Why do we think that the energy density of false vacuum/scalar field doesn’t dilute with the expansion?

I have a question about inflation theory.

Why do we think that the energy density of false vacuum/scalar field doesn’t dilute with the expansion?

Could it be that the energy density does dilute with the expansion?

Is it the nature of a scalar field to maintain its energy density even with the expansion? or Do we think that energy density doesn’t dilute with the expansion so that it will have negative pressure which cause expansion?

• – Kosm
Commented Jul 16, 2018 at 20:52
• energy density (almost) doesn't dilute BECAUSE it has negative pressure. Look at the equations. A scalar field can have negative pressure if its potential energy dominates over kinetic energy for some period of time
– Kosm
Commented Jul 20, 2018 at 2:38

The energy density does dilute during inflation: it has to, or else inflation would never end. The energy density is given by $$\rho = \frac{1}{2}\dot{\phi}^2+ V(\phi)$$ where $V(\phi)$ is some potential energy function. As long as $V(\phi)$ is not strictly constant, the inflaton field will evolve according to $$\dot{\rho} = \dot{\phi}\,(\ddot{\phi} + V'(\phi)).$$ If the total energy of the inflaton was conserved, then indeed we'd have $\dot{\rho} = 0$, but if we look at the Klein-Gordon equation for a scalar field in the Friedmann-Robertson-Walker background, $$\ddot{\phi} - 3H\dot{\phi} + V'(\phi) = 0,$$ we see that there is a drag term, proportional to the Hubble parameter, $3H\dot{\phi}$, that draws energy out of the inflaton as it evolves in its potential.
We therefore have $\dot{\rho} = -3H\dot{\phi}^2$, which is nonzero (and negative) as long as $\phi$ is evolving.
EDIT: I should add that the universe can still accelerate even if $\dot{\rho} \neq 0$. To see this, note that $\ddot{a} = a(\dot{H} + H^2) > 0$ which leads to the following bound, $$\frac{|\dot{H}|}{H^2} < 1.$$