Monday, August 14, 2017

cosmology - If the nucleation bubble radius is greater than the deSitter radius, does that make the de Sitter space stable?


In our de Sitter phase, the cosmological constant is tiny. $10^{-123}M_P^4$. Suppose there is another phase with a lower vacuum energy. Is de Sitter phase still stable? The tunneling bubble radius has to exceed the de Sitter radius. Suppose a metastable decay to such a bubble happened. Take that final state, and evolve back in time. It's unlikely to tunnel back because of exponential suppression factors. Light cones are dragged outward in expanding de Sitter at such radii, so, by causality, the bubble radius has to keep shrinking back in time until at least the de Sitter radius. This contradicts our earlier assumption.


What about engineering a phase transition? Form a small bubble and stuff it with enough matter in the new phase with sufficient interior pressure to keep the bubble from shrinking. It collapses to a black hole if the radius R is much greater than $M_P^2/T$, which is much less than the de Sitter phase. The black hole then evaporates.



Even if the cosmological constant in the new phase is large and negative, the tunneling radius still has to be larger than the de Sitter radius because of the hyperbolic geometry of AdS means the volume of the new phase is only proportional to the domain wall area?


If we assume there is a large matter density in the de Sitter phase, it has to be very very large to make the tunneling radius smaller than the de Sitter radius.


Is this deSitter phase stable and not metastable?




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