Cosmological Scalar Field Density Dilution

  • #1
Samama Fahim
52
4
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.
 
Last edited:
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  • #4
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.
 
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