Question

Equations (2-38) relate momentum and total energy in two frames. Show that they make sense in the non-relativistic limit.

Short answer

In non-relativistic collisions, the system’s mass does not change, which is shown in the following solution with the help of relations between momentum and energy formula.

Step-by-step solution

State the relation of energy and momentum in two frames.

The relations between momentum and energy in two frames are expressed as,

$$\begin{gathered} p{\text{'}}_x={\gamma }_v\left[p_x-\frac{V}{C}\left(\frac{E}{C}\right)\right] \\ E\text{'}={\gamma }_v\left[E-v{p}_x\right] \end{gathered}$$

Express the relativistic momentum expression in the classical limit.

Classically,$\frac{\mathrm{v}}{\mathrm{c}}<<\hspace{0.33em}1\hspace{0.33em}\Rightarrow \hspace{0.33em}\mathrm{\gamma \_v}\hspace{0.33em}\approx \hspace{0.33em}1\hspace{0.33em}\&\hspace{0.33em}\mathrm{E}\hspace{0.33em}\cong \hspace{0.33em}{\mathrm{mc}}^2$, the momentum expression becomes,

$$\begin{gathered} p_x^{\text{'}}\hspace{0.33em}=\left[p_x-\hspace{0.33em}\frac{v}{c}\left(\frac{m{c}^2}{c}\right)\right] \\ {p}_x^{\text{'}}\hspace{0.33em}=\hspace{0.33em}{p}_x\hspace{0.33em}-\hspace{0.33em}mv \end{gathered}$$

which is momentum in classical Galilean transformation. Similarly, the energy equation will approximate to,

$$\begin{gathered} E\text{'}=E-v{p}_x \\ \frac{E\text{'}}{c}=\frac{E}{c}-\frac{v}{c}{p}_x \\ \frac{E\text{'}}{c}=\frac{E}{c} \\ m\text{'}c=mc\because \hspace{0.33em}\frac{v}{c}<<\hspace{0.33em}1 \end{gathered}$$

This implies that mass remains the same and is valid for the non-relativistic frame. Therefore, the relativistic momentum and energy equations are valid for classical limits too.