# User:Tohline/Appendix/Ramblings/SphericalWaveEquation

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# Playing With Spherical Wave Equation

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The traditional presentation of the (spherically symmetric) adiabatic wave equation focuses on fractional radial displacements, $~x \equiv \delta r/r_0$, 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 $~p(r_0) \equiv P_1/P_0$, $~d(r_0) \equiv \rho_1/\rho_0$ and $~x(r_0) \equiv r_1/r_0$, for various characteristic eigenfrequencies, $~\omega$:

 Linearized Equation of Continuity $r_0 \frac{dx}{dr_0} = - 3 x - d ,$ Linearized Euler + Poisson Equations $\frac{P_0}{\rho_0} \frac{dp}{dr_0} = (4x + p)g_0 + \omega^2 r_0 x ,$ Linearized Adiabatic Form of the First Law of Thermodynamics $p = \gamma_\mathrm{g} d \, .$

### First Effort

Let's switch from the perturbation variable, $~p$, to an enthalpy-related variable,

 $~W$ $~\equiv$ $~\frac{P_1}{\rho_0} = \biggl(\frac{P_0}{\rho_0}\biggr) p \, .$

The second expression then becomes,

 $~x(4g_0 + \omega^2 r_0)$ $~=$ $~\frac{P_0}{\rho_0} \frac{d}{dr_0}\biggl(\frac{W\rho_0}{P_0}\biggr) - \biggl(\frac{g_0 \rho_0}{P_0}\biggr)W$ $~=$ $~\frac{dW}{dr_0} + \frac{W}{\rho_0} \frac{d\rho_0}{dr_0} - \frac{W }{P_0} \frac{dP_0}{dr_0} - \biggl(\frac{g_0 \rho_0}{P_0}\biggr)W$ $~=$ $~\frac{dW}{dr_0}+ \frac{W}{\rho_0} \frac{d\rho_0}{dr_0} \, .$

Taking the derivative of this expression with respect to $~r_0$ gives,

 $~\frac{dx}{dr_0}$ $~=$ $~\frac{d}{dr_0}\biggl\{ (4g_0 + \omega^2 r_0)^{-1}\biggl[\frac{dW}{dr_0} + \frac{W}{\rho_0} \frac{d\rho_0}{dr_0} \biggr] \biggr\}$ $~=$ $~ (4g_0 + \omega^2 r_0)^{-1}\frac{d}{dr_0} \biggl[\frac{dW}{dr_0} + \frac{W}{\rho_0} \frac{d\rho_0}{dr_0} \biggr] +\biggl[\frac{dW}{dr_0} + \frac{W}{\rho_0} \frac{d\rho_0}{dr_0} \biggr]\frac{d}{dr_0} (4g_0 + \omega^2 r_0)^{-1}$ $~=$ $~ (4g_0 + \omega^2 r_0)^{-1} \biggl\{ \frac{d^2W}{dr^2_0} + \frac{d}{dr_0}\biggl[\frac{W}{\rho_0} \frac{d\rho_0}{dr_0} \biggr] \biggr\} -(4g_0 + \omega^2 r_0)^{-2}\biggl[\frac{dW}{dr_0} + \frac{W}{\rho_0} \frac{d\rho_0}{dr_0} \biggr] \biggl\{ 4\frac{dg_0}{dr_0} + \omega^2 \biggr\}$ $~\Rightarrow~~~~ (4g_0 + \omega^2 r_0)^{2} \biggl[ \frac{dx}{dr_0} \biggr]$ $~=$ $~ (4g_0 + \omega^2 r_0)\biggl\{ \frac{d^2W}{dr^2_0} + \frac{d}{dr_0}\biggl[\frac{W}{\rho_0} \frac{d\rho_0}{dr_0} \biggr] \biggr\} -\biggl[\frac{dW}{dr_0} + \frac{W}{\rho_0} \frac{d\rho_0}{dr_0} \biggr] \biggl\{ 4\frac{dg_0}{dr_0} + \omega^2 \biggr\} \, .$

