Question

(a) Repeat Prob. 12.2 (a) using the (incorrect) definition $\mathbf{p}=\mathrm{m}\mathbf{u}$, but with the (correct) Einstein velocity addition rule. Notice that if momentum (so defined) is conserved in S, it is not conserved inlocalid="1654750932476" $\overline{\mathit{S}}$. Assume all motion is along the x axis.

(b) Now do the same using the correct definition,localid="1654750939709" $\mathbf{p}\mathbf{=}\mathrm{m}\mathbf{\eta }$ . Notice that if momentum (so defined) is conserved in S, it is automatically also conserved inlocalid="1654750943454" $\overline{\mathit{S}}$. [Hint: Use Eq. 12.43 to transform the proper velocity.] What must you assume about relativistic energy?

Short answer

(a) It is proved that the momentum is not conserved in$\overline{S}$frame.

(b) It is proved that the momentum is also conserved in $\overline{S}$frame.

Step-by-step solution

Expression for the conservation of momentum:

Write the expression for the conservation of momentum.

${\mathit{m}}_{\mathbf{A}}{\mathit{u}}_{\mathbf{A}}\mathbf{+}{\mathit{m}}_{\mathbf{B}}{\mathit{u}}_{\mathbf{B}}\mathbf{=}{\mathit{m}}_{\mathbf{c}}{\mathit{u}}_{\mathbf{c}}\mathbf{+}{\mathit{m}}_{\mathbf{D}}{\mathit{u}}_{\mathbf{D}}$ …… (1)

Here, $m_A$is the mass of particle A, $u_A$is the initial velocity of the particle, $m_B$is the mass of particle B, $u_B$is the initial velocity of particle B,$m_C$ is the mass of particle C, $u_C$is the initial velocity of particle C, is the mass of particle D and$u_D$ is the initial velocity of particle D.

Determine that the momentum is not conserved in S :

(a)

Using the Einstein velocity addition rule in the S frame, write the initial velocities of all the particles.

$$\begin{gathered} u_A=\frac{{\overline{u}}_A+v}{1+{\displaystyle \frac{{\overline{u}}_{A}v}{c^2}}} \\ {u}_B=\frac{{\overline{u}}_B+v}{1+{\displaystyle \frac{{\overline{u}}_{B}v}{c^2}}} \\ {u}_C=\frac{{\overline{u}}_C+v}{1+{\displaystyle \frac{{\overline{u}}_{C}v}{c^2}}} \\ {u}_D=\frac{{\overline{u}}_D+v}{1+{\displaystyle \frac{{\overline{u}}_{D}v}{c^2}}} \end{gathered}$$

Substitute all the above values in equation (1).

$m_A\left(\frac{\begin{array}{c}\\ {\overline{u}}_A+v\end{array}}{1+{\displaystyle \frac{{\overline{u}}_{A}v}{c^2}}}\right)+m_{\mathit{B}}\left(\frac{\begin{array}{c}\\ {\overline{u}}_B+v\end{array}}{1+{\displaystyle \frac{{\overline{u}}_{B}v}{c^2}}}\right)=m_C\left(\frac{\begin{array}{c}\begin{array}{c}\\ {\overline{u}}_C+v\end{array}\end{array}}{1+{\displaystyle \frac{{\overline{u}}_{C}v}{c^2}}}\right)+m_D\left(\frac{\begin{array}{c}\\ {\overline{u}}_D+v\end{array}}{1+{\displaystyle \frac{{\overline{u}}_{D}v}{c^2}}}\right)$

If the masses of all particles are equal, the initial velocities of all the particles will be,

$$\begin{gathered} u_A=-u_B=v \\ {u}_C=u_D=0 \end{gathered}$$

As the above condition indicates that it is a symmetric, completely inelastic collision in the S frame, the momentum is clearly conserved in the S frame.

