Doppler effect (A-level only)

What you'll learn

How relative motion between a light source and an observer changes the observed frequency and wavelength.
How to use the Doppler equation $\frac{\Delta \lambda}{\lambda} = \frac{v}{c}$ for speeds much less than the speed of light.
How to calculate the velocity of receding galaxies and quasars using cosmological redshift.
How to determine the orbital speed of spectroscopic binary stars from the periodic shifting of their spectral lines.

The Doppler Effect for Light

You likely already know the Doppler effect from everyday life: when an ambulance drives past you, the pitch of its siren sounds higher as it approaches and lower as it drives away. The exact same principle applies to electromagnetic waves, including visible light and radio waves.

When a source of waves moves relative to an observer:

If the source moves towards the observer, the waves are compressed. The observed wavelength decreases and the frequency increases. For visible light, the spectrum shifts towards the blue end.
If the source moves away from the observer, the waves are stretched out. The observed wavelength increases and the frequency decreases. For visible light, the spectrum shifts towards the red end.

Diagram showing laboratory, redshifted, and blueshifted spectra

Definition

Redshift and Blueshift

Redshift ( $z$ ) is the fractional increase in wavelength (or decrease in frequency) due to the source moving away from the observer.

Blueshift is the corresponding decrease in wavelength when a source approaches the observer.

Astrophysicists measure this by looking at absorption lines in stellar spectra. Elements in a star's atmosphere absorb specific, known wavelengths of light. By comparing the position of these dark lines to a stationary laboratory sample, we can instantly tell if a star or galaxy is moving towards or away from us.

The Doppler Equations

To calculate the exact speed of the source, AQA provides two versions of the Doppler equation. For a source moving with relative velocity $v$ , emitting light of original frequency $f$ and original wavelength $λ\lambda$ :

\frac{\Delta f}{f} = \frac{v}{c}

\frac{\Delta \lambda}{\lambda} = -\frac{v}{c}

Where:

$Δf\Delta f$ is the change in frequency ( $Hz\text{Hz}$ )
$f$ is the true (laboratory) frequency ( $Hz\text{Hz}$ )
$Δλ\Delta \lambda$ is the change in wavelength ( $m\text{m}$ )
$λ\lambda$ is the true (laboratory) wavelength ( $m\text{m}$ )
$v$ is the relative velocity of the source ( $s−1\text{m s}^{-1}$ )
$c$ is the speed of light in a vacuum ( $3.00 \times 10^8 \text{ m s}^{-1}$ )
$z$ is the redshift parameter (a dimensionless number)

Common Mistake

The negative sign convention

In the AQA specification, the formula is written as $\frac{\Delta \lambda}{\lambda} = -\frac{v}{c}$ .

The negative sign simply accounts for vector directions. If a source is moving towards you (approach velocity is positive), the wavelength decreases ( $Δλ\Delta \lambda$ is negative), keeping the equation balanced.

In exam calculations, it is almost always safer to calculate the magnitude of the shift using $\frac{\Delta \lambda}{\lambda} = \frac{v}{c}$ and then explicitly state whether the object is moving towards or away based on whether it is blueshifted or redshifted.

The Crucial Condition: $\ll c$

The equations above are approximations. They rely on classical physics and only work accurately if the relative velocity of the source is much less than the speed of light ( $\ll c$ ).

If an object is moving at a significant fraction of the speed of light (e.g., $v > 0.1 c$ ), special relativity takes over. Time dilation means the simple classical formula breaks down, and a more complex relativistic Doppler formula (which you don't need to know for A-level) must be used.

Key Idea

When to use the simple formula

Always use $Δλλ=vc\frac{\Delta \lambda}{\lambda} = \frac{v}{c}$ unless the question explicitly asks you to discuss its limitations. If a multi-part question asks you why your calculated value for a distant quasar might be inaccurate, the answer is always that the $\ll c$ condition is not met, so relativistic effects must be considered.

Example

Calculating the recession velocity of a galaxy

A distant galaxy emits a prominent hydrogen absorption line. In a laboratory on Earth, this line has a wavelength of $\text{ nm}$ . When observed from the galaxy, the line is detected at $\text{ nm}$ .

