PlanetPhysics/Stefan Boltzamann Law

Stefan-Boltzmann law
The Stefan-Boltzmann law, also known as Stefan's law, states that the total energy radiated per unit surface area of a black body in unit time, P is directly proportional to the fourth power of the black body's Thermodynamic temperature T (also called absolute temperature):

$$ P(T) = \epsilon \sigma T^{4} $$

The irradiance P has dimensions of power density (energy per time per square distance), and the SI units of measure are joules per second per square meter, or equivalently, watts per square meter. The SI unit for absolute temperature T is the kelvin. e is the emissivity of the blackbody; if it is a perfect blackbody e = 1.

The constant of proportionality $$\sigma$$, called the Stefan-Boltzmann constant or Stefan's constant, is non-fundamental in the sense that it derives from other known constants of nature. The value of the constant is

$$   \sigma=\frac{2\pi^5 k^4}{15c^2h^3}= 5.670 400 \times 10^{-8} \,\, [\textrm{W\,s}^{-1}\textrm{m}^{-2}\textrm{K}^{-4}].$$

Thus at 100 K the energy flux density is 5.67 W/m2, at 1000 K 56.7 kW/m2, etc.

The law was discovered experimentally by Jo\v{z}ef Stefan (1835-1893) in 1879 and derived theoretically, using thermodynamics, by Ludwig Boltzmann (1844-1906) in 1884. Boltzmann treated a certain ideal heat engine with the light as a working matter instead of the gas. This law is the only physical law of nature named after a Slovene physicist. The law is valid only for ideal black objects, the perfect radiators, called black bodies. Stefan published this law on March 20 in the article \"Uber die Beziehung zwischen der W\"armestrahlung und der Temperatur (On the relationship between thermal radiation and temperature) in the Bulletins from the sessions of the Vienna Academy of Sciences.

Derivation
The Stefan-Boltzmann law can be derived by integrating over all wavelengths the spectral intensity of a black body as given by Planck's radiation law.

$$ P(T)=\int_0^\infty I(\lambda,T) d\lambda$$

where $$I(\lambda,T)$$ is the amount of energy emitted by a black body at temperature T per unit surface per unit time per unit solid angle. The equation for $$I$$ comes From Planck's radiation law and is given as

$$ I(\lambda,T) = \frac{2 \pi c^2 h}{\lambda^5} \, \frac{1}{e^{hc/ \lambda kT} - 1} $$

which leaves us to integrate

$$ P(T)= 2 \pi c^2 h \int_0^\infty \, \frac{d\lambda}{\lambda^5(e^{hc/ \lambda kT} - 1)} $$

using u substitution by setting

$$u = \frac{hc}{\lambda kT} $$

so that

$$ du = -\frac{hc}{\lambda^2 kT} d\lambda $$ $$ d\lambda = -\frac{\lambda^2 kT}{hc} du$$

with the limits of integration changing to

$$ \lambda \, -> \infty, u \, -> 0 $$ $$ \lambda \, -> 0, u \, -> \infty $$

substituting this into the integral and yields

$$ P(T)= -2 \pi c^2 h \int_\infty^0 \, \frac{\lambda^2 kT du}{\lambda^5 hc(e^{u} - 1)} $$

simplifying a little and switching the limits of integration to get rid of the minus sign

$$ P(T)= 2 \pi c kT \int_0^\infty \, \frac{du}{\lambda^3(e^{u} - 1)} du $$

making sure we convert all $$\lambda's$$ to $$u's$$, use

$$ \lambda^3 = \left( \frac{hc}{u kT} \right )^3 $$

leaving us with

$$ P(T)= 2 \pi c kT \left( \frac{kT}{hc} \right )^3 \int_0^\infty \, \frac{u^3 du}{(e^{u} - 1)}$$

The theory needed to analytically solve this integral is beyond this article. Even looking up this integral in a table takes a few moments because the solution, given in [3], is defined as

$$ \int_0^{\infty} \frac{u^{n-1} dx}{e^u - 1} = \Gamma (n) \zeta (n) $$

where $$\Gamma (n) $$ is the gamma function and $$\zeta (n)$$ is the Riemann zeta function.

The values of the gamma function are simple for integers

$$ \Gamma (n) = (n - 1)! $$

so for the case of $$n = 4$$

$$ \Gamma (4) = (3)! = 3 \cdot 2 \cdot 1 = 6 $$

The values of the Riemann zeta function are more involved, but for even integers, we can use the theorem given on PlanetMath,

which lets us get $$\zeta(4)$$ from

$$\zeta(n)=\frac{2^{n-1}|B_n|\pi^n}{n!}$$

We still need $$B_n$$, the Bernoulli number, for $$n = 4$$, we can get this from [5], PlanetMath,

$$ B_4 = -\frac{1}{30} $$

so

$$\zeta(n)=\frac{2^{3}\pi^n}{30 \cdot 4!} = \frac{\pi^4}{90}$$

Finally, the integral solution is

$$ \int_0^{\infty} \frac{u^{n-1} dx}{e^u - 1} = \Gamma (n) \zeta (n) = 6 \cdot \frac{\pi^4}{90} = \frac{\pi^4}{15} $$

gathering all the constants from the original integral, we are left with

$$ P(T)= 2 \pi c kT \left( \frac{kT}{hc} \right )^3 \frac{\pi^4}{15}$$

simplifying yields the Stefan-Boltzmann law

$$ P(T) = \frac{2 \pi^5 k^4}{15 h^3 c^2}T^4 $$