Supplement 1.2: Solving Maxwell's Equations for Electromagnetic Waves (2/3)

Solving the wave equation

We derived the wave equations which characterise electromagnetic fields in vacuum. They combine the second spatial and temporal derivatives of the electric and magnetic field:

Δ \vec{E} = ε_{o} μ_{o} \frac{\partial^{2} \vec{E}}{\partial t^{2}}

Δ \vec{B} = ε_{o} μ_{o} \frac{\partial^{2} \vec{B}}{\partial t^{2}}

We examine the electric field vector $\vec{E}$ of the wave at a position $\vec{r} = (x, y, z)$ in space. The wave is assumed to propagate in a direction given by the unit vector $\vec{a}$ . The wave equation for the electric field is then solved by the following equation:

\vec{E} = {\vec{E}}_{o} f (\vec{a} \cdot \vec{r} - \frac{1}{\sqrt{ε_{o} μ_{o}}} t)

where ${\vec{E}}_{o}$ is a constant vector having the same orientation as $\vec{E}$ and f is a second differentiable function.

With the first Maxwell equation for the divergence of the electric field it follows:

\nabla \cdot \vec{E} = \vec{a} \cdot {\vec{E}}_{o} \frac{\partial}{\partial r} f (\vec{a} \cdot \vec{r} - \frac{1}{\sqrt{ε_{o} μ_{o}}} t) = 0

and therefore: $\vec{a} \cdot {\vec{E}}_{o} = 0$

The scalar product of two vectors vanishes if the vectors are orthogonal to each other. Hence, ${\vec{E}}_{o}$ and also $\vec{E}$ are orthogonal to the direction the wave travels.

In the third Maxwell equation

\nabla \times \vec{E} = - \frac{\partial \vec{B}}{\partial t}

the vector $\nabla \times \vec{E}$ is orthogonal to $\vec{E}$ . Since the time derivative does not change the orientation of $\vec{B}$ , it is evident that $\vec{E}$ and $\vec{B}$ are orthogonal to each other, and both are orthogonal to the direction the wave travels: electromagnetic waves are transverse waves.

Plane monochromatic waves

In chapter 1, section electromagnetic waves on page 2, the E-field and the B-field of a plane monochromatic wave travelling into direction x (ignoring the vector character of $\vec{E}$ and $\vec{B}$ ) is written as: