Stefan-Boltzmann Law

The total power radiated per unit area by a black-body at all frequencies is given by

$$P_{\rm tot}(T) = \int_0^\infty P(\omega) \, d\omega = \frac{\hbar... ...{2}} \int_0^\infty \frac{\omega^{3}\,d\omega} {\exp(\hbar\,\omega/k_B\,T) - 1},$$ (5.478)

or

$$P_{\rm tot}(T) = \frac{k_B^{\,4}\, T^{4} }{4\pi^{2}\, c^{2}\,\hbar^{3}} \int_0^\infty \frac{\eta^{3}\,d\eta}{{\rm e}^\eta -1},$$ (5.479)

where $\eta = \hbar\,\omega/k_B \,T$. The previous integral can be looked up in standard reference books on integrals. In fact,

$$\int_0^\infty \frac{\eta^{3}\,d\eta}{{\rm e}^\eta -1} = \frac{\pi^{4}}{15}.$$ (5.480)

Thus, the total power radiated per unit area by a black-body is

$$P_{\rm tot}(T) = \frac{\pi^{2}}{60} \frac{k_B^{\,4}}{c^{2}\, \hbar^{3}} \,T^{4} = \sigma \,T^{4}.$$ (5.481)

This $T^{4}$ dependence of the radiated power is called the Stefan-Boltzmann law, after Josef Stefan, who first obtained it experimentally 1877, and Ludwig Boltzmann, who first derived it theoretically in 1884. The parameter

$$\sigma = \frac{\pi^{2}}{60} \frac{k_B^{\,4}}{c^{2}\, \hbar^{3}} = 5.67\times 10^{-8}\,\,{\rm W} \,{\rm m}^{-2} \,{\rm K}^{\,-4},$$ (5.482)

is known as the Stefan-Boltzmann constant.

We can use the Stefan-Boltzmann law to estimate the temperature of the Earth from first principles. The Sun is a ball of glowing gas of radius $R_\odot\simeq 7\times 10^{\,5}$ km and surface temperature $T_\odot\simeq 5770$ K. Its luminosity is

$$L_\odot = 4\pi\, R_\odot^{\,2} \, \sigma\, T_\odot^{\,4},$$ (5.483)

according to the Stefan-Boltzmann law. The Earth is a globe of radius $R_\oplus\simeq 6000$ km located an average distance $r_\oplus\simeq 1.5\times 10^8$ km from the Sun. The Earth intercepts an amount of energy

$$P_\oplus=L_\odot\,\frac{ \pi \,R_\oplus^{\,2}/r_\oplus^{\,2}}{4\pi}$$ (5.484)

per second from the Sun's radiative output; that is, the power output of the Sun reduced by the ratio of the solid angle subtended by the Earth at the Sun to the total solid angle $4\pi$. The Earth absorbs this energy, and then re-radiates it at longer wavelengths. The luminosity of the Earth is

$$L_\oplus = 4\pi\, R_\oplus^{\,2} \, \sigma\, T_\oplus^{\,4},$$ (5.485)

according to the Stefan-Boltzmann law, where $T_\oplus$ is the average temperature of the Earth's surface. Here, we are ignoring any surface temperature variations between polar and equatorial regions, or between day and night. In a steady state, the luminosity of the Earth must balance the radiative power input from the Sun, so, equating $L_\oplus$ and $P_\oplus$, we arrive at

$$T_\oplus = \left(\frac{R_\odot}{2\,r_\oplus}\right)^{1/2} T_\odot.$$ (5.486)

Remarkably, the ratio of the Earth's surface temperature to that of the Sun depends only on the Earth-Sun distance and the solar radius. The previous expression yields $T_\oplus\simeq 279$ K or $6^\circ$ C (or $43^\circ$ F). This is slightly on the cold side, by a few degrees, because of the greenhouse action of the Earth's atmosphere, which was neglected in our calculation. Nevertheless, it is quite encouraging that such a crude calculation comes so close to the correct answer.