A radar transmitter$\mathit{T}$isfixedto a reference frame ${\mathit{S}}^{\mathbf{\text{'}}}$that is moving to the right with speed v relative to reference frame $\mathit{S}$(Fig. 37-33). A mechanical timer (essentially a clock) in frame ${\mathit{S}}^{\mathbf{\text{'}}}$, having a period ${\mathit{\tau }}_{\mathbf{\circ }}$ (measured in$S^{\text{'}}$ ), causes transmitter $\mathit{T}$to emit timed radar pulses, which travel at the speed of light and are received by $\mathit{R}$, a receiver fixed in frame $\mathit{S}$. (a) What is the period $\mathit{\tau }$of the timer as detected by observer $\mathit{A}$, who is fixed in frame $\mathit{S}$? (b) Show that at receiver $\mathit{R}$ the time interval between pulses arriving from$\mathit{T}$ is not $\mathit{\tau }$or ${\mathit{\tau }}_{\mathbf{o}}$, but

${\mathit{\tau }}_{\mathbf{R}}\mathbf{=}{\mathit{\tau }}_{\mathbf{o}}\sqrt{\frac{\mathbf{c}\mathbf{+}\mathbf{v}}{\mathbf{c}\mathbf{-}\mathbf{v}}}$

(c) Explain why receiver $\mathit{R}$and observer $\mathit{A}$, who are in the same reference frame, measure a different period for the transmitter. (Hint: A clock and a radar pulse are not the same thing.)

Time dilation, in special relativity, the "slowing down" of a clock as determined by an observer in relative motion relative to that clock.

Doppler effect, the apparent difference between the frequency at which sound or light waves leave the source and the frequency at which they reach the observer, caused by the relative motion of the observer and the wave source.

Suppose two consecutive events occur at the same location; the time interval measured in the same inertial reference frame is called proper time. And the time interval measured in any other reference frame relative to this frame will be longer than the proper time. The formula given below is used to determine the time interval in another frame.

$∆t=\frac{∆t_{\circ }}{\sqrt{1-{\displaystyle \raisebox{1ex}{$u^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}}}$

Where $∆t_o$is the proper time, and $∆t$is the interval measured by an observer moving with a relative speed $u$.

Here, the time period measured by the observer in frame $S$ is,

$\tau =\frac{{\tau }_{\circ }}{\sqrt{1-{\displaystyle \raisebox{1ex}{$u^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}}}$

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 ${\lambda }_{\circ }$. And the detected wavelength $\lambda$is related to the proper wavelength as

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

Where, $\beta$is the speed parameter $\left(v/c\right)$.

The wavelength and frequency are related by,

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

Inserting this in above expression, we get

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

The frequency and time period are related by $f=2\pi /\tau$.Inserting this in above equation and we get the expression as,

localid="1663135596103" ${\tau }_R={\tau }_o\sqrt{\frac{1+\beta }{1-\beta }}$

$$\begin{gathered} ={\tau }_{\circ }\sqrt{\frac{1+{\displaystyle \raisebox{1ex}{$v$}\!\left/ \!\raisebox{-1ex}{$c$}\right.}}{1-{\displaystyle \raisebox{1ex}{$v$}\!\left/ \!\raisebox{-1ex}{$c$}\right.}}} \\ {\tau }_R={\tau }_o\sqrt{\frac{c+v}{c-v}} \end{gathered}$$

Thus, the expression for time period measured by the detector is derived.

The doppler shift comprises of two phenomena: the time dilation of the moving transmitter and time differences in two travelling periodic pulses. The observer in frame measures purely time dilations effect whereas the receiver measures the doppler shift of the periodic waves. Hence the time period measured are different.