Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

During inflation the metric is de-Sitter so $dt^2-d\underline{X}^2 $.

I know that the eqn.motion governing GW's from inflation (tensor perturbations) is


derived from varying $$S^{(2)}~=~\int \frac{a^{2}(t)}{2}~\partial^{\mu}h ~\partial_{\mu}h ~d^{4}x.$$

See paper 1 and paper 2. Surely this isn't valid during inflation because it isn't de-Sitter. I have read lots of papers/lectures where quantum fluctuations are found by deriving equations of motion from a metric with an $a(t)$ and then Fourier decomposed. Surely during inflation the metric can't contain an $a(t)$. So why do all these lectures/papers use it?

share|cite|improve this question
Related question by OP: – Qmechanic Mar 12 '13 at 23:18

De Sitter space is a special case of the Robertson-Walker spacetime. If you want, you can even work out what the coordinate transformation is.

If your question isn't answered by this, could you please amend it to make it clearer exactly where your hangup is?

share|cite|improve this answer
ok sorry, just to be clear could the metric $dt^2 - a^2 dx^2$ describe inflation? I am just getting confused because the above derivation seems to use that mtric in a perturbed form but surely that is only valid in flat space-time after inflation? – user21119 Mar 12 '13 at 22:20
I just don;t understand how the above eqn can be valid for perturbations created during inflation because de-sitter space is similar to minkowski and yet that derivation seemed to involved a non de-sitter metric (at least to my understanding). – user21119 Mar 12 '13 at 22:24
There is a simple coordinate transformation from the metric you are using to an FRW metric with a particular $a$ function. They are physically the same. – Michael Brown Mar 13 '13 at 0:56
@user21119: yes. There is a coordinate system that covers a patch of De Sitter space in which the metric has that form. – Jerry Schirmer Mar 13 '13 at 15:20

Your Answer


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