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 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

145 submissions , 122 unreviewed
3,930 questions , 1,398 unanswered
4,852 answers , 20,624 comments
1,470 users with positive rep
501 active unimported users
More ...

Examples for p-form gauge fields

+ 3 like - 0 dislike
247 views

I don't completely understand the notion "p-form". Can you give me examples of 1-form, 2-form and 3-form gauge fields? What kind of p-form is e.g. the Higgs field, the electromagnetic four-potential, etc.?

This post imported from StackExchange Physics at 2015-10-03 21:47 (UTC), posted by SE-user LCF
asked Oct 2, 2015 in Theoretical Physics by LCF (15 points) [ no revision ]
Higgs, photons, W-Bosons are charge carriers, not gauge fields. Their description in QFT is achieved by means of gauge theories on fibre bundles, which in turn make use of tensor calculus on manifold, where $p$-form formalism takes place.

This post imported from StackExchange Physics at 2015-10-03 21:47 (UTC), posted by SE-user Gennaro Tedesco
Sorry, I meant not the gauge bosons but the corresponding gauge fields such as the electromagnetic four-potential. Why do we have a 3-form for Higgs and not a 2-form?

This post imported from StackExchange Physics at 2015-10-03 21:47 (UTC), posted by SE-user LCF

2 Answers

+ 3 like - 0 dislike

One way to understand it is that the "reason" why ordinary gauge fields are given by 1-forms is that the term in the action of a charged point particle that gives, under variation, the Lorentz force of a background electromagnetic field on the particle, is simply the integral of that 1-form over the worldline of the particle.

Now as a particle = 0-brane is replaced by a p-brane, then the direct analog of the gauge coupling term is the integral of a (p+1)-form field over the worldvolume of the p-brane.

A famous example of a 2-form gauge field is the Kalb-Ramond B-field that the fundamental string (a 1-brane) is charged under. Then there is the 3-form C-field.under which the M2-brane is charged. 

For more see also at Higher Prequantum Geometry I: The need for prequantization.

answered Oct 7, 2015 by Urs Schreiber (5,775 points) [ no revision ]
+ 2 like - 0 dislike

Not sure whether I understood the intention of your question correctly.

In electrodynamics you usually use a 1-form $A=A_{\mu}dx^{\mu}$ to write down an action \begin{equation*} S = \int F \wedge \star F \end{equation*} with the 2-form field strength $F=dA$, which the gives the known (vacuum) Maxwell equations.

Scalar fields $\phi$ (such as the Higgs) are described by 0-forms or alternatively by their dual $(d-2)$-forms in $d$ dimensions.

In string theory one finds cousins of the electromagnetic field $A$, but they turn out to be $p$-forms, e.g. in type IIB we have the R-R-fields $C_0, C_2,C_4$,...

psm

This post imported from StackExchange Physics at 2015-10-03 21:47 (UTC), posted by SE-user psm
answered Oct 2, 2015 by psm (55 points) [ no revision ]
Thanks a lot! However, unfortunately I am not used to your notations. Your action is equivalent to $S=\int F_{\mu \nu} F^{\mu \nu}$ or not? What exactly means wedge product followed by Hodge operator? And $dA$ means $d_{\mu}A_{\nu}-d_{\nu}A_{\mu}$?

This post imported from StackExchange Physics at 2015-10-03 21:47 (UTC), posted by SE-user LCF
And what do you mean with "dual field", what is that physically?

This post imported from StackExchange Physics at 2015-10-03 21:47 (UTC), posted by SE-user LCF
Yes, up to some factor $F \wedge \star F \sim d^dx F_{\mu\nu}F^{\mu\nu}$. In $d$ dimensions, $\star F$ is by definition a $(d-2)$-form if $F$ is a 2-form. Then, the wedge product can be applied as in the case of wedging a p- and a q-form. And $dA= \partial_{\mu}A_{\nu}dx^{\mu}\wedge dx^{\nu}$. (See definition of exterior derivative.)

This post imported from StackExchange Physics at 2015-10-03 21:47 (UTC), posted by SE-user psm
Concerning the dual field: Take a p-form potential $C_p$. Its field strength tensor is then $F=dC_p$, a $(p+1)$-form. Thus, in $d$ dimensions $\tilde{F} \equiv \star dC_p$ is a $(d-p-1)$-form, and locally there exists a $(d-p-2)$-form $\tilde{C}_{d-p-2}$ s.t. $\tilde{F}=d \tilde{C}$. We call $\tilde{C}$ the dual field to $C$. Physically they should be equivalent...

This post imported from StackExchange Physics at 2015-10-03 21:47 (UTC), posted by SE-user psm
Thanks again! I am just wondering where these dual fields come from, because up to know I never met them in QFT. When the dual field is a (d-2)-form if the field is a 2-form, how can they then be physically equivalent?

This post imported from StackExchange Physics at 2015-10-03 21:47 (UTC), posted by SE-user LCF
You're welcome! Careful, the dual field of a 2-form is a $(d-4)$-form. One has to figure this out by dualising the field strength tensor. (See my last comment.) To see, why these should be physically equivalent: i) One can rewrite the action $S[C_p]$ into $S[\tilde{C}_{d-p-2}]$ and they're structurally identical. In particular, the Bianchi identity & the eq. of motion for $C$ become the e.o.m. & Bianchi id. for $\tilde{C}$. ii) The on-shell degrees of freedom for a $p$-form and a $(d-p-2)$-form gauge field are identical.

This post imported from StackExchange Physics at 2015-10-03 21:47 (UTC), posted by SE-user psm
Notice however, that I'm not sure whether the physical equivalence holds true in all field theories involving $p$-form gauge fields. But it should be true at least for abelian gauge fields with Lagrangians involving kinetic terms as above. Maybe someone else can provide more details...

This post imported from StackExchange Physics at 2015-10-03 21:47 (UTC), posted by SE-user psm

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$\hbar$ysi$\varnothing$sOverflow
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
...