The big idea: A star radiates like a black body. Two laws read its light: one gives its temperature, the other its total power.
Metal in a furnace: Watch metal in a furnace: as it heats it glows from dull red to orange to brilliant white. Its colour betrays its temperature. A star does the same.
The two laws
- The colour of the peak tells you how hot the surface is — that is Wien's law
- The total power it pours out depends on its size and temperature — that is the Stefan-Boltzmann law
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New words, plainly
- Black body = an ideal object that absorbs all radiation falling on it and glows with a characteristic spectrum set only by its temperature. A star is a good approximation.
- Peak wavelength λmax = the wavelength at which the star radiates most intensely — the top of its curve.
- Luminosity L = the total power the star radiates, in watts (W). (Brightness as seen from Earth is something else — that fades with distance.)
Wien's law — the COLOUR
- Looks at where the peak is (the colour)
- Hotter star → peak at a shorter wavelength (bluer)
- Gives you the surface temperature T
Stefan-Boltzmann — the POWER
- Looks at the total power radiated (luminosity L)
- Depends on the area and on T⁴ — a tiny T rise is huge
- Links luminosity, radius and temperature together
Hotter = bluer AND brighter: As a star gets hotter two things happen at once on its curve:
- the peak slides to a shorter wavelength (it looks bluer) — Wien, - the whole curve rises (it radiates far more power) — Stefan-Boltzmann.
So a blue star is both hotter and, size for size, much more luminous than a red one.
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Measure where the spectrum peaks, λmax, and Wien's law gives you the star's surface temperature straight away. Shorter peak ⇒ hotter star.
- peak wavelength — where the star's radiation is most intense (m)
- surface (absolute) temperature of the star (K)
- Wien's constant, 2.9 × 10⁻³ m K (given)
Watch the units: Put λmax in metres (a wavelength of 480 nm is 480 × 10⁻⁹ m) and T comes out in kelvin (K).
Rearrange the given law for whichever you need:
- temperature from the peak: T = 2.9 × 10⁻³ ÷ λmax - peak from the temperature: λmax = 2.9 × 10⁻³ ÷ T
A star's black-body spectrum peaks at a wavelength of λmax = 480 nm. Determine the surface temperature of the star. (Wien's constant = 2.9 × 10⁻³ m K.)
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The Stefan-Boltzmann law gives the total power a star radiates. For a star the radiating area is the surface of a sphere, A = 4πR², so the law connects luminosity, radius and temperature in one equation.
- luminosity — the total power the star radiates (W)
- Stefan-Boltzmann constant, 5.67 × 10⁻⁸ W m⁻² K⁻⁴ (given)
- surface area of the star; for a sphere A = 4πR² (m²)
- surface (absolute) temperature of the star (K)
The T⁴ is the key: Because the temperature is raised to the fourth power, it dominates. Double the temperature and the luminosity jumps by 2⁴ = 16 times (radius unchanged).
The radius only enters as R² (double R → 4× the luminosity).
For comparing two stars, use the proportional form L ∝ R² T⁴ — the constants cancel.
| If you change… | What happens to luminosity L |
|---|---|
| Double the radius R (T fixed) | Area is 4πR², so L goes up 4× (L ∝ R²) |
| Double the temperature T (R fixed) | L goes up 16× — because of the T⁴ |
| Double both R and T | L goes up 4 × 16 = 64× |
Comparing two stars — the ratio trick: When a question gives one star's values and asks about a second, divide the two Stefan-Boltzmann equations so σ and 4π cancel:
LB ÷ LA = (RB ÷ RA)² × (TB ÷ TA)⁴
Then rearrange for the radius ratio: RB ÷ RA = √(LB ÷ LA) ÷ (TB ÷ TA)².
From telescope data to a star's radius
Work in solar units: Divide by the Sun's values and σ and 4π cancel: L/L☉ = (R/R☉)² × (T/T☉)⁴. A star of 0.9 R☉ at 5200 K: L/L☉ = 0.9² × (5200/5800)⁴ = 0.81 × 0.65 = 0.52, about half the Sun's.
Star Y has 4 times the luminosity and 2 times the surface temperature of Star X. Determine the ratio of Star Y's radius to Star X's radius.
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How this is tested — the two laws appear on both papers, almost always together:
Paper 1A
- A one-step MCQ — read off which star is hotter from its peak.
- Or find a radius ratio of two stars from their luminosities and temperatures (L ∝ R²T⁴).
Paper 2
- Show that the surface temperature is about a stated value from the spectrum peak (Wien).
- Then estimate the radius by combining the laws (often via b = L/4πd² to get L first).
The classic trap: Forgetting the T⁴. When you take the ratio, the temperature ratio is raised to the fourth power and the radius ratio to the second — and λmax must be in metres for Wien.
Star X has a black-body peak at λmax = 580 nm and Star Y has a peak at λmax = 290 nm; Star Y is also 4 times as luminous as Star X. (a) Determine the surface temperature of each star. (b) Determine the ratio of Star Y's radius to Star X's radius. (Wien's constant = 2.9 × 10⁻³ m K.)
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