Appendix/Ramblings/SphericalWaveEquation

Playing With Spherical Wave Equation

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 $r_0\frac{dx}{d{r}_0}=-3x-d,$ Linearized Euler + Poisson Equations $\frac{P_0}{{\rho }_0}\frac{dp}{d{r}_0}=(4x+p)g_0+{\omega }^2{r}_{0}x,$ 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,

$W$

$\equiv$

$\frac{P_1}{{\rho }_0}=\left(\frac{P_0}{{\rho }_0}\right)p.$

The second expression then becomes,

$x(4g_0+{\omega }^2{r}_0)$	$=$	$\frac{P_0}{{\rho }_0}\frac{d}{d{r}_0}\left(\frac{W{\rho }_0}{P_0}\right)-\left(\frac{g_0{\rho }_0}{P_0}\right)W$
	$=$	$\frac{dW}{d{r}_0}+\frac{W}{{\rho }_0}\frac{d{\rho }_0}{d{r}_0}-\frac{W}{P_0}\frac{d{P}_0}{d{r}_0}-\left(\frac{g_0{\rho }_0}{P_0}\right)W$
	$=$	$\frac{dW}{d{r}_0}+\frac{W}{{\rho }_0}\frac{d{\rho }_0}{d{r}_0}.$

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

$\frac{dx}{d{r}_0}$	$=$	$\frac{d}{d{r}_0}\{(4g_0+{\omega }^2{r}_0{)}^{-1}[\frac{dW}{d{r}_0}+\frac{W}{{\rho }_0}\frac{d{\rho }_0}{d{r}_0}]\}$
	$=$	$(4g_0+{\omega }^2{r}_0{)}^{-1}\frac{d}{d{r}_0}[\frac{dW}{d{r}_0}+\frac{W}{{\rho }_0}\frac{d{\rho }_0}{d{r}_0}]+[\frac{dW}{d{r}_0}+\frac{W}{{\rho }_0}\frac{d{\rho }_0}{d{r}_0}]\frac{d}{d{r}_0}(4g_0+{\omega }^2{r}_0{)}^{-1}$
	$=$	$(4g_0+{\omega }^2{r}_0{)}^{-1}\{\frac{d^{2}W}{d{r}_0^2}+\frac{d}{d{r}_0}\left[\frac{W}{{\rho }_0}\frac{d{\rho }_0}{d{r}_0}\right]\}-(4g_0+{\omega }^2{r}_0{)}^{-2}[\frac{dW}{d{r}_0}+\frac{W}{{\rho }_0}\frac{d{\rho }_0}{d{r}_0}]\{4\frac{d{g}_0}{d{r}_0}+{\omega }^2\}$
$\Rightarrow (4g_0+{\omega }^2{r}_0{)}^2[\frac{dx}{d{r}_0}]$	$=$	$(4g_0+{\omega }^2{r}_0)\{\frac{d^{2}W}{d{r}_0^2}+\frac{d}{d{r}_0}[\frac{W}{{\rho }_0}\frac{d{\rho }_0}{d{r}_0}\left]\right\}-[\frac{dW}{d{r}_0}+\frac{W}{{\rho }_0}\frac{d{\rho }_0}{d{r}_0}]\{4\frac{d{g}_0}{d{r}_0}+{\omega }^2\}.$

Hence, the linearized equation of continuity becomes,

$-(4g_0+{\omega }^2{r}_0{)}^2(\frac{W{\rho }_0}{{\gamma }_g{r}_0{P}_0})$	$=$	$(4g_0+{\omega }^2{r}_0{)}^2[\frac{dx}{d{r}_0}]+\frac{3(4g_0+{\omega }^2{r}_0)}{r_0}[(4g_0+{\omega }^2{r}_0)x]$
	$=$	$(4g_0+{\omega }^2{r}_0)\{\frac{d^{2}W}{d{r}_0^2}+\frac{d}{d{r}_0}[\frac{W}{{\rho }_0}\frac{d{\rho }_0}{d{r}_0}\left]\right\}-[\frac{dW}{d{r}_0}+\frac{W}{{\rho }_0}\frac{d{\rho }_0}{d{r}_0}]\{4\frac{d{g}_0}{d{r}_0}+{\omega }^2\}$
		$+\frac{3(4g_0+{\omega }^2{r}_0)}{r_0}[\frac{dW}{d{r}_0}+\frac{W}{{\rho }_0}\frac{d{\rho }_0}{d{r}_0}]$

Second Effort

Direct Approach

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

$W$

$\equiv$

$\frac{P_1}{{\rho }_0{\overline{\sigma }}^2}=\left(\frac{P_0}{{\rho }_0{\overline{\sigma }}^2}\right)p,$

where,

${\overline{\sigma }}^2\equiv \frac{4g_0}{r_0}+{\omega }^2.$

Note, as well, that,

$g_0$	$=$	$-\frac{1}{{\rho }_0}\cdot \frac{d{P}_0}{d{r}_0}$
$\Rightarrow {\overline{\sigma }}^2$	$=$	${\omega }^2-\frac{4}{{\rho }_0{r}_0}\cdot \frac{d{P}_0}{d{r}_0}$
	$=$	${\omega }^2-\frac{4P_0}{{\rho }_0{r}_0^2}\cdot \frac{d\ln P_0}{d\ln r_0}$
$\Rightarrow \frac{{\rho }_0{\overline{\sigma }}^2{r}_0^2}{P_0}$	$=$	$\frac{{\rho }_0{r}_0^2}{P_0}\cdot {\omega }^2-4\frac{d\ln P_0}{d\ln r_0}$

The second expression then becomes,

$x{r}_0{\overline{\sigma }}^2$	$=$	$\frac{P_0}{{\rho }_0}\frac{d}{d{r}_0}\left(\frac{W{\rho }_0{\overline{\sigma }}^2}{P_0}\right)-\left(\frac{g_0{\rho }_0{\overline{\sigma }}^2}{P_0}\right)W$
	$=$	${\overline{\sigma }}^2\cdot \frac{dW}{d{r}_0}+W\left[\frac{P_0}{{\rho }_0}\frac{d}{d{r}_0}\right(\frac{{\rho }_0{\overline{\sigma }}^2}{P_0})-(\frac{g_0{\rho }_0{\overline{\sigma }}^2}{P_0}\left)\right]$
$\Rightarrow x{r}_0$	$=$	$\frac{dW}{d{r}_0}+\frac{W}{{\rho }_0{\overline{\sigma }}^2}\left[P_0\frac{d}{d{r}_0}\right(\frac{{\rho }_0{\overline{\sigma }}^2}{P_0})+(\frac{{\rho }_0{\overline{\sigma }}^2}{P_0}\left)\frac{d{P}_0}{d{r}_0}\right]$
	$=$	$\frac{dW}{d{r}_0}+W\left[\frac{d\ln ({\rho }_0{\overline{\sigma }}^2)}{d{r}_0}\right].$

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

$\frac{dx}{d{r}_0}$	$=$	$\frac{d}{d{r}_0}\left\{\frac{1}{r_0}\right[\frac{dW}{d{r}_0}+W\cdot \frac{d\ln ({\rho }_0{\overline{\sigma }}^2)}{d{r}_0}\left]\right\}$
$\Rightarrow r_0\frac{dx}{d{r}_0}$	$=$	$\frac{d}{d{r}_0}[\frac{dW}{d{r}_0}+W\cdot \frac{d\ln ({\rho }_0{\overline{\sigma }}^2)}{d{r}_0}]-\frac{1}{r_0}[\frac{dW}{d{r}_0}+W\cdot \frac{d\ln ({\rho }_0{\overline{\sigma }}^2)}{d{r}_0}]$
	$=$	$\frac{d^{2}W}{d{r}_0^2}+\frac{dW}{d{r}_0}[\frac{d\ln ({\rho }_0{\overline{\sigma }}^2)}{d{r}_0}-\frac{1}{r_0}]+W\{\frac{d^2\ln ({\rho }_0{\overline{\sigma }}^2)}{d{r}_0^2}-\frac{1}{r_0}[\frac{d\ln ({\rho }_0{\overline{\sigma }}^2)}{d{r}_0}\left]\right\}.$

