Question

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

Answer 1

Let us consider the Maxwell's equations in free space (in Gauss units)

$$\nabla \cdot \vec{E} = \rho\\ \nabla \cdot \vec{H} = 0\\ \nabla \times \vec{E} = - \frac{1}{c} \frac{\partial \vec{H}}{\partial t}\\ \nabla \times \vec{H} = \frac{1}{c} \frac{\partial \vec{E}}{\partial t} + \frac{4 \pi}{c} \vec{j}$$

After calculating the curl from the left side of the third equation, we obtain:

$$\nabla \times (\nabla \times \vec{E}) = \nabla(\nabla \cdot \vec{E}) - \Delta \vec{E} = \nabla \rho - \Delta \vec{E}$$

Performing this procedure over the right side of the same equation and taking into account that partial derivative over time commutes with nable operator, we obtain:

$$\nabla \times \left( - \frac{1}{c} \frac{\partial \vec{H}}{\partial t} \right) = - \frac{1}{c} \frac{\partial}{\partial t} \left( \frac{1}{c} \frac{\partial \vec{E}}{\partial t} + \frac{4 \pi}{c} \vec{j} \right) = - \frac{1}{c^2} \frac{\partial^2 \vec{E}}{\partial t^2} - \frac{4 \pi}{c^2} \frac{\partial \vec{j}}{\partial t}$$

Finally, putting the expressions together, we deduce:

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

In order to get the equation for z-component, one should project this expression onto the z-axis:

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