Saturday, 22 September 2018

quantum mechanics - 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?



Answer



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 $c$.


No comments:

Post a Comment

Understanding Stagnation point in pitot fluid

What is stagnation point in fluid mechanics. At the open end of the pitot tube the velocity of the fluid becomes zero.But that should result...