Hence, the linearized continuity equation gives,

$-\left(\frac{W{\rho }_0{\overline{\sigma }}^2}{{\gamma }_g{P}_0}\right)$	$=$	$r_0\frac{dx}{d{r}_0}+3x$
	$=$	$\frac{d^{2}W}{d{r}_0^2}+\frac{dW}{d{r}_0}[\frac{d\ln ({\rho }_0{\overline{\sigma }}^2)}{d{r}_0}-\frac{1}{r_0}]+W\{\frac{d^2\ln ({\rho }_0{\overline{\sigma }}^2)}{d{r}_0^2}-\frac{1}{r_0}[\frac{d\ln ({\rho }_0{\overline{\sigma }}^2)}{d{r}_0}\left]\right\}$
		$+\frac{3}{r_0}[\frac{dW}{d{r}_0}+W\cdot \frac{d\ln ({\rho }_0{\overline{\sigma }}^2)}{d{r}_0}]$
	$=$	$\frac{d^{2}W}{d{r}_0^2}+\frac{dW}{d{r}_0}[\frac{d\ln ({\rho }_0{\overline{\sigma }}^2)}{d{r}_0}+\frac{2}{r_0}]+W\{\frac{d^2\ln ({\rho }_0{\overline{\sigma }}^2)}{d{r}_0^2}+\frac{2}{r_0}[\frac{d\ln ({\rho }_0{\overline{\sigma }}^2)}{d{r}_0}\left]\right\}$
$\Rightarrow 0$	$=$	$\frac{d^{2}W}{d{r}_0^2}+\frac{dW}{d{r}_0}[\frac{d\ln ({\rho }_0{\overline{\sigma }}^2)}{d{r}_0}+\frac{2}{r_0}]+W\{\frac{d^2\ln ({\rho }_0{\overline{\sigma }}^2)}{d{r}_0^2}+\frac{2}{r_0}[\frac{d\ln ({\rho }_0{\overline{\sigma }}^2)}{d{r}_0}]+(\frac{{\rho }_0{\overline{\sigma }}^2}{{\gamma }_g{P}_0}\left)\right\}.$

Playing Around

Multiply thru by $r_0^2$:

$0$

$=$

$r_0^2\cdot \frac{d^{2}W}{d{r}_0^2}+r_0\cdot \frac{dW}{d{r}_0}[\frac{d\ln ({\rho }_0{\overline{\sigma }}^2)}{d\ln r_0}+2]+W\{r_0^2\cdot \frac{d^2\ln ({\rho }_0{\overline{\sigma }}^2)}{d{r}_0^2}+2[\frac{d\ln ({\rho }_0{\overline{\sigma }}^2)}{d\ln r_0}]+(\frac{{\rho }_0{\overline{\sigma }}^2{r}_0^2}{{\gamma }_g{P}_0}\left)\right\}$

Now,

$r_0\cdot \frac{d}{d{r}_0}\left[\frac{d\ln ({\rho }_0{\overline{\sigma }}^2)}{d\ln r_0}\right]$	$=$	$r_0\cdot \frac{d}{d{r}_0}[r_0\cdot \frac{d\ln ({\rho }_0{\overline{\sigma }}^2)}{d{r}_0}]$
	$=$	$\frac{d\ln ({\rho }_0{\overline{\sigma }}^2)}{d\ln r_0}+r_0^2\cdot \frac{d^2\ln ({\rho }_0{\overline{\sigma }}^2)}{d{r}_0^2}$
$\Rightarrow r_0^2\cdot \frac{d^2\ln ({\rho }_0{\overline{\sigma }}^2)}{d{r}_0^2}$	$=$	$r_0\cdot \frac{d}{d{r}_0}\left[\frac{d\ln ({\rho }_0{\overline{\sigma }}^2)}{d\ln r_0}\right]-\frac{d\ln ({\rho }_0{\overline{\sigma }}^2)}{d\ln r_0}$

