# Is eternal inflation Lorentz invariant?

Start without general relativity. Consider a metastable vacuum over good ol'-fashioned Minkowski space. It decays. A bubble forms and the domain wall expands. The domain wall is timelike, and accelerates, driven by the difference in pressure between both phases. Its speed approaches the speed of light asymptotically. Far away, it looks like an expanding light cone, but it stops looking so around the moment of decay. Roughly speaking, this breaks Lorentz covariance, and this picks out an approximate frame of reference. It might be argued that the metastable phase prior to decay is Lorentz invariant. If so, there is only one Lorentz invariant measure over the space of future-oriented unit 4-velocity vectors up to a multiplicative factor. This space has infinite measure, and the typical decay velocity of a decay would have to be infinite. Nonsense. The point to note is we have to specify the initial conditions of the metastable vacuum. That can't be infinitely in the past, as the lifetime, while exponentially long, is still finite. A Lorentz invariant initial boundary can only be an expanding light cone, or a hyperboloid. However, both have infinite volumes, and to regulate, we need to cut off these surfaces at some finite radius. This radius has to be less than the decay lifetime. This cutoff roughly picks out a frame of reference. At any rate, affine parameters on light cones are tricky. Boost them enough, and you can get an exponential change in the affine parameter, and in this boosted frame, the likelihood of crossing a bubble nucleation approaches one in probability.

Turn on general relativity. Does eternal inflation break Lorentz invariance?

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Why do you say your example breaks Lorentz covariance? – John Rennie Aug 6 '12 at 15:21