Bose-Einstein distribution for photons

[TheCanadian]

When computing the probability distribution of bosons, why is A = 1 for photons? Does this not imply that photons will have an increasingly high probability of being present as E approaches 0? What is the significance of such a situation?

 

[Charles Link]

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.

 

[TheCanadian]

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.

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?

 

[Charles Link]

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?

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.

 

[Henryk]

Some corrections. The link in the sentence

When computing the probability distribution of bosons

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.
Why is probability incorrect? well, for Bose-Einstein distribution, the value can be larger than 1 if E is sufficiently small. Probability cannot be larger than one, ever.

is A = 1 for photons? Does this not imply that photons will have an increasingly high probability of being present as E approaches 0?

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.

I am wondering if there is any minimum to a photon's energy (i.e. frequency).

This is actually a very interesting question.
Let's consider a one dimensional case, think of a cavity of length L. Then the lowest mode of electromagnetic wave (photon) will have the wavelength of 2L, That gives the energy ## E = \frac{hc}{2L}##. This is the lowest energy of a photon confined in one dimension to a cavity of length L. In 3D, if the cavity is a cube of the same length, the minimum photon energy is ##\sqrt 3## times larger.
So, as L goes to infinity, the minimum energy for a photon goes to zero.
However, if you confined photons to a finite volume, the minimum energy is inversely proportional to the dimension of the volume.
This leads to observable effect, such as Casimir effect

Henryk