^2W}{dr^2_0} + \frac{d}{dr_0}\biggl[\frac{W}{\rho_0} \frac{d\rho_0}{dr_0} \biggr] \biggr\}
-\biggl[\frac{dW}{dr_0} + \frac{W}{\rho_0} \frac{d\rho_0}{dr_0} \biggr]
\biggl\{ 4\frac{dg_0}{dr_0} + \omega^2 
\biggr\} \, .
</math>
  </td>
</tr>
</table>
</div>

Hence, the linearized equation of continuity becomes,

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

<tr>
  <td align="right">
<math>~- (4g_0 + \omega^2 r_0)^{2}\biggl(\frac{W\rho_0}{\gamma_g r_0P_0}\biggr) </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(4g_0 + \omega^2 r_0)^{2} \biggl[ \frac{dx}{dr_0} \biggr] +\frac{3 (4g_0 + \omega^2 r_0)}{r_0} \biggl[ (4g_0 + \omega^2 r_0)x \biggr] </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(4g_0 + \omega^2 r_0)\biggl\{
\frac{d^2W}{dr^2_0} + \frac{d}{dr_0}\biggl[\frac{W}{\rho_0} \frac{d\rho_0}{dr_0} \biggr] \biggr\}
-\biggl[\frac{dW}{dr_0} + \frac{W}{\rho_0} \frac{d\rho_0}{dr_0} \biggr]
\biggl\{ 4\frac{dg_0}{dr_0} + \omega^2 
\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+\frac{3 (4g_0 + \omega^2 r_0)}{r_0} \biggl[ \frac{dW}{dr_0}+ \frac{W}{\rho_0} \frac{d\rho_0}{dr_0} \biggr] 
</math>
  </td>
</tr>
</table>
</div>



===Second Effort===
====Direct Approach====
Let's switch from the perturbation variable, <math>~p</math>, to an enthalpy-related variable,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~W</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\frac{P_1}{\rho_0 {\bar\sigma}^2} = \biggl(\frac{P_0}{\rho_0 {\bar\sigma}^2}\biggr) p \, ,</math>
  </td>
</tr>
</table>
</div>
where,
<div align="center">
<math>~{\bar\sigma}^2 \equiv \frac{4g_0}{r_0} + \omega^2 \, .</math>
</div>

Note, as well, that,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~g_0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- \frac{1}{\rho_0} \cdot \frac{dP_0}{dr_0}</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~~ {\bar\sigma}^2 </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\omega^2
-\frac{4}{\rho_0 r_0} \cdot \frac{dP_0}{dr_0} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\omega^2
-\frac{4P_0}{\rho_0 r_0^2} \cdot \frac{d\ln P_0}{d\ln r_0} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~~ \frac{\rho_0 {\bar\sigma}^2 r_0^2}{P_0} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{\rho_0 r_0^2}{P_0} \cdot\omega^2- 4\frac{d\ln P_0}{d\ln r_0} 
</math>
  </td>
</tr>
</table>
</div>

The second expression then becomes,

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

<tr>
  <td align="right">
<math>~xr_0{\bar\sigma}^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{P_0}{\rho_0} \frac{d}{dr_0}\biggl(\frac{W\rho_0 {\bar\sigma}^2}{P_0}\biggr)  - \biggl(\frac{g_0 \rho_0 {\bar\sigma}^2}{P_0}\biggr)W</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
{\bar\sigma}^2 \cdot \frac{dW}{dr_0}  
+ W \biggl[ \frac{P_0}{\rho_0} \frac{d}{dr_0}\biggl(\frac{\rho_0 {\bar\sigma}^2}{P_0}\biggr)  
- \biggl(\frac{g_0 \rho_0 {\bar\sigma}^2}{P_0}\biggr)\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~~ xr_0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{dW}{dr_0}  
+ \frac{W}{\rho_0{\bar\sigma}^2} \biggl[ P_0 \frac{d}{dr_0}\biggl(\frac{\rho_0 {\bar\sigma}^2}{P_0}\biggr)  
+ \biggl(\frac{\rho_0 {\bar\sigma}^2}{P_0}\biggr)\frac{dP_0}{dr_0} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{dW}{dr_0}  
+ W \biggl[ \frac{d \ln(\rho_0 {\bar\sigma}^2)}{dr_0}  \biggr] \, .
</math>
  </td>
</tr>
</table>
</div>

Taking the derivative of this expression with respect to <math>~r_0</math> gives,

