物理

{栖岸计划}热力统计学大赏

$Maxwell$速度分布律、$Boltzmann$分布

$Maxwell 速度分布律$

$\vec{v}(v_x, v_y, v_z)$

$dp = A \cdot e^{-\frac{m v^2}{2kT}} dv_x dy dv_z$

$Boltzmann$分布律

$p \propto e^{-\frac{E}{kT}}, \quad f(E) = A e^{-\frac{E}{kT}}$

$Maxwell$符合 $Boltzmann$ 分布

$x, y, z \text{ 方向等权}$

$\int_{-\infty}^{\infty} A^{\frac{1}{3}} \cdot e^{-\frac{m v_x^2}{2kT}} dv_x = 1$

利用 $Gauss$ 积分：

$A = \left( \frac{m}{2\pi kT} \right)^{\frac{3}{2}}$

$dp = \left( \frac{m}{2\pi kT} \right)^{\frac{3}{2}} e^{-\frac{m}{2kT}(v_x^2 + v_y^2 + v_z^2)} dv_x dv_y dv_z$

$f(v) = \left( \frac{m}{2\pi kT} \right)^{\frac{3}{2}} e^{-\frac{m}{2kT} v^2}, \quad v^2 = v_x^2 + v_y^2 + v_z^2$

$dp = f(v) \cdot 4\pi v^2 dv$

在 $ v $ 速度相空间的单位体积中：

$\text{由 } dx \cdot dp = h_0 \text{为相格体积}$

在量子力学中作普朗克常数

$\text{能级 } \varepsilon →\varepsilon + d\varepsilon \sim \omega$

$\text{状态数 } \sim \frac{\omega_0}{h_0} $

$\text{能级简并数 } \sim \sum \frac{\omega_0}{h_0}$

$p = \sqrt{2m\varepsilon}, \quad dp = \frac{m}{\sqrt{2m\varepsilon}} d\varepsilon$

$d\omega = \sum dp_x \cdot dx \cdot dp_y \cdot dy \cdot dp_z \cdot dz = \sum dp_x \cdot dp_y \cdot dp_z \cdot dx \cdot dy \cdot dz$

$= 4\pi p^2 dp \cdot V$

三维相格 $\sim h_0^3$

$\text{能级简并数 } \sim \frac{d\omega}{h_0^3} = \frac{4\pi p^2 dp \cdot V}{h_0^3}$

使用 $Gauss$ 公式推导 $Maxwell$ 速率分布

$Gauss's formula$

$\int_0^\infty e^{-a x^2} dx = \frac{1}{2} \sqrt{\frac{\pi}{a}} \quad (\text{当 } n=0)$

只考虑单方向，由 $-\infty \to \infty$，加系数 2：

$2A^{\frac{1}{3}} \int_0^\infty e^{-\frac{m v_x^2}{2kT}} dv_x = 1$

$A^{\frac{1}{3}} \sqrt{\frac{\pi}{\frac{m}{2kT}}} = 1 \Rightarrow A = \left( \frac{m}{2\pi kT} \right)^{\frac{3}{2}}$

用 $ v_x, v_y, v_z $ 的独立性，出得 Gauss 积分：

$\frac{d G_n(a)}{da} = G_{n+2}(a)$

$\bar{V} = \int_0^\infty f(v) \cdot 4\pi v^2 dv \sim \int_0^\infty v dp$

$\bar{V} = \int_0^\infty f(v) \cdot 4\pi v^3 dv$

$\bar{V} = 4\pi \left( \frac{m}{2\pi kT} \right)^{\frac{3}{2}} \int_0^\infty e^{-\frac{m}{2kT} v^2} v^3 dv$

$G_4(\frac{m}{2kT})=\frac{3}{8} \sqrt{\frac{π}{a^2}}=\frac{3}{8} \sqrt{\frac{\pi}{(\frac{m}{2kT}^2}} =\frac{3}{8} (\frac{2kT}{m})^{\frac{3}{2}} \sqrt{π}$

$\bar{V} = 4\pi \left( \frac{m}{2\pi kT} \right)^{\frac{3}{2}} \cdot \frac{3}{2} \left( \frac{2kT}{m} \right)^{\frac{3}{2}} \sqrt{\pi}$

$= \frac{3}{2} \pi^{\frac{3}{2}} \left( \frac{1}{\pi} \right)^{\frac{3}{2}} \cdot \frac{2kT}{m} = \frac{3}{2} \cdot \frac{2kT}{m}$

