Playground
Interactive glowing blackbody sphere: adjust temperature and surface area to see radiated power with color-shifting glow and a T^4 power curve.
Variables
| Symbol | Name | SI | Dimension | Range |
|---|---|---|---|---|
| Radiated poweroutput Total power radiated by the blackbody surface | W | M·L²·T⁻³ | 0 – 100000 | |
| Stefan-Boltzmann constant Fundamental constant relating temperature to radiated power | W/(m²·K⁴) | M·T⁻³·Θ⁻⁴ | 5.670374419e-8 – 5.670374419e-8 | |
| Surface area Total radiating surface area of the body | m² | L² | 0.001 – 100 | |
| Surface temperature Absolute temperature of the radiating surface | K | Θ | 100 – 10000 |
Deep dive
Derivation
Boltzmann (1884) derived P = σAT⁴ by combining thermodynamics with Maxwell's radiation pressure (p = u/3). From the first law applied to a photon gas in a cavity: u = aT⁴ where a = 4σ/c. Integrating the Planck spectrum over all wavelengths gives σ = 2π⁵k_B⁴/(15h³c²) = 5.670374419×10⁻⁸ W·m⁻²·K⁻⁴.
Experimental verification
Stefan inferred the T⁴ law (1879) from Tyndall's data on platinum wire radiation. Lummer and Pringsheim (1899) confirmed it using blackbody cavities. Modern pyrometers and bolometers verify the law to high precision for calibration standards.
Common misconceptions
- All objects radiate, not just hot ones — even an ice cube emits thermal radiation, just far less than a furnace
- The law gives total power across all wavelengths, not the peak wavelength (Wien's law does that)
- Real objects emit less than a blackbody by a factor of emissivity ε — shiny metals can have ε as low as 0.03
Real-world applications
- Stellar luminosity and surface temperature measurement in astrophysics
- Infrared thermography and thermal imaging cameras
- Earth's energy balance and climate modeling
- Industrial furnace and kiln design
Worked examples
Radiating blackbody sphere
Given:
- sigma:
- 5.670374419e-8
- r:
- 0.1
- T:
- 1000
Find: P
Solution
P = σAT⁴ = 5.670374419×10⁻⁸ × 0.1257 × 10¹² = 7126 W ≈ 7.13 kW
Sun's total luminosity
Given:
- sigma:
- 5.670374419e-8
- R_sun:
- 695700000
- T_sun:
- 5778
Find: P
Solution
P = σ(4πR²)T⁴ = 5.670374419×10⁻⁸ × 6.079×10¹⁸ × 1.114×10¹⁵ = 3.85×10²⁶ W
Scenarios
What if…
- scenario:
- What if the Sun's temperature doubled?
- answer:
- Luminosity would increase by 2⁴ = 16 times, from 3.85×10²⁶ W to ~6.2×10²⁷ W. Earth's equilibrium temperature would rise by a factor of 2⁴ᐟ⁴ ≈ 1.41, from ~288 K to ~406 K (133°C). Life as we know it would be impossible.
- scenario:
- What if emissivity is 0.5 instead of 1?
- answer:
- Radiated power halves. P = εσAT⁴, so a grey body with ε = 0.5 emits half the power of a blackbody at the same temperature. Polished metals (ε ~ 0.03-0.1) radiate far less — this is why thermos flasks use reflective coatings.
Limiting cases
- condition:
- T → 0
- result:
- P → 0
- explanation:
- At absolute zero, no thermal radiation is emitted.
- condition:
- T doubled
- result:
- P increases 16×
- explanation:
- The T⁴ dependence means small temperature increases cause dramatic power increases.
- condition:
- A → 0 (point source)
- result:
- P → 0
- explanation:
- Zero surface area means no radiation, regardless of temperature.
Context
Josef Stefan · 1879
Stefan deduced the T⁴ law experimentally in 1879. Boltzmann derived it theoretically in 1884 from thermodynamics and Maxwell's electrodynamics.
Hook
How can astronomers measure a star's temperature from billions of kilometers away?
A blackbody sphere of radius 0.1 m is at 1000 K. Find the total power it radiates (σ = 5.670374419×10⁻⁸ W·m⁻²·K⁻⁴).
Dimensions: [P] = [σ][A][T⁴] → M·L²·T⁻³ = (M·T⁻³·Θ⁻⁴)(L²)(Θ⁴) = M·L²·T⁻³ ✓
Validity: Exact for ideal blackbodies. For real objects, multiply by emissivity ε (0 < ε ≤ 1). Assumes thermal equilibrium and isotropic radiation.