Question

Using equations (2-20), show that

${{\mathit{y}}^{\mathbf{\text{'}}}}_{\mathbf{u}}\mathbf{=}\left(1-\frac{u_{x}v}{c^2}\right){\mathit{y}}_{\mathbf{v}}\mathbf{,}{\mathit{y}}_{\mathbf{u}}$

Short answer

The required equation ${\gamma }_u=\left(1-\frac{u_{x}v}{c^2}\right){\gamma }_g{\gamma }_v$is obtained.

Step-by-step solution

Write the given data from the question

The equation 2.20 is given as,

$$\begin{gathered} u_x^{\text{'}}=\frac{u_x-v}{1-{\displaystyle \frac{u_{x}v}{c^2}}} \\ {u}_y^{\text{'}}=\frac{u_y}{{\gamma }_v\left(1-{\displaystyle \frac{u_{x}v}{c^2}}\right)} \\ {u}_z^{\text{'}}=\frac{u_z}{{\gamma }_v\left(1-\frac{u_{x}v}{c^2}\right)} \end{gathered}$$

Determine the equation to prove the equation

The expression to calculate the velocity transformation for velocity of object in direction is given as follows.

${{\mathit{u}}^{\mathbf{\text{'}}}}_{\mathbf{x}}\mathbf{=}\frac{{\mathbf{u}}_{\mathbf{x}}\mathbf{-}\mathbf{v}}{\mathbf{1}\mathbf{-}\frac{{\mathbf{u}}_{\mathbf{x}}\mathbf{v}}{{\mathbf{c}}^{\mathbf{2}}}}$ ...(i)

Here, is the velocity component in the x direction,v is the velocity of frame ${\mathit{S}}^{\mathbf{\text{'}}}$relative to S, and c is the velocity of the light.

The expression to calculate the velocity transformation for velocity of object iny direction is given as follows.

${\mathit{u}}_{\mathbf{y}}^{\mathbf{\text{'}}}\mathbf{=}\frac{{\mathbf{u}}_{\mathbf{y}}}{{\mathbf{\gamma }}_{\mathbf{v}}\mathbf{(}\mathbf{1}\mathbf{-}{\displaystyle \frac{u_{x}v}{c^2}}\mathbf{)}}$ ...(ii)

The expression to calculate the velocity transformation for velocity of object in z

direction is given as follows.

${\mathbf{u}}_{\mathbf{z}}^{\mathbf{\text{'}}}\mathbf{=}\frac{\mathbf{1}}{{\mathit{\gamma }}_{\mathbf{v}}\left(1-{\displaystyle \frac{u_{x}v}{c^2}}\right)}$ ...(iii)

Prove the equation γu'=(1-uxvc2)γvγu.

The Lorentz factor is written out for the velocity $u^{\text{'}}$as,

$$\begin{gathered} {\gamma }_{u^{\text{'}}}=\frac{1}{\sqrt{1-{\left({\displaystyle \frac{u^{\text{'}}}{c}}\right)}^2}} \\ {\gamma }_{v^{\text{'}}}=\frac{1}{\sqrt{1-{\frac{u^{\text{'}}}{c^2}}^2}} \end{gathered}$$

The velocity$u^{{\text{'}}^2}$is expand as,

$$\begin{gathered} {\gamma }_{u\text{'}}=\frac{1}{\sqrt{1-{\displaystyle \frac{u_x^{\text{'}2}+u_y^{\text{'}2}+u_z^{\text{'}2}}{c^2}}}} \\ {\gamma }_{u\text{'}}=\frac{1}{\sqrt{1-{\displaystyle \frac{1}{c^2}\left(u_x^{\text{'}2}+u_y^{\text{'}2}+u_z^{\text{'}2}\right)}}} \end{gathered}$$

Substitute $\frac{u_x-v}{1-\frac{u_{x}V}{c^2}}$for $u_x^{\text{'}}$, $\frac{u_y}{{\gamma }_v\left(1-\frac{u_{x}V}{c^2}\right)}$for ${u^{\text{'}}}_y$and $\frac{u_z}{{\gamma }_v(\left(1-\frac{u_{x}V}{c^2}\right)}$for $u_z^{\text{'}}$into above equation.

$$\begin{gathered} {\gamma }_{u\text{'}}=\frac{1}{\sqrt{1-{\displaystyle \frac{1}{c^2}}}\left[{\left({\displaystyle \frac{u_x-v}{1-{\displaystyle \frac{u_{x}v}{c^2}}}}\right)}^2+{\left({\displaystyle \frac{u_y}{{\gamma }_v\left(1-{\displaystyle \frac{u_{x}v}{c^2}}\right)}}\right)}^2+{\left({\displaystyle \frac{u_z}{{\gamma }_v\left(1-{\displaystyle \frac{u_{x}v}{c^2}}\right)}}\right)}^2\right]} \\ {\gamma }_{u\text{'}}=\frac{1}{\sqrt{1-{\displaystyle \frac{1}{c^2}}}\left[\frac{{\left(u_x-v\right)}^2}{{\left(1-{\displaystyle \frac{u_{x}v}{c^2}}\right)}^2}+\frac{u_y^2}{{\lambda }_v^2{\left(1-{\displaystyle \frac{u_{x}v}{c^2}}\right)}^2}+\frac{u_y^2}{{\lambda }_v^2{\left(1-{\displaystyle \frac{u_{x}v}{c^2}}\right)}^2}\right]} \end{gathered}$$