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

<tr>
  <td align="right">
<math>~\frac{dx}{dr_0}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{d}{dr_0}\biggl\{\frac{1}{r_0}\biggl[ \frac{dW}{dr_0}  
+ W \cdot \frac{d \ln(\rho_0 {\bar\sigma}^2)}{dr_0}  \biggr] \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~~r_0 \frac{dx}{dr_0}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{d}{dr_0}\biggl[ \frac{dW}{dr_0}  
+ W \cdot \frac{d \ln(\rho_0 {\bar\sigma}^2)}{dr_0}  \biggr] 
- \frac{1}{r_0}\biggl[ \frac{dW}{dr_0}  
+ W \cdot \frac{d \ln(\rho_0 {\bar\sigma}^2)}{dr_0}  \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{d^2W}{dr_0^2}  
+ \frac{dW}{dr_0} \biggl[ \frac{d \ln(\rho_0 {\bar\sigma}^2)}{dr_0} -\frac{1}{r_0}\biggr] 
+ W \biggl\{ \frac{d^2 \ln(\rho_0 {\bar\sigma}^2)}{dr_0^2}
- \frac{1}{r_0}\biggl[  \frac{d \ln(\rho_0 {\bar\sigma}^2)}{dr_0}  \biggr]\biggr\} \, .
</math>
  </td>
</tr>
</table>
</div>

Hence, the linearized continuity equation gives,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~- \biggl(\frac{W\rho_0 {\bar\sigma}^2}{\gamma_gP_0} \biggr)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~r_0 \frac{dx}{dr_0} +3x</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{d^2W}{dr_0^2}  
+ \frac{dW}{dr_0} \biggl[ \frac{d \ln(\rho_0 {\bar\sigma}^2)}{dr_0} -\frac{1}{r_0}\biggr] 
+ W \biggl\{ \frac{d^2 \ln(\rho_0 {\bar\sigma}^2)}{dr_0^2}
- \frac{1}{r_0}\biggl[  \frac{d \ln(\rho_0 {\bar\sigma}^2)}{dr_0}  \biggr]\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+\frac{3}{r_0}\biggl[
\frac{dW}{dr_0}  
+ W \cdot\frac{d \ln(\rho_0 {\bar\sigma}^2)}{dr_0}  
\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{d^2W}{dr_0^2}  
+ \frac{dW}{dr_0} \biggl[ \frac{d \ln(\rho_0 {\bar\sigma}^2)}{dr_0} +\frac{2}{r_0}\biggr] 
+ W \biggl\{ \frac{d^2 \ln(\rho_0 {\bar\sigma}^2)}{dr_0^2}
+ \frac{2}{r_0}\biggl[  \frac{d \ln(\rho_0 {\bar\sigma}^2)}{dr_0}  \biggr]\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~~ 0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{d^2W}{dr_0^2}  
+ \frac{dW}{dr_0} \biggl[ \frac{d \ln(\rho_0 {\bar\sigma}^2)}{dr_0} +\frac{2}{r_0}\biggr] 
+ W \biggl\{ \frac{d^2 \ln(\rho_0 {\bar\sigma}^2)}{dr_0^2}
+ \frac{2}{r_0}\biggl[  \frac{d \ln(\rho_0 {\bar\sigma}^2)}{dr_0}  \biggr] + \biggl(\frac{\rho_0 {\bar\sigma}^2}{\gamma_gP_0} \biggr)\biggr\} \, .
</math>
  </td>
</tr>
</table>
</div>

====Playing Around====
Multiply thru by <math>~r_0^2</math>:

<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_0^2 \cdot \frac{d^2W}{dr_0^2}  
+ r_0 \cdot \frac{dW}{dr_0} \biggl[ \frac{d \ln(\rho_0 {\bar\sigma}^2)}{d\ln r_0} +2 \biggr] 
+ W \biggl\{ r_0^2 \cdot \frac{d^2 \ln(\rho_0 {\bar\sigma}^2)}{dr_0^2}
+ 2\biggl[  \frac{d \ln(\rho_0 {\bar\sigma}^2)}{d\ln r_0}  \biggr] + \biggl(\frac{\rho_0 {\bar\sigma}^2 r_0^2 }{\gamma_gP_0} \biggr)\biggr\} 
</math>
  </td>
</tr>
</table>
</div>

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

<tr>
  <td align="right">
<math>~r_0 \cdot \frac{d}{dr_0} \biggl[\frac{d\ln (\rho_0 {\bar\sigma}^2)}{d\ln r_0} \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~r_0 \cdot \frac{d}{dr_0} \biggl[r_0 \cdot \frac{d\ln (\rho_0 {\bar\sigma}^2)}{dr_0} \biggr]</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{d\ln (\rho_0 {\bar\sigma}^2)}{d\ln r_0} 
+ r_0^2 \cdot \frac{d^2\ln (\rho_0 {\bar\sigma}^2)}{dr_0^2} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~~
r_0^2 \cdot \frac{d^2\ln (\rho_0 {\bar\sigma}^2)}{dr_0^2} 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
r_0 \cdot \frac{d}{dr_0} \biggl[\frac{d\ln (\rho_0 {\bar\sigma}^2)}{d\ln r_0} \biggr]
- \frac{d\ln (\rho_0 {\bar\sigma}^2)}{d\ln r_0} 
</math>
  </td>
</tr>
</table>
</div>


