# Seeming violation-wave travelling faster than speed of light

Consider the basic relation $$E=\sqrt{(pc)^2+(mc^2)^2}.$$

Every particle possesses a wave nature and it depends on the situation in which one among the two is perceptible...

Consider a particle with rest mass $$m$$. If we consider the speed of De-Broglie Waves, as usual for a wave $$v_{wave}=\nu \lambda.$$ And since we are taking relativistic effects into account, let's write $$\lambda =\frac{h}{\gamma mv}$$ where $$\gamma$$ denotes the Lorentz factor $$\gamma =1/\sqrt{1-(v/c)^2}$$, and $$v$$ the speed of the particle. Now clearly the energy of the wave could be written as $$E=h \nu$$. And for the particle, Energy is equal to $$\gamma mc^2$$. So clearly $$h \nu =\gamma mc^2.$$ Now plugging into $$v_{wave}=\nu \lambda$$, we get $$v_{wave}=\frac{\gamma mc^2}{h}\frac{h}{\gamma mv},$$ or $$v_{wave}=\frac{c^2}{v}.$$ Doesn't this seem to go against what we know, that the velocity of the wave is less than or equal to $$c$$?

So can anyone point out what's the mistake here? Does this have anything to do with phase or group velocity?

What you have calculated is the phase velocity, $$v_p$$, of the de Broglie wave associated with the particle. The phase velocity can be greater than $$c$$, and indeed it is always greater than $$c$$.
The velocity of the particle is the group velocity, $$v_g$$, and as you have demonstrated the two are linked by:
$$v_p v_g = c^2$$
The group velocity must always be less than $$c$$ and that implies the phase veocity must always be greater than $$c$$.