Cosmological Redshift vs the Doppler Effect: How to Tell

A hydrogen spectral line has a wavelength of \(656.3\ \text{nm}\) when measured in a laboratory. Light from a distant galaxy shows the same line at \(662.9\ \text{nm}\). Meanwhile, a siren moving away from you also produces a longer observed wavelength.

Are these two examples of the same Doppler effect?

Before reading on, predict whether you could calculate the galaxy’s redshift using the sound Doppler equation.

You can’t. Both observations involve changes in wavelength and frequency, but their physical explanations are different. The ordinary sound Doppler effect depends on the motion of a source and observer relative to a medium, such as air. Cosmological redshift occurs because the expansion of the universe stretches the wavelength of light as it travels through space.

Understanding that difference will help you choose the correct model rather than simply looking for the word redshift and reaching for a familiar equation.

01The three effects you need to distinguish

There are actually three related ideas worth separating:

1. The Doppler effect for sound.
2. The ordinary Doppler effect for light caused by relative motion.
3. Cosmological redshift caused by the expansion of the universe.

All three can change an observed wavelength or frequency. However, they do not use the same physical model.

The most important distinction is the first one.

Sound travels through air. Its Doppler equation must account for the motion of the source and observer relative to that air.

Light does not need air or another material medium. Its ordinary Doppler shift is described using special relativity, while cosmological redshift describes the stretching of light during cosmic expansion.

Let’s build each model separately.

02The ordinary sound Doppler effect: why the medium matters

Imagine a train approaching a stationary observer while sounding its horn.

The horn produces sound waves at a constant frequency. However, the train moves forward between successive emissions.

This means each new wavefront is produced slightly closer to the observer than the previous one. The wavefronts become compressed in front of the train.

The observer therefore receives more wavefronts each second and hears a higher frequency.

There are two separate ways motion can affect the observed frequency:

• Moving the source changes the spacing between successive wavefronts.
• Moving the observer changes how quickly they encounter those wavefronts.

This is why source velocity and observer velocity appear in different parts of the sound Doppler equation.

The correct reference frame for sound

Sound travels at a particular speed relative to its medium.

For sound travelling through air, we use:

\[
c_{\text{air}} = \text{speed of sound relative to air}
\]

At approximately \(20^\circ\text{C}\), a useful value is \(343\ \text{m s}^{-1}\). In worked questions, use the speed supplied in the question.

We also need two velocities:

• \(u_s\): the source’s velocity relative to the air.
• \(u_o\): the observer’s velocity relative to the air.

Both velocities refer to their components along the line joining the source and observer.

Use the following sign convention:

The sound Doppler equation is:

\[
\boxed{
f_{\text{obs}} =
f_{\text{emit}}
\frac{c_{\text{air}}+u_o}
{c_{\text{air}}-u_s}
}
\]

Where:

• \(f_{\text{obs}}\) is the observed frequency in hertz (Hz).
• \(f_{\text{emit}}\) is the emitted frequency in hertz (Hz).
• \(c_{\text{air}}\) is the speed of sound relative to air in \(\text{m s}^{-1}\).
• \(u_o\) is the signed observer velocity relative to air in \(\text{m s}^{-1}\).
• \(u_s\) is the signed source velocity relative to air in \(\text{m s}^{-1}\).

The equation assumes straight-line motion along the line of sight and a uniform medium.

Notice that these are not simply the source and observer’s speeds relative to the ground.

If there is wind, you must account for the motion of the air before substituting velocities.

For a more detailed explanation of the sound equation and its sign conventions, see tutorgum’s guides to the Doppler effect and choosing the correct Doppler signs.

What happens when there is wind?

Consider a speaker and listener that are both stationary relative to the ground. A steady wind blows from the speaker towards the listener.

You might predict that the wind increases the observed frequency because it carries sound towards the listener faster.

That prediction is incorrect.

In the reference frame of the air, both the speaker and listener move together in the opposite direction to the wind. Their velocities relative to the air produce equal and opposite corrections in the Doppler equation.

The observed frequency remains equal to the emitted frequency.

The wind changes the speed of sound relative to the ground, but a steady, uniform wind does not by itself change the frequency received by a stationary listener.

This is a useful reminder: the medium is not just a detail in the sound Doppler equation. It determines the reference frame in which the velocities must be measured.

