9 minute lesson · Physics · Original foundation material
Before you start: Surface area of a sphere; Powers and ratios; Temperature in kelvin.
A star's luminosity describes the power it emits. The flux at an observer describes power arriving per unit area. Keeping these quantities separate explains why a powerful distant star can look faint.
Spread one luminosity over a sphere
Imagine a source radiating equally in every direction, with no intervening absorption. Its luminosity L crosses every surrounding sphere. At distance d the sphere has area 4πd², so the received flux is F = L/(4πd²).
For the same source, doubling d reduces F to one quarter. No energy has been destroyed in this model; the same outgoing power is shared across a larger area. L is measured in watts and F in watts per square metre.
Connect size and temperature
An ideal blackbody surface emits power σT⁴ per unit area, where T is absolute temperature in kelvin and σ is the Stefan–Boltzmann constant. For a spherical blackbody of radius R, L = 4πR²σT⁴.
When comparing two such stars, L2/L1 = (R2/R1)² × (T2/T1)⁴. At unchanged radius, doubling temperature multiplies luminosity by 16. This blackbody model approximates a star's overall radiation; detailed spectra contain additional features.
A worked example
Star B has twice the luminosity of star A and is four times farther from us. Ignoring absorption, find FB/FA.
- Write flux as F = L/(4πd²) for each star.
- Divide the expressions: FB/FA = (LB/LA) × (dA/dB)².
- Insert the given ratios: FB/FA = 2 × (1/4)² = 2/16.
- Simplify to 1/8. Star B delivers one eighth of A's flux despite its larger luminosity.
FB/FA = 1/8
Check your understanding
Two spherical blackbodies have the same temperature, but B has three times A's radius. What is LB/LA?
Show the explanation
9. The temperature factor is 1 and the area factor is (3/1)² = 9.