Question

Calculate the ratio of (a) energy to momentum for a photon, (b) kinetic energy to momentum for a relativistic massive object of speed u, and (e) total energy to momentum for a relativistic massive object. (d) There is a qualitative difference between the ratio in part (a) and the other two. What is it? (e) What are the ratios of kinetic and total energy to momentum for an extremelyrelativistic massive object, for which$\text{u}\cong c\text{?}\underset{\delta x\to 0}{lim}$ What about the qualitative difference now?

Short answer

(a)de-Broglie relation result is c

(b) kinetic energy and momentum formulas$(1-\frac{1}{\gamma })\frac{{\mathbf{c}}^2}{u}$

(c)Instead of kinetic energy as in section b is$\frac{{\text{c}}^{\text{2}}}{\text{u}}$

(d)Unlike massless particles, the ratios in parts (b) and (c) are affected by the velocity of the particle u.

(e)Massive particles act similarly to massless particles, resulting in the same c ratio.

Step-by-step solution

:de-Broglie relation.

We have E = pc for massless particles, which means that the ratio of energy to momentum will be equal to c. However, if you start with the energy relation and combine it with the de-Broglie relation, you'll get the same result.

$\text{E}=\text{hf}=\frac{\text{hc}}{\text{λ}}=\text{pc}$……………..(1)

Calculate the relation between energy and momentum (a)

Therefore the ratio of the energy and momentum from equation (1) is

$\frac{\text{E}}{\text{p}}=\text{c}$

Relation between the kinetic energy and momentum (b)

Now, apply the relativistic kinetic energy and momentum formulas.

$\begin{array}{c}\frac{\text{KE}}{\text{p}}=\frac{(\gamma -\text{1}){\text{mc}}^{\text{2}}}{\gamma \text{mu}}\\ =(\text{1}-\frac{\text{1}}{\gamma })\frac{{\text{c}}^{\text{2}}}{\text{u}}\end{array}$

Here, $\gamma =\frac{1}{\sqrt{1-u^2/c^2}}istheLorentzfactor.$

Total energy to momentum relation(c)

Instead of kinetic energy as in section (b), the total energy is now used.

$\begin{array}{c}\frac{\text{E}}{\text{p}}=\frac{(\gamma -\text{1}){\text{mc}}^{\text{2}}+{\text{mc}}^{\text{2}}}{\gamma \text{mu}}\\ =\frac{{\text{mc}}^{\text{2}}}{\gamma \text{mu}}\\ =\frac{{\text{c}}^{\text{2}}}{\text{u}}\end{array}$

Difference between part (a) and other parts, i.e., (b) and (c)(d)

The ratios in parts (b) and (c) are affected by the particle's velocity u, whereas the ratio in part (a) for massless particles is constant.

Because the speed of light is constant for massless particles in part (c).

Ratios for an extremely relativistic particle(e)

For an extremely relativistic object, we may take the limit of the expressions in parts(b) and (c) as $u\to c$, and therefore, $\gamma \to \infty$.

$\begin{array}{l}\frac{\text{KE}}{\text{p}}=\text{c}\\ \frac{\text{E}}{\text{p}}=\text{c}\end{array}$

As we approach c, the internal energy of the relativistic decreases, thus the two ratios are equal. Where all of the energy is in the form of kinetic energy, and large particles act more like massless particles, with about the same c/c ratio.

Result.

(a)

de-Broglie relation result c

(b)

kinetic energy and momentum formulas $(1-\frac{1}{\gamma })\frac{{\mathbf{c}}^2}{u}$.

(c)

Instead of kinetic energy as in section b is$\frac{{\text{c}}^{\text{2}}}{\text{u}}$

(d)

Unlike massless particles, the ratios in parts b) and c) are affected by the velocity of the particle u.

(e)

Massive particles act similarly to massless particles, resulting in the same c ratio