# Phase-alignment in EM-waves

In a physics-book I read that "it should be noted that the [electric and magnetic] fields [in a moving EM-wave] alternate at first, 90 degrees out of phase with each other. As they travel, they gradually come into phase". Is this true ?? I've never read or heard this before.

• This is surely a duplicate, but I can't find one. The statement is correct. – garyp Oct 12 '16 at 2:13
• Context please! What do you mean by a moving EM wave? Moving from where to where and generated by what? – ProfRob Oct 12 '16 at 6:45
• -1. Which physics book? You should not "quote" an authority without giving the reference for it. Context is also important. – sammy gerbil Feb 28 '17 at 4:01

Consulting the late J.D. Jackson's "Classical Electrodynamics, 3rd edition", page 411, we find the following expressions for the fields due to a sinusoidally excited, center-fed, linear antenna that is small compared to a wavelength:

$$\mathbf{H} = \frac{ck^2}{4\pi}\left(\mathbf{n} \times \mathbf{p}\right)\frac{e^{ikr}}{r}\left(1 - \frac{1}{ikr}\right)$$

$$\mathbf{E} = \frac{1}{4\pi\epsilon_0}\left\{k^2\left(\mathbf{n} \times \mathbf{p}\right)\times \mathbf{n}\frac{e^{ikr}}{r} + \left[3\mathbf{n}(\mathbf{n}\cdot \mathbf{p}) - \mathbf{p}\right]\left(\frac{1}{r^3} - \frac{ik}{r^2}\right)e^{ikr}\right\}$$

We note that there are magnetic and electric terms that decay as $1/r$ and $1/r^2$ as well as an electric term that decays as $1/r^3$ where $r$ is the distance from the antenna.

The $1/r$ terms are in phase (always) and thus transport energy away from the antenna; these are the radiation terms and they dominate in the far-field:

$$\mathbf{H} = \frac{ck^2}{4\pi}\left(\mathbf{n} \times \mathbf{p}\right)\frac{e^{ikr}}{r}$$

$$\mathbf{E} = \frac{1}{4\pi\epsilon_0}k^2\left(\mathbf{n} \times \mathbf{p}\right)\times \mathbf{n}\frac{e^{ikr}}{r} = Z_0 \mathbf{H} \times \mathbf{n}$$

In the region near the antenna (near-field), the dominant terms approach

$$\mathbf{H} = \frac{i \omega}{4\pi}\left(\mathbf{n} \times \mathbf{p}\right)\frac{1}{r^2}$$

$$\mathbf{E} = \frac{1}{4\pi\epsilon_0}\left[3\mathbf{n}(\mathbf{n}\cdot \mathbf{p}) - \mathbf{p}\right]\frac{1}{r^3}$$

which are out of phase (reactive) and thus energy flows back and forth between these fields and the antenna.

In summary, it is true that the dominant fields near the antenna are out of phase but these aren't the fields that propagate to the far-field since these fields decay as the inverse square and the inverse cube of the distance from the antenna.

• Thank you for this nice demonstration that the electromagnetic "near fields" at an antenna are not propagating waves as opposed to the "far fields". This is the reason why the terms "near field" and "far field" at an antenna have been introduced. – freecharly Oct 12 '16 at 14:49

The electric and magnetic fields in a linearly polarized sinusoidal plane electromagnetic wave in vacuum or in a simple dielectric are always in phase, but their planes of oscillation in space are orthogonal to each other. The mentioned statement "they gradually come into phase" seems not to hold for such a wave.

• Are you talking about the phase relations of electrical and magnetic fields at an antenna? – freecharly Oct 12 '16 at 1:10
• @AlfredCentauri I don't see the reference to a plane wave. I think the question is about fields generated by a source (e.g., antenna). Hence the mention of "at first". – garyp Oct 12 '16 at 2:08
• The question does not mention plane waves. @PERFESSERCREEK-WATER is correct. The $E$ and $B$ fields near a source are 90 degrees out of phase, but after some distance they are in phase. – garyp Oct 12 '16 at 2:10
• @Alfred Centauri - If the question had been better formulated a lot of unnecessary guesswork could have been avoided. The OP reads a book but fails to mention that the radiation from an antenna (e.g. dipole) is considered. It is clear that the near-field of an antenna is different from freely propagating waves at a larger distance. – freecharly Oct 12 '16 at 4:45
• @AlfredCentauri Sorry. I didn't read carefully enough, and I thought the comment was directed toward the question, not this answer. This answer is factually correct, but does not address the question. But the question is very poorly constructed, begging for misinterpretation. – garyp Oct 12 '16 at 11:58