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

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Since mass can be given to particles via the interaction with the Higgs Field could there be a "Charger Field" that supplies particles with charge? Possibly this would require two different "charger bosons" one for + and one for -.

share|cite|improve this question
Related: – Qmechanic Aug 11 '12 at 21:11

Though charge and mass are fundamentally different concepts, one can cook up an interaction like

$L = |d\phi - |\alpha|^2 A \phi|^2+\lambda(|\alpha|^2-c^2)^2 + |d_B \alpha|^2 + dA^2 + dB^2$.

Here $\alpha$ and $\phi$ are complex scalars and $A$ and $B$ are $U(1)$ gauge fields. $|-|$ denotes complex magnitude and $d_B$ is the covariant derivative for $\alpha$ under $B$.

There are two gauge symmetries

$\phi \rightarrow e^{i\theta}\phi$ with corresponding transformation for $A$, and $B$ fixed


$\alpha \rightarrow e^{i\zeta}\alpha$, with corresponding transformation for $B$, and $A$ fixed

At large $\lambda$, the second gauge symmetry is broken, causing $B$ to gain mass and $\phi$ to gain charge $c^2$, the vev of $|\alpha|^2$, which acts as a charge (for $\phi$ in the $A$ field) in the above Lagrangian. The phase of $\alpha$ is then left over as a real scalar : the Higgs boson.

Experts, please let me know if this is hogwash. This is not something I've ever seen in the literature.

share|cite|improve this answer
the difference is that the gauge transformations here cannot close a circle, you can't have a naturally compact U(1), and this model is also nonrenormalizable since the interaction is dimension 5. This really doesn't work with compact gauge groups. – Ron Maimon Aug 11 '12 at 20:56
What do you mean cannot close a circle? Are you saying that there are more gauge transformations than the ones generated by the two I've listed? Renormalizability is a good point. – Ryan Thorngren Aug 11 '12 at 20:59
The U(1) cannot be compact if there are different charged particles with non-quantized charged relationship in the presence of the field $\alpha$. The ratio of charges has to stay an integer, and then your coupling is unnatural. – Ron Maimon Aug 11 '12 at 21:04
I don't understand what you mean by compactness. To me U(1) is compact. The charge relationship here is a bit funky, but I don't think there is any (Dirac) quantization condition as long as the principal bundles for $A$ and $B$ are both trivial. How do you see such a constraint arising? – Ryan Thorngren Aug 11 '12 at 21:07
Yes, that's it. But if you rescale the field "A" to absorb $\alpha$, it can stay U(1), just with a space-time dependent radius for the circle. This is not inconsistent, just nonrenormalizable. That's fine, so your model is ok, it is just that it is not natural to do it in the way you did, because the ratios of charges of different particles have to stay integers. +1 for the answer, it's probably the most interesting example. You don't need to ask for help from "experts"--- it's not hogwash--- it's an interesting coupling one should consider. – Ron Maimon Aug 11 '12 at 22:26

The standard model of particle physics describes three forces of nature - electromagnetism, the weak interaction and the strong nuclear force. Each interaction is described by a so-called gauge theory.  That is, additional symmetries are added to the theory for each force.  One symmetry you may be familiar with from classical physics is rotational invariance - you can rotate a system, and Newton's laws still work.  So, classical physics is invariant under rotations.  The group of rotations has a name, O(N).  This also includes reflections.  N is the number of dimensions.  Another symmetry you may know of is Lorentz invariance - this includes rotations and translations, along with boosts.  Boosts are the familiar Lorentz transformations of time and space that leave the speed of light invariant.  Since special relativity is formulated in four dimensional spacetime, we need four dimensions.  However, we distinguish between one of the dimensions and the rest, which is time.  So, the Lorentz group is SO(3,1).  Why an S?  This stands for special, meaning that this group doesn't have all of the symmetries of O(3,1), which has too many to be an actual description of the world.

So, gauge symmetries are a bit different.  They're essentially symmetries of the theory that are redundant.  While in the classical version of the theory one can disregard them (you can use the electric and magnetic field instead of the 4-potential), they are necessary for consistency in the quantum theory.  So, each force has one of these gauge symmetries.  

So, when you impose gauge symmetries of a theory, you get an interesting result - they end up requiring that some particles have charge, that electric and magnetic fields exist, that anti-particles exist, what quantities are conserved, what interactions can and cannot occur, and many other things.

So, we can see that many features of the Standard Model, such as charge, exist because of the local gauge symmetries that exist in the theory.  This isn't the case for mass, which is why we have the Higgs mechanism.  However, we don't need a Higgs mechanism for charge, because it emerges out of the gauge symmetries.

And also see Ron's post, showing that such a thing wouldn't even be consistent.

share|cite|improve this answer
It wouldn't necessarily be inconsistent, just nonrenormalizable, and requiring that the charges change proportionally. This is not fatal--- this type of coupling is expected in GUT's near the string scale. But it's not interpretable as $\phi$ "giving charge" to a field, rather as $\phi$ proportionately changing the charge of a field. – Ron Maimon Aug 11 '12 at 22:23
Oh, okay. Thanks for correcting that. – Mark M Aug 11 '12 at 22:25

You can't have charger fields giving charge to particles individually, (except for the case of noncompact U(1) gauge group described by user404143), because charges are discrete not continuous. The closest you can come is through the nonrenormalizable interaction

$$ \phi \mathrm{tr}(F_{\mu\nu}F^{\mu\nu})$$

Which, by altering the VEV of $\phi$ alters the charges of all the particles coupled to the gauge field in $F$ by a proportional amount. You can rescale the A field to do it in the particle part of the action, but this leads to all charges changing proportionally to the fundamental charge.

The only case where this fails is the noncompact U(1) (QED without charge quantization, which is exceptional in many regards, in particular it remains renormalizable with a mass term). In this case, you can do it by arbitrarily changing the charge using $\phi$.

The models are never renormalizable, since by power counting, the minimal coupling is already at the limit of dimension allowed, and any variation in the constants will just make it non-renormalizable. The reason is obvious from the kinetic coupling--- kinetic terms are those that define the dimension of the field, and making the kinetic term coefficient a field, making a nonlinear sigma-model, always takes you away from renormalizable theories except in dimension 2 where fields are dimensionless.

In 2d you can probably do this, there is no barrier to a term of the form above, the field $\phi$ is dimensionless. But in this case, the gauge field is non-propagting, and you have confinement always, so it would be a field dependent Regge slope, not a field dependent charge.

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.