Calculate the velocity of the galaxy relative to Earth and state its direction.

Identify the true wavelength $λ\lambda$ and calculate the shift $Δλ\Delta \lambda$ .

\lambda = 656.28 \text{ nm}

\Delta \lambda = 660.12 - 656.28 = 3.84 \text{ nm}

State the Doppler equation for magnitude.

\frac{\Delta \lambda}{\lambda} = \frac{v}{c}

Rearrange for $v$ and substitute the values (you can leave both wavelengths in $nm\text{nm}$ because the units cancel in the ratio).

\left( \frac{\Delta \lambda}{\lambda} \right)

\times 10^8 \times \left( \frac{3.84}{656.28} \right)

\times 10^6 \text{ m s}^{-1}

State final answer to an appropriate number of significant figures, with the direction. The observed wavelength is longer (redshifted), so the galaxy is moving away. $\times 10^6 \text{ m s}^{-1}$ , moving away from Earth.

Galaxies and Quasars

The Doppler effect is our primary tool for studying the large-scale structure of the universe. When astronomers observe distant galaxies, almost all of them exhibit redshift. This cosmological redshift tells us that the universe is expanding.

Quasars (quasi-stellar radio sources) are among the most distant and luminous objects in the universe. Because they are so far away, the expansion of the universe carries them away from us at tremendous speeds.

Quasars display huge redshifts, often with $z$ values greater than $1$ .
A $z$ value greater than $1$ implies $Δλ>λ\Delta \lambda > \lambda$ , which would absurdly suggest $v > c$ if you strictly applied the classical $z = v / c$ formula!
This confirms that for quasars, the classical approximation $\ll c$ is invalid, and relativistic formulas are required.

Spectroscopic Binary Stars

A binary star system consists of two stars orbiting their common centre of mass. Often, these stars are too far away from Earth to be resolved individually by a telescope—they just look like a single point of light. However, we can prove it's a binary system using the Doppler effect.

If we view the system in the plane of its orbit (edge-on), the stars will alternately move towards us and away from us as they complete their orbits.

Binary star system edge-on showing redshift and blueshift

If both stars are roughly the same brightness and have similar spectral lines, we will see the spectral lines split into two and then recombine periodically:

As Star A moves towards us, its lines blueshift.
At the same time, Star B moves away from us, so its lines redshift.
The separation between the lines reaches a maximum when the stars are moving directly along our line of sight (one straight towards us, one straight away).
When the stars are moving perpendicular to our line of sight (across the sky), there is zero relative velocity towards/away from us, so the lines merge back into their rest positions.

Hint

What if one star is much brighter?

If one star completely outshines the other, you won't see splitting lines. Instead, you will see a single set of spectral lines smoothly oscillating back and forth between a redshift and a blueshift as the bright star completes its orbit.

Example

Orbital speed of a binary star

A spectroscopic binary system consists of two identical stars in a circular orbit. An absorption line, known to have a laboratory wavelength of $\text{ nm}$ , is observed to periodically split. The maximum recorded wavelength for this line during the cycle is $\text{ nm}$ .

Calculate the orbital speed of the stars.

Identify the maximum wavelength shift, $Δλ\Delta \lambda$ . This occurs when the star is moving directly away from us at its full orbital speed.

\Delta \lambda = 580.058 - 580.000 = 0.058 \text{ nm}

Use the Doppler equation to find the orbital velocity $v$ .

\frac{\Delta \lambda}{\lambda} = \frac{v}{c}

Rearrange and calculate.

\left( \frac{\Delta \lambda}{\lambda} \right)

\times 10^8 \times \left( \frac{0.058}{580.000} \right)

\times 10^4 \text{ m s}^{-1}

State the final answer. The orbital speed is $3.0 \times 10^4 \text{ m s}^{-1}$ (or $\text{ km s}^{-1}$ ).

Common Mistake

Confusing maximum shift with total line separation

In binary star questions, an exam might tell you the maximum difference in wavelength between the two split lines (the gap between Star A's blueshift and Star B's redshift). If the stars are identical, this gap represents $\times \Delta \lambda$ . You must halve this value to find the true $Δλ\Delta \lambda$ for one star before calculating its orbital speed!

