- #1
jstrunk
- 55
- 2
In Quantum Mechanics Concepts and Applications by Zettili the following formulas are used
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?
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?