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Are the electric and magnetic fields of an electromagnetic wave perpendicular to each other even in the near-field zone or in the intermediate zone (when the radiation zone approximation is not valid)?

The example given by ProfRob is nice but I would like to know whether E and B fields are always perpendicular for any arbitrary radiation source when the radiation zone approximation is not valid.

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  • $\begingroup$ If you want to know whether or not the E and B field are perpendicular then no as the near field includes static fields aswell. However the "far field " approximation, is just another way of finding terms that behave like 1/r, which IS radiation. And the far field approximation predicts that yes, the radiative components are perpendicular no matter how close you go $\endgroup$ Mar 12 at 14:12
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    $\begingroup$ Please clarify what radiating source you are asking about. $\endgroup$
    – ProfRob
    Mar 12 at 14:34

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Do some degree, the answer depends on exactly how you define "electromagnetic wave" and "radiation source".

Jackson (Third Edition) works out the exact expressions for the ${\bf E}$ and ${\bf B}$ fields from a radiating source in eqs. 9.18, making the sole approximation that the source is small compared to the wavelength of radiation and distance from the source under consideration. (The distance under consideration can be either larger or smaller than the wavelength, so this encompasses both the near and far zones.) He finds that the ${\bf E}$ and ${\bf B}$ fields are indeed always exactly perpendicular (although in the near zone, the ${\bf E}$ field does have a radial component, unlike in the far zone where it becomes transverse).

In full generality, where the wavelength, the spatial extent of the source, and the distance under consideration are all arbitrary, the ${\bf E}$ and ${\bf B}$ fields don't need to be perpendicular. This is essentially just a completely general solution to Maxwell's equations, and the Lorentz-covariant scalar field ${\bf E} \cdot {\bf B}$ certainly isn't required to be zero in general. The fact that you mention the "near-field and intermediate zones" in your question suggests that you have some specific assumptions in mind about the nature of the radiation sources, wavelengths, and/or spatial region under consideration, but we can't answer your question without more details about what those assumptions are.

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  • $\begingroup$ thanks, @tparker Why do you say that $\vec E$ and $\vec B$ are scalar fields? I did not understand that point. Could you elaborate? Did you mean to say that $\vec E\cdot\vec B\neq 0$? $\endgroup$ Mar 14 at 15:59
  • $\begingroup$ @Solidification Yes, sorry, that was a typo. Corrected. $\endgroup$
    – tparker
    Mar 14 at 22:43
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I will assume you mean the fields generated by an oscillating electric dipole.

The E- and B-fields are $$ {\bf E} = \frac{j}{2\pi \epsilon_0} kp_0 \cos\theta\left(1 - \frac{j}{kr}\right) \frac{e^{j(\omega t -kr)}}{r^2}\ {\bf \hat{r}} - \frac{k^2}{4\pi \epsilon_0} p_0 \sin \theta \left(1 - \frac{j}{kr} - \frac{1}{k^2 r^2}\right) \frac{e^{j(\omega t -kr)}}{r}\ {\bf \hat{\theta}}$$ $${\bf B} = -\frac{k^2}{4\pi\epsilon_0 c} p_0 \sin \theta\left(1 - \frac{j}{kr}\right)\frac{e^{j(\omega t -kr)}}{r}\ {\bf \hat{\phi}}\ ,$$ where $k$ is the magnitude of the wave vector, $p_0$ is the electric dipole moment amplitude and $\omega$ is the angular frequency.

As you can see, the scalar product of these two fields is always zero and the E- and B-fields are perpendicular, whatever the value of $r$ and $\theta$.

However, you want to know about more general fields. Well we could show that they are not always perpendicular by considering a trivial counterexample.

Take the dipole mentioned above and surround it with a small oscillating current loop - an oscillating magnetic dipole, with fields $$ {\bf E} = \frac{\mu_0}{4\pi} k^2 m_0 c\sin\theta \left(1 + \frac{j}{kr}\right) \frac{e^{j(\omega t -kr)}}{r}\, {\bf \hat{\phi}} $$ $${\bf B} = -j\frac{\mu_0}{2\pi} k m_0\cos\theta\left(1 +\frac{j}{kr} \right) \frac{e^{j(\omega t -kr)}}{r^2}\, {\bf \hat{r}} - \frac{\mu_0}{4\pi} k^2 m_0\sin\theta \left(1+ \frac{j}{kr} -\frac{1}{k^2r^2} \right)\frac{e^{j(\omega t -kr)}}{r}\, {\bf \hat{\theta}} $$ Since the solutions of Maxwell's equations superpose, then we can take the scalar product of the total fields to be $${\bf E}\cdot{\bf B} = \frac{\mu_0 }{4\pi^2 \epsilon_0} k^2 p_0 m_0 \cos^2\theta \left(1 +\frac{1}{k^2 r^2}\right) \frac{e^{2j(\omega t-kr)}}{r^4} + \frac{\mu_0 }{16\pi^2 \epsilon_0} k^4 p_0 m_0 \sin^2\theta \left(\left(1-\frac{1}{k^2r^2}\right)^2 +\frac{1}{k^2 r^2}\right) \frac{e^{2j(\omega t-kr)}}{r^2} - \frac{\mu_0 }{16\pi^2 \epsilon_0} k^4 p_0 m_0 \sin^2\theta \left(1 +\frac{1}{k^2 r^2}\right) \frac{e^{2j(\omega t-kr)}}{r^2}$$

In the limit where $kr \gg 1$ then indeed the scalar product approaches zero. This is the radiation field limit where the waves approximate to plane waves and the E- and B-fields must be perpendicular. However for smaller $r$ values, this isn't true. The scalar product depends on $r$ and $\theta$ and is generally non-zero.