03Ordinary Doppler shift for light

Now replace the train’s horn with a star emitting electromagnetic radiation.

There is one major difference.

Light does not require a material medium to travel. In a vacuum, its speed is:

\[
c=3.00\times10^8\ \text{m s}^{-1}
\]

This speed is measured locally by observers in inertial reference frames.

Consequently, you cannot use the sound Doppler equation with the speed of light substituted for the speed of sound.

The ordinary Doppler effect for light is explained using special relativity.

If a light source moves directly away from an observer, the observed wavelength increases. If it approaches, the observed wavelength decreases.

This is called redshift or blueshift, respectively.

For purely radial relative motion, the relativistic Doppler relationship is:

\[
1+z_D=
\frac{\lambda_{\text{obs}}}{\lambda_{\text{emit}}}
=
\sqrt{\frac{1+\beta}{1-\beta}}
\]

Where:

• \(z_D\) is the relativistic Doppler redshift.
• \(\lambda_{\text{obs}}\) is the observed wavelength in metres.
• \(\lambda_{\text{emit}}\) is the emitted wavelength in metres.
• \(\beta=v_r/c\) is the signed radial velocity as a fraction of the speed of light.
• \(v_r\) is positive when the source is receding and negative when it is approaching.
• \(c\) is the speed of light in a vacuum.

This equation assumes ordinary relative motion in flat spacetime, without cosmological expansion or gravitational redshift.

For relatively small speeds compared with light, it simplifies to:

\[
z_D\approx\frac{v_r}{c}
\]

The approximation is useful for understanding small light Doppler shifts, but it is not the general cosmological redshift equation.

Extension: The exact relativistic light Doppler equation is included to distinguish the physical models. Do not assume that it is a required HSC calculation in your school’s Physics program.

04Cosmological redshift: space stretches the wavelength

Here is where the explanation changes.

Imagine two distant galaxies that are not moving significantly relative to the surrounding expanding universe. Light travels from one galaxy towards the other.

As the universe expands during the journey, the wavelength of that light increases.

This is called cosmological redshift.

The important point is that the wavelength stretches as space expands. We do not need to explain the effect by treating the light as a sound wave travelling through a moving medium.

Picture the expansion

Imagine drawing a wave across a stretchy grid.

Initially, the grid lines are close together, and the wave crests are separated by a particular distance.

Now stretch the grid. The distance between the crests increases as the grid expands.

This is a useful way to picture cosmological redshift.

The analogy has a limit: the universe is not a rubber sheet expanding into a surrounding room. The grid represents the changing distances between locations in space.

The cosmological redshift equation

Cosmological redshift is defined by:

\[
\boxed{
z=\frac{\lambda_{\text{obs}}-\lambda_{\text{emit}}}
{\lambda_{\text{emit}}}
}
\]

Rearranging:

\[
\boxed{
1+z=\frac{\lambda_{\text{obs}}}
{\lambda_{\text{emit}}}
}
\]

Where:

• \(z\) is the cosmological redshift.
• \(\lambda_{\text{obs}}\) is the wavelength measured by the observer.
• \(\lambda_{\text{emit}}\) is the wavelength when the light was emitted.

The redshift is dimensionless because it is a ratio of two wavelengths.

Its sign tells us what happened:

A redshift does not necessarily mean that the light looks visibly red. The wavelength may shift beyond the visible spectrum into infrared.

The scale factor connection

Cosmologists describe the expansion of the universe using a quantity called the scale factor, \(a(t)\).

It describes how distances between locations that follow the overall cosmic expansion change with time.

For light travelling between two such locations:

\[
\boxed{
1+z=
\frac{a(t_{\text{obs}})}
{a(t_{\text{emit}})}
}
\]

Here:

• \(a(t_{\text{obs}})\) is the scale factor when the light is observed.
• \(a(t_{\text{emit}})\) is the scale factor when the light was emitted.

If the universe expands during the light’s journey, the scale factor increases. The observed wavelength is therefore longer than the emitted wavelength.

Because the local speed of light remains \(c\), a longer wavelength corresponds to a lower observed frequency:

\[
f_{\text{obs}}=
\frac{f_{\text{emit}}}{1+z}
\]

This is the central mechanism of cosmological redshift.

