Black-body radiators and how to read a black-body radiation curve

Wien's displacement law, λ_(max)T=2.898×10⁻³ m K, to find surface temperatures

The Stefan–Boltzmann law, L=σAT⁴, for the luminosity (power output) of a star

Intensity and the inverse-square law, I=L/4πd²

Distances: trigonometric parallax, the parsec and the light-year, and standard candles

The Hertzsprung–Russell (HR) diagram and how it shows the life cycle of stars of different masses

1. A black body absorbs all radiation falling on it and emits a continuous spectrum whose shape depends only on its temperature, and stars are close to black bodies.

2. Wien: λ_(max)T=2.898×10⁻³ m K; hotter stars peak at shorter wavelengths (bluer). T is in kelvin.

3. Stefan–Boltzmann: L=σAT⁴ with A=4πr²; doubling the temperature multiplies the power by 16.

4. Inverse-square law: I=L/4πd². Measure I; know d and you find L; know L (a standard candle) and you find d.

5. Parallax: a nearby star shifts against distant stars as the Earth orbits; d=r/tan p. A star at 1 parsec has p = 1 arcsecond.

6. HR diagram: luminosity against temperature (hot on the left). Main sequence, red giants, supergiants and white dwarfs sit in distinct regions; a star's mass decides its path and its end.

Three equations and two diagrams unlock almost everything we know about stars. The colour of a star gives its temperature (Wien), temperature and size give its power output (Stefan–Boltzmann), and power and distance give how bright it looks from Earth (inverse-square law). All three rest on one idealisation.

Every object above absolute zero emits electromagnetic radiation. A black body is the perfect emitter as well as the perfect absorber: no object at the same temperature emits more at any wavelength.

The spectrum is continuous (all wavelengths are present), but the power is not spread evenly: the curve of intensity against wavelength rises steeply to a peak at λ_(max) and falls away more gradually at long wavelengths.

Stars are a good approximation to black bodies, so their continuous spectra can be used to find their surface temperatures.

Reading a black-body curve

Hotter → curve higher at every wavelength. A hotter body emits more at every wavelength, so the curves never cross.

Hotter → peak at a shorter wavelength. The peak moves from the infrared (cool stars) through the visible to the ultraviolet (very hot stars). This is why cool stars look red and hot stars look blue-white.

Area under the curve = total power emitted per unit area, which is proportional to T⁴.

The Sun's peak, near 501 nm, lies in the visible region; a filament lamp at about 3000 K peaks in the infrared, so most of its output is heat, not light.

λ_(max) is the wavelength at which the intensity is greatest (m) and T the surface temperature (K). Wien's law is in the formulae list. Since λ_(max)∝1/T, doubling the temperature halves the peak wavelength. The temperature must be in kelvin (T/K=θ/°C+273).

Measure the spectrum of the star, find the wavelength at the peak, and divide 2.898×10⁻³ m K by it.

This gives the surface temperature: the temperature of the layer the light escapes from, not the much hotter core.

WORKED EXAMPLE · QUESTION

(a) The spectrum of a blue-white star has its peak intensity at a wavelength of 380 nm. Calculate the surface temperature of the star. (b) A red dwarf star has a surface temperature of 3200 K. Calculate the wavelength at which its emission peaks and state the region of the electromagnetic spectrum it lies in.

SOLUTION

(a) Rearrange Wien's law, with the wavelength in metres: T=2.898×10⁻³÷380×10⁻⁹= 7.63 × 10³ K

(b) λ_(max)=2.898×10⁻³÷3200=9.06×10⁻⁷ m (906 nm). This is longer than 700 nm (the visible range is about 400–700 nm), so the peak is in the infrared. The star still emits visible light, mostly at the red end, so it looks red.

L is the luminosity (W), σ the Stefan–Boltzmann constant =5.67×10⁻⁸ W m⁻² K⁻⁴, A the surface area (m²) and T the surface temperature (K). The Stefan–Boltzmann law is in the formulae list. For a spherical star of radius r, A=4πr², so L=4πr²σT⁴; the area formula is not in the list (No — learn it).

L∝r² and L∝T⁴. Temperature has the stronger effect: doubling T multiplies L by 2⁴=16; doubling r multiplies it by 4.

A cool star can still be very luminous if it is very large (a red giant); a hot star can be dim if it is very small (a white dwarf).

Astronomers often quote luminosity in units of the Sun's luminosity, L_⊙=3.83×10²⁶ W (radius R_⊙=6.96×10⁸ m, surface temperature ≈ 5780 K). These are not on the data list: a question will give them.

WORKED EXAMPLE · QUESTION

A star has a radius of 1.7 × 10⁹ m and a surface temperature of 8500 K. Calculate its luminosity and express it as a multiple of the Sun's luminosity, 3.83 × 10²⁶ W.

SOLUTION

A=4π(1.7×10⁹)²=3.63×10¹⁹ m²

L=σAT⁴=5.67×10⁻⁸×3.63×10¹⁹×8500⁴=1.07×10²⁸ W≈ 1.1 × 10²⁸ W

1.07×10²⁸÷3.83×10²⁶= 28 L☉. The radius is given to 2 s.f., so the luminosity is given to 2 s.f.

WORKED EXAMPLE · QUESTION

A red giant with a surface temperature of 3600 K and a white dwarf with a surface temperature of 18 000 K have the same luminosity. Show that the radius of the red giant is 25 times the radius of the white dwarf, and explain why equal luminosity does not mean that the two stars look equally bright from Earth.

SOLUTION

Equal luminosity, so 4πr₁²σT₁⁴=4πr₂²σT₂⁴; cancel 4πσ: r₁/r₂=(T₂/T₁)²=(18 000/3600)²=5²= 25

How bright a star looks depends on the intensity at Earth, I=L/4πd², so equal luminosities look equally bright only if the stars are at the same distance.

Ratio methods are quick and avoid big powers of ten: write the equation for both stars and divide.

The star's power spreads out over the surface of a sphere of radius d, area 4πd². The equation assumes nothing absorbs the radiation on the way (no dust or gas in between).

Measured on Earth: the intensity I (with a detector of known area) and the spectrum (so λ_(max) and T).

Then: if d is known, L=4πd²I; and with T from Wien, the radius follows from L=4πr²σT⁴.

Example: the Sun's luminosity spread over a sphere of radius 1 AU = 1.50 × 10¹¹ m gives about 1.35 × 10³ W m⁻² above the Earth's atmosphere.

WORKED EXAMPLE · QUESTION

A star is 2.6 × 10¹⁷ m from Earth. The intensity of its radiation at the Earth is 2.2 × 10⁻⁹ W m⁻², and its spectrum peaks at 420 nm. Calculate (a) its luminosity, (b) its surface temperature and (c) its radius. (d) The Sun's radius is 6.96 × 10⁸ m. Suggest which type of star this is.

SOLUTION

(a) L=4πd²I=4π×(2.6×10¹⁷)²×2.2×10⁻⁹=1.87×10²⁷ W≈ 1.9 × 10²⁷ W

(b) T=2.898×10⁻³÷420×10⁻⁹= 6900 K

(c) r=√(L/4πσT⁴)=√((1.87×10²⁷)/(4π×5.67×10⁻⁸×6900⁴))=1.08×10⁹ m≈ 1.1 × 10⁹ m

(d) About 1.5 times the Sun's radius, about 4.9 times its luminosity and a little hotter: a main sequence star somewhat more massive than the Sun.

Keep the unrounded luminosity in your calculator for (c); the square root halves the effect of any rounding, but the fourth power of T magnifies slips in T.

Hold a finger at arm's length and close each eye in turn: the finger jumps against the background. Trigonometric parallax is the same effect with the Earth's orbit as the baseline. As the Earth moves round its orbit, a nearby star appears to shift its position against the background of much more distant stars (which barely move).

Measure the star's angular position, then again six months later, when the Earth is on the other side of the Sun, 2 AU away. Half the total angular shift is the parallax angle p.

The parallax angle is the angle at the star subtended by the radius of the Earth's orbit, r = 1 AU, so tan p=r/d.

d is the distance to the star (m), r the radius of the Earth's orbit = 1 AU = 1.50 × 10¹¹ m and p the parallax angle. This is not in the formulae list (No — learn it (trigonometry)). Parallax angles are tiny, always less than 1 arcsecond (1 arcsecond = 1/3600 of a degree), so tan p≈p in radians.

Because d∝1/p: d in parsecs = 1 ÷ (p in arcseconds). A parallax of 0.5 arcseconds means 2 pc.

Limit: for distant stars the angle is too small to measure reliably against the uncertainty, so parallax works only for relatively nearby stars (in our part of the galaxy).

Telescopes in space, above the blurring atmosphere, extend the range.

WORKED EXAMPLE · QUESTION

The parallax angle of a star is measured as 0.29 arcseconds. 1 AU = 1.50 × 10¹¹ m and 1 light-year = 9.46 × 10¹⁵ m. Calculate the distance to the star (a) in parsecs, (b) in metres, (c) in light-years.

SOLUTION

(a) d=1÷0.29= 3.4 pc

(b) Convert the angle: p=0.29÷3600=8.06×10⁻⁵ degrees =1.41×10⁻⁶ rad. Then d=1.50×10¹¹÷tan(1.41×10⁻⁶)=1.07×10¹⁷ m≈ 1.1 × 10¹⁷ m

(c) 1.07×10¹⁷÷9.46×10¹⁵= 11 ly

Check: 3.45 pc×3.09×10¹⁶ m=1.07×10¹⁷ m, the same as (b).

Measure the intensity I at Earth; with L known, rearrange the inverse-square law: d=√(L/4πI).

Cepheid variable stars pulsate, and their luminosity is linked to their period of pulsation: measure the period, deduce L. The link was calibrated using Cepheids whose distance was known by parallax. They work out to other nearby galaxies.

Type Ia supernovae (exploding white dwarfs) all reach about the same peak luminosity, billions of times the Sun's. They are visible across billions of light-years.

Each method is calibrated by the one before it (parallax → Cepheids → supernovae), so uncertainties build up as distances grow.

WORKED EXAMPLE · QUESTION

A type Ia supernova in a distant galaxy is observed. At its brightest it has a luminosity of 2.0 × 10³⁶ W, and the peak intensity measured at Earth is 7.4 × 10⁻¹⁶ W m⁻². (a) Calculate the distance to the galaxy in metres and in light-years (1 ly = 9.46 × 10¹⁵ m). (b) Explain why parallax could not be used for this galaxy.

SOLUTION

(a) d=√((2.0×10³⁶)/(4π×7.4×10⁻¹⁶))=1.47×10²⁵ m≈ 1.5 × 10²⁵ m; 1.47×10²⁵÷9.46×10¹⁵= 1.6 × 10⁹ ly

(b) At this distance the parallax angle (about 2 × 10⁻⁹ arcseconds) would be far too small to measure.

A Hertzsprung–Russell (HR) diagram is a plot of luminosity (vertical, logarithmic scale, often in units of L_⊙) against surface temperature (horizontal, logarithmic, with temperature decreasing from left to right). Stars do not fall randomly: most lie on a diagonal band, the main sequence, and the rest gather in a few distinct groups. Because L=4πr²σT⁴, every point on the diagram also has a radius: lines of constant radius run diagonally, and stars get bigger towards the top right.

A star forms when a cloud of gas and dust (a nebula) collapses under gravity. The protostar heats up until hydrogen fusion begins in its core and it joins the main sequence. Where it joins depends on its mass.

