The Doppler Effect For Light

OCR H556 mapping. This lesson covers Module 5.5.3 specification points on the Doppler effect for light: the change in observed wavelength and frequency when source and observer are in relative motion along the line of sight; the non-relativistic low-velocity formula $\Delta\lambda/\lambda \approx \Delta f/f \approx v/c$Δλ/λ≈Δf/f≈v/c for $v \ll c$v≪c; the convention that red shift indicates recession and blue shift indicates approach; and applications to stellar, binary and galactic motion. It is the foundation for Hubble's law (lesson 8) and for all of observational cosmology.

You have met the Doppler effect already, in Module 4 (Waves) of OCR A-Level Physics A: the change in observed frequency of a wave when source and observer are in relative motion. A fire engine siren sounds high-pitched as it approaches and low-pitched as it recedes. The same effect happens to light — but with cosmological consequences that shaped twentieth-century physics.

This lesson develops the Doppler formula for electromagnetic radiation, distinguishes red shift from blue shift, and prepares the ground for Lessons 8 and 9, where we shall use the Doppler effect to measure the expansion of the universe itself.

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The Doppler effect, briefly recalled

For sound waves, the classical Doppler formula in your Module 4 notes is

$$f' = f \times \dfrac{v_\text{sound} \pm v_\text{observer}}{v_\text{sound} \mp v_\text{source}}$$f′=f×vsound​∓vsource​vsound​±vobserver​​

The details depend on who is moving and in which direction. For sound waves through air, the medium itself defines a preferred frame, and the full formula treats source and observer asymmetrically.

For light, the situation is different. There is no medium for light to travel through — no "aether" — as the Michelson-Morley experiment (1887) demonstrated. Light propagates through vacuum at the speed $c$c in every inertial frame. The full relativistic Doppler formula relates emitted and observed frequencies via

$$\dfrac{f'}{f} = \sqrt{\dfrac{1 - \beta}{1 + \beta}}, \qquad \beta = \dfrac{v}{c}$$ff′​=1+β1−β​​,β=cv​

where $v$v is the relative velocity along the line of sight (positive for recession). This is the formula an astronomer uses for objects moving at a significant fraction of $c$c. For the non-relativistic case ($v \ll c$v≪c), which is what OCR A-Level Physics A requires, we can simplify it considerably.

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The low-velocity approximation

Expanding the relativistic formula for small $\beta$β using a binomial expansion:

$$\begin{aligned} \dfrac{f'}{f} &= (1 - \beta)^{1/2}(1 + \beta)^{-1/2} \\ &\approx \left(1 - \dfrac{\beta}{2}\right)\left(1 - \dfrac{\beta}{2}\right) \\ &\approx 1 - \beta + O(\beta^{2}) \end{aligned}$$ff′​​=(1−β)1/2(1+β)−1/2≈(1−2β​)(1−2β​)≈1−β+O(β2)​

To first order in $\beta$β, the change in frequency is

$$\Delta f = f' - f \approx -f\beta = -\dfrac{fv}{c}$$Δf=f′−f≈−fβ=−cfv​

or equivalently

$$\dfrac{\Delta f}{f} \approx -\dfrac{v}{c}$$fΔf​≈−cv​

The negative sign reflects the convention that $v > 0$v>0 means the source is receding, and that a receding source appears to have a lower frequency (red shift). For an approaching source ($v < 0$v<0), the frequency increases (blue shift).

Because frequency and wavelength are related by $c = f\lambda$c=fλ in vacuum, an increase in wavelength corresponds to a decrease in frequency. The fractional shifts have opposite signs:

$$\dfrac{\Delta\lambda}{\lambda} = -\dfrac{\Delta f}{f} \approx \dfrac{v}{c}$$λΔλ​=−fΔf​≈cv​

The OCR formula you are expected to use is

$$\boxed{\dfrac{\Delta\lambda}{\lambda} \approx \dfrac{\Delta f}{f} \approx \dfrac{v}{c}}$$λΔλ​≈fΔf​≈cv​​

Here $v$v is the radial velocity of the source (along the line of sight), positive for recession. The approximation is valid when $v \ll c$v≪c. For essentially all stellar motions within our Galaxy, and for all but the most distant galaxies, this approximation is excellent.

