- #1

jstrunk

- 55

- 2

for phase and group velocities.

[itex]

{\rm{ }}{v_{ph}} = \frac{w}{k} = \frac{{E\left( p \right)}}{{p}}{\rm{ }}\\

{\rm{ }}{v_g} = \frac{{dw}}{{dk}}{\rm{ = }}\frac{{dE\left( p \right)}}{{dp}}{\rm{ }}

[/itex]

In the low velocity classical case for a free particle, E means Kinetic Energy. In the

relativistic case, E means Kinetic plus Rest Energy, according to the book. I suspect

that we should use Kinetic Energy on both cases. You seem to get the same results when

it matters whether you use Kinetic or Kinetic plus Rest in the relativistic case, so it

makes more sense to be consistent and use the Kinetic in both cases.

Is there a real reason why you have to use Kinetic Energy classically but Kinetic plus Rest

relativisticly?