next up previous contents
Next: Rotating Disk Up: Physical Background Previous: Jeans Instability   Contents


Gravitational Instability of Thin Disk

Figure 2.3: Thin disk.
\begin{figure}\centering\leavevmode
\epsfxsize =.45\columnwidth \epsfbox{eps/thin_disk.ps}\end{figure}

Disks are common in the Universe. Spiral and barred spiral galaxies have disks where stars are formed. In more small scale, gas and dust disks are often found around protostars. Moreover, such a disk may become a proto-planetary disk. It is valuable to study how the self-gravity works in such thin structures. Here, we assume a thin disk extending in $x$- and $y$-directions whose surface density is equal to $\sigma=\int_{-\infty}^\infty \rho dz$, in other word the density is written using the Dirac's delta function $\delta(z)$ as

\begin{displaymath}
\rho(x,y,z)=\sigma(x,y)\delta(z).
\end{displaymath} (2.55)

Integrating along the $z$-direction basic equations (2.45), (2.46), and (2.47), the linearized basic equations for the thin disk are as follows:
\begin{displaymath}
\frac{\partial \delta \sigma}{\partial t}+\sigma_0\frac{\partial \delta u}{\partial x}=0,
\end{displaymath} (2.56)


\begin{displaymath}
\sigma_0\frac{\partial \delta u}{\partial t}=-c_{is}^2\frac{...
...}{\partial x}-\sigma_0\frac{\partial \delta \phi}{\partial x},
\end{displaymath} (2.57)


\begin{displaymath}
\frac{\partial^2 \delta \phi}{\partial x^2}+\frac{\partial^2 \delta \phi}{\partial z^2}=4\pi G \delta \sigma \delta(z),
\end{displaymath} (2.58)

where we assumed $\sigma=\sigma_0+\delta \sigma$, $u=\delta u$, $\phi=\phi_0+\delta \phi$ and took the first order terms (those contain only one $\delta$).

Outside the disk, the rhs of equation (2.58) is equal to zero. It reduces to the Laplace equation

\begin{displaymath}
\frac{\partial^2 \delta \phi}{\partial x^2}+\frac{\partial^2 \delta \phi}{\partial z^2}=0.
\end{displaymath} (2.59)

Taking a plane wave of
\begin{displaymath}
\delta X(x,t)=\delta A \exp(i\omega t - i kx),
\end{displaymath} (2.60)

equation (2.59) is reduced to
\begin{displaymath}
\frac{\partial^2 \delta \phi}{\partial z^2}-k^2 \delta \phi =0.
\end{displaymath} (2.61)

This has a solution which does not diverge at the infinity $z=\pm \infty$ as
\begin{displaymath}
\delta \phi=\delta\phi(z=0) \exp(-k\vert z\vert).
\end{displaymath} (2.62)

On the other hand, integrating equation (2.58) from $z=-0$ to $z=+0$ or in other word, applying the Gauss' theorem to the region containing the $z=0$ surface, it is shown that the gravity $\delta g_z=-\partial \delta \phi/\partial z$ has a jump crossing the $z=0$ surface as
\begin{displaymath}
\left. \frac{\partial \delta \phi}{\partial z}\right\vert _{...
...lta \phi}{\partial z}\right\vert _{z=-0}=4\pi G \delta \sigma.
\end{displaymath} (2.63)

Equations (2.62) and (2.63) lead a final form of the potential as
\begin{displaymath}
\delta \phi=-\frac{2\pi G \delta \sigma}{k}\exp(-k\vert z\vert).
\end{displaymath} (2.64)

Putting this to equations (2.57), and using equations (2.56) and (2.57), we obtain the dispersion relation for the gravitational instability in a thin disk as
\begin{displaymath}
\omega^2=c_{is}^2k^2 -2\pi G \sigma_0 k.
\end{displaymath} (2.65)

This reduces to the dispersion relation of the sound wave for the short wave $k\gg 2\pi G \sigma_0/c_{is}^2$. While for a longer wave than $\lambda_{cr}=c_{is}^2/G\sigma_0$, an exponential growth of $\delta \sigma$ is expected. The dispersion relation is shown in Fig.2.2.



Subsections
next up previous contents
Next: Rotating Disk Up: Physical Background Previous: Jeans Instability   Contents
Kohji Tomisaka 2007-07-08