My name is Alexander FufaeV and here I write about:
Bose Distribution: The Occupation Probability of Bosons
Important Formula
What do the formula symbols mean?
Probability
$$ P $$ Unit $$ - $$
The occupation probability indicates with which probability \(P\) a state with energy \( W \) at temperature \( T \) is occupied by a boson (for example a photon or phonon).
Energy
$$ W $$ Unit $$ \mathrm{J} = \mathrm{Nm} = \frac{ \mathrm{kg} \, \mathrm{m^2} }{ \mathrm{s}^2 } $$
Energy state which can be occupied by a boson, for example by a photon or \(^4\text{He}\) atom.
Chemical potential
$$ \mu $$ Unit $$ \mathrm{J} $$
Chemical potential indicates the change of internal energy when the particle number of the boson gas changes. At \(T = 0 \) the chemical potential corresponds to the Fermi energy.
Temperature
$$ T $$ Unit $$ \mathrm{K} $$
Absolute temperature of the boson gas (for example a photon gas).
Boltzmann Constant
$$ k_{\text B} $$ Unit $$ \frac{\mathrm J}{\mathrm K} = \frac{\mathrm{kg} \,\mathrm{m}^2}{\mathrm{s}^2 \, \mathrm{K}} $$Boltzmann constant is a physical constant from many-particle physics and has the following exact value:
$$ k_{\text B} ~=~ 1.380 \, 649 ~\cdot~ 10^{-23} \, \frac{\mathrm{J}}{\mathrm{K}} $$
The Bose distribution (also called Bose-Einstein distribution) indicates the occupation probability \( P(W) \) that a boson in a system takes on the energy \( W \). The occupation probability depends on the temperature \( T \) of the system.
$$ \begin{align} P(W) ~=~ \frac{1}{\mathrm{e}^{ \frac{ W - \mu }{ k_{\text B} \, T}} ~-~ 1} \end{align} $$
The crucial difference to the Boltzmann distribution is the presence of the term "-1" in the denominator, which ensures that the probability of occupying energy states does not become infinitely large when the temperature approaches zero. This leads to Bose-Einstein condensation, in which bosons can occupy the same energetic ground state at very low temperatures.