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?


1 Answer 1


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 $c$.


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