Why are the $\mathbf E$ and $\mathbf B$ fields of an electromagnetic wave mutually perpendicular?

Question

Why are the wave number $\mathbf k$ and the electric and magnetic fields $\mathbf E$ and $\mathbf B$ are perpendicular to each other?

I know it but I haven't thought about it deeply.

How can I prove this conclusion mathematically?

Answer 1

They're not. There's plenty of situations where the $\mathbf E$ and $\mathbf B$ fields are not orthogonal to each other or (where the latter can be defined at all) to the wavevector $\mathbf k$. Notable examples include tightly-focused gaussian beams, waveguides, and spherical waves, but it's plenty easy (and a good exercise) to cook up examples using superpositions of two different plane waves.

On the other hand, "morally" speaking, i.e. in a decidedly hand-wavy sense, the property very often still mostly holds, in the sense that if your field looks similar enough to a plane wave to have a reasonably-well-defined propagation direction, at least inside some confined region, then the electric and magnetic fields will be often be mostly orthogonal to each other and to the propagation direction. However, a strict result along those lines cannot be shown - the only hard zeros there come at the PDE level of the Maxwell equations.

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What is true is the fact that if you have a plane wave with spatial and temporal dependence $$ \mathbf E(\mathbf r,t) = \mathrm{Re}\mathopen{}\left[\mathbf E_0e^{i(\mathbf k\cdot\mathbf r-\omega t)}\right]\mathclose{} \quad \text{and} \quad \mathbf B(\mathbf r,t) = \mathrm{Re}\mathopen{}\left[\mathbf B_0e^{i(\mathbf k\cdot\mathbf r-\omega t)}\right]\mathclose{}, $$ then the nabla operator can be substituted as $\nabla \mapsto i\mathbf k$, because $$ \nabla\cdot \mathbf C(\mathbf r,t) = \mathrm{Re}\mathopen{}\left[i\mathbf k \cdot \mathbf C_0e^{i(\mathbf k\cdot\mathbf r-\omega t)}\right]\mathclose{} \quad \text{and} \quad \nabla\times \mathbf C(\mathbf r,t) = \mathrm{Re}\mathopen{}\left[i\mathbf k\times \mathbf C_0e^{i(\mathbf k\cdot\mathbf r-\omega t)}\right]\mathclose{}, $$ (where $\mathbf C=\mathbf E,\mathbf B$); similarly, time derivatives can be replaced via $\frac{\partial}{\partial t} \mapsto -i\omega$. From this you can recover the Maxwell equations in vacuum in the form \begin{align} i\mathbf k \cdot \mathbf E_0 & = 0 &&& i\mathbf k \cdot \mathbf B_0 & = 0 \\ i\mathbf k \times\mathbf E_0 & = i\omega \mathbf B_0 &&& i\mathbf k \times\mathbf B_0 & = -\frac{i\omega}{c^2} \mathbf E_0. \end{align} These then directly imply that $(\mathbf E_0,\mathbf B_0,\mathbf k)$ are a right-handed orthogonal triad.

But again, as noted above: this property, and its proof, is only strictly valid for plane waves.

Answer 2

There are many situations in which a longitudinal component of E or B can exist. We're used to thinking in terms of plane waves in vacuum. But plane waves do not exist in nature. If you try to form a finite beam, for example, you'll find a small longitudinal component exists, even in free space. And in waveguides it is easy to calculate and visualize the longitudinal component of E or B. But, if you do assume plane waves in vacuum, then $\nabla \times B$ is proportional to $k \times B$ and similarly for E. Plug these into the 2 Maxwell curl equations and you get an orthogonal triple after Fourier transforming time.

Added. Here are some citations to work showing examples of longitudinal fields in finite beams in free space. A beam with a nonuniform transverse intensity profile necessarily has a longitudinal component. In other cases the longitudinal components are caused by focusing or collimation.

• Contribution of longitudinal electric field of a gaussian beam to second harmonic generation. S.R. Mishra and K.C. Rustagi. Opt. Commun. 74, 419 (1990)

• Gaussian laser beams with radial polarization. K.T. McDonald (2000)

• Graphical study of Laguerre-Gaussian beam modes. R.K. Arora and Z. Lu. IEE Proc.-Microw. Antennas Propag. 141, 145 (1994)

A quick google search will turn up many more. The effect is used a lot in laser physics.

