- Einstein condensation Density
*face*Fermi and Bose gases*face*Lecture120 min.

##### Fermi and Bose gases

Thermal and Statistical Physics 2020Fermi level fermion boson Bose gas Bose-Einstein condensate ideal gas statistical mechanics phase transition

These lecture notes from week 7 of Thermal and Statistical Physics apply the grand canonical ensemble to fermion and bosons ideal gasses. They include a few small group activities.*face*Phase transformations*face*Lecture120 min.

##### Phase transformations

Thermal and Statistical Physics 2020phase transformation Clausius-Clapeyron mean field theory thermodynamics

These lecture notes from the ninth week of Thermal and Statistical Physics cover phase transformations, the Clausius-Clapeyron relation, mean field theory and more. They include a number of small group activities.*assignment*Surface temperature of the Earth*assignment*Homework##### Surface temperature of the Earth

Temperature Radiation Thermal and Statistical Physics 2020 Calculate the temperature of the surface of the Earth on the assumption that as a black body in thermal equilibrium it reradiates as much thermal radiation as it receives from the Sun. Assume also that the surface of the Earth is a constant temperature over the day-night cycle. Use the sun's surface temperature \(T_{\odot}=5800\text{K}\); and the sun's radius \(R_{\odot}=7\times 10^{10}\text{cm}\); and the Earth-Sun distance of \(1.5\times 10^{13}\text{cm}\).*face*Thermal radiation and Planck distribution*face*Lecture120 min.

##### Thermal radiation and Planck distribution

Thermal and Statistical Physics 2020Planck distribution blackbody radiation photon statistical mechanics

These notes from the fourth week of Thermal and Statistical Physics cover blackbody radiation and the Planck distribution. They include a number of small group activities.*assignment*Radiation in an empty box*assignment*Homework##### Radiation in an empty box

Thermal physics Radiation Free energy Thermal and Statistical Physics 2020As discussed in class, we can consider a black body as a large box with a small hole in it. If we treat the large box a metal cube with side length \(L\) and metal walls, the frequency of each normal mode will be given by: \begin{align} \omega_{n_xn_yn_z} &= \frac{\pi c}{L}\sqrt{n_x^2 + n_y^2 + n_z^2} \end{align} where each of \(n_x\), \(n_y\), and \(n_z\) will have positive integer values. This simply comes from the fact that a half wavelength must fit in the box. There is an additional quantum number for polarization, which has two possible values, but does not affect the frequency.

**Note that in this problem I'm using different boundary conditions from what I use in class. It is worth learning to work with either set of quantum numbers.**Each normal mode is a harmonic oscillator, with energy eigenstates \(E_n = n\hbar\omega\) where we will not include the zero-point energy \(\frac12\hbar\omega\), since that energy cannot be extracted from the box. (See the Casimir effect for an example where the zero point energy of photon modes does have an effect.)- Note
- This is a slight approximation, as the boundary conditions for light are a bit more complicated. However, for large \(n\) values this gives the correct result.

Show that the free energy is given by \begin{align} F &= 8\pi \frac{V(kT)^4}{h^3c^3} \int_0^\infty \ln\left(1-e^{-\xi}\right)\xi^2d\xi \\ &= -\frac{8\pi^5}{45} \frac{V(kT)^4}{h^3c^3} \\ &= -\frac{\pi^2}{45} \frac{V(kT)^4}{\hbar^3c^3} \end{align} provided the box is big enough that \(\frac{\hbar c}{LkT}\ll 1\). Note that you may end up with a slightly different dimensionless integral that numerically evaluates to the same result, which would be fine. I also do not expect you to solve this definite integral analytically, a numerical confirmation is fine.

**However, you must manipulate your integral until it is dimensionless and has all the dimensionful quantities removed from it!**Show that the entropy of this box full of photons at temperature \(T\) is \begin{align} S &= \frac{32\pi^5}{45} k V \left(\frac{kT}{hc}\right)^3 \\ &= \frac{4\pi^2}{45} k V \left(\frac{kT}{\hbar c}\right)^3 \end{align}

Show that the internal energy of this box full of photons at temperature \(T\) is \begin{align} \frac{U}{V} &= \frac{8\pi^5}{15}\frac{(kT)^4}{h^3c^3} \\ &= \frac{\pi^2}{15}\frac{(kT)^4}{\hbar^3c^3} \end{align}

*assignment*Potential energy of gas in gravitational field*assignment*Homework##### Potential energy of gas in gravitational field

Potential energy Heat capacity Thermal and Statistical Physics 2020 Consider a column of atoms each of mass \(M\) at temperature \(T\) in a uniform gravitational field \(g\). Find the thermal average potential energy per atom. The thermal average kinetic energy is independent of height. Find the total heat capacity per atom. The total heat capacity is the sum of contributions from the kinetic energy and from the potential energy. Take the zero of the gravitational energy at the bottom \(h=0\) of the column. Integrate from \(h=0\) to \(h=\infty\).*You may assume the gas is ideal.**assignment*Pressure of thermal radiation*assignment*Homework##### Pressure of thermal radiation

Thermal radiation Pressure Thermal and Statistical Physics 2020(modified from K&K 4.6) We discussed in class that \begin{align} p &= -\left(\frac{\partial F}{\partial V}\right)_T \end{align} Use this relationship to show that

\begin{align} p &= -\sum_j \langle n_j\rangle\hbar \left(\frac{d\omega_j}{dV}\right), \end{align} where \(\langle n_j\rangle\) is the number of photons in the mode \(j\);

Solve for the relationship between pressure and internal energy.

*assignment*Pressure and entropy of a degenerate Fermi gas*assignment*Homework##### Pressure and entropy of a degenerate Fermi gas

Fermi gas Pressure Entropy Thermal and Statistical Physics 2020Show that a Fermi electron gas in the ground state exerts a pressure \begin{align} p = \frac{\left(3\pi^2\right)^{\frac23}}{5} \frac{\hbar^2}{m}\left(\frac{N}{V}\right)^{\frac53} \end{align} In a uniform decrease of the volume of a cube every orbital has its energy raised: The energy of each orbital is proportional to \(\frac1{L^2}\) or to \(\frac1{V^{\frac23}}\).

Find an expression for the entropy of a Fermi electron gas in the region \(kT\ll \varepsilon_F\). Notice that \(S\rightarrow 0\) as \(T\rightarrow 0\).

*face*Quantum Reference Sheet*face*Boltzmann probabilities and Helmholtz*face*Lecture120 min.

##### Boltzmann probabilities and Helmholtz

Thermal and Statistical Physics 2020ideal gas entropy canonical ensemble Boltzmann probability Helmholtz free energy statistical mechanics

These notes, from the third week of Thermal and Statistical Physics cover the canonical ensemble and Helmholtz free energy. They include a number of small group activities.-
Thermal and Statistical Physics 2020
**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.