Doppler Effect in Astronomy Revision Notes | A Level AQA Physics

1. The Doppler Effect and Why Astronomers Care

Picture an ambulance siren: as it rushes towards you the pitch sounds higher, and as it moves away the pitch drops. That shift in frequency is the Doppler effect, and it happens with light just as it does with sound. For astronomers, this is not a curiosity — it is the primary tool for measuring how fast stars, galaxies, and quasars are moving relative to Earth.

When a light source moves away from us, its emitted wavelengths are stretched towards the red end of the spectrum — a redshift. When it moves towards us, wavelengths are compressed towards the blue end — a blueshift. By comparing the observed wavelength of a known spectral line with its laboratory (rest) wavelength, we can calculate the source's velocity with remarkable precision.

The AQA specification requires you to apply this to optical and radio frequencies, to binary star systems, and to galaxies and quasars. Each application uses the same pair of equations, so mastering those equations first unlocks everything else.

2. The Core Equations: Fractional Shift and Redshift Parameter

Two equations sit at the heart of this topic. The first links the fractional change in frequency to the source's speed; the second defines the redshift parameter $z$ in terms of wavelength. Both apply only when $v \ll c$ — that is, when the source is moving at a speed much less than the speed of light ($3.0 \times 10^{8}$ m s$^{-1}$).

Note that $\Delta f = f_{\text{observed}} - f_{\text{emitted}}$ carries a sign: it is positive when the source approaches (observed frequency is higher) and negative when it recedes (observed frequency is lower). In the equation $\Delta f / f = v / c$, $v$ also carries a sign — positive for approach, negative for recession — consistent with the sign of $\Delta f$. In practice, many students find it clearest to calculate the magnitude of the shift and then state the direction separately based on whether the observed frequency is higher or lower than the rest frequency.

The sign convention here catches students out every year. In the AQA formula $z = -v/c$, $v$ is defined as positive when the source is approaching and negative when it is receding. When a galaxy recedes, $\lambda_{\text{observed}} > \lambda_{\text{emitted}}$, so $\Delta\lambda$ is positive and $z$ is positive. Substituting into $z = -v/c$ gives a negative value of $v$, which in this convention correctly indicates recession. To find the recessional speed as a positive magnitude, use $|v| = zc$. Some textbooks define $v$ as the recessional speed (always positive) and drop the minus sign — AQA uses the form above, so stick with it.

These equations apply equally to optical light and to radio waves — the physics is identical because both are electromagnetic radiation travelling at $c$. Radio astronomers routinely measure the 21 cm hydrogen line and compare its observed wavelength with the known rest wavelength to determine the velocity of interstellar gas clouds and distant galaxies.

3. Worked Calculations Using the Doppler Equations

Theory is only useful once you can execute it under exam conditions. The two worked examples below cover both equation forms — one using frequency, one using wavelength — so you can see exactly how to set out your working for full marks.

Worked example: Calculating recessional speed from a frequency shift

A distant galaxy emits a radio signal at a rest frequency of $1.420 \times 10^{9}$ Hz. An observer on Earth measures the frequency as $1.413 \times 10^{9}$ Hz. Calculate the recessional speed of the galaxy. [3 marks]

1. $f = 1.420 \times 10^{9}$ Hz (rest frequency) $f_{\text{obs}} = 1.413 \times 10^{9}$ Hz (observed frequency) $c = 3.0 \times 10^{8}$ m s$^{-1}$ — Always write down what you are given before substituting. Note the observed frequency is lower than the rest frequency — this is a redshift, so the galaxy is receding. [B1]
2. $\Delta f = f_{\text{obs}} - f_{\text{emitted}} = 1.413 \times 10^{9} - 1.420 \times 10^{9} = -0.007 \times 10^{9}$ Hz $= -7.0 \times 10^{6}$ Hz — The negative sign confirms recession. For the speed calculation, use the magnitude.
3. $\frac{\Delta f}{f} = \frac{v}{c}$ so $v = c \times \frac{|\Delta f|}{f}$ [M1] — Using the magnitude here gives the speed; the direction (recession) is already established from the sign of $\Delta f$.
4. $v = 3.0 \times 10^{8} \times \frac{7.0 \times 10^{6}}{1.420 \times 10^{9}}$ $v = 3.0 \times 10^{8} \times 4.930 \times 10^{-3}$ $v = 1.479 \times 10^{6}$ m s$^{-1}$ [A1] — Check: $7.0 \times 10^{6} \div 1.420 \times 10^{9} = 4.930 \times 10^{-3}$. Then $3.0 \times 10^{8} \times 4.930 \times 10^{-3} = 1.479 \times 10^{6}$ m s$^{-1}$. This is much less than $c$, confirming the $v \ll c$ condition is satisfied.

