Wave Equations

From Eq.3 and Eq.6

$${\rm rot} {\bf E} = - \mu \mu_0 \frac{\partial {\bf H}}{\partial t}$$ (13)

From Eqs.4, 5, 7

$${\rm rot} {\bf H} = \sigma {\bf E} + \epsilon \epsilon_0 \frac{\partial {\bf E}}{\partial t}$$ (14)

If we apply $\nabla \times$ to Eq.13 and $\partial /\partial t$ to Eq.14 and using the relation

$$\nabla \times \nabla \times {\bf E} = {\rm rot} \left\vert \begin{array}{ccc} {\bf i} & {\bf j} & {\bf ... ...rtial E_y}{\partial x} - \frac{\partial E_x}{\partial y} \end{array}\right\vert$$

$$= {\bf i} \left[ \frac{\partial^2 E_y}{\partial x \partial y} - \fr... ...frac{\partial ^2 E_z}{\partial x \partial z}\right] + {\bf j}[..] + {\bf k}[..]$$

$$= {\bf i}\frac{\partial }{\partial x}\left( \frac{\partial E_x}{\pa... ...rtial y} {\rm div}{\bf E} + {\bf k}\frac{\partial }{\partial z}{\rm div}{\bf E}$$

$$- {\bf i}\nabla^2 E_x - {\bf j}\nabla^2 E_y - {\bf k}\nabla^2 E_z$$

$$= {\rm grad}({\rm div}{\bf E}) - \nabla^2 {\bf E}$$ (15)

If we assume $\rho =0$, ${\rm div}{\bf E} = 0$ then

$$\nabla^2 {\bf E} = \sigma \mu \mu_0 \frac{\partial {\bf E}}{... ...0 \epsilon \epsilon_0 \frac{\partial^2 {\bf E}}{\partial t^2}$$ (16)

In the same way we can get

$$\nabla^2 {\bf H} = \sigma \mu \mu_0 \frac{\partial {\bf H}}{... ...0 \epsilon \epsilon_0 \frac{\partial^2 {\bf H}}{\partial t^2}$$ (17)

In the case that the electric field is plane wave

$${\bf E} = {\bf E}_0 e^{i({\bf k}\cdot{\bf r} - \omega t)}$$ (18)

then

$$k^2 = i\sigma \mu \mu_0 \omega + \mu \mu_0 \epsilon \epsilon_0 \omega^2$$ (19)

In vacuum, $\epsilon =1, \mu =1, \sigma = 0$ and $c / \nu = \lambda, c/\omega = \lambda/(2\pi ), k = 2\pi/ \lambda = \omega /c$ then

$$c = 1/\sqrt{\mu_0 \epsilon_0 }$$ (20)

If we define the comlex optical index $\tilde{n}$ by

$$k = \frac{2\pi}{\lambda } = \frac{\omega}{v} = \frac{\tilde{n}\omega}{c}$$ (21)

$$\tilde{n}^2 = k^2 c^2 /\omega^2 = \mu \epsilon + i \frac{\sigma \mu }{\epsilon_0 \omega }$$ (22)

¹

If we take divergence of electic filed in the form of plane wave.

$${\rm div} {\bf E} = \left( {\bf i}\frac{\partial }{\partial x} + {\bf i}\frac{\partia... ...\bf i} + E_{0y}{\bf j} + E_{0z}{\bf k} )e^{i(k_x x + k_y y + k_z z - \omega t)}$$ (28)

$$= i ( k_x E_{0x} + k_y E_{0y} + k_z E_{0z} ) = i{\bf k}\cdot{\bf E}_0$$ (29)

$$= 0$$ (30)

Then the electric field is transverse wave.

If the magnetic-flux density ${\bf B}$ is written as

$${\bf B} = \frac{{\bf k}\times {\bf E}_0 }{\omega} e^{i({\bf k}\cdot{\bf r} - \omega t)},$$ (31)

the magnetic-flux density satisfies Eq.3, because

$${\rm rot} {\bf E} = {\bf i}(\frac{\partial E_z }{\partial y} - \frac{\partial E_y}{\partial z}) + ..$$ (32)

$$= {\bf i}(ik_y E_z - ik_z E_y) + ..$$ (33)

$$= i{\bf k}\times {\bf E}$$ (34)

$$- \frac{\partial {\bf B}}{\partial t} = - \frac{{\bf k}\times {\bf E}_0}{\omega} (-i \omega) e^{i({\bf k}\cdot{\bf r} - \omega t)}$$ (35)

$$= i{\bf k}\times {\bf E}$$ (36)

In vacuum

$$\frac{\vert{\bf E}_0 \vert}{\vert{\bf H}_0\vert} = \frac{\mu... ...u_0} \omega} = \sqrt{\frac{\mu_0}{\epsilon_0}} = 376.7 \Omega$$ (37)