Why cosmological redshift is not simply ordinary Doppler motion

For nearby galaxies, cosmological redshift can often be approximated using:

\[
v\approx cz
\]

This is a low-redshift approximation, assuming the galaxy’s motion is dominated by the overall cosmic expansion.

It is useful because it allows astronomers to express small redshifts as approximate recession velocities.

However, at larger redshifts, this simple relationship cannot be used as a general conversion from redshift to velocity.

The expansion history of the universe matters. There is no single global, special-relativistic source velocity that explains every cosmological redshift.

The distinction is important:

• Ordinary light Doppler shift: relative motion between source and observer changes the observed frequency.
• Cosmological redshift: the expansion of space stretches the wavelength during the light’s journey.

Both can produce a redshift, but they describe different physical situations.

05How astronomers measured cosmological redshift

The distinction became important when astronomers began investigating the motion of distant galaxies.

The original problem was straightforward: were the faint, spiral-shaped objects seen through telescopes part of our own galaxy, or were they separate galaxies? And were they moving towards or away from us?

Astronomers needed a way to measure motion across enormous distances.

Step 1: Collect light from a galaxy

A telescope gathers light from a distant galaxy using its primary mirror or lens.

The collected light is focused towards an instrument called a spectrograph.

A narrow slit directs the light into the spectrograph, where a prism or diffraction grating separates it into its different wavelengths.

A photographic plate or modern detector records the resulting spectrum.

Step 2: Compare spectral lines with laboratory measurements

Elements absorb and emit light at characteristic wavelengths.

For example, hydrogen produces identifiable spectral lines. Scientists can measure these wavelengths in laboratories and use them as reference points.

When astronomers examine light from a galaxy, they look for the same pattern of spectral lines.

If the galaxy’s light has shifted towards longer wavelengths, the lines appear displaced towards the red end of the spectrum.

If the lines shift towards shorter wavelengths, they appear displaced towards the blue end.

The key is to compare the positions of known lines, not simply judge the overall colour of a galaxy.

NASA explains this measurement process in its guide to spectroscopy and spectral lines.

Step 3: What did the early observations show?

In 1912, astronomer Vesto Slipher measured the radial velocity of the Andromeda nebula using its spectrum.

The result indicated that Andromeda was approaching the Milky Way, producing a blueshift.

Further observations showed that most of the spiral nebulae being studied had redshifted spectra.

This was an important observation, but it did not yet establish that the universe was expanding.

There was another problem.

A spectrum could provide information about an object’s radial motion under a Doppler interpretation, but it did not independently tell astronomers how far away the object was.

Step 4: Add independent distance measurements

Edwin Hubble and his collaborators combined spectral velocity measurements with independent estimates of galaxy distances.

One important method involved Cepheid variable stars.

These stars change in brightness over time. By measuring their periods of variation, astronomers could use the relationship between a Cepheid’s period and its intrinsic luminosity to estimate its distance.

The improved distance information allowed astronomers to compare galaxy distance with radial velocity.

In 1929, Hubble published the relationship between distance and radial velocity for a sample of galaxies. This became known as Hubble’s law.

The key historical development was therefore not simply a better spectrograph. It was combining two different measurements:

1. Spectral shifts provided information about radial motion.
2. Independent distance measurements showed how the observed shifts varied with distance.

The relationship became important evidence for the expansion of the universe.

NASA’s historical timeline of Hubble’s expansion measurements covers Slipher’s early observations, Hubble’s Cepheid measurements, and the 1929 distance-velocity relationship.

07A reliable method for distinguishing the effects

When a question mentions a frequency shift, wavelength change, or redshift, use this decision process.

Step 1: Identify the wave

Is the question about sound or electromagnetic radiation?

• Sound requires a medium.
• Light does not require a medium.

This immediately rules out using the sound Doppler equation for light.

Step 2: Identify what causes the change

For sound, ask whether the source or observer moves relative to the air.

For ordinary light Doppler shift, consider the relative radial motion of the source and observer.

For cosmological redshift, consider whether the light has travelled through an expanding universe.

Step 3: Choose the appropriate relationship

Step 4: Check your interpretation

A positive redshift tells you that the observed wavelength is longer than the emitted wavelength.

It does not, by itself, prove why that happened.

