Stefan-Boltzmann Law

Equivalent forms

P = \varepsilon \sigma A T^4

\sigma = \frac{2\pi^5 k_B^4}{15 h^3 c^2} \approx 5.6704 \times 10^{-8}\,\text{W·m}^{-2}\text{K}^{-4}

One constant connects the kitchen oven, the Sun, and the cosmic microwave background.

Symbol	Name	SI unit	Dimension	Range
$j$	Radiant exitanceoutput Total power emitted per unit area	W/m²	M·T⁻³	0 – 1000000000000
$T$	Temperature Absolute temperature of the surface	K	Θ	1 – 10000000
$\sigma$	Stefan-Boltzmann constant Universal radiation constant	W·m⁻²·K⁻⁴	M·T⁻³·Θ⁻⁴	5.670374419e-8 – 5.670374419e-8

Unit systems

Symbol	SI	CGS	Imperial
$j$	W/m²	erg·s⁻¹·cm⁻²	BTU/(h·ft²)
$T$	K	K	°R
$\sigma$	W·m⁻²·K⁻⁴	—	—

Where it holds

Exact for ideal blackbodies. For real surfaces, multiply by emissivity

\varepsilon (T)

∈ [0,1]. Net radiative exchange between two bodies at

T_{1}

,

T_{2} is \varepsilon \sigma (T_{1}^{4} - T_{2}^{4})

. Breaks down at quantum optics scales (few photons, non-thermal sources).

Dimensional analysis

$[\sigma T^{4}] = (W\cdot m^{-2}\cdot K^{-4})\cdot K^{4} = W/m^{2} \checkmark$ (power per area)

Discovery

Josef Stefan & Ludwig Boltzmann · 1879

Stefan inferred T⁴ empirically from experiments. Boltzmann derived it thermodynamically in 1884 by treating radiation as a photon gas with pressure P = u/3. A rare case where thermodynamics nailed a result decades before quantum mechanics explained it.

Try this

Why does doubling a star's temperature make it 16 times brighter?

A blackbody radiates more intensely as it heats up — but the power scales as the fourth power of temperature. Why such a steep dependence, and what fixes the constant?

Research status: stable

Real-world applications

Climate science — Earth's equilibrium T set by balancing solar input $vs \sigma T^{4}$ output.
Stellar physics — luminosity $L = 4\pi R^{2}\sigma T^{4}$ classifies stars (H-R diagram).
Industrial furnaces — radiative heat transfer dominates above $\sim 600 ^{\circ}C$ .
Spacecraft thermal control — radiators sized $via \sigma T^{4} to$ dump waste heat to space.

Common misconceptions

$T^{4}$ comes from geometry — actually from integrating the Planck spectrum over all frequencies.
Stefan-Boltzmann gives net heat transfer — only emission; net transfer requires subtracting absorbed radiation from surroundings.
All bodies radiate equally well — emissivity matters: polished metals have $\varepsilon \sim 0.05$ , soot $\varepsilon \sim 0.95$ .

Experimental verification

Cavity-radiation experiments since the 1880s confirm

T^{4} to

high precision. Stellar luminosities derived from Stefan-Boltzmann match independent distance/parallax measurements. CMB total energy density

= aT^{4}

confirmed cosmologically.

Derivation

Integrate Planck's law:

j = \int _{0}^\infty \pi B(\lambda

,

T) d\lambda

.

Substituting

x = hc/(\lambda k_B T)

gives

j = (2\pi ^{5}k_B^{4}/(15h^{3}c^{2}))\cdot T^{4}

.

The integral

\int x^{3}/(e

ˣ-

1) dx = \pi ^{4}/15

supplies the numerical factor.

Limiting cases

T = 300\,\mathrm{K}

(room temperature)⟶

j \approx 459\,\mathrm{W/m}^{2}

Why people in a cold room feel chilled — net radiative loss from skin to walls.

T = 5778\,\mathrm{K}

(solar surface)⟶

j \approx 6.3 \times 10^{7} W/m^{2}

Sun's surface emits

63\,\mathrm{MW/m}^{2}

— multiplied by surface area gives total luminosity

3.8 \times 10^{26} W

.

T \to 0

⟶

j \to 0

Absolute zero emits no radiation (classically). Real materials with quantum zero-point fluctuations have tiny residual emission.

What if…

What if Sun's T were

2\times

higher (11556 K)?

Luminosity grows $16\times$ . Earth's equilibrium $T = 255 \times 2 = 510\,\mathrm{K}$ — boiling. Oceans evaporate; Venus-like greenhouse.

What

if \sigma

were

10\times

smaller?

Hot bodies cool $10\times$ slower. Stars would burn $10\times$ longer; Earth's IR balance breaks — climate dynamics radically change.

1

Solar luminosity

Given ·

T:: 5778
\sigma:: 5.670374419e-8

Find · L_sun (using

R_sun = 6.96e_{8} m)

Steps

$j = \sigma T^{4} = 5.67e-8 \times (5778)^{4} \approx 6.32e_{7} W/m^{2}$
Area: $4\pi \times (6.96e_{8})^{2} \approx 6.09e18\,\mathrm{m}^{2}$
$L = j \times A \approx 3.85e26\,\mathrm{W}$ — matches measured solar luminosity.

Result ·

L \approx 3.85 \times 10^{26} W

2

Earth's equilibrium temperature (no atmosphere)

Given ·

\sigma:: 5.670374419e-8

Find · T_eq

Steps

Solar flux at Earth: $S = 1361\,\mathrm{W/m}^{2}$ . Absorbed (after 30% albedo): $(1-0.3)\cdot S\cdot \pi R^{2}$ distributed over $4\pi R^{2}$ .
$Set \sigma T^{4} = (1-\alpha )S/4 \to T^{4} = 0.7 \times 1361 / (4\times 5.67e-8)$
$T = 255\,\mathrm{K}$ . Earth is warmer (288 K) due to greenhouse trapping.

Result ·

T_eq \approx 255\,\mathrm{K} (-18 ^{\circ}C)

Planck Radiation Law→Wien's Displacement Law→Mass-Energy Equivalence→