Multiply by ${\gamma }_v^2$in the $u_x^{\text{'}}$part to simplify the above equation,

role="math" localid="1659332643154" $$\begin{gathered} {\gamma }_{u\text{'}}=\frac{1}{\sqrt{1-{\displaystyle \frac{1}{c^2}\left[\frac{{\gamma }_v^2{\left(u_x-v\right)}^2}{{\gamma }_v^2{\left(1-{\displaystyle \frac{u_x-v}{c^2}}\right)}^2}+\frac{u_y^2}{{\gamma }_v^2{\left(1-\frac{u_x-v}{c^2}\right)}^2}+\frac{u_z^2}{{\gamma }_v^2{\left(1-\frac{u_x-v}{c^2}\right)}^2}\right]}}} \\ {\gamma }_{u\text{'}}=\frac{1}{\sqrt{1-{\displaystyle \frac{1}{c^2}}\left[{\displaystyle \frac{{\gamma }_v^2{\left(u_x-v\right)}^2+u_y^2+u_z^2}{{\gamma }_v^2{\left(1-\frac{u_{x}v}{c^2}\right)}^2}}\right]}} \\ {\gamma }_{u\text{'}}=\frac{1}{\sqrt{1-\left[{\displaystyle \frac{{\gamma }_v^2{\left(u_x-v\right)}^2+u_y^2+u_z^2}{c^2{\gamma }_v^2{\left(1-\frac{u_{x}v}{c^2}\right)}^2}}\right]}} \\ {\gamma }_{u\text{'}}=\frac{1}{\sqrt{{\displaystyle \frac{c^2{\gamma }_v^2{\left(1-{\displaystyle \frac{u_{x}v}{c^2}}\right)}^2-\left[{\gamma }_v^2{\left(u_x-v\right)}^2+u_y^2+u_z^2\right]}{c^2{\gamma }_v^2{\left(1-{\displaystyle \frac{u_{x}v}{c^2}}\right)}^2}}}} \end{gathered}$$

Solve further as,

$$\begin{gathered} {\gamma }_{u\text{'}}=\frac{1}{{\displaystyle \frac{1}{{\gamma }_v\left(1-{\displaystyle \frac{u_{x}v}{c^2}}\right)}}\sqrt{{\displaystyle \frac{c^2{\gamma }_v^2{\left(1-{\displaystyle \frac{u_{x}v}{c^2}}\right)}^2\left[{\gamma }_v^2{\left(u_x-v\right)}^2+u_y^2+u_z^2\right]}{c^2}}}} \\ {\gamma }_{u\text{'}}=\frac{{\gamma }_v\left(1-{\displaystyle \frac{u_{x}v}{c^2}}\right)}{{\displaystyle \sqrt{\frac{c^2{\gamma }_v^2{\left(1-{\displaystyle \frac{u_{x}v}{c^2}}\right)}^2-\left[{\gamma }_v^2{\left(u_x-v\right)}^2+u_y^2+u_z^2\right]}{c^2}}}} \\ {\gamma }_{u\text{'}}=\frac{{\gamma }_v\left(1-{\displaystyle \frac{u_{x}v}{c^2}}\right)}{{\displaystyle \sqrt{\frac{c^2{\gamma }_v^2{\left(1+{\displaystyle \frac{{u^2}_x{v}^2}{c^4}-2\frac{u_{x}v}{c}}\right)}^2-\left[{\gamma }_v^2\left({u^2}_x+v^2-2u_{x}v\right)+u_y^2+u_z^2\right]}{c^2}}}} \\ {\gamma }_{u\text{'}}=\frac{{\gamma }_v\left(1-{\displaystyle \frac{u_{x}v}{c^2}}\right)}{{\displaystyle \sqrt{\frac{{\gamma }_v^2\left[c^2\left(1+{\displaystyle \frac{{u^2}_x{v}^2}{c^4}-2\frac{u_{x}v}{c}}\right)-\left({u^2}_x+v^2-2u_{x}v\right)\right]}{c^2}}}} \end{gathered}$$

