Đạo hàm phân đoạn, dao động năng lượng và nhiệt dung
Đã giới thiệu| English | Ý nghĩa |
|---|---|
| heat capacity/hiːt kəˈpæsɪti/ | The heat needed to raise a body’s temperature by one kelvin. |
| free energy/friː ˈenədʒi/ | The quantity minus kT ln Z, minimised at equilibrium. |
| partition function/pɑːˈtɪʃn ˈfʌŋkʃn/ | The canonical sum of Boltzmann weights over microstates, used to normalise their probabilities and determine equilibrium thermodynamic properties. |
Small heat capacity 热容 — hot and cold
- A two-level system absorbs little heat when it is very cold (it cannot reach the upper level) and again when very hot (both levels are already equally filled). Its heat capacity peaks in between.
- GRE statistical mechanics tests $Z$ and its derivatives: mean energy, free energy 自由能, entropy, heat capacity and energy fluctuations.
The partition function 配分函数
- $Z = \sum_i g_ie^{-\beta E_i}$, $\beta = 1/k_BT$. Level probability $g_ie^{-\beta E_i}/Z$; $U = -\partial\ln Z/\partial\beta$ with the spectrum held fixed.
- Independent distinguishable subsystems: $Z_{\text{tot}} = z^N$, $U_{\text{tot}} = Nu$.
Ground level $g = 1$ at 0; excited level $g = 2$ at $\Delta = k_BT\ln2$. Find $Z$.
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
$Z = 1 + 2e^{-\ln 2} = 1 + 2 \times 1/2 = 2$.
Free energy, entropy, heat capacity
- $F = -k_BT\ln Z$; $S = k_B(\ln Z + \beta U)$; $C_V = (\partial U/\partial T)_V$.
- Shifting every level by a constant $\varepsilon$ shifts $U$ and $F$ by $\varepsilon$ and leaves probabilities, $S$ and $C$ unchanged.
Adding a constant $\varepsilon$ to every energy level changes
Độ khó 1/5 · 100 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
U shifts by ε; the others do not.
Fluctuations
- $\text{Var}(E) = \langle E^2\rangle - U^2$ and $C_V = \dfrac{\text{Var}(E)}{k_BT^2}$, so $C_V \ge 0$. Relative fluctuations shrink like $1/\sqrt N$.
Levels $0$ and $\Delta$ with $\Delta/k_BT = \ln3$: $z = \tfrac43$, $p_{\text{exc}} = \tfrac14$, $U = \Delta/4$, $\text{Var} = \tfrac{3}{16}\Delta^2$, so $C/k_B = \tfrac{3}{16}(\ln3)^2 \approx 0.226$.
At very high temperature, the two-level energy variance is $k\Delta^2$. Find $k$.
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
At very high temperature each level has probability 1/2: $\langle E^2\rangle - \langle E\rangle^2 = \Delta^2/2 - \Delta^2/4 = \Delta^2/4$.
The two-level system
- $x = \Delta/k_BT$: $p_{\text{exc}} = \dfrac{1}{1 + e^x}$, $\dfrac{C}{k_B} = \dfrac{x^2e^x}{(1 + e^x)^2}$.
- Low $T$: everything freezes out. High $T$: $p_{\text{exc}} \to \tfrac12$, $U \to \Delta/2$, $S \to k_B\ln2$, $C \to 0$.
Cold limit or hot limit?
Recall the low- and high-temperature limits of a two-level system.
Two nondegenerate levels with $\Delta/k_BT = \ln 2$. Excited-state probability (three decimal places)?
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
The excited weight is $e^{-\ln2} = 1/2$ against 1 for the ground state, so $p = (1/2)/(1 + 1/2) = 1/3$.
The heat capacity of a two-level system grows without limit as the temperature rises.
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The heat capacity of a two-level system peaks (the Schottky anomaly) and then falls to zero once both levels are equally filled.
Recap
- Include degeneracies in $Z$; differentiate with the levels fixed.
- $F = -kT\ln Z$; $C_V = \text{Var}(E)/kT^2$.
- A finite spectrum gives a heat-capacity peak, vanishing at both ends.
- heat capacity: The heat needed to raise a body’s temperature by one kelvin.
- free energy: The quantity minus kT ln Z, minimised at equilibrium.