The discussion centers around the Bose-Einstein distribution for photons, particularly addressing the implications of the parameter A being equal to 1 for photons, the behavior of photon occupancy as energy approaches zero, and the existence of a minimum energy for photons in various confinement scenarios. The scope includes theoretical considerations and conceptual clarifications related to statistical mechanics and quantum physics.
Participants express differing views on the implications of A = 1 for photons and the nature of photon occupancy at low energies. There is no consensus on the existence of a minimum energy for photons, with various perspectives presented regarding confinement and energy states.
Limitations include unresolved assumptions about the behavior of photons at low energies and the dependence of conclusions on the specific definitions of energy states and confinement scenarios.
Charles Link said:You can do a Taylor expansion for ## E_p << kT ## and it says the mean occupancy for this case (for each photon mode) is ## kT/E_p ##. This implies you will get multiple photons in the same mode if ## kT>> E_p ## where ## E_p ## is the photon energy.
When you compute the energy from these states, you multiply the occupancy by the energy. There is also one other factor included in this integration though, and that is the density of states in k-space, where ## E=\hbar c k ##. That density is proportional to ## k^2 ## and goes to zero as ## k=2 \pi/\lambda ## goes to zero, i.e. for long wavelengths.TheCanadian said:But why? Why is it more probable to find a state occupied by a photon as ##E_p## goes to 0? I suppose I am wondering if there is any minimum to a photon's energy (i.e. frequency). In this case, the mean occupancy is ##kT/E_p## but can't ##E_p## be an arbitrarily small quantum of energy?
is to a hyperphysics page. The explanation on that page requires a correction. The Bose-Einstein distribution is not the 'probability' of a particle having the energy but the thermodynamic average number of particles having the energy E.TheCanadian said:When computing the probability distribution of bosons
It means that as we approach zero energy, the number of photons in the given energy state approaches to infinity, i.e. infinite number of photons at zero energy.TheCanadian said:is A = 1 for photons? Does this not imply that photons will have an increasingly high probability of being present as E approaches 0?
This is actually a very interesting question.TheCanadian said:I am wondering if there is any minimum to a photon's energy (i.e. frequency).