Effective occupation number of photon

  • #1

Main Question or Discussion Point

Hello everyone:

Here is the problem:

Under thermal equilibrium, photon's number can be described as the
photonic density of state (PDOS) * occupation number(ON).
Also, the photon's flux can be described as
PDOS * ON * effective particle velocity into certain direction ( V)

The occupation number is determined by the source, typically it will be a Boltzmann distribution characterized by the temperature and chemical potential.

When a beam scattering by an interface between media 1 and media 2, only part of incident energy can go through into the second media. If we continuously tracing the energy of the photon flux, we may can write the following equation:

PDOS1* ON(IN) * V1 = .PDOS2 * ON(T) * V2 + PDOS1 * ON(R) * V1

, where IN, T, R means incidence, transmission and reflection respectively.

This means the occupation number of photon in the media 2 will be different to the one in media 1. Is such an effective occupation number is meaningful and this consideration is correct? Does this mean that once the reflection happens the occupation number of photon in media 2 will never reach the thermodynamic limit?
 

Answers and Replies

  • #2
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This means the occupation number of photon in the media 2 will be different to the one in media 1.
You have the effect of different reflection, but you also have the different velocity. Why are you sure they won't cancel?
 
  • #3
You have the effect of different reflection, but you also have the different velocity. Why are you sure they won't cancel?
If you write down the full equals, you will find that will not cancelled. The ON(T) in the second media will be a function of refractive index.
 

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