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

205 submissions , 163 unreviewed
5,047 questions , 2,200 unanswered
5,345 answers , 22,709 comments
1,470 users with positive rep
816 active unimported users
More ...

  Why is there no dynamics in 3D general relativity?

+ 7 like - 0 dislike
1207 views

I heard a couple of times that there is no dynamics in 3D (2+1) GR, that it's something like a topological theory. I got the argument in the 2D case (the metric is conformally flat, Einstein equations trivially satisfied and the action is just a topological number) but I don't get how it is still true or partially similar with one more space dimension.

This post imported from StackExchange Physics at 2014-04-04 16:30 (UCT), posted by SE-user toot
asked May 24, 2012 in Theoretical Physics by toot (445 points) [ no revision ]

1 Answer

+ 7 like - 0 dislike

The absence of physical excitations in 3 dimensions has a simple reason: the Riemann tensor may be fully expressed via the Ricci tensor. Because the Ricci tensor vanishes in the vacuum due to Einstein's equations, the Riemann tensor vanishes (whenever the equations of motion are imposed), too: the vacuum has to be flat (no nontrivial Schwarzschild-like curved vacuum solutions). So there can't be any gravitational waves, there are no gravitons (quanta of gravitational waves). In other words, Ricci flatness implies flatness.

Counting components of tensors

The reason why the Riemann tensor is fully determined by the Ricci tensor is not hard to see. The Riemann tensor is $R_{abcd}$ but it is antisymmetric in $ab$ and in $cd$ and symmetric under exchange of the index pairs $ab$, $cd$. In 3 dimensions, one may dualize the antisymmetric index pairs $ab$ and $cd$ to simple indices $e,f$ using the antisymmetric $\epsilon_{abe}$ tensor and the Riemann tensor is symmetric in these new consolidated $e,f$ indices so it has 6 components, just like the Ricci tensor $R_{gh}$.

Because the Riemann tensor may always be written in terms of the Ricci tensor and because they have the same number of components at each point in $D=3$, it must be true that the opposite relationship has to exist, too. It is $$ R_{abcd} = \alpha(R_{ac}g_{bd} - R_{bc}g_{ad} - R_{ad}g_{bc} + R_{bd}g_{ac} )+\beta R(g_{ac}g_{bd}-g_{ad}g_{bc}) $$ I leave it as a homework to calculate the right values of $\alpha,\beta$ from the condition that the $ac$-contraction of the object above produces $R_{bd}$, as expected from the Ricci tensor's definition.

Counting polarizations of gravitons (or linearized gravitational waves)

An alternative way to prove that there are no physical polarizations in $D=3$ is to count them using the usual formula. The physical polarizations in $D$ dimensions are the traceless symmetric tensor in $(D-2)$ dimensions. For $D=3$, you have $D-2=1$ so only the symmetric tensor only has a unique component, e.g. $h_{22}$, and the traceless condition eliminates this last condition, too. So just like you have 2 physical graviton polarizations in $D=4$ and 44 polarizations in $D=11$, to mention two examples, there are 0 of them in $D=3$. The general number is $(D-2)(D-1)/2-1$.

In 2 dimensions, the whole Riemann tensor may be expressed in terms of the Ricci scalar curvature $R$ (best the Ricci tensor itself is $R_{ab}=Rg_{ab}/2$) which is directly imprinted to the component $R_{1212}$ etc.: Einstein's equations become vacuous in 2D. The number of components of the gravitational field is formally $(-1)$ in $D=2$; the local dynamics of the gravitational sector is not only vacuous but it imposes constraints on the remaining matter, too.

Other effects of gravity in 3D

While there are no gravitational waves in 3 dimensions, it doesn't mean that there are absolutely no gravitational effects. One may create point masses. Their gravitational field creates a vacuum that is Riemann-flat almost everywhere but creates a deficit angle.

Approximately equivalent theories

Due to the absence of local excitations, this is formally a topological theory and there are maps to other topological theories in 3D, especially the Chern-Simons theory with a gauge group. However, this equivalence only holds in some perturbative approximations and under extra assumptions, and for most purposes, it is a vacuous relationship, anyway.

This post imported from StackExchange Physics at 2014-04-04 16:30 (UCT), posted by SE-user Luboš Motl
answered May 24, 2012 by Luboš Motl (10,278 points) [ no revision ]
Thanks for the very detailed explanation Lubos, should have counted the components of the Riemann and the Ricci tensors.

This post imported from StackExchange Physics at 2014-04-04 16:30 (UCT), posted by SE-user toot
I like this nice explanation too :-)

This post imported from StackExchange Physics at 2014-04-04 16:30 (UCT), posted by SE-user Dilaton
In the expression for the Riemann tensor in terms of Ricci tensor, $\alpha=1,\beta=-1$, right?

This post imported from StackExchange Physics at 2014-04-04 16:30 (UCT), posted by SE-user c.p.

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$\hbar$ysicsOve$\varnothing$flow
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
...