User:Tohline/Appendix/Ramblings/SphericalWaveEquation
Playing With Spherical Wave Equation
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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>:
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Linearized Linearized Linearized |
First Effort
Let's switch from the perturbation variable, <math>~p</math>, to an enthalpy-related variable,
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<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,
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<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> |
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<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> |
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<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,
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<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> |
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<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> |
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<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> |
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<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,
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<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> |
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<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> |
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<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,
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<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,
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<math>~g_0</math> |
<math>~=</math> |
<math>~- \frac{1}{\rho_0} \cdot \frac{dP_0}{dr_0}</math> |
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<math>~\Rightarrow ~~~~ {\bar\sigma}^2 </math> |
<math>~=</math> |
<math>~\omega^2 -\frac{4}{\rho_0 r_0} \cdot \frac{dP_0}{dr_0} </math> |
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<math>~=</math> |
<math>~\omega^2 -\frac{4P_0}{\rho_0 r_0^2} \cdot \frac{d\ln P_0}{d\ln r_0} </math> |
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<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,
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<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> |
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<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> |
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<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> |
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<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,
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<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> |
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<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> |
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<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,
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<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> |
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<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> |
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<math>~ +\frac{3}{r_0}\biggl[ \frac{dW}{dr_0} + W \cdot\frac{d \ln(\rho_0 {\bar\sigma}^2)}{dr_0} \biggr] </math> |
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<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> |
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<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>:
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<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,
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<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> |
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<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> |
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<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,
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<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> |
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<math>~=</math> |
<math>~ \frac{dW}{d\ln r_0} + r_0^2 \cdot \frac{d^2W}{dr_0^2} </math> |
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<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> |
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<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> |
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<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> |
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<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,
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<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,
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<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> |
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<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> |
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<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,
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<math>~u dy</math> |
<math>~=</math> |
<math>~d \ln(\rho_0 {\bar\sigma}^2) \, .</math> |
See Also
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© 2014 - 2021 by Joel E. Tohline |