Solve further as,

$$\begin{gathered} {\gamma }_{u\text{'}}=\frac{{\gamma }_u\left(1-{\displaystyle \frac{u_{x}v}{c^2}}\right)}{\sqrt{{\displaystyle \frac{{\gamma }_v^2\left[c^2+{\displaystyle \frac{u_x^2{v}^2}{c^2}}-2u_{x}v-u_x^2-v^2+2u_{x}v\right]-u_y^2-u_z^2}{c^2}}}} \\ {\gamma }_{u\text{'}}=\frac{{\gamma }_u\left(1-{\displaystyle \frac{u_{x}v}{c^2}}\right)}{\sqrt{{\displaystyle \frac{{\gamma }_v^2\left[c^2+{\displaystyle \frac{u_x^2{v}^2}{c^2}}-{u^2}_x{v}^2\right]-u_y^2-u_z^2}{c^2}}}} \\ {\gamma }_{u\text{'}}=\frac{{\gamma }_u\left(1-{\displaystyle \frac{u_{x}v}{c^2}}\right)}{\sqrt{{\displaystyle \frac{{\gamma }_v^2\left[c^2-v^2+u_x^2\left({\displaystyle \frac{v^2}{c^2}}-1\right)\right]-u_y^2-u_z^2}{c^2}}}} \\ {\gamma }_{u\text{'}}=\frac{{\gamma }_v\left(1-{\displaystyle \frac{u_{x}v}{c^2}}\right)}{\sqrt{{\displaystyle \frac{{\gamma }_v^2\left[\left(c^2-v^2\right)-y_v^2{u}_x^2\left({\displaystyle 1-\frac{v^2}{c^2}}\right)\right]-u_y^2-u_z^2}{c^2}}}} \end{gathered}$$

Solve further as,

${\gamma }_{u\text{'}}=\frac{{\gamma }_v\left(1-{\displaystyle \frac{u_{x}v}{c^2}}\right)}{\sqrt{{\displaystyle \frac{c^2{y}_v^2\left(1-\frac{v^2}{c^2}\right)-y_v^2{u}_x^2\left(1-\frac{v^2}{c^2}\right)u_y^2-u_z^2}{c^2}}}}$

Substitute $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\sqrt{1-{\left(v/c\right)}^2}$}\right.$for ${\gamma }_v$into above equation.

$$\begin{gathered} {\gamma }_{u\text{'}}=\frac{{\gamma }_v\left(1-{\displaystyle \frac{u_{x}v}{c^2}}\right)}{{\displaystyle \frac{\sqrt{c^2{\displaystyle \frac{1}{{\left(\sqrt{1-{\left({\displaystyle \frac{v}{c}}\right)}^2}\right)}^2}}}\left(1-{\displaystyle \frac{v^2}{c^2}}\right)-{\displaystyle \frac{1}{{\left(\sqrt{1-{\left({\displaystyle \frac{v}{c}}\right)}^2}\right)}^2}}{u}_x^2\left(1-{\displaystyle \frac{v^2}{c^2}}\right)-u_y^2-u_z^2}{c^2}}} \\ {\gamma }_{u\text{'}}=\frac{{\gamma }_v\left(1-\frac{u_{x}v}{c^2}\right)}{\sqrt{{\displaystyle \frac{c^2\left(1-{\displaystyle \frac{v^2}{c^2}}\right)-u_x^2{\displaystyle \frac{\left(1-{\displaystyle \frac{v^2}{c^2}}\right)}{1-{\left({\displaystyle \frac{v}{c}}\right)}^2}}-u_y^2-u_z^2}{c^2}}}} \\ {\gamma }_{u\text{'}}=\frac{{\gamma }_v\left(1-\frac{u_{x}v}{c^2}\right)}{\sqrt{{\displaystyle \frac{c^2-u_x^2+u_y^2+u_z^2}{c^2}}}} \\ {\gamma }_{u\text{'}}=\frac{{\gamma }_v\left(1-\frac{u_{x}v}{c^2}\right)}{\sqrt{{\displaystyle \frac{c^2-\left(u_x^2+u_y^2+u_z^2\right)}{c^2}}}} \end{gathered}$$

Substitute$u_x^2+u_y^2+u_z^2$for $u^2$into above equation.

$$\begin{gathered} {\gamma }_{u\text{'}}=\frac{{\gamma }_v\left(1-{\displaystyle \frac{u_{x}v}{c^2}}\right)}{\sqrt{{\displaystyle \frac{c^2-u^2}{c^2}}}} \\ {\gamma }_{u\text{'}}=\frac{{\gamma }_v\left(1-{\displaystyle \frac{u_{x}v}{c^2}}\right)}{\sqrt{{\displaystyle 1-\frac{u^2}{c^2}}}} \end{gathered}$$

The root mean is equal to Lorentz factor for the velocityu through,

$$\begin{gathered} {\gamma }_{u\text{'}}={\gamma }_v\left(1-\frac{u_{x}v}{c^2}\right){\gamma }_u \\ {\gamma }_{u\text{'}}=\left(1-\frac{u_{x}v}{c^2}\right){\gamma }_u{\gamma }_v \end{gathered}$$

Hence the required equation${\gamma }_{u\text{'}}=\left(1-\frac{u_{x}v}{c^2}\right){\gamma }_u{\gamma }_v$ is obtained.