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

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.

• Try substituting those solutions into one of the two Maxwell equations that relate $\vec E$ and $\vec B$. – G. Smith Jan 23 '20 at 20:16
• Yeah, I had that feeling but then I get for example for the curl of the electric field : $\nabla \times \vec{E_0}e^{i(\vec{k} \cdot \vec{r} - wt)} = - \omega \vec{B(\vec{r}, t)}$ But I don't know how to calculate this curl ? – Mathieu Rousseau Jan 23 '20 at 20:22
• I don't know how to calculate this curl Use $\vec k\cdot\vec r=k_xx+k_yy+k_zz$ and take the appropriate partial derivatives. – G. Smith Jan 23 '20 at 21:00

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 $$||\vec{E}|| = c||\vec{B}||\, .$$
$$\vec{E}=c \hspace{0.1cm} \vec{k} \times \vec{B}$$ $$\vec{B}=\frac{1}{c} \vec{k} \times \vec{E}$$
$\vec{k} is unitary, I dont know how to write unitary vectors here.Take modules in any equation and u will get ur relation. Edit: These relations are deduced for sure in the Griffiths book for Electromagnetism. • I dont know how to write unitary vectors Write \hat{k} to get$\hat{k}\$. In English this is called a “unit vector”. – G. Smith Jan 23 '20 at 20:57