# Cosmology Lecture 14

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

### Jeans Instability Continued

So far we’ve written down 3 equations: the Continuity Equation, Euler’s Equation, and Poisson’s Equation. From these we derived a dispersion relation for a static medium. We evaluated two wavelength regimes for this relation and found that for wavelengths longer than the Jeans wavelength, we had exponential collapse due to self-gravity, and for wavelengths shorter than the Jeans wavelength, we have normal, stable oscillation.

Basically, we are seeing a balance between pressure outward and gravity inward. Collapse occurs when gravity wins out over pressure. Consider a sphere of density $\rho$ , and a piece of volume $V$ at distance $R$ . Then the force of gravity on that piece is:

${F_{grav} \over V}\sim {GM\rho \over R^{2}}\sim G\rho ^{2}R\,\!$ since $M\sim \rho R^{3}$ . The pressure force on that volume is:

${F_{pres} \over V}\sim \nabla P\sim v_{s}^{2}\nabla \rho \sim v_{s}^{2}{\rho \over R}\,\!$ Comparing these forces, we find that gravity wins if:

$G\rho ^{2}R>v_{s}^{2}{\rho \over R}\,\!$ Solving for $R$ :

$R>{\sqrt {v_{s}^{2} \over G\rho }}\sim \lambda _{J}\,\!$ So this is a quick-and-dirty way of deriving the Jeans length scale. The timescale for collapse is also a balance between gravity and pressure. The time for gravitational free fall is given by:

{\begin{aligned}R&\sim a_{grav}t_{ff}^{2}\sim {GM \over R^{2}}t_{ff}^{2}\sim G\rho Rt_{ff}^{2}\\t_{ff}&\sim {1 \over {\sqrt {G\rho }}}\\\end{aligned}}\,\! Thus, the free fall time is independent of distance. The pressure timescale is the sound crossing time, given by:

$t_{sound}\sim {R \over v_{s}}\,\!$ We will have gravitational collapse if $t_{ff} , which means:

${\begin{matrix}{1 \over {\sqrt {G\rho }}}<{R \over v_{s}}&\Rightarrow &R>{\sqrt {v_{s}^{2} \over G\rho }}\end{matrix}}\,\!$ So again we get out the Jeans wavelength.

### Gravitational Instability in an Expanding Fluid

Density is obviously affected by expansion. To zeroth order:

{\begin{aligned}\rho _{0}&=\rho _{0}(t)={\rho _{0}(t_{today}) \over a^{3}}\\{\partial \rho _{0} \over \partial t}&=-3{{\dot {a}} \over a}\rho _{0}\\\end{aligned}}\,\! where $a$ is our familiar scale factor. To examine first order perturbations, we define the density field $\delta$ , where $\rho _{1}\equiv \rho _{0}\delta$ , so that:

$\delta ={\rho _{1} \over \rho _{0}}={\rho -\rho _{0} \over \rho _{0}}\,\!$ ${\delta ={\partial \rho \over \rho _{0}}}\,\!$ Our velocities are also changed by expansion:

${\vec {v}}_{0}=H{\vec {r}}={{\dot {a}} \over a}{\vec {r}}\,\!$ where ${\vec {r}}$ is in physical coordinates. This velocity is purely from Hubble expansion. Note that:

$\nabla \cdot {\vec {v}}_{0}={{\dot {a}} \over a}(\nabla \cdot {\vec {r}})=3{{\dot {a}} \over a}\,\!$ We’ll use ${\vec {v}}_{1}$ to describe the peculiar velocity (motion with respect to the expansion). Now we need to solve our fluid equations using these parameters. The continuity equation, to zeroth order, says:

{\begin{aligned}{\partial \rho _{0} \over \partial t}+\rho _{0}(\nabla \cdot {\vec {v}}_{0})&=0\\{\partial \rho _{0} \over \partial t}+3{{\dot {a}} \over a}\rho _{0}&=0\\\rho _{0}(t)&\propto a^{-3}\\\end{aligned}}\,\! To first order:

{\begin{aligned}{\partial \rho _{1} \over \partial t}+\rho _{0}(\nabla \cdot {\vec {v}}_{1})+\nabla (\rho _{1}{\vec {v}}_{1})&=0\\\rho _{0}{\dot {\delta }}+{\dot {\rho }}_{0}\delta +\rho _{0}(\nabla \cdot {\vec {v}}_{1})+\rho _{0}{\vec {v}}_{0}(\nabla \cdot \delta )+\rho _{0}\delta (\nabla \cdot {\vec {v}}_{0})&=0\\\end{aligned}}\,\! Using that ${\dot {\rho }}_{0}\delta =-3{{\dot {a}} \over a}\rho _{0}\delta$ , and $\nabla \cdot {\vec {v}}_{0}=3{{\dot {a}} \over a}$ :

${{\dot {\delta }}+\nabla \cdot {\vec {v}}_{1}+{{\dot {a}} \over a}({\vec {r}}\cdot \nabla )\delta =0}\,\!$ Euler’s equation gives us, to zeroth order:

${\partial {\vec {v}}_{0} \over \partial t}+({\vec {v}}_{0}\cdot \nabla ){\vec {v}}_{0}=-\nabla \cdot \phi _{0}\neq 0\,\!$ So there is no Jeans swindle in this case. Expanding this to first order, we get:

${\partial {\vec {v}}_{1} \over \partial t}+({\vec {v}}_{0}\cdot \nabla ){\vec {v}}_{1}+({\vec {v}}_{1}\cdot \nabla ){\vec {v}}_{0}=-{v_{s}^{2} \over \rho _{0}}\nabla \rho _{1}-\nabla \phi _{1}\,\!$ To figure out what $({\vec {v}}_{1}\cdot \nabla ){\vec {v}}_{0}$ is, let’s expand it:

{\begin{aligned}({\vec {v}}_{1}\cdot \nabla ){\vec {v}}_{0}&=H(v_{1x}{\hat {x}}+v_{1y}{\hat {y}}+v_{1z}{\hat {z}})({\hat {x}}\partial _{x}+{\hat {y}}\partial _{y}+{\hat {z}}\partial _{z}){\vec {r}}\\&=H{\vec {v}}_{1}={{\dot {a}} \over a}{\vec {v}}_{1}\\\end{aligned}}\,\! Thus our equation above becomes:

${{\partial {\vec {v}}_{1} \over \partial t}{{\dot {a}} \over a}({\vec {r}}\cdot \nabla ){\vec {v}}_{1}+{{\dot {a}} \over a}{\vec {v}}_{1}=-v_{s}^{2}\nabla \rho -\nabla \pi _{1}}\,\!$ Finally, we have Poisson’s Equation:

${\nabla ^{2}\phi =4\pi G\rho _{1}=4\pi G\rho _{0}\delta }\,\!$ 