Exam technique

In the exam

Always check your wavelength units. While the ratio $Δλ/λ\Delta \lambda / \lambda$ is dimensionless (so both can stay in $nm\text{nm}$ ), make sure you don't accidentally mix $nm\text{nm}$ for one and $m\text{m}$ for the other.
Read the question carefully to see if it provides the observed wavelength or the change in wavelength ( $Δλ\Delta \lambda$ ).
If a calculation gives a velocity greater than about $\times 10^7 \text{ m s}^{-1}$ (which is $0.1 c$ ), be prepared for a 1-mark follow-up question asking why the value might not be perfectly accurate. The magic phrase is "the formula assumes $\ll c$ ".
For binary stars, orbital speed $v$ is calculated from the maximum Doppler shift, which occurs when the velocity vector points directly along the observer's line of sight.

Self review

Check yourself

What happens to the observed frequency of light when the source is moving towards the observer?
In the Doppler equation, what does the symbol $z$ represent, and what is its unit?
Why can the simple formula $Δf/f=v/c\Delta f / f = v / c$ not be used for distant quasars?
In a spectroscopic binary system viewed edge-on, at what point in the orbit do the spectral lines show zero Doppler shift?

Doppler effect (A-level only)

What you'll learn

How relative motion between a light source and an observer changes the observed frequency and wavelength.
How to use the Doppler equation $\frac{\Delta \lambda}{\lambda} = \frac{v}{c}$ for speeds much less than the speed of light.
How to calculate the velocity of receding galaxies and quasars using cosmological redshift.
How to determine the orbital speed of spectroscopic binary stars from the periodic shifting of their spectral lines.

The Doppler Effect for Light

When a source of waves moves relative to an observer:

If the source moves towards the observer, the waves are compressed. The observed wavelength decreases and the frequency increases. For visible light, the spectrum shifts towards the blue end.
If the source moves away from the observer, the waves are stretched out. The observed wavelength increases and the frequency decreases. For visible light, the spectrum shifts towards the red end.

Diagram showing laboratory, redshifted, and blueshifted spectra

Definition

Redshift and Blueshift

Redshift ( $z$ ) is the fractional increase in wavelength (or decrease in frequency) due to the source moving away from the observer.

Blueshift is the corresponding decrease in wavelength when a source approaches the observer.

The Doppler Equations

\frac{\Delta f}{f} = \frac{v}{c}

\frac{\Delta \lambda}{\lambda} = -\frac{v}{c}

Where:

$Δf\Delta f$ is the change in frequency ( $Hz\text{Hz}$ )
$f$ is the true (laboratory) frequency ( $Hz\text{Hz}$ )
$Δλ\Delta \lambda$ is the change in wavelength ( $m\text{m}$ )
$λ\lambda$ is the true (laboratory) wavelength ( $m\text{m}$ )
$v$ is the relative velocity of the source ( $s−1\text{m s}^{-1}$ )
$c$ is the speed of light in a vacuum ( $3.00 \times 10^8 \text{ m s}^{-1}$ )
$z$ is the redshift parameter (a dimensionless number)

Common Mistake

The negative sign convention

In the AQA specification, the formula is written as $\frac{\Delta \lambda}{\lambda} = -\frac{v}{c}$ .

The Crucial Condition: $\ll c$

The equations above are approximations. They rely on classical physics and only work accurately if the relative velocity of the source is much less than the speed of light ( $\ll c$ ).

Key Idea

When to use the simple formula

Example

Calculating the recession velocity of a galaxy

Calculate the velocity of the galaxy relative to Earth and state its direction.

Identify the true wavelength $λ\lambda$ and calculate the shift $Δλ\Delta \lambda$ .

\lambda = 656.28 \text{ nm}

\Delta \lambda = 660.12 - 656.28 = 3.84 \text{ nm}

State the Doppler equation for magnitude.

\frac{\Delta \lambda}{\lambda} = \frac{v}{c}

Rearrange for $v$ and substitute the values (you can leave both wavelengths in $nm\text{nm}$ because the units cancel in the ratio).

