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How can contracting dimensions lead to cosmological inflation?

+ 4 like - 0 dislike
20 views

Using the Kasner metric, given by

$$ ds^2 = -dt^2 + \sum_{j=1}^D t^{2p_j}(dx^j) $$

it is possible to not only describe the cosmological expansion of some space directions (the ones with positive Kasner exponents $p_j$, but this metric allows for some dimensions to contract too, those have negative $p_j$. The two Kasner conditions

$$ \sum_{j=1}^{D-1} p_j = 1 $$

and

$$ \sum_{j=1}^{D-1} (p_j)^2 = 1 $$

say that there have to be contracting and expanding dimensions at the same time, as the $p_j$ can not all have the same sign.

In a comment I have read, that in models with for example 3 expanding and $n>1$ contracting dimennsions, the contracting dimensions drive the inflation in the other directions by leading their expansion to accelerate without a cosmological constant. This is interesting and about this I'd like to learn some more.

So can somebody a bit more explicitely explain how such inflation models work? For example what exactly would the vacuum energy from a physics point of view be in this case? Up to now I only heard about inflation models where the vacuum energy density is the potential energy of some inflaton field(s) in a little bit more detail.

asked Jun 12, 2013 in Theoretical Physics by Dilaton (4,175 points) [ revision history ]
Which dimensions have negative $p_j$? Knowing a little bit about the global topology might help with the vacuum energy question.

This post imported from StackExchange Physics at 2014-03-09 16:17 (UCT), posted by SE-user Chay Paterson
@ChayPaterson there are for example 3 expanding spatial dimensions and $n>1$ contracting dimensions.

This post imported from StackExchange Physics at 2014-03-09 16:17 (UCT), posted by SE-user Dilaton
Sure. Do we know if the contracting dimensions are compact or noncompact?

This post imported from StackExchange Physics at 2014-03-09 16:17 (UCT), posted by SE-user Chay Paterson
@ChayPaterson about that there is unfortunately no information, I rather thought they should be compact ?

This post imported from StackExchange Physics at 2014-03-09 16:17 (UCT), posted by SE-user Dilaton
Oh, ok. Well, my thinking was that the cosmological constant has a different meaning depending on if you are looking at just the 3+1 dimensional slice of the spacetime or the full D+1 dimensional spacetime: on the 3+1 dimensional slice, it might still be interpretable as the potential energy of a scalar field, but I think that will depend on whether or not you can separate out the dynamics of the compact dimensions. (Disclaimer: I'm not an expert)

This post imported from StackExchange Physics at 2014-03-09 16:17 (UCT), posted by SE-user Chay Paterson
@ChayPaterson this is about the full D+1 dimensional spacetime, even though the metric can be partitioned into the expanding and contracting dimensions, they are not "warped" or something (if I rememeber correctly what "wharped" means).

This post imported from StackExchange Physics at 2014-03-09 16:17 (UCT), posted by SE-user Dilaton
OK: the full D+1 dimensional spacetime doesn't have a vacuum energy.

This post imported from StackExchange Physics at 2014-03-09 16:17 (UCT), posted by SE-user Chay Paterson

1 Answer

+ 1 like - 0 dislike

You seem to be talking about inflation and expansion as if they were the same thing; they aren't. A Kasner metric has expansion and contraction, but it doesn't have anything like inflation. Inflation is exponential and is driven by a scalar field; the Kasner metrics are vacuum solutions and their behavior isn't exponential.

[...]what exactly would the vacuum energy from a physics point of view be in this case?

There is no vacuum energy in a Kasner metric; the Kasner metrics are vacuum solutions, i.e., solutions of the Einstein field equations with zero stress-energy tensor and zero cosmological constant.

[...] the contracting dimensions drive the inflation in the other directions by leading their expansion to accelerate without a cosmological constant.

This is something that has to happen because it's a vacuum solution. In a vacuum solution, the Ricci tensor vanishes. The interpretation of a vanishing Ricci tensor is that the only gravitational forces are tidal in character, as opposed to the kind of gravitational forces you get from a source that's present in that the region of space where you're measuring the curvature. One way to distinguish a tidal from a non-tidal force is that if you release a cloud of test particles in a purely tidal field, the volume of the cloud is conserved. If you want to conserve volume, you can't have expansion along all axes.

This post imported from StackExchange Physics at 2014-03-09 16:17 (UCT), posted by SE-user Ben Crowell
answered Jun 12, 2013 by Ben Crowell (1,070 points) [ no revision ]

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