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 -.
Asked
Active
Viewed 804 times
3 Answers
4
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.
Mark M
- 1,485
3
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
and
$\alpha \rightarrow e^{i\zeta}\alpha$, 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.
Ryan Thorngren
- 8,254
2
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.
