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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?

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    $\begingroup$ A simple plausibility argument is the following. The only scalar invariants of the electromagnetic field under a Lorentz boost are $P=E^2-B^2$ and $Q=\textbf{E}\cdot\textbf{B}$. The actual structure of a plane wave is that $P=0$ and $Q=0$, and the invariance of these quantities guarantees that a wave that has these properties in one frame has them in another as well. We expect that a plane wave is still a valid plane wave under a Lorentz boost. If a wave had $P\ne0$ or $Q\ne0$, then it's not clear what sane structure we would require that would be Lorentz-invariant. $\endgroup$ – Ben Crowell Apr 18 '18 at 22:53
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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.


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.

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  • $\begingroup$ This rather off-topic discussion should continue in chat. (Actually, there's no point in continuing it at all, but if you insist on dragging it on, then do it on chat.) $\endgroup$ – Emilio Pisanty Apr 20 '18 at 18:48
  • $\begingroup$ OK, I'll give it a try. $\endgroup$ – freecharly Apr 20 '18 at 19:11
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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.

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

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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.

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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.

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  • $\begingroup$ The electric and magnetic field are not always perpendicular, even in vacuum; counter-examples are trivial to cook up using superpositions of plane waves. The plane-wave proof is fine, but it doesn't generalize to arbitrary fields. $\endgroup$ – Emilio Pisanty Apr 18 '18 at 21:21
  • $\begingroup$ Are you talking about superposition of waves travelling in different directions? $\endgroup$ – FGSUZ Apr 19 '18 at 20:27
  • $\begingroup$ Yes, obviously. Other examples are slightly more involved (see here for more details) but anything from a simple laser pointer onwards will exhibit some amount of non-orthogonality between $\mathbf E$, $\mathbf B$, and (to whatever extent it can be defined at all) $\mathbf k$. $\endgroup$ – Emilio Pisanty Apr 19 '18 at 20:51
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    $\begingroup$ Okay now I get you. I hadn't thought about that case, I was thinking about the simplest case: "ordinary simple plane waves", as that's what people think about when they hear "light" or "wave", but thank you for the note. $\endgroup$ – FGSUZ Apr 20 '18 at 20:47
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    $\begingroup$ @EmilioPisanty - To claim that plane waves are "unphysical" is simply ridiculous. All physicist know that a mathematical plane wave would extend over infinite space and doesn't exist as such in reality. But it is a good approximations in many situations . Most physical models are only valid approximation in certain situations. This doesn't mean that they are "unphysical". $\endgroup$ – freecharly Apr 22 '18 at 15:32
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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.

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    $\begingroup$ For clarity, this follows strictly from the Maxwell equations and the plane-wave assumption; without the latter, it fails. The (mis-)assumption that the orthogonality property will hold in other contexts is a particularly pernicious misconception that should be given as few opportunities to grow as possible. $\endgroup$ – Emilio Pisanty Apr 18 '18 at 21:45
  • $\begingroup$ @EmilioPisanty - Dear Emilio, this seems a bit of wisecracking. It was rather obvious that the OP was asking about plane electromagnetic waves. One usually learns first about plane electromagnetic waves including that their electric and magnetic fields are perpendicular to each other and to the wave vector. I gave the correct answer and also stated that it was for plane waves. In your answer, you gave the exactly the same explanation. But you didn't provide a proof for your claim that it doesn't hold for other electromagnetic waves. $\endgroup$ – freecharly Apr 19 '18 at 3:19
  • $\begingroup$ As I said above, explicit examples are in the linked in my answer. And as I said, the misconception is extremely widespread and very hard to get rid of, because of the sheer number of resources that don't make it clear that it's a plane-wave-only result and by extension imply that it holds universally (as this answer does). OP might actually understand that restriction (though given the phrasing I find that claim extremely dubious) but many more people than just OP will see this thread. $\endgroup$ – Emilio Pisanty Apr 19 '18 at 6:46
  • $\begingroup$ @EmilioPisanty - Dear Emilio, my answer doesn't suggest any misconceptions. It explicitly states that it is valid for the assumption of a plane wave, as you also did in your answer. Also the OP most likely was referring to a plane wave in posing this question. I don't know any scholarly text that spreads your supposed "widespread misconception". $\endgroup$ – freecharly Apr 19 '18 at 11:14
  • $\begingroup$ I would appreciate if those luminaries who are voting down this obviously correct derivation of the fact that $\vec E$, $\vec B$, and $\vec k$ of a plane EM wave are mutually perpendicular would also dare to explain why they are doing this. $\endgroup$ – freecharly Apr 20 '18 at 14:57

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