# Bose-Einstein distribution for photons

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In summary: A = 1 for photons and if this implies an increasingly high probability as E approaches 0. The answer is that this is due to the Bose-Einstein distribution, which gives the average number of photons in a given energy state. As E approaches 0, the number of photons in the given state approaches infinity. However, there is a minimum energy for photons, which is inversely proportional to the size of the volume they are confined in. This leads to observable effects such as the Casimir effect.
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?

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.

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?

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.

Some corrections. The link in the sentence
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

## 1. What is the Bose-Einstein distribution for photons?

The Bose-Einstein distribution for photons is a statistical distribution that describes the probability of finding photons in different energy states at a given temperature. It is based on the principles of quantum mechanics and is used to understand the behavior of photons in a system.

## 2. How is the Bose-Einstein distribution different from other statistical distributions?

The Bose-Einstein distribution is different from other statistical distributions because it takes into account the quantum nature of particles, specifically the fact that photons are indistinguishable and can occupy the same energy state. This leads to the formation of a Bose-Einstein condensate at low temperatures.

## 3. What are the key parameters in the Bose-Einstein distribution for photons?

The key parameters in the Bose-Einstein distribution for photons are the temperature of the system and the energy of the photons. These parameters determine the shape of the distribution and the number of photons in each energy state.

## 4. How does the Bose-Einstein distribution change with temperature?

As the temperature of the system decreases, the Bose-Einstein distribution shifts towards lower energy states, leading to an increase in the number of photons in the lowest energy state. At low temperatures, the distribution approaches a step function, indicating the formation of a Bose-Einstein condensate.

## 5. What applications does the Bose-Einstein distribution for photons have?

The Bose-Einstein distribution for photons has various applications, including in the study of laser physics, photonics, and quantum optics. It is also used in the development of new technologies such as Bose-Einstein condensate-based sensors and quantum computing devices.

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