- #1

travelingscienceman

- 5

- 1

## Homework Statement

The general goal of the problem is to derive some useful identities involving the slow-roll parameters during inflation.

For part a show that:

$$\frac {d} {d\eta} (\frac {1} {aH})= \epsilon - 1$$

## Homework Equations

$$\epsilon \equiv \frac {d} {dt} (\frac {1} {H})= \frac {-\dot H} {aH^2}$$

$$H=\frac {\dot a}{a}$$

$$\eta\approx\frac{-1}{aH}$$

## The Attempt at a Solution

First I put everything into the scale factor a and its derivatives:

$$-\dot H = \frac{\dot a^2 - \ddot a a}{a^2}$$

$$\frac{1}{a H^2}=\frac{a}{\dot a^2}$$

$$\frac {-\dot H} {aH^2} = \frac{\dot a^2 - \ddot a a}{a\dot a^2}=\frac{1}{a}-\frac{\ddot a}{\dot a^2}=\epsilon$$

Now this is where I am struggling.

He uses a derivative with respect to time and one with respect to conformal time (##\eta##). Coupled with the fact that

$$\frac {d} {d\eta} (\frac {1} {aH})=\frac {d} {d\eta} (-\eta) =-1=\epsilon - 1$$

Which means that ##\epsilon=0## which is not correct. Instead of substituting I could do the following:

$$\frac {d} {d\eta} (\frac {1} {aH})=\frac {d} {d\eta} (\frac {1} {\dot a})=\frac {-1} {\ddot a}=\epsilon - 1$$

Rearranging we get

$$\epsilon=1-\frac {1} {\ddot a}$$

And I am not sure this is correct either. I cannot see any way to make the RHS equal to epsilon.

Any help showing where I might have gone wrong would be appreciated and thanks in advance for taking a look at this!