Hence, the linearized equation of continuity becomes,

 $~- (4g_0 + \omega^2 r_0)^{2}\biggl(\frac{W\rho_0}{\gamma_g r_0P_0}\biggr)$ $~=$ $~(4g_0 + \omega^2 r_0)^{2} \biggl[ \frac{dx}{dr_0} \biggr] +\frac{3 (4g_0 + \omega^2 r_0)}{r_0} \biggl[ (4g_0 + \omega^2 r_0)x \biggr]$ $~=$ $~ (4g_0 + \omega^2 r_0)\biggl\{ \frac{d^2W}{dr^2_0} + \frac{d}{dr_0}\biggl[\frac{W}{\rho_0} \frac{d\rho_0}{dr_0} \biggr] \biggr\} -\biggl[\frac{dW}{dr_0} + \frac{W}{\rho_0} \frac{d\rho_0}{dr_0} \biggr] \biggl\{ 4\frac{dg_0}{dr_0} + \omega^2 \biggr\}$ $~ +\frac{3 (4g_0 + \omega^2 r_0)}{r_0} \biggl[ \frac{dW}{dr_0}+ \frac{W}{\rho_0} \frac{d\rho_0}{dr_0} \biggr]$

### Second Effort

#### Direct Approach

Let's switch from the perturbation variable, $~p$, to an enthalpy-related variable,

 $~W$ $~\equiv$ $~\frac{P_1}{\rho_0 {\bar\sigma}^2} = \biggl(\frac{P_0}{\rho_0 {\bar\sigma}^2}\biggr) p \, ,$

where,

$~{\bar\sigma}^2 \equiv \frac{4g_0}{r_0} + \omega^2 \, .$

Note, as well, that,

 $~g_0$ $~=$ $~- \frac{1}{\rho_0} \cdot \frac{dP_0}{dr_0}$ $~\Rightarrow ~~~~ {\bar\sigma}^2$ $~=$ $~\omega^2 -\frac{4}{\rho_0 r_0} \cdot \frac{dP_0}{dr_0}$ $~=$ $~\omega^2 -\frac{4P_0}{\rho_0 r_0^2} \cdot \frac{d\ln P_0}{d\ln r_0}$ $~\Rightarrow ~~~~ \frac{\rho_0 {\bar\sigma}^2 r_0^2}{P_0}$ $~=$ $~ \frac{\rho_0 r_0^2}{P_0} \cdot\omega^2- 4\frac{d\ln P_0}{d\ln r_0}$

The second expression then becomes,

 $~xr_0{\bar\sigma}^2$ $~=$ $~\frac{P_0}{\rho_0} \frac{d}{dr_0}\biggl(\frac{W\rho_0 {\bar\sigma}^2}{P_0}\biggr) - \biggl(\frac{g_0 \rho_0 {\bar\sigma}^2}{P_0}\biggr)W$ $~=$ $~ {\bar\sigma}^2 \cdot \frac{dW}{dr_0} + W \biggl[ \frac{P_0}{\rho_0} \frac{d}{dr_0}\biggl(\frac{\rho_0 {\bar\sigma}^2}{P_0}\biggr) - \biggl(\frac{g_0 \rho_0 {\bar\sigma}^2}{P_0}\biggr)\biggr]$ $~\Rightarrow ~~~~ xr_0$ $~=$ $~ \frac{dW}{dr_0} + \frac{W}{\rho_0{\bar\sigma}^2} \biggl[ P_0 \frac{d}{dr_0}\biggl(\frac{\rho_0 {\bar\sigma}^2}{P_0}\biggr) + \biggl(\frac{\rho_0 {\bar\sigma}^2}{P_0}\biggr)\frac{dP_0}{dr_0} \biggr]$ $~=$ $~ \frac{dW}{dr_0} + W \biggl[ \frac{d \ln(\rho_0 {\bar\sigma}^2)}{dr_0} \biggr] \, .$

Taking the derivative of this expression with respect to $~r_0$ gives,

 $~\frac{dx}{dr_0}$ $~=$ $~ \frac{d}{dr_0}\biggl\{\frac{1}{r_0}\biggl[ \frac{dW}{dr_0} + W \cdot \frac{d \ln(\rho_0 {\bar\sigma}^2)}{dr_0} \biggr] \biggr\}$ $~\Rightarrow~~~~r_0 \frac{dx}{dr_0}$ $~=$ $~ \frac{d}{dr_0}\biggl[ \frac{dW}{dr_0} + W \cdot \frac{d \ln(\rho_0 {\bar\sigma}^2)}{dr_0} \biggr] - \frac{1}{r_0}\biggl[ \frac{dW}{dr_0} + W \cdot \frac{d \ln(\rho_0 {\bar\sigma}^2)}{dr_0} \biggr]$ $~=$ $~ \frac{d^2W}{dr_0^2} + \frac{dW}{dr_0} \biggl[ \frac{d \ln(\rho_0 {\bar\sigma}^2)}{dr_0} -\frac{1}{r_0}\biggr] + W \biggl\{ \frac{d^2 \ln(\rho_0 {\bar\sigma}^2)}{dr_0^2} - \frac{1}{r_0}\biggl[ \frac{d \ln(\rho_0 {\bar\sigma}^2)}{dr_0} \biggr]\biggr\} \, .$

