91Ó°ÊÓ

The kinetic energy of an object whose mass is and moving with a velocity of is given by $\mathit{K}\mathit{E}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{m}{\mathit{v}}^{\mathbf{2}}\mathbf{=}\frac{{\mathbf{p}}^{\mathbf{2}}}{\mathbf{2}\mathbf{m}}$. The kinetic energy of a photon is given by $\mathit{K}\mathit{E}\mathbf{=}\mathit{p}\mathit{c}$, where is the momentum and is the velocity of light in vacuum.

(a)

It is given that the momentum of photon and the massive object are same, also the velocity of the massive object is very slow than . So, the required ratio can be obtained as follows:

$$\begin{gathered} \frac{K{E}_o}{K{E}_p}=\frac{\frac{p}{2mc}}{pc} \\ \frac{K{E}_o}{K{E}_p}=\frac{p}{2mc} \end{gathered}$$

Thus, the ratio is $\frac{K{E}_o}{K{E}_p}=\frac{p}{2mc}$.

(b)

The relativistically correct formula of kinetic energy for the massive object can be obtained as follows:

localid="1659335218499" $$\begin{gathered} \frac{\mathit{K}{\mathit{E}}_{\mathit{o}}}{\mathit{K}{\mathit{E}}_{\mathit{p}}}\mathit{=}E\mathit{-}m{c}^{\mathit{2}} \\ \mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{=}\sqrt{{\mathit{p}}^{\mathit{2}}{\mathit{c}}^{\mathit{2}}\mathit{+}{\mathit{m}}^{\mathit{2}}{\mathit{c}}^{\mathit{4}}}\mathit{-}m{c}^{\mathit{2}} \end{gathered}$$

So, the required ratio can be obtained as follows:

localid="1659335276834" $$\begin{gathered} \frac{K{E}_o}{K{E}_p}=\frac{\sqrt{p^2{c}^2+m^2{c}^4}-m{c}^2}{pc} \\ \frac{K{E}_o}{K{E}_p}=\sqrt{1+\frac{m^2{c}^2}{p^2}}-\frac{mc}{p} \end{gathered}$$

Thus, the ratio is $\frac{K{E}_o}{K{E}_p}=\sqrt{1+\frac{m^2{c}^2}{p^2}}-\frac{mc}{p}$.

(c)

From part (b), the ratio is $\frac{K{E}_o}{K{E}_p}=\sqrt{1+\frac{m^2{c}^2}{p^2}}-\frac{mc}{p}$. In the low-speed limit, we have localid="1659335557009" $p≪mc$. So, the ratio can be simplified as follows:

$$\begin{gathered} \frac{K{E}_o}{K{E}_p}=\sqrt{1+\frac{m^2{c}^2}{p^2}}-\frac{mc}{p} \\ =\frac{mc}{p}\left(\sqrt{\frac{p^2}{m^2{c}^2}+1}-1\right) \end{gathered}$$

Now, we have $p≪mc$, so, the binomial expansion of ${\left(\frac{p^2}{m^2{c}^2}+1\right)}^{\frac{1}{2}}$is approximate equal

to $1+\frac{p^2}{2m^2{c}^2}$. So, the ratio becomes:

$$\begin{gathered} \frac{K{E}_o}{K{E}_p}≈\frac{mc}{p}\left(1+\frac{p^2}{2m^2{c}^2}-1\right) \\ ≈\frac{mc}{p}\left(\frac{p^2}{2m^2{c}^2}\right) \\ ≈\frac{p}{2mc} \end{gathered}$$

This is the ratio we got in part (a). So, it is proved that the relativistic formula approaches classical limit at low speeds.

Now, for high-speed limit, we have localid="1659336113888" $p≫mc$, which means $\frac{mc}{p}$is negligible or zero. So, the ratio becomes 1.

Thus, the low-speed limit of the ratio of part (a) agrees with part (b) and the high-speed limit is 1.

The kinetic energy of a massive object is given by $KE=\sqrt{p^2{c}^2+m^2{c}^4}-m{c}^2$. In very high-speed limit, we have $p≫mc⇒pc≫m{c}^2$. So, $m{c}^2$can be ignored in the above equation. Therefore, the kinetic energy of the massive object only left with pc .

Thus, at very high speed, the kinetic energy of massive object approaches pc .