Chapter 1

I. Thermodynamic Systems, States, and Processes

1. Thermodynamic Systems

• Isolated system: No exchange of matter or energy with the surroundings.
• Closed system: No exchange of matter, but has an exchange of energy.
• Open system: Has exchange of both matter and energy.

2. State of a System

Isolated and closed systems will eventually reach an equilibrium state: Thermodynamic equilibrium (macroscopic quantities do not change).

Systems deviating from equilibrium but not moving far can be divided into several local equilibrium states. If there is a steady "flow", it is in a quasi-steady state (non-equilibrium state).

Independent macroscopic variables depend on the degrees of freedom of the equilibrium state: State parameters → Characteristic functions → State functions.

1. Particle number: $N$
2. Volume: $V$
3. Temperature: $T$ (Equation of state)
4. Pressure: $P$
5. Entropy: $S$ (Entropy differential representation)
6. Internal energy: $E$ (Energy equation), $C_V = \left(\frac{\partial E}{\partial T}\right)_V$
7. Enthalpy: $H = E + PV$, $C_P = \left(\frac{\partial H}{\partial T}\right)_P$
8. Helmholtz free energy: $F = E - TS$
9. Gibbs free energy: $G = E + PV - TS$
10. Chemical potential: $\mu = \left(\frac{\partial G}{\partial N}\right)_{T,P} = \left(\frac{\partial F}{\partial N}\right)_{T,V} = \left(\frac{\partial H}{\partial N}\right)_{S,P} = \left(\frac{\partial E}{\partial N}\right)_{S,V}$

Quantities can also be classified as:

3. Thermodynamic Processes

A thermodynamic process describes the transition from one equilibrium state to another equilibrium state.

• Work: Transfer of ordered motion energy: $dW = \sum Y_i dy_i$
• Heat: Transfer of disordered motion energy: $dQ = T dS$ (for a reversible process)

4. First Law of Thermodynamics

From Green's formula, we can obtain the Maxwell relations.

5. Second Law of Thermodynamics

Reversible process: $dS = \frac{dQ}{T}$. Irreversible process: $dS > \frac{dQ}{T}$.

If the thermodynamic process is adiabatic ($dQ = 0$), then $dS \ge 0$.

II. Equilibrium Distribution of Nearly Independent Systems

Macroscopic quantities (state parameters, state functions) are the statistical average of microscopic quantities.

• $\varepsilon_i$: Energy of one particle on the $i$-th energy level.
• $g_i$: Degeneracy of the $i$-th energy level.
• $n_i$: Number of particles on the $i$-th energy level; $\{n_i\}$: Particle number distribution over all energy levels.
• $W\{n_i\} = \prod W_i$: Number of microscopic states corresponding to the distribution $\{n_i\}$.
• Constraints: $\sum n_i = N$, $\sum \varepsilon_i n_i = E$.

Equal probability hypothesis:

Most probable distribution method: If $P\{n_i\}_m \gg P_{\text{other}}$, take $\{n_i\}_m$ as the equilibrium distribution.

1. Localized Particles: Boltzmann Distribution

From $\frac{\partial}{\partial n_i} \ln W_{Bol} + \alpha \frac{\partial}{\partial n_i} (N - \sum n_i) + \beta \frac{\partial}{\partial n_i} (E - \sum n_i \varepsilon_i) = 0$

We get the Boltzmann distribution:

2. Non-localized Particles: Bose-Einstein Distribution

From $\frac{\partial}{\partial n_i} \ln W_{BE} + \alpha \frac{\partial}{\partial n_i} (N - \sum n_i) + \beta \frac{\partial}{\partial n_i} (E - \sum n_i \varepsilon_i) = 0$

We get the Bose-Einstein distribution:

3. Non-localized Particles: Fermi-Dirac Distribution

From $\frac{\partial}{\partial n_i} \ln W_{FD} + \alpha \frac{\partial}{\partial n_i} (N - \sum n_i) + \beta \frac{\partial}{\partial n_i} (E - \sum n_i \varepsilon_i) = 0$

We get the Fermi-Dirac distribution:

4. Classical Limit Condition

When $e^\alpha \gg 1$ or $\frac{n_i}{g_i} \ll 1$ or $\left(\frac{V}{N}\right)^{\frac{1}{3}} \gg \frac{h}{\sqrt{2\pi m kT}}$:

From $\frac{\partial}{\partial n_i} \ln W_S + \alpha \frac{\partial}{\partial n_i} (N - \sum n_i) + \beta \frac{\partial}{\partial n_i} (E - \sum n_i \varepsilon_i) = 0$, we get:

(This does not consider particle indistinguishability).

