光速及折射率表达式推导

光速及折射率表达式推导

Sunfove 632 2021-12-11

在没有自由电荷和传导电流的情况下,麦克斯韦方程组表达式如下:

D=0\nabla \bullet \mathbf{D} = 0

B=0\nabla \bullet \mathbf{B} = 0

×E=Bt\nabla \times \mathbf{E} = - \frac{\partial\mathbf{B}}{\partial t}

×H=Dt\nabla \times \mathbf{H} = - \frac{\partial\mathbf{D}}{\partial t}

其中,

D=εE=ε0εrE\mathbf{D} = \varepsilon\mathbf{E} = \varepsilon_{0}\varepsilon_{r}\mathbf{E}

B=μH=μ0μrH\mathbf{B} = \mu\mathbf{H} = \mu_{0}\mu_{r}\mathbf{H}

对$$\nabla \times \mathbf{E} = - \frac{\partial\mathbf{B}}{\partial
t}$$两边取旋度,可得:

×(×E)=×Bt=t(×B)=εμ2Et2\nabla \times \left( \nabla \times \mathbf{E} \right) = - \nabla \times \frac{\partial\mathbf{B}}{\partial t} = - \frac{\partial}{\partial t}\left( \nabla \times \mathbf{B} \right) = - \text{εμ}\frac{\partial^{2}\mathbf{E}}{\partial t^{2}}

利用矢量分析公式和

E=1εD=0\nabla \bullet \mathbf{E}\mathbf{=}\frac{\mathbf{1}}{\varepsilon}\nabla \bullet \mathbf{D} =0

两遍取旋度,可得:

×(×E)=(E)2E=2E\nabla \times \left( \nabla \times \mathbf{E} \right)\mathbf{=}\nabla\left( \nabla \bullet \mathbf{E} \right)\mathbf{-}\nabla^{2}\mathbf{E} = \nabla^{2}\mathbf{E}

由此可得:

2Eεμ2Et2=0\nabla^{2}\mathbf{E}\mathbf{-}\text{εμ}\frac{\partial^{2}\mathbf{E}}{\partial t^{2}} = 0

2Ev22Et2=0\nabla^{2}\mathbf{E}\mathbf{-}v^{2}\frac{\partial^{2}\mathbf{E}}{\partial t^{2}} = 0

其中,v表示电磁波传播速度,电磁波真空中的传播速度为

c=1ε0μ0=2.997924×108 m/sc = \frac{1}{\sqrt{\varepsilon_{0}\mu_{0}}} = 2.997924 \times 10^{8}\ m/s

折射率表达式为:

n=cv=εrμrn = \frac{c}{v} = {\sqrt{\varepsilon_{r}\mu_{r}}}