<!--
SECOND
-->

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

<tr>
  <td align="right">
<math>~r_0 \cdot \frac{d}{dr_0} \biggl[\frac{dW}{d\ln r_0} \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~r_0 \cdot \frac{d}{dr_0} \biggl[r_0 \cdot \frac{dW}{dr_0} \biggr]</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{dW}{d\ln r_0} 
+ r_0^2 \cdot \frac{d^2W}{dr_0^2} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~~
r_0^2 \cdot \frac{d^2W}{dr_0^2} 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{d}{d\ln r_0} \biggl[\frac{dW}{d\ln r_0} \biggr] - \frac{dW}{d\ln r_0} 
</math>
  </td>
</tr>
</table>
</div>



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

<tr>
  <td align="right">
<math>~ \Rightarrow ~~~~ 0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{d}{d\ln r_0} \biggl[\frac{dW}{d\ln r_0} \biggr] - \frac{dW}{d\ln r_0}  
+ r_0 \cdot \frac{dW}{dr_0} \biggl[ \frac{d \ln(\rho_0 {\bar\sigma}^2)}{d\ln r_0} +2 \biggr] 
+ W \biggl\{ r_0 \cdot \frac{d}{dr_0} \biggl[\frac{d\ln (\rho_0 {\bar\sigma}^2)}{d\ln r_0} \biggr]
+ \frac{d\ln (\rho_0 {\bar\sigma}^2)}{d\ln r_0} 
+ \biggl(\frac{\rho_0 {\bar\sigma}^2 r_0^2 }{\gamma_gP_0} \biggr)\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{d}{d\ln r_0} \biggl[\frac{dW}{d\ln r_0} \biggr] 
+ \frac{dW}{d\ln r_0} \biggl[ \frac{d \ln(\rho_0 {\bar\sigma}^2)}{d\ln r_0} +1 \biggr] 
+ W \biggl\{ \frac{d}{d\ln r_0} \biggl[\frac{d\ln (\rho_0 {\bar\sigma}^2)}{d\ln r_0} \biggr]
+ \frac{d\ln (\rho_0 {\bar\sigma}^2)}{d\ln r_0} 
+ \biggl(\frac{\rho_0 {\bar\sigma}^2 r_0^2 }{\gamma_gP_0} \biggr)\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{d}{d\ln r_0} \biggl[\frac{dW}{d\ln r_0} \biggr] 
+ \frac{dW}{d\ln r_0} \biggl[ u +1 \biggr] 
+ W \biggl\{ \frac{du}{d\ln r_0} 
+ u 
+ \biggl(\frac{\rho_0 {\bar\sigma}^2 r_0^2 }{\gamma_gP_0} \biggr)\biggr\} \, ,
</math>
  </td>
</tr>
</table>
</div>
where,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~u</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\frac{d \ln(\rho_0 {\bar\sigma}^2)}{d\ln r_0}  \, .</math>
  </td>
</tr>
</table>
</div>

Let,
<div align="center">
<math>~y \equiv \ln r_0 </math>
&nbsp; &nbsp; &nbsp; <math>~\Rightarrow</math> &nbsp; &nbsp; &nbsp;
<math>~r_0 = e^{y} \, . </math>
</div>

Then we have,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{d^2W}{dy^2} 
+ \frac{dW}{dy} \biggl[ u +1 \biggr] 
+ W \biggl\{ \frac{du}{dy} 
+ u 
+ \biggl(\frac{\rho_0 {\bar\sigma}^2 e^{2y} }{\gamma_gP_0} \biggr)\biggr\} \, .
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{d^2W}{dy^2} + \frac{dW}{dy} \biggl[ u +1 \biggr] + W \biggl\{ \frac{du}{dy} + u 
+ \biggl[\frac{\rho_0 r_0^2}{P_0} \cdot\omega^2- 4\frac{d\ln P_0}{d\ln r_0}  \biggr] \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{d^2W}{dy^2} + \frac{dW}{dy} \biggl[ u +1 \biggr] + W \biggl\{ \frac{d(u- P_0^4)}{dy} + u 
+ \frac{\rho_0 r_0^2}{P_0} \cdot\omega^2\biggr\}
</math>
  </td>
</tr>
</table>
</div>

Therefore, it must also be the case that,

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


<tr>
  <td align="right">
<math>~u dy</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~d \ln(\rho_0 {\bar\sigma}^2) \, .</math>
  </td>
</tr>
</table>
</div>

=See Also=
* [[SSC/Perturbations#Spherically_Symmetric_Configurations_.28Stability_.E2.80.94_Part_II.29|Traditional Presentation of Adiabatic Wave Equation]]


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