Jeans instablity

不考虑宇宙膨胀的影响和冷暗物质的内能,气体达到完全的维里化。

假设一团气体有初始条件:

$$ \begin{cases} \rho_0=const\\ T_0=const\\ \vec{V_0}=0\\ \vec{P_0}=const\\ \end{cases} $$

则它们满足

$$ \begin{cases} \frac{\partial \rho}{\partial t}+\nabla\cdot(\rho\vec{V})=0\qquad连续性方程\\ \frac{\partial \vec{V}}{\partial t}+(\vec{V}\cdot\nabla)\vec{V}=-\frac{1}{\rho}\nabla\vec{P}-\nabla\phi\qquad运动学方程\\\ \nabla^2\phi=4\pi G\rho\qquad泊松方程\\ \vec{P}=\frac{k}{\mu}\rho T=c_s^2\rho\qquad理想气体 \end{cases} $$

取扰动

$$ \begin{cases} \rho=\rho_0+\rho_1\\ \vec{P}=\vec{P_0}+\vec{P_1}\\ \vec{V}=0+\vec{V_1}\\ \phi=\phi_0+\phi_1 \end{cases} $$

连续性方程可写为

$$ \frac{\partial \rho_1}{\partial t}+\nabla\cdot(\rho_1\vec{V_1})+\rho_0\nabla\cdot\vec{V_1}=0 $$

略去高阶小量即

$$ \frac{\partial \rho_1}{\partial t}+\rho_0\nabla\cdot\vec{V_1}=0\tag{1} $$

对运动学方程

$$ \frac{\partial(\vec{V_0}+\vec{V_1})}{\partial t}[(\vec{V_0}+\vec{V_1})\cdot\nabla](\vec{V_0}+\vec{V_1})=-\frac{1}{\rho_0+\rho_1}\nabla(\vec{P_0}+\vec{P_1})-\nabla(\phi_0+\phi_1)\\ [\frac{\partial\vec{V_0}}{\partial t}+\frac{1}{\rho_1+\rho_0}\nabla\vec{P_0}+\nabla\phi_0+\vec{V_0}\cdot\nabla\vec{V_0}]+[\frac{\partial\vec{V_1}}{\partial t}+\vec{V_1}\cdot\nabla\vec{V_1}+\frac{1}{\rho_0+\rho_1}\nabla\vec{P_1}+\nabla\phi_1]=0 $$

化简得

$$ \frac{\partial\vec{V_1}}{\partial t}+\frac{1}{\rho_0+\rho_1}\nabla\vec{P_1}+\nabla\phi_1=0\tag{2} $$

对泊松方程

$$ \nabla^{2}\phi_1=4\pi G\rho_1\tag{3} $$

理想气体方程

$$ \vec{P_1}=c_s^2\rho_1\tag{4} $$

对(1)求导

$$ \frac{\partial^2\rho_1}{\partial t^2}+\frac{\partial}{\partial t}(\rho_0\nabla\cdot\vec{V_1})=\frac{\partial^2\rho_1}{\partial t^2}+\rho_0\nabla\cdot(\frac{\partial\vec{V_1}}{\partial t})=0\tag{5} $$

对(2)求散度,略去二阶小量

$$ \nabla\cdot(\frac{\partial\vec{V_1}}{\partial t})+\nabla^2\phi_1=0\tag{6} $$

(3)(4)(5)代入(6)

$$ \frac{\partial^2\rho_1}{\partial t^2}-c_s^2\nabla^2\rho_1=4\pi G\rho_1\rho_0 $$

作傅里叶变换(假设时空模独立)

$$ \rho_1(\vec{r},t)=\frac{\rho_1(t)}{(2\pi)^\frac{3}{2}}\iiint\rho_1(\vec{r})e^{-i\vec{r}\cdot\vec{k}}dk^3 $$

代入上式

$$ \frac{d^2\rho_1(t)}{dt^2}+c_s^2k^2\rho_1(t)=4\pi G\rho_1(t)\rho_0 $$

可得

$$ \rho_1(t)=Ae^{-i\omega t} $$

其中

$$ \omega=\sqrt{c_s^2k^2-4\pi G\rho_0} $$

讨论:

当$\omega=0$时

$$ k_J=\frac{2\sqrt{\pi G\rho_0}}{c_s}\\ \lambda_J=\frac{2\pi}{k_s}=\sqrt{\frac{\pi}{G\rho_0}}c_s\\ M_J=\frac{4}{3}\pi\lambda_J^3\rho_J $$

$\lambda_J$是金斯波长,$M_J$是金斯质量

引力的作用时标$\tau_引\sim(G\rho_0)^{\frac{1}{2}}$

压力的作用时标$\tau_压\sim\frac{\lambda_J}{c_s}\sim(G\rho_0)^{\frac{1}{2}}$

当$\omega>0$时

微扰会以振荡的形式传播出去

如果考虑宇宙膨胀,振荡会衰减

此时$\tau_压<\tau_引$

当$\omega$s是复数

扰动增大,气体会塌缩