Hence, the linearized continuity equation gives,

 $~- \biggl(\frac{W\rho_0 {\bar\sigma}^2}{\gamma_gP_0} \biggr)$ $~=$ $~r_0 \frac{dx}{dr_0} +3x$ $~=$ $~ \frac{d^2W}{dr_0^2} + \frac{dW}{dr_0} \biggl[ \frac{d \ln(\rho_0 {\bar\sigma}^2)}{dr_0} -\frac{1}{r_0}\biggr] + W \biggl\{ \frac{d^2 \ln(\rho_0 {\bar\sigma}^2)}{dr_0^2} - \frac{1}{r_0}\biggl[ \frac{d \ln(\rho_0 {\bar\sigma}^2)}{dr_0} \biggr]\biggr\}$ $~ +\frac{3}{r_0}\biggl[ \frac{dW}{dr_0} + W \cdot\frac{d \ln(\rho_0 {\bar\sigma}^2)}{dr_0} \biggr]$ $~=$ $~ \frac{d^2W}{dr_0^2} + \frac{dW}{dr_0} \biggl[ \frac{d \ln(\rho_0 {\bar\sigma}^2)}{dr_0} +\frac{2}{r_0}\biggr] + W \biggl\{ \frac{d^2 \ln(\rho_0 {\bar\sigma}^2)}{dr_0^2} + \frac{2}{r_0}\biggl[ \frac{d \ln(\rho_0 {\bar\sigma}^2)}{dr_0} \biggr]\biggr\}$ $~\Rightarrow~~~~ 0$ $~=$ $~ \frac{d^2W}{dr_0^2} + \frac{dW}{dr_0} \biggl[ \frac{d \ln(\rho_0 {\bar\sigma}^2)}{dr_0} +\frac{2}{r_0}\biggr] + W \biggl\{ \frac{d^2 \ln(\rho_0 {\bar\sigma}^2)}{dr_0^2} + \frac{2}{r_0}\biggl[ \frac{d \ln(\rho_0 {\bar\sigma}^2)}{dr_0} \biggr] + \biggl(\frac{\rho_0 {\bar\sigma}^2}{\gamma_gP_0} \biggr)\biggr\} \, .$

#### Playing Around

Multiply thru by $~r_0^2$:

 $~ 0$ $~=$ $~ r_0^2 \cdot \frac{d^2W}{dr_0^2} + r_0 \cdot \frac{dW}{dr_0} \biggl[ \frac{d \ln(\rho_0 {\bar\sigma}^2)}{d\ln r_0} +2 \biggr] + W \biggl\{ r_0^2 \cdot \frac{d^2 \ln(\rho_0 {\bar\sigma}^2)}{dr_0^2} + 2\biggl[ \frac{d \ln(\rho_0 {\bar\sigma}^2)}{d\ln r_0} \biggr] + \biggl(\frac{\rho_0 {\bar\sigma}^2 r_0^2 }{\gamma_gP_0} \biggr)\biggr\}$

Now,

 $~r_0 \cdot \frac{d}{dr_0} \biggl[\frac{d\ln (\rho_0 {\bar\sigma}^2)}{d\ln r_0} \biggr]$ $~=$ $~r_0 \cdot \frac{d}{dr_0} \biggl[r_0 \cdot \frac{d\ln (\rho_0 {\bar\sigma}^2)}{dr_0} \biggr]$ $~=$ $~ \frac{d\ln (\rho_0 {\bar\sigma}^2)}{d\ln r_0} + r_0^2 \cdot \frac{d^2\ln (\rho_0 {\bar\sigma}^2)}{dr_0^2}$ $~\Rightarrow ~~~~ r_0^2 \cdot \frac{d^2\ln (\rho_0 {\bar\sigma}^2)}{dr_0^2}$ $~=$ $~ r_0 \cdot \frac{d}{dr_0} \biggl[\frac{d\ln (\rho_0 {\bar\sigma}^2)}{d\ln r_0} \biggr] - \frac{d\ln (\rho_0 {\bar\sigma}^2)}{d\ln r_0}$

