- #1

Airton Rampim

- 6

- 1

$$ \rho_T = \intop_0^{\infty}\rho(\nu, T)\mathrm{d}\nu = \frac{8\pi h}{c^3} \intop_0^\infty \frac{\nu^3}{\exp\left(\frac{h\nu}{k T}\right) - 1}\mathrm{d} \nu $$

$$ \rho_T = \frac{8\pi h}{c^3}\left(\frac{k T}{h}\right)^4 \underbrace{\intop_0^\infty \frac{u^3}{\exp\left(u\right) - 1}\mathrm{d} u}_{\zeta(4)\Gamma(4) = \pi^4/15} = \frac{8\pi^5 k^4}{15 h^3 c^3}T^4 $$

At equilibrium, these modes will have the same energy of the blackbody that is proportional to the fourth power of the temperature. This is the same result predicted by Stefan-Boltzmann law for the radiance of a black body

$$ R_T = \intop_0^\infty R(\nu, T)\mathrm{d}\nu = \sigma T^4, $$

where ##R(\nu, T)\mathrm{d}\nu## is the radiance emitted by frequencies between ##\nu## and ##\nu + \mathrm{d}\nu## and ##\sigma## is the Stefan-Boltzmann constant. My question is how can I find the following relation

$$ R(\nu, T)\mathrm{d}\nu = \frac{c}{4} \rho(\nu, T)\mathrm{d}\nu? $$