Chapter 26, Problem 96P

(a)

To determine

The time elapse between the emission of wavefronts by the source, in observer’s frame.

Answer to Problem 96P

The time elapse between the emission of wavefronts by the source, in observer’s frame is $\underset{\_}{T_{\text{s}}{}^{\prime }=\frac{\gamma }{f_{\text{s}}}}$.

Explanation of Solution

Write the expression for the proper time interval.

  $\Delta t_0=T_{\text{s}}$                                                                                                              (I)

Here, $\Delta t_0$ is the proper time interval, $T_{\text{s}}$ is the period of wavefront emission.

Write the expression for the period of the wavefront emission.

  $T_{\text{s}}=\frac{1}{f_{\text{s}}}$                                                                                                               (II)

Here, $f_{\text{s}}$ is the frequency of emission.

Use equation (II) in (I).

  $\Delta t_0=\frac{1}{f_{\text{s}}}$                                                                                                             (III)

Write the expression for the time elapse between the emission of wavefronts by the source, in observer’s frame.

  $T_{\text{s}}{}^{\prime }=\Delta t$                                                                                                              (IV)

Here, $T_{\text{s}}{}^{\prime }$ is the time elapse between the emission of wavefronts by the source, in observer’s frame, and $\Delta t$ is the time interval in observer’s frame.

Write the expression for the time interval corresponding to the proper time interval, in observer’s frame.

  $\Delta t=\gamma \Delta t_0$                                                                                                             (V)

Here, $\gamma$ is the Lorentz factor.

Use equation (III) in (V).

  $\Delta t=\frac{\gamma }{f_{\text{s}}}$                                                                                                                (VI)

Use equation (VI) in (IV).

  $T_{\text{s}}{}^{\prime }=\frac{\gamma }{f_{\text{s}}}$                                                                                                               (VII)

Conclusion:

Therefore, the time elapse between the emission of wavefronts by the source, in observer’s frame is $\underset{\_}{T_{\text{s}}{}^{\prime }=\frac{\gamma }{f_{\text{s}}}}$.

(b)

To determine

The time $T_{\text{o}}$ between the arrival of the two wavefronts as measured by the observer.

Answer to Problem 96P

The time $T_{\text{o}}$ between the arrival of the two wavefronts as measured by the observer is $\underset{\_}{T_{\text{o}}=\frac{\gamma }{f_{\text{s}}}\left(1+\frac{v}{c}\right)}$.

Explanation of Solution

Write the expression for the time that the first wavefront described in the problem statement arrives at the observer.

  $t_1=\frac{d}{c}$                                                                                                                (VIII)

Here, $t_{\text{1}}$ is the time that the first wavefront arrives at the observer, $d$ is the distance between the source and the observer, and $c$ is the speed of light.

Write the expression for the time of travel of the second wavefront as it arrives at the observer.

  $t_2=T_{\text{s}}{}^{\prime }+\frac{d+v{T}_{\text{s}}{}^{\prime }}{c}$                                                                                                   (IX)

Here, $t_2$ is the time of travel of the second wavefront as it arrives at the observer.

Write the expression for time $T_{\text{o}}$ between the arrival of the two wavefronts as measured by the observer.

  $T_{\text{o}}=t_{\text{2}}-t_1$                                                                                                         (X)

Use equation (VIII) and (IX) in (X).

  $T_{\text{o}}=T_{\text{s}}{}^{\prime }+\frac{d+v{T}_{\text{s}}{}^{\prime }}{c}-\frac{d}{c}$                                                                                       (XI)

Use equation (VII) in (XI).

  $\begin{array}{c}{T}_{\text{o}}=\frac{\gamma }{f_{\text{s}}}+\frac{d}{c}+\frac{v\gamma }{c{f}_{\text{s}}}-\frac{d}{c}\\ =\frac{\gamma }{f_{\text{s}}}\left(1+\frac{v}{c}\right)\end{array}$                                                                                         (XII)

Conclusion:

Therefore, the time $T_{\text{o}}$ between the arrival of the two wavefronts as measured by the observer is $\underset{\_}{T_{\text{o}}=\frac{\gamma }{f_{\text{s}}}\left(1+\frac{v}{c}\right)}$.

(c)

To determine

To show that the frequency detected by the observer is $f_{\text{o}}=f_{\text{s}}\sqrt{\frac{1-v/c}{1+v/c}}$.

Answer to Problem 96P

The frequency detected by the observer is $f_{\text{o}}=f_{\text{s}}\sqrt{\frac{1-v/c}{1+v/c}}$.

Explanation of Solution

Write the expression for the frequency detected by the observer.

  $f_{\text{o}}=\frac{1}{T_{\text{o}}}$                                                                                                            (XIII)

Use equation (XII) in (XIII).

  $\begin{array}{c}{f}_{\text{o}}=\frac{1}{\frac{\gamma }{f_{\text{s}}}\left(1+\frac{v}{c}\right)}\\ =f_{\text{s}}\frac{1}{\gamma \left(1+v/c\right)}\end{array}$                                                                                          (XIV)

Write the expression of Lorentz constant.

  $\gamma =\frac{1}{\sqrt{1-v^2/c^2}}$                                                                                                 (XV)

Use equation (XV) in (XIV) and simplify.

  $\begin{array}{c}{f}_{\text{o}}=f_{\text{s}}\frac{\sqrt{1-v^2/c^2}}{\left(1+v/c\right)}\\ =f_{\text{s}}\sqrt{\frac{1-v^2/c^2}{{\left(1+v/c\right)}^2}}\\ =f_{\text{s}}\sqrt{\frac{\left(1+v/c\right)\left(1-v/c\right)}{{\left(1+v/c\right)}^2}}\\ =f_{\text{s}}\sqrt{\frac{1-v/c}{1+v/c}}\end{array}$

Conclusion:

Therefore, the frequency detected by the observer is $f_{\text{o}}=f_{\text{s}}\sqrt{\frac{1-v/c}{1+v/c}}$.