Last page you learned that a receding source stretches its waves. Lovely — but a star is not an ambulance. You cannot stand next to it with a stopwatch. All you get is a thin smear of light. And yet, from that smear alone, we can tell you how fast a galaxy 50 million light-years away is running from us, to within a few kilometres per second. The trick is that atoms leave fingerprints in starlight, and those fingerprints get shifted.
📘 What you need to know
For a light source with v << c, the Doppler shift is Δf/f = Δλ/λ ≈ Δv/c
Δλ = λ0 − λ, where λ0 is the observed wavelength and λ the reference (lab) wavelength
Δv = vs − vo. If the observer is stationary, Δv = v, the source’s speed
The Doppler shift Δλ/λ is a ratio, so it has no units — and the wavelengths need not be converted to metres
Rearranged: v = cΔλ / λ
Atoms give spectral lines. Comparing a lab spectrum with a star’s tells you which way it is moving
When a source of light moves slowly compared with light itself, the fractional change in frequency, the fractional change in wavelength, and the fraction of the speed of light all come out equal:
Doppler shift for light (non-relativistic)Δf / f = Δλ / λ ≈ Δv / cvalid only when v << c
Read the symbols carefully, because half the marks on this topic are lost right here:
λ — the reference wavelength, measured in a laboratory on Earth (m)
λ0 — the observed wavelength, measured from the moving source (m)
Δλ = λ0 − λ — the change in wavelength (m)
Δv — the relative velocity of source and observer along the line joining them (m s−1)
c = 3.00 × 108 m s−1, given in the data booklet
The relative velocity is the difference of the two velocities:
Relative velocityΔv = vs − voobserver stationary (vo = 0) → Δv = vs = v
Almost always in an exam the observer is us, sitting still on Earth, so Δv is simply the speed of the star or galaxy. Rearranging for that speed:
The one you will actually usev = c Δλ / λ
No units, no conversions. Δλ/λ is a wavelength divided by a wavelength, so the units cancel. Leave both in nanometres if you like — the answer for v still comes out in m s−1, because c brought the units with it.
Now the condition everyone skips: v << c. The ≈ sign is not decoration. Finish a calculation, get v = 1.2 × 108 m s−1, and you should feel uneasy — that is 40% of the speed of light, and this equation has no business being used there. Divide your answer by c and glance at it. If it’s a percent or two, you’re safe.
Spectral lines: the fingerprints
A hot star glows across all visible wavelengths. But the cooler gas in its atmosphere absorbs particular wavelengths, unique to each element, leaving dark lines in the spectrum. Hydrogen always absorbs at the same wavelengths — in a lab, in the Sun, in a quasar.
So measure those lines in a distant galaxy, compare them with a lab source, and any shift is due to motion.
The pattern of lines is identical, which is how we know it’s the same element. Only its position has slid. All lines shifted right → red-shifted → the galaxy is receding.
Which way is it moving?
Everything hinges on comparing λ0 with λ. Zoom in on a single line:
Flip the picture — put the observed line on the left — and every arrow reverses: Δλ negative, blue-shift, approaching.
Observation
Δλ
Δf
Name
Motion
λ0 > λ (longer)
Positive
Negative
Red-shift
Receding from Earth
λ0 < λ (shorter)
Negative
Positive
Blue-shift
Approaching Earth
λ0 = λ
Zero
Zero
No shift
No relative motion
Spot the sign trap in that table. A red-shift means Δλ is positive but Δf is negative — longer wavelength is lower frequency. So don’t chase the minus sign through the algebra. Find the size of v from the magnitudes, then look at whether λ0 got bigger or smaller and say in words which way it moves. That earns the mark every time.
A galaxy seen edge-on
A favourite exam question. A spiral galaxy rotates. Seen edge-on, one limb of the disc is swinging towards us and the other is swinging away — so the same spectral line arrives blue-shifted from one side and red-shifted from the other.
The whole galaxy may also be receding. Averaging the two shifts removes that, leaving just the rotational speed.
🔭 Every Doppler-for-light calculation
Identify λ and λ0. The lab or reference value is λ; the one from the star is λ0.
Find Δλ = λ0 − λ. Don’t convert nm to m — it cancels.
Use v = cΔλ/λ to get the speed.
Check v/c is small. If it isn’t, the equation didn’t apply.
