Figure 37-21 shows one of four star cruisers that are in a race. As each cruiser passes the starting line, a shuttle craft leaves the cruiser and races toward the finish line. You, judging the race, are stationary relative to the starting and finish lines. The speeds v_c of the cruisers relative to you and the speeds of the shuttle craft relative to their respective starships are, in that order, (1) 0.70c, 0.40c; (2) 0.40c, 0.70c; (3) 0.20c, 0.90c; (4) 0.50c, 0.60c. (a) Rank the shuttle craft according to their speeds relative to you, greatest first. (b) Rank the shuttle craft according to the distances their pilots measure from the starting line to the finish line, greatest first. (c) Each starship sends a signal to its shuttle craft at a certain frequency f₀ as measured on board the starship. Rank the shuttle craft according to the frequencies they detect, greatest first.

Suppose an object is moving with the velocity${\mathit{u}}^{\mathbf{\text{'}}}$ with respect to S’ frame, which is moving with the velocity with respect to frame S,then the velocity of the object with respect to S frame is

role="math" localid="1663075755325" $\mathit{u}\mathbf{=}\frac{{\mathbf{u}}^{\mathbf{\text{'}}}\mathbf{+}\mathbf{v}}{\mathbf{1}\mathbf{+}\frac{{\mathbf{u}}^{\mathbf{\text{'}}}\mathbf{v}}{{\mathbf{c}}^{\mathbf{2}}}}$

The length of an object measured in the object’s inertial reference frame is called proper length. When the length of this object is measured in any other inertial frame moving at relative speed is always shorter than the proper length. This is known as length contraction.

$\mathit{L}\mathbf{=}{\mathit{L}}_{\mathbf{0}}\sqrt{\mathbf{1}\mathbf{-}{\mathbf{\beta }}^{\mathbf{2}}}\mathbf{=}\frac{{\mathbf{L}}_{\mathbf{0}}}{\mathbf{\gamma }}$

Here, the proper length between starting and the finish line is measured from the stationary frame. And the speed parameter is

.role="math" localid="1663076128794" $\mathit{\beta }\mathbf{=}\frac{\mathbf{v}}{\mathbf{c}}$

The four sets of the speed of cruiser relative to you $\left(v_c\right)$and the speed of shuttle relative to cruiser $\left(v_s\right)$in respective order are-

i. $0.7c,0.40c$

ii. $0.4c,0.70c$

iii. $0.20c,0.90c$

iv.$0.50c,0.60c$

In this case, the star cruiser is moving at speed $v_c$and the shuttle is moving at a speed $v_s$relative to the star cruiser. Therefore, the velocity of the shuttle relative to the stationary frame is

$u_{s,rel}=\frac{v_s+V_c}{1+\frac{V_s{V}_c}{c^2}}$

For $V_c=0.70c,V_s=0.40c$, the speed of the shuttle relative to the stationary frame is

$$\begin{gathered} u_1=\frac{0.70c+0.40c}{1+{\displaystyle \frac{0.70c+0.40c}{c^2}}} \\ =\frac{1.1c}{1.28} \\ =0.86c \end{gathered}$$

For$V_c=0.40c,V_s=0.70c$, the speed of the shuttle relative to the stationary frame is

$$\begin{gathered} u_2=\frac{0.40c+0.70c}{1+{\displaystyle \frac{0.40c+0.70c}{c^2}}} \\ =\frac{1.1c}{1.28} \\ =0.86c \end{gathered}$$

For$V_c=0.20c,V_s=0.90c$ , the speed of the shuttle relative to the stationary frame is

$$\begin{gathered} u_3=\frac{0.20c+0.90c}{1+{\displaystyle \frac{0.20c+0.90c}{c^2}}} \\ =\frac{1.1c}{1.18} \\ =0.93c \end{gathered}$$

For$V_c=0.50c,V_s=0.60c$, the speed of the shuttle relative to the stationary frame is

$$\begin{gathered} u_4=\frac{0.50c+0.60c}{1+{\displaystyle \frac{0.50c+0.60c}{c^2}}} \\ =\frac{1.1c}{1.30} \\ =0.85c \end{gathered}$$

The relative velocities of each shuttle are ranked as follows.

