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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This is a good question, it can be resolved by considering the thermal nature of deSitter and the continuation to a sphere--- you just thermally mix the two vacua, you go back and forth. The second part of the question is not so sensible, though. What are you doing with the bubble radius? The pressure is not up to you to control, it's determined by the new vacuum energy and the instanton structure. – Ron Maimon Aug 23 '12 at 6:15
To what extent can we be confident we're not sitting in some metastable vacuum that might yet decay (eg. we're in the symmetric phase of a Higgs potential which will fall apart once the universe cools enough - or alternatively, dark energy is some slow-rolling scalar field that will eventually reach its minimum)? – James Aug 30 '12 at 12:45

1 Answer

No, de Sitter is still metastable. Our minds evolved to visualize flat Euclidean geometry in 2D and 3D only and has problems visualizing higher dimensional highly warped Lorentzian geometries, which is why so much confusion exists. To aid our visual intuition, we have to "pretend" and project out a few extra dimensions, while imagining curved spacetime as a curved 2D surface embedded within a flat 3D Euclidean space which really transforms as a 2+1D Lorentzian space.

To aid this intuition, embed 3+1D de Sitter space within 4+1D Minkowski space with Cartesian coordinates T, W, X, Y and Z. Only the first is timelike. de Sitter space is the induced surface satisfying $-T^2 + W^2 + X^2 + Y^2 + Z^2 = 1/k^2$.

The solution that is tunneled to corresponds to cutting off de Sitter space to $W\leqslant -\sqrt{1/k^2 -c^2M_P^4/T_w^2}$ where $T_w$ is the domain wall tension, and attaching a flat Minkowski space at $W=\sqrt{1/k^2 -c^2M_P^4/T_w^2}$ bounded by its intersection with de Sitter.

Let $r^2 = X^2+Y^2+Z^2$, $Z=r\cos\theta$, $X=r\sin\theta \cos\phi$, $Y=r\sin\theta \sin\phi$.

In the coordinate system $(\tau=\sinh^{-1}(kT)/k,r,\Omega)$, the minimum bubble radius is $cM_P^2/T_w$, but there is inside-outside reversal for de Sitter space so that the remaining de Sitter space outside is compact with size $cM_P^2/T_w$ as well. This is what naive Wick rotation and the Coleman-de Luccia analysis gives.

Expanding de Sitter coordinates only cover a patch of half of de Sitter space corresponding to $W+T > 0$. Let $kt=\ln (k(W+T))$. In this coordinate system, $-(e^{kt}/k+\sqrt{1/k^2 -c^2M_P^4/T_w^2})^2+(1/k^2 -c^2M_P^4/T_w^2)+r^2 =1/k^2$. As $t\to -\infty$, the bubble radius approaches $1/k$ from the outside. In this coordinate system, it looks that going back in time, the wall asymptotically approaches the de Sitter radius but never gets smaller than that. This is just a coordinate singularity.

There is a tunneling from expanding de Sitter space everywhere to this solution. The tunneling doesn't happen at $T=0$ as suggested by Wick rotation, but for a small tiny positive value for $W+T$ to accommodate the bubble wall thickness just outside the de Sitter radius in the expanding de Sitter coordinates. Wick rotation gets it wrong because the Killing vector in the expanding de Sitter coordinates isn't globally timelike, and also, the Euclidean instanton solution doesn't apply for this steady state with no time reversal invariance in this coordinate system for t.

It's not true that the tunneling radius has to be around $T/k^2M_P^2$. That's what the naive Minkowski analysis gives.

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