2
$\begingroup$

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.

Improve this question
$\endgroup$
  • 1
    $\begingroup$ Try substituting those solutions into one of the two Maxwell equations that relate $\vec E$ and $\vec B$. $\endgroup$ – G. Smith Jan 23 '20 at 20:16
  • $\begingroup$ 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 ? $\endgroup$ – Mathieu Rousseau Jan 23 '20 at 20:22
  • 2
    $\begingroup$ 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. $\endgroup$ – G. Smith Jan 23 '20 at 21:00
2
$\begingroup$

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

Improve this answer
$\endgroup$
1
$\begingroup$

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.

Improve this answer
$\endgroup$
  • $\begingroup$ I dont know how to write unitary vectors Write \hat{k} to get $\hat{k}$. In English this is called a “unit vector”. $\endgroup$ – G. Smith Jan 23 '20 at 20:57
  • $\begingroup$ Thanks @G.Smith $\endgroup$ – Isaac Domínguez Larrañaga Jan 24 '20 at 12:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.