Physics › Astrophysics › The Doppler effect and red shift
The Doppler effect and red shift
Motion along the line of sight shifts every spectral line by the same fraction: red for a receding source, blue for an approaching one. For speeds well below c the shift measured against laboratory wavelengths gives that speed, through z = Δλ/λ and Δf/f = v/c. A spectroscopic binary's lines swing red and blue once per orbit.
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The Doppler effect and Hubble's law, part 1 of 2. Part 2 is Hubble's law and the Big Bang.
Builds on Progressive waves and Black-body radiation and spectral classes.
IN THIS TOPIC
- Use Δf/f = v/c and z = Δλ/λ with the sign of each shift defined, and say where the approximation stops being safe.
- Interpret the periodic Doppler shift of a spectroscopic binary seen in the plane of its orbit.
COMMON MISCONCEPTION
A galaxy with a red shift of z = 2 is receding at twice the speed of light.
z = Δλ/λ and Δf/f = v/c are small-shift approximations, so zc is a speed only while it stays far below c. A red shift of 2 does not mean 2c; past about z = 0.1 the relativistic treatment takes over and the simple ratio is an estimate at best.
Sound first
You already know the Doppler effect by ear. The pitch of a siren drops the moment it passes you. While the source approaches, each wavefront is emitted a little closer to you than the one before, so the fronts arrive bunched together and the frequency you hear is raised. Once it recedes they arrive stretched apart, and the pitch falls. For a source moving at speed vs through sound travelling at v, the observed frequency is
for approach, and a receding source flips the sign to v + vs so the pitch drops. Nothing about the source itself has changed. Because the size of the shift depends on the source's speed, measuring the shift gives the speed, and the same idea works for a passing siren and a receding galaxy.
Motion, measured from light
Light shows the same effect, with no medium needed. A source moving away stretches every wavelength it emits towards the red; a source approaching squeezes them towards the blue. For speeds much less than c, the size of the fractional shift equals the speed as a fraction of light's:
Neither equation can be used until Δ is defined. Take Δf as the observed frequency minus the rest frequency, and Δλ as the observed wavelength minus the rest wavelength. A source receding at speed v then gives Δλ/λ = +v/c and Δf/f = −v/c, because receding light arrives stretched and therefore lower in frequency. Approach reverses both signs. The data booklet prints Δf/f = −Δλ/λ = v/c, which agrees with all of that provided its v counts as positive for a source coming towards you.
Safer still, work with magnitudes and settle the direction physically. Away means red, towards means blue, and the size of the shift gives the speed. The quantity z is the red shift, positive for recession by construction. Both equations hold only while v is far below c, so treat anything past z of about 0.1 as an estimate and say so.
The measurement is possible at all because spectra carry absorption lines at exactly known laboratory wavelengths. Every line in the pattern shifts by the same fraction of its wavelength, and that shift is what is measured.
Describe the drawing
Two spectrum strips, each a continuum with dark absorption lines cut into it. The upper strip carries the hydrogen lines measured in the laboratory; the lower carries the same lines from a receding galaxy, every one of them sitting at a longer wavelength. Dashed leaders join each laboratory line to its shifted partner, and the stretch is the same factor 1 plus z for all four.
WORKED EXAMPLE
A galaxy's speed from one line
Hydrogen's 656.3 nm line arrives from a galaxy at 662.9 nm. Find the galaxy's velocity.
Δλ = 662.9 − 656.3 = 6.6 nm, so z = 6.6 / 656.3 = 0.010.
v = zc = 0.010 × 3.0 × 108 = 3.0 × 106 m s−1, receding, since the shift is towards longer wavelength.
The nanometres never needed converting, because z is a ratio and any wavelength unit cancels. Always state the direction as part of the answer. The sign of the shift is what gives the direction of motion.
Binary stars: the wobble in the lines
Many stars come in orbiting pairs too close for any telescope to split. The spectrum reveals them. Watch a binary in the plane of its orbit and each star alternately approaches and recedes along the line of sight, so its spectral lines swing blue, then red, then blue again, once per orbit. The result is a wavelength that oscillates about the rest value.
Two readings follow directly. The period of the oscillation is the orbital period of the pair, and the maximum shift gives the orbital speed through z = Δλ/λ. Tilt the orbit out of the line of sight and the measured shifts shrink, so the clean textbook case specifies the plane of the orbit.
Hubble's law, and what it implies has moved to the next part of this lesson.
What happened next, an OCR section has moved to the next part of this lesson.
ASSESSMENT FOCUS
- z = Δλ/λ uses the laboratory wavelength on the bottom, and both formulas hold only for v much less than c. Quote both facts when the question probes understanding.
- State the direction with every shift. Longer wavelength means receding, shorter means approaching; a magnitude on its own does not say which.
- If you write a signed answer, define Δ first. With Δf as observed minus rest, recession gives a negative Δf and a positive Δλ. Substituting magnitudes and naming the direction in words avoids the error altogether.
- The binary-star answer is a sentence about geometry. Seen in the plane of the orbit, each star alternately approaches and recedes, so its lines oscillate red and blue with the orbital period.
- Read the two numbers off a binary's wavelength graph separately. The time for one full cycle is the orbital period; the largest departure from the rest wavelength is the one to put into z = Δλ/λ, and halving the peak-to-peak swing is what gets you there.
CHECK YOURSELF
One star of a spectroscopic binary shows a spectral line whose wavelength runs from 588.87 nm up to 589.13 nm and back once every 4.0 days, about a rest wavelength of 589.00 nm. Find the orbital period and the star's orbital speed, and say what a tilt of the orbit away from edge-on would do to the measurement.
Show a hint
The period of the wavelength cycle is the orbital period. For the speed, take the largest departure from rest, not the full swing.
Show the answer
The wavelength completes one cycle in 4.0 days, so the orbital period is 4.0 days.
The largest departure from rest is 589.13 − 589.00 = 0.13 nm, so z = 0.13/589.00 = 2.2 × 10−4.
v = zc = 2.2 × 10−4 × 3.0 × 108 = 6.6 × 104 m s−1, about 66 km s−1.
Tilting the orbit puts part of the star's velocity across the line of sight, where it shifts nothing. The measured shifts shrink and the speed comes out too small, which is why the textbook case specifies the plane of the orbit.
Away means red, towards means blue, and the fractional shift is the speed over c.
A spectroscopic binary's lines swing red and blue once per orbit: the period of the swing is the orbital period, and its size is the orbital speed.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
Or read them with their mark schemes on the Doppler effect and red shift questions page.
USED LATER IN
CHECK YOUR PROGRESS
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- Use Δf/f = v/c and z = Δλ/λ with the sign of each shift defined, and say where the approximation stops being safe.
- Interpret the periodic Doppler shift of a spectroscopic binary seen in the plane of its orbit.
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