Proof that $||\vec{E}|| = c||\vec{B}||$ for electromagnetic waves from maxwells equations in vacuum

Question

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.

Answer 1

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 $$||\vec{E}|| = c||\vec{B}||\, .$$

Answer 2

The relations between the electric field and the magnetic field for the solutions of Maxwell equations in vacuum are (wave planes):

$$\vec{E}=c \hspace{0.1cm} \vec{k} \times \vec{B}$$ $$\vec{B}=\frac{1}{c} \vec{k} \times \vec{E}$$

$\vec{k} is unitary, I dont know how to write unitary vectors here.Take modules in any equation and u will get ur relation.

Edit: These relations are deduced for sure in the Griffiths book for Electromagnetism.