# Why are electric and magnetic fields perpendicular in an electromagnetic wave?

I want to know why electric and magnetic fields are perpendicular in an electromagnetic wave and what would happen if they weren't.

• Mar 5, 2016 at 19:34

This can be explained by Maxwell's equations. For which one needs to know the basics of vector calculus, gradient, divergence and curl.

When there are no charges:

\begin{aligned} c\nabla\times \boldsymbol E(\boldsymbol r,t) + \partial_t \boldsymbol B(\boldsymbol r,t) = 0\\ \nabla\cdot \boldsymbol B(\boldsymbol r,t) = 0 \\ c\nabla\times \boldsymbol B(\boldsymbol r,t) - \partial_t \boldsymbol E(\boldsymbol r,t) = 0\\ \nabla\cdot \boldsymbol E(\boldsymbol r, t) = 0 \end{aligned}

Where, $\boldsymbol E$ and $\boldsymbol B$ are the electric field strength and the magnetic induction, respectively and $c$ is the speed of light in free space.

The spatial and time periodicity of the radiation be utilized to write Maxwell's equations in Fourier transformed form:

\begin{aligned} c\boldsymbol q\times \boldsymbol E(\boldsymbol q,\omega)− \omega B(\boldsymbol q,\omega) = 0 \\ \boldsymbol q \cdot \boldsymbol B(\boldsymbol q, \omega) = 0 \\ c\boldsymbol q\times \boldsymbol B(\boldsymbol q,\omega)+\omega \boldsymbol E(\boldsymbol q,\omega) = 0\\ \boldsymbol q \cdot \boldsymbol E(\boldsymbol q, \omega) = 0 \end{aligned} where $\boldsymbol q$ is a wave vector.

From the third equation we get $$\omega \boldsymbol B(\boldsymbol q, \omega)=c\boldsymbol q\times \boldsymbol E(\boldsymbol q, \omega)$$

Now we take the scalar product with $\omega\boldsymbol E(\boldsymbol q, \omega)\cdot \boldsymbol B(\boldsymbol q, \omega) = c\omega\boldsymbol E(\boldsymbol q, \omega)\cdot \boldsymbol q\times \boldsymbol E(\boldsymbol q, \omega)$ but from the first equation we know that $i\boldsymbol q \cdot \boldsymbol E(\boldsymbol q, \omega) = 0$

Therefore, $$\boldsymbol E(\boldsymbol q, \omega)\cdot \boldsymbol B(\boldsymbol q, \omega) = 0$$

For the scalar product between two vectors to be zero either one of them is the zero vector or they are perpendicular to each other.

Therefore, the electric and magnetic fields are perpendicular.

Also, for a propagating EM wave, the $\boldsymbol E$ and $\boldsymbol B$ fields are always perpendicular in a homogenous, linear, anisotropic medium. This type of media includes many things like air, water, glass (without stress or tempering). However, in inhomogenous, non-linear, or isotropic media, the $\boldsymbol E$ and $\boldsymbol B$ fields may not be perpendicular, e.g. in a crystal (which is isotropic).

Although this explains the math of your question, I would also add that its just God's design that these two energies are perpendicular to each other :)

• I think that you meant "isotropic" where you said "anisotropic", and "anisotropic" where you said "isotropic".
– user93237
Mar 5, 2016 at 19:35
• Oh Ya! Pardon my mistake. The words got flipped Mar 5, 2016 at 19:40
• This answer's conclusions are incorrect. The result as proved is true for plane-wave states of radiation, but that is the full extent of the result, and in general $\boldsymbol E$ and $\boldsymbol B$ are not perpendicular; see e.g. this thread for examples and an explanation. Sep 13, 2021 at 12:28