For a distant galaxy, the distance-redshift relationship provides evidence for cosmological expansion. For a nearby star, a changing spectral shift may instead indicate ordinary orbital motion.

Also remember that galaxies can have local motions relative to the overall expansion. These are called peculiar velocities.

The measured redshift can therefore contain contributions from more than one physical effect.

08Questions and solutions

The following questions move from straightforward redshift calculations to more challenging situations involving reference frames and competing explanations.

All numerical scenarios are illustrative.

Question 1

A galaxy produces an absorption line with a laboratory wavelength of \(486.1\ \text{nm}\).

An astronomer measures the same line at \(495.8\ \text{nm}\).

Calculate the redshift and determine whether the galaxy’s light is redshifted or blueshifted.

Solution 1

The galaxy has a redshift of approximately \(\boxed{0.0200}\).

The redshift equation is:

\[
z=
\frac{\lambda_{\text{obs}}-\lambda_{\text{emit}}}
{\lambda_{\text{emit}}}
\]

Substituting:

\[
\begin{aligned}
z&=\frac{495.8-486.1}{486.1}\\
&=\frac{9.7}{486.1}\\
&=0.01995
\end{aligned}
\]

Therefore:

\[
\boxed{z=+0.0200}
\]

The positive sign indicates redshift because the observed wavelength is longer than the laboratory wavelength.

The important distinction is that redshift describes the measured wavelength change. You still need to consider the physical cause before interpreting it as cosmological expansion.

Question 2

A train-mounted siren is positioned west of a stationary observer.

The train travels east at \(4.0\ \text{m s}^{-1}\) relative to the ground. A steady wind also blows east at \(12\ \text{m s}^{-1}\).

The siren emits a frequency of \(680\ \text{Hz}\).

The speed of sound relative to the air is \(340\ \text{m s}^{-1}\).

Calculate the frequency heard by the observer.

Explain why substituting the train’s ground velocity directly into the sound Doppler equation would be incorrect.

Solution 2

The observer hears a frequency of approximately \(\boxed{688\ \text{Hz}}\).

The important step is converting both velocities into the reference frame of the air.

Step 1: Calculate the source velocity relative to the air.

The train moves east at \(4.0\ \text{m s}^{-1}\), while the air moves east at \(12\ \text{m s}^{-1}\).

Therefore:

\[
u_s=4.0-12=-8.0\ \text{m s}^{-1}
\]

The negative sign means the source moves west relative to the air, away from the observer.

Step 2: Calculate the observer velocity relative to the air.

The observer is stationary relative to the ground.

Therefore:

\[
u_{\text{observer, air}}=0-12
\]

\[
=-12\ \text{m s}^{-1}
\]

The observer moves west relative to the air, towards the source.

Using our sign convention:

\[
u_o=+12\ \text{m s}^{-1}
\]

Step 3: Substitute into the sound Doppler equation.

\[
f_{\text{obs}}=
f_{\text{emit}}
\frac{c_{\text{air}}+u_o}
{c_{\text{air}}-u_s}
\]

\[
\begin{aligned}
f_{\text{obs}}
&=680\left(\frac{340+12}{340-(-8)}\right)\\
&=680\left(\frac{352}{348}\right)\\
&=687.82\ \text{Hz}
\end{aligned}
\]

Step 4: State the answer.

\[
\boxed{f_{\text{obs}}\approx688\ \text{Hz}}
\]

The observer hears a higher frequency than the emitted frequency.

This is consistent with the train approaching the stationary observer in the ground frame.

The tempting mistake is to use \(4.0\ \text{m s}^{-1}\) as the source velocity while keeping \(340\ \text{m s}^{-1}\) as the sound speed. Those values belong to different reference frames.

The sound equation requires both source and observer velocities relative to the air.

Question 3

A star emits light with a particular spectral line at \(500.0\ \text{nm}\).

An astronomer observes the same line at \(550.0\ \text{nm}\).

Assume the shift is entirely due to the star’s radial motion through otherwise non-expanding, flat spacetime. Ignore gravitational effects.

Using the exact relativistic light Doppler relationship, calculate the star’s recession speed as a fraction of the speed of light.

Explain why this calculation is different from calculating the cosmological redshift of a distant galaxy.

Solution 3

The star is receding at approximately \(\boxed{0.0950c}\), or \(2.85\times10^7\ \text{m s}^{-1}\).

