I am currently working on the topic of inflation.

It seems that at the stage of inflation, the universe can be described as a de Sitter space. In such a space, all spacetime diffeomorphisms are preserved. (That is something I don't really understand but I keep reading that so I admit it for now.)

Now, I read that, in order to give the FLRW universe that we know today, the time diffeomorphisms are broken and therefore there is a Goldstone boson associated with this symmetry breaking. I also read somewhere else that it is not the inflaton field that is concerned by this broken symmetry but rather the fluctuation field $\delta \phi$ defined as $\phi(x,t) = \overline{\phi_0} (t) + \delta \phi (x,t)$.

  • What is the Goldstone boson ? It is not the inflaton field...
  • Does it makes sense to say that only time diffs are broken, since space and time coordinates are not so evidently separated ? I know that the background field $\overline{\phi_0}$ gives a clock but I'm still confused.
  • Why is FLRW not invariant under time diffs ? Because of the evolution of $a(t)$ ?
  • I thought that in order to apply Goldstone's theorem, the broken symmetry had to be continuous. When we speak about broken time diffs, is it a continuous symmetry ? (Maybe because it's only shift in time ?)

Sorry if it is not clear, as you can see I'm confused about this topic. Thank to anyone who can provide some answer, even partial.

EDIT : Found in 0905.3746, section 2.

In inflation there is a physical clock that controls when inflation ends. This means that time translations are spontaneously broken, and that therefore there is a Goldstone boson associated with this symmetry breaking. As usual, the Lagrangian of this Goldstone boson is highly constrained by the symmetries of the problem, in this case the fact that the spacetime is approximately de Sitter space, in the sense that $|\dot{H}| / H^2 \ll 1$. The Goldstone boson, that we can call $\pi$, can be thought of as being equivalent, in standard models of inflation driven by a scalar field, to the perturbations in the scalar field $\delta \phi$.

I believe it makes the things a bit more clear, at least for my first question.

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    $\begingroup$ Hey, this is very interesting, I did not see it as it was asked ... +1 $\endgroup$ – Dilaton Aug 22 '13 at 16:02
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    $\begingroup$ Thanks. I keep asking people about that, hopefully I'll get some convincing answer. If you are interested they are many other aspects about that. Essentially, this is used in the development of an effective field theory for inflation, arxiv.org/abs/0709.0293 :) They try to find the most general lagrangian on symmetries considerations, but for that they need to pick up a specific gauge, breaking a symmetry, leading to a GB boson (that's the part I don't understand). This can be restored using the so-called Stuckelberg trick, making the GB apparent. $\endgroup$ – AnSy Aug 23 '13 at 8:21

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