State the direction in words, from whether λ0 was bigger or smaller.
WE 1
A hydrogen line has a wavelength of 486.1 nm when measured in a laboratory. The same line, in light from a distant star, is observed at 486.9 nm. Calculate the speed of the star relative to Earth, state its direction of motion, and confirm the equation was valid.
Step 1 — identify the wavelengths
λ = 486.1 nm (lab), λ₀ = 486.9 nm (observed)
Step 2 — find ΔλΔλ = 486.9 − 486.1 = 0.8 nmStep 3 — use v = cΔλ/λ, leaving both in nmv = (3.00 × 10⁸) × 0.8 / 486.1v = 4.9 × 10⁵ m s⁻¹Step 4 — direction, and the validity check
λ₀ is longer → red-shift → the star is receding.
v/c = 0.0016, comfortably << 1.
Under 500 km s⁻¹, from a shift of less than a nanometre. Spectroscopy is astonishingly sensitive.
WE 2
A spectral line emitted at 6.000 × 10¹⁴ Hz is observed from a distant object at 6.012 × 10¹⁴ Hz. Determine the speed of the object relative to Earth and state whether it is approaching or receding.
Step 1 — find the change in frequencyΔf = f₀ − f = (6.012 − 6.000) × 10¹⁴ = 1.2 × 10¹² HzStep 2 — use Δf/f ≈ v/c, rearrangedv = cΔf/f = (3.00 × 10⁸) × (1.2 × 10¹²) / (6.000 × 10¹⁴)v = 6.0 × 10⁵ m s⁻¹Step 3 — which way?
The observed frequency is higher → shorter wavelength → blue-shift.
approaching EarthCareful: here Δf is positive, yet a blue-shift has Δλ negative. The two forms carry opposite signs. Decide the direction from the physics, not the sign.
WE 3
A galaxy is observed edge-on. A spectral line with a reference wavelength of 486.13 nm is measured at 486.25 nm from the left-hand limb and 486.01 nm from the right-hand limb. (a) State and explain which limb is moving towards Earth. (b) Calculate the rotational speed of the galaxy.
(a) which limb approaches?
The right limb reads 486.01 nm, shorter than the reference 486.13 nm, so it is blue-shifted.
the right-hand limb(b) Step 1 — average the two shiftsΔλ = (486.25 − 486.01) / 2 = 0.12 nmStep 2 — apply v = cΔλ/λv = (3.00 × 10⁸) × 0.12 / 486.13v = 7.4 × 10⁴ m s⁻¹ = 74 km s⁻¹Halving the difference between the limbs cancels any motion of the galaxy as a whole, leaving pure rotation.
💡 Top tips
Δλ = observed − reference. Get that round the wrong way and the direction flips.
Leave wavelengths in nanometres. The ratio cancels the units and saves you a conversion error.
Always check v/c at the end. This equation is only valid for v << c.
State the direction in words from whether the wavelength grew or shrank. Don’t rely on a stray minus sign.
Edge-on galaxy? Halve the difference between the two limbs to get the rotational speed.
c is in the data booklet. You never need to memorise it.
⚠ Common mistakes
Computing Δλ as reference − observed, reversing the direction of motion
Dividing by λ0 instead of the referenceλ
Wasting time converting nm to m — the units cancel in Δλ/λ
Using this equation when v comes out as a large fraction of c. It is not valid there
Assuming Δλ and Δf have the same sign — a red-shift has Δλ > 0 but Δf < 0
Forgetting to halve the limb difference when finding a rotational speed
Quick recap: For a light source with v << c, Δf/f = Δλ/λ ≈ Δv/c, where Δλ = λ0 − λ and the shift is a unitless ratio. Rearranged, v = cΔλ/λ. A longer observed wavelength is a red-shift (receding); a shorter one is a blue-shift (approaching). Spectral lines give the reference, and an edge-on galaxy shows both shifts at once.
Point a telescope anywhere you like and something odd emerges: almost every galaxy is red-shifted. Not half of them, as you would expect if they were milling about at random. Nearly all. And the further away one is, the faster it seems to flee. That single observation rewrote our picture of the universe, and it is the next page: galactic redshift.
Red-shift calculations not shifting for you?
Book a free meeting and we’ll work through Δλ/λ, spectral lines and past-paper Doppler questions together.
