0
$\begingroup$

My notes have stated that the relations between direction of electric and magnetic field is as follows: $$ \frac \omega k \vec B_0 = \hat{\vec k} \times \vec E_0$$ where $\hat{\vec k}$ is the unit vector in the direction of $\vec k$.

My question is how do we know that it is $\hat{\vec k} \times \vec E_0$ and not $\vec E_0 \times \hat{\vec k}$ (which will obviously give the negative of the previous one).

$\endgroup$
  • 2
    $\begingroup$ Because the Maxwell equations say so (because of our relative choice of units for $E$ and $B$). $\endgroup$ – Sebastian Riese May 9 '18 at 15:16
  • $\begingroup$ It is the convention which is used. $\endgroup$ – Farcher May 9 '18 at 15:20
1
$\begingroup$

As pointed out by Sebastian Riese in the comments, this follows from Maxwell's equations, and specifically Faraday's Law. Suppose $\vec{B}$ and $\vec{E}$ are plane-polarized plane waves traveling in the same direction with the same frequency: $$ \vec{E} = \vec{E}_0 \cos \left[ \vec{k} \cdot \vec{r} - \omega t\right]\\ \vec{B} = \vec{B}_0 \cos \left[ \vec{k} \cdot \vec{r} - \omega t\right] $$ where $\vec{E}_0 = E_{0x} \hat{\imath} + E_{0y} \hat{\jmath} + E_{0z} \hat{k}$ is a constant vector, and $\vec{B}_0$ is defined similarly. If you write out the $x$-, $y$-, and $z$-components of Faraday's Law $\vec{\nabla} \times \vec{E} = - \partial \vec{B}/\partial t$, you will find that the components of the vectors must satisfy $$ \omega B_{0x} = k_y E_{0z} - k_z E_{0y} \\ \omega B_{0y} = k_z E_{0x} - k_x E_{0z} \\ \omega B_{0x} = k_x E_{0y} - k_y E_{0x} $$ which can be summarized by the equation $\omega \vec{B}_0 = \vec{k} \times \vec{E}$.

You can also apply the other three of Maxwell's equations to the above solution to find out other information about $\vec{E}_0$, $\vec{B}_0$, and $\vec{k}$. Specifically, Gauss's Laws for electric fields and for magnetic fields yield $$ \vec{k} \cdot \vec{E}_0 = \vec{k} \cdot \vec{B}_0 = 0 $$ while Ampere's Law yields $$ \frac{\omega}{c^2} \vec{E}_0 = - \vec{k} \times \vec{B}_0. $$

$\endgroup$

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.