Answer: $v \approx 1.5 \times 10^{6}$ m s$^{-1}$ (recessional speed, galaxy moving away from Earth)

Worked example: Calculating speed from a wavelength shift using z

The hydrogen-alpha line has a rest wavelength of 656.3 nm. In the spectrum of a quasar, this line is observed at 722.0 nm. Calculate the redshift parameter $z$ and the recessional speed of the quasar. [4 marks]

1. $\lambda_{\text{emitted}} = 656.3$ nm $\lambda_{\text{observed}} = 722.0$ nm $c = 3.0 \times 10^{8}$ m s$^{-1}$ — Units are consistent (both in nm), so no conversion is needed for the ratio.
2. $\Delta \lambda = \lambda_{\text{observed}} - \lambda_{\text{emitted}} = 722.0 - 656.3 = 65.7$ nm — Positive $\Delta\lambda$ confirms the line has shifted to a longer wavelength — a redshift.
3. $z = \frac{\Delta \lambda}{\lambda} = \frac{65.7}{656.3} = 0.1001$ — $65.7 \div 656.3 = 0.1001$ (to 4 s.f.). This is a substantial redshift, typical of a moderately distant quasar. [A1]
4. $z = -\frac{v}{c}$ so $v = -zc$ $v = -0.1001 \times 3.0 \times 10^{8}$ $v = -3.003 \times 10^{7}$ m s$^{-1}$ — The negative value of $v$ in this convention indicates recession. The speed is $3.0 \times 10^{7}$ m s$^{-1}$, which is 10% of $c$ — borderline for the $v \ll c$ approximation, but within the scope of AQA questions. [A1]

Answer: $z = 0.100$; recessional speed $\approx 3.0 \times 10^{7}$ m s$^{-1}$

4. Binary Stars Viewed in the Plane of Orbit

Now that the equations are established, we can apply them to one of the most elegant uses of the Doppler effect in astronomy: determining the properties of binary star systems. A binary star consists of two stars orbiting their common centre of mass. When we observe the system edge-on — that is, in the plane of the orbit — each star periodically moves towards us and then away from us as it completes its orbit.

This produces a periodic oscillation in the observed wavelength of spectral lines from each star. When a star moves towards Earth, its lines are blueshifted; half an orbit later, the same star moves away and its lines are redshifted. By measuring the maximum shift and the period of oscillation, we can extract the orbital speed and — with the period — the orbital radius.

The key assumption here is that the orbit is viewed in the plane of orbit. If the orbit were face-on (perpendicular to our line of sight), there would be no component of velocity along our line of sight and no Doppler shift would be observed. AQA questions will specify this geometry, so watch for it in the question stem.

Worked example: Binary star: finding orbital speed and separation

Two stars in a binary system are observed in the plane of their orbit. Star A has a rest wavelength spectral line at 486.1 nm. The observed wavelength oscillates between 483.7 nm and 488.5 nm with a period of $T = 2.0 \times 10^{7}$ s. Assuming both stars have equal mass and orbit their common centre of mass, calculate: (a) the orbital speed of Star A, and (b) the orbital radius of Star A. [5 marks]

