24.5

The Doppler Effect and Electromagnetic Waves

Section 16.9 presents a discussion of the Doppler effect that sound waves exhibit when either the source of a sound wave, the observer of the wave, or both are moving with respect to the medium of propagation (e.g., air). This effect is one in which the observed sound frequency is greater or smaller than the frequency emitted by the source. A different Doppler effect arises when the source moves than when the observer moves.
Electromagnetic waves also can exhibit a Doppler effect, but it differs from that for sound waves for two reasons. First, sound waves require a medium such as air in which to propagate. In the Doppler effect for sound, it is the motion (of the source, the observer, and the waves themselves) relative to this medium that is important. In the Doppler effect for electromagnetic waves, motion relative to a medium plays no role, because the waves do not require a medium in which to propagate. They can travel in a vacuum. Second, in the equations for the Doppler effect in Section 16.9, the speed of sound plays an important role, and it depends on the reference frame relative to which it is measured. For example, the speed of sound with respect to moving air is different than it is with respect to stationary air. As we will see in Section 28.2, electromagnetic waves behave in a different way. The speed at which they travel has the same value, whether it is measured relative to a stationary observer or relative to one moving at a constant velocity. For these two reasons, the same Doppler effect arises for electromagnetic waves when either the source or the observer of the waves moves; only the relative motion of the source and the observer with respect to one another is important.
When electromagnetic waves and the source and the observer of the waves all travel along the same line in a vacuum (or in air, to a good degree of approximation), the single equation that specifies the Doppler effect is

| (24.6)

In this expression, fo is the observed frequency, and fs is the frequency emitted by the source. The symbol vrel stands for the speed of the source and the observer relative to one another, and c is the speed of light in a vacuum. Equation 24.6 applies only if vrel is very small compared to c—that is, if . It is essential to realize that vrel is the relative speed of the source and the observer. Thus, if the source is moving due east at a speed of 28 m/s with respect to the earth, while the observer is moving due east at a speed of 22 m/s, the value for vrel is . Because vrel is the relative speed, it has no algebraic sign. The direction of the relative motion is taken into account by choosing the plus or minus sign in Equation 24.6. The plus sign is used when the source and the observer come together, and the minus sign is used when they move apart. For example, suppose the source is traveling at a velocity of 28 m/s due east. The observer, traveling ahead of the source, is moving at a slower velocity of 22 m/s due east. The source and observer are coming together (the source is catching up with the observer), and the plus sign is used in Equation 24.6. Suppose, on the other hand, that the velocity of the observer is 34 m/s due east (greater than the source velocity of 28 m/s due east). Now the source and observer are moving apart (the observer is pulling away from the source) and the minus sign is used in Equation 24.6. Example 6 illustrates one familiar use of the Doppler effect for electromagnetic waves.

Problem solving insight The Doppler effect for electromagnetic waves depends on the speed vrel of the observer and the source of the waves relative to one another. In general, do not use the speed of the observer or of the source with respect to the ground in Equation 24.6.

A N A L Y Z I N G M U L T I P L E - C O N C E P T P R O B L E M S

| Example 6 | | | Radar Guns and the Doppler Effect

Police use radar guns and the Doppler effect to catch speeders. Figure 24-17 illustrates a moving car approaching a stationary police car. A radar gun emits an electromagnetic wave that reflects from the oncoming car. The reflected wave returns to the police car with a frequency (measured by on-board equipment) that is different from the emitted frequency. One such radar gun emits a wave whose frequency is . When the speed of the car is 39 m/s and the approach is essentially head-on, what is the difference between the frequency of the wave returning to the police car and that emitted by the radar gun?

Figure 24-17 A radar gun in the police car emits an electromagnetic wave that reflects from the moving car. The frequencies of the emitted and reflected waves are different because of the Doppler effect.

The physics of rader speed traps.