Using the Einstein velocity addition rule in $\overline{S}$frame, write the initial velocities of all the particles.

$$\begin{gathered} u_A=0 \\ {\overline{u}}_B=\frac{-2u}{1+{\displaystyle \frac{u^2}{c^2}}} \\ {u}_C=-u \\ {u}_D=-u \end{gathered}$$

Substitute all the above values in equation (1).

$m_A\left(0\right)+m_B\left(\frac{\begin{array}{c}\\ -2u\end{array}}{1+{\displaystyle \frac{u^2}{c^2}}}\right)=m_C\left(-u\right)+m_D\left(-u\right)$

As all the masses are equal then,

$$\begin{gathered} m\left(0\right)+m\left(\frac{\begin{array}{c}\\ -2u\end{array}}{1+{\displaystyle \frac{u^2}{c^2}}}\right)=m\left(-u\right)+m\left(-u\right) \\ 0+m\left(\frac{\begin{array}{c}\\ -2u\end{array}}{1+{\displaystyle \frac{u^2}{c^2}}}\right)=-2mu \\ m\left(\frac{\begin{array}{c}\\ -2u\end{array}}{1+{\displaystyle \frac{u^2}{c^2}}}\right)\ne -2mu \end{gathered}$$

Therefore, it is proved that the momentum is not conserved in $\overline{S}$frame.

Determine that the momentum is conserved in S:

(b)

Write the expression for the conservation of momentum for proper velocity.

${\mathit{m}}_{\mathbf{A}}{\mathit{\eta }}_{\mathbf{A}}\mathbf{+}{\mathit{m}}_{\mathbf{B}}{\mathit{\eta }}_{\mathbf{B}}\mathbf{=}{\mathit{m}}_{\mathbf{c}}{\mathit{\eta }}_{\mathbf{c}}\mathbf{+}{\mathit{m}}_{\mathbf{D}}{\mathit{\eta }}_{\mathbf{D}}$ …… (2)

Using the Lorentz inverse transformation, write the initial velocities of all the particles.

$$\begin{gathered} {\eta }_A=\gamma \left({\overline{\eta }}_A+\beta {\overline{\eta }}_A^0\right) \\ {\eta }_B=\gamma \left({\overline{\eta }}_B+\beta {\overline{\eta }}_B^0\right) \\ {\eta }_C=\gamma \left({\overline{\eta }}_C+\beta {\overline{\eta }}_C^0\right) \\ {\eta }_D=\gamma \left({\overline{\eta }}_D+\beta {\overline{\eta }}_D^0\right) \end{gathered}$$

Substitute all the above values in equation (2).

$$\begin{gathered} m_A\left[\gamma \left({\overline{\eta }}_A+\beta {\overline{\eta }}_A^0\right)\right]+m_B\left[\gamma \left({\overline{\eta }}_B+\beta {\overline{\eta }}_B^0\right)\right]=m_C\left[\gamma \left({\overline{\eta }}_C+\beta {\overline{\eta }}_C^0\right)\right]+m_D\left[\gamma \left({\overline{\eta }}_D+\beta {\overline{\eta }}_D^0\right)\right] \\ {m}_A{\overline{\eta }}_A+m_B{\overline{\eta }}_B+\beta \left({\overline{\eta }}_A^0+{\overline{\eta }}_B^0\right)=m_C{\overline{\eta }}_C+m_D{\overline{\eta }}_D+\beta \left({\overline{\eta }}_C^0+{\overline{\eta }}_D^0\right) \end{gathered}$$

As it is known by the relativistic energy.

$P^0=m\eta =\frac{E^0}{c}$

So, if energy is conserved in$\overline{S}$frame, the energy of the particle will also be conserved. Hence,

${\overline{E}}_A+{\overline{E}}_B={\overline{E}}_0+{\overline{E}}_D$

So, the momentum will also be conserved. Hence,

$m_A{\overline{\eta }}_A+m_{\mathit{B}}{\overline{\eta }}_B=m_A{\overline{\eta }}_C+m_D{\overline{\eta }}_D$

Therefore, it is proved that the momentum is also conserved in$\overline{S}$frame.