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The usual representation of a free electromagnetic wave in vacuum looks like this:

The blue parts are the local electric field, while the green parts are the local magnetic field.

The circularly polarized wave is also another standard representation of an EM travelling wave, but it is a bit less well known.

Here are the wave functions describing the above usual linearly polarized wave:

\begin{align} \vec{\boldsymbol{E}} &= \vec{\boldsymbol{a}} \sin{(k \, x - \omega \, t + \phi)}, \tag{1} \\[12pt] \vec{\boldsymbol{B}} &= \vec{\boldsymbol{b}} \sin{(k \, x - \omega \, t + \phi)}, \tag{2} \end{align} where $\vec{\boldsymbol{a}}$ and $\vec{\boldsymbol{b}}$ are two orthogonal constant polarisation vectors, transverse to the propagation (the $x$ axis here).

Now, I believe that I remember (I'm not sure!) that there's a special superposition of travelling waves that gives a non-standard representation of the travelling EM plane wave, with the magnetic parts shifted by a quarter of wavelength relative to the electric parts. I'm unable to find which superposition exactly can do this, but I know that this superposition (if it exists !) is actually very simple.

I'm NOT talking about standard standing wave here, and I'm not talking about the circularly polarized travelling wave neither. However, the superposition I'm looking for may be related to such waves, I don't know.

So somebody could tell which superposition of waves can shift the magnetic crests (as seen above) by one quarter of a wavelength, relative to the electric crests ?

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  • $\begingroup$ Cham, have you moved on in your thoughts? $\endgroup$
    – HolgerFiedler
    Jun 27, 2021 at 8:10

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I was looking for a travelling electromagnetic wave of the following form, in vacuum :

Well, apparently this is impossible by Maxwell equations in vacuum : \begin{align}\tag{1} \boldsymbol{\nabla} \times \boldsymbol{E} &= \frac{\partial \, \boldsymbol{B}}{\partial \, t}, &\boldsymbol{\nabla} \times \boldsymbol{B} &= -\,\frac{\partial \, \boldsymbol{E}}{\partial \, t}. \end{align} The picture above implies two wave functions of the following form ($\boldsymbol{B}$ shifted by a quarter of a wavelenght) : \begin{align}\tag{2} \boldsymbol{E}(t, x) &= \boldsymbol{a} \sin{(k \, x - \omega \, t + \phi)}, \\[12pt] \boldsymbol{B}(t, x) &= \boldsymbol{b} \cos{(k \, x - \omega \, t + \phi)}.\tag{3} \end{align} These vectors cannot satisfy the equations (1), since the derivative of a $\sin{\theta}$ cannot be equal to the derivative of a $\cos{\theta}$ !

If we try a superposition of the $\sin{\theta}$ and the $\cos{\theta}$, we could easily find a solution to equations $(1)$, but this lead us to the standard circularly polarized travelling plane wave (or more generally a wave with elliptical polarization).

So I conclude that there is no solution to Maxwell equations which describes a travelling wave in vacuum of the type shown on the picture above.

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  • $\begingroup$ Yeah, pretty much. Well handled. $\endgroup$
    – Emilio Pisanty
    Feb 15, 2016 at 3:22
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    $\begingroup$ Note also that your argument only rules out travelling waves of that form. However, you can prepare a standing-wave arrangement with fields of the form \begin{align} \mathbf E &= \mathbf{a} \sin(kx)\cos(\omega t)+\mathbf{b} \cos(kx)\sin(\omega t)\\ \mathbf{B}&= \mathbf{a} \sin(kx)\sin(\omega t)+\mathbf{b} \cos(kx)\cos(\omega t), \end{align} which looks like what you want at $t=0$ (i.e. you prepare that as an initial condition) but it does not propagate like you were hoping it would. $\endgroup$
    – Emilio Pisanty
    Feb 15, 2016 at 12:51
  • $\begingroup$ @EmilioPisanty, the functions you wrote aren't a solution to maxwell equations. $\endgroup$
    – Cham
    Feb 15, 2016 at 13:41
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    $\begingroup$ probably not, but (depending on whether you're on SI or cgs) fiddling with the signs will give you solutions. $\endgroup$
    – Emilio Pisanty
    Feb 15, 2016 at 13:42
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    $\begingroup$ OK. It was your due diligence to work these out, but the full solutions are \begin{align}\mathbf E&=\hat{\mathbf y}\left(E_1 \cos(kx)\cos(\omega t)+E_2 \cos(kx)\sin(\omega t)\right) \\ \mathbf B&=\frac{\hat{\mathbf z}}{c}\left(-E_1 \sin(kx)\sin(\omega t)+E_2 \sin(kx)\cos(\omega t)\right).\end{align} The point is that a standing wave oscillates between all-electric and all-magnetic, so you can manufacture the green and blue fields ($E_1$ and $E_2$) independently and then superpose them. $\endgroup$
    – Emilio Pisanty
    Feb 15, 2016 at 14:20
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What you are asking about is the near field of a modulated EM radiation. When an antenna generator accelerates the electrons from one side of the antenna rod to the opposite side, the electron's magnetic dipole moments get aligned and by this there is in sum a varying magnetic field induced around the rod. As you know, a varying magnetic field induces an electric field. Depending from the frequency and the power of the antenna generator the induced from the varying magnetic field electric field in parts couldn't go back to the antenna rod, stay in space and induces the next following magnetic field. Here is the picture for the EM radiation of an antenna:

enter image description here

And here is one of the two possible pictures for the induction process for a direction only:

enter image description here

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  • $\begingroup$ The second picture is exactly what I'm looking for. How do you get this configuration from a superposition of the monochromatic plane waves I described in my question ? $\endgroup$
    – Cham
    Feb 14, 2016 at 20:55
  • $\begingroup$ @Cham independent.academia.edu/HolgerFiedler $\endgroup$
    – HolgerFiedler
    Feb 14, 2016 at 20:59
  • $\begingroup$ Please, could you point to a specific document ? This is not answering the question above. $\endgroup$
    – Cham
    Feb 14, 2016 at 21:17

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