Wednesday, November 27, 2013

classical photons

the flux density of a blackbody would have form

S(\omega) \propto \omega^3\exp\left(-\frac{\hbar\omega}{k_BT}\right)

if the photons were classical distinguishable particles. can you derive Wien's displacement law and the Stefan-Boltzmann law and point out the differences to the Planck-spectrum?

1 comment:

1. the differences between the true result for quantum mechanical indistinguishable particles and classical distinguishable ones are very tiny: in fact, one recovers the Stefan-Boltzmann law by integration:

\int\mathrm{d}\omega\:S(\omega) = \int_0^\infty\mathrm{d}\omega\:\omega^3\exp(-\lambda\omega)

with $\lambda = \hbar/(k_BT)$, which yields using this trick:

\ldots=
-\frac{\mathrm{d}^3}{\mathrm{d}\lambda^3}\int_0^\infty\mathrm{d}\omega\:\exp(-\lambda\omega) =
-\frac{\mathrm{d}^3}{\mathrm{d}\lambda^3}\frac{1}{\lambda} = \frac{6}{\lambda^4} \propto T^4,

which gives the same scaling of the total energy with temperature, only with a different numerical prefactor, $6$ instead of $\pi^4/15\simeq6.49$. the Wien-law can be derived by looking for the maximum:

\frac{\mathrm{d}S}{\mathrm{d}\omega} = \omega^2\exp(-\lambda\omega)\left(3-\lambda\omega\right) = 0,

which yields $\omega_\mathrm{max}=3/\lambda$ instead of $\simeq2.82/\lambda$.