By applying the relativistic velocity transformation to the left side of equation (2 - 23) and using the algebraic identity derived in Exercise 67, verify equation (2 - 23).

Consider a algebraic identity is${\gamma }_{u\text{'}}=\left(1-\frac{u_{x}v}{c^2}\right){\gamma }_v{\gamma }_u$.

Write the formula of algebraic equation of relativistic velocity transformation. $\mathbf{\sum }_{\mathbf{i}}{\mathit{\gamma }}_{{\mathbf{u}}_{\mathbf{i}}^{\mathbf{\text{'}}}}{\mathit{m}}_{\mathbf{i}}{\mathit{u}}_{\mathbf{i}}^{\mathbf{\text{'}}}\mathbf{=}{\mathbf{\gamma }}_{\mathbf{u}}\mathbf{\sum }_{\mathbf{i}}{\mathit{\gamma }}_{{\mathbf{u}}_{\mathbf{i}}^{\mathbf{\text{'}}}}{\mathit{m}}_{\mathbf{i}}{\mathit{u}}_{\mathbf{i}}\mathbf{-}\mathit{u}{\mathbf{\gamma }}_{\mathbf{u}}\mathbf{\sum }_{\mathbf{i}}{\mathit{\gamma }}_{\mathbf{u}\mathbf{i}}{\mathit{m}}_{\mathbf{i}}$ …… (1)

Here, ${\mathit{\gamma }}_{{\mathbf{u}}_{\mathbf{i}}^{\mathbf{\text{'}}}}$is relativistic velocity transformation, ${\mathbf{m}}_{\mathbf{i}}$is mass and ${\mathbf{u}}_{\mathbf{i}}$ is velocity transformation.

As reference of equation 2.23.

Determine the algebraic equation of relativistic velocity transformation.

Consider the LHS of the equation (1).

$\sum _i{\gamma }_{u_i^{\text{'}}}{m}_i{u}_i^{\text{'}}$

First we solve it without summation.

$$\begin{gathered} {\gamma }_{\mathrm{u}\text{'}}{m}_{\mathrm{i}}u\text{'}=\left(1-\frac{u_{x}v}{c^2}\right){\gamma }_{\mathrm{u}}{\gamma }_{\mathrm{u}}m·\frac{u_x-v}{\left(1-\frac{u_{x}v}{c^2}\right)} \\ ={\gamma }_{\mathrm{u}}{\gamma }_{\mathrm{u}}m{u}_x-{\gamma }_{\mathrm{u}}{\gamma }_{\mathrm{u}}vm \end{gathered}$$

Now summing over whole system’s particles:

$\sum _i{\gamma }_{u_i^{\text{'}}}{m}_i{u}_i^{\text{'}}={\gamma }_u\sum _i{\gamma }_{u_i^{\text{'}}}{m}_i{u}_i-u{\gamma }_u\sum _i{\gamma }_{ui}{m}_i$

Therefore, the value of algebraic equation of relativistic velocity transformation is $\sum _{\mathrm{i}}{\gamma }_{{\mathrm{u}}_{\mathrm{i}}^{\text{'}}}{m}_{\mathrm{i}}{u}_{\mathrm{i}}^{\text{'}}={\mathrm{\gamma }}_{\mathrm{u}}\sum _{\mathrm{i}}{\gamma }_{{\mathrm{u}}_{\mathrm{i}}^{\text{'}}}{m}_{\mathrm{i}}{u}_{\mathrm{i}}-u{\mathrm{\gamma }}_{\mathrm{u}}\sum _{\mathrm{i}}{\gamma }_{\mathrm{ui}}{m}_{\mathrm{i}}$.