I Inflation lecture from Guth : why does d(ρ)/dt = 0 ?

1. Jul 24, 2017

DoobleD

I watched a video lecture from Alan Guth on inflation (undergrad level), and there is something in it I don't understand.

He first presents the inflation scalar field, Φ, which has an energy density associated with it, V(Φ). V(Φ) is stuck at a local minimum, at Φ = 0 (what is called a "false vaccum") :

He then invoke an equation, derived earlier in the course, which relates the derivative with respect to time of the mass density of the Universe, ρ, to its pressure p :

Then, at around 9:43 prof. Guth says that the left-hand side of the equation, d(ρ)/dt, is 0, because the scalar field is stuck at the false vaccum value. Leading to :

Where u is the energy density of the Universe.

Here is what bugs me : ρ represents the mass density of the Universe, not the energy density of the false vaccum, V(Φ). I get that V(Φ) is constant, but why does that implies that the mass density ρ is also constant ?

2. Jul 24, 2017

DoobleD

I realize that if the false vaccum energy density is constant, the mass density associated with it, V(Φ)/c2, is also constant. But what about the mass density from the primordial matter in the Universe ? This part shouldn't be constant.

BUT, actually, a bit later in the lecture, Guth says that we assume the Universe is dominated by the false vaccum. So, in the equation for d(ρ)/dt, maybe he is simply making the approximation that there is no matter in the Universe, so that ρ is pretty much only the masse density corresponding of the false vaccum energy density ?

Last edited: Jul 24, 2017
3. Jul 24, 2017

Orodruin

Staff Emeritus
Inflation would quickly dilute any other contribution to the energy density. You can also solve the Friedmann equations with several different components, but if you have a dark energy component, it will dilute much slower than a matter or radiation component. One of the points of inflation is that when it ends you have essentially gotten rid of all other components and you can start to repopulate them by reheating.

4. Jul 24, 2017

DoobleD

Thank you, that answers the question.