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According to this answer I understood that particles with mass also move at speed of light but interaction with higgs field make this movement zigzag. So average speed is below speed of light. Is this true?

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Alberto
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    That answer is a just-so story. Nowhere in the actual technical description of the Higgs mechanism are particles zigzagging or bouncing around - there is no classical point-like particle at all in quantum field theory. This question isn't really well-defined. – ACuriousMind Apr 11 '16 at 14:54

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I'm not 100% sure of your level so just as a heads up, I put some comments in parentheses that are meant to give technical caveats. If they don't make sense just ignore everything in parentheses, the zeroth order answer

You can look at it that way, but actually it turns out to be much more complicated to understand what's going on in detail. (Basically you can think of a massive particle as a massless particle scattering off of a background Higgs field, but in order to compute how the massive particle propagates you need to sum over an infinite number of different quantum scattering amplitudes each describing one describing different ways the particle could interact with the background, and the full infinite sum will produce the same behavior we'll get in a much simpler way below. In short this gives you a vivid picture but to actually use the picture to make accurate statements is actually quite subtle and requires knowledge of quantum mechanics and is not generally the way anyone thinks about it in practice because it's just way more complicated than necessary)

A simpler way is to think of the relationship between energy and momentum for a relativistic particle \begin{equation} E^2= m^2 c^4 + p^2 c^2 \end{equation} The net result of, say, the electron interacting with the Higgs field (to leading order) is to make $m$ nonzero in this relationship.

(Or more precisely, the result is to redress the value of $m$ if you want to allow the electron to have some part of its mass not come from the Higgs, although that's a dangerous game since giving the electron a bare mass violates the chiral gauge symmetry of the weak interactions).

This relationship between energy and momentum in turn leads to the particle moving at a speed less than $c$, which you can see for example by using \begin{equation} v=\frac{p}{E}=\frac{p}{\sqrt{p^2 c^2 + m^2 c^4}} \end{equation} As you can see for $m=0$ the particle moves at $c$ whereas for nonzero (real) $m$ the particle moves at $v<c$.

(This result can also be obtained by taking the relationship between energy and momentum into a relationship between frequency and wave number using the $E=\hbar \omega$ and $p=\hbar k$ and then computing the group velocity $\partial \omega/\partial k$)

Andrew
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