Does Group Velocity Equal Particle Velocity in Relativistic Physics?

[CoreyJKelly]

Homework Statement

"Use relativistic expressions for total energy and momentum to verify that the group velocity vg of a matter wave equals the velocity v of the associated particle."

Homework Equations

E$$^{2}$$ = (pc)$$^{2}$$ + (mc$$^{2}$$)$$^{2}$$

p = hbar * k

E = hbar * $$\omega$$

vg = $$\frac{\partial \omega}{\partial k}$$

The Attempt at a Solution

So I know how to do this, and I've seen the solution written out in many places (the Wikipedia article for 'group velocity', for example), but the final step of these derivations only works if the lorentz factor has a plus sign instead of a minus sign. My answer has a plus sign, and I'm not sure how i can justify writing it as the relativistic velocity... any ideas?

 

[BigJDubs]

The Wikipedia article does this: \frac{dE}{dk} = \frac{d}{dk} \sqrt{p^{2}c^{2} + m^{2}c^{4}} = \frac{pc^{2}}{\sqrt{p^{2}c^{2} + m^{2}c^{4}}}v_{g} = \frac{d\omega}{dk} = \frac{dE}{dk}/\hbar = \frac{pc^{2}}{\hbar \sqrt{p^{2}c^{2} + m^{2}c^{4}}} = \frac{pc}{\hbar \sqrt{1 + (mc^{2}/pc)^{2}}} = \frac{pc}{\hbar \gamma} = v_{p}The only difference in my derivation is that the last step says vg = \frac{pc}{\hbar \gamma} = -v_{p}. Where did the plus sign go? EDIT: I think I understand what happened now, but if anyone can provide more of an explanation that would be great. Thanks!