Exam tip. OCR gives the formula $\Delta\lambda/\lambda = \Delta f/f = v/c$Δλ/λ=Δf/f=v/c on the data sheet. What you must supply is the physical interpretation, the sign convention, and the algebra to go from observed wavelength shifts to source velocities. The discriminator move between Mid and Stronger is naming the convention ($v > 0$v>0 recession → red shift) explicitly.

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Red shift and blue shift

When a source moves away from the observer, the emitted wavefronts are "stretched out" behind the receding source. The observed wavelength is longer — shifted towards the red end of the spectrum. This is called a red shift.

When a source moves towards the observer, the wavefronts are "piled up" in front of it. The observed wavelength is shorter — shifted towards the blue end. This is called a blue shift.

The key conventions:

• Red shift → $\Delta\lambda > 0$Δλ>0, $v > 0$v>0 (receding).
• Blue shift → $\Delta\lambda < 0$Δλ<0, $v < 0$v<0 (approaching).

Astronomers report red shifts as positive $z = \Delta\lambda/\lambda$z=Δλ/λ values and blue shifts as negative. For small velocities, $z \approx v/c$z≈v/c.

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Worked example 1: velocity of a star from a spectral line

A hydrogen absorption line has laboratory wavelength $\lambda = 656.3$λ=656.3 nm (the H-$\alpha$α line in the Balmer series). In a particular star's spectrum this line is observed at $\lambda' = 656.8$λ′=656.8 nm. Find the star's radial velocity and direction.

Step 1. Compute $\Delta\lambda$Δλ:

$$\Delta\lambda = \lambda' - \lambda = 656.8 - 656.3 = 0.5\,\text{nm}$$Δλ=λ′−λ=656.8−656.3=0.5nm

Step 2. Apply the Doppler formula $v = c\,\Delta\lambda/\lambda$v=cΔλ/λ:

$$\begin{aligned} v &= (3.00 \times 10^{8}\,\text{m s}^{-1}) \times \dfrac{0.5}{656.3} \\ &= (3.00 \times 10^{8}) \times 7.62 \times 10^{-4} \\ &= 2.29 \times 10^{5}\,\text{m s}^{-1} \end{aligned}$$v​=(3.00×108m s−1)×656.30.5​=(3.00×108)×7.62×10−4=2.29×105m s−1​

This is about 230 km s$^{-1}$−1. Since $\Delta\lambda > 0$Δλ>0, the line is red-shifted — the star is moving away from us at $\approx 230$≈230 km s$^{-1}$−1 along the line of sight. The fractional shift is $0.5/656.3 \approx 0.076\%$0.5/656.3≈0.076%, comfortably in the non-relativistic regime ($v/c \approx 7.6 \times 10^{-4}$v/c≈7.6×10−4), so the simple formula is justified.

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Worked example 2: an approaching galaxy

A particular emission line in the Andromeda galaxy (M31) has laboratory wavelength $\lambda = 486.1$λ=486.1 nm (the H-$\beta$β line). The observed wavelength is $\lambda' = 485.6$λ′=485.6 nm. What is the radial velocity of Andromeda?

$$\begin{aligned} \Delta\lambda &= 485.6 - 486.1 = -0.5\,\text{nm} \\ v &= c \times \dfrac{\Delta\lambda}{\lambda} \\ &= (3.00 \times 10^{8}) \times \dfrac{-0.5}{486.1} \\ &\approx -3.09 \times 10^{5}\,\text{m s}^{-1} \end{aligned}$$Δλv​=485.6−486.1=−0.5nm=c×λΔλ​=(3.00×108)×486.1−0.5​≈−3.09×105m s−1​

The velocity is negative, meaning the galaxy is approaching us at $\approx 310$≈310 km s$^{-1}$−1. Andromeda is indeed one of the very few galaxies blue-shifted relative to the Milky Way — it is gravitationally bound to our Local Group and on a collision course with the Milky Way, expected to merge in about 4.5 billion years. Most galaxies, at greater distances, are red-shifted because of the cosmological expansion of space (see lesson 8) — and the cosmological red shift, strictly, is not a Doppler effect at all (more on this in the misconceptions section).