First, calculate the Doppler redshift:

\[
z_D=\frac{550.0-500.0}{500.0}
\]

\[
z_D=0.100
\]

Therefore:

\[
1+z_D=1.10
\]

For a receding source, the exact relativistic light Doppler equation is:

\[
1+z_D=
\sqrt{\frac{1+\beta}{1-\beta}}
\]

Step 1: Square both sides.

\[
(1+z_D)^2=
\frac{1+\beta}{1-\beta}
\]

Step 2: Rearrange to isolate \(\beta\).

\[
\beta=
\frac{(1+z_D)^2-1}
{(1+z_D)^2+1}
\]

Step 3: Substitute the redshift.

\[
\begin{aligned}
\beta
&=\frac{1.10^2-1}{1.10^2+1}\\
&=\frac{1.21-1}{1.21+1}\\
&=\frac{0.21}{2.21}\\
&=0.0950
\end{aligned}
\]

Since \(\beta=v_r/c\):

\[
v_r=0.0950c
\]

Using \(c=3.00\times10^8\ \text{m s}^{-1}\):

\[
\boxed{v_r=2.85\times10^7\ \text{m s}^{-1}}
\]

The distinction is in the physical interpretation.

Here, the shift is attributed entirely to the star’s relative radial motion. The special-relativistic Doppler equation describes that motion.

For cosmological redshift, the same wavelength ratio would instead indicate:

\[
\frac{a(t_{\text{obs}})}
{a(t_{\text{emit}})}
=1.10
\]

This describes the expansion of space between emission and observation.

The measured spectral shift alone does not identify which mechanism produced it. The physical context and additional observations are needed.

Question 4

Two astronomical observations are made.

Observation A: A nearby binary star has spectral lines that repeatedly shift towards longer wavelengths, then shorter wavelengths, over its orbital cycle. The pattern repeats over time.

Observation B: A distant galaxy has several known absorption lines, all shifted towards longer wavelengths by approximately the same fractional amount. The measured redshift remains stable over repeated observations.

Explain:

1. Which observation is evidence of ordinary Doppler motion caused by orbital movement?
2. Which observation is consistent with cosmological redshift?
3. Why is Observation B not, by itself, sufficient to prove that the universe is expanding?

Solution 4

Observation A is evidence of ordinary orbital Doppler motion, while Observation B is consistent with cosmological redshift.

However, Observation B alone does not establish that the shift is caused by cosmic expansion.

1. Observation A: orbital Doppler motion

As the binary star moves through its orbit, its radial velocity relative to the observer changes.

When the star moves away, its spectral lines shift towards longer wavelengths.

When it moves towards the observer, its spectral lines shift towards shorter wavelengths.

The repeated reversal is consistent with ordinary Doppler motion caused by orbital movement.

The expansion of the universe would not produce this periodic red-to-blue reversal over a star’s orbital cycle.

2. Observation B: cosmological redshift

The distant galaxy shows a consistent shift towards longer wavelengths.

The fact that multiple known absorption lines are displaced by the same fractional amount supports the interpretation that the galaxy’s light has been redshifted.

This is consistent with cosmological expansion.

3. Why one galaxy is not enough

A galaxy moving away from us through ordinary relative motion can also produce a redshift.

Therefore, a positive redshift alone does not prove that space expanded during the light’s journey.

Astronomers need additional evidence, particularly the relationship between galaxy distance and redshift across many galaxies.

When increasingly distant galaxies generally show greater redshifts, the overall pattern provides evidence for an expanding universe.

The important distinction is between measuring a redshift and establishing its physical cause.

Question 5

A distant galaxy emits a spectral line with a wavelength of \(600.0\ \text{nm}\).

An observer on Earth measures the wavelength as \(599.4\ \text{nm}\).

The galaxy is therefore observed to be slightly blueshifted.

However, an independent measurement establishes that the galaxy has a peculiar velocity of \(900\ \text{km s}^{-1}\) towards Earth relative to its local comoving reference frame.

Assume:

• The peculiar motion is purely radial.
• The speed of light is \(3.00\times10^5\ \text{km s}^{-1}\).
• Gravitational redshift and other effects are negligible.
• The cosmological and peculiar-motion wavelength factors combine multiplicatively.

Determine the cosmological redshift of the galaxy.

