# What are the consequences in high-energy of the non-interaction of the Higgs Field?

At high-energies when the Higgs field won't affect (interact with) particles, when the symmetry breaking won't occur, what would be $\rm W\pm$ or $\rm Z^{0}$ bosons speed if they would then have a $0$ rest mass? What about Fermions?

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Related: physics.stackexchange.com/q/31395/2451 and links therein. – Qmechanic Aug 18 '13 at 7:30

In the unbroken phase of the Standard Model, all particles are massless. Thus, they can only move with the speed of light, in order to fulfill the mass-energy-relation $$E^2 = \vec p^2 c^2 + m^2 c^4 = \vec p^2 c^2 \big|_{m = 0}$$ This is true for photons and gluons in our broken world, but would be true for $W^\pm, Z^0$ (or more correctly for the $W^{i}, B$ bosons of $SU(2) \times U(1)$ as there would be no need to form linear combinations of them) and for all fermions as well.

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Originally, you have 3 massless gauge bosons $W_x, W_y, W_z$ for the symmetry $SU(2)_L$ and one massless gauge boson $X$ for the symmetry $U(1)_Y$.

With the higgs mechanism, you have linear combinations of $W_z$ and $X$ which give the massless photon and the massive $Z_0$, while the $W \pm = W_x \pm iW_y$ acquire a mass, too.

So, all these particles : photon, $Z_0, W \pm$, are not the original massless gauge photons of $SU(2)_L$ and $U(1)_Y$.

While the mass of the photon is always zero, even at high energy, there is, of course, a variation of the mass of the $Z_0, W \pm$, due to the energy scale (all masses of massive particles depend on the energy scale, like coupling constants).

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Quantum corrections to the mass should always be proportional to the mass itself. Thus, as long as the bosons are massless, no mass should be "generated" through self-energy contributions! (Your post did not make that clear enough, imo) – Neuneck Jul 19 '13 at 6:42
There are thermal masses $m^2 \sim g T^2$ if the temperature is nonzero, which it must be to get into the unbroken phase to begin with. – Michael Brown Jul 19 '13 at 6:57