声子stefan-boltzmann系数

声子stefan-bolzamann系数

黑体边界

给定温度边界，声子半球辐射热流为： \[ \begin{aligned} E(T) &= \sum_p \int_0^{2\pi} \mathrm{d}\phi \int_0^{\pi/2} \cos\theta\sin\theta\mathrm{\mathrm{d}\theta} \int_0^{\omega_m} \hbar\omega\frac{D(\omega)}{4\pi}f^0(T)v_g\mathrm{d}\omega \\ &= \sum_p \frac{1}{4} \int_0^{\omega_m}v_g\hbar\omega D(\omega)f^0(T)\mathrm{d}\omega \\ &= \sum_p \frac{1}{4} \int_0^{\omega_m}v_g\hbar\omega \frac{k^2}{2\pi^2v_g} \frac{1}{\exp(\hbar\omega/k_BT - 1)}\mathrm{d}\omega \end{aligned} \] 引入两个近似

• 灰体近似：假设声子的性质和频率无关
• 线性色散：假设声子的群速度为常数，这就是Debye模型中的近似，此时声子相速度和群速度相等

在这两个近似下，辐射热流可以写成： \[ \begin{aligned} E(T) &= \sum_p \frac{1}{4} \int_0^{\omega_m}v_g\hbar\omega \frac{k^2}{2\pi^2v_g} \frac{1}{\exp(\hbar\omega/k_BT - 1)}\mathrm{d}\omega \\ &= \frac{3}{8\pi^2v_g^2}\int_0^{\omega_m}\frac{\hbar\omega^3}{\exp(\hbar\omega/k_BT - 1)} \mathrm{d}\omega \\ \end{aligned} \] 令 \[ y = \hbar\omega/k_BT \] 于是 \[ \begin{aligned} E(T) &= \frac{3}{8\pi^2v_g^2}\int_0^{\omega_m}\frac{\hbar\omega^3}{\exp(\hbar\omega/k_BT - 1)} \mathrm{d}\omega \\ &= \frac{3(k_BT)^4}{8\pi^2v_g^2\hbar^3}\int_0^{\frac{\hbar\omega_m}{k_BT}}\frac{y^3}{\exp(y)-1} \mathrm{d}y \\ &= T^4 \frac{3k_B^4}{8\pi^2v_g^2\hbar^3}\int_0^{\frac{\hbar\omega_m}{k_BT}}\frac{y^3}{\exp(y)-1} \mathrm{d}y\\ &= \sigma T^4\\ \text{其中}\\ \sigma &= \frac{3k_B^4}{8\pi^2v_g^2\hbar^3}\int_0^{\frac{\hbar\omega_m}{k_BT}}\frac{y^3}{\exp(y)-1} \mathrm{d}y \end{aligned} \] 在小温差近似下，可以认为整个求解域内的\(\sigma\)保持不变。

关于比热容

在小温差近似下，有 \[ \hbar \omega D(\omega)\left(f^{0}(T)-f^{0}\left(T_{r e f}\right)\right) \approx C(\omega)\left(T-T_{r e f}\right) \] 同时，在数学上，有 \[ T - T_{ref} \approx 4 T_{ref}^3 \left(T - T_{ref}\right) \] 于是

\[ \begin{aligned} q_{x}^{+}-q_{r e f} &=\int_{0}^{2 \pi} \mathrm{d} \phi \int_{0}^{\pi / 2} \cos \theta \sin \theta \mathrm{d} \theta \int_{0}^{\omega_{m}} \hbar \omega \frac{D(\omega)}{4 \pi}\left(f_{\omega}^{0}\left(T_{h}\right)-f_{\omega}^{0}\left(T_{r e f}\right)\right) v_{g} \mathrm{~d} \omega \\ & \approx \int_{0}^{2 \pi} \mathrm{d} \phi \int_{0}^{\pi / 2} \cos \theta \sin \theta \mathrm{d} \theta \int_{0}^{\omega_{m}} \frac{C(\omega)\left(T_{h}-T_{r e f}\right)}{4 \pi} v_{g} \mathrm{~d} \omega \\ &=\frac{1}{4} \int_{0}^{\omega_{m}} C(\omega)\left(T_{h}-T_{r e f}\right) v_{g} \mathrm{~d} \omega \\ \end{aligned} \] 灰体近似下，考虑相当于参考温度的情况，有 \[ \begin{aligned} q_{x}^{+}-q_{r e f} &\approx \frac{1}{4} \int_{0}^{\omega_{m}} C(\omega)\left(T_{h}-T_{r e f}\right) v_{g} \mathrm{~d} \omega \\ &= \frac{1}{4}C(T_h-T_{ref})v_g \\ &= \sigma(T_h^4 - T_{ref}^4) \\ &\approx 4 \sigma T_{ref}^3 \left(T - T_{ref}\right) \end{aligned} \] 于是得到了\(\sigma\)和比热容的关系： \[ \frac{Cv_g}{16T_{ref}^3} = \sigma \]

关于内部声子散射

当地散射功率密度和局域温度的关系为： \[ d Q_{\mathrm{e}} / d V=\int_{0}^{\omega_{D}} \frac{4}{l_w} e_{ \omega}(T) d \omega = 4\sigma T^4 /l_{ave} \] 这个推导有点复杂..需要借助辐射传递方程，还有一些平均自由程取法的问题，下一篇里再系统整理一下罢. stefan-boltzmann系数的形式是声子蒙特卡洛统计的基础，再下一篇再整理一下..