The last page said a hotter object radiates more — but how much more? The Stefan–Boltzmann law makes it startlingly precise: a black body’s total power depends on the fourth power of its absolute temperature. Nudge the temperature up a little and the power output rockets. It’s the rule that turns a star’s temperature and size into its luminosity.
📚 What you need to know
The total power (luminosity) radiated by a black body is L = σAT4.
σ = the Stefan–Boltzmann constant = 5.67 × 10−8 W m−2 K−4.
A = surface area (m2); T = absolute (surface) temperature — kelvin, always.
For a star (a sphere): A = 4πr2, so L = 4πr2σT4.
T4 dependence: double the temperature → 16× the power. Radiated power is hugely sensitive to temperature.
Combine with b = L/4πd2 to link temperature, size and distance to how bright a star looks.
Power that scales with T⁴
The heart of the law is that little exponent. Radiated power doesn’t just grow with temperature — it grows with temperature to the fourth power. That makes it dramatically sensitive: a small change in temperature produces an enormous change in power output.
Radiated power against temperature is a steep fourth-power curve. Double the temperature and the power doesn’t double — it goes up sixteenfold.
Let the fourth power sink in. Two stars the same size, one twice as hot as the other: the hotter one isn’t twice as luminous, it’s 24 = 16 times as luminous. Triple the temperature and it’s 34 = 81 times. Temperature utterly dominates a body’s power output.
The law, factor by factor
Written out, the law multiplies three things — a fixed constant, the surface area, and the temperature raised to the fourth:
Stefan–Boltzmann lawL = σAT4for a star: A = 4πr2 → L = 4πr2σT4
A star of radius r and surface temperature T radiates a total power L. Its surface area is A = 4πr2.
There are two routes to a big luminosity: be large (a big surface area A) or be hot (a high T). Both matter — a giant cool star and a small blazing one can rival each other — but because T enters to the fourth power and r only to the second, temperature is the far stronger lever.
Same size, different temperature
Hold the size fixed and let only temperature change, and the T4 effect stands out starkly:
Two stars of equal radius: the hotter one (twice the temperature) radiates sixteen times the power. It also looks whiter-blue — a preview of the next law.
This is why hot blue-white stars are such fierce beacons. Even a modest, ordinary-sized star will pour out staggering power if its surface runs hot — the fourth power sees to that. Their colour, meanwhile, hints at that temperature, which is exactly what Wien’s law (coming next) turns into a measurement.
Worked examples
WE 1
The Sun has radius 7.0 × 108 m and surface temperature 5800 K. Find its luminosity. (σ = 5.67 × 10−8 W m−2 K−4)
Surface area: A = 4πr² = 4π(7.0 × 108)²= 6.2 × 1018 m²Luminosity: L = σAT⁴= 5.67 × 10−8 × 6.2 × 1018 × 58004L ≈ 3.9 × 1026 WMatches the Sun’s known output — close to 4 × 1026 joules every second.
WE 2
A star has the same radius as the Sun but twice its surface temperature. How does its luminosity compare?
Same radius → same area A, so L ∝ T⁴.Twice the temperature → 2⁴ = 16.16× as luminous as the SunThe fourth power makes temperature the dominant factor — a modest rise means a huge jump in power.
WE 3
A star has luminosity 1.0 × 1028 W and surface temperature 10 000 K. Find its radius. (σ = 5.67 × 10−8)
L = 4πr²σT⁴, so r = √(L / (4πσT⁴))= √(1.0 × 1028 ÷ (4π × 5.67 × 10−8 × (10 000)4))= √(1.4 × 1018)r ≈ 1.2 × 109 mAbout 1.7 solar radii — a bigger, hotter and far more luminous star than the Sun.
🔧 Using L = σAT⁴
T in kelvin, area A in m2 (for a star, A = 4πr2).
Power:L = σAT4 — don’t forget to raise T to the fourth.
Rearrange for r, T or A as needed.
Ratios:L ∝ T4 (and ∝ A ∝ r2). ×2 T → ×16 L.
Combine with b = L/4πd2 for how bright it looks from distance d.
temperature T
raise to T4
× σ × A
=
luminosity L
Quick recap: A black body’s total radiated power is L = σAT4, with σ = 5.67 × 10−8 W m−2 K−4 and T in kelvin. For a star A = 4πr2, giving L = 4πr2σT4. Because power goes as T4, doubling the temperature multiplies the power by 16 — temperature dominates. Pair it with b = L/4πd2 to reach apparent brightness.
💡 Top tips
Fourth power, in kelvin. The two classic slips are dropping the exponent and using °C.
Surface, not circle. A star’s area is A = 4πr2, not πr2.
Ratios are quick.L ∝ T4: ×2 T → ×16; ×3 T → ×81.
Size counts too.L ∝ r2: double the radius → four times the luminosity.
Chain the two laws. Get L here, then b = L/4πd2 for brightness.
⚠ Common mistakes
Forgetting the fourth power — using T or T2 instead of T4
Using °C instead of kelvin for the temperature
Using πr2 (a circle) instead of 4πr2 (a sphere’s surface) for A
Dropping σ or its power of ten (5.67 × 10−8)
Confusing L (total power) with b (power per m2) — that’s the brightness law
This law gives a star’s total power from its temperature and size. But notice we keep needing that surface temperature — so how do we actually measure it, for something we can never visit? The answer is hiding in the star’s colour: the peak of its radiation curve. Next: Wien’s displacement law, which reads a body’s temperature straight off the wavelength it shines brightest at.
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