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

New features!

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

123 submissions , 104 unreviewed
3,547 questions , 1,198 unanswered
4,552 answers , 19,366 comments
1,470 users with positive rep
411 active unimported users
More ...

Torsion and gauge invariant EM kinetic term

+ 6 like - 0 dislike
28 views

Everytime I hear about adding torsion to GR, something struggles me: how do you create a kinetic term for the electromagnetic field that is still gauge-invariant? One of the consequences of torsion is that covariant derivatives no more commute on scalar functions, so by looking at the $F_{\mu\nu} = \nabla_{\mu}A_{\nu}-\nabla_{\nu}A_{\mu}$ (constrained to minimal coupling by the equivalence principle) and we perform a gauge transformation on it $A_{\mu} \to A_{\mu} + \partial_{\mu}\Lambda$ we get $F_{\mu\nu} \to F_{\mu\nu} + T^{\rho}_{\mu\nu}\partial_{\rho}\Lambda$ where $T^{\rho}_{\mu\nu}$ is the torsion tensor. So how can we handle torsion in non pure-gravity?

This post has been migrated from (A51.SE)
asked Jan 11, 2012 in Theoretical Physics by toot (445 points) [ no revision ]
retagged Mar 7, 2014 by dimension10

2 Answers

+ 4 like - 0 dislike

Your definition of $F_{\mu\nu}$ is strange. Assume the relevant $U(1)$-bundle is trivial, then $A$ is a 1-form on the base. The curvature $F=dA$ is independent of the metric. In coordinates you still use the formula without covariant derivatives: $F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu$.

This post has been migrated from (A51.SE)
answered Jan 11, 2012 by Pavel Safronov (1,115 points) [ no revision ]
+ 3 like - 0 dislike

The gauge invariance problem of the Maxwell field in the presence of torsion has been known for many years. There is no known perfect solution for this problem.

The importance of having the gauge invariance of the Maxwell action is that it implies charge conservation which is very well established experimentally.

One type of suggestions (please see Sabbata ) is through nonminimal coupling of the maxwell field to torsion. There are other suggestions restricting the types of connections.

However, the "solution" by Benn Dereli and Tucker Phys Lett. B. 96B 100-104 (1980) reviewed is a recent paper by Socolovsky(section 35) seems more appealing as it does not involve any change in the geometric structure of the Maxwell theory coupled to an Einstein-Cartan background, nor waives the minimal coupling. Here the Maxwell field is defined to be the exterior derivative of the vector potential (as in Pavel's answer), and the Maxwell action has its "flat space" form except for the curved space-time measure, which is manifestly gauge invariant. However, the independent variables are taken to be the tetrad components of the gauge potential $A_a =e_a^{\mu} A_{\mu}$ rather the space-time components. In this solution, the gauge noninvariance is manifested in the solution of the field equations for the torsion itself being proportional to the field's non gauge invariant spin density tensor. However, the Lorentz generators being the space integrals of the spin density zero components are still gauge invariant.

This post has been migrated from (A51.SE)
answered Jan 12, 2012 by David Bar Moshe (3,505 points) [ no revision ]
Just to see whether I understand. The problem arises if you take a connection $\nabla:TM\rightarrow TM\otimes\Omega^1$ on the tangent bundle, use the metric to get a connection on the cotangent bundle $\nabla':\Omega^1\rightarrow\Omega^1\otimes\Omega^1$, and then define $F = \nabla' A$, where you wedge the 1-forms in the end?

This post has been migrated from (A51.SE)
Firstly you don't need a metric to get a connection on the cotangent bundle. Isomorphism of vector spaces induces isomorphism of dual vector spaces hence if you can parallel transport vectors you can parallel transport convectors too. Hence an affine connection yields a covariant exterior derivative operator which equals the usual exterior derivative iff the torsion vanishes. Secondly I also don't understand why we need this covariant exterior derivative in the action instead of using the ordinary exterior derivative. Perhaps it is needed because otherwise the torsion has no physical meaning.

This post has been migrated from (A51.SE)
@Pavel, Sorry for the late response, I am afraid I didn't understand the question. Nevertheless, here are some more details , which I hope will be helpful. When formulated by means of the Vector potential, the Maxwell Lagrangian has the same form as on a Riemannian manifold. However, when written in terms of the tetrad components of the gauge field (which are locally $u(1)$ valued sections of the frame bundle, it becomes dependent on the spin connection and as a consequence one gets the correct Maxwell equations $\nabla *F = 0$ from its variation with respect to the tetrad components $A_a$.

This post has been migrated from (A51.SE)

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:
$\varnothing\hbar$ysicsOverflow
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).
To avoid this verification in future, please log in or register.




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

Your rights
...