# Does Electromagnetic wave propagation really use $E$ to create $B$

I was studying how electromagnetic waves (EM waves) were generated and propagated. I was shown a dipole antenna with an AC source sloshing charges back and forth.

To my understanding, when the charges are accelerated their field lines change and "update" the existing field lines, effectively creating a pulse of electric field lines, shown below (credit to Alfred Centauri here) which race off to infinity As this was a current in the wire, it also would create a matching magnetic field, which also races off to infinity on it's own right? So with these two elements we have completed the electric and magnetic part of the EM wave.

However, the textbook also talks about how there are changing electric fields which induce magnetic fields (and vice versa). Where does that happen? I don't think that the electric field needs anything to "induce" the front of it right? Don't the electric and magnetic fields just keep on propagating because they're field lines?

So my question is, how does this "induction" occur and how does it produce the next segment of the wave?

Magnetic fields do not induce electric fields and neither do electric fields induce magnetic fields. They are simultaneously induced by time varying currents and by time varying charges that are the sources and causes of the fields. This has been well-known since 1908 .

It has been since erroneously renamed as "Jefimenko's equations", see. https://en.wikipedia.org/wiki/Jefimenko%27s_equations.

These equations have been well-known to engineers since the 1930's when Stratton and Chu published them in "Diffraction Theory of Electromagnetic Waves" Phys. Rev. July 1, 1939 vol. 56.

Professor McDonald wrote a fascinating note  on the subject and it should be compulsory reading to all EE/Phys majors while replacing hundreds of textbooks used in EM teaching.

Here they are reproduced in the form McDonald wrote them:

$$\textbf{E} = \int \frac{[\rho]\hat{\textbf{n}}}{R^2}d\textbf{x}' +\frac{1}{c}\int \frac{([\textbf {J}]\cdot \hat{\textbf{n}})\hat{\textbf{n}}+([\textbf {J}] \times \hat{\textbf{n}}) \times \hat{\textbf{n}}}{R^2}d\textbf{x}' + \frac{1}{c^2}\int \frac{([\dot {\textbf {J}}] \times \hat{\textbf{n}}) \times \hat{\textbf{n}}}{R}d\textbf{x}'$$

$$\textbf{B} = \frac{1}{c}\int \frac{[\textbf {J}]\times \hat{\textbf{n}}}{R^2}d\textbf{x}' + \frac{1}{c^2}\int \frac{[\dot {\textbf {J}}] \times \hat{\textbf{n}}}{R}d\textbf{x}'$$

In these equations the square brackets mean taking the retarded time $$[f]=f(t-R/c)$$. Notice that both the electric and magnetic fields are explicitly expressed as the retarded functions of the charge density $$\rho$$ and current density $$\textbf{J}$$. In this interpretation the two "$$\textrm{curl}$$" Maxwell equations are restrictions as the two fields my develop but one does not induce the other, only sources create fields.

The beauty of Ignatowsky's equations is that they explicitly and separately show the reactive fields and the propagating fields. The reactive or static fields are the ones that have $$1/R^2$$ dependence and they are induced by charges and currents while the propagating fields have $$1/R$$ in the respective integrands and they induced by the time rate variation of the current density. Furthermore it is also manifestly displayed that the propagating fields are perpendicular to each other and also to the direction of propagation, i.e., the propagating field is a transverse wave and they are induced by the time rate variation of the current density.

Since a current is a moving charge, $$\textbf{J}=\rho \textbf{v}$$, the time rate of the current is accelerating charge, hence the propagating terms are induced by accelerating charges.

: Ignatowsky: Ann. d. Physik 23, 875 (1907); 25, 99 (1908).