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Worked example 3: rotation of a distant galaxy

A spiral galaxy is observed face-on along the line of sight (well, edge-on relative to the line of sight, so that we see both edges of the disc rotating). The H-$\alpha$α line (laboratory $\lambda = 656.3$λ=656.3 nm) is measured at the two ends of the galaxy's disc. On the western edge it appears at $656.7$656.7 nm; on the eastern edge at $656.9$656.9 nm.

(a) What is the average radial velocity of the galaxy?

(b) What is its rotational velocity?

(a) Average $\lambda' = (656.7 + 656.9)/2 = 656.8$λ′=(656.7+656.9)/2=656.8 nm. Average $\Delta\lambda = 0.5$Δλ=0.5 nm.

$$v_\text{avg} = c \times \dfrac{0.5}{656.3} = 2.29 \times 10^{5}\,\text{m s}^{-1}$$vavg​=c×656.30.5​=2.29×105m s−1

The galaxy as a whole is receding at $\approx 229$≈229 km s$^{-1}$−1.

(b) The difference between the edges is $0.2$0.2 nm, split as $0.1$0.1 nm on either side of the average. The line-of-sight velocity difference is:

$$v_\text{rot} = c \times \dfrac{0.1}{656.3} = 4.57 \times 10^{4}\,\text{m s}^{-1}$$vrot​=c×656.30.1​=4.57×104m s−1

So one edge is rotating towards us at $\approx 46$≈46 km s$^{-1}$−1 relative to the centre, and the other away at $\approx 46$≈46 km s$^{-1}$−1. This is exactly the kind of rotation-curve measurement that, accumulated over many spiral galaxies and extended out to large radii, ultimately led to the discovery of dark matter — but that is beyond the OCR specification.

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Applications: measuring stellar and galactic motion

The Doppler effect for light is the single most important tool astronomers use to measure the motion of distant objects. Among its applications:

1. Radial velocities of stars. Stellar motions within our Galaxy are measured routinely from the Doppler shifts of their absorption lines, mapping the kinematic structure of the Milky Way.
2. Spectroscopic binaries. If a star has an unseen companion, its spectral lines shift periodically as the two orbit a common centre of mass. Half the masses of all measured stellar masses come from this technique.
3. Exoplanet detection (radial-velocity method). A star with an orbiting planet wobbles slightly as the planet tugs it. The wobble produces tiny periodic Doppler shifts in the star's spectral lines, revealing the planet's existence and (with assumptions about inclination) estimating its mass. Responsible for the first $\sim 1000$∼1000 planet discoveries.
4. Galactic rotation curves. Doppler shifts across a galaxy's disc reveal how fast it is rotating at each radius, the basic data set used to infer dark matter.
5. Cosmological red shifts. For distant galaxies and quasars, the Doppler shift (combined with Hubble's law) becomes a distance measure.
6. Radar speed guns. The same principle in reverse: a radar gun emits a known frequency, bounces it off a moving car, and measures the Doppler-shifted reflection to deduce the car's speed.

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Worked example 4: exoplanet detection

A Sun-like star is monitored for Doppler wobbles. Over one orbital period its line centre shifts between extreme values $\lambda_\text{min} = 500.0000$λmin​=500.0000 nm and $\lambda_\text{max} = 500.0002$λmax​=500.0002 nm. Calculate the star's radial velocity amplitude.

$$\begin{aligned} \Delta\lambda_\text{amp} &= \dfrac{\lambda_\text{max} - \lambda_\text{min}}{2} = 10^{-4}\,\text{nm} = 10^{-13}\,\text{m} \\ v_\text{amp} &= c \times \dfrac{\Delta\lambda_\text{amp}}{\lambda} \\ &= (3.00 \times 10^{8}) \times \dfrac{10^{-13}}{5.00 \times 10^{-7}} \\ &= 60\,\text{m s}^{-1} \end{aligned}$$Δλamp​vamp​​=2λmax​−λmin​​=10−4nm=10−13m=c×λΔλamp​​=(3.00×108)×5.00×10−710−13​=60m s−1​

The star is wobbling by about 60 m s$^{-1}$−1 around the system's centre of mass. This is typical of a giant planet orbiting a Sun-like star at $\approx 1$≈1 AU. The Doppler precision required — wavelength shifts of a few parts in $10^{8}$108 — is achieved with specialised echelle spectrographs such as HARPS and ESPRESSO, which can resolve radial velocities down to a few centimetres per second. The first planet found this way around a Sun-like star, 51 Pegasi b in 1995, induced a $\approx 56$≈56 m s$^{-1}$−1 wobble.

