A light beam moves at an angle$\mathit{\theta }$ with the x-axis as seen from frame S. Using the relativistic velocity transformation, find the components of its velocity when viewed from frame $\mathit{S}\mathbf{\text{'}}$. From these, verify explicitly that its speed is c.

The angle at which light beam is moving is$\theta$ withx axis.

The expression to calculate the velocity transformation for velocity of object inx 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 role="math" localid="1659325003700" ${\mathit{S}}^{\mathbf{\text{'}}}$relative to S, and c is the velocity of the light.

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

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

The expression to calculate the Lorentz factor is given by,

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

Since the light beam is moving with the x axis, so the component of velocity in x and y direction is given by,

$$\begin{gathered} u_x=ccos\theta \\ {u}_y=csin\theta \end{gathered}$$

Calculate the component of velocity in x direction relative to frame $S^{\text{'}}$.

Substitute $c\cos \theta$for $u_x$into equation (i).

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

Calculate the component of velocity in y direction relative to frame $S^{\text{'}}$.

Substitute $cs\mathrm{in}\mathrm{\theta }$for $u_y$into equation (ii).

$$\begin{gathered} u_y^{\text{'}}=\frac{c\sin \theta }{{\gamma }_v\left(1-{\displaystyle \frac{\left(c\cos \theta \right)v}{c^2}}\right)} \\ {u}_y^{\text{'}}=\frac{c\sin \theta }{{\gamma }_v\left(1-{\displaystyle \frac{\left(c\cos \theta \right)v}{c^2}}\right)} \\ {u}_y^{\text{'}}=\frac{c\sin \theta }{{\gamma }_v\left(1-{\displaystyle \frac{v\cos \theta }{c}}\right)} \\ {u}_y^{\text{'}}=\frac{c^2\sin \theta }{{\gamma }_v\left(c-v\cos \theta \right)} \end{gathered}$$

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

$\frac{c\left(c\cos \theta -v\right)}{c-v\cos \theta }$and $\frac{c^2\sin \theta }{Y_v\left(c-v\cos \theta \right)}$respectively.

Take the sum of the squares of the velocity component to prove the total velocity is equal to light velocity.

$(u^{\text{'}}{)}^2=(u_x^{\text{'}}{)}^2+(u_y^{\text{'}}{)}^2$

Substitute $\frac{c\left(c\cos \theta -v\right)}{c-v\cos \theta }$for ${\left({u^{\text{'}}}_x\right)}^2$and $\frac{c^2\sin \theta }{Y_v\left(c-v\cos \theta \right)}$for ${\left({u^{\text{'}}}_y\right)}^2$into above equation.

$$\begin{gathered} (u^{\text{'}}{)}^2={\left(\frac{c(c\cos \theta -v)}{c-v\cos \theta }\right)}^2{\left(\frac{c^{2}sin\theta }{{\gamma }_v(c-vcos\theta )}\right)}^2 \\ (u^{\text{'}}{)}^2={\left(\frac{c}{c-v\cos \theta }\right)}^2\left[{(c-vcos\theta )}^2+\frac{c2sin\theta }{\gamma 2}\right] \end{gathered}$$_v

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

$$\begin{gathered} {\left(u^{\text{'}}\right)}^2={\left(\frac{c}{c-v\cos \theta }\right)}^2\left[{\left(c\cos \theta -v\right)}^2+\frac{c^2{\sin }^2\theta }{{\left({\displaystyle \frac{1}{\sqrt{1-\left({\displaystyle \frac{v}{c}}\right)}}}\right)}^2}\right] \\ {\left(u^{\text{'}}\right)}^2={\left(\frac{c}{c-v\cos \theta }\right)}^2\left\{c^2{\cos }^2\theta +v^2-2vc\cos \theta +\left[1-{\left(\frac{v}{c}\right)}^2\right]c^2{\sin }^2\theta \right\} \\ {\left(u^{\text{'}}\right)}^2={\left(\frac{c}{c-v\cos \theta }\right)}^2\left\{c^2+v^2-2vc\cos \theta -v^2{\sin }^2\theta \right\} \end{gathered}$$

Solve further as,

localid="1659327155880" $$\begin{gathered} (u^{\text{'}}{)}^2={\left(\frac{c}{c-v\cos \theta }\right)}^2[c^2-2vccos\theta -v^2(1-si{n}^2\theta )\left] \\ \right(u^{\text{'}}{)}^2={\left(\frac{c}{c-v\cos \theta }\right)}^2(c^2-2(v\left)\right(ccos\theta )-v^{2}co{s}^2\theta ) \\ (u^{\text{'}}{)}^2=\frac{c}{{\left(c-v\cos \theta \right)}^2}(c-vcos\theta {)}^2 \\ (u^{\text{'}}{)}^2=\frac{c^2}{{\left(c-v\cos \theta \right)}^2}(c-vcos\theta {)}^2 \end{gathered}$$

Solve further as,

$\begin{array}{l}{\left(u^{\text{'}}\right)}^2=c^2\\ u^{\text{'}}=\sqrt{c^2}\\ u^{\text{'}}=c\end{array}$

Therefore, it is proved that, the total velocity of the beam is equal to the speed of the light.