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Einstein condensation temperature Starting from the density of free particle orbitals per unit energy range \begin{align} \mathcal{D}(\varepsilon) = \frac{V}{4\pi^2}\left(\frac{2M}{\hbar^2}\right)^{\frac32}\varepsilon^{\frac12} \end{align} show that the lowest temperature at which the total number of atoms in excited states is equal to the total number of atoms is \begin{align} T_E &= \frac1{k_B} \frac{\hbar^2}{2M} \left( \frac{N}{V} \frac{4\pi^2}{\int_0^\infty\frac{\sqrt{\xi}}{e^\xi-1}d\xi} \right)^{\frac23} T_E &= \end{align} The infinite sum may be numerically evaluated to be 2.612. Note that the number derived by integrating over the density of states, since the density of states includes all the states except the ground state.

Note: This problem is solved in the text itself. I intend to discuss Bose-Einstein condensation in class, but will not derive this result.


120 min.

Fermi and Bose gases
These lecture notes from week 7 of https://paradigms.oregonstate.edu/courses/ph441 apply the grand canonical ensemble to fermion and bosons ideal gasses. They include a few small group activities.
Consider one particle confined to a cube of side \(L\); the concentration in effect is \(n=L^{-3}\). Find the kinetic energy of the particle when in the ground state. There will be a value of the concentration for which this zero-point quantum kinetic energy is equal to the temperature \(kT\). (At this concentration the occupancy of the lowest orbital is of the order of unity; the lowest orbital always has a higher occupancy than any other orbital.) Show that the concentration \(n_0\) thus defined is equal to the quantum concentration \(n_Q\) defined by (63): \begin{equation} n_Q \equiv \left(\frac{MkT}{2\pi\hbar^2}\right)^{\frac32} \end{equation} within a factor of the order of unity.