# Slow-roll parameter in Inflation

What is the definition of slow-roll parameter? What equation that these parameters are showed up?

I read a book "Modern Cosmology" by Scott Dodelson. In chapter 6, it is just given the parameter without mention what are these parameters actually saying and how cosmologists know the field roll slowly from those parameters.

• Just to say this is not a precise science at the moment. It is more a case of trying out models and seeing whether they appear to be plausible. In this sense the cosmologists don't "know" what the parameter may have done (slowly or otherwise), since they don't know that the model is correct. But in order for the model to have any chance of plausibility, certain energy features have to be very flat as a function of some parameter, making the evolution slow enough to allow a large expansion during the time the field evolves. Commented Feb 1 at 13:12

In order to produce an accelerating expansion (desirable for inflation and dark energy models) using a scalar field, that field must be slowly rolling. The two slow-roll parameters are defined as \begin{align} \epsilon &= \frac{1}{\kappa^2} \left(\frac{V_{, \phi}}{V} \right)^2, \\ \eta &= \frac{V_{, \phi \phi}}{\kappa^2 \phi}, \end{align} where $$\kappa = 8 \pi G$$, $$\phi$$ is the scalar field and $$V$$ its potential.

When $$\epsilon \ll 1$$ and $$|\eta| \ll 1$$, the field is rolling slowly enough for acceleration to occur.

More on the slow roll formalism can be found in this work by Liddle, Parsons and Barrow: https://arxiv.org/abs/astro-ph/9408015.

• Actually, I don't have any idea why it's called "slowly rolling".Can you give an intuitive explanation? Commented Jan 20, 2021 at 0:46
• @Adika essentially the time derivative of the field wrt the potential must be much smaller than the expansion rate of the Universe. Commented Jan 20, 2021 at 9:59

Cosmologists introduce a scalar field, named inflation, to provide vacuum energy for an inflation phase. In a homogeneous universe, the field equation for the inflation $$\phi$$ looks like: $$\ddot{\phi}+3H\dot{\phi}+\frac{d}{d\phi}V(\phi)=0$$ This equation looks like a ball rolling in a potential field $$V(\phi)$$, with a friction term proportional to $$3H$$. When we are able to trap $$\phi$$ in some local minimum, we can solve the Friedmann equations to arrive at a de Sitter solution $$a\propto e^{Ht}$$. What will happen if the field $$\phi$$ cannot be trapped? It will roll downhill, but it is not a big problem, since there is the friction term here. The friction is going to slow down the "little ball" until $$\dot{\phi}^2\ll V(\phi)$$, and then we may state that the universe is in an asymptotic de Sitter state. One lucky thing for us, is that the slow roll solution is an attractor for for most potentials, in other words, the evolution of the system will converge into a slow roll state whatever the initial conditions are. Hope this could help you build an intuitive picture.

• I have just noticed it is a question 3 years ago...
– wei
Commented Dec 27, 2023 at 9:17