Modern Cosmology by Dodelson: Problem 6.12 a

[travelingscienceman]

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!

 

[Dick]

Try using that $\frac {d} {d\eta} = a \frac{d}{dt}$. So you've got $a \frac{d}{dt} (\frac{1}{a} \frac{1}{H})$. It's pretty straightforward from there.

 

[travelingscienceman]

That helps but I am still having trouble. With $\epsilon=\frac{-\dot H}{a H^2}$ I just end up going in circles. But if I work with $\epsilon=\frac{d}{dt}\frac{1}{H}$ I get the following:

$$\epsilon=\frac{d}{dt}\frac{1}{H}=\frac{\dot a^2 - \ddot a a}{\dot a^2}$$
$$\frac{d}{d\eta}\frac{1}{\dot a}=a\frac{d}{dt}\frac{1}{\dot a}=\frac{-a}{\ddot a}=\epsilon-1$$
$$1+\frac{a}{\ddot a}=\epsilon=\frac{\dot a^2 - \ddot a a}{\dot a^2}$$
$$\dot a^2 \ddot a -a \dot a^2=\ddot a \dot a^2 - \ddot a^2 a$$
$$\dot a =\ddot a$$

Which is better but not right.

 

[TSny]

From looking inside Dodelson's text at Amazon, it appears that the author uses a dot to denote differentiation with respect to $\eta$. But, it looks like you are sometimes interpreting the dot as differentiation with respect to $t$ (for example, when you write $H =\large \frac{\dot a}{a}$).

 

[Dick]

From looking inside Dodelson's text at Amazon, it appears that the author uses a dot to denote differentiation with respect to $\eta$. But, it looks like you are sometimes interpreting the dot as differentiation with respect to $t$ (for example, when you write $H =\large \frac{\dot a}{a}$).

Ah, that explains some inconsistencies. I assumed dot was differentiation by $t$ from that definition of $H$.

 

[travelingscienceman]

Thank you both. Today with a clear head I figured it out.

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

And so

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

Which made things much easier.