Question

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

Answer 1

The Maxwell's equations in free space can be written as follows:

$$\nabla \times \vec{E} = - \frac{1}{c} \frac{\partial \vec{H}}{\partial t}, \quad (1) \\ \nabla \times \vec{H} = \frac{1}{c} \frac{\partial \vec{E}}{\partial t} + \frac{4 \pi}{c} \vec{j}, \quad (2)\\ \nabla \cdot \vec{E} = \rho, \quad (3) \quad\quad \nabla \cdot \vec{H} = 0. \quad (4)$$

Taking the curl operation from the first equation, we obtain:

$$\nabla \times (\nabla \times \vec{E}) = \nabla \times \left( - \frac{1}{c} \frac{\partial \vec{H}}{\partial t} \right)$$

This expression can be simplified as follows:

$$\nabla(\nabla \cdot \vec{E}) - \Delta \vec{E} = - \frac{1}{c} \frac{\partial }{\partial t} \left( \nabla \times \vec{H} \right)$$

After substitution of the corresponding expressions from the 2nd and 3rd Maxwell's equations, we derive:

$$\nabla \rho - \Delta \vec{E} = - \frac{1}{c^2} \frac{\partial^2 \vec{E}}{\partial t^2} - \frac{4 \pi}{c^2} \frac{\partial \vec{j}}{\partial t}$$

This expression can be re-written in the following way:

$$\Delta \vec{E} - \frac{1}{c^2} \frac{\partial^2 \vec{E}}{\partial t^2} = \nabla \rho + \frac{4 \pi}{c^2} \frac{\partial \vec{j}}{\partial t}$$

In order to derive the expression for the z-component of the electric field vector, one should project this expression onto the z-axis:

$$\Delta E_z - \frac{1}{c^2} \frac{\partial^2 E_z}{\partial t^2} = \frac{\partial \rho}{\partial z} + \frac{4 \pi}{c^2} \frac{\partial j_z}{\partial t},$$

which is the answer on the question.