: Kirk T. McDonald, The relation between expressions for time-dependent electromagnetic fields given by Jefimenko and by Panofsky and Phillips, American Journal of Physics 65 (11) (1997), 1074-1076

• Where is the current and charge density in an EM wave? How could there be current if it's just moving electric and magnetic waves. Do you mean the current density during the generation and then during propagation nothing more happens? – John Hon Feb 4 '19 at 6:14
• There is no EM wave in the currents or charges, the fields and EM waves are in the vacuum (ether), the space between the charges/currents but the waves/fields are generated or induced by the currents and charges. This is similar to the way a single point charge has an electric field around it, or a static current with a magnetic. As you move the charge/current its field will move with it. This is what is described by Ignatowsky's equations explicitly and they even show explicitly how time varying currents cause radiation. – hyportnex Feb 4 '19 at 14:12

In an electromagnetic wave, changing electric fields induce changing magnetic fields, which turn induce changing electric fields. This procedure self-propagates at the speed of light (since it is light) across space until it is absorbed or reflected.

• Firstly, thank you, but I still cannot visualize what that would look like (all animations just look like the E and B fields just travel straight) and secondly, how would you explain the fact that the front of a wave is always the same if a new bit is becoming the new front? – John Hon Feb 3 '19 at 6:55

Outside the source (i.e., where $$\rho=0$$ and $$\vec{j}=0$$) we have the Maxwell equations stating that $$\partial\vec{E}/\partial t\propto\text{rot}(\vec{B})$$ and $$\partial\vec{B}/\partial t\propto\text{rot}(\vec{E})$$. Change in time of one field is induced by the other, and these are quantitative relationships.

Of course, we can obtain from them a wave equation for each field and supply these with the right boundary conditions, and the field "coupling" may become less visible. But we should not forget the original Maxwell equations with non zero charge density and current, as well as that the total force acting on something (a receiver antenna, for example) is generally determined with both fields: $$\vec{E}$$ and $$\vec{B}$$. In other words, the coupling is at least inavoidable at the source and at the "sink".

The picture you show is qualitative. The implicit and important thing in it is a wave character of perturbation propagation followed from the exact quantitative relationships ;-)

There is a point of view, that all the accelerated electrons in the antenna rod are producing a macroscopic EM wave. That is fine for the understanding of how radio translation works.

But it’s worth to digg a little deeper. The only possibility to induce a EM radiation is the acceleration of subatomic particles, mostly electrons. Even the thermic radiation, which emits every body with temperature above 0 Kelvin, is induced by these accelerated particles.

The deeper reason for a radio wave is the synchronized acceleration of a hugh number of electrons and the acceleration of these electrons on the surface of the antenna rod. This skin effect is a needed condition because the photons from electrons, accelerated inside the rod, get dissipated at once and get transformed step by step to infrared radiation.

Each electron emits his own EM radiation. The quanta of this radiation are called photons. If one agree to this point, than it is obvious that the radio wave in a microscopic view is the combination of all the emitted photons from the synchronized acceleration of the surface electrons of the rod.

Hertz was the first who observed that the radio waves have aelectric and a magnetic field component. Since the radio wave consists of synchronized photons it is legitime to conclude that each photons also consists of a magnetic and an electric field component.

However, the textbook also talks about how there are changing electric fields which induce magnetic fields (and vice versa). Where does that happen?

The field components (E and B) of the photons are oscillating and changing their sign periodically. This is the nature of the photon. The representation of this phenomenon is drawn as follows:

Closer to the antenna - in the near field -, the EM wave looks like this:

From this sketch it is seen the induction of the magnetic field component form the electric and the electric from the magnetic.

• Thank you, but can you explain exactly how at a point in space, the wave induces the next "front" of the wave? – John Hon Feb 3 '19 at 8:41
• @JohnHon unfortunately no, my imagination ends by the point that a quanta of energy transforms from E to B and back, moving with c. – HolgerFiedler Feb 3 '19 at 8:48
• @JohnHon The only think I see is, that electron (proton,...) has an electric field AND a magnetic dipole. Seems all these subatomic particles contain the “ingredients” for the photon emission with it E and B. – HolgerFiedler Feb 3 '19 at 10:44