Sign up ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

I was browsing through the hep-th arXiv and came across this article:

Spontaneous Lorentz Violation in Gauge Theories. A. P. Balachandran, S. Vaidya. arXiv:1302.3406 [hep-th]. (Submitted on 14 Feb 2013)

The authors say that Lorentz symmetry is spontaneously broken in QED and also in non-abelian gauge theories by infrared effects. However as far as I know the observed world is Lorentz invariant. Can someone kindly explain to me what is going on? Thanks for any help!

share|cite|improve this question
Huh interesting, +1. I hope somebody will give an answer here ... – Dilaton Feb 18 '13 at 11:18
They enlarge the gauge group to a "Sky" group that doesn't commute with Lorentz transformations and then superselect based on the new group. So of course they break Lorentz invariance. My problem is motivation. Why do they think what they are doing is physical? I really don't understand... – Michael Brown Feb 18 '13 at 14:19

1 Answer 1

up vote 10 down vote accepted

The possibility of spontaneous Lorentz symmetry violation due to the infrared problem of the Dirac-Maxwell equation was conjectured a long time ago by Frohlich, Morchio and Strocchi, in references [1,2] mentioned in the given Balachandran and Vaidya article.

In perturbative QED, we usually assume that the scattering states are free eigenstates of the number operator. But, due to the masslessness of the photon, this assumption is not true due to the infinite range of the electromagnetic force. This fact gives rise to the existence of nonfree asymptotic states for example an electron surrounded by an indefinite number of soft photons. This state can be represented by a photon coherent state (eigenstate of the annihilation operators of the physical polarizations). In fact the inclusion of these ‘dressed” states solves the infrared divergence problem in perturbation theory.

Now for large gauge transformations which do not vanish at infinity, one can prove that the charge density of such a charged state is conserved at every direction. This is a direct consequence of the Gauss law. (please, see an explanation in the following Wikipedia page).

This implies that a charged vacuum can assume any charge distribution at infinity, which is the sign of a spontaneous Lorentz symmetry breakdown since a Lorentz boost modifies the charge distribution at infinity). The only way that this violation does not take place is when the total charge (of the universe) is zero.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.