When $\Delta \varepsilon \ll kT$, we can use continuous integration:

(This does not consider energy quantization, but considers the uncertainty principle).

In a $2\gamma$-dimensional phase space, $\varepsilon = \varepsilon(q_1, q_2 \dots q_\gamma, p_1, p_2 \dots p_\gamma)$:

III. Macroscopic Quantities in Boltzmann Systems

Partition function (Characteristic function):

Under classical limit conditions:

Using the partition function, we can derive the macroscopic state functions:

• Number of particles: $N = \sum g_i e^{-\alpha - \beta \varepsilon_i} = e^{-\alpha} z \implies \alpha = \ln \frac{z}{N}$
• Internal energy: $E = \sum \varepsilon_i g_i e^{-\alpha - \beta \varepsilon_i} = -N \frac{\partial \ln z}{\partial \beta}$

Differential relations:

In semi-classical distribution, $z = z(\beta, V)$:

In Boltzmann distribution, $z = z(\beta, V)$:

1. Ideal gas of monoatomic molecules: semi-classical distribution, classical limit

• 3D: $$ z = \int_0^\infty e^{-\beta \varepsilon} \frac{2\pi V (2m)^{3/2}}{h^3} \sqrt{\varepsilon} d\varepsilon = V \left(\frac{2\pi m}{\beta h^2}\right)^{3/2} $$ $$ E = -N \frac{\partial}{\partial \beta} \ln z = \frac{3}{2} NkT $$ $$ C_V = \left(\frac{\partial E}{\partial T}\right)_V = \frac{3}{2} Nk $$ $$ P = \frac{N}{\beta} \frac{\partial \ln z}{\partial V} = \frac{NkT}{V} = \frac{2E}{3V} $$
• 2D: $$ z = \int_0^\infty e^{-\beta \varepsilon} \frac{2\pi A m}{h^2} d\varepsilon = \frac{2\pi A m}{\beta h^2} $$ $$ E = NkT, \quad C_V = Nk, \quad P = \frac{NkT}{A} = \frac{E}{A} $$
• 1D: $$ z = \int_0^\infty e^{-\beta \varepsilon} \frac{L \sqrt{2m}}{h} \frac{1}{\sqrt{\varepsilon}} d\varepsilon = \frac{L \sqrt{2\pi m}}{\sqrt{\beta} h} $$ $$ E = \frac{1}{2} NkT, \quad C_V = \frac{1}{2} Nk, \quad P = \frac{NkT}{L} = \frac{2E}{L} $$

2. Ideal gas of monoatomic molecules: semi-classical distribution, ultra-relativistic limit

• 3D: $$ z = \int_0^\infty e^{-\beta \varepsilon} \frac{4\pi V}{(hc)^3} \varepsilon^2 d\varepsilon = \frac{8\pi V}{(\beta hc)^3} $$ $$ E = 3NkT, \quad C_V = 3Nk, \quad P = \frac{NkT}{V} = \frac{E}{3V} $$
• 2D: $$ z = \int_0^\infty e^{-\beta \varepsilon} \frac{2\pi A}{(hc)^2} \varepsilon d\varepsilon = \frac{2\pi A}{(\beta hc)^2} $$ $$ E = 2NkT, \quad C_V = 2Nk, \quad P = \frac{E}{2A} $$
• 1D: $$ z = \int_0^\infty e^{-\beta \varepsilon} \frac{2L}{hc} d\varepsilon = \frac{2L}{\beta hc} $$ $$ E = NkT, \quad C_V = Nk, \quad P = \frac{E}{L} $$

3. Ideal gas of diatomic molecules: semi-classical distribution, classical limit

① Translation: Same as monoatomic molecules.

