# Understanding what the Bose-Einstein distribution

I'm currently studying Kittel's Solid State Physics and in his chapter on Phonon heat capacity, we need to first calculate the total energy $$U$$. Phonons have energy $$E_n = (n+1/2)\hbar\omega$$ and he first calculated the average energy $$\langle E\rangle$$ and using the Boltzmann factor, he showed: $$\langle E\rangle = \dfrac{1}{2}\hbar\omega+\hbar\omega\dfrac{1}{e^{\hbar\omega/k_BT}-1}=\dfrac{1}{2}\hbar\omega+\hbar\omega\langle n \rangle$$ so then we must have $$\langle n \rangle=\dfrac{1}{e^{\hbar\omega/k_BT}-1}$$ I recognise this as the Bose-Einstein but I'm surprised to see this as being interpreted as an average of the number of states. I always thought this was a probabilistic distribution and in fact, Kittel does seem to use this as a probability since he later writes: $$U=\int d\omega\ \hbar\omega D(\omega)\langle n \rangle$$ where $$D(\omega)$$ is the density of state. In this expression $$D$$ already accounts for the number of photons so $$\langle n \rangle$$ must be some probability weight? I'm sure something is flawed in my understanding so any help is appreciated!

• <n> isn't an average number of states. It's the average number of phonons in a state with frequency $\omega$ at temperature T. – Samuel Weir Nov 1 '18 at 5:27

I wouldn't say that $$D\left(\omega\right)$$ accounts for the number of phonons; I'd say it accounts for the number of vibrational (phonon) modes at a given energy. A phonon is a unit of energy in a vibrational mode. Then, $$\left$$ is the average number of phonons in a given vibrational mode (what Samuel Weir said).
So the integrand is the energy of a phonon $$\hbar \omega$$ times the average number of those phonons in a mode $$\left$$ times the density of modes at that energy $$D\left(\omega\right)$$.