We just saw that the speed of a nonrelativistic particle of mass m is not equal to its phase velocity. Show that the particle speed is equal to the group velocity.

Strategy We can use the relation for the group velocity in Equation (5.28) or (5.31). Either should work, and using both equations will be instructive.

$u_{ gr }=\frac{d \omega}{d k}$ (5.28)

$u_{ gr }=\frac{d E}{d p}=\frac{p c^{2}}{E}$ (5.31)

Question

We just saw that the speed of a nonrelativistic particle of mass m is not equal to its phase velocity. Show that the particle speed is equal to the group velocity.

Strategy We can use the relation for the group velocity in Equation (5.28) or (5.31). Either should work, and using both equations will be instructive.

[latex]u_{ gr }=\frac{d \omega}{d k}[/latex] (5.28)

[latex]u_{ gr }=\frac{d E}{d p}=\frac{p c^{2}}{E}[/latex] (5.31)

Accepted Answer

First, we look at Equation (5.31) for our nonrelativistic particle:

[latex]u_{ gr }=\frac{p c^{2}}{E}=\frac{m v c^{2}}{m c^{2}}=v[/latex] (5.35)

In order to use Equation (5.28) we utilize the results in Equations (5.29) and (5.30) for ω and k.

[latex]E=h f=\hbar(2 \pi f)=\hbar \omega[/latex] (5.29)

[latex]u_{ gr }=\frac{d E}{d p}=\frac{p c^{2}}{E}[/latex] (5.31)

[latex]u_{ gr }=\frac{d \omega}{d k}=\frac{d(E / \hbar)}{d(p / \hbar)}=\frac{d E}{d p}=\frac{d}{d p} \frac{p^{2}}{2 m}=\frac{2 p}{2 m}=v[/latex]

We agree that the particle, when acting as a wave, moves with the group velocity, not the phase velocity.

Question 5.5: We just saw that the speed of a nonrelativistic particle of ...

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