# How electromagnetic waves are created?

Macroscopically, electromagnetic waves are produced by a changing dipole or an oscillating charged particle as shown below:

In this case, the frequency of the electromagnetic radiation is equal to the frequency of oscillation.

However, we also know that electromagnetic radiation is produced by simply accelerating a charged particle like below:

I have a few questions:

1. What frequency will the radiation of the second animation be? For simplicity, lets say it undergoes the exact same acceleration that the charge in the first animation does from t=0 to t=T/4, but then abruptly stops accelerating (so essentially its just the first quarter of a sinusoid acceleration).
2. If a charged particle is then hit by the radiation emitted by the second animation, would it just feel a force in a single direction, rather than an oscillatory force?
3. In terms of the actual photons, it would seem the first animation emits only photons with the same frequency as the particle's oscillation. What about the second animation? As far as the particle knows for t=0 to t=T/4 it is under the exact same acceleration as the first animation, so would it would emit the exact same photons for that period?
• I love the second animation. Can I steal it / where is it from? – Arthur Aug 25 '17 at 7:04
• I love it too! It's not mine, found it first on this caltech site: tapir.caltech.edu/~teviet/Waves/empulse.html – jaydnul Aug 25 '17 at 15:58
• the 2nd may be a little misleading, with this "fixed" radiating potentials, and only 1 ripple... a physicist may be able to improve on it? – Olivier Dulac Aug 25 '17 at 16:07

First of all one should have clear the difference between classical electromagnetic theory, which completely describes synchrotron radiation with the potentials generated by a classical moving charged particle, and the quantum frame for EM which consists of photons building up the classical radiation.

What frequency will the radiation of the second animation be?

Synchrotron radiation has a spectrum of frequencies as seen in the lecture referenced above:

It is a continuum of photon energies emitted. At the quantