Class Information

The three laws of thermodynamics; thermodynamic potentials; ideal and non-ideal gases; phase transitions; chemical equilibrium. Introduction to equilibrium statistical mechanics; statistical definition of entropy; applications to fluids, magnetic systems, the ideal quantum gas. Prior to Fall 2017, this course was named “Thermodynamics and Kinetic Theory”.

Textbook: Schroeder

Probability

$$ P(n, r) = \frac{n!}{(n-r)!}$$
Permutation: Choose $r$ objects from a set of $n$ total objects. Order matters. Gives the number of possible arrangements

  • Note that permutation also refers to a different but related concept where $P = n!$
    • The number of arrangements that a set can be placed in
    • Note that this is just a specific case of the above where $r=n$

$$ P(n, r) = \frac{n!}{(n-r)!r!}$$
Permutation: Choose $r$ objects from a set of $n$ total objects. Order does not matters. Gives the number of possible arrangements

Product Rule

Find the probability that $n$ molecules are in the left half of a box with the $N$ total molecules

Total number of configurations of the system where $n$ molecules out of $N$ are in a specific half
$$ C(N, n) = \frac{N!}{n!(N-n)!}$$

For one single particle
$$ P_0 = \frac{1}{2}$$

For the system to have $n$ particles to be in a specific state
$$ P_n = (P_0)^n$$

The probability for a system to have $n$ particles in the left half of the box:
$$ P = P_n C(N, n) = (P_0)^n \frac{N!}{(N-n)!n!}$$

Alternatively, $C(N, n)$ gives the total number of configurations that lead to molecules in the left half the box and $2^N$ is the total number of configurations possible:
$$ P = \frac{C(N, n)}{2^N}$$

Boltzmann Statistics

Partition Function

  • At any given time, $t$, a system of $N$ particles can be described by 6 variables: $x, y, z, p_x, p_y, p_z$

  • These 6 variables form a 6-D phase space

  • Microcanonical Ensemble: Each microstate with energy between $E$ and $E+ dE$ are equally probable

  • Canonical Ensemble: Each microstate with a particular energy is not equally probable, but follows Boltzmann-Statistics

    • Heat can be exchanged
    • Temperature are constant since there is a heat and particle resevoir connected to the system

  • Grand Canonical Ensemble: Each microstate with a particular energy is not equally probable

    • Chemical potential, $mu$, tells how much energy is needed to add/create a particle to the system
    • Heat and particles can be exchanged
    • Temperature and $mu$ are constant since there is a heat and particle resevoir connected to the system

  • The partition function gives the number of available system states

Canonical Partition Function:
$$ Z = \sum_i \exp(-\frac{\epsilon_i}{k_B T}) = \sum_\epsilon g(\epsilon) \exp(-\frac{\epsilon}{k_B T})$$

$$ F= -k_B T \ln(Z)$$
$$ U = -k_B T^2 \pdv{}{T} \left(\ln(Z)\right)$$
$$ U = \overline{E} N$$
$$ \mu = -k_B T \left(\pdv{\ln(Z)}{N}\right)_{V, T}$$

The probability:

  • $N(s)$ is the number of particles in state $s$
    $$ P(s) = N(s)/ N$$
    $$ P(s) = \frac{1}{Z} e^{E(s)/(k_B T)}$$

Paramagnet

Average magnetic dipole moment, $\mu$
$$ \mu = \sum_s \mu(s) P(s)$$

Total Magnetization
$$ M = N\overline{\mu}$$

Maxwell Speed Distribution

The partition function can be rewritten as an integral over the phase space as follows:

$$ Z = \frac{1}{(2\pi\hbar)^3}\int d^3 r \int d^3 p e^{-(E(p)/(k_B T)} $$
where the $(2\pi\hbar)^3$ is a quantum volume factor, $d^3 r$ is a 3D positional integral, and $E(p)$ means energy is expressed in terms of momentum

Then the probability as a function of momentum, $P(p)$:
$$P(p) = \frac{1}{Z} e^{-E(p)/(k_B T)}$$

The distribution of momentum, $D(p)$ is:
$$ D(p) = P(p)\ g(p)$$

where $g(p)$ is the density of states:
$$D(p) \ dp = P(p) \ g(p) \ dp = P(p) \ \frac{V \ d^3 p}{(2\pi\hbar)^3} = P(p) \ dp \frac{4\pi p^2 V}{(2\pi\hbar)^3} $$
$$ \implies D(p) = P(p) \ \frac{4\pi p^2 V}{(2\pi\hbar)^3} $$

$$D(p) = \frac{4\pi p^2 V}{(2\pi\hbar)^3} \frac{1}{Z} e^{-E(p)/(k_B T)}$$

Partition Function for Composite Systems

For $N$ distinguisable, non-interacting particles:
$$ Z_{total} = Z_1Z_2 Z_3…Z_N$$
where $Z_{tota}$ is the partition function for the entire system, and $Z_i$ is the individual single particle partition function

For $N$ indistinguisable, non-interacting particles:
$$ Z_{total} = \frac{Z_1^N}{N!}$$

  • where $Z_1$ is the single particle partition function
  • Only applies when $Z_1$ » N
    • When the number of available single-particle states is much greater than the number of particles
    • Equivalently this formula only applies when
      $$ \frac{V}{N} » V_{Q}$$

$$ V_Q = l_Q^3 = \left(\frac{h}{\sqrt{2\pi mk_BT}}\right)^3$$

Quantum Statistics

Grand Canonical Partition Function

  • $\mu VT $ Ensemble
  • Thermal resevoir and particle resevoir
    • Heat and particles can be exchanged
    • $T$ is constant
      $$ \tilde{Z} = \sum_N \sum_i \exp(\frac{\mu N - \epsilon_i N}{k_B T}) $$
      alternatively
      $$ \tilde{Z} = \exp(\frac{PV}{k_B T}) $$

