Starting from Maxwell-equations in vacuum :
$$
\nabla \cdot \vec{E} = 0
$$
$$
\nabla \times \vec{E} = – \frac{\partial \vec{B}}{\partial t}
$$
$$
\nabla \cdot \vec{B} = 0
$$
$$
\nabla \times \vec{B} = \frac{1}{c^2}\frac{\partial \vec{E}}{\partial t}
$$
We can show the existence of electromagnetic waves (using the identity $\nabla \times \nabla \times \vec{F} = \nabla(\nabla \cdot F) – \nabla^2 \vec{F})$ :
$$
\frac{\partial^2 \vec{E}}{\partial t^2} = c^2 \nabla^2 \vec{E}
$$
$$
\frac{\partial^2 \vec{B}}{\partial t^2} = c^2 \nabla^2 \vec{B}
$$
The solutions to these equations are the following for plane waves (using $\mathbb{C}$ notation) :
$$
\vec{E}(\vec{r}, t) = \vec{E_0}e^{i(\vec{k} \cdot \vec{r} – wt)}
$$
$$
\vec{B}(\vec{r}, t) = \vec{B_0}e^{i(\vec{k} \cdot \vec{r} – wt)}
$$
We can show (using divergence of the electric and magnetic field in vacuum) that these waves form an orthonormal basis $(\vec{E}, \vec{B}, \vec{k})$
However, I'm looking for a proof that :
$$
||\vec{E}|| = c||\vec{B}||
$$
I've looked everywhere, in Griffith electrodynamics, in my books (Berkeley vol. II and III) but I've found nothing.
Best Answer
The short proof is as follows. Your second Maxwell equation in vacuum states among others that $$ k||\vec{E}|| = \omega||\vec{B}|| \, .$$ Since $\omega = kc$ it follows that $$||\vec{E}|| = c||\vec{B}||\, .$$