You fire a light signal at $\mathbf{60}\mathbf{°}$north of west (a) Find the velocity component of this, signal according to an observer moving eastward relative to you at half the speed of light. From them. determine the magnitude and direction of the light signal's velocity according to this other observer. (b) Find the component according to a different observer, moving westward relative to you at half the speed of light.

The observer is moving eastward related to the half of the speed of the light.

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,${\mathit{u}}_{\mathbf{x}}$ 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 in y direction is given as follows.

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

The expression to calculate the Lorentz factor is given by,

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

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

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

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

Substitute $c\mathrm{sin\theta }$ for $u_y$ into equation (ii).

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

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

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

Hence the component of velocity of light beam in x and y directions are $\frac{-c\cos \theta -v}{1+\frac{v\cos \theta }{c}}$ and $\frac{c\sqrt{1-{\left(\frac{v}{c}\right)}^2}\sin \theta }{\left(1+\frac{v\cos \theta }{c}\right)}$respectively.

Since the observer is moving to right, the velocity,$v=+0.5c$

Calculate the x component of the velocity,

Substitute ${60}^{°}$for $\theta$ and +0.5 c for v into equation (iii).

role="math" localid="1659329161430" $$\begin{gathered} u_x^{\text{'}}=\frac{-c\cos 60°-0.5c}{1+\frac{0.5c\cos {60}^{°}}{c}} \\ {u}_x^{\text{'}}=\frac{-0.5c-0.5c}{1+0.25} \\ {u}_x^{\text{'}}=\frac{-1c}{1.25} \\ {u}_x^{\text{'}}=-0.8c \end{gathered}$$

Calculate the component of the velocity,

Substitute $60°$for $\theta$ and 0.5c for v into equation (iv).

$$\begin{gathered} u_y^{\text{'}}=\frac{c\sqrt{1-{\left(\frac{0.5c}{c}\right)}^2\sin 60°}}{\left(1+\frac{0.5c\cos 60°}{\left(1+0.5\times 0.5\right)}\right)} \\ {u}_y^{\text{'}}=\frac{c\sqrt{1-0.25}\left(0.866\right)}{\left(1+0.5\times 0.5\right)} \\ {u}_y^{\text{'}}=\frac{c\left(0.866\right)\left(0.866\right)}{\left(1+0.25\right)} \\ {u}_y^{\text{'}}=0.6c \end{gathered}$$

So, the velocities for the light beam for the moving eastward observer are.

$(u_x^{\text{'}},u_y^{\text{'}})=(-0.8c,0.6c)$.

Since the observer is moving to left, the velocity,$V=-0.5c$

Calculate the x component of the velocity,

Substitute $60°$ for $\theta$ and $-0.5c$ for v into equation (iii).

$$\begin{gathered} u_x^{\text{'}}=\frac{-ccos60°-(-0.5c)}{1+\frac{(-0.5c)\cos 60°}{c}} \\ {u}_x^{\text{'}}=\frac{-0.5c+0.5c}{1-0.25} \\ {u}_x^{\text{'}}=\frac{0}{0.75} \\ {u}_x^{\text{'}}==0 \end{gathered}$$

Calculate the y component of the velocity,

Substitute $60°$ for $\theta$ and -0.5c for v into equation (iv).

$$\begin{gathered} u_y^{\text{'}}=\frac{c\sqrt{1-{\left(\frac{-0.5c}{c}\right)}^2}\sin 60°}{\left(1+\frac{\left(-0.5c\right)\cos 60°}{c}\right)} \\ {u}_y^{\text{'}}=\frac{c\sqrt{1-0.25}\left(0.866\right)}{\left(1-0.5\times 0.5\right)} \\ {u}_y^{\text{'}}=\frac{c\left(0-.866\right)\left(0.866\right)}{0.75} \\ {u}_y^{\text{'}}=c \end{gathered}$$

So, the velocities for the light beam for the moving westward observer are $(u_x^{\text{'}},u_y^{\text{'}})=(0,c)$.