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Worked example 5: getting to the limit of the formula

Use the binomial expansion to estimate how good the non-relativistic formula is for various source velocities. Express the second-order correction as a percentage.

The full relativistic wavelength relation gives

$$\dfrac{\lambda'}{\lambda} = \sqrt{\dfrac{1+\beta}{1-\beta}} \approx 1 + \beta + \dfrac{\beta^{2}}{2} + O(\beta^{3})$$λλ′​=1−β1+β​​≈1+β+2β2​+O(β3)

So $\Delta\lambda/\lambda \approx \beta + \beta^{2}/2 = v/c + (v/c)^{2}/2$Δλ/λ≈β+β2/2=v/c+(v/c)2/2. The fractional error of using the first-order formula is

$$\dfrac{\text{error}}{\Delta\lambda/\lambda} \approx \dfrac{\beta^{2}/2}{\beta} = \dfrac{\beta}{2} = \dfrac{v}{2c}$$Δλ/λerror​≈ββ2/2​=2β​=2cv​

$v$v	$\beta = v/c$β=v/c	Fractional error
$300$300 km s$^{-1}$−1 (galactic star)	$10^{-3}$10−3	$0.05\%$0.05%
$3000$3000 km s$^{-1}$−1 (nearby galaxy)	$10^{-2}$10−2	$0.5\%$0.5%
$30\,000$30000 km s$^{-1}$−1 ($z \approx 0.1$z≈0.1)	$0.1$0.1	$5\%$5%
$0.5c$0.5c (high-$z$z quasar)	$0.5$0.5	$25\%$25%

So $\Delta\lambda/\lambda = v/c$Δλ/λ=v/c is excellent for galactic stars and nearby galaxies. It starts to break down at $z \approx 0.1$z≈0.1 (a few hundred Mpc), and is wholly inadequate for the most distant quasars. OCR A-Level questions stay safely in the non-relativistic regime; if a question mentions quasars or $v > 0.1c$v>0.1c, expect a clear instruction about which approximation to use.

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Synoptic Links

• Waves (Module 4). The Doppler effect for sound is in Module 4.4. The mathematical form there is not the same as for light — sound has a preferred frame (the air), light does not. The conceptual seed is the same (wavefronts piling up or stretching out), but the formulae differ. A common synoptic question gives you a sound Doppler problem and a light Doppler problem and asks why one is reciprocal in $v$v and the other not.
• Special relativity (Module 6 extension or beyond A-Level). The relativistic Doppler formula is a direct consequence of the Lorentz transformation. The fact that no medium-based formula reproduces it is one of the foundational experimental tests of special relativity.
• Hubble's law (lesson 8). Doppler shifts of distant galaxies, plotted against distance, give a straight line of slope $H_{0}$H0​. This is the empirical discovery that started modern cosmology. Note however that for cosmological distances the red shift is not strictly a Doppler effect — it is expansion of space stretching wavelengths in transit (a subtlety we explore in the misconceptions).
• Stellar evolution (lesson 6). Pulsational variable stars (Cepheids, RR Lyrae) show periodic Doppler shifts of their photospheres as they breathe in and out — a direct line-of-sight measurement of their pulsation velocity.
• Quantum physics (Module 4.4-4.5). The spectral lines themselves come from quantum transitions in atoms. The Doppler-shifted H-$\alpha$α Balmer line is a direct application of the Bohr model: $n = 3 \to n = 2$n=3→n=2 in hydrogen gives $\lambda = 656.3$λ=656.3 nm. Doppler astronomy is unthinkable without atomic spectroscopy.
• Module 5.4 (gravitational fields). Combining the Doppler method with Newton's law of gravitation lets you derive the mass function of binary stars from radial-velocity amplitudes — extending the orbital-mechanics calculations of Module 5.4 into observational practice.

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Specimen question modelled on the OCR H556 paper format

Question (10 marks): The H-$\alpha$α Balmer line of hydrogen has rest wavelength $\lambda_{0} = 656.28$λ0​=656.28 nm. A distant galaxy's spectrum shows this line at $\lambda = 658.91$λ=658.91 nm.