Modern Cosmology Dodelson Problem 6.12b (Inflation)

[travelingscienceman]

Homework Statement

Show that $4\pi G(\dot \phi)^2=\epsilon a^2 H^2$

Homework Equations

Over dots mean derivative with respect to $\eta$.
$$\frac{1}{a}\frac{d}{d\eta}=\frac{d}{dt}$$
$$H=\frac{\dot a}{a^2}$$
$$\epsilon=\frac{-\dot H}{aH^2}$$
$$(\frac{\dot a}{a^2})^2=(\frac{da/dt}{a})^2=\frac{8\pi G\rho}{3}$$
$$\frac{\ddot a}{a^3}=\frac{d^2 a/dt^2}{a}=-\frac{4\pi G}{3}(\rho-P)$$
$$\rho=\frac{1}{2}(\frac{d\phi}{dt})^2 + V(\phi)$$
$$P=\frac{1}{2}(\frac{d\phi}{dt})^2 - V(\phi)$$
$$\frac{d\phi}{dt}=\frac{\dot\phi}{a}$$

The Attempt at a Solution

First I plug $\epsilon$ into the main equation so I get $4\pi G(\dot phi)^2=-a \dot H$. Then I find
$$\dot H=\frac{\ddot a a-2\dot a^2}{a^3}$$
$$-a\dot H=\frac{2\dot a^2-\ddot a a}{a^2}=2a^2(\frac{\dot a}{a^2})^2-a^2\frac{\ddot a}{a^3}$$
$$=a^2(\frac{16\pi G}{3}\rho+\frac{4\pi G}{3}(\rho+3P))$$
$$=a^2\frac{4\pi G}{3}(4\rho +\rho+3P)=a^2\frac{4\pi G}{3}(5\rho +3P)$$
$$=a^2\frac{4\pi G}{3}(\frac{5}{2}\frac{\dot\phi^2}{a^2}+5V+\frac{3}{2}\frac{\dot\phi^2}{a^2}-3V)$$
$$=a^2\frac{4\pi G}{3}(4\frac{\dot\phi^2}{a^2}+2V)$$

Which is where I have problems as I cannot simplify things further to get the desired $4\pi G(\dot \phi)^2$. If the 2V was gone and there was one less factor of $\dot \phi^2$ then I would get the right answer, but I keep coming to this same equation. Any help would be appreciated. Thanks in advance!

 

[TSny]

$$ \frac{\ddot a}{a^3}=\frac{d^2 a/dt^2}{a}=-\frac{4\pi G}{3}(\rho-P)$$

$\large \frac{\ddot a}{a^3} \neq \frac{d^2 a/dt^2}{a}$ You will need to be careful when relating $\large \frac{d^2 a}{dt^2}$ to $\ddot a$.

Also, $\large \frac{d^2 a/dt^2}{a}$ $\neq -\frac{4\pi G}{3}(\rho-P)$. Looks like a typographical error here. Shouldn't the $-P$ be $+3P$?

 

[travelingscienceman]

Thanks for your help once again! I was able to solve it after reevaluating $\frac{d^2 a}{dt^2}$ and got $\frac{\ddot a a-\dot a^2}{a^3}$.

And yeah it's 3P, messed up typing it in.

 

[TSny]

I was able to solve it after reevaluating $\frac{d^2 a}{dt^2}$ and got $\frac{\ddot a a-\dot a^2}{a^3}$.

Looks good.