② Vibration:

• 1D: $$ z^\nu = \sum_{n=0}^\infty e^{-\beta(n+\frac{1}{2})h\nu} \cdot 1 = \frac{e^{-\frac{1}{2}\beta h\nu}}{1 - e^{-\beta h\nu}} $$ $$ E^\nu = -N \frac{\partial}{\partial \beta} \ln z^\nu = N\left[\frac{1}{2}h\nu + \frac{h\nu}{e^{\beta h\nu} - 1}\right] $$ $$ C_V^\nu = \left(\frac{\partial E^\nu}{\partial T}\right)_V = Nk \frac{x^2 e^x}{(e^x - 1)^2}, \quad x = \frac{\Theta^\nu}{T}, \quad \Theta^\nu = \frac{h\nu}{k} = \frac{\Delta\varepsilon^\nu}{k} $$ $$ \begin{cases} T \ll \Theta^\nu \text{ (room temp)}: C_V^\nu \approx Nk \left(\frac{\Theta^\nu}{T}\right)^2 e^{-\frac{\Theta^\nu}{T}} \\ T \gg \Theta^\nu: C_V^\nu \approx Nk \end{cases} \quad P = 0 $$
• 2D: $$ z^\nu = \sum_{n=0}^\infty e^{-\beta(n+1)h\nu} \cdot (n+1) = \frac{e^{\beta h\nu}}{(e^{\beta h\nu}-1)^2} $$ $$ E^\nu = -N \frac{\partial}{\partial \beta} \ln z^\nu = N\left[\frac{2h\nu}{1 - e^{-\beta h\nu}} - h\nu\right] $$ $$ C_V^\nu = \left(\frac{\partial E^\nu}{\partial T}\right)_V = 2Nk \left(\frac{\Theta^\nu}{T}\right)^2 \frac{e^{\frac{\Theta^\nu}{T}}}{(1 - e^{-\frac{\Theta^\nu}{T}})^2} $$ $$ \begin{cases} T \ll \Theta^\nu: C_V^\nu \approx 2Nk \left(\frac{\Theta^\nu}{T}\right)^2 e^{-\frac{\Theta^\nu}{T}} \\ T \gg \Theta^\nu: C_V^\nu \approx 2Nk \end{cases} \quad P = 0 $$

③ Rotation:

When $T \ll \Theta^r$:

When $T \gg \Theta^r$ (room temperature):

4. Paramagnetism of solids: Boltzmann distribution, spin

Magnetic moment:

IV. Macroscopic Quantities in Bose Systems and Fermi Systems

Grand partition function (Characteristic function):

Under classical limit condition:

(where $+$ is for Fermi, $-$ is for Bose).

• Particle number: $N = \sum \frac{g_i}{e^{\alpha+\beta\varepsilon_i} \pm 1} = -\frac{\partial \Phi}{\partial \alpha}$
• Internal energy: $E = \sum \frac{\varepsilon_i g_i}{e^{\alpha+\beta\varepsilon_i} \pm 1} = -\frac{\partial \Phi}{\partial \beta}$

Similar to the Boltzmann derivations, we can get:

From the full differential equation of the open system $dE = TdS + \sum Y_k dy_k + \mu dN$ and the full differential formula of $\Phi$, we can get $\alpha = -\frac{\mu}{kT}$.

1. Strongly Degenerate Bose Gas: Bose Distribution, Classical Limit

We know $n_i = \frac{g_i}{e^{\frac{\varepsilon_i-\mu}{kT}} - 1} \ge 0$. Take $\varepsilon_0 = 0$, then $\mu < 0$.

Since $N = \sum n_i = \text{constant}$, we know that as $T \downarrow \implies \mu \uparrow$. When $T = T_c$, $\mu = 0$.

• 3D: $$ \Phi = -\sum g_i \ln(1 - e^{-\alpha - \beta\varepsilon_i}) \approx -g_0 \ln(1 - e^{-\alpha}) - CV \int_0^\infty \sqrt{\varepsilon} \ln(1 - e^{-\alpha-\beta\varepsilon}) d\varepsilon $$ $$ N = -\frac{\partial \Phi}{\partial \alpha} = CV \int_0^\infty \frac{\sqrt{\varepsilon} d\varepsilon}{e^{\alpha+\beta\varepsilon} - 1} $$

  When $T \le T_c$, $\alpha = \mu = 0$, this leads to Bose-Einstein Condensation (BEC).

