(a) In Ex. 12.6 we found how velocities in thex direction transform when you go from Sto $\overline{\mathit{S}}$. Derive the analogous formulas for velocities in the y and z directions.

(b) A spotlight is mounted on a boat so that its beam makes an angle$\overline{\mathit{\theta }}$ with the deck (Fig. 12.20). If this boat is then set in motion at speedv, what angle$\mathit{\theta }$ does an individual photon trajectory make with the deck, according to an observer on the dock? What angle does the beam (illuminated, say, by a light fog) make? Compare Prob. 12.10.

Let’s assumeS be the frame of reference of a boat,$S\text{'}$be the frame of the observer, $dy$be the distance in frame S and $dt$be the time interval.

Write the expression for the Lorentz transformation equation based on the assumed data.

$\mathit{d}\mathit{t}\mathbf{\text{'}}\mathbf{=}\mathit{\gamma }\left(\begin{array}{c}dt-\frac{v}{c^2}dx\end{array}\right)$

Here, v is the velocity of the particle, x is the displacement, and c is the speed of light.

(a)

Express the equation when the particles move with respect to the frame along the y-direction.

$$\begin{gathered} \frac{\mathrm{dy}\text{'}}{\mathrm{dt}\text{'}}=\frac{\mathrm{dy}}{\gamma \left(\begin{array}{c}\mathrm{dt}-{\displaystyle \frac{\mathrm{v}}{{\mathrm{c}}^2}}\mathrm{dx}\end{array}\right)} \\ \frac{\mathrm{dy}\text{'}}{\mathrm{dt}\text{'}}=\frac{\mathrm{dy}}{\gamma dt\left(\begin{array}{c}1-{\displaystyle \frac{\mathrm{v}}{{\mathrm{c}}^2}}{\displaystyle \frac{\mathrm{dx}}{\mathrm{dt}}}\end{array}\right)} \end{gathered}$$ …… (1)

Write the expression for the instantaneous speed of a particle along the x-direction.

$u_x=\frac{dx}{dt}$

Substitute$\frac{dy\text{'}}{dt\text{'}}=u{\text{'}}_y,dy=u_y$and $\frac{dx}{dt}=u_x$ in equation (1) to form the analogous formula for velocity in the y-direction.

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

Substitute $\frac{dy\text{'}}{dt\text{'}}=u{\text{'}}_{\mathrm{z}},dy=u_{\mathrm{z}}$and $\frac{dx}{dt}=u_x$in equation (1) to form the analogous formula for velocity in the z-direction.

$u{\text{'}}_{\mathrm{z}}=\frac{u_{\mathrm{z}}}{\gamma \left(\begin{array}{c}1-{\displaystyle \frac{v}{c^2}}{u}_x\end{array}\right)}$

Therefore, the analogous formulas for velocities in the y and z directions are $u{\text{'}}_y=\frac{u_y}{\gamma \left(\begin{array}{c}1-{\displaystyle \frac{v}{c^2}}{u}_x\end{array}\right)}$and $u{\text{'}}_{\mathrm{z}}=\frac{u_{\mathrm{z}}}{\gamma \left(\begin{array}{c}1-{\displaystyle \frac{v}{c^2}}{u}_x\end{array}\right)}$respectively.

(b)

Write the expression for the angle made by the observer on the deck.

$\tan \theta \text{'}=\frac{-u{\text{'}}_y}{u{\text{'}}_x}$ …… (2)

Here,$\mathrm{u}{\text{'}}_{\mathrm{x}}$is the instantaneous speed of a particle along the x-direction with respect to $S\text{'}$frame of reference, which is given as:

$u{\text{'}}_x=\frac{dx\text{'}}{dt}$ …… (3)

Here, $dx\text{'}$is the distance moved by boat in $S\text{'}$frame of reference.

Express the equation when the particles move with respect to the frame along the x-direction using the Lorentz transformation equation.

$$\begin{gathered} dx\text{'}=\gamma \left(dx-vdt\right) \\ dt\text{'}=\gamma \left(\begin{array}{c}dt-\frac{v}{c^2}dx\end{array}\right) \end{gathered}$$

Substitute$dx\text{'}=\gamma \left(dx-vdt\right),dt\text{'}=\gamma \left(\begin{array}{c}dt-\frac{v}{c^2}dx\end{array}\right)$and$u_x=dx$in equation (3).

$$\begin{gathered} u{\text{'}}_x=\frac{\gamma \left(d-vdt\right)}{\gamma \left(dt-{\displaystyle \frac{v}{c^2}}dx\right)} \\ u{\text{'}}_x=\frac{u_x-v}{1-{\displaystyle \frac{v}{c^2}}{u}_x} \end{gathered}$$

Substitute $u{\text{'}}_x=\frac{u_x-v}{1-{\displaystyle \frac{v}{c^2}}{u}_x}$and$u{\text{'}}_y=\frac{u_y}{\gamma \left(1-{\displaystyle \frac{v}{c^2}}{u}_x\right)}$in equation (2).

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

Here, $u_y=c\sin \theta$and$u_x=-c\cos \theta$

Hence, the above equation becomes,

$$\begin{gathered} \tan \theta \text{'}=\frac{-\left(\begin{array}{c}c\sin \theta \end{array}\right)}{\gamma \left(\begin{array}{c}-c\cos \theta -v\end{array}\right)} \\ \tan \theta \text{'}=\frac{\sin \theta }{\gamma \left(\begin{array}{c}\cos \theta +{\displaystyle \frac{v}{c}}\end{array}\right)} \end{gathered}$$

From problem 12.10, the result was obtained as$\tan \theta \text{'}=\gamma \frac{\sin \theta \text{'}}{\cos \theta \text{'}}$ which is the only case of velocity.Still, in this problem, the speeds are sensitive to the transformation of distances and times, which is the only reason there is no universal rule for translating angles. One must know whether it’s an angle made by a velocity vector or a position vector.

As the beam is made with many different photons at one instant time, it transforms the same way the mast does.

Therefore, the angle of an individual photon on the dock and on the beam is$\tan \theta \text{'}\frac{\sin \theta }{\gamma \left(\begin{array}{c}\cos \theta +{\displaystyle \frac{v}{c}}\end{array}\right)}$.