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title: Thermal Physics
tags: 上課筆記

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# Thermal Physics
:::info
**thermodynamics：** 大尺度
the arrow of time 時間有一定的演化方向
:::

:::info
**statistical mechanics：** 小尺度
time-reversal-invariant 時間反轉不變
:::

${N_A}$ = $6.02 \times 10^{23}$
$\Omega(狀態數量) = C_{n}^{r}$

#### e.g. 估計$10^{23}!$的數量級：
$ln(10^{23}!) \approx 10^{23}ln(10^{23})-10^{23} \approx 5.2 \times 10^{24}$

==Stiring's formula==
$ln(N!) \approx Nln(N) - N$ $\rightarrow$ $\{N | N \gg 1\}$

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_Temperature_ T

$T_A >T_B$：未達熱平衡

**Thermal Equilibrium**
允許兩物體能量交換 $\rightarrow$ 熱接觸thermal contact
達到熱平衡的過程：thermolization 熱化

### 熱力學第零定律

若物體A和B達熱平衡，B、C亦達平衡，則A、C必達平衡。

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溫度絕對定義：
* Carnot engine
* Boltzmann's definition

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micro表示法: (name, energy)
macro表示法: (energy, amount)

#### Statistical definition of temperature

1. Postulate of equal a priori probability： For a macrostate in equilibrium all microstates corresponding to that macrostate (equilibrium state) are equally likely.
$\rightarrow$ The probability of finding the system in any one of the microstates = $\frac{1}{\Omega}$, where $\Omega \equiv$ number of accessible states

2. The observed macrostate is the one with the most microstates.

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$E_{tot} = E_1 +E_2 = const$
$\Omega_{tot}(E_{tot}) = \Omega_{1}(E_{1}) \cdot \Omega_{2}(E_{2})$
:::info
$\Omega_{1}(E_{1}) \cdot \Omega_{2}(E_{2})其實就是聯集的意思。$
:::

When the equilibrium is established, we must have：

$\displaystyle \frac{d \Omega_{tot}}{dE_1} = 0$

$\rightarrow i.e.\displaystyle \frac{d}{dE_1}[\Omega_{1}(E_{1}) \cdot \Omega_{2}(E_{2})] = 0$

$\displaystyle\rightarrow \Omega_2\frac{d \Omega_{1}(E_1)}{dE_1} + \Omega_1\frac{d \Omega_{2}(E_2)}{dE_2} \frac{d E_2}{d E_1} = 0$

$\displaystyle\because E_1 + E_2 = const$, $\displaystyle\therefore \frac{d E_2}{d E_1} = -1$

$\displaystyle\rightarrow \Omega_2\frac{d \Omega_{1}}{dE_1} = \Omega_1\frac{d \Omega_{2}}{dE_2}$

$\displaystyle\therefore \frac{1}{\Omega_1}\frac{d \Omega_{1}}{dE_1} = \frac{1}{\Omega_2}\frac{d \Omega_{2}}{dE_2}$

$\displaystyle i.e. \frac{d \ln \Omega_1}{d E_1} = \frac{d \ln \Omega_2}{d E_2}$

According to the $0^{th}$ law, we can define the temperature as
$\displaystyle\frac{1}{k_BT} \equiv \frac{d \ln \Omega}{d E}$

Boltzmann constant $k_B = 1.38 \times 10^{-23} J/K$

**The Boltzmann distribution**

$\mathcal{P}(E)$ is probability

$\varepsilon + E_R = E$

$\because E \gg \varepsilon$

$i.e.E - \varepsilon \gg \varepsilon$

$\therefore \Omega(\varepsilon) \ll \Omega(E - \varepsilon)$

$\mathcal{P}(\varepsilon) \propto \Omega(\varepsilon)\Omega(E - \varepsilon)$

$i.e.\mathcal{P}(\varepsilon) \propto \Omega(E - \varepsilon)$

#### Taylor expand $\displaystyle\ln\Omega(E - \varepsilon)$ around $\varepsilon = 0$ up to $1^{st}$ order：

$\displaystyle\ln\Omega(E - \varepsilon) = \displaystyle\ln\Omega(E) - \frac{d\ln\Omega(E)}{dE}\varepsilon + ...$ (因為$\varepsilon$很小，高階項可忽略)

$\displaystyle\therefore\ln\Omega(E - \varepsilon) \approx \ln\Omega(E) - \frac{\varepsilon}{k_BT}$

$\rightarrow \Omega(E - \varepsilon) = \Omega(E)e^{- \frac{\varepsilon}{k_BT}}$

* Remark：
1. $\displaystyle e^{-\frac{\varepsilon}{k_BT}}$ is called the Boltzmann factor.

2. Significant probability to reach equilibrium if $\varepsilon <k_BT$.
Negligible probability to reach equilibrium if $\varepsilon > k_BT$.

3. if $\varepsilon = E_r$ , then $\displaystyle\mathcal{P}(E_r) = \frac{e^{-E_r/k_BT}}{\displaystyle\sum_i e^{-E_i/k_BT}}$
$E_r$ is random one.

4. $\displaystyle\beta \equiv \frac{1}{k_BT} = \frac{d\ln\Omega}{dE}$

$\therefore \mathcal{P}(E_r) = \frac{e^{-\beta E_r}}{\displaystyle\sum_i e^{-\beta E_i}}$