## Symmetry of filled and vacant orbitals

• 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 2020
1. Show 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 $\frac1{V^{\frac23}}$.

2. 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 2020

The 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?

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 Lecture

120 min.

##### Chemical potential and Gibbs distribution
Thermal and Statistical Physics 2020

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 Lecture

120 min.

##### Fermi and Bose gases
Thermal and Statistical Physics 2020

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 Lecture

120 min.

##### Ideal Gas
Thermal and Statistical Physics 2020

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 identity

The 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 2020

The 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 $$\left<\left(\varepsilon-\langle\varepsilon\rangle\right)^2\right> = k_BT^2\left(\frac{\partial U}{\partial T}\right)_{V}$$ 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 Lecture

5 min.

##### Energy and Entropy review
Thermal and Statistical Physics 2020 (3 years)

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.