Tell me more ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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):

enter image description here

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

1 Answer

Hints:

  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 $a=a(t)$, $$ \frac{2\dot{a}\dot{h}}{a} + d_{\mu}d^{\mu}h~=~0. $$

  4. Finally, Fourier transform the three spatial directions to get eq. (92).

share|improve this answer
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 at 18:07

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.