Chapter 2, Problem 89P

(a)

To determine

The number of signals received by Frank at rate $f^{\prime }$ and the time at which Frank detects Mary’s turnaround.

Answer to Problem 89P

Frank receives half the number of signals sent by Mary and the time at which Frank detects is $\frac{L}{v}+\frac{L}{c}$ .

Explanation of Solution

The number of signals received by Frank at the rate $f^{\prime }$ is $fL/\gamma v$ . It is half of the number of signals sent by Mary at the same rate.

Write the expression for detected time.

  $t=\frac{n}{f^{\prime }}$        (I)

Here, $n$ is the number of signals, $f^{\prime }$ is the frequency and $t$ is the detected time.

Substitute $fL/\gamma v$ for $n$ and $f\sqrt{\left(1-\beta \right)/\left(1+B\right)}$ for $f^{\prime }$ in equation (I).

  $\begin{array}{c}t=\frac{fL/\gamma v}{f\sqrt{\left(1-\beta \right)/\left(1+B\right)}}\\ =\frac{L\sqrt{1+\beta }}{\gamma v\sqrt{1-\beta }}\end{array}$

Substitute $\frac{1}{\sqrt{1-{\beta }^2}}$ for $\gamma$ in above equation and simplify.

  $\begin{array}{c}t=\frac{L\left(1+\beta \right)}{v}\\ =\frac{L}{v}+\frac{L}{c}\end{array}$

Conclusion:

Thus, Frank receives half the number of signals sent by Mary and the time at which Frank detects is $\frac{L}{v}+\frac{L}{c}$ .

(b)

To determine

The number of signals received by Mary at rate $f^{\prime }$ and the time at which Mary turns around.

Answer to Problem 89P

The number of signals received by Mary at rate $f^{\prime }$ is $fL\left(1-\beta \right)/v$ and the time at which Mary turns around is $L/\gamma v$ .

Explanation of Solution

Write the expression for number of signals received by Mary.

  $n^{\prime }=f^{\prime }\frac{T^{\prime }}{2}$        (II)

Here, $n^{\prime }$ is the number of signals received, $f^{\prime }$ is the frequency and $T^{\prime }$ is the total time.

Substitute $f\sqrt{\left(1-\beta \right)/\left(1+B\right)}$ for $f^{\prime }$ and $2L/\gamma v$ for $T^{\prime }$ in equation (II).

  $\begin{array}{c}{n}^{\prime }=\left(f\sqrt{\left(1-\beta \right)/\left(1+B\right)}\right)\left(\frac{2L/\gamma v}{2}\right)\\ =\frac{fL\left(1-\beta \right)}{v}\end{array}$

The turnaround time is half of the total time is equal to $L/\gamma v$.

Conclusion:

Thus, the number of signals received by Mary at rate $f^{\prime }$ is $fL\left(1-\beta \right)/v$ and the time at which Mary turns around is $L/\gamma v$ .

(c)

To determine

The time for the remainder of trip, number of signal received at rate $f^{″}$ , total number of signal received and Mary’s age from perspective of Frank.

Answer to Problem 89P

The time for the remainder of trip is $L/v-L/c$, number of signal received at rate $f^{″}$ is $fL/\gamma v$, total number of signal received is $2fL/\gamma v$ and Mary’s age from perspective of Frank is $2L/\gamma v$.

Explanation of Solution

Write the expression for remainder time of trip.

  $t_2=T-t_1$        (III)

Here, $T$ is the total time, $t_1$ is the time of detecting Mary’s turnaround by Frank and $t_2$ is the time for remainder of trip.

Substitute $2L/v$ for $T$ and $L/v+L/c$ for $t_1$ in equation (III).

  $\begin{array}{c}{t}_2=2L/v-\left(L/v+L/c\right)\\ =L/v-L/c\end{array}$

Write the expression for number of signals received.

  $\begin{array}{c}n=f^{″}{t}_2\\ =f\sqrt{\frac{1+\beta }{1-\beta }}\left(L/v-L/c\right)\\ =\frac{fL}{\gamma v}\end{array}$

Write the expression for total number of signals.

  $\begin{array}{c}N=\frac{fL}{\gamma v}+\frac{fL}{\gamma v}\\ =\frac{2fL}{\gamma v}\end{array}$

Here, N is the total number of signals.

Write the expression for Mary’s age.

  $\begin{array}{c}A=\frac{N}{f}\\ =\frac{2L}{\gamma v}\end{array}$

Here, A is the Mary’s age.

Conclusion:

Thus, the time for the remainder of trip is $L/v-L/c$, number of signal received at rate $f^{″}$ is $fL/\gamma v$, total number of signal received is $2fL/\gamma v$ and Mary’s age from perspective of Frank is $2L/\gamma v$.

(d)

To determine

The time for the remainder of trip, number of signal received at rate $f^{″}$ , total number of signal received and Frank’s age from perspective of Mary.

Answer to Problem 89P

The time for the remainder of trip is $L/\gamma v$, number of signal received at rate $f^{″}$ is $fL\left(1+\beta \right)/v$ , total number of signal received is $2fL/v$ and Frank’s age from perspective of Mary is $2L/v$ .

Explanation of Solution

Write the expression for remainder time of trip.

  ${t^{\prime }}_2=T^{\prime }-{t^{\prime }}_1$        (IV)

Here, $T^{\prime }$ is the total time, ${t^{\prime }}_1$ is the time of detecting Mary’s turnaround by Mary and ${t^{\prime }}_2$ is the time for remainder of trip.

Substitute $2L/\gamma v$ for $T$ and $L/\gamma v$ for $t_1$ in equation (III).

  $\begin{array}{c}{t^{\prime }}_2=2L/\gamma v-L/\gamma v\\ =L/\gamma v\end{array}$

Write the expression for number of signals received.

  $\begin{array}{c}n=f^{″}{t}_2\\ =f\sqrt{\frac{1+\beta }{1-\beta }}\cdot \frac{L}{\gamma v}\\ =\frac{fL}{v}\left(1+\beta \right)\end{array}$

Write the expression for total number of signals.

  $\begin{array}{c}N=\frac{fL}{v}\left(1-\beta +1+\beta \right)\\ =\frac{2fL}{v}\end{array}$

Here, N is the total number of signals.

Write the expression for Frank’s age.

  $\begin{array}{c}A=\frac{N}{f}\\ =\frac{2L}{v}\end{array}$

Here, A is the Frank’s age.

Conclusion:

Thus, time for the remainder of trip is $L/\gamma v$, number of signal received at rate $f^{″}$ is $fL\left(1+\beta \right)/v$ , total number of signal received is $2fL/v$ and Frank’s age from perspective of Mary is $2L/v$ .