I Cosmological Scalar Field Density Dilution

[Samama Fahim]

Consider a homogeneous free scalar field $\phi$ of mass m which has a
potential

$$V(\phi) = \frac{1}{2}m^2\phi^2$$

Show that, for $m ≫ H$, the scalar field undergoes oscillations with
frequency given by $m$ and that its energy density dilutes as
$a^{−3}$.

This is from Modern Cosmology, Scott Dodelson, Chapter 6.

For the part "Show that its energy density dilutes as $a^{−3}$", following is my attempt:

In the equation $\frac{\partial \rho}{\partial t} = -3H(P+\rho)$, put $P = \frac{1}{2} \dot{\phi}^2-V(\phi)$ and $\rho=\frac{1}{2} \dot{\phi}^2+V(\phi)$ to get

$$\frac{\partial \rho}{\partial t} = -3\frac{\dot{a}}{a}\dot{\phi}^2,$$

where $H = \dot{a}/a$. I am not sure how to proceed or proceed or whether this is the correct approach. Should I use Friedmann's equation instead? But it involves densities of other species as well, and there is no assumption here whether one species dominates. Or Should I convert $d\rho/dt$ to $d \rho/ da$?

Kindly provide a hint as to how to proceed.

 

[ergospherical]

See section II, Turner M.S. 1983,
https://journals.aps.org/prd/abstract/10.1103/PhysRevD.28.1243

 

[Samama Fahim]

See section II, Turner M.S. 1983,
https://journals.aps.org/prd/abstract/10.1103/PhysRevD.28.1243

It needs subscription.

 

[ergospherical]

Sorry, going through my replies and must have missed this at the time...
The paper goes into some detail but you can proceed in a more simple way.

Recall from the Klein Gordon equation for the inflation,$$\ddot{\phi} + 3H\dot{\phi} + V' = 0$$The key assumption is to neglect the Hubble friction, $3H\dot{\phi} \ll 1$, i.e. you have an underdamped oscillator when $m$ is large. We are given that the potential is quadratic (this would also apply for small oscillations around any quadratic minimum), so you have $V' = m^2\phi$ and the equation of motion $\ddot{\phi} + m^2 \phi = 0$ is just SHM with frequency $m$. So write, for example,$$\phi = \phi_0 \cos{m t}$$For the next part, you have the right idea,$$\dot{\rho} = -3H(\rho + P) = -3H \dot{\phi}^2 $$Finally you just look at the time average. Note $\langle \dot{\phi}^2 \rangle = \frac{1}{2} m^2 \phi_0^2 = \langle \rho \rangle$, so you just get$$d \ (\log \langle \rho \rangle) = -3 \ d (\log a)$$which then spits out the $\rho \sim a^{-3}$ dependence after inflation ends.