(adsbygoogle = window.adsbygoogle || []).push({}); Are spontaneous and stimulated emission selected by a Boltzmann's statistics ?

Consider 2 levels(m,n) oscillators in thermal equilibrium with Einstein's coefficients Amn (spontaneous emission), Bmn (stimulated emission), Bnm (absorption) and r(f) the energy density at the frequency f (black body). The ratio of probality for spontaneous emission upon all emission is :

Amn/(Amn+r.Bmn)=1-exp(-hf/kT)=(sum (0 to hf) exp(-E/kT)dE)/(sum (0 to infinity) exp(-E/kT)dE)

That's the statistical part in a Boltzmann's statistics corresponding to photons with energy less than hf. This is actually the condition which defines the spontaneous emission.

In the same way, the ratio of probality for stimulated emission upon all emission is :

rBmn/(Amn+r.Bmn)=exp(-hf/kT)=(sum (hf to infinity) exp(-E/kT)dE)/(sum (0 to infinity) exp(-E/kT)dE)

That's the statistical part in a Boltzmann's statistics corresponding to photons with energy more than hf. This is actually the condition which defines the stimulated emission.

Does it mean that the statistical distribution of spontaneous and stimulated emission in thermal equilibrium obeys to a Boltzmann's statistics ?

Is it really self consistent with the previous theory of black body proposed by Einstein with its coefficients ?

**Physics Forums | Science Articles, Homework Help, Discussion**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Do spontaneous and stimulated emission obey to a Boltzmann's statistics ?

**Physics Forums | Science Articles, Homework Help, Discussion**