How does the graph of V(φ) support the concept of slow-roll inflation?

[robertjford80]

This is a screenshot from one of susskind's cosmology lectures.

it shows a graph of V(φ). As you can see it slowly rules down then, bam, it rolls really fast down hill. I thought inflation erased all the information before it happened. So how does he know there is a gradual decline before a fast decline?

 

[Chalnoth]

This is a screenshot from one of susskind's cosmology lectures.

it shows a graph of V(φ). As you can see it slowly rules down then, bam, it rolls really fast down hill. I thought inflation erased all the information before it happened. So how does he know there is a gradual decline before a fast decline?

Well, yes, inflation erases pretty much all information from before, but leaves its own imprint upon the resulting universe. That said, this is just a heuristic model of inflation used to explain the key concepts. There are many others. The current best-fit actually seems to be simply a potential given by:

V(\phi) = \alpha \phi^2.

Here \alpha is a constant that gives the overall scale of the potential. This shouldn't be too much of a surprise since most any potential will look like a harmonic potential near its minimum, and it is pretty much only the behavior of the potential near the end of inflation that is detectable. So even if we had a complicated potential like the one shown in that graph, it might still come out just looking like a harmonic potential.

 

[clamtrox]

Well, yes, inflation erases pretty much all information from before, but leaves its own imprint upon the resulting universe. That said, this is just a heuristic model of inflation used to explain the key concepts. There are many others. The current best-fit actually seems to be simply a potential given by:

V(\phi) = \alpha \phi^2.

Here \alpha is a constant that gives the overall scale of the potential. This shouldn't be too much of a surprise since most any potential will look like a harmonic potential near its minimum, and it is pretty much only the behavior of the potential near the end of inflation that is detectable. So even if we had a complicated potential like the one shown in that graph, it might still come out just looking like a harmonic potential.

I don't think that's true -- although I'm not an expert in inflation so please educate me if I'm wrong. But in the usual models, doesn't inflation end right away when the slow roll conditions are broken? In a potential like that, after you start oscillating in the minimum, the slow roll conditions do not hold so the universe is no longer inflating. In fact, I seem to remember that the oscillating inflaton behaves like regular matter, satisfying \rho \sim a^{-3}. If the oscillating inflaton is coupled to regular standard model particles, then it can decay into them, heating up the universe and starting the regular big bang expansion.

 

[Chalnoth]

I don't think that's true -- although I'm not an expert in inflation so please educate me if I'm wrong. But in the usual models, doesn't inflation end right away when the slow roll conditions are broken? In a potential like that, after you start oscillating in the minimum, the slow roll conditions do not hold so the universe is no longer inflating. In fact, I seem to remember that the oscillating inflaton behaves like regular matter, satisfying \rho \sim a^{-3}. If the oscillating inflaton is coupled to regular standard model particles, then it can decay into them, heating up the universe and starting the regular big bang expansion.

Right, but that end is never instantaneous, in any inflation model. The slow roll conditions are, after all, a statement that the change in the field value is very slow compared to the expansion. Such conditions cannot be broken instantaneously, because they involve one continuous value becoming significant in size compared to another. So inflation always has an at least somewhat gradual end.

I believe in realistic inflation models the end of inflation has to be fairly rapid, but it is never instantaneous.