Difference between revisions of "User:Tohline/Appendix/Ramblings/SphericalWaveEquation"

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(→‎Assembling the Key Relations: Work toward establishing eigenvalue problem in terms of W)
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===First Effort===
Let's switch from the perturbation variable, <math>~p</math>, to an enthalpy-related variable,
Let's switch from the perturbation variable, <math>~p</math>, to an enthalpy-related variable,
<div align="center">
<div align="center">
Line 277: Line 278:
<math>~
<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]  
+\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===
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>
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>
</math>
   </td>
   </td>

Revision as of 02:32, 15 May 2016

Playing With Spherical Wave Equation

Whitworth's (1981) Isothermal Free-Energy Surface
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The 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 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>:

Linearized
Equation of Continuity
<math> r_0 \frac{dx}{dr_0} = - 3 x - d , </math>

Linearized
Euler + Poisson Equations
<math> \frac{P_0}{\rho_0} \frac{dp}{dr_0} = (4x + p)g_0 + \omega^2 r_0 x , </math>

Linearized
Adiabatic Form of the
First Law of Thermodynamics

<math> p = \gamma_\mathrm{g} d \, . </math>


First Effort

Let's switch from the perturbation variable, <math>~p</math>, to an enthalpy-related variable,

<math>~W</math>

<math>~\equiv</math>

<math>~\frac{P_1}{\rho_0} = \biggl(\frac{P_0}{\rho_0}\biggr) p \, .</math>

The second expression then becomes,

<math>~x(4g_0 + \omega^2 r_0)</math>

<math>~=</math>

<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>

 

<math>~=</math>

<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>

 

<math>~=</math>

<math>~\frac{dW}{dr_0}+ \frac{W}{\rho_0} \frac{d\rho_0}{dr_0} \, . </math>

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

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

<math>~=</math>

<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>

 

<math>~=</math>

<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>

 

<math>~=</math>

<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>

<math>~\Rightarrow~~~~ (4g_0 + \omega^2 r_0)^{2} \biggl[ \frac{dx}{dr_0} \biggr] </math>

<math>~=</math>

<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>

Hence, the linearized equation of continuity becomes,

<math>~- (4g_0 + \omega^2 r_0)^{2}\biggl(\frac{W\rho_0}{\gamma_g r_0P_0}\biggr) </math>

<math>~=</math>

<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>

 

<math>~=</math>

<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>

 

 

<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>


Second Effort

Let's switch from the perturbation variable, <math>~p</math>, to an enthalpy-related variable,

<math>~W</math>

<math>~\equiv</math>

<math>~\frac{P_1}{\rho_0 {\bar\sigma}^2} = \biggl(\frac{P_0}{\rho_0 {\bar\sigma}^2}\biggr) p \, ,</math>

where,

<math>~{\bar\sigma}^2 \equiv \frac{4g_0}{r_0} + \omega^2 \, .</math>

The second expression then becomes,

<math>~xr_0{\bar\sigma}^2</math>

<math>~=</math>

<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>

 

<math>~=</math>

<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>

<math>~\Rightarrow ~~~~ xr_0</math>

<math>~=</math>

<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>

 

<math>~=</math>

<math>~ \frac{dW}{dr_0} + W \biggl[ \frac{d \ln(\rho_0 {\bar\sigma}^2)}{dr_0} \biggr] \, . </math>

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

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

<math>~=</math>

<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>

<math>~\Rightarrow~~~~r_0 \frac{dx}{dr_0}</math>

<math>~=</math>

<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>

 

<math>~=</math>

<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>

See Also


Whitworth's (1981) Isothermal Free-Energy Surface

© 2014 - 2021 by Joel E. Tohline
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