Spectral lines, the Doppler shift and line broadening

Builds on: Sections 4.6 and S4
Reading time: 40 min
Exercises: 4
Olympiad problems: 1

In 1814 Joseph von Fraunhofer spread sunlight into a long spectrum and found it crossed by hundreds of fine dark lines. Two close lines in the yellow, at 589.0 and 589.6 nm, appear at the same wavelengths as the bright yellow light given out by salt in a flame, so they come from sodium. In 1868 astronomers watching an eclipse found a bright yellow line at 587.6 nm in the glowing gas at the edge of the Sun that matched no element known on Earth. It was named helium, after the Greek word for the Sun, and was found on Earth only in 1895. In this section we see how such lines form, how they sort the stars by temperature, and what the shape and position of a single line tell us about the star that made it.

4.9.1How lines form

Before you readMake a guess first

A cloud of thin hydrogen gas is heated and seen against a dark sky. What does its spectrum look like?

An atom can absorb or give out a photon only if its energy matches a jump between two of the atom's levels, as Section S4 showed for hydrogen. Each element has its own set of levels, and so its own set of wavelengths, which identify it as surely as a fingerprint. Three kinds of spectrum follow from this, first set out by Gustav Kirchhoff in 1859.

A hot, dense body, such as a lamp filament or the deep layers of a star, gives a continuous spectrum, light at every wavelength, close to the blackbody curve of Section 4.6.
A thin, hot gas seen against a dark background gives an emission-line spectrum, bright lines at the wavelengths of its atoms. The red glow of hydrogen in nebulae such as the Orion Nebula is one example.
A thin gas in front of a hotter source of continuous light absorbs at its own wavelengths and gives an absorption-line spectrum, a continuous spectrum crossed by dark lines.

The spectrum of a star is of the third kind. The deep layers of the star give continuous light, and the cooler gas of its outer layers absorbs at its own wavelengths. The light absorbed is given out again, but in all directions, so less of it travels on towards us at those wavelengths and the lines look dark.

4.9.2Spectral classes

The lines in a star's spectrum depend far more on its temperature than on what it is made of, since almost all stars are mostly hydrogen and helium. In a cool star the electrons of hydrogen sit in the lowest level, $n = 1$ , and cannot absorb the Balmer lines, which start from $n = 2$ . In a very hot star most hydrogen atoms have lost their electron altogether. The Balmer lines are therefore strongest in between, near 10 000 K. Other lines peak at other temperatures, those of ionised helium in the hottest stars and those of molecules such as titanium oxide in the coolest, which are the only stars cool enough for molecules to survive.

Astronomers sort stars into spectral classes by their lines. In order of falling temperature the classes are O, B, A, F, G, K and M, each divided into ten subclasses from 0 to 9, so the Sun, of class G2, is a little cooler than a G0 star.

Class	Temperature (K)	Strongest lines	Example
O	above 30 000	ionised helium	Alnitak, O9.5
B	10 000 to 30 000	neutral helium	Spica, B1
A	7500 to 10 000	hydrogen, the Balmer lines	Vega, A0
F	6000 to 7500	hydrogen, ionised calcium	Procyon, F5
G	5200 to 6000	ionised calcium, metals	the Sun, G2
K	3700 to 5200	neutral metals	Arcturus, K2
M	below 3700	titanium oxide	Betelgeuse, M2

A Roman numeral after the class gives the luminosity class, from I for supergiants through III for giants to V for ordinary stars like the Sun, which are called dwarfs. A giant has a thin, spread-out atmosphere, in which atoms collide less often than in the dense atmosphere of a dwarf, so its lines are narrower. The full type of Betelgeuse is M2 Ib, a cool supergiant, and that of the Sun is G2 V.

Quick checkThe peak of Procyon's light

Procyon, of class F5, has a surface temperature of about 6550 K. At what wavelength does its spectrum peak?

4.9.3The position of a line

A line from a moving star is shifted by the Doppler effect of (S.30). Comparing the wavelength of the line with its , measured in the laboratory, gives the star's radial velocity, the part of its velocity along our line of sight. Every line in the spectrum is shifted by the same fraction of its wavelength, which is how astronomers know that a shift is due to motion and not to some other cause.

Quick checkA shifted Hα line

The Hα line, of rest wavelength 656.28 nm, is seen at 656.50 nm in a star's spectrum. What is the star's radial velocity?

km/s

4.9.4The width of a line

A line is never infinitely sharp. The atoms in a star's atmosphere move in all directions with speeds set by the temperature (Section S4), and each atom absorbs at a wavelength shifted by its own Doppler shift. Together they spread the line over a range of wavelengths, an effect called thermal broadening. Its width depends on the typical speed of the atoms, $\sqrt{k T / m}$ , so hot gas and light atoms give wide lines.