  $$ N = CV(kT_c)^{\frac{3}{2}} \frac{\sqrt{\pi}}{2} \zeta\left(\frac{3}{2}\right), \quad T_c = \frac{h^2}{2\pi mk} \left(\frac{N}{2.612 V g}\right)^{2/3} $$ $$ N_{\varepsilon \neq 0}(T) = N \left(\frac{T}{T_c}\right)^{3/2} $$ $$ N_{\varepsilon=0}(T) = N \left[1 - \left(\frac{T}{T_c}\right)^{3/2}\right] $$ $$ E = -\frac{\partial \Phi}{\partial \beta} = 0.770 NkT \left(\frac{T}{T_c}\right)^{3/2} \quad \text{(using numerical integration)} $$ $$ C_V = \left(\frac{\partial E}{\partial T}\right)_V = 1.925 Nk \left(\frac{T}{T_c}\right)^{3/2} $$ $$ P = \frac{1}{\beta} \frac{\partial \Phi}{\partial V} = 0.514 \frac{NkT}{V} \left(\frac{T}{T_c}\right)^{3/2} = \frac{2E}{3V} $$
• 2D / 1D:

  The integral form of $N$ does not converge, meaning there is no BEC.

2. Strongly Degenerate Photon Gas: Bose Distribution, Ultra-relativistic Limit

Photon rest mass is 0. Under any condition, a photon gas is a strongly degenerate Bose gas. Spin degeneracy $g=2$. Particle number is not conserved, so $\alpha=0$.

• 3D: $$ g(\varepsilon) d\varepsilon = \frac{8\pi V}{(hc)^3} \varepsilon^2 d\varepsilon $$ $$ E(\nu, T) d\nu = h\nu n(\nu) d\nu = \frac{8\pi V}{c^3} \frac{h\nu^3}{e^{\beta h\nu} - 1} d\nu $$ $$ \begin{cases} \frac{h\nu}{kT} \ll 1: E(\nu, T) d\nu \approx \frac{8\pi V}{c^3} kT \nu^2 d\nu \\ \frac{h\nu}{kT} \gg 1: E(\nu, T) d\nu \approx \frac{8\pi V}{c^3} h\nu^3 e^{-\beta h\nu} d\nu \end{cases} $$ $$ E(\lambda, T) d\lambda = \frac{hc}{\lambda} n(\lambda) d\lambda = \frac{8\pi hc V}{\lambda^5} \frac{d\lambda}{e^{\beta hc / \lambda} - 1} $$

  From $\frac{\partial E(\lambda, T)}{\partial \lambda} = 0$ we get Wien's Displacement Law: $\lambda_m T = \frac{hc}{4.965 k}$.

  $$ \Phi(\beta, V) \sim T^4 \quad N \sim T^3 \quad E \sim T^4 \quad J = \frac{1}{4} c \frac{E}{V} \text{ (Stefan-Boltzmann law)} $$ $$ C_V \sim T^3 \quad P = \frac{E}{3V} $$
• 2D: $$ g(\varepsilon) d\varepsilon = \frac{4\pi A}{(hc)^2} \varepsilon d\varepsilon $$ $$ E(\nu, T) d\nu = \frac{4\pi A}{c^2} \frac{h\nu^2}{e^{\beta h\nu} - 1} d\nu $$ $$ \Phi(\beta, V) \sim T^3 \quad N \sim T^2 \quad E \sim T^3 \quad J = \frac{1}{\pi} c \frac{E}{A} $$ $$ C_V \sim T^2 \quad P = \frac{E}{2A} $$
• 1D: $$ g(\varepsilon) d\varepsilon = \frac{4L}{hc} d\varepsilon $$ $$ E(\nu, T) d\nu = \frac{4L}{c} \frac{h\nu}{e^{\beta h\nu} - 1} d\nu $$ $$ \Phi(\beta, V) \sim T^2 \quad N \sim T \quad E \sim T^2 \quad C_V \sim T \quad P = \frac{E}{L} $$

3. Strongly Degenerate Phonon Gas: Bose Distribution, Ultra-relativistic Limit

The phonon quantum number can take any integer under simplification, so the phonon gas is a strongly degenerate Bose gas. $g_l=1, g_t=2$. Vibrational state is not conserved, so $\alpha=0$.

