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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
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up vote 2 down vote accepted

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
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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
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