ʻATENISI INSTITUTE
An institute for critical education in the South Pacific
Faikava
The classical Doppler effect
The classical Doppler effect is that waves as seen by a observer have a different frequency or wavelength than those as emitted from a source due to their relative movement. In the classic model it is important to distinguish between the source standing still and the observer moving (as measured compared to the medium which carries the waves), and the opposite. This is typically valid for sound waves, which are carried by the air.
In the formulas below c = phase speed of wave, u = speed of source, v = speed of observer, f = frequency, which is 1 hertz as the source produces 1 pulse every second.
The source (left) and observer (right) are standing still: u = 0, v = 0, and c = wave speed has a particular value. The observer hears the sound at the same frequency as the source emitted it, 1 Hz in this case, as the formula for the observed frequency = f × (c - v) ÷ (c + u) = f × (c - 0) ÷ (c + 0) = f
The observer moves towards the stationary source: u = 0, v = -½c (approaching velocities are negative), so that the observed frequency = f × (c + ½c) ÷ (c + 0) = 3/2 f. The moving observer sees 12 waves coming along in the time the stationary observer sees only 8, so the waves seem to be compressed and the sound has a higher tone. Or in case of light waves, the spectrum shifted towards the blue.
With a source approaching with u = -½c the observed frequency = f * (c - 0) ÷ (c - ½c) = 2f. The observer in this case sees 12 waves coming along in the time only 6 pass with the source standing still. Still a much higher tone than the previous case, even though the relative speeds of source and observer are the same.
The observer moves away from the stationary source: u = 0, v = +½c, so that the observed frequency = f * (c - ½c) ÷ (c + 0) = 1/2 f. That is: the waves pass along twice as slow, or seem to be extended and the sound has a lower tone. For lightwaves the spectrum is shifted towards the red. If the observer would move away faster than c it would never receive any wave, or overtake those already emitted.
When the source is moving away from the observer, the latter again sees a lower tone or redshifted spectrum, but since this time: u = +½c, v = 0, the observed frequency = f * (c - 0) ÷ (c + ½c) = 2/3 f, so that 8 waves will pass the observer in the time the source produces 12. In addition the source may move away faster than c, and still the observer will receive the waves one day.
The formula works for observers with any speed. But with a source going from subsonic to supersonic approach: u = -c, problems arise. Then the observed frequency = f * (c - 0) ÷ (c - c) = ∞. All waves pass the observer at the same time as a shockwave causing a sonic boom. After that the trailing waves will reach the observer at half frequency.