Use Stefan's law and Wien's displacement law to classify stars by luminosity, temperature, spectral class, and position on the Hertzsprung-Russell diagram.
AQA A-Level Physics · Astrophysics Option
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Use L = 4πr²σT⁴ (Stefan-Boltzmann law)
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Apply Wien's displacement law λ_max = b/T
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Interpret the Hertzsprung-Russell (HR) diagram
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Recall spectral classes O B A F G K M and their properties
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Distinguish absolute and apparent magnitude
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Use the distance-magnitude formula
Luminosity: Stefan-Boltzmann Law
A star radiates approximately as a black body. Its total power output (luminosity L) depends on its radius r and surface temperature T:
The Sun: L_☉ = 3.85 × 10²⁶ W, T_☉ ≈ 5778 K, r_☉ = 6.96 × 10⁸ m. Stars are often compared to solar values: L = n × L_☉, T in K.
Luminosity depends on T to the fourth power — a small increase in temperature produces a very large increase in luminosity.
Wien's Displacement Law
The peak wavelength of a star's black-body spectrum is inversely proportional to its temperature:
λ_max = b / T
Where:
λ_max = peak wavelength (m)
b = Wien's constant = 2.898 × 10⁻³ m K
T = surface temperature (K)
Star temperature
λ_max
Colour
30 000 K (O-type)
97 nm (UV)
Blue-white
10 000 K (A-type)
290 nm (near-UV)
White
5 800 K (G-type, Sun)
500 nm
Yellow
3 500 K (M-type)
828 nm (near-IR)
Red
Wien's law allows astronomers to determine a star's surface temperature from the colour of its light (or the peak wavelength of its spectrum).
Spectral Classification: OBAFGKM
Stars are classified into spectral types based on their surface temperature and the absorption lines in their spectrum:
Class
Temperature (K)
Colour
Example
O
>30 000
Blue
Mintaka
B
10 000–30 000
Blue-white
Rigel, Spica
A
7 500–10 000
White
Sirius, Vega
F
6 000–7 500
Yellow-white
Procyon
G
5 200–6 000
Yellow
Sun, Alpha Centauri A
K
3 700–5 200
Orange
Arcturus, Aldebaran
M
<3 700
Red
Betelgeuse, Proxima Centauri
Mnemonic: Oh Be A Fine Girl/Guy, Kiss Me. The spectral type is determined by which elements show absorption lines (hydrogen Balmer series, helium, metal lines) — these depend on temperature.
The Hertzsprung-Russell Diagram
The HR diagram plots luminosity (y-axis, log scale) against surface temperature (x-axis, decreasing to the right, log scale). Stars cluster into distinct regions:
Main sequence: A diagonal band from top-left (hot, luminous) to bottom-right (cool, dim). The Sun sits roughly in the middle. ~90% of stars are on the main sequence.
Red giants: Upper right — cool but luminous (large radius)
Supergiants: Top of diagram — extremely luminous
White dwarfs: Lower left — hot but dim (very small radius)
A star's position on the HR diagram reveals its size. Along lines of constant T, higher L means larger radius (from L = 4πr²σT⁴). Lines of constant radius slope diagonally across the HR diagram.
Apparent and Absolute Magnitude
Apparent magnitude m: how bright a star appears from Earth (depends on distance and luminosity). Scale: brighter stars have smaller (or more negative) magnitudes.
Absolute magnitude M: the apparent magnitude a star would have at exactly 10 parsecs (pc) from Earth. Depends only on luminosity.
m − M = 5 log₁₀(d/10) (d in parsecs)
This quantity (m − M) is called the distance modulus.
1 parsec (pc) = 3.086 × 10¹⁶ m = 3.26 light-years. The Sun's absolute magnitude is M = +4.83; its apparent magnitude is m = −26.7 (so close it appears far brighter than any other star).
Example 1: Star luminosity from Stefan's law
A star has surface temperature 12 000 K and radius 3.0 × 10⁹ m. Calculate its luminosity. (σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴)
r_A = 42 r_☉ — Star A is a red giant, 42 times larger than the Sun despite being cooler
Example 4: Distance from apparent and absolute magnitude
A star has apparent magnitude m = 8.5 and absolute magnitude M = 3.5. Calculate its distance in parsecs.
1 m − M = 5 log₁₀(d/10) → 8.5 − 3.5 = 5.0 = 5 log₁₀(d/10)
2 log₁₀(d/10) = 1.0 → d/10 = 10¹ = 10
3 d = 100 pc
d = 100 parsecs
Q1. A star has twice the radius of the Sun and the same surface temperature. What is its luminosity compared to the Sun?
Q2. The peak wavelength of a star's spectrum is 600 nm. What is its surface temperature? (b = 2.90 × 10⁻³ m K)
Q3. Where on the HR diagram would you find a white dwarf?
Q4. Which spectral class is the hottest?
Q5. A star has absolute magnitude M = 0 and apparent magnitude m = 5. What is its distance?
Challenge 1. Betelgeuse has surface temperature T = 3600 K and luminosity L = 1.26 × 10³¹ W. Calculate its radius in solar radii. (σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴; r_☉ = 6.96 × 10⁸ m)
L = 4πr²σT⁴ → r² = L/(4πσT⁴)
T⁴ = (3600)⁴ = 1.680 × 10¹⁴ K⁴
4πσT⁴ = 4π × 5.67 × 10⁻⁸ × 1.680 × 10¹⁴ = 1.196 × 10⁸ W/m²
r² = 1.26 × 10³¹ / 1.196 × 10⁸ = 1.053 × 10²³ m²
r = √(1.053 × 10²³) = 3.24 × 10¹¹ m
r/r_☉ = 3.24 × 10¹¹ / 6.96 × 10⁸ = 466 solar radii
Betelgeuse is ~466 times the radius of the Sun — a red supergiant that would engulf Mercury, Venus, Earth, and Mars if placed at the Sun's position.
Challenge 2. Two stars A and B have the same luminosity. Star A has temperature 3000 K and star B has temperature 6000 K. Calculate the ratio of their radii r_A/r_B. What types of stars might these be?
L_A = L_B → 4πr_A²σT_A⁴ = 4πr_B²σT_B⁴
r_A²/r_B² = T_B⁴/T_A⁴ = (T_B/T_A)⁴ = (6000/3000)⁴ = 2⁴ = 16
r_A/r_B = 4
Star A (T = 3000 K, large) is likely a red giant or sub-giant — cool but large.
Star B (T = 6000 K, smaller) is likely a G-type main sequence star like the Sun.
For the same luminosity, the cooler star must be 4× larger in radius to compensate for its lower surface brightness.
Challenge 3. A Cepheid variable star has a period of 10 days. Using the known period-luminosity relationship for Cepheids (assume L = 1000 L_☉ for a 10-day Cepheid), its apparent magnitude is measured as m = 10. Calculate (a) its absolute magnitude, and (b) its distance in parsecs. (L_☉ = 3.85 × 10²⁶ W; M_☉ = +4.83)
(b) m − M = 5 log₁₀(d/10): 10 − (−2.7) = 12.7 = 5 log₁₀(d/10)
log₁₀(d/10) = 2.54 → d/10 = 10^2.54 = 346.7
d = 3467 pc ≈ 3.5 kpc
This shows the power of Cepheid variables as "standard candles" for measuring distances to nearby galaxies.