Stefan-Boltzmann Law

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

$$P_{\rm tot}(T) = \int_0^\infty P(\omega) d\omega = \frac{\hbar... ...,2}} \int_0^\infty \frac{\omega^{ 3} d\omega} {\exp(\hbar \omega/k T) - 1},$$ (8.125)

or

$$P_{\rm tot}(T) = \frac{k^{ 4} T^{ 4} }{4\pi^{ 2} c^{ 2} \hbar^{ 3}} \int_0^\infty \frac{\eta^{ 3} d\eta}{\exp\eta -1},$$ (8.126)

where $\eta = \hbar \omega/k T$ . The previous integral can easily be looked up in standard mathematical tables. In fact,

$$\int_0^\infty \frac{\eta^{ 3} d\eta}{\exp\eta -1} = \frac{\pi^{ 4}}{15}.$$ (8.127)

(See Exercise 2.) Thus, the total power radiated per unit area by a black-body is

$$P_{\rm tot}(T) = \frac{\pi^{ 2}}{60} \frac{k^{ 4}}{c^{ 2} \hbar^{ 3}} T^{ 4} = \sigma T^{ 4}.$$ (8.128)

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

$$\sigma = \frac{\pi^{ 2}}{60} \frac{k^{ 4}}{c^{ 2} \hbar^{ 3}} = 5.67\times 10^{-8} {\rm W} {\rm m}^{-2} {\rm K}^{ -4},$$ (8.129)

is called 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

$$L_\odot = 4\pi R_\odot^{ 2} \sigma T_\odot^{ 4},$$ (8.130)

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

$$P_\oplus=L_\odot \frac{ \pi R_\oplus^{ 2}/r_\oplus^{ 2}}{4\pi}$$ (8.131)

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

$$L_\oplus = 4\pi R_\oplus^{ 2} \sigma T_\oplus^{ 4},$$ (8.132)

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 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

$$T_\oplus = \left(\frac{R_\odot}{2 r_\oplus}\right)^{1/2} T_\odot.$$ (8.133)

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.