Question

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.

Short answer

The component of velocity of light beam in x and y directions are$\frac{c(c\cos \theta -v)}{c-v\cos \theta }$ and$\frac{c^2\sin \theta }{{\gamma }_v(c-vcos\theta )}$ respectively. It is also verified that the total velocity of the beam is equal to the speed of the light.

Step-by-step solution

Write the given data from the question.

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

Determine the formulas to calculate the component of velocity of light beam relative to S'.

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,

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

Calculate the component of velocity of light beam relative to S'.

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.