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  Cosmic bubble collision

+ 2 like - 0 dislike
1708 views

I'm reading this review right now. The claim seems to be that when you have an expanding "false vacuum", finite size bubbles form due to phase transition (cause by finite action instantons) (which, in itself might then evolve, etc). My questions is, if I imagine these bubbles to be moving away from each other due to exponential expansion of the false vacuum, then how will they come near each other and collide?

This post has been migrated from (A51.SE)
asked Oct 6, 2011 in Theoretical Physics by Siva (720 points) [ no revision ]
retagged Apr 19, 2014 by dimension10
I would've also liked to tag it under cosmology, multiverse and string-landscape (some might consider those tags controversial), but wasn't able to since I didn't have enough points.

This post has been migrated from (A51.SE)
May I ask why you are using a famous person's name instead of your own, given that you are a physics graduate student, and presumably interested in building a positive reputation for yourself in the international community? It leaves a bad taste in my mouth, because it is as though you are claiming to be God's gift to physics (or at least this site).

This post has been migrated from (A51.SE)
Oh... Don't read so much into it. It was a nickname I picked up online in homage to Feynman. That was a long time ago. It stuck and a lot of my online friends know me by that nickname. If you notice, it's a single word, like a nickname and not "Dick Feynman". I don't see why it has to be an issue. I'm not claiming to be Feynman.

This post has been migrated from (A51.SE)

1 Answer

+ 5 like - 0 dislike

The size of these bubbles is growing nearly by the speed of light. So if the boundaries of two such bubbles are close enough to each other, the exponential expansion of the parent space in between them, even if this expansion exists, is negligible relatively to the shrinking distance between the bubbles due to their growth. That's why the bubbles collide after a finite time.

Of course, if the bubbles are far enough, e.g. (or i.e.) behind the horizons of each other, their growth isn't enough to overcome the exponentially growing separation and they cannot collide.

This post has been migrated from (A51.SE)
answered Oct 6, 2011 by Luboš Motl (10,278 points) [ no revision ]
So you're saying that the linear(with near light speed) growth of the bubbles will trump the exponential expansion of the false vacuum (which is sublinear, for small distances). Is it obvious that the bubbles should expand at near light speed?

This post has been migrated from (A51.SE)
Dear @Siva, concerning the first point, one really has to be careful about the emotional meaning of words. If "something" grows exponentially, it doesn't mean that it will beat "anything" that grows linearly because we're comparing different quantities here. The shrinking of the space between the bubbles goes like $1-t$ where $t$ is time, that's the linear one because of the size growth, but it's corrected to $\exp(ct)-t$ because the distance inflates by cosmology. Still, $\exp(ct)-t$ crosses zero and goes negative for some values of $c,t$. The word "exponential" doesn't mean that it wins.

This post has been migrated from (A51.SE)
Concerning the second point, the near-luminal expansion is obvious from the maths of the bubbles. A bubble is described by an instanton localized in the Euclidean spacetime, near $x^2+y^2+z^2+t_E^2\leq a^2$ or so. By Wick rotation, you may see that the transition is concentrated near the light cone $x^2+y^2+z^2-t^2\leq a^2$, so the point where the Universe switches is expanding nearly by the speed of light. Alternatively, the huge energy released from the difference of energy densities (in the bulk, difference between the two phases) is invested to accelerate the wall between the two phases.

This post has been migrated from (A51.SE)
The energy released is $V(phase_1)-V(phase_2)$ times $volume$ while the "rest mass" of the domain wall separating the two phases only grows as the $area$ which is much smaller for large radii, $r^2

This post has been migrated from (A51.SE)

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