Chapter 2, Problem 39P

To determine

Frequency of the signal received in beginning $\left(f\prime \right)$ and end $\left(f\prime \prime \right)$ of the trip and time period of Frank and Marry in receiving these frequencies.

Answer to Problem 39P

frequency of the signal received in beginning is $f\prime =f_0\sqrt{\frac{\left(1-\beta \right)}{\left(1+\beta \right)}}$ and end of the trip is $f\prime =f_0\sqrt{\frac{\left(1+\beta \right)}{\left(1-\beta \right)}}$ and time period of Frank in starting is $t_1=\frac{L}{v}+\frac{L}{c}$ and at end of the trip is $t_2=\frac{L}{v}-\frac{L}{c}$ and for Marry in starting time period is $t_1^{\prime }=\frac{L}{\gamma v}$ and at end of the trip is $t_2^{\prime }=\frac{L}{\gamma v}$, in receiving these frequencies.

Explanation of Solution

Initially the signal sent by Marry will be received by the Frank. The source emitting n waves in a time interval T, here Marry going away with velocity (v) from frank that is source is moving away from observer

Write the equation to get the distance travelled by a signal. Thus,

$d=cT+vT$        (I)

Write the equation to get the wavelength of waves emitting n waves in a time interval T and travelling a distance d in that time interval. Thus,

$\lambda \prime =\frac{cT+vT}{n}$        (II)

Write the equation to show the relation between frequency and wavelength. Thus,

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

Where, c is the speed of the light, f is the frequency of the waves, and $\lambda$ is the wavelength of the waves.

Substitute the value of wavelength from equation (II) in above equation. Thus,

$\begin{array}{l}f\prime =\frac{c}{\frac{cT+vT}{n}}\\ =\frac{cn}{cT+vT}\end{array}$        (III)

Write the equation to get the total number of the signals in time interval $T_0^{\prime }$. Thus,

$n=f_0{T}_0^{\prime }$        (IV)

Where, $T_0^{\prime }=\frac{T}{\gamma }$, T is the time measured in the observer’s (Frank) clock and $\gamma =\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$ is the relativistic factor.

Substituting the value of n from equation (IV) in equation (III). Thus

$\begin{array}{l}f\prime =\frac{c{f}_0{T}_0^{\prime }}{cT+vT}\\ \end{array}$

Substitute the $T_0^{\prime }=\frac{T}{\gamma }$ and $\sqrt{1-\frac{v^2}{c^2}}=\frac{1}{\gamma }$ in above equation

$\begin{array}{l}f\prime =\frac{c{f}_0\left(\frac{T}{\gamma }\right)}{cT+vT}\\ =\sqrt{1-\frac{v^2}{c^2}}\left(\frac{c{f}_0}{c+v}\right)\\ =\sqrt{1-\frac{v^2}{c^2}}\left(\frac{f_0}{1+\frac{v}{c}}\right)\\ =\sqrt{1-{\beta }^2}\left(\frac{f_0}{1+\beta }\right)\\ \end{array}$

Because $\beta =\frac{v}{c}$, simplify the above equation, thus,

$\begin{array}{l}f\prime =\sqrt{\left(1-\beta \right)\left(1+\beta \right)}\left(\frac{f_0}{\sqrt{{\left(1+\beta \right)}^2}}\right)\\ f\prime =f_0\sqrt{\frac{\left(1-\beta \right)}{\left(1+\beta \right)}}\end{array}$        (V)

Initially Marry and Frank receives same frequency.

At the end of the trip Marry going away from Frank thus, $\beta$ changes to $\beta$ in the above expression. Thus,

$\begin{array}{l}f\prime \prime =f_0\sqrt{\frac{\left(1-\left(-\beta \right)\right)}{\left(1+\left(-\beta \right)\right)}}\\ =f_0\sqrt{\frac{\left(1+\beta \right)}{\left(1-\beta \right)}}\end{array}$        (VI)

Thus, marry initially receives signals at a rate of $f\prime$ with $t_1^{\prime }=\frac{L}{\gamma v}$ and at end of the trip signals will be received at a rate of  $f\prime \prime$ for $t_2^{\prime }=\frac{L}{\gamma v}$.

And, Frank initially receives signals at a rate of $f\prime \prime$ with $t_1=\frac{L}{v}+\frac{L}{c}$ and at end of the trip signals will be received at a rate of $f\prime \prime$ for $t_2=\frac{L}{v}-\frac{L}{c}$.

Where, L is the distance of the Marry’s Journey from earth to star

Conclusion:

Therefore, frequency of the signal received in beginning is $f\prime =f_0\sqrt{\frac{\left(1-\beta \right)}{\left(1+\beta \right)}}$ and end of the trip is $f\prime =f_0\sqrt{\frac{\left(1+\beta \right)}{\left(1-\beta \right)}}$ and time period of Frank in starting is $t_1=\frac{L}{v}+\frac{L}{c}$ and at end of the trip is $t_2=\frac{L}{v}-\frac{L}{c}$ and for Marry in starting time period is $t_1^{\prime }=\frac{L}{\gamma v}$ and at end of the trip is $t_2^{\prime }=\frac{L}{\gamma v}$, in receiving these frequencies