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
5,082 questions , 2,232 unanswered
5,353 answers , 22,789 comments
1,470 users with positive rep
820 active unimported users
More ...

  Relationship between local and global scaling (Weyl) symmetry

+ 3 like - 0 dislike
1123 views

Theorem 5.1 on page 80 of this paper says that

Assuming that the matter fields satisfy their equations of motion, the matter field action is locally Weyl invariant if and only if the corresponding energy-momentum tensor is traceless.

If I have understood this right, Weyl invariance reduces to scale invariance in flat spacetime, which is basically what I am up to at the moment.

My question now is:

If one has Weyl or scale invariance in accordance with this theorem, does this mean that global Weyl or scale invariance (such that the scaling parameter does not depend on position) is guaranteed too? If not why not?

asked Nov 27, 2012 in Theoretical Physics by Dilaton (6,240 points) [ revision history ]
Dear Dilaton, the Weyl invariance is scaling invariance but the coefficient of scaling must be allowed to depend on the position, otherwise it's not Weyl. The global scaling is always a subgroup of the Weyl symmetry if the latter exists, so the scaling is guaranteed if Weyl holds. So Weyl is always local. Above 2 spacetime dimensions, it's somewhat hard or unnatural to find Weyl-invariant theories. In 2 dimensions, the 2-derivative kinetic terms are naturally Weyl-symmetric because the scaling of $g^{\mu\nu}$ contracting $\partial_\mu$ and $\partial_\mu$ cancels the transformation of $\sqrt g$

This post imported from StackExchange Physics at 2014-03-12 15:39 (UCT), posted by SE-user Luboš Motl
Thanks @LubošMotl for this helpful and enlightening comment. Is the vanishing of the trace of the energy-momentum tensor a neccessary and sufficient contition for global scale invariance to hold then?

This post imported from StackExchange Physics at 2014-03-12 15:39 (UCT), posted by SE-user Dilaton
Actually, from the proof in that paper it would appear that what you propose is true. They use an arbitrary conformal scaling of the metric so it should be true globally.

This post imported from StackExchange Physics at 2014-03-12 15:39 (UCT), posted by SE-user levitopher

1 Answer

+ 3 like - 0 dislike

I do not think that local Weyl (conformal) invariance implies global invariance. The fields are only defined over local open sets $U_i$, so if the action is invariant under a constant conformal transformation (which does not depend on $x\in U_i$)

$\phi\to \Omega_i^2 \phi$

this does not imply anything about the global properties of $\Omega:M\to \mathbb{R}$, only that $\Omega|U_i$ is constant. There is something called a "conformal connection" which I don't know very much about, but I think the short answer is that if you restrict the fields to satisfy some specific symmetry group $G$ (i.e. they are part of a $G$-bundle and $G$ is some group of conformal transformations) then the fields will always be conformally invariant, locally and globally.

This post imported from StackExchange Physics at 2014-03-12 15:39 (UCT), posted by SE-user levitopher
answered Nov 27, 2012 by levitopher (160 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$\hbar$ys$\varnothing$csOverflow
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
...