Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

206 submissions , 164 unreviewed
5,103 questions , 2,249 unanswered
5,355 answers , 22,798 comments
1,470 users with positive rep
820 active unimported users
More ...

  How can contracting dimensions lead to cosmological inflation?

+ 4 like - 0 dislike
1502 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 (6,240 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 ]

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ysicsOverfl$\varnothing$w
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...