On the main sequence the outward pressure from the hot core balances the inward pull of gravity, and the star stays in nearly the same place on the diagram for most of its life. More massive stars use their fuel much faster, so they have shorter lives.

When the hydrogen in the core runs out, the core contracts and heats, the outer layers expand and cool, and the star moves up and to the right on the diagram.

Low-mass stars (up to about 8 solar masses, including the Sun)

Main sequence → red giant (helium fusion begins in the core; the outer layers swell to perhaps a hundred times their former radius).

The outer layers drift away as a planetary nebula, exposing the hot core, which is seen briefly at the top left of the diagram.

The core becomes a white dwarf: no fusion, very dense, about Earth-sized. It cools and fades, moving down and to the right along a line of almost constant radius.

High-mass stars (more than about 8 solar masses)

Main sequence (high and to the left) → red supergiant, fusing heavier and heavier elements in shells around the core, up to iron.

When the core can release no more energy by fusion it collapses in a fraction of a second and the star explodes as a supernova, briefly outshining a whole galaxy.

The remnant core becomes a neutron star (core about 1.4 to 3 solar masses: incredibly dense, made almost entirely of neutrons) or, if more than about 3 solar masses, a black hole, whose gravity is so strong that not even light can escape.

WORKED EXAMPLE · QUESTION

A star of 20 solar masses and a star of 1 solar mass form from the same nebula. Describe and compare how the two stars will change during their lives, referring to their positions on a Hertzsprung–Russell diagram.

SOLUTION

A full answer makes these points, linking each stage to the diagram:

Both stars join the main sequence when hydrogen fusion starts in the core; the 20 solar mass star joins at the top left (hotter, far more luminous), the 1 solar mass star lower down, near the middle.

The more massive star uses its hydrogen much faster, so it leaves the main sequence first (after millions rather than billions of years).

When core hydrogen runs out, both stars expand and their surfaces cool: they move to the right. The Sun-like star becomes a red giant, the massive star a red supergiant near the top right.

The 1 solar mass star sheds its outer layers as a planetary nebula; its core becomes a white dwarf at the bottom left, which cools and fades.

The 20 solar mass star fuses heavier elements until its core collapses, and it explodes as a supernova.

Its remnant core becomes a neutron star or, if the core mass is more than about 3 solar masses, a black hole; neither appears on the HR diagram.

Photons and line spectra (Unit 2)

The dark absorption lines on a star's continuous spectrum identify the elements in its outer layers; Wien's law uses the continuous spectrum underneath them. Intensity as power per unit area was first met with waves.

Thermal physics and 5.3 Nuclear fusion

Wien's law and the Stefan–Boltzmann law need absolute temperatures in kelvin, the same scale used for the ideal gas. The energy a star radiates as luminosity comes from fusion in its core; the mass of the star decides how fast it is used up.

7 Gravitation and 5.9 Cosmology

Gravity drives the collapse that forms a star and ends its life; standard candles give the distances that, with redshift, reveal the expanding universe.

1. What is meant by a black body?

2. A star has a surface temperature of 10 000 K. At what wavelength does its emission peak?

3. Star X has twice the radius and twice the surface temperature of star Y. How many times more luminous is X?

4. A star has a parallax angle of 0.05 arcseconds. How far away is it in parsecs, and why would this measurement be harder than for a star at 2 pc?

5. Where on an HR diagram are white dwarfs, and what does their position tell you about their size?

1. An idealised object that absorbs all the electromagnetic radiation falling on it; its emitted spectrum depends only on its temperature.

2. λ_(max)=2.898×10⁻³÷10 000= 2.90 × 10⁻⁷ m (290 nm), in the ultraviolet.

3. L∝r²T⁴=2²×2⁴= 64 times.

4. d=1÷0.05= 20 pc. The angle is ten times smaller than for a star at 2 pc, so the same angular uncertainty is a much larger percentage of it.

5. Bottom left: hot but dim. For a high temperature to give a low luminosity in L=4πr²σT⁴, the radius must be very small, about the size of the Earth.