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

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(→‎Assembling the Key Relations: Finished deriving W-based eigenvalue problem)
(→‎Second Effort: Just playing around with 2nd-order ODE, trying to simplify)
 
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===Second Effort===
===Second Effort===
====Direct Approach====
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 307: Line 308:
<div align="center">
<div align="center">
<math>~{\bar\sigma}^2 \equiv \frac{4g_0}{r_0} + \omega^2 \, .</math>
<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>
</div>


Line 512: Line 573:
+ \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\} \, .
+ \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>
</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>
   </td>
</tr>
</tr>

Latest revision as of 03:58, 23 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

Direct Approach

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>

Note, as well, that,

<math>~g_0</math>

<math>~=</math>

<math>~- \frac{1}{\rho_0} \cdot \frac{dP_0}{dr_0}</math>

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

<math>~=</math>

<math>~\omega^2 -\frac{4}{\rho_0 r_0} \cdot \frac{dP_0}{dr_0} </math>

 

<math>~=</math>

<math>~\omega^2 -\frac{4P_0}{\rho_0 r_0^2} \cdot \frac{d\ln P_0}{d\ln r_0} </math>

<math>~\Rightarrow ~~~~ \frac{\rho_0 {\bar\sigma}^2 r_0^2}{P_0} </math>

<math>~=</math>

<math>~ \frac{\rho_0 r_0^2}{P_0} \cdot\omega^2- 4\frac{d\ln P_0}{d\ln r_0} </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>

Hence, the linearized continuity equation gives,

<math>~- \biggl(\frac{W\rho_0 {\bar\sigma}^2}{\gamma_gP_0} \biggr)</math>

<math>~=</math>

<math>~r_0 \frac{dx}{dr_0} +3x</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>

 

 

<math>~ +\frac{3}{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{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>

<math>~\Rightarrow~~~~ 0</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{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>

Playing Around

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

<math>~ 0</math>

<math>~=</math>

<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>

Now,

<math>~r_0 \cdot \frac{d}{dr_0} \biggl[\frac{d\ln (\rho_0 {\bar\sigma}^2)}{d\ln r_0} \biggr]</math>

<math>~=</math>

<math>~r_0 \cdot \frac{d}{dr_0} \biggl[r_0 \cdot \frac{d\ln (\rho_0 {\bar\sigma}^2)}{dr_0} \biggr]</math>

 

<math>~=</math>

<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>

<math>~\Rightarrow ~~~~ r_0^2 \cdot \frac{d^2\ln (\rho_0 {\bar\sigma}^2)}{dr_0^2} </math>

<math>~=</math>

<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>


Also,

<math>~r_0 \cdot \frac{d}{dr_0} \biggl[\frac{dW}{d\ln r_0} \biggr]</math>

<math>~=</math>

<math>~r_0 \cdot \frac{d}{dr_0} \biggl[r_0 \cdot \frac{dW}{dr_0} \biggr]</math>

 

<math>~=</math>

<math>~ \frac{dW}{d\ln r_0} + r_0^2 \cdot \frac{d^2W}{dr_0^2} </math>

<math>~\Rightarrow ~~~~ r_0^2 \cdot \frac{d^2W}{dr_0^2} </math>

<math>~=</math>

<math>~ \frac{d}{d\ln r_0} \biggl[\frac{dW}{d\ln r_0} \biggr] - \frac{dW}{d\ln r_0} </math>


<math>~ \Rightarrow ~~~~ 0</math>

<math>~=</math>

<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>

 

<math>~=</math>

<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>

 

<math>~=</math>

<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>

where,

<math>~u</math>

<math>~\equiv</math>

<math>~\frac{d \ln(\rho_0 {\bar\sigma}^2)}{d\ln r_0} \, .</math>

Let,

<math>~y \equiv \ln r_0 </math>       <math>~\Rightarrow</math>       <math>~r_0 = e^{y} \, . </math>

Then we have,

<math>~0</math>

<math>~=</math>

<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>

 

<math>~=</math>

<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>

 

<math>~=</math>

<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>

Therefore, it must also be the case that,

<math>~u dy</math>

<math>~=</math>

<math>~d \ln(\rho_0 {\bar\sigma}^2) \, .</math>

See Also


Whitworth's (1981) Isothermal Free-Energy Surface

© 2014 - 2021 by Joel E. Tohline
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Recommended citation:   Tohline, Joel E. (2021), The Structure, Stability, & Dynamics of Self-Gravitating Fluids, a (MediaWiki-based) Vistrails.org publication, https://www.vistrails.org/index.php/User:Tohline/citation