Question: A light beam moves in the xy plane and has an x component of velocity of ${\mathit{u}}_{\mathbf{x}}$ (a) In terms of ${\mathit{u}}_{\mathbf{x}}$, and c, what is its y component? (b) Using equations (2-20a) and (2-20b). Calculate its velocity components in a frame moving in the x direction at speed$\mathit{v}\mathbf{=}{\mathit{u}}_{\mathbf{x}}$ and comment on your result.

The x component of the light beam moving in the xy plane is $u_x$.

The equation 2.20a is given by,

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

The equation 2.20b is given by,

$u_x^{\text{'}}=\frac{u_y}{{\gamma }_v\left(1-{\displaystyle \frac{u_{x}v}{c^2}}\right)}$

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

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

Here, is the velocity component in the direction, is the velocity of frame relative to , and is the velocity of the light.

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

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

Here,${\mathbf{u}}_{\mathbf{y}}$ is the velocity component in the y direction and${\mathbf{\gamma }}_{\mathbf{v}}$ is the Lorentz factor.

The expression to calculate the Lorentz factor is given by,

${\mathbf{\gamma }}_{\mathbf{v}}\mathbf{=}\frac{\mathbf{1}}{\sqrt{\mathbf{1}\mathbf{-}{\left({\displaystyle \frac{v}{c}}\right)}^{\mathbf{2}}}}$

Since the light beam is moving in the xy plane, therefore the velocity is related by the sum of the square.

$$\begin{gathered} u_x^2+u_y^2=c^2 \\ {u}_y^2=c^2-u_x^2 \\ {u}_y=\sqrt{c^2-u_x^2} \end{gathered}$$

Hence the velocity component in y direction, in terms of c and $u_x\mathrm{is}{\mathrm{u}}_{\mathrm{y}}=\sqrt{{\mathrm{c}}^2-{\mathrm{u}}_{\mathrm{x}}^2}$.

$u_y\mathrm{and}{\mathrm{u}}_{\mathrm{x}}$Calculate the component of velocity of the light beam in x direction,

Substitute$u_x$for into v equation (i)

$$\begin{gathered} u_x^{\text{'}}=\frac{u_x-u_x}{1-{\displaystyle \frac{u_x{u}_x}{c}}} \\ {u}_x^{\text{'}}=\frac{0}{1-{\displaystyle \frac{u_x{u}_x}{c}}} \\ {u}_x^{\text{'}}=0 \end{gathered}$$

Calculate the component of velocity of the light beam in y direction,

Substitute$1/\sqrt{1-{\left(\frac{v}{c}\right)}^2}$into equation (ii).

$u_y^{\text{'}}=\frac{u_v}{{\displaystyle \frac{1}{\sqrt{1-{\displaystyle \frac{v^2}{c^2}}}}}\left(1-{\displaystyle \frac{u_{x}v}{c^2}}\right)}$

Substitute$\sqrt{c^2-u_x^2}$for $u_y\mathrm{and}{\mathrm{u}}_{\mathrm{x}}$for v into above equation.

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

Solve further as,

$$\begin{gathered} u_y^{\text{'}}=\frac{c\sqrt{1-\frac{u_x^2}{c^2}}}{\sqrt{1-\frac{u_x^2}{c^2}}} \\ {u}_y^{\text{'}}=c \end{gathered}$$

Since the velocity in the x direction is zero. Therefore, all the velocity would be in y direction. The light always travels with the velocity c. Therefore, the result of velocity is x and y are direction correct.

Hence the velocity component of the light beam in x and y directions are 0 and c respectively.