Pulsating stars and Cepheids

Builds on: Sections 4.3, 4.6, 8.4 and D8
Reading time: 40 min
Exercises: 3
Olympiad problems: 2

In 1784, a year after he timed Algol's eclipses, John Goodricke noticed that δ Cephei, a fourth-magnitude star in Cepheus, brightens and fades every 5.37 days, rising quickly and sinking slowly. Unlike Algol, it is a single star swelling and shrinking. In 1912 Henrietta Leavitt, studying photographs of the Small Magellanic Cloud at Harvard, found that the brighter such stars had the longer periods. All the stars in the Cloud are at nearly the same distance, so the brighter ones really were more luminous. A Cepheid's period therefore tells us its luminosity, and comparing that with how bright it looks gives its distance. In 1923 Edwin Hubble found Cepheids in the Andromeda Nebula and showed that it lies far outside our Galaxy. In this section we see why these stars pulsate and how they measure the Universe.

8.7.1Why some stars pulsate

Before you readMake a guess first

A pulsating star is at its brightest during its cycle. Where is it in its swelling and shrinking?

A star is a ball of gas held up by its pressure against its gravity ((S.24)). Squeeze it, and the pressure grows and pushes it back out, past its balance, so it can ring like a bell. In most stars the ringing dies away, because the gas loses the energy of each squeeze as heat. In some it is kept going by a valve. Deep below the surface lies a layer where helium is partly ionised. When this layer is squeezed it becomes more opaque instead of less, and traps the heat flowing out from below. The trapped heat raises the pressure and pushes the layer out, where it becomes clear again and lets the heat escape, so it falls back and the cycle repeats. This valve works only when the layer sits at the right depth, which happens in a narrow band of temperatures on the Hertzsprung-Russell diagram called the instability strip. Cepheids are yellow supergiants in the strip, and RR Lyrae stars are old, low-mass stars where it crosses the horizontal branch.

A pressure wave crosses a star in about the time the star would take to fall in on itself under its own gravity, so the period depends mainly on the star's mean density.

ResultThe period of a pulsating star(8.11)

P \approx \frac{1}{\sqrt{G \bar{ρ}}}, \bar{ρ} = \frac{3 M}{4 π R^{3}}

$P$	the period of pulsation
$\bar{ρ}$	the mean density of the star
$M, R$	the mass and the radius of the star

This is an estimate to within a factor of order 1. A larger star of similar mass has a lower density and a longer period, and it is also more luminous, which is why the period and the luminosity are linked.

Figure 8.7A Cepheid through its cycle

Phase

Watch

Brightness
Surface moving
Size and temperature

The figure shows a simple model of a pulsating star with numbers close to δ Cephei, a mean radius of 44 solar radii and a period of 5.37 days. On the left the star swells and shrinks between the dotted circles, with its change of size drawn eight times larger than it is, and it is lighter when it is hotter. On the right are its brightness, the outward speed of its surface and its radius over two cycles. The radius changes by only 4%, and the brightness follows the temperature, so the star is brightest just after it is smallest, while its surface moves outwards at its fastest. The spectrum shows this motion as a Doppler shift, so in a real Cepheid the velocity curve is close to a mirror image of the light curve.

Quick checkThe period of δ Cephei

δ Cephei has about 4.5 solar masses and a radius of 44 solar radii. Estimate its period from (8.11).

days

8.7.2The period-luminosity relation

The more luminous a Cepheid, the longer its period. For the classical Cepheids, young massive stars like δ Cephei, the relation is a straight line between the absolute magnitude and the logarithm of the period. Measuring a Cepheid's period and its mean apparent magnitude, and correcting for extinction ((4.14)), gives its distance by the distance modulus ((4.4)).

ResultThe Cepheid period-luminosity relation(8.12)

M_{V} = - 2.43 (\log P - 1) - 4.05

$M_{V}$	the mean absolute visual magnitude of a classical Cepheid
$P$	its period, in days

This calibration rests on parallaxes measured by the Hubble Space Telescope (Benedict and others, 2007), with a scatter of about 0.2 magnitudes. In the infrared, as in the K band of Section D2, the scatter is smaller and extinction matters less. The older type II Cepheids are about 1.5 magnitudes fainter at the same period, and RR Lyrae stars, with periods under a day, all have

M_{V} \approx + 0.6

. A problem will give the relation to use.

Example 8.13A Cepheid in the Large Magellanic Cloud

A Cepheid in the Large Magellanic Cloud has a period of 30.0 days and a mean visual magnitude of 13.50, with an extinction of 0.20 magnitudes. How far away is it?

Solution

By (8.12), $M_{V} = - 2.43 \times (1.477 - 1) - 4.05 = - 5.21$ . The distance modulus corrected for extinction is $13.50 + 5.21 - 0.20 = 18.51$ , so $d = 10^{(18.51 + 5) / 5} = 10^{4.702}$ pc.

AnswerThe Cepheid is about 50 kpc away, the distance of the Large Magellanic Cloud.

8.7.3The radius from the light and the velocity

The velocity curve tells us how fast the surface moves, so adding up the speed over time gives how much the radius changes between two moments. The light and the colour give the ratio of the radii at those moments, since by the Stefan-Boltzmann law the luminosity goes as $R^{2} T^{4}$ and the colour gives $T$ . The change and the ratio together fix the radius itself. This is the Baade-Wesselink method, and with the angular size from interferometry or from the flux, it gives the distance as well.