$= \frac{3kT}{m}$

$\bar{v}_x^2 + \bar{v}_y^2 + \bar{v}_z^2 = \frac{3kT}{m}$

$\bar{v}_x^2 = \bar{v}_y^2 = \bar{v}_z^2 = \frac{kT}{m}$

$\frac{1}{2} m \bar{v}^2 = \frac{3}{2} kT$

$\frac{1}{2} m \langle \bar{v}^2 \rangle = \frac{3}{2}kT$

$\bar{v}^2 = v^2 \int dp = v^2 \int 4\pi v^2 dv \cdot f(v) = 4\pi \int f(v) v^4 dv$

$= 4\pi \int \left( \frac{m}{2\pi kT} \right)^{\frac{3}{2}} e^{-\frac{m}{2kT} v^2} v^4 dv$

$= 4\pi \left( \frac{m}{2\pi kT} \right)^{\frac{3}{2}} \int e^{-\frac{m}{2kT} v^2} v^4 dv$

$G_4(\frac{m}{2kT}) = \frac{3}{8} \sqrt{\frac{π}{a^2}}=\frac{3}{8} \sqrt{\frac{π}{(\frac{m}{2kT}^2}}= \frac{3}{8}(\frac{2kT}{m})^{\frac{3}{2}} \sqrt{π}$

$\bar{v}^2 = 4\pi \left( \frac{m}{2\pi kT} \right)^{\frac{3}{2}} \cdot \frac{3}{2} \left( \frac{2kT}{m} \right)^{\frac{3}{2}} \sqrt{\pi}$

$= \frac{3}{2} \pi^{\frac{3}{2}} \left( \frac{1}{\pi} \right)^{\frac{3}{2}} \cdot \frac{2kT}{m} = \frac{3}{2} \cdot \frac{2kT}{m}$

$= \frac{3kT}{m}$

扩散流流密度

$J = \frac{dN}{ds dt}$  扩散流率

$dp = 4\pi v^2 \cdot f(v) dv \cdot \frac{d\Omega}{4\pi} = 4\pi v^2 \cdot f(v) dv \cdot \frac{\sin\theta d\theta}{2}$

$J = \frac{dN}{ds dt} = \frac{1}{ds dt} \int_0^\infty n dp \cdot ds \cdot v dt \cos\theta$

$dN = n dp \cdot [v \cos\theta dt \times ds]$

$J = \frac{1}{ds dt} \int_0^\infty \int_0^{\frac{\pi}{2}} n dp \cdot v \cos\theta \cdot ds \cdot dt$

$= \int_0^\infty \int_0^{\frac{\pi}{2}} n v \cos\theta \cdot f(v) \cdot n dv \cdot \int_0^{\frac{\pi}{2}} \sin\theta \cos\theta d\theta$

$= \int_0^\infty n^2 v^3 \left( \frac{m}{2\pi kT} \right)^{\frac{3}{2}} e^{-\frac{m}{2kT} v^2} dv \cdot \frac{1}{2}$

$= n \bar{v} \int_0^\infty v^3 \left( \frac{m}{2\pi kT} \right)^{\frac{3}{2}} e^{-\frac{m}{2kT} v^2} dv$

$G_3 = -\frac{\partial G_1}{\partial a} = \pi \left( \frac{1}{2a^2} \right)$

$J = n \bar{v}\frac{m}{2\pi kT}^{\frac{3}{2}} (\frac{1}{2}\frac{m}{2kT})^2= \frac{1}{2} n(\frac{2kT}{m})^{\frac{1}{2}}$

$= \frac{1}{2} n \left( \frac{8kT}{m\pi} \right)^{\frac{1}{2}} = \frac{1}{2} n \left( \frac{8kT}{m\pi} \right)^{\frac{1}{2}}$

$= \frac{1}{2} n \left( \frac{8kT}{m\pi} \right)^{\frac{1}{2}} \Rightarrow J = \frac{1}{2} n \bar{v}$

Boltzmann 分布率、状态数、统计系统

 Boltzmann 分布率

$f(\varepsilon) = A e^{-\frac{\varepsilon}{kT}}, \quad \varepsilon = \frac{p^2}{2m} + V(r)$

$dx \cdot dp = h_0$为相格体积

在量子力学中作普朗克常数

能级 $\varepsilon \to \varepsilon + d\varepsilon \sim d\omega$

状态数 $\sim \frac{\omega_0}{h_0}$

能级简并数 $\sim \sum \frac{\omega_0}{h_0}$

$p = \sqrt{2m\varepsilon}, \quad dp = \frac{m}{\sqrt{2m\varepsilon}} d\varepsilon$