\left( \frac{\Delta \lambda}{\lambda} \right)

\times 10^8 \times \left( \frac{3.84}{656.28} \right)

\times 10^6 \text{ m s}^{-1}

State final answer to an appropriate number of significant figures, with the direction. The observed wavelength is longer (redshifted), so the galaxy is moving away. $\times 10^6 \text{ m s}^{-1}$ , moving away from Earth.

Galaxies and Quasars

Quasars display huge redshifts, often with $z$ values greater than $1$ .
A $z$ value greater than $1$ implies $Δλ>λ\Delta \lambda > \lambda$ , which would absurdly suggest $v > c$ if you strictly applied the classical $z = v / c$ formula!
This confirms that for quasars, the classical approximation $\ll c$ is invalid, and relativistic formulas are required.

Spectroscopic Binary Stars

If we view the system in the plane of its orbit (edge-on), the stars will alternately move towards us and away from us as they complete their orbits.

Binary star system edge-on showing redshift and blueshift

If both stars are roughly the same brightness and have similar spectral lines, we will see the spectral lines split into two and then recombine periodically:

As Star A moves towards us, its lines blueshift.
At the same time, Star B moves away from us, so its lines redshift.
The separation between the lines reaches a maximum when the stars are moving directly along our line of sight (one straight towards us, one straight away).
When the stars are moving perpendicular to our line of sight (across the sky), there is zero relative velocity towards/away from us, so the lines merge back into their rest positions.

Hint

What if one star is much brighter?

Example

Orbital speed of a binary star

Calculate the orbital speed of the stars.

Identify the maximum wavelength shift, $Δλ\Delta \lambda$ . This occurs when the star is moving directly away from us at its full orbital speed.

\Delta \lambda = 580.058 - 580.000 = 0.058 \text{ nm}

Use the Doppler equation to find the orbital velocity $v$ .

\frac{\Delta \lambda}{\lambda} = \frac{v}{c}

Rearrange and calculate.

\left( \frac{\Delta \lambda}{\lambda} \right)

\times 10^8 \times \left( \frac{0.058}{580.000} \right)

\times 10^4 \text{ m s}^{-1}

State the final answer. The orbital speed is $3.0 \times 10^4 \text{ m s}^{-1}$ (or $\text{ km s}^{-1}$ ).

Common Mistake

Confusing maximum shift with total line separation

Exam technique

In the exam

Always check your wavelength units. While the ratio $Δλ/λ\Delta \lambda / \lambda$ is dimensionless (so both can stay in $nm\text{nm}$ ), make sure you don't accidentally mix $nm\text{nm}$ for one and $m\text{m}$ for the other.
Read the question carefully to see if it provides the observed wavelength or the change in wavelength ( $Δλ\Delta \lambda$ ).
If a calculation gives a velocity greater than about $\times 10^7 \text{ m s}^{-1}$ (which is $0.1 c$ ), be prepared for a 1-mark follow-up question asking why the value might not be perfectly accurate. The magic phrase is "the formula assumes $\ll c$ ".
For binary stars, orbital speed $v$ is calculated from the maximum Doppler shift, which occurs when the velocity vector points directly along the observer's line of sight.

Self review

Check yourself

What happens to the observed frequency of light when the source is moving towards the observer?
In the Doppler equation, what does the symbol $z$ represent, and what is its unit?
Why can the simple formula $Δf/f=v/c\Delta f / f = v / c$ not be used for distant quasars?
In a spectroscopic binary system viewed edge-on, at what point in the orbit do the spectral lines show zero Doppler shift?

Doppler effect (A-level only)

What you'll learn

The Doppler Effect for Light

The Doppler Equations

The Crucial Condition: v≪cv \ll cv≪c

Galaxies and Quasars

Spectroscopic Binary Stars

Doppler effect (A-level only) Revision Guide

Doppler effect (A-level only)

What you'll learn

The Doppler Effect for Light

The Doppler Equations

The Crucial Condition: v≪cv \ll cv≪c

Galaxies and Quasars

Spectroscopic Binary Stars

Doppler effect (A-level only) Revision Guide

The Crucial Condition: $\ll c$

The Crucial Condition: $\ll c$