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).

  • 2
    $\begingroup$ Because the Maxwell equations say so (because of our relative choice of units for $E$ and $B$). $\endgroup$ May 9, 2018 at 15:16
  • $\begingroup$ It is the convention which is used. $\endgroup$
    – Farcher
    May 9, 2018 at 15:20

1 Answer 1


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 $$ \frac{\omega}{c^2} \vec{E}_0 = - \vec{k} \times \vec{B}_0. $$


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