We already can do this in materials--- it is called "superconductivity". The phenomenon for photons was understood before the phenomenon in the weak interactions, and the description of superconductivity by Landau, and the Bardeen Cooper Schriefer model for fermionic paired condensates, was the inspiration for Nambu's fermion vacuum condensate idea, and Brout and Englert's later point-particle superconducting Higgs mechanism.

Photons do not go slower in a superconductor, they do not go at all. Superconductors don't have photon excitations at all, and if you have an electric and magnetic field in the superconductor trying to propagate, the fields decay away exponentially.

It is certain that we won't be able to do this in vacuum, because we know all the fields around us are stable. In order to make an instability in the field, we have to alter the fundamental constants in such a way that a charged field makes a superconducting condensate. In order to alter the constants, we would need a certain energy density per unit volume which is going to be practically infinite for the purposes of engineering.

But the condensed matter analog, the superconductor, is a perfect analog, and we understand the dynamics of what would happen in this situation simply by examining what happens in a superconductor, and extrapolating to the situation where the material doesn't break Einstein's relativity invariance with respect to constant motion.

The Higgs field is a scalar field, $h$, so as far as the Lorentz symmetry goes, it is allowed and expect to interact with any other field. Whatever is the Lorentz-invariant term in the Lagrangian for other fields may be multiplied by $h$ to get an allowed interaction term, often a renormalizable one.

In particular, the Higgs field interacts with all the fermionic fields via the so-called Yukawa interaction, schematically $y\cdot h\bar\psi \psi$, where $\psi$ is a fermion field. The (classically) dimensionless Yukawa couplings $y$ may be large or small. The top quark Yukawa coupling is very large, of order one, which makes the top quark heavy. On the contrary, the Yukawa couplings with the neutrinos is (almost) zero which keeps the neutrinos massless.

In the previous paragraph, I ignored the fact that the Higgs is really a component of the Higgs doublet and the interaction terms has to be gauge-invariant. So a doublet must be contracted with another doublet, and so on. The vev produces masses for one component of the doublet only. There are interactions with $h$ (doublet) and its complex conjugate that make all quarks massive. However, it's not possible to write down the simplest interaction that would give the neutrinos masses in the simplest model. In the simplest model, the neutrinos are left-handed and it's exactly the left-handed part of the neutrinos that can't get the masses. But the neutrinos are still interacting with the Higgs field.

The gauge invariance in fact *dictates* that the Higgs field has to interact with other fields, the gauge fields. Because the Higgs field carries no color charges like quarks, it's neutral under $SU(3)_{QCD}$ which means that it doesn't have any interactions with gluons.

However, the Higgs field has nonzero charges under the electroweak $SU(2)_W$ because the Higgs field is a doublet; and under the hypercharge $U(1)_Y$ part of the electroweak gauge symmetry. These charges of the Higgs fields mean that the derivatives in its kinetic term have to be replaced by the covariant derivatives and this produces interactions with the electroweak gauge fields.

In general, that makes the gauge bosons for $SU(2)\times U(1)$ massive, too, when the Higgs gets a vacuum expectation value. However, the vev is such that the classical vacuum configuration is invariant under the electromagnetic $U(1)_{em}$ generated by the electric charge $Q = (1/2)Y+T_3$. So under this particular generator, the component of the Higgs field that has a nonzero vev is neutral which is why the Higgs field doesn't directly interact with the corresponding gauge field, the electromagnetic field, and that's why the photon stays massless and the electromagnetic force remains a long-range force. The other three generators of the electroweak group produce gauge bosons $W^\pm, Z^0$ if an orthonormal basis is chosen which is why these three gauge bosons are massive due to the Higgs mechanism.

## Best Answer

The full answer is unknown, i.e. nobody can tell you why the electron is only 0.00484... times as heavy as the muon. In fact, all interactions of massive particles (leptons) with the Higgs are of the same form (called a 'Yukawa interaction'), but for every particle, there is a different constant of proportionality ('Yukawa coupling constant'). For completeness, you can find the mathematical description here. (For the quarks, this is essentially the 'CKM matrix'.)

So the problem of different masses is equivalent to asking why some particles interact more strongly or weakly with the Higgs, although that doesn't help you much.

So the take-home message is: there is no accepted theory of physics that predicts the masses of particles in the Standard Model. If you find such a theory (or a partial theory), you can be sure to win a Nobel Prize, so good luck ;)