# Varying an action (cosmological perturbation theory)

I am stuck varying an action, trying to get an equation of motion. (Going from eq. 91 to eq. 92 in the image.) This is the action

$$S~=~\int d^{4}x \frac{a^{2}(t)}{2}(\dot{h}^{2}-(\nabla h)^2).$$

And this is the solution,

$$\ddot{h} + 2 \frac{\dot{a}}{a}\dot{h} - \nabla^{2}h~=~0.$$

This is what I get

$$\partial_{0}(a^{2}\partial_{0}h)-\partial_{0}(a^{2}\nabla h)-\nabla(a^{2}\partial_{0}h)+\nabla^{2}(ha^{2})~=~0.$$

I don't really see my mistake, perhaps I am missing something. (dot represents $\partial_{0}$)

It is this problem (see Lectures on the Theory of Cosmological Perturbations, by Brandenburger):

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Comment to the question (v1): How do you get the second and third term with mixed temporal and spatial derivatives? – Qmechanic Mar 10 '13 at 17:41
Cross-posted from math.stackexchange.com/q/325481/11127 – Qmechanic Apr 23 '13 at 20:18

1. The Lagrangian density in the $(+,-,-,-)$ convention is $${\cal L}~=~\frac{a^2}{2}d_{\mu}h ~d^{\mu}h.$$
2. The corresponding Euler-Lagrange equation (by varying the action $S[h]=\int \!d^4x ~{\cal L}$ wrt. the field $h$) is $$d_{\mu}(a^2 ~d^{\mu}h)~=~0.$$
3. Or equivalently, under the assumption that $a=a(t)$, $$\frac{2\dot{a}\dot{h}}{a} + d_{\mu}d^{\mu}h~=~0.$$
thx of course, just to check does the Fourier transform mean that one should sub in $h=h(t)e^{ikx}$ or is it $h=h(t)e^{ikx}+h(t)^{*}e^{-ikx}$ ? I always get confused with doing this, how would one simplify it to get the desired result? – user21119 Mar 10 '13 at 18:07