Also,

$r_0\cdot \frac{d}{d{r}_0}\left[\frac{dW}{d\ln r_0}\right]$	$=$	$r_0\cdot \frac{d}{d{r}_0}[r_0\cdot \frac{dW}{d{r}_0}]$
	$=$	$\frac{dW}{d\ln r_0}+r_0^2\cdot \frac{d^{2}W}{d{r}_0^2}$
$\Rightarrow r_0^2\cdot \frac{d^{2}W}{d{r}_0^2}$	$=$	$\frac{d}{d\ln r_0}\left[\frac{dW}{d\ln r_0}\right]-\frac{dW}{d\ln r_0}$

$\Rightarrow 0$	$=$	$\frac{d}{d\ln r_0}\left[\frac{dW}{d\ln r_0}\right]-\frac{dW}{d\ln r_0}+r_0\cdot \frac{dW}{d{r}_0}[\frac{d\ln ({\rho }_0{\overline{\sigma }}^2)}{d\ln r_0}+2]+W\{r_0\cdot \frac{d}{d{r}_0}[\frac{d\ln ({\rho }_0{\overline{\sigma }}^2)}{d\ln r_0}]+\frac{d\ln ({\rho }_0{\overline{\sigma }}^2)}{d\ln r_0}+(\frac{{\rho }_0{\overline{\sigma }}^2{r}_0^2}{{\gamma }_g{P}_0}\left)\right\}$
	$=$	$\frac{d}{d\ln r_0}\left[\frac{dW}{d\ln r_0}\right]+\frac{dW}{d\ln r_0}[\frac{d\ln ({\rho }_0{\overline{\sigma }}^2)}{d\ln r_0}+1]+W\left\{\frac{d}{d\ln r_0}\right[\frac{d\ln ({\rho }_0{\overline{\sigma }}^2)}{d\ln r_0}]+\frac{d\ln ({\rho }_0{\overline{\sigma }}^2)}{d\ln r_0}+(\frac{{\rho }_0{\overline{\sigma }}^2{r}_0^2}{{\gamma }_g{P}_0}\left)\right\}$
	$=$	$\frac{d}{d\ln r_0}\left[\frac{dW}{d\ln r_0}\right]+\frac{dW}{d\ln r_0}[u+1]+W\{\frac{du}{d\ln r_0}+u+(\frac{{\rho }_0{\overline{\sigma }}^2{r}_0^2}{{\gamma }_g{P}_0}\left)\right\},$

where,

$u$

$\equiv$

$\frac{d\ln ({\rho }_0{\overline{\sigma }}^2)}{d\ln r_0}.$

Let,

$y\equiv \ln r_0$       $\Rightarrow$       $r_0=e^y.$

Then we have,

$0$	$=$	$\frac{d^{2}W}{d{y}^2}+\frac{dW}{dy}[u+1]+W\{\frac{du}{dy}+u+(\frac{{\rho }_0{\overline{\sigma }}^2{e}^{2y}}{{\gamma }_g{P}_0}\left)\right\}.$
	$=$	$\frac{d^{2}W}{d{y}^2}+\frac{dW}{dy}[u+1]+W\{\frac{du}{dy}+u+[\frac{{\rho }_0{r}_0^2}{P_0}\cdot {\omega }^2-4\frac{d\ln P_0}{d\ln r_0}\left]\right\}$
	$=$	$\frac{d^{2}W}{d{y}^2}+\frac{dW}{dy}[u+1]+W\{\frac{d(u-P_0^4)}{dy}+u+\frac{{\rho }_0{r}_0^2}{P_0}\cdot {\omega }^2\}$

Therefore, it must also be the case that,

$udy$

$=$

$d\ln ({\rho }_0{\overline{\sigma }}^2).$

See Also

• Traditional Presentation of Adiabatic Wave Equation