Question

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).

Short answer

The value of algebraic equation of relativistic velocity transformation is $\sum _i{\gamma }_{u_i^{\text{'}}}{m}_i{u}_i^{\text{'}}={\gamma }_u\sum _i{\gamma }_{u_i^{\text{'}}}{m}_i{u}_i-u{\gamma }_u\sum _i{\gamma }_{ui}{m}_i$.

Step-by-step solution

Write the given data from the question.

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

Determine the formula of algebraic equation of relativistic velocity transformation.

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.

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