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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, ~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


Whitworth's (1981) Isothermal Free-Energy Surface

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