物理

(栖岸计划)黑体辐射

$G(T) = \frac{dE}{ds \cdot dt} = σ \cdot T^4 = \int_0^\infty g(\nu, t) \, d\nu$

$u(\nu, T) d\nu = n(\nu, T) \cdot f(\nu, t) d\nu \cdot h\nu$

$ G \cdot T $ 圈出

光子泄漏（光子通量）：

$dN = c \cos\theta , dt \cdot ds \cdot \frac{\sin\theta , d\theta \times 2\pi}{4\pi} \cdot n dp$

$= c \cos\theta , dt \cdot ds \cdot \frac{\sin\theta , d\theta}{2} \cdot n(\nu, T) f(\nu, T) d\nu$

$g(\nu, t) , d\nu = \int_0^{\frac{\pi}{2}} h\nu \cdot \frac{dN}{ds , dt}$

$= \int_0^{\frac{\pi}{2}} h\nu \cdot c \cos\theta \cdot \frac{\sin\theta , d\theta}{2} \cdot n(\nu, T) f(\nu, T) d\theta $

$= \frac{1}{4} c \, u(\nu, T) d\nu$

波矢空间与简并度：

$N\nu d\nu = \frac{4\pi k^2 \, dk}{(2\pi)^3} \times 2 \quad \text{(每个模式两个偏振)}$

$k = \frac{\omega}{c} = \frac{2\pi \nu}{c}, \quad \beta = V$

$N\nu d\nu = \Theta \cdot 8\pi \cdot \frac{d\nu}{c^3} \cdot V$

Bose 分布：

$\frac{a_i}{\omega_i} = \frac{1}{e^{\beta \omega_i} - 1}, \quad \varepsilon = m_c h\nu, \quad m = 0,1,2,\dots $

$P_m = A(\nu) \cdot e^{-\frac{h\nu}{kT} m} \quad \text{(Boltzmann 分布)}$

平均粒子数（Bose-Einstein 分布）：

$\bar{n}(\nu) = \sum_{m=0}^\infty m \cdot P_m = \frac{e^{-\frac{h\nu}{kT}}}{1 - e^{-\frac{h\nu}{kT}}} = \frac{1}{e^{\frac{h\nu}{kT}} - 1}$

$\Rightarrow \bar{n}(\nu) = \frac{1}{e^{\frac{h\nu}{kT}} - 1}$

温度下总粒子数密度：

$n(T) = \int f(\nu, T) \, d\nu = \frac{N(\nu) \, d\nu}{V} \cdot \bar{n}(\nu)$

$= \frac{8\pi \nu^2}{c^3} \cdot \frac{1}{e^{\frac{h\nu}{kT}} - 1} d\nu $

 黑体能量密度：

$u(\nu, T) d\nu = n(\nu, T) \cdot f(\nu, t) d\nu \cdot h\nu$

$= \frac{N(\nu) d\nu}{V} \cdot \bar{n}(\nu) \cdot h\nu$

$= \frac{8\pi \nu^2 d\nu}{c^3} \cdot \frac{h\nu}{e^{\frac{h\nu}{kT}} - 1}$

$g(\nu, t) d\nu = \frac{1}{4} c u(\nu, T) d\nu$

$\int_0^\infty g(\nu, t) d\nu = \frac{1}{4} c \int_0^\infty u(\nu, T) d\nu$

$= \frac{1}{4} c \cdot \frac{8\pi h}{c^3} \int_0^\infty \frac{\nu^3}{e^{\frac{h\nu}{kT}} - 1} \, d\nu$

令 $ x = \frac{h\nu}{kT} $，则 $ \nu = \frac{kT}{h} x $，$ d\nu = \frac{kT}{h} dx $

$= \frac{1}{4} c \cdot \frac{8\pi h}{c^3} \left( \frac{kT}{h} \right)^4 \int_0^\infty \frac{x^3}{e^x - 1} dx$

$= \frac{2\pi k^4 T^4}{c^2 h^3} \int_0^\infty \frac{x^3}{e^x - 1}dx$

关键积分：$ \int_0^\infty \frac{x^3}{e^x - 1} dx $

$\frac{1}{e^x - 1} = \frac{e^{-x}}{1 - e^{-x}} = \sum_{n=1}^\infty e^{-nx}$

$\Rightarrow \int_0^\infty \frac{x^3}{e^x - 1} dx = \int_0^\infty x^3 \sum_{n=1}^\infty e^{-nx} dx = \sum_{n=1}^\infty \int_0^\infty x^3 e^{-nx} dx$

令 $ u = nx $，则：

$\int_0^\infty x^3 e^{-nx} dx = \frac{1}{n^4} \int_0^\infty u^3 e^{-u} du = \frac{\Gamma(4)}{n^4} = \frac{6}{n^4}$

$\Rightarrow \int_0^\infty \frac{x^3}{e^x - 1} dx = 6 \sum_{n=1}^\infty \frac{1}{n^4} = 6 \zeta(4)$

已知 $ \zeta(4) = \frac{\pi^4}{90} $，所以：

$\int_0^\infty \frac{x^3}{e^x - 1} dx = 6 \cdot \frac{\pi^4}{90} = \frac{\pi^4}{15}$

$G(T) = \frac{2\pi k^4 T^4}{c^2 h^3} \cdot \frac{\pi^4}{15} = \left( \frac{2\pi^5 k^4}{15 c^2 h^3} \right) T^4$

$\Rightarrow G(T) = \sigma T^4, \quad \sigma = \frac{2\pi^5 k^4}{15 c^2 h^3}$