1. Maximum observed wavelength: 488.5 nm (redshifted, star moving away) Minimum observed wavelength: 483.7 nm (blueshifted, star approaching) Rest wavelength: 486.1 nm $\Delta\lambda_{\text{max}} = 488.5 - 486.1 = 2.4$ nm — Use the maximum shift from the rest wavelength — this corresponds to the full orbital speed directed along the line of sight. Either the red or blue extreme gives the same speed magnitude.
2. $\frac{\Delta\lambda}{\lambda} = \frac{v}{c}$ $v = c \times \frac{\Delta\lambda}{\lambda} = 3.0 \times 10^{8} \times \frac{2.4}{486.1}$ — Rearranging $z = \Delta\lambda / \lambda = -v/c$ for speed magnitude: $v = c \times |\Delta\lambda| / \lambda$.
3. $\frac{2.4}{486.1} = 4.938 \times 10^{-3}$ $v = 3.0 \times 10^{8} \times 4.938 \times 10^{-3} = 1.481 \times 10^{6}$ m s$^{-1}$ — Check: $2.4 \div 486.1 = 0.004938$. Then $3.0 \times 10^{8} \times 0.004938 = 1.481 \times 10^{6}$ m s$^{-1}$. This is well below $c$, confirming the approximation is valid.
4. $r = \frac{vT}{2\pi}$ $r = \frac{1.481 \times 10^{6} \times 2.0 \times 10^{7}}{2\pi}$ $r = \frac{2.962 \times 10^{13}}{6.2832}$ — The orbital circumference is $2\pi r$, and the star travels this distance in one period $T$, so $v = 2\pi r / T$ rearranges to $r = vT / 2\pi$. [M1]
5. $r = \frac{2.962 \times 10^{13}}{6.2832} = 4.71 \times 10^{12}$ m — $2.962 \times 10^{13} \div 6.2832 = 4.71 \times 10^{12}$ m. For equal-mass stars, the separation between the two stars is $2r = 9.4 \times 10^{12}$ m.

Answer: (a) Orbital speed of Star A: $v \approx 1.5 \times 10^{6}$ m s$^{-1}$. (b) Orbital radius of Star A: $r \approx 4.7 \times 10^{12}$ m. Separation between stars: $\approx 9.4 \times 10^{12}$ m.

5. Galaxies, Quasars, and the Expanding Universe

The same Doppler equations that work for nearby stars apply to the most distant objects in the observable universe. When astronomers measure the spectra of galaxies, they find that almost all of them show redshifts — their spectral lines are shifted to longer wavelengths compared with laboratory values. This is direct evidence that galaxies are receding from us.

More distant galaxies show larger redshifts, indicating higher recessional speeds. This relationship — known as Hubble's law — underpins the model of an expanding universe. For AQA, you need to be able to calculate recessional speeds from galaxy spectra using the Doppler equations, and to interpret what those speeds imply about the universe's expansion.

Quasars (quasi-stellar objects) are the most luminous persistent sources in the universe. They are active galactic nuclei — supermassive black holes accreting matter at enormous rates — and they can outshine entire galaxies. Their defining observational feature is an extremely large redshift: values of $z > 1$ are common, and some quasars have $z > 6$. At these values, the $v \ll c$ approximation breaks down and relativistic corrections are needed, but AQA questions will keep you within the non-relativistic regime.

The large redshifts of quasars tell us two things simultaneously: they are receding at high speeds, and because light takes time to travel, we are seeing them as they were billions of years ago. Quasars are therefore both the most distant and the most ancient objects we can observe — windows into the early universe.

The chart above illustrates the enormous range of redshift values encountered in astronomy. A nearby star moving at a few tens of km s$^{-1}$ produces a barely detectable shift, while a quasar at $z = 1.5$ is receding at a substantial fraction of $c$. This range is why the Doppler equations must be applied carefully, always checking the $v \ll c$ condition.

6. Applying the Equations to Radio Frequencies

The Doppler equations are not restricted to visible light. Radio astronomers use exactly the same expressions for radio-frequency observations. The most important radio spectral line in astronomy is the 21 cm hydrogen line, emitted at a frequency of $1.420 \times 10^{9}$ Hz by neutral hydrogen atoms. Because hydrogen is the most abundant element in the universe, this line is detectable across enormous distances.

By mapping the Doppler shift of the 21 cm line across different regions of the sky, radio astronomers can measure the rotation of our own galaxy and the motion of hydrogen clouds in other galaxies. The physics is identical to the optical case: compare the observed frequency with the rest frequency, apply $\Delta f / f = v / c$, and extract the velocity.

One practical advantage of radio observations is that radio waves pass through interstellar dust that would block optical light. This means radio Doppler measurements can probe regions of the galaxy that are optically obscured — making radio astronomy an indispensable complement to optical work.

7. Exam Technique and Common Mistakes

The Doppler effect questions in AQA papers follow predictable patterns. Most marks are awarded for correct substitution, consistent units, and a clearly stated final answer with units. The following guidance targets the specific errors that cost students marks most frequently.

For extended-response questions asking you to evaluate the use of Doppler measurements in astronomy, structure your answer as: observation (what is measured) → calculation (how speed is found) → interpretation (what the speed tells us about the object or the universe) → limitation (where the $v \ll c$ approximation breaks down or where other factors complicate the measurement). This four-part structure reliably covers the marking points.