Reasoning As the car moves toward the wave emitted by the radar gun, the car intercepts a greater number of wave crests per second than it would if it were stationary. Thus, the car “observes” a wave frequency fo that is greater than the frequency fs emitted by the radar gun. The car then reflects the wave back toward the police car. In effect, the car becomes a moving source of radar waves that are emitted with a frequency fo. Since the car is moving toward the police car, the reflected crests arrive at the police car with a frequency that is even greater than fo. Thus, there are two frequency changes due to the Doppler effect, one associated with fo and the other with . We will employ Equation 24.6 to determine these frequencies.
Knowns and Unknowns The data for this problem are listed below:

Description | Symbol | Value | Comment
Frequency emitted by radar gun | fs |  |
Relative speed between moving car and police car | vrel | 39m/s | The relative speed is just that of the moving car, since the police car is stationary.
Unknown Variable |  |  |
Frequency difference |  | ? | Subscript “o” indicates that is the frequency detected by the “observer” (the police car).

Modeling the Problem

| Step 1 Frequency Difference
| The desired frequency difference is , where fs is the frequency emitted by the radar gun and is known; is the frequency of the wave returning to the police car after being reflected from the moving car and is not known. However, and the frequency fo observed by the moving car are related by Equation 24.6:

where vrel is the relative speed between the moving car and the stationary police car. We have chosen the + sign in Equation 24.6, since the moving car and the police car are coming together. With this expression for , the desired frequency difference can be written as shown in Equation 1 at the right. The relative speed vrel and the frequency fs are known; the frequency fo will be determined in Step 2.

| (1)

| Step 2 Frequency of Wave “Observed” by Moving Car
| The radar gun sends out a wave whose frequency is fs. As the moving car approaches this wave, the car “observes” a frequency fo that is greater than fs. The relation between these frequencies is given by Equation 24.6:

Again, the + sign has been chosen in Equation 24.6, since the moving car and the police car are coming together. All the variables on the right side of this equation are known, and we substitute this expression for fo into Equation 1, as indicated at the right.

Solution Algebraically combining the results of the two steps above, we obtain

Multiplying and combining terms in this equation yields

The value of vrel/c is . This is an extremely small number when compared to the number 2, so the term is very nearly equal to 2. Thus, we find that the difference between the frequency of the wave returning to the police car and that emitted by the radar gun is

Related Homework: Problems 32 and 46

The Doppler effect of electromagnetic waves provides a powerful tool for astronomers. For instance, Example 10 in Chapter 5 discusses how astronomers have identified a supermassive black hole at the center of galaxy M87 by using the Hubble space telescope. They focused the telescope on regions to either side of the center of the galaxy (see Figure 5-15). From the light emitted by these two regions, they were able to use the Doppler effect to determine that one side is moving away from the earth, while the other side is moving toward the earth. In other words, the galaxy is rotating. The speeds of recession and approach enabled astronomers to determine the rotational speed of the galaxy, and Example 10 in Chapter 5 shows how the value for this speed leads to the identification of the black hole. Astronomers routinely study the Doppler effect of the light that reaches the earth from distant parts of the universe. From such studies, they have determined the speeds at which distant light-emitting objects are receding from the earth.

The physics of astronomy and the Doppler effect.

Check Your Understanding 3

The drawing shows three situations A, B, and C in which an observer and a source of electromagnetic waves are moving along the same line. In each case the source emits a wave of the same frequency. The arrows in each situation denote velocity vectors relative to the ground and have magnitudes of either v or 2v. Rank the frequencies of the observed waves in descending order (largest first) according to magnitude. (The answer is given at the end of the book.)

Background: The electromagnetic wave frequency that is observed is not necessarily the same as the frequency emitted by the source because of the Doppler effect. The observed frequency depends on the speed of the source and the observer relative to one another.
For similar questions (including calculational counterparts), consult Self-Assessment Test 24.1, which is described at the end of Section 24.6.

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