Probability
$$ P(n) = \frac{1}{\tilde{Z}} e^{-n(\epsilon-\mu)/(k_B T)} $$

Average Number of Particles in a State (Distribution Functions)

Let $\overline{n}$ denote the average number of particles in a state

For quantum particles
$$ \bar{n} = \sum_n n P(n)$$
where $n$ is an integer

Let $x = (\epsilon - \mu)/(k_B T)$
$$ \bar{n} = \sum_n n \frac{e^{-nx}}{\tilde{Z}} = -\frac{1}{\tilde{Z}} \sum_n \pdv{}{x} e^{-n x} = -\frac{1}{\tilde{Z}} \pdv{\tilde{Z}}{x}$$

or alternatively:
$$ \bar{n} = k_B T\frac{1}{\tilde{Z}} \pdv{\tilde{Z}}{\mu}$$

  • you can use these formulas to find the distribution function for Fermions or Bosons
    • Fermions: Particles cannot occupy the same state
      • For some fermions, like electrons, two particles can have the energy, but they still have different states (spin up and spin down)
    • Bosons: Any number of particles can occupy the same state/energy

Compare with Boltzmann:
Probablity for single particle
$$ P_{boltz}(n) = \frac{1}{Z_1} e^{-\epsilon/(k_B T)}$$

If we have $N$ independent, distinguishable particles:
$$ \overline{n_{boltz}} = NP(s) = \frac{N}{Z_1}e^{-\epsilon/(k_B T)}$$

Fermi-Dirac Statistics

$$ N = \int_0^\infty g(\epsilon) \overline{n_{FD}}(\epsilon) \ d\epsilon $$
$$ U = \int_0^\infty g(\epsilon) \ \epsilon \ \overline{n_{FD}}(\epsilon) \ d\epsilon$$
$$ g(\epsilon) = \frac{\pi (8m)^{3/2}}{2(2\pi\hbar)^3} V \sqrt{\epsilon}$$

  • Degenerate Gas: All states below $\epsilon_F$ are occupied and all states above are unoccupied
    • Occurs when $T = 0$
      $$ \epsilon_F = \mu $$
    • Note that degeneracy here is completely different than the same word used to describe the number of states with the same energy (similar to Density of State)

Density of States, Derivation for Fermi Energy, and Derivation for Number of Particles

Number of particles at a given energy, $\epsilon$
$$ N = \sum_i n p_i = \sum_i 2 \overline{n}$$
where $n$ is an integer

The $2$ is from the two spins that an electron has
$$ \overline{n_{FD}} = \frac{1}{e^{(\epsilon - \mu)/(k_B T)} + 1}$$

$$ N = \sum_i \frac{2}{e^{(\epsilon - \mu)/(k_B T)} + 1} $$
Convert to an integral by integrating over all possible states (over the 6D phase space variables):
$$ N = \frac{2}{(2\pi\hbar)^3} \int dV \int d^3p \frac{1}{e^{(\epsilon - \mu)/(k_B T)} + 1} $$
We divide by $(2\pi\hbar)^3$, the quantum unit of volume

$$ N = \frac{2}{(2\pi\hbar)^3} V \int d^3p \frac{1}{e^{(\epsilon - \mu)/(k_B T)} + 1} $$
$$ \int d^3 p = \int d\Omega \int_0^\infty dp \ p^2 = 4\pi \int_0^\infty dp \ p^2$$
$$ \implies N = \frac{4\pi 2}{(2\pi\hbar)^3} V \int_0^\infty dp \ p^2 \frac{1}{e^{(\epsilon - \mu)/(k_B T)} + 1} $$

Convert from momentum $p$ to energy, $\epsilon$
$$ \epsilon = \frac{p^2}{2m} \implies p^2 = 2m \epsilon \implies p = \sqrt{2\pi\epsilon} \implies dp = \frac{\sqrt{2m}}{2\epsilon} \ d\epsilon$$

$$ N = \frac{8\pi}{(2\pi\hbar)^3} \int_0^\infty \frac{\sqrt{2m}}{2\sqrt{\epsilon}} \ d\epsilon \ (2m\epsilon) \frac{1}{e^{(\epsilon-\mu)/(k_B T)} + 1}$$
$$ N = \frac{8\pi}{(2\pi\hbar)^3} \int_0^\infty \frac{(2m)^{3/2} \sqrt{\epsilon}}{2} \frac{1}{e^{(\epsilon-\mu)/(k_B T)} + 1}$$

  • We also have $N$ in terms of the density of states: $g(\epsilon)$
    • The density of states gives the number of single-particle states per energy

$$ N = \int_0^\infty \ \bar{n}(\epsilon) g(\epsilon) \ d\epsilon$$

This implies
$$ g(\epsilon) = \frac{8\pi}{(2\pi\hbar)^3}\frac{(2m)^{3/2}\sqrt{\epsilon}}{2} =\frac{\pi}{(2\pi\hbar)^3}\frac{(8m)^{3/2}\sqrt{\epsilon}}{2} $$

We arrive at the same density of states (DOS) from before in the Fermi Dirac Statistics Section