① Einstein Model: $\nu = \nu_E$

② Debye Model: $\nu \le \nu_D$

• 3D: $$ g(\varepsilon) d\varepsilon = \frac{d\Omega_l + d\Omega_t}{h^3} = \frac{4\pi V}{h^3} \left(\frac{1}{v_l^3} + \frac{2}{v_t^3}\right) \varepsilon^2 d\varepsilon $$

  From $\int_0^{\nu_D} g(\nu) d\nu = \int_0^{\nu_D} B\nu^2 d\nu = 3N$ we get $\nu_D = \left[\frac{9N}{4\pi V(\frac{1}{v_l^3} + \frac{2}{v_t^3})}\right]^{1/3}$ and $B = \frac{9N}{\nu_D^3}$.

  $$ E = \int_0^{h\nu_D} \frac{\varepsilon g(\varepsilon)}{e^{\beta\varepsilon} - 1} d\varepsilon = \frac{9N}{(h\nu_D)^3} \int_0^{h\nu_D} \frac{\varepsilon^3}{e^{\beta\varepsilon} - 1} d\varepsilon $$

  Let $\Theta^D = \frac{h\nu_D}{k}$:

  $$ \begin{cases} T \ll \Theta^D: C_V \sim T^3 \\ T \gg \Theta^D: C_V \approx 3Nk \end{cases} \quad P = \frac{E}{3V} $$
• 2D: $$ g(\varepsilon) d\varepsilon = \frac{d\Omega_l + d\Omega_t}{h^2} = \frac{2\pi A}{h^2} \left(\frac{1}{v_l^2} + \frac{1}{v_t^2}\right) \varepsilon d\varepsilon $$

  From $\int_0^{\nu_D} g(\nu) d\nu = 3N$ we get $\nu_D = \left[\frac{3N}{\pi A(\frac{1}{v_l^2} + \frac{1}{v_t^2})}\right]^{1/2}$ and $B = \frac{6N}{\nu_D^2}$.

  $$ E = \frac{6N}{(h\nu_D)^2} \int_0^{h\nu_D} \frac{\varepsilon^2}{e^{\beta\varepsilon} - 1} d\varepsilon $$ $$ \begin{cases} T \ll \Theta^D: C_V \sim T^2 \\ T \gg \Theta^D: C_V \approx 3Nk \end{cases} \quad P = \frac{E}{2A} $$
• 1D: $$ g(\varepsilon) d\varepsilon = \frac{d\Omega_l + d\Omega_t}{h} = \frac{2L}{h} \left(\frac{1}{v_l} + \frac{1}{v_t}\right) d\varepsilon $$

  From $\int_0^{\nu_D} g(\nu) d\nu = 3N$ we get $\nu_D = \frac{3N}{2L(\frac{1}{v_l} + \frac{1}{v_t})}$ and $B = \frac{3N}{\nu_D}$.

  $$ E = \frac{3N}{h\nu_D} \int_0^{h\nu_D} \frac{\varepsilon}{e^{\beta\varepsilon} - 1} d\varepsilon $$ $$ \begin{cases} T \ll \Theta^D: C_V \sim T \\ T \gg \Theta^D: C_V \approx 3Nk \end{cases} \quad P = \frac{E}{L} $$

Note: Sommerfeld Expansion

Calculate the integral of $f(\varepsilon) \varphi(\varepsilon)$ under $0 < T \ll T_F$:

Proof: using integration by parts $I = -[g(\varepsilon) f(\varepsilon)]_0^\infty - \int_0^\infty g(\varepsilon) \frac{df}{d\varepsilon} d\varepsilon$. Since $f$ is non-zero only near $\varepsilon = \mu$, expand $g(\varepsilon)$ near $\mu$ and substitute. Let $x = \frac{\varepsilon - \mu}{kT}$. Since $T \ll \frac{\mu}{k}$, the lower bound approaches $-\infty$, and since $\frac{\partial f}{\partial x}$ is an even function, we integrate over the whole axis to get the expansion formula.

