What happens to sound and light waves when this is no longer true? Assuming that the medium is at rest, we have three possible situations:

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1 2.9 Doppler Effect So far, we have ignored the motion of both the wave source and any observers measuring the wave. In effect, it has been assumed that both the source and observer are at rest with respect to each other and the medium i.e. in some inertial reference frame. What happens to sound and light waves when this is no longer true? Assuming that the medium is at rest, we have three possible situations: The wave source moves with respect to the observer; the observer is at rest with respect to the medium. The wave source is at rest with respect to the medium; the observer moves with respect to the medium. Both the wave source and observer move with respect to the medium. Let us consider the various situations in sound and light in turn Doppler Effect in Sound Waves We are aware that as an emergency vehicle approaches us with its siren going, the pitch of the sound waves appear to be higher than the pitch when the vehicle is receding. What is happening to the frequency of the sound waves? Source moving, Observer stationary (wrt Medium) Consider two stationary observers monitoring a single sound wave emitted from a moving source (figures 40 and ). A source of sound waves moves away from Pablo and Nancy at a steady speed v s (i.e. this is speed of the source, not the waves). 31 Knight, Figures (a), (b), page

2 Suppose that the source emits a sound wave of speed v, frequency ν 0 and wavelength λ 0 relative to the source s frame. The period T of the sound wave is thus T = 1/ν 0 relative to the source s frame. The positions of the source at times t = 0, T, 2T and 3T are shown in figure 40. Consider the system for times 0 t 3T. What does the source see? At t = 0, source emits a circular wave front, centred on its position at t = 0. At t = T, source emits a circular wave front, centred on its position at t = T. At t = 2T, source emits a circular wave front, centred on its position at t = 2T. At t = 3T, source emits a circular wave front, centred on its position at t = 3T. Note: after a wave crest has been emitted, its motion is governed by the medium the motion of the source cannot effect a wave that has been already emitted. What do both observers see? Pablo: observes the source to be receding. Nancy: observes the source to be approaching. We try to relate ν ± to ν 0 and v s. Consider the situation at t = 3T : 71

3 Figure 40: Motion of source in observers frame. 72

4 Figure 41: Wave fronts at time 3T. 73

5 Wave front emitted at t = 0 has a radius 3λ 0, and just reaches both observers (figure 41). Wave front emitted at t = T has a radius 2λ 0. Wave front emitted at t = 2T has a radius λ 0. Pablo: observes the wave fronts to be spread out. Suppose he measures a frequency ν and wavelength λ for the wave. Then λ = v ν > λ 0, ν = v λ < ν 0. (2.37) Nancy: observes the wave fronts to be bunched up. Suppose she measures a frequency ν + and wavelength λ + for the wave. Then λ + = v ν + < λ 0, ν + = v λ + > ν 0. (2.38) In (2.37) or (2.38), these changes in wavelength or frequency measured when the source moves relative to the observer is called the Doppler effect. Nancy observes the three wave fronts to be bunched up into a distance d from the source (see figure 41). She measures a wavelength λ + = d/3 for the sound wave. The source has moved a distance x source = v s t in a time t = 3T, i.e. x source = 3v s T. (2.39) The first wave front has moved a distance x wave t = 3T, i.e. = v t in a time x wave = 3vT. (2.40) 74

6 Thus, using (2.39) and (2.40), we see that d = x wave x source = 3 (v v s ) T. (2.41) Therefore, Nancy measures a wavelength for the sound wave. λ + = d 3 = 3 (v v s) T 3 = (v v s ) T (2.42) Substituting for (2.42) in (2.38), we get ν + = v λ + = v/ν 0 (v v s ) = ν 0 (1 v s /v). (2.43) By analysing Pablo s viewpoint, we can get a similar equation to (2.43). Equation (2.43) for Nancy, and the corresponding expression for Pablo, form the general expressions Doppler effect: approaching source, stationary observer: ν + = ν 0 1 v s /v. (2.44) Doppler effect: receding source, stationary observer: ν = ν v s /v. (2.45) Example: after bothering her young, a mother hawk screeches as she dives at you. You recall from biology class that female hawks screech at 825 Hz, but you hear the screech at 925 Hz. How fast is the hawk approaching? Solution: the mother hawk s frequency is altered by the Doppler effect. As she approaches you, you will measure a frequency ν + given by (2.44) (v = 343 m s 1 ): 75

7 ( ν 0 ν + = 1 v s /v v s = v 1 ν ) 0 ν ( + = ) = 37.1 m s Source stationary, Observer moving (wrt Medium) In the above example, suppose that the source and Nancy were interchanged. Nancy would claim that the source is moving towards her at a speed v s. Intuitively, you may think that she could apply (2.44) to find ν +. Note: this is not the case! In the above case: the observer is stationary with respect to the medium. the source is moving with respect to the medium (i.e. approaching the observer). In this case: the source is stationary with respect to the medium. the observer is is moving relative to the medium. This fundamental difference is very important the Doppler effect depends not only upon how the source and observer move relative to each other, but also how they move relative to the medium. If an observer travels at a speed v 0 relative to the medium/stationary wave source, then it can be shown that Doppler effect: approaching observer, stationary source: ( ν + = 1 + v 0 v 76 ) ν 0. (2.46)

8 Doppler effect: receding observer, stationary source: ( ν = 1 v 0 v ) ν 0. (2.47) Source moving, Observer moving (wrt Medium) In the case where both the wave source and observer are moving with respect to the medium, we can generalise (2.44) (2.47) quite easily: Replacing the ν 0 in (2.44) with ν + from (2.46). Replacing the ν 0 in (2.45) with ν from (2.47). We can therefore write down the equations Doppler effect: observer approaching source, both moving: ( ) v + v0 ν + = ν 0 v v s. (2.48) Doppler effect: observer receding from source, both moving: ( ) v v0 ν = ν 0 v + v s. (2.49) Doppler Effect in Light Waves The Doppler effect is also observed for electromagnetic waves. The underlying reason for the effect in light is the same as for that in sound, i.e. Source of light receding from an observer: the observer measures a wavelength λ > λ 0. In the visible spectrum, the observer sees the wavelength shifted towards the red end of the spectrum i.e. to waves of longer wavelength. This effect is called the red shift. 77

9 Source of light approaches an observer: the observer measures a wavelength λ + < λ 0. In the visible spectrum, the observer sees the wavelength shifted towards the blue end of the spectrum i.e. to waves of shorter wavelength. This effect is called the blue shift. Note: a fundamental problem with the analogy with sound is that for light waves traveling in a vacuum, we do not have a medium. From special relativity, we state, without proof, the following equations for the Doppler effect in light waves: for a source of speed v s relative to the observer, Doppler effect for receding source: λ red = Doppler effect for approaching source: λ blue = 1 + v s /c 1 v s /c λ 0. (2.50) 1 v s /c 1 + v s /c λ 0. (2.51) Example: hydrogen atoms in the laboratory emit red light with wavelength 656 nm. In the light from a distant galaxy, the same spectral line is observed to have a wavelength of 695 nm. What is the speed of this galaxy with respect to the earth? Solution: so the wavelength of the light from the distant galaxy is longer than that measured at rest with respect to the observer thus we can conclude that since the light is red-shifted, the galaxy is moving away from us. From (2.50), we have that 78

10 i.e. λ red = ( ) 1 ( ) 1 + v s /c 1 v s /c λ 0 v s = c 1 + λ2 red λ 2 red 1 λ 2 0 λ 2 0, ) 1 ( ) v s = c ( = 0.058c = m s 1, 2 i.e. the galaxy is receding from us at approx 6% of the speed of light. 79