TD 5：电解质中的Drude模型，数值近似，脉宽，群速度，光缆

• 电解质中的运动方程：

$$\frac{d^2\vec x}{dt^2}m_e = -e\vec E-\frac {m_e}\tau\frac{d\vec x}{dt}-m_e\omega_0^2\vec x$$

• 解得：

$$\vec x = -\frac{e}{m_e(\omega_0^2-\omega^2+i\omega /\tau)}\vec E$$

• 极化：

$$\vec P = -\sum_j f_jNe\vec x = \frac{e^2Nf_j}{m_e}\frac{1}{\omega_{0j}^2-\omega_j^2+i\omega_j /\tau_j}\vec E\\\vec D = \varepsilon_0\vec E+\vec P = (1+\sum_j\frac{e^2N}{m_e\varepsilon_0}\frac{f_j}{\omega_{0j}^2-\omega_j^2+i\omega_j /\tau_j})\varepsilon_0\vec E$$

• 复介电常数：

$$\varepsilon = 1+\sum_j\frac{e^2N}{m_e\varepsilon_{0}}\frac{f_j}{\omega_{0j}^2-\omega_j^2+i\omega_j /\tau_j} = 1+\sum_j\frac{\omega_{pj}^2f_j}{\omega_{0j}^2-\omega_j^2+i\omega_j /\tau}$$

• $f_j$是占比

• 使用波长表示的复介电常数：

$$\omega = \frac{2\pi c}{\lambda}\Rightarrow \varepsilon = 1+\sum_j\frac{\lambda_{0j}^2\lambda_j^2f_j}{\lambda_{j}^2\lambda_{pj}^2-\lambda_{0j}^2\lambda_{pj}^2+A(\lambda)}$$

• 式中，$A(\lambda)$是吸收项，考虑到纯净二氧化硅在该光线范围内吸收较弱，忽略

$$\varepsilon = 1+\sum_j\frac{\frac{\lambda_{0j}^2}{\lambda_{pj}^2}\lambda_j^2f_j}{\lambda_{j}^2-\lambda_{0j}^2}$$

• 可以求出三个$\lambda_{0j} = 68nm,116nm,9896nm$

求群速度以求时间

• 群速度:

$$v_g = \frac{d\omega}{dk} = \frac c{\frac{d(\omega n(\omega))}{d\omega}} = c\cdot\frac 1{n(\omega)+\omega\frac{dn(\omega)}{d\omega}} =c\cdot\frac 1{n(\omega)-\lambda\frac{dn(\lambda)}{d\lambda}} =\frac{c}{N(\lambda)}$$

• $d\omega = -\frac{2\pi c}{\lambda^2}d\lambda$，$N(\lambda) = n-\lambda\frac{n(\lambda)}{\lambda}$
• 到达时间：$t = L/v_g = \frac{LN}{c}$

求脉宽

• 研究时间与距离的微分关系：

$$\frac{dt}{d\lambda} = \frac LC\frac{dN}{d\lambda}\Rightarrow\Delta t = \frac LC\frac{d(n-\lambda\frac{dn(\lambda)}{d\lambda})}{d\lambda}\Delta\lambda = \frac LC(-\lambda\frac{d^2n}{d\lambda^2})\Delta\lambda$$

• 已知: $T = 10^{-12}s,\ \ \Delta \nu = 1/T = 10^{12}Hz,\ \ \Delta\lambda = \frac{\lambda^2}{c}\Delta\nu = 2.7nm$
• 后续部分记为讨论，参考：