Thus for any arbitrary distribution of charge and current sources, the E- and B-fields will not be perpendicular. However, if you go to the radiation field limit and at a distance much larger than the source size, then the superposed fields from a single source will approximate to plane waves with perpendicular E- and B-fields.

NB. It is trivial to show that the E- and B-fields of plane waves arriving from different directions from different sources are not necessarily perpendicular.

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    $\begingroup$ These look like the fields for an idealized point dipole (though I might be wrong.) But for an idealized point dipole, the distinction between near and far field gets blurry because the size of the "antenna" is zero and so the distance to the antenna is always large compared to the antenna size. So if my statements about are correct, I'm not sure this answers the question about what happens in the near field. $\endgroup$ Mar 12 at 15:01
  • $\begingroup$ @ProfRob Can we conclude this for any source? Could it be that in this special example E and B are perpendicular but not in general? $\endgroup$ Mar 12 at 15:20
  • $\begingroup$ @MichaelSeifert as I said, it assumes an oscillating electric dipole and is not a general formula. Near and far field terms are dominant dependent on their relative sizes. Every textbook I have seen refers to the terms depending on $r^{-1}$ as the far field and the terms depending on $r^{-2}$ or $r^{-3}$ as the near field. The distinction between them is whether $kr \gg 1$ or $\ll 1$. The size of the dipole does not appear in these formulae because it is assumed to be negligible. I do not claim this to be a general answer for all time-dependent currents/charge densities. $\endgroup$
    – ProfRob
    Mar 12 at 16:08
  • $\begingroup$ Why are you assuming that the OP means the fields generated by an oscillating electric dipole? That seems pretty arbitrary. $\endgroup$
    – tparker
    Mar 13 at 17:40
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    $\begingroup$ @Solidification Probably because it addressed a restricted case of the oscillating dipole. Then you clarified your Q to confirm you wanted something more general. Then it was downvoted before I had chance to address that. And now I have. $\endgroup$
    – ProfRob
    Mar 14 at 16:05
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Imagine the following setup. We have a capacitor whose $E$ field is oriented in the $z$ direction. If you put AC voltage on that capacitor with frequency $\omega$ it will radiate (it's like an electric dipole). Now put a solenoid concentric with that capacitor (either inside the capacitor or around it or something) so that its $B$ field is in the $z$ direction. If you put an AC voltage on that solenoid with frequenecy $\omega$ it will also radiate (it's like a magnetic dipole).

If you now drive the capacitor and the solenoid simultaneously you will have some weird radiator with some weird radiation pattern in the far field, but in the near field, we can see that $E$ and $B$ are parallel.

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If we are only considering the radiative effects of the EM field, we can prove that in general they are always perpendicular to eachother. In general this is not true for the total EM field, but for the radiative components it is always true!

This can be proved by analysing the dot products between the radiative components of jefimenkos equations, for the electric and magnetic field.

The radiative components of jefimenkos equations are -

$\vec{E} = \frac{1}{4\pi\epsilon_0}\int-\frac{1}{|\vec{r}-\vec{r'}|}\frac{1}{c^2} \frac{\partial \vec{J_{tr}}}{\partial t} d^3r'$

$\vec{B} =\frac{-\mu_{0}}{4\pi}\int \frac{\vec{r}-\vec{r'}}{|\vec{r}-\vec{r'}|^3}×\frac{1}{c} \frac{\partial \vec{J_{tr}}}{\partial t} d^3r'$

From here it is seen that $\vec{E} \cdot \vec{B} = 0$ as the difference in the integrand is a cross product, so by definition, the contribution to the integral at each point in space must be perpendicular

(I believe the logic involving analysing the integrants itself to determine the outcome of the dot product is correct, please tell me if I am wrong though)

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    $\begingroup$ I don't think the logic about the integrals does hold, at least not without additional assumptions. As an example, we could be in the "far field" of two different sources in two different directions, such that each source "looks like" a plane wave. This should be a case that Jefimenko's equations can handle. In that case we have $\vec{E} \cdot \vec{B} = \vec{E}_1 \cdot \vec{B}_2 + \vec{E}_2 \cdot \vec{B}_1$ (where "1" and "2" stand for the two plane waves). Depending on the directions of propagation of the two plane waves, this quantity need not vanish. ... $\endgroup$ Mar 12 at 16:31
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    $\begingroup$ ... Most likely there are additional assumptions that need to be imposed on the integrals (something like $\vec{J}_{tr}$ is localized near the origin) to make the argument you want to make. $\endgroup$ Mar 12 at 16:32
  • $\begingroup$ Can you give me any examples of 2 plane waves superposing such that the dot product is non zero? I can't picture a scenario that doesn't. I have been able to prove in the past that if the 2 waves move in the same direction, no matter the polarisation, the dot product is zero, but not for waves in multiple directions $\endgroup$ Mar 12 at 17:09
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    $\begingroup$ Wave 1: $\vec{E}_1 \propto \hat{x}$, $\vec{B}_1 \propto \hat{y}$, propagating in the $z$-direction. Wave 2: $\vec{E}_2 \propto \hat{z}$, $\vec{B}_2 \propto \hat{x}$, propagating in the $y$-direction. $\vec{E}_1 \cdot \vec{B}_2 \neq 0$, and $\vec{E}_2 \cdot \vec{B}_1 = 0$, so their sum is non-zero. $\endgroup$ Mar 12 at 18:24
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    $\begingroup$ I don't think this reasoning works, because when you take the dot product of the two integrals and then move the dot product inside the integrals, you get a double integral with two different dummy integration variables. So in general the integrand is a triple product of vectors evaluated at different points in space, and there's no reason to expect that the cross product of the time derivative of $J$ at two different points in spacetime would be zero. $\endgroup$
    – tparker
    Mar 13 at 17:48

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