Question

Using Maxwell's equation in free space,derive the wave equations for the z-component of electric field vector.

Answer 1

1) Take $\bf{\nabla} \times$ of Maxwell-Faraday differential equation:

$$\bf{\nabla} \times[\bf{\nabla} \times \bf{E}]=\bf{\nabla} \times[-\frac{\it{\partial} \bf{B}}{\it{\partial t}}]=-\frac{\it{\partial}}{\it{\partial t}}[\bf{\nabla}\times \bf{B}].$$

2) It's easy to see that in the square brackets of the last right part above there is Ampere's circuit law in the differential form:

$$\bf{\nabla} \times[\bf{\nabla} \times \bf{E}]=\it{-\frac{\partial}{\partial t}[\mu_0 \epsilon_0 \frac{\partial \bf{E}}{\partial t}]=-\it{\mu_0 \epsilon_0 \frac{\partial^2 \bf{E}}{\partial t^2}}}.$$

3) It can be shown for any vector that $\bf{\nabla} \times[\bf{\nabla} \times \bf{E}]=\bf{\nabla}(\bf{\nabla}\cdot \bf{E})-\bf{\nabla}^2\bf{E}.$ But we derive the wave equation in free

space, that is why charge density is 0 and $\bf{\nabla} \times[\bf{\nabla} \times \bf{E}]$ becomes simply $-\bf{\nabla}^2\bf{E}.$ Use this result in the equation above:

$$\bf{\nabla} \times[\bf{\nabla} \times \bf{E}]=-\bf{\nabla}^2\bf{E}=\it{-\mu_0 \epsilon_0 \frac{\partial^2 \bf{E}}{\partial t^2}}.$$

4) Let's polarize our wave in z-direction so that x- and y-components were 0. The wave equation above is written in the vector form. Now written in the scalar form for the z-component it will look like

$$-\bf{\nabla}^2 \it{E_z}=-\mu_0 \epsilon_0 \frac{\partial^2 E_z}{\partial t^2}.$$