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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.

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Which dimensions have negative $p_j$? Knowing a little bit about the global topology might help with the vacuum energy question. –  Chay Paterson Jun 12 '13 at 21:22
    
@ChayPaterson there are for example 3 expanding spatial dimensions and $n>1$ contracting dimensions. –  Dilaton Jun 12 '13 at 21:26
    
Sure. Do we know if the contracting dimensions are compact or noncompact? –  Chay Paterson Jun 12 '13 at 21:28
    
@ChayPaterson about that there is unfortunately no information, I rather thought they should be compact ? –  Dilaton Jun 12 '13 at 21:32
    
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) –  Chay Paterson Jun 12 '13 at 21:45

1 Answer 1

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.

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