# Cosmology Lecture 14

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### 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 ${\displaystyle \rho }$, and a piece of volume ${\displaystyle V}$ at distance ${\displaystyle R}$. Then the force of gravity on that piece is:

${\displaystyle {F_{grav} \over V}\sim {GM\rho \over R^{2}}\sim G\rho ^{2}R\,\!}$

since ${\displaystyle M\sim \rho R^{3}}$. The pressure force on that volume is:

${\displaystyle {F_{pres} \over V}\sim \nabla P\sim v_{s}^{2}\nabla \rho \sim v_{s}^{2}{\rho \over R}\,\!}$

Comparing these forces, we find that gravity wins if:

${\displaystyle G\rho ^{2}R>v_{s}^{2}{\rho \over R}\,\!}$

Solving for ${\displaystyle R}$:

${\displaystyle R>{\sqrt {v_{s}^{2} \over G\rho }}\sim \lambda _{J}\,\!}$

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:

{\displaystyle {\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:

${\displaystyle t_{sound}\sim {R \over v_{s}}\,\!}$

We will have gravitational collapse if ${\displaystyle t_{ff}, which means:

${\displaystyle {\begin{matrix}{1 \over {\sqrt {G\rho }}}<{R \over v_{s}}&\Rightarrow &R>{\sqrt {v_{s}^{2} \over G\rho }}\end{matrix}}\,\!}$

So again we get out the Jeans wavelength.

### Gravitational Instability in an Expanding Fluid

Density is obviously affected by expansion. To zeroth order:

{\displaystyle {\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 ${\displaystyle a}$ is our familiar scale factor. To examine first order perturbations, we define the density field ${\displaystyle \delta }$, where ${\displaystyle \rho _{1}\equiv \rho _{0}\delta }$, so that:

${\displaystyle \delta ={\rho _{1} \over \rho _{0}}={\rho -\rho _{0} \over \rho _{0}}\,\!}$
${\displaystyle {\delta ={\partial \rho \over \rho _{0}}}\,\!}$

Our velocities are also changed by expansion:

${\displaystyle {\vec {v}}_{0}=H{\vec {r}}={{\dot {a}} \over a}{\vec {r}}\,\!}$

where ${\displaystyle {\vec {r}}}$ is in physical coordinates. This velocity is purely from Hubble expansion. Note that:

${\displaystyle \nabla \cdot {\vec {v}}_{0}={{\dot {a}} \over a}(\nabla \cdot {\vec {r}})=3{{\dot {a}} \over a}\,\!}$

We’ll use ${\displaystyle {\vec {v}}_{1}}$ 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:

{\displaystyle {\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:

{\displaystyle {\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 ${\displaystyle {\dot {\rho }}_{0}\delta =-3{{\dot {a}} \over a}\rho _{0}\delta }$, and ${\displaystyle \nabla \cdot {\vec {v}}_{0}=3{{\dot {a}} \over a}}$:

${\displaystyle {{\dot {\delta }}+\nabla \cdot {\vec {v}}_{1}+{{\dot {a}} \over a}({\vec {r}}\cdot \nabla )\delta =0}\,\!}$

Euler’s equation gives us, to zeroth order:

${\displaystyle {\partial {\vec {v}}_{0} \over \partial t}+({\vec {v}}_{0}\cdot \nabla ){\vec {v}}_{0}=-\nabla \cdot \phi _{0}\neq 0\,\!}$

So there is no Jeans swindle in this case. Expanding this to first order, we get:

${\displaystyle {\partial {\vec {v}}_{1} \over \partial t}+({\vec {v}}_{0}\cdot \nabla ){\vec {v}}_{1}+({\vec {v}}_{1}\cdot \nabla ){\vec {v}}_{0}=-{v_{s}^{2} \over \rho _{0}}\nabla \rho _{1}-\nabla \phi _{1}\,\!}$

To figure out what ${\displaystyle ({\vec {v}}_{1}\cdot \nabla ){\vec {v}}_{0}}$ is, let’s expand it:

{\displaystyle {\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:

${\displaystyle {{\partial {\vec {v}}_{1} \over \partial t}{{\dot {a}} \over a}({\vec {r}}\cdot \nabla ){\vec {v}}_{1}+{{\dot {a}} \over a}{\vec {v}}_{1}=-v_{s}^{2}\nabla \rho -\nabla \pi _{1}}\,\!}$

Finally, we have Poisson’s Equation:

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