# 5.7: Doppler Effect for Light

As discussed in the chapter on sound, if a source of sound and a listener are moving farther apart, the listener encounters fewer cycles of a wave in each second, and therefore lower frequency, than if their separation remains constant. For the same reason, the listener detects a higher frequency if the source and listener are getting closer. The resulting Doppler shift in detected frequency occurs for any form of wave. For sound waves, however, the equations for the Doppler shift differ markedly depending on whether it is the source, the observer, or the air, which is moving. Light requires no medium, and the Doppler shift for light traveling in vacuum depends only on the relative speed of the observer and source.

# The Relativistic Doppler Effect

Suppose an observer in *S* sees light from a source in S'S′ moving away at velocity *v* (Figure). The wavelength of the light could be measured within S'S′—for example, by using a mirror to set up standing waves and measuring the distance between nodes. These distances are proper lengths with S'S′ as their rest frame, and change by a factor 1−v2/c2−−−−−−−−√1−v2/c2 when measured in the observer’s frame *S*, where the ruler measuring the wavelength in S'S′ is seen as moving.

*S*sees the wavelength measured in S'.S′. to be shorter by a factor 1−v2/c2−−−−−−−√.1−v2/c2. (b) Because the observer sees the source moving away within

*S*, the wave pattern reaching the observer in

*S*is also stretched by the factor(cΔt+vΔt)/(cΔt)=1+v/c.(cΔt+vΔt)/(cΔt)=1+v/c.</figcaption> </figure>

If the source were stationary in *S*, the observer would see a length cΔtcΔt of the wave pattern in time Δt.Δt. But because of the motion of S'S′relative to *S*, considered solely within *S*, the observer sees the wave pattern, and therefore the wavelength, stretched out by a factor of

cΔtperiod+vΔtperiodcΔtperiod=1+vccΔtperiod+vΔtperiodcΔtperiod=1+vc

as illustrated in (b) of Figure. The overall increase from both effects gives

λobs=λsrc(1+vc)11−v2c2−−−−−−√=λsrc(1+vc)1(1+vc)(1−vc)−−−−−−−−−−−−−√=λsrc(1+vc)(1−vc)−−−−−−−⎷λobs=λsrc(1+vc)11−v2c2=λsrc(1+vc)1(1+vc)(1−vc)=λsrc(1+vc)(1−vc)

where λsrcλsrc is the wavelength of the light seen by the source in S'S′ and λobsλobs is the wavelength that the observer detects within *S*.

# Red Shifts and Blue Shifts

The observed wavelength λobsλobs of electromagnetic radiation is longer (called a “red shift”) than that emitted by the source when the source moves away from the observer. Similarly, the wavelength is shorter (called a “blue shift”) when the source moves toward the observer. The amount of change is determined by

λobs=λs1+vc1−vc−−−−−√λobs=λs1+vc1−vc

where λsλs is the wavelength in the frame of reference of the source, and *v* is the relative velocity of the two frames *S* and S'.S′. The velocity *v* is positive for motion away from an observer and negative for motion toward an observer. In terms of source frequency and observed frequency, this equation can be written as

fobs=fs1−vc1+vc−−−−−√.fobs=fs1−vc1+vc.

Notice that the signs are different from those of the wavelength equation.

Calculating a Doppler ShiftSuppose a galaxy is moving away from Earth at a speed 0.825*c*. It emits radio waves with a wavelength of

0.525 m. What wavelength would we detect on Earth?

StrategyBecause the galaxy is moving at a relativistic speed, we must determine the Doppler shift of the radio waves using the relativistic Doppler shift instead of the classical Doppler shift.

Solution

- Identify the knowns: u=0.825c;λs=0.525m.u=0.825c;λs=0.525m.
- Identify the unknown: λobs.λobs.
- Express the answer as an equation:
λobs=λs1+vc1−vc−−−−−√.λobs=λs1+vc1−vc.

- Do the calculation:
λobs=λs1+vc1−vc−−−−√=(0.525m)1+0.825cc1−0.825cc−−−−−−√=1.70m.λobs=λs1+vc1−vc=(0.525m)1+0.825cc1−0.825cc=1.70m.

SignificanceBecause the galaxy is moving away from Earth, we expect the wavelengths of radiation it emits to be redshifted. The wavelength we calculated is 1.70 m, which is redshifted from the original wavelength of 0.525 m. You will see in Particle Physics and Cosmology that detecting redshifted radiation led to present-day understanding of the origin and evolution of the universe.

**Check Your Understanding** Suppose a space probe moves away from Earth at a speed 0.350*c*. It sends a radio-wave message back to Earth at a frequency of 1.50 GHz. At what frequency is the message received on Earth?

The relativistic Doppler effect has applications ranging from Doppler radar storm monitoring to providing information on the motion and distance of stars. We describe some of these applications in the exercises.

# Summary

- An observer of electromagnetic radiation sees relativistic Doppler effects if the source of the radiation is moving relative to the observer. The wavelength of the radiation is longer (called a red shift) than that emitted by the source when the source moves away from the observer and shorter (called a blue shift) when the source moves toward the observer. The shifted wavelength is described by the equation:
λobs=λs1+vc1−vc−−−−−√.λobs=λs1+vc1−vc.

*v*is the relative velocity of the source to the observer.

# Conceptual Questions

Explain the meaning of the terms “red shift” and “blue shift” as they relate to the relativistic Doppler effect.

What happens to the relativistic Doppler effect when relative velocity is zero? Is this the expected result?

Is the relativistic Doppler effect consistent with the classical Doppler effect in the respect that λobsλobs is larger for motion away?

All galaxies farther away than about 50×106ly50×106ly exhibit a red shift in their emitted light that is proportional to distance, with those farther and farther away having progressively greater red shifts. What does this imply, assuming that the only source of red shift is relative motion?

# Problems

A highway patrol officer uses a device that measures the speed of vehicles by bouncing radar off them and measuring the Doppler shift. The outgoing radar has a frequency of 100 GHz and the returning echo has a frequency 15.0 kHz higher. What is the velocity of the vehicle? Note that there are two Doppler shifts in echoes. Be certain not to round off until the end of the problem, because the effect is small.