$d\omega = \sum dp_x \cdot dx \cdot dp_y \cdot dy \cdot dp_z \cdot dz = \sum dp_x \cdot dp_y \cdot dp_z \cdot dx \cdot dy \cdot dz$

$= 4\pi p^2 dp \cdot V$

三维相格 $\sim h_0^3$

能级简并数 $\sim \frac{d\omega}{h_0^3} = \frac{4\pi p^2 dp \cdot V}{h_0^3}$

状态数 $\Omega$ 为系统状态数（$Maxwell-Boltzmann$系统）

$\Omega = \frac{N!}{\prod_i a_i!} \cdot \prod_i (w_i)^{a_i}$

$\ln \Omega ≈ \ln N! - \sum \ln a_i! + \sum \ln w_i \cdot a_i$

Stirling 近似：$\ln M! \approx M \ln M - M$, $M \gg 1$

$\ln \Omega \approx \sum (\ln N - 1) N - \sum (\ln a_i - 1) a_i + \sum a_i \ln w_i$

$\delta \ln \Omega = \sum [-\ln a_i \cdot \delta a_i + \ln w_i \cdot \delta a_i]$

约束条件：

$\sum a_i = N \Rightarrow \sum \delta a_i = 0$①

$\sum \varepsilon_i a_i = E \Rightarrow \sum \varepsilon_i \delta a_i = 0$②

引入 Lagrange 数乘：

$\text{(①)} + \alpha \text{(②)} + \beta \text{(③)} = 0$

$\sum [-\ln a_i + \ln w_i - \alpha - \beta \varepsilon_i] \delta a_i = 0$

$\ln \frac{w_i}{a_i} = \alpha + \beta \varepsilon_i$

$\frac{a_i}{w_i} = e^{-\alpha} e^{-\beta \varepsilon_i}$

$p_i = \frac{a_i}{w_i} = A e^{-\frac{\varepsilon_i}{kT}}$

推导出 $Boltzmann$ 分布

 Bose 系统

$\Omega = \prod_i \frac{(w_i + a_i - 1)!}{a_i! (w_i - 1)!}$

$\ln \Omega \approx \sum [\ln(w_i + a_i - 1)! - \ln a_i! - \ln(w_i - 1)!]$

$\delta \ln \Omega \approx \sum [\ln(w_i + a_i) \delta a_i - \ln a_i \cdot \delta a_i - \ln(w_i - 1) \cdot \delta a_i] = 0$

$\sum [\ln(w_i + a_i) - \ln a_i - \ln(w_i - 1)] \delta a_i = 0$

$\sum [\ln(w_i + a_i) - \ln a_i - \ln(w_i - 1)] \delta a_i = 0$

约束：

$\sum \delta a_i = 0$

$\sum \varepsilon_i \delta a_i = 0$

$\text{(①)} + \alpha \text{(②)} + \beta \text{(③)} = 0$

$\sum [\ln(w_i + a_i) - \ln a_i - \alpha - \beta \varepsilon_i] \delta a_i = 0$

$\ln \frac{w_i + a_i}{a_i} = \alpha + \beta \varepsilon_i$

$\frac{a_i}{w_i} = \frac{1}{e^{\alpha + \beta \varepsilon_i} - 1}$

$Bose$分布：

$p_i = \frac{1}{e^{\alpha + \beta \varepsilon_i} - 1}$

$Fermi$系统

$\Omega = \prod_i \frac{w_i!}{(w_i - a_i)! a_i!}$

$\ln \Omega = \sum \ln w_i! - \sum \ln(w_i - a_i)! - \sum \ln a_i!$

$\delta \ln \Omega = \sum \ln w_i \cdot \delta a_i - \sum \ln(w_i - a_i) \cdot \delta a_i - \sum \ln a_i \cdot \delta a_i$

$= \sum [\ln w_i - \ln(w_i - a_i) - \ln a_i] \delta a_i$

约束：

$\sum \delta a_i = 0$

$\sum \varepsilon_i \delta a_i = 0$

$\text{(①)} + \alpha \text{(②)} + \beta \text{(③)} = 0$

$\sum [\ln w_i - \ln(w_i - a_i) - \ln a_i - \alpha - \beta \varepsilon_i] \delta a_i =0$

$\ln \frac{w_i - a_i}{a_i} = \alpha + \beta \varepsilon_i$

$\frac{w_i - a_i}{a_i} = e^{\alpha + \beta \varepsilon_i}$

$p_i = \frac{1}{e^{\alpha + \beta \varepsilon_i} +1}$