ResultThe Baade-Wesselink method(8.13)

\begin{aligned} \frac{R_{1}}{R_{2}} & = \sqrt{\frac{F_{1}}{F_{2}}} {(\frac{T_{2}}{T_{1}})}^{2} \\ R_{2} - R_{1} & = \int_{t_{1}}^{t_{2}} v d t \end{aligned}

$R_{1}, R_{2}$	the radii of the star at two moments
$F_{1}, F_{2}$	the bolometric fluxes received at those moments
$T_{1}, T_{2}$	the temperatures then, from the colour or the peak of the spectrum
$v$	the outward speed of the surface

The measured radial velocity is the speed of the surface as seen from Earth, averaged over the disc of the star, and real work multiplies it by a factor of about 1.3 and takes off the speed of the star as a whole. Olympiad problems usually leave these out.

Example 8.14How large a Cepheid isfinish it

A Cepheid is 1.40 times as bright at maximum light as at minimum, and its temperature is 1.15 times as high. Between the two moments its surface moves inwards by $3.0 \times 10^{9}$ m. What is its radius at minimum light?

The start

By (8.13), with 1 for maximum and 2 for minimum, $R_{1} / R_{2} = \sqrt{1.40} / {1.15}^{2} = 1.1832 / 1.3225 = 0.8947$ , and $R_{2} - R_{1} = 3.0 \times 10^{9}$ m.

Your stepNow solve for

R_{2}

solar radii

Common errorsSection 8.7

The period-luminosity relation gives the mean absolute magnitude, so it must be compared with the mean apparent magnitude, not the brightest.
Classical Cepheids, type II Cepheids and RR Lyrae stars follow different relations. Before 1952 the two kinds of Cepheid were mixed up, and all distances beyond the Milky Way came out about half as large as they are.
A pulsating star is not brightest when it is largest. Its brightness follows its temperature, which peaks just after it is smallest.
Extinction makes a Cepheid look fainter than it is, so leaving it out puts the Cepheid too far away.

ExercisesSection 8.73 questions

8.7.1
What is the mean absolute visual magnitude of a classical Cepheid with a period of 3.00 days?
mag
8.7.2
An RR Lyrae star, with $M_{V} = + 0.6$ , has a mean visual magnitude of 15.6, with no extinction. How far away is it?
kpc
8.7.3
What keeps a Cepheid pulsating instead of settling down?

Olympiad problemsPast IOAA problems, restated2 problems

2016The pulsations of β Doradus20 points

Adapted from IOAA 2016 (Bhubaneswar), theory problem 6. We have restated it in our own words and kept the original data. The solution was last checked on 21 October 2026.

The Cepheid β Doradus pulsates with a period of 9.84 days. Assume that it is brightest when most contracted, with radius $R_{1}$ , and faintest when most expanded, with radius $R_{2}$ , that it stays spherical and that it radiates as a blackbody. Its bolometric magnitude varies from 3.46 to 4.08. Doppler measurements show that the surface moves in or out at an average speed of 12.8 km/s. The peak of its spectrum moves between 531.0 nm and 649.1 nm. Take the Sun's luminosity as $3.826 \times 10^{26}$ W.

(a) Find the ratio $R_{1} / R_{2}$ . [7 points]

(b) Find $R_{1}$ and $R_{2}$ in metres. [3 points]

(d) Find the distance of the star, in parsecs. [5 points]

I solved it without the solution

Answer(a) $R_{1} / R_{2} = 0.890$ . (b) $R_{1} = 4.42 \times 10^{10}$ m and $R_{2} = 4.96 \times 10^{10}$ m. (c) $F_{2} = 5.88 \times 10^{- 10}$ W/m². (d) About 315 pc. The official solution uses $- 26.72$ for the Sun, close to its visual magnitude, in place of its bolometric magnitude, which gives $6.51 \times 10^{- 10}$ W/m² and 298 pc.

2019The changing amplitude of an RR Lyrae star20 points

Adapted from IOAA 2019 (Keszthely), theory problem 8. We have restated it in our own words and kept the original data. The solution was last checked on 21 October 2026.

Some RR Lyrae stars show a slow, repeating change in the amplitude of their pulsation, called the Blazhko effect. An RR Lyrae star is observed at $λ_{1} = 500$ nm and $λ_{2} = 2000$ nm, which come from different layers of the star. Take the light at each wavelength as set by the radius and the blackbody intensity of its layer, and use Wien's approximation, $F (λ, T) \propto λ^{- 5} \exp (- C_{b} / (λ T))$ , with $C_{b} = h c / k \approx 0.0144$ m K.

(a) The temperature of each layer varies between $T_{1} = 6000$ K and $T_{2} = 7400$ K. Ignoring changes of radius, find the ratio of the amplitudes in magnitudes at the two wavelengths. [5 points]

(b) Find the peak-to-peak amplitude at $λ_{1}$ . [3 points]

(c) Ignoring changes of temperature, what does a change of radius between $0.9 ⟨ R ⟩$ and $1.05 ⟨ R ⟩$ add to the amplitude at any wavelength? [3 points]

(d) At the smallest amplitude of the Blazhko cycle the temperature varies only between 6100 K and 6900 K, while the radius changes little. Taking the temperatures of (a) for the largest amplitude, by how much does the amplitude change over the Blazhko cycle at each wavelength? [5 points]

(e) Which statements are true? There may be more than one. [4 points] (A) The Blazhko effect is easier to see in the infrared. (B) The changes of temperature dominate the visual light curve. (C) Ignoring changes of radius, the amplitude is inversely proportional to the wavelength. (D) Observations in several colours do not help in understanding the Blazhko effect.

I solved it without the solution

Answer(a) 4. (b) About 1.0 magnitude. (c) 0.33 magnitudes. (d) About 0.4 magnitudes at 500 nm and 0.1 at 2000 nm. (e) B and C.