4. Strongly Degenerate Fermi Gas: Fermi Distribution, Classical Limit

① At $T = 0K$:

• 3D: $$ N_0 = \frac{2}{3} \cdot \frac{2\pi V \cdot 2(2m)^{3/2}}{h^3} \cdot \mu_0^{3/2} \implies \mu_0 = \frac{h^2}{2m} \left(\frac{3N_0}{8\pi V}\right)^{2/3} $$ $$ E_0 = \int_0^{\mu_0} \varepsilon g(\varepsilon) d\varepsilon = \frac{3}{5} N_0 \mu_0 $$ $$ P_0 = -\frac{\partial E_0}{\partial V} = \frac{2E_0}{3V} $$
• 2D: $$ N_0 = \frac{2\pi A \cdot 2 m}{h^2} \mu_0 \implies \mu_0 = \frac{h^2}{2m} \frac{N_0}{2\pi A} $$ $$ E_0 = \int_0^{\mu_0} \varepsilon g(\varepsilon) d\varepsilon = \frac{1}{2} N_0 \mu_0, \quad P_0 = -\frac{\partial E_0}{\partial A} = \frac{E_0}{A} $$
• 1D: $$ N_0 = 2 \cdot \frac{L \sqrt{2m}}{h} \cdot \mu_0^{1/2} \implies \mu_0 = \frac{h^2}{2m} \left(\frac{N_0}{2L}\right)^2 $$ $$ E_0 = \int_0^{\mu_0} \varepsilon g(\varepsilon) d\varepsilon = \frac{1}{3} N_0 \mu_0, \quad P_0 = -\frac{\partial E_0}{\partial L} = \frac{2E_0}{L} $$

② When $0 < T \ll \frac{\mu_0}{k} = T_F$:

• 3D: $$ \Phi(\alpha, \beta, V) = CV \int_0^\infty \ln(1 + e^{-\alpha-\beta\varepsilon}) \varepsilon^{1/2} d\varepsilon $$ $$ N = -\frac{\partial \Phi}{\partial \alpha} = \frac{2}{3} CV \mu^{3/2} \left[1 + \frac{\pi^2}{8} \left(\frac{kT}{\mu}\right)^2 + \dots\right] $$

  Solving for chemical potential:

  $$ \mu = \mu_0 \left[1 - \frac{1}{12} \left(\frac{\pi kT}{\mu_0}\right)^2 - \dots\right] $$ $$ E = E_0 \left[1 + \frac{5}{12} \left(\frac{\pi kT}{\mu_0}\right)^2 - \dots\right] $$ $$ C_V \approx Nk \frac{\pi^2}{2} \cdot \frac{kT}{\mu_0} $$ $$ S = k\left(\Phi - \beta \frac{\partial \Phi}{\partial \beta} - \alpha \frac{\partial \Phi}{\partial \alpha}\right) = \frac{1}{2\mu_0} N \pi^2 k^2 T \dots $$

5. Strongly Degenerate Fermi Gas: Fermi Distribution, Ultra-relativistic Limit

At $T = 0K, S_0 = 0$:

• 3D: $$ N_0 = \frac{8\pi V}{(hc)^3} \cdot \frac{1}{3} \mu_0^3 \implies \mu_0 = \frac{hc}{2} \left(\frac{3N_0}{\pi V}\right)^{1/3} $$ $$ E_0 = \frac{3}{4} N_0 \mu_0, \quad P_0 = \frac{E_0}{3V} $$
• 2D: $$ N_0 = \frac{4\pi A}{(hc)^2} \cdot \frac{1}{2} \mu_0^2 \implies \mu_0 = \frac{hc}{2\pi} \left(\frac{N_0}{A}\right)^{1/2} $$ $$ E_0 = \frac{2}{3} N_0 \mu_0, \quad P_0 = \frac{E_0}{2A} $$
• 1D: $$ N_0 = \frac{4L}{hc} \cdot \mu_0 \implies \mu_0 = \frac{hc}{4} \frac{N_0}{L} $$ $$ E_0 = \frac{1}{2} N_0 \mu_0, \quad P_0 = \frac{E_0}{L} $$

Energy scaling and pressure relations (from Schrödinger equation):

$$ E \sim p^L $$

For a 3D infinite potential well, solving the Schrödinger equation $\mathcal{H}\Psi = E\Psi$ gives boundary conditions containing $V^{-1/3}$. In general, for an $n$-dimensional infinite potential well:

(Assuming $H = E + PV$ holds).