The Einstein-Hilbert Lagrangian (along with a scalar field) in FRW spacetime reads:
\begin{equation}
\mathcal{L} = – \frac{1}{8 \pi G} (3 a \dot{a}^2 – 3 k a + \Lambda a^3) + \frac{1}{2} \dot{\phi}^2 a^3 – \mathcal{V}(\phi) a^3.
\end{equation}
Let's determine the Euler-Lagrange EoM w.r.t. the variable $a$. First, the canonical conjugate momentum
$$ \frac{\partial \mathcal{L}}{\partial\dot{a}} = -\frac{6 a \dot{a}}{8\pi G} $$
Then the time derivative of the conjugate momentum
$$\frac{{\rm d} }{{\rm d}t} \frac{\partial \mathcal{L}}{\partial\dot{a}} = -\frac{6 \dot{a}^2 + 6 a \ddot{a}}{8\pi G} $$
Thus the EoM reads
$$\frac{{\rm d} }{{\rm d}t} \frac{\partial \mathcal{L}}{\partial\dot{a}} = \frac{\partial\mathcal{L}}{\partial a} \implies -\frac{6 \dot{a}^2 + 6 a \ddot{a}}{8\pi G} = – \frac{-3 k + 3 \Lambda a^2}{8\pi G} + 3 a^2 \left(\frac{1}{2}\dot{\phi}^2 – \mathcal{V}(\phi)\right) $$
Simplifying we get
$$ \frac{\ddot{a}}{a} + \frac{\dot{a}^2}{a^2} + \frac{k}{2 a^2} – \frac{\Lambda}{2} = – 4 \pi G \left(\frac{1}{2}\dot{\phi}^2 – \mathcal{V}(\phi)\right) $$
Somehow I am not getting the correct factor in front of the $H^2$ term. I believe the correct EoM is
\begin{equation}
2 \frac{\ddot{a}}{a} + \frac{\dot{a}^2}{a^2} + \frac{k}{a^2} = \Lambda – 8 \pi G \Big( \frac{1}{2} \, \dot{\phi}^2 – \mathcal{V}(\phi) \Big).
\end{equation}
Can someone point out to me where I am going wrong in the derivation?
Best Answer
I think you are missing the term $-\frac{3\dot{a}^2}{8 \pi G}$ when you compute $\frac{\partial\mathcal{L}}{\partial a}$.