Answer 3

For a single, monochromatic plane wave in an electrically neutral, uniform medium, $\vec{E}=\partial_t \vec{A}$ and $\vec{B} = \vec{\nabla} \times \vec{A}$ are perpendicular.

In general, e.g. for a superposition of such waves, this is not true.

Answer 4

I think this must be probably asked already. However I couldn't find it easily, so I'll try to briefly explain.

The two main ideas are:

1. Everything related to electromagnetism is governed by Maxwell's equations.
2. $\vec{E}$ and $\vec{B}$ are not always perpendicular, but they are perpendicular in vaccuum and air (which are the most frequent cases).

The way to prove this is precisely using Maxwell's equations. They are

$$ \begin{array}{ccc} \vec{\nabla}\cdot\vec{E}=\dfrac{\rho}{\varepsilon_0} & \ & \vec{\nabla}\cdot\vec{B}=0 \\ \vec{\nabla}\times\vec{E}=-\dfrac{\partial \vec{B}}{\partial t} & \ & \vec{\nabla}\times\vec{B}=\mu_0\vec{J}+\mu_0\varepsilon_0 \dfrac{\partial\vec{E}}{\partial t} \end{array}$$

But, if you are in vaccuum, where there is no charge nor current, then $J=0,\ \rho=0$, and hence

$$ \begin{array}{ccc} \vec{\nabla}\cdot\vec{E}=0 & \ & \vec{\nabla}\cdot\vec{B}=0 \\ \vec{\nabla}\times\vec{E}=-\dfrac{\partial \vec{B}}{\partial t} & \ & \vec{\nabla}\times\vec{B}=\mu_0\varepsilon_0 \dfrac{\partial\vec{E}}{\partial t} \end{array}$$

Notice they have become beautifully simmilar for both fields.

From here you can show that both $E$ and $B$ satisfy the wave equation. ... but the wave equation has the solution of plane waves in vaccuum. For a plane wave, you still have

$$ \begin{array}{ccc} \vec{\nabla}\times\vec{E}=-\dfrac{\partial \vec{B}}{\partial t} & \ & \vec{\nabla}\times\vec{B}=\mu_0\varepsilon_0 \dfrac{\partial\vec{E}}{\partial t} \end{array}$$

For example, the x component of the first equation is

$$ \dfrac{\partial E_z}{\partial y}-\dfrac{\partial E_y}{\partial z}=-\dfrac{\partial B_x}{\partial t}$$

and, if you replace $\vec{E}=\vec{E_0}e^{i(\omega t - k_x x - k_y y - k_z z)}$, then,

$$ k_y E_z - k_Z E_y = \omega B_x$$

Doing the same for teh rest of components, you see that

$$\vec{s}\times \vec{E} = \frac{\omega}{k} \vec{B}$$

So $B$ is perpendicular to both the direction of propagation $\vec{s}$ and the electric field. The same results can be obtained by substituting on the second equation.

Answer 5

This follows directly from Maxwell's equations. Assuming a plane wave solution of these equations $$\vec E=\vec E_0 \exp(i\vec k \vec r-i\omega t)\tag 1$$ Gauss's law in free space becomes $$\nabla \vec E=i\vec k ·\vec E=0 \tag 2$$ This means that the electric field vector $\vec E$ is perpendicular to the wave vector $\vec k$. On the other hand, the Faraday-Maxwell equation yields $$\nabla \times \vec E=-\frac {\partial \vec B}{\partial t}=i\vec k \times \vec E=i\omega\vec B \tag 3$$ This means that the magnetic field vector $\vec B$ is perpendicular to both the wave vector $\vec k$ and the electric field vector $\vec E$.

Note following comments: This derivation that $\vec E$, $\vec B$ and $\vec k$ are mutually perpendicular holds only for a plane EM wave, as stated in the assumption of the first sentence. My answer doesn't imply or is intended to suggest that this relation holds for all electromagnetic waves. There are plenty examples to the contrary.