$u_3>u_1=u_2>u_4$

For part (1), the shuttle moves at a speed relative to you. The distance measured by the pilot of this shuttle is

$\begin{array}{l}{L}_1=L_0\sqrt{1-{u^2}_1/c^2}\\ =L_0\sqrt{1-{\left(0.86\right)}^2}\\ =0.51L_0\end{array}$

For part (2), the shuttle moves at a speed relative to you. The distance measured by the pilot of this shuttle is

$\begin{array}{l}{L}_2=L_0\sqrt{1-{u^2}_2/c^2}\\ =L_0\sqrt{1-{\left(0.85\right)}^2}\\ =0.51L_0\end{array}$

For part (3), the shuttle moves at a speed relative to you. The distance measured by the pilot of this shuttle is

$\begin{array}{l}{L}_3=L_0\sqrt{1-{u^2}_3/c^2}\\ =L_0\sqrt{1-{\left(0.93\right)}^2}\\ =0.37L_0\end{array}$

For part (4), the shuttle moves at a speed relative to you. The distance measured by the pilot of this shuttle is

$\begin{array}{l}{L}_4=L_0\sqrt{1-{u^2}_4/c^2}\\ =L_0\sqrt{1-{\left(0.85\right)}^2}\\ =0.53L_0\end{array}$

Hence the length measured by pilots of each shuttle is as follows.

$L_4>L_1=L_2>L_3$

In astronomy applications, the velocities of galaxies are estimated using Doppler shifts. Doppler shift is the difference between the observed and proper wavelength of light. The wavelength measured in the rest frame of the source is called proper wavelength${\lambda }_0$ . And the detected wavelength $\lambda$is related to the proper wavelength as

$\lambda ={\lambda }_0\sqrt{\frac{1+\beta }{1-\beta }}$

where$\beta$ is the speed parameter ($v/c$).The wavelength ad frequency is related by

$\lambda =\frac{c}{f}$

Inserting this in the above expression, we get

$$\begin{gathered} \frac{c}{f}=\frac{c}{f_0}\sqrt{\frac{1+\beta }{1-\beta }} \\ f=f_0\sqrt{\frac{1+\beta }{1-\beta }} \end{gathered}$$

This formula is valid when the source-detector separation is increasing. For decreasing separation, the signs of$\beta$ will flip.

For part (1), the shuttle moves at a speed role="math" localid="1663078209263" $u_1$relative to you. The frequency detected by the shuttle pilot is

$$\begin{gathered} f_1=f_0\sqrt{\frac{1-0.86}{1+0.86}} \\ =f_0\sqrt{\frac{0.14}{1.86}} \\ =0.27f_0 \end{gathered}$$

For part (2), the shuttle moves at a speed $u_2$relative to you. The frequency detected by the shuttle pilot is

$$\begin{gathered} f_2=f_0\sqrt{\frac{1-0.86}{1+0.86}} \\ =f_0\sqrt{\frac{0.14}{1.86}} \\ =0.27f_0 \end{gathered}$$

For part (3), the shuttle moves at a speed $u_3$relative to you. The frequency detected by the shuttle pilot is

$$\begin{gathered} f_3=f_0\sqrt{\frac{1-0.93}{1+0.93}} \\ =f_0\sqrt{\frac{0.07}{1.93}} \\ =0.19f_0 \end{gathered}$$

For part (4), the shuttle moves at a speed $u_4$relative to you. The frequency detected by the shuttle pilot is

$$\begin{gathered} f_4=f_0\sqrt{\frac{1-0.85}{1+0.85}} \\ =f_0\sqrt{\frac{0.15}{1.85}} \\ =0.29f_0 \end{gathered}$$

Hence, the frequency detected by each shuttle pilot is ranked as follows.

$f_4>f_1=f_2>f_3$