- Symmetry Orbitals
*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 \(T\rightarrow 0\).

*assignment*Active transport*assignment*Homework##### Active transport

Active transport Concentration Chemical potential Thermal and Statistical Physics 2020The concentration of potassium \(\text{K}^+\) ions in the internal sap of a plant cell (for example, a fresh water alga) may exceed by a factor of \(10^4\) the concentration of \(\text{K}^+\) ions in the pond water in which the cell is growing. The chemical potential of the \(\text{K}^+\) ions is higher in the sap because their concentration \(n\) is higher there. Estimate the difference in chemical potential at \(300\text{K}\) and show that it is equivalent to a voltage of \(0.24\text{V}\) across the cell wall. Take \(\mu\) as for an ideal gas. Because the values of the chemical potential are different, the ions in the cell and in the pond are not in diffusive equilibrium. The plant cell membrane is highly impermeable to the passive leakage of ions through it. Important questions in cell physics include these: How is the high concentration of ions built up within the cell? How is metabolic energy applied to energize the active ion transport?

- David adds
- You might wonder why it is even remotely plausible to consider the ions in solution as an ideal gas. The key idea here is that the ideal gas entropy incorporates the entropy due to position dependence, and thus due to concentration. Since concentration is what differs between the cell and the pond, the ideal gas entropy describes this pretty effectively. In contrast to the concentration dependence, the temperature-dependence of the ideal gas chemical potential will not be so great.

*assignment*Derivative of Fermi-Dirac function*assignment*Homework##### Derivative of Fermi-Dirac function

Fermi-Dirac function Thermal and Statistical Physics 2020**Derivative of Fermi-Dirac function**Show that the magnitude of the slope of the Fermi-Direc function \(f\) evaluated at the Fermi level \(\varepsilon =\mu\) is inversely proportional to its temperature. This means that at lower temperatures the Fermi-Dirac function becomes dramatically steeper.*face*Chemical potential and Gibbs distribution*face*Lecture120 min.

##### Chemical potential and Gibbs distribution

Thermal and Statistical Physics 2020chemical potential Gibbs distribution grand canonical ensemble statistical mechanics

These notes from the fifth week of Thermal and Statistical Physics cover the grand canonical ensemble. They include several small group activities.*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*Ideal Gas*face*Lecture120 min.

##### Ideal Gas

Thermal and Statistical Physics 2020ideal gas particle in a box grand canonical ensemble chemical potential statistical mechanics

These notes from week 6 of Thermal and Statistical Physics cover the ideal gas from a grand canonical standpoint starting with the solutions to a particle in a three-dimensional box. They include a number of small group activities.*assignment*Energy, Entropy, and Probabilities*assignment*Homework##### Energy, Entropy, and Probabilities

Energy Entropy Probabilities Thermodynamic identityThe goal of this problem is to show that once we have maximized the entropy and found the microstate probabilities in terms of a Lagrange multiplier \(\beta\), we can prove that \(\beta=\frac1{kT}\) based on the statistical definitions of energy and entropy and the thermodynamic definition of temperature embodied in the thermodynamic identity.

The internal energy and entropy are each defined as a weighted average over microstates: \begin{align} U &= \sum_i E_i P_i & S &= -k_B\sum_i P_i \ln P_i \end{align}: We saw in clase that the probability of each microstate can be given in terms of a Lagrange multiplier \(\beta\) as \begin{align} P_i &= \frac{e^{-\beta E_i}}{Z} & Z &= \sum_i e^{-\beta E_i} \end{align} Put these probabilities into the above weighted averages in order to relate \(U\) and \(S\) to \(\beta\). Then make use of the thermodynamic identity \begin{align} dU = TdS - pdV \end{align} to show that \(\beta = \frac1{kT}\).

*assignment*Energy, Entropy, and Probabilities*assignment*Homework##### Energy, Entropy, and Probabilities

Thermal and Statistical Physics 2020The goal of this problem is to show that once we have maximized the entropy and found the microstate probabilities in terms of a Lagrange multiplier \(\beta\), we can prove that \(\beta=\frac1{kT}\) based on the statistical definitions of energy and entropy and the thermodynamic definition of temperature embodied in the thermodynamic identity.

The internal energy and entropy are each defined as a weighted average over microstates: \begin{align} U &= \sum_i E_i P_i & S &= -k_B\sum_i P_i \ln P_i \end{align} We saw in clase that the probability of each microstate can be given in terms of a Lagrange multiplier \(\beta\) as \begin{align} P_i &= \frac{e^{-\beta E_i}}{Z} & Z &= \sum_i e^{-\beta E_i} \end{align} Put these probabilities into the above weighted averages in order to relate \(U\) and \(S\) to \(\beta\). Then make use of the thermodynamic identity \begin{align} dU = TdS - pdV \end{align} to show that \(\beta = \frac1{kT}\).

*assignment*Energy fluctuations*assignment*Homework##### Energy fluctuations

energy Boltzmann factor statistical mechanics heat capacity Thermal and Statistical Physics 2020 Consider a system of fixed volume in thermal contact with a resevoir. Show that the mean square fluctuations in the energy of the system is \begin{equation} \left<\left(\varepsilon-\langle\varepsilon\rangle\right)^2\right> = k_BT^2\left(\frac{\partial U}{\partial T}\right)_{V} \end{equation} Here \(U\) is the conventional symbol for \(\langle\varepsilon\rangle\).*Hint:*Use the partition function \(Z\) to relate \(\left(\frac{\partial U}{\partial T}\right)_V\) to the mean square fluctuation. Also, multiply out the term \((\cdots)^2\).*face*Energy and Entropy review*face*Lecture5 min.

##### Energy and Entropy review

Thermal and Statistical Physics 2020 (3 years)thermodynamics statistical mechanics

This very quick lecture reviews the content taught in Energy and Entropy, and is the first content in Thermal and Statistical Physics.-
Thermal and Statistical Physics 2020
Show that
\begin{align}
f(\mu+\delta) &= 1 - f(\mu-\delta)
\end{align}
This means that the probability that an orbital above the
Fermi level is occupied is equal to the probability an orbital
the same distance below the Fermi level
*being empty*. These unoccupied orbitals are called*holes*.