Then explain how the galaxy can have a positive cosmological redshift while its total observed spectrum is blueshifted.

Solution 5

The galaxy has a positive cosmological redshift of approximately:

\[
\boxed{z_{\text{cos}}\approx+0.00200}
\]

This means the expansion of the universe stretches the light, even though the total observed wavelength is slightly shorter than the emitted wavelength.

The key is that the galaxy’s local motion towards Earth produces a blueshift that partially cancels the cosmological redshift.

Step 1: Calculate the total observed redshift.

\[
z_{\text{obs}}=
\frac{\lambda_{\text{obs}}-\lambda_{\text{emit}}}
{\lambda_{\text{emit}}}
\]

Substitute:

\[
\begin{aligned}
z_{\text{obs}}
&=\frac{599.4-600.0}{600.0}\\
&=\frac{-0.6}{600.0}\\
&=-0.0010
\end{aligned}
\]

Therefore:

\[
1+z_{\text{obs}}=0.9990
\]

The negative value confirms that the total observed shift is a blueshift.

Step 2: Calculate the dimensionless peculiar velocity.

The galaxy approaches Earth at \(900\ \text{km s}^{-1}\).

Its speed as a fraction of the speed of light is:

\[
\beta=\frac{v}{c}
\]

\[
\begin{aligned}
\beta&=\frac{900}{300\,000}\\
&=0.0030
\end{aligned}
\]

Step 3: Calculate the peculiar Doppler factor.

For an approaching source, the relativistic wavelength factor is:

\[
1+z_{\text{pec}}=
\sqrt{\frac{1-\beta}{1+\beta}}
\]

Substituting:

\[
\begin{aligned}
1+z_{\text{pec}}
&=\sqrt{\frac{1-0.0030}{1+0.0030}}\\
&=\sqrt{\frac{0.997}{1.003}}\\
&\approx0.9970045
\end{aligned}
\]

This is less than one, as expected for an approaching source.

Step 4: Combine the two effects.

When separate wavelength shifts occur, their wavelength ratios multiply:

\[
1+z_{\text{obs}}=
(1+z_{\text{cos}})
(1+z_{\text{pec}})
\]

Rearranging:

\[
1+z_{\text{cos}}=
\frac{1+z_{\text{obs}}}
{1+z_{\text{pec}}}
\]

Substituting:

\[
\begin{aligned}
1+z_{\text{cos}}
&=\frac{0.9990}{0.9970045}\\
&\approx1.0020015
\end{aligned}
\]

Therefore:

\[
\boxed{z_{\text{cos}}\approx+0.00200}
\]

Step 5: Interpret the result.

The two effects act in opposite directions:

• Cosmological expansion increases the wavelength by approximately \(0.2\%\).
• The galaxy’s local approach towards Earth decreases the wavelength by approximately \(0.3\%\).

The combined result is a net wavelength decrease of approximately \(0.1\%\).

This is why the observer measures a blueshift even though the cosmological contribution is a redshift.

The tempting approach is to assume that an observed blueshift means the universe is contracting, or that the galaxy must have a negative cosmological redshift.

Neither conclusion follows.

The measured spectrum contains the combined effects of expansion and local motion. Separating those contributions is necessary to identify the underlying cosmological behaviour.

This is also why the factors \(1+z\) multiply rather than the redshift values simply adding exactly. For small shifts, the redshift values can be approximately added, but the multiplicative relationship is the more general method.

09What to study next: Hubble’s law

You now have three different models for understanding frequency and wavelength shifts.

The sound Doppler effect depends on source and observer velocities relative to a medium.

The ordinary Doppler effect for light depends on relative radial motion and special relativity.

Cosmological redshift occurs because the expansion of the universe stretches the wavelength of light.

The next useful step is understanding Hubble’s law and how astronomers use the relationship between galaxy distance and redshift to investigate the expansion of the universe.

That is where individual spectral measurements become evidence about the universe as a whole.

For NSW students, remember that the Physics Stage 6 (2017) syllabus continues for Year 12 during 2026. The Physics 11-12 Syllabus (2025) begins implementation for Year 11 in 2027, with Year 12 implementation commencing in Term 4 of 2027. Check your school’s program for the assessed scope of the cosmological and relativistic extensions discussed here. See NESA’s syllabus implementation information.