LocusBlog

The Doppler effect, and why a moving source is not a moving observer

The Doppler effect is the change in the frequency you hear when a source of sound and a listener move relative to each other. For sound travelling through still air it is:

$$f_{\text{obs}} = f_s\,\frac{v_{\text{sound}} \pm v_o}{v_{\text{sound}} \mp v_s}$$

$f_s$ is the frequency the source emits, $v_o$ is the observer’s speed, $v_s$ is the source’s speed, and $v_{\text{sound}} = 343$ m/s is the speed of sound in air near room temperature. Both speeds are measured relative to the air, and both motions are along the line joining source and observer. The formula also needs $v_s < v_{\text{sound}}$: at or above the speed of sound the denominator goes to zero or negative, and what happens instead is a shock wave.

Getting the signs right

There are two ambiguous signs and four cases, and that is where most of the wrong answers live. One rule settles all of them: motion toward raises the pitch. Pick the sign in each slot so that the whole fraction gets larger when the motion is toward, and smaller when it is away.

An observer moving toward the source adds in the numerator, $v_{\text{sound}} + v_o$. A source moving toward the observer subtracts in the denominator, $v_{\text{sound}} - v_s$. Both make the fraction bigger, and bigger is a higher pitch. Receding flips each one: subtract in the numerator, add in the denominator.

The denominator is the sign people reverse, because the arithmetic runs against the intuition. Approaching means minus there, not plus.

Two ways to close the same gap

Take a 500 Hz siren and a closing speed of 30 m/s, arranged two ways.

The source moves at 30 m/s toward an observer standing still:

$$f_{\text{obs}} = 500 \times \frac{343}{343 - 30} = 500 \times \frac{343}{313} = 547.9\ \text{Hz}$$

The observer moves at 30 m/s toward a source standing still:

$$f_{\text{obs}} = 500 \times \frac{343 + 30}{343} = 500 \times \frac{373}{343} = 543.7\ \text{Hz}$$

Same relative speed, same air, same siren. The answers differ by 4.2 Hz. If the Doppler shift depended only on how fast the gap was closing, those two numbers would be identical, and they are not.

Build both from wavefronts and the reason appears. The siren emits one compression every $T = 1/500 = 0.002$ s. In the case where the source moves, it advances $30 \times 0.002 = 0.06$ m between emissions, so each front starts 0.06 m closer to the observer than the last one did. The wavelength in the air ahead of the source is genuinely shorter:

$$\lambda' = 343 \times 0.002 - 0.06 = 0.686 - 0.06 = 0.626\ \text{m}$$

The observer is at rest in the air, so those fronts still sweep past at 343 m/s, and $f = 343/0.626 = 547.9$ Hz. That matches the formula, which it has to, because the formula is this argument with the algebra done in advance.

Now the case where the observer moves. The source is at rest in the air, so the air carries ordinary wavelengths, $\lambda = 343/500 = 0.686$ m, unsqueezed. The observer runs into them at a closing speed of $343 + 30 = 373$ m/s, so $f = 373/0.686 = 543.7$ Hz. Again the formula agrees.

Two different mechanisms, then. A moving source changes the wavelength in the medium. A moving observer changes the rate at which unchanged wavelengths are met. The air is what breaks the symmetry: it is a medium, it defines a rest frame, and “who is moving” is a real question with a real answer rather than a matter of bookkeeping.

The size of the gap is easy to pin down. Write $x = v/v_{\text{sound}}$, so $x = 30/343 = 0.0875$. The moving-observer answer is exactly $f_s(1 + x)$. The moving-source answer is $f_s/(1 - x) = f_s(1 + x + x^2 + \cdots)$. They agree to first order and part company at order $x^2$, which is $500 \times 0.0875^2 = 3.8$ Hz, most of the 4.2 Hz actually measured. So at walking or driving speeds the asymmetry is a rounding error, and at aircraft speeds it is not. Run the same pair at 200 m/s and the moving source gives 1199 Hz while the moving observer gives 792 Hz.

One more thing falls out of that expansion. Since $1/(1-x) > 1+x$ and $1/(1+x) > 1-x$, the moving-source answer is the higher of the two no matter which way the motion runs. Approaching, 547.9 Hz beats 543.7 Hz. Receding, 459.8 Hz beats 456.3 Hz. The bias has a sign, and it always points the same way.

When both move, both signs apply at once. A source at 30 m/s and an observer at 20 m/s, approaching each other, give $500 \times 363/313 = 579.9$ Hz. Practice on the Doppler effect skill mixes the four sign cases deliberately, which is the only way to find out whether the rule has stuck.

Where the asymmetry bites

The common failure is treating the two cases as interchangeable and reaching for whichever version of the formula is remembered. Use the moving-observer form on a 30 m/s train horn and you report 543.7 Hz instead of 547.9 Hz. That is 13 cents flat, small enough to survive a physics problem set and get marked wrong, and small enough that nobody’s ear files a complaint. Do the same for a low pass at 200 m/s and you report 792 Hz for a tone that is actually near 1199 Hz, off by more than a musical fifth.

The other failure is louder. Flip the denominator sign for an approaching source and the formula returns $500 \times 343/373 = 459.8$ Hz, a pitch that has dropped. Nothing in the world sounds like that. If the arithmetic says an approaching siren gets deeper, the sign is wrong, not the siren, and this is the cheapest check available: decide whether the pitch should go up or down before computing anything, then confirm the number moved that way.

Light does not work this way

There is no medium for light in a vacuum, so there is nothing to be at rest with respect to, and the shift cannot depend on anything but the relative velocity of source and observer. For a source and observer separating along the line of sight at speed $v$, with $\beta = v/c$, the relativistic result is:

$$f_{\text{obs}} = f_s\sqrt{\frac{1 - \beta}{1 + \beta}}$$

Swap which one you call the mover and $\beta$ is unchanged, so the formula is unchanged. That is the whole difference. The sound formula has two speeds in it because there are two things that can move with respect to the air; the light formula has one because there is only relative motion to speak of. At $\beta = 0.1$ the received frequency is $0.9045 f_s$, and the same square root, read backwards, is what turns a galaxy’s spectral lines into a redshift.

The simulations browser has the wave and sound problems set up so you can change a speed and watch which way the frequency moves, and the speed, intensity and decibels skill covers the $v_{\text{sound}} = 343$ m/s that every one of these calculations leans on.

References