Question

From $\mathit{p}\mathbf{=}{\mathit{\gamma }}_{\mathbf{u}}\mathit{m}\mathit{u}$ (i.e., ${\mathit{p}}_{\mathbf{x}}\mathbf{=}{\mathit{\gamma }}_{\mathbf{u}}\mathit{m}{\mathit{u}}_{\mathbf{x}}\mathbf{,}{\mathit{p}}_{\mathbf{y}}\mathbf{=}{\mathit{\gamma }}_{\mathbf{u}}\mathit{m}{\mathit{u}}_{\mathbf{y}}$ and ${\mathit{p}}_{\mathbf{z}}\mathbf{=}{\mathit{\gamma }}_{\mathbf{u}}\mathit{m}{\mathit{u}}_{\mathbf{z}}$ ), the relativistic velocity transformation $\mathbf{(}\mathbf{2}\mathbf{-}\mathbf{20}\mathbf{)}$, and the identity ${\mathit{\gamma }}_{{\mathbf{u}}^{\mathbf{\text{'}}}}\mathbf{=}\left(1-u_{x}v/c^2\right){\mathit{\gamma }}_{\mathbf{v}}{\mathit{\gamma }}_{\mathbf{u}}$ show that ${\mathit{p}}_{\mathbf{y}}^{\mathbf{\text{'}}}\mathbf{=}{\mathit{p}}_{\mathbf{y}}$ and ${\mathit{p}}_{\mathbf{z}}^{\mathbf{\text{'}}}\mathbf{=}{\mathit{p}}_{\mathbf{z}}$.

Short answer

It is shown that both frames’ y and z component momentum are equal by using the relativistic velocity and the given identity for ${\gamma }_{w^{\text{'}}}$.

Step-by-step solution

 State the relativistic velocity transformation equations.

The relativistic velocity transformation between two frames $S\&S^{\text{'}}$is expressed as,

$u_x^{\text{'}}=\frac{u_x-v}{1-\frac{u_{x}v}{c^2}}$

$u_y^{\text{'}}=\frac{u_y}{{\gamma }_r\left(1-\frac{u_x{\eta }^3}{c^2}\right)}$

$u_z^{\text{'}}=\frac{u_z}{y_{\mathrm{v}}\left(1-\begin{array}{c}{u}_{x}v\\ c^2\end{array}\right)}$

Insert the above equations in the corresponding momentum equation.

The relativistic momentum in $S^{\text{'}}$frame for the y-component is, $p_y^{\text{'}}={\gamma }_{u}m{u}_y^{\text{'}}$

Inserting the expression for $u_y^{\text{'}}$and the given identity for ${\gamma }_{w^{\text{'}}}$in the above equation,

$p_y^{\text{'}}=\left[\left(1-\frac{u_{x}v}{c^2}\right){\gamma }_v{\gamma }_n\right]\left(m\right)\left[\frac{u_y}{{\gamma }_v\left(1-\frac{u_{x}v}{c^2}\right)}\right]$

$={\gamma }_{u}m{u}_y$

$p_y^{\text{'}}=p_y$

Similarly, for the z-component in the frame $S^{\text{'}}$,

$p_z^{\text{'}}={\gamma }_{u^m}mm{u}_z^{\text{'}}$

$=\left[\left(1-\frac{u_{c}v}{c^2}\right){\gamma }_v{\gamma }_{iu}\right]\left(m\right)\left[\frac{u_z}{{\gamma }_v\left(1-\frac{u_{x}v}{c^2}\right)}\right]$

$={\gamma }_{u}m{u}_z$

$p_z^{\text{'}}=p_z$

Thus, unlike relativistic x- component momentum, the y and z component momentum of both frames is equal.