ResultWidths of a spectral line(4.17)

\begin{aligned} Δ λ_{th} & = \frac{λ_{0}}{c} \sqrt{\frac{8 \ln 2 k T}{m}} \\ Δ λ_{rot} & = \frac{2 λ_{0} v \sin i}{c} \end{aligned}

$Δ λ_{th}$	the thermal width, the full width of the line at half its depth
$T, m$	the temperature of the gas and the mass of one of its atoms
$Δ λ_{rot}$	the full width that rotation gives the line, from edge to edge
$v \sin i$	the speed of the star's equator times the sine of the angle $i$ between its axis and our line of sight

The factor

8 \ln 2

comes from the bell-shaped spread of the speeds of the atoms. A rotating star has one edge coming towards us and the other moving away, so its lines are spread from

- v \sin i

+ v \sin i

. A star seen pole-on, with

i = 0

, shows no rotational broadening however fast it spins. Collisions between atoms widen the lines further, more in the dense atmospheres of dwarfs than in giants.

Figure 4.4One line of hydrogen in a star's spectruminteractive

Radial velocityTemperatureRotation

Centre of the line
Width from the heat of the gas
Width from rotation

The figure shows the brightness of a star's spectrum near the Hα line of hydrogen, as a fraction of the brightness on either side, with the line drawn heavy. The dashed line R marks the rest wavelength, 656.28 nm, and C is the centre of the line. The sliders set the star's radial velocity, the temperature of its gas and its rotation

v \sin i

. The line takes about the same amount of light out of the spectrum at every setting, so a wider line is shallower. A gives a narrow line only 0.047 nm wide, shifts the line 0.26 nm to the red without changing its shape, and , a speed common among hot stars, spreads the line almost 0.9 nm wide.

Example 4.17Thermal width in the Sun

What is the thermal width of the Hα line in gas at 5772 K, the temperature of the Sun's surface? The mass of a hydrogen atom is $1.674 \times 10^{- 27}$ kg.

Solution

By (4.17),

\begin{aligned} Δ λ_{th} & = \frac{656.28 nm}{2.998 \times 10^{8}} \sqrt{\frac{8 \ln 2 \times 1.381 \times 10^{- 23} \times 5772}{1.674 \times 10^{- 27}}} \\ = 0.036 nm . \end{aligned}

AnswerThe thermal width is about 0.036 nm, which matches a spread of speeds of 16 km/s. An iron line at the same wavelength would be

\sqrt{55.8 / 1.008} = 7.4

times narrower, since iron atoms are 55 times as heavy and move more slowly.

Example 4.18The width the Sun's rotation givesfinish it

The Sun's equator moves at about 2.0 km/s. Seen from far away in the plane of its equator, how wide would the rotation make the Hα line?

The start

Seen in the plane of the equator, $i = 90 °$ , so $v \sin i = 2.0$ km/s.

Your stepNow use (4.17).

Common errorsSection 4.9

A Doppler shift gives only the radial velocity. A star moving across our line of sight shows no shift at all, however fast it moves.
Dark lines do not mean that a star's gas is cold. They mean only that the outer layers are cooler than the layers below them.
The thermal width goes as $\sqrt{T / m}$ , so a gas four times as hot gives lines only twice as wide.

ExercisesSection 4.94 questions

4.9.1
The Hβ line, of rest wavelength 486.13 nm, is seen at 485.90 nm. What is the star's radial velocity? A star coming towards us has a negative radial velocity.
km/s
4.9.2
Rotation widens the Hα line of a star to 0.20 nm from edge to edge. What is its $v \sin i$ ?
km/s
4.9.3
The gas in one star's atmosphere is four times as hot as in another's. How many times as wide are its lines from thermal broadening alone?
times
4.9.4
In which spectral class are the Balmer lines of hydrogen strongest?

Olympiad problemsPast IOAA problems, restated1 problem

2023A line from a magnetic white dwarf5 points

Adapted from IOAA 2023 (Katowice), theory problem Q2. We have restated it in our own words and kept the original data. The solution was last checked on 20 October 2026.

An emission line of wavelength 600 nm is seen in the spectrum of a white dwarf. Assume that it comes from free electrons, moving much slower than light, that circle in the star's magnetic field.

(a) Find the magnetic flux density $B$ of the field.

(b) Estimate the wavelength of another line whose discovery would confirm that the lines come from the particles of a plasma circling in the field.

I solved it without the solution

AnswerThe field is about $2 \times 10^{4}$ T, hundreds of millions of times stronger than Earth's, and a line from protons should appear near 1.1 mm.