Stefan-Boltzmann Law

Equivalent forms

j^* = \sigma T^4

P = \epsilon \sigma A T^4

P_{\text{net}} = \sigma A (T^4 - T_{\text{env}}^4)

A T⁴ dependence — double the temperature and you radiate sixteen times the power. This single exponent shapes stellar evolution and climate science.

Symbol	Name	SI unit	Dimension	Range
$P$	Radiated poweroutput Total power radiated by the blackbody surface	W	M·L²·T⁻³	0 – 100000
$sigma$	Stefan-Boltzmann constant Fundamental constant relating temperature to radiated power	W/(m²·K⁴)	M·T⁻³·Θ⁻⁴	5.670374419e-8 – 5.670374419e-8
$A$	Surface area Total radiating surface area of the body	m²	L²	0.001 – 100
$T$	Surface temperature Absolute temperature of the radiating surface	K	Θ	100 – 10000

Unit systems

Symbol	SI	CGS	Imperial
$P$	W	erg/s	BTU/h
$sigma$	W/(m²·K⁴)	erg/(s·cm²·K⁴)	BTU/(h·ft²·°R⁴)
$A$	m²	cm²	ft²
$T$	K	K	°R

Where it holds

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

\varepsilon (0

<

\varepsilon \le 1)

. Assumes thermal equilibrium and isotropic radiation.

Dimensional analysis

$[P] = [\sigma ][$ A $][T^{4}] \to M\cdot L^{2}\cdot T^{-3} = (M\cdot T^{-3}\cdot \Theta ^{-4})(L^{2})(\Theta ^{4}) = M\cdot L^{2}\cdot T^{-3} \checkmark$

Discovery

Josef Stefan · 1879

Stefan deduced the T⁴ law experimentally in 1879. Boltzmann derived it theoretically in 1884 from thermodynamics and Maxwell's electrodynamics.

Try this

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).

Research status: stable

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

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 $\varepsilon$ — shiny metals can have $\varepsilon as low$ as 0.03

Experimental verification

Stefan inferred the

T^{4} 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.

Derivation

Boltzmann (1884) derived

P = \sigma AT^{4} by

combining thermodynamics with Maxwell's radiation pressure

(p = u/3)

.

From the first law applied to a photon gas in a cavity:

u = aT^{4}

where

a = 4\sigma /c

.

Integrating the Planck spectrum over all wavelengths gives

\sigma = 2\pi ^{5}k_B^{4}/(15h^{3}c^{2}) = 5.670374419\times 10^{-8} W\cdot m^{-2}\cdot K^{-4}

.

Limiting cases

T \to 0

⟶

P \to 0

At absolute zero, no thermal radiation is emitted.

T doubled⟶ P increases

16\times

The

T^{4}

dependence means small temperature increases cause dramatic power increases.

A \to 0

(point source)⟶

P \to 0

Zero surface area means no radiation, regardless of temperature.

What if…

What if the Sun's temperature doubled?

Luminosity would increase by $2^{4} = 16$ times, from $3.85\times 10^{26} W$ $to \sim 6.2\times 10^{27} W$ . Earth's equilibrium temperature would rise by a factor of $2^{4}$ ᐟ $^{4} \approx 1.41$ , from $\sim 288\,\mathrm{K}$ $to \sim 406\,\mathrm{K} (133^{\circ}C)$ . Life as we know it would be impossible.

What if emissivity is 0.5 instead of 1?

Radiated power halves. $P = \varepsilon \sigma AT^{4}$ , so a grey body with $\varepsilon = 0.5$ emits half the power of a blackbody at the same temperature. Polished metals $(\varepsilon \sim 0.03-0.1)$ radiate far less — this is why thermos flasks use reflective coatings.

1

Radiating blackbody sphere

Given ·

sigma:: 5.670374419e-8
r:: 0.1
T:: 1000

Find · P

Steps

Surface area: $A = 4\pi r^{2} = 4\pi (0.1)^{2} = 0.1257\,\mathrm{m}^{2}$
$T^{4} = (1000)^{4} = 10^{12} K^{4}$
$P = 5.670374419\times 10^{-8} \times 0.1257 \times 10^{12} = 7126\,\mathrm{W}$

Result ·

P = \sigma AT^{4} = 5.670374419\times 10^{-8} \times 0.1257 \times 10^{12} = 7126\,\mathrm{W} \approx 7.13\,\mathrm{kW}

2

Sun's total luminosity

Given ·

sigma:: 5.670374419e-8
R sun:: 695700000
T sun:: 5778

Find · P

Steps

$A = 4\pi R^{2} = 4\pi (6.957\times 10^{8})^{2} = 6.079\times 10^{18} m^{2}$
$T^{4} = (5778)^{4} = 1.114\times 10^{15} K^{4}$
$P = 5.670374419\times 10^{-8} \times 6.079\times 10^{18} \times 1.114\times 10^{15} \approx 3.85\times 10^{26} W$ (solar luminosity)

Result ·

P = \sigma (4\pi R^{2})T^{4} = 5.670374419\times 10^{-8} \times 6.079\times 10^{18} \times 1.114\times 10^{15} = 3.85\times 10^{26} W

Fourier's Law of Heat Conduction→Carnot Efficiency→Maxwell-Boltzmann Speed Distribution→