TD 5：电解质中的Drude模型，数值近似，脉宽，群速度，光缆
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：
