# Curvature perturbation

1. Nov 27, 2009

### Rick89

Hi,
I am having some problems understanding this concept, I hope you can help.
I studied on Hobson, Efstathiou and Lasenby, in chapter 16 on Inflationary cosmology that in cosmological perturbation theory we need to express quantities in a gauge invariant way, very clear so far. The problem there is: we have a scalar field perturbation $\phi (t) -> \phi_0 (t) +\delta\phi (t,x)$ and we assume a perturbed metric from the flat background FRW which is $ds^2=(1+2\Phi)dt^2 -(1-2\Phi) R(t)^2 (dx^2+dy^2+dz^2)$ ,where $\Phi$ is the small perturbation.
They set about to derive a gauge invariant related to these quantities and derive what they call the curvature perturbation:
$\Phi+ H\delta \phi / \phi_0 (t)'$
where $\phi_0'$ denotes the time derivative of $\phi_0$ and H is the Hubble parameter. They then work in terms of this curvature perturbation saying that it does not depend on the coordinates used for the perturbation and so it can be done. I would like to understand how to obtain and how to understand that curvature perturb. That piece of the text is very unclear in my opinion, everything else in the book is very clear and well written, at least for undergrads... Thanx

Last edited: Nov 27, 2009
2. Nov 28, 2009

### Rick89

Thanx anyway

3. Nov 29, 2009

### clamtrox

Basically, the curvature perturbation is proportional to the curvature scalar of a constant-time hypersurface in comoving coordinates. Usually it's defined as

$$4 \nabla^2 \mathcal{R} = R^{(3)}.$$

4. Nov 29, 2009

### Rick89

You said in a constant time hypersurface, but if we perturbed the metric (to solve for the perturbed scalar field) how do we know that the time coord we are choosing does not give us gauge dependent effects?

5. Nov 30, 2009

### clamtrox

It's not gauge-independent in the general case, but transforms as

$$\mathcal{R} \rightarrow \mathcal{R} + H \delta t$$

However, you'll find that for adiabatic perturbations the different gauges coincide as long as the expansion of the universe is not exactly exponential (in which case everything with $$\delta t^{-1}$$ in it will diverge).

There's a short (but slightly longer) explanation in arxiv, in http://arxiv.org/abs/astro-ph/0003278

6. Nov 30, 2009

### cristo

Staff Emeritus
$\mathcal{R}$ doesn't transform like that; it is a gauge invariant quantity. To see this,

$$\mathcal{R}=\psi+\frac{\mathcal{H}}{\varphi_0'}\delta\varphi\to\psi+\mathcal{H}\delta\eta+\frac{\mathcal{H}}{\varphi_0'}\left(\delta\varphi-\varphi_0'\delta\eta\right)$$

where the last equality comes from using the gauge transformations (5) and (6) in the above paper. This gives

$$\mathcal{R}\to\psi+\frac{\mathcal{H}}{\varphi_0'}\delta\varphi=\mathcal{R}$$

so R is gauge invariant.

7. Dec 1, 2009

### clamtrox

That R is the curvature perturbation evaluated in the uniform density hypersurface, which is defined in equation [7] on the paper I posted. You're right in that it's gauge-independent, but that's only because you define it by transforming the generic curvature perturbation into a certain gauge before calculating it.

I guess I should have mentioned that :)

8. Dec 1, 2009

### Rick89

Hi, thanks. I also found this article which I think is very very clear about that (it's where the concept was first introduced...)

Gauge-invariant cosmological perturbations
J. M. Bardeen, Phys. Rev. D 22, 1882 (1980).

9. Dec 1, 2009

### cristo

Staff Emeritus
No, R is the curvature perturbation on comoving hypersurfaces. The curvature perturbation on uniform density hypersurfaces, $\zeta$ is in general different, though the two do coincide on large scales.

Absolutely, if you can follow Bardeen's 1980 paper, then that's the best source for this topic, but it is pretty complicated (which is why there have been several reviews written). I imagine (to answer your first question) the reason things weren't done rigorously in the text book you mention is that to do so would require first writing down the metric with general scalar perturbations (the one you quote is already in a gauge-- the longitudinal, or newtonian, gauge), and then considering the gauge transformations for each quantity, before constructing the gauge invariant variables. If you want details, I suggest you look at some of the reviews (or maybe even some more specific textbooks).

10. Dec 1, 2009

### Rick89

Yes, it is pretty difficult. Can you suggest some reviews? Thanx