2

I am stuck varying an action. This is the action $$S=\int\mathrm d^{4}x \frac{a^{2}(t)}{2}(\dot{h}^{2}-(\nabla h)^2)$$

And this is the solution, $\ddot{h} + 2 \frac{\dot{a}}{a}\dot{h} - \nabla^{2}h $.

This is what I get $$\partial_{0}(a^{2}\partial_{0}h)-\partial_{0}(a^{2}\nabla h)-\nabla(a^{2}\partial_{0}h)+\nabla^{2}(ha^{2})$$ 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)

$\quad$To quadratic order in the fluctuating fields, the action separates into separate terms involving $h_+$ and $h_x$. Each term is of the form $$S^{(2)}=\int\mathrm d^4x\dfrac{a^2}{2}\left[h'^2-(\nabla h)^2\right]\,,\tag{91}$$lading to the equation of motion $$h_k''+2\dfrac{a'}{a}h_k'+k^2h_k=0\,.\tag{92}$$ The variable in terms of which the action $\text{(91)}$ has canonical kinetic term is $$\mu_k\equiv ah_k\,,\tag{93}$$ and its equation of motion is $$\mu_k''+\left(k^2+\dfrac{a''}{a}\right)\mu_k=0\,.\tag{94}$$

Going from step 91 to 92

  • 1
    The solution to which problem? – joriki Mar 09 '13 at 13:47
  • The solution is what I have found in papers to varying the action w.r.t h, but I can't get it I only get the last equation which is wrong and I don't understand why. –  Mar 09 '13 at 14:22
  • "varying the action" isn't a well-defined problem that I recognize. In case you mean "forming the first variation of the action", that would have to be a functional; but what you call the solution isn't a functional. – joriki Mar 09 '13 at 14:26
  • I am going to post an image of the paper that might clear things up. –  Mar 09 '13 at 14:44
  • Aha, you're trying to derive an equation of motion. Note that this is a mathematics site, not a physics site, so to reach a lot of potential answerers you should explain any physics that you're presupposing, e.g. how the variation of the action is related to the equation of motion. Also you might want to explain how the $k$s in the image you posted relate to the differential operators in your question. – joriki Mar 09 '13 at 15:04
  • sorry, the k's are just a fourier transform, il just post on the physics one. –  Mar 09 '13 at 15:30
  • Cross-posted to http://physics.stackexchange.com/q/56444/2451 – Qmechanic Apr 23 '13 at 20:18

1 Answers1

2

Varying the action we find the Euler-Lagrange equation of motion $$\frac{\partial}{\partial x^\mu} \frac{\partial \mathcal{L}}{\partial (\partial_\mu h)} = \frac{\partial \mathcal{L}}{\partial h}$$ or $$\frac{\partial}{\partial t} \frac{\partial \mathcal{L}}{\partial \dot h} - \frac{\partial}{\partial x^i} \frac{\partial \mathcal{L}}{\partial (\partial_i h)} = \frac{\partial \mathcal{L}}{\partial h},$$ where $$\mathcal{L} = \frac{1}{2} a(t)^2 \left(\dot h^2 - (\nabla h)^2\right)$$ is the Lagrangian (also called the Lagrangian density). Boundary terms can typically arise but are here ignored on physical grounds.

We find $$\begin{eqnarray*} \frac{\partial}{\partial t}\frac{\partial \mathcal{L}}{\partial \dot h} &=& \frac{\partial}{\partial t}(a^2 \dot h) \\ &=& a^2\ddot h + 2 a \dot a \dot h \\ \frac{\partial}{\partial x^i}\frac{\partial \mathcal{L}}{\partial (\partial_i h)} &=& \frac{\partial}{\partial x^i}(a^2 \partial_i h) \\ &=& a^2 \nabla^2 h \\ \frac{\partial \mathcal{L}}{\partial h} &=& 0. \end{eqnarray*}$$ Therefore, $$a^2 \ddot h + 2 a\dot a \dot h - a^2 \nabla^2 h = 0.$$ Dividing by $a^2$ we find the claimed result, $$\ddot h + 2 \frac{\dot a}{a} \dot h - \nabla^2 h = 0.$$

user26872
  • 19,465