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Sound Wave

If the self-gravity is ignorable, equations(2.35)

\begin{displaymath}
\frac{\partial \delta \rho}{\partial t}+\rho_0 \frac{\partial \delta u}{\partial x}=0,
\end{displaymath} (2.40)

and equations(2.37)
\begin{displaymath}
\rho_0 \frac{\partial \delta u}{\partial t}= -c_{is}^2\frac{\partial \delta \rho}{\partial x},
\end{displaymath} (2.41)

where we assumed the gas is isothermal, these two equations describe the propagation and growth of perturbations. If the gas acts adiabatically, replace $c_{is}$ with $c_s$.

Making $\partial /\partial x\times $(2.40) and $\partial /\partial t\times$(2.41) vanishes $\delta \rho$ and we obtain

\begin{displaymath}
\frac{\partial^2 \delta u}{\partial t^2}-c_{is}^2\frac{\partial^2 \delta u}{\partial x^2}=0.
\end{displaymath} (2.42)

Since this leads to
$\displaystyle \frac{\partial \delta u}{\partial t}-c_{is}\frac{\partial \delta u}{\partial x}$ $\textstyle =$ $\displaystyle 0,$ (2.43)
$\displaystyle \frac{\partial \delta u}{\partial t}+c_{is}\frac{\partial \delta u}{\partial x}$ $\textstyle =$ $\displaystyle 0,$ (2.44)

equation (2.42) has a solution that a wave propagates with a phase velocity of $\pm c_s$. Since the displacement ( $\propto \delta u$) is parallel to the propagation direction $x$, and the restoring force comes from the pressure, this seems the sound wave.



Subsections
next up previous contents
Next: Problem Up: Gravitational Instability Previous: Linear Analysis   Contents
Kohji Tomisaka 2007-07-08