# The Mass Terms

The standard model would be much more simple and beautiful without the mass terms. But we have them, and a complete theory has to describe them.

### Should the mass terms be explained?

At the current moment the model presented here is unable to predict the mass terms. And it seems not very probable that future research will change this situation – the point is that there will be some freedom in the definition of the model, some freedom of choice of the material properties, and this freedom means that not all parameters will be fixed. It seems quite reasonable to expect that most of the mass terms do not have some more fundamental explanation, but are simple material paramters.

But there are a few qualitative properties which require some explanation. In particular, one characteristic property of the mass terms is that the three neutrinos have much smaller masses than the other leptons and the quarks. This large difference in the mass terms is something which requires explanation.

The same can be said about the other large mass difference – that the weak gauge bosons have quite large masses and, on the other hand, gluons and photons are massless (or have extremely small masses).

## The masses of the gauge bosons

### The explanation of the zero masses of gluons

If a particle does not have mass, this may be explained by a symmetry which would be destroyed by the introduction of a mass term. This is the case for gauge fields: If there is gauge symmetry, the gauge bosons cannot have a nonzero mass. They have to be massless.

In the standard model, the three gauge fields, or, more accurate, the three fields described with the formalism of gauge theory, are not all massless: Only the gluons (the gauge bosons of the strong interaction) and the photon are massless. Instead, the W- and Z-bosons which are used to describe the weak interaction have masses.

This is an interesting difference which cries for explanation. And here our model allows to give a quite simple and natural explanation in terms of gauge symmetry.

The question is how the continuous gauge symmetry becomes represented at the fundamental level, on the lattice. The problem is that the lattice model should be in principle only an approximation. But an approximation possibly violates the symmetry of the continuous problem.

In the case of strong interactions, this does not happen. The strong interaction is described by a Wilson gauge field in our lattice model. But such Wilson lattice gauge fields have an exact gauge symmetry even in the lattice model. Thus, in this case we have an exact gauge symmetry in our lattice model. And, in agreement with the general theory of gauge fields, this leads to massless gauge bosons of the corresponding gauge field. That means, the gluons should be massless.

### The explanation why the weak bosons are massive

The situation is different for weak interactions which are represented on the lattice as fields which describe deformations of the lattice. These fields give, in the continuous limit, some gauge-like fields. In particular, they interact with the fermions in exactly the same way (with the same interaction Lagrangian) as true gauge fields. But this is only an approximation, and there is no exact gauge symmetry in this theory.

But once there is no exact gauge symmetry on the lattice, there is no reason to expect that the corresponding bosons are massless. So, we can expect that they have non-zero masses.

Unfortunately, this is all we can tell at the current moment. We have no rule how to compute the masses themself. Nonetheless, the qualitative correspondence is a nice one: If we have gauge symmetry on the lattice, we have massless gauge bosons, and if we have no gauge symmtry on the lattice, the gauge bosons have masses.

### The case of the photon

But what about the photon? The photon is described as a combination of a strong and weak bosons. So why is it massless? One would expect that a combination of a massless and a massive particle will have mass.

Thus, the zero mass of the photon seems to be a problem. But there seems to be also a solution of this problem. This solution requires some more research, it is, at the current moment, more an idea how to solve the problem than really a solution.

Anyway, the idea is that it is the exact gauge symmetry of the U(1)B part which causes the zero mass of the photon. The Wilson gauge fields on the lattice which have exact gauge symmetry form the group SU(3)c⊗ U(1)B. This group is the same as the group of massless gauge fields in the standard model, which is SU(3)c⊗ U(1)EM. It only acts in a slightly different way.

Now, there is a good reason why the gauge field cannot act like U(1)B. In this case, the ground state (the "Dirac sea") would be charged, and there are good reasons to require that the ground state has to be neutral. The gauge action of U(1)EM has this property.

This suggests the following picture: Because of the principle that the ground state has to be neutral, the U(1) gauge field cannot act like U(1)B. But this does not mean that its fundamental symmetry simply disappears, but, instead, means that the action of the symmetry on fermions becomes distorted. So we obtain U(1)EM as a distorted version of the action of U(1)B.

But if this is only a distorted action of a symmetry, it means that we have some exact (even if distorted) U(1) gauge symmetry. And, as a consequence, the photon mass should be zero.

## The fermion masses

There is not much we can explain at the current moment about the fermion masses, except the remarkable fact that the neutrino masses are much smaller than the masses of the other fermions. And, slightly generalizing the argument, also why leptons have lower masses than quarks and the lower (down, strange, bottom) quarks have lower masses than the upper (up, charmed, top) quarks. But these are also only qualitative explanations, no explicit computations of these masses.

### Why there are three acoustic phonons

The fermions are some sort of phonons of the lattice. "Phonons" are a known quantum effects of sound waves in a similar way as photons are quantum effects of EM waves. One can find a lot of information about them in every book on quantum condensed matter theory.

One characteristic property of the phonons is that they have something similar to masses. And there is the remarkable fact that in a crystal lattice there are three phonons which have zero masses, the so-called "acoustic phonons". There are also other phonons, which have masses, and which are named "optical phonons".

And even more interesting for our purpose is the explanation in condensed matter theory for this fact. The point is that there are oscillations associated with translations: The whole lattice oscillates as a whole. Because there is translational symmetry, and the internal relations between neighbour atoms remain unchanged if the whole lattice oscillates in the same way, these oscillations have low energy for low frequencies, and, in the limit of zero frequency, they need no energy at all. Thus, they have no "mass gap", or, in other words, these phonons are massless. Such oscillations of the lattice as a whole are possible in all three directions of space, thus, we obtain three such massless phonons.

There are other modes of oscillations. For example, atoms on even positions may oscillate in counterphase with atoms on odd positions. Here we have no such effect that the energy goes to zero because of translational symmetry of the lattice. The oscillations of this type always deform the lattice, and this needs energy. And, because of quantization, there remains some minimal energy which is necessary to obtain a nontrivial oscillation of this type – in other words, a nontrivial mass term.

### The association of neutrinos with translations

So, if we think that particles are quantum effects of oscillations of some lattice, then we should expect some particles which are massless or have at least much lower masses, which are the analog of the "acoustic phonons". And these (almost) massless particles should be associated with translations.

But we already have an association of the neutrinos with translations. We have postulated this association for completely different reasons, not thinking at all about "acoustic phonons". The reason was that the particles associated with translations will not interact with any gauge fields, and that the particles of the SM which do not interact with any gauge fields are the right-handed neutrinos.

Thus, the association of the neutrinos with translations already exists: The direction of translations are three directions inside the sector of the right-handed neutrinos.

But what follows exactly for the mass terms? Here, a complication occurs, namely that the neutrinos define only translations of the lattice of cells, but not of the material between them. This makes the association with translations non-ideal. As a consequence, once the translations of the cells describe only an approximate symmetry, we do not obtain ideal mass zero for the neutrinos. All we obtain is that their masses should be smaller than the masses of the other fermions.

### Relative masses of different types of fermions

We can use a similar thinking to decide about the closeness to translations of different particles to decide about their relative masses. The principle is similar: The particles more closely associated with translations will have smaller masses.

This works nicely. The translations are located in the leptonic sector, in its lower part. Thus, the principle would predict that leptons have lower masses than quarks, and that lower particles have lower masses than upper particles. This is exactly what we observe in each generation: The particle with lowest mass is the neutrino, followed by the lepton, followed by the lower quark, and with the upper quark as the particle with highest mass in each generation.