Question

Classically and nonrelativistically, we say that the energy$\mathbf{\text{E}}$of a massive free particle is just its kinetic energy. (a) With this assumption, show that the classical particle velocity${\mathbf{\text{v}}}_{\mathbf{\text{particle}}}$is$\mathbf{\text{2E}}\mathbf{/}\mathbf{\text{p}}$. (b) Show that this velocity and that of the matter wave differ by a factor of 2. (c) In reality, a massive object also has internal energy, no matter how slowly it moves, and its total energy$\mathbf{\text{E}}$is$\mathit{\gamma }{\mathbf{\text{mc}}}^{\mathbf{\text{2}}}$, where_{$\mathit{\gamma }\mathbf{=}\mathbf{\text{1}}\mathbf{/}\sqrt{\mathbf{\text{1-}}{\mathbf{(}{\mathbf{\text{v}}}_{\mathbf{\text{particle}}}\mathbf{/}\mathbf{\text{c}}\mathbf{)}}^{\mathbf{\text{2}}}}$.}Show that${\mathbf{\text{v}}}_{\mathbf{\text{particle}}}$is${\mathbf{\text{pc}}}^{\mathbf{\text{2}}}\mathbf{/}\mathbf{\text{E}}$and that_{${\mathbf{\text{v}}}_{\mathbf{\text{wave}}}$}is${\mathbf{\text{c}}}^{\mathbf{\text{2}}}\mathbf{/}{\mathbf{\text{v}}}_{\mathbf{\text{particle}}}$Is there anything wrong with it_{${\mathbf{\text{v}}}_{\mathbf{\text{wave}}}$}? (The issue is discussed further in Chapter 6.)

Short answer

a. The classical particle velocity ${\text{v}}_{\text{particle}}$is$2E/\text{p}$ has proved by relating kinetic energy and momentum.

b. The velocity and the matter-wave differ by a factor of $\text{2}$ by relating wave velocity with the result of component a).

c. By relating wave velocity to particle velocity it is found that the value of particle velocity is distinct from the value of wave velocity.

Step-by-step solution

Concept Introduction

The kinetic energy of the particle in the classical or nonrelativistic picture is_{$\mathbf{\text{KE}}\mathbf{=}\frac{\mathbf{\text{1}}}{\mathbf{\text{2}}}{\mathbf{\text{mv}}}_{\mathbf{\text{particle}}}^{\mathbf{\text{2}}}$and for a}particle, in the relativistic approach, the kinetic energy can b expressed as $\text{KE}=\gamma {\text{mc}}^{\text{2}}$.

Ratio between kinetic energy and momentum(a)

We could derive the ratio between the kinetic energy (total energy of a free particle) and the momentumusing classicalreasoning.

_{$\text{KE}=\frac{\text{1}}{\text{2}}{\text{mv}}_{\text{particle}}^{\text{2}}$}

$\text{p}={\text{mv}}_{\text{particle}}$

$\text{E}=\frac{{\text{pv}}_{\text{particle}}}{\text{2}}$

${\text{v}}_{\text{particle}}=\frac{\text{2E}}{\text{p}}$

Wave velocity(b)

The wave velocity is discovered to be,

_{${\text{v}}_{\text{wave}}=\frac{\text{E}}{\text{p}}$}

As a result, when we compare this to the result of component (a), we can see that the two velocitiesdiffer by a factor of_$\text{two}$.

Calculations for particle Velocity(c)

We can use the relativistic correction to repeat the particle velocity (_{${\text{v}}_{\text{particle}}$)} calculations that we did in part (a).

_{$\begin{array}{l}\text{E}={\gamma }_v{\text{mc}}^{\text{2}}\frac{{\partial }^2\Omega }{\partial v^2}\\ \text{p}={\gamma }_v{\text{mv}}_{\text{particle}}\end{array}$}

_{$\frac{\text{E}}{\text{p}}=\frac{{\gamma }_v{\text{mc}}^{\text{2}}}{{\gamma }_v{\text{mv}}_{\text{particle}}}$}

_{${\text{v}}_{\text{particle}}=\frac{{\text{pc}}^2}{\text{E}}$}

This value of ${\text{v}}_{\text{particle}}$is still distinct from the value of ${\text{v}}_{\text{wave}}$that we previously discussed. We can substitute ${\text{v}}_{\text{wave}}$from part (b), to relate the wave velocity to the particle velocity.

.$\begin{array}{l}{\text{v}}_{\text{particle}}=\frac{{\text{c}}^{\text{2}}}{{\text{v}}_{\text{wave}}}\\ {\text{v}}_{\text{wave}}=\frac{{\text{c}}^{\text{2}}}{{\text{v}}_{\text{particle}}}\end{array}$

This velocity, known as the phase velocity, conveys no information, it can travel faster than the speed of light.

The group velocity, where the energy and momentum are carried through, is the wave velocity that is not allowed to have speeds larger than.