Also,

 $~r_0 \cdot \frac{d}{dr_0} \biggl[\frac{dW}{d\ln r_0} \biggr]$ $~=$ $~r_0 \cdot \frac{d}{dr_0} \biggl[r_0 \cdot \frac{dW}{dr_0} \biggr]$ $~=$ $~ \frac{dW}{d\ln r_0} + r_0^2 \cdot \frac{d^2W}{dr_0^2}$ $~\Rightarrow ~~~~ r_0^2 \cdot \frac{d^2W}{dr_0^2}$ $~=$ $~ \frac{d}{d\ln r_0} \biggl[\frac{dW}{d\ln r_0} \biggr] - \frac{dW}{d\ln r_0}$

 $~ \Rightarrow ~~~~ 0$ $~=$ $~ \frac{d}{d\ln r_0} \biggl[\frac{dW}{d\ln r_0} \biggr] - \frac{dW}{d\ln r_0} + r_0 \cdot \frac{dW}{dr_0} \biggl[ \frac{d \ln(\rho_0 {\bar\sigma}^2)}{d\ln r_0} +2 \biggr] + W \biggl\{ r_0 \cdot \frac{d}{dr_0} \biggl[\frac{d\ln (\rho_0 {\bar\sigma}^2)}{d\ln r_0} \biggr] + \frac{d\ln (\rho_0 {\bar\sigma}^2)}{d\ln r_0} + \biggl(\frac{\rho_0 {\bar\sigma}^2 r_0^2 }{\gamma_gP_0} \biggr)\biggr\}$ $~=$ $~ \frac{d}{d\ln r_0} \biggl[\frac{dW}{d\ln r_0} \biggr] + \frac{dW}{d\ln r_0} \biggl[ \frac{d \ln(\rho_0 {\bar\sigma}^2)}{d\ln r_0} +1 \biggr] + W \biggl\{ \frac{d}{d\ln r_0} \biggl[\frac{d\ln (\rho_0 {\bar\sigma}^2)}{d\ln r_0} \biggr] + \frac{d\ln (\rho_0 {\bar\sigma}^2)}{d\ln r_0} + \biggl(\frac{\rho_0 {\bar\sigma}^2 r_0^2 }{\gamma_gP_0} \biggr)\biggr\}$ $~=$ $~ \frac{d}{d\ln r_0} \biggl[\frac{dW}{d\ln r_0} \biggr] + \frac{dW}{d\ln r_0} \biggl[ u +1 \biggr] + W \biggl\{ \frac{du}{d\ln r_0} + u + \biggl(\frac{\rho_0 {\bar\sigma}^2 r_0^2 }{\gamma_gP_0} \biggr)\biggr\} \, ,$

where,

 $~u$ $~\equiv$ $~\frac{d \ln(\rho_0 {\bar\sigma}^2)}{d\ln r_0} \, .$

Let,

$~y \equiv \ln r_0$       $~\Rightarrow$       $~r_0 = e^{y} \, .$

Then we have,

 $~0$ $~=$ $~ \frac{d^2W}{dy^2} + \frac{dW}{dy} \biggl[ u +1 \biggr] + W \biggl\{ \frac{du}{dy} + u + \biggl(\frac{\rho_0 {\bar\sigma}^2 e^{2y} }{\gamma_gP_0} \biggr)\biggr\} \, .$ $~=$ $~ \frac{d^2W}{dy^2} + \frac{dW}{dy} \biggl[ u +1 \biggr] + W \biggl\{ \frac{du}{dy} + u + \biggl[\frac{\rho_0 r_0^2}{P_0} \cdot\omega^2- 4\frac{d\ln P_0}{d\ln r_0} \biggr] \biggr\}$ $~=$ $~ \frac{d^2W}{dy^2} + \frac{dW}{dy} \biggl[ u +1 \biggr] + W \biggl\{ \frac{d(u- P_0^4)}{dy} + u + \frac{\rho_0 r_0^2}{P_0} \cdot\omega^2\biggr\}$

Therefore, it must also be the case that,

 $~u dy$ $~=$ $~d \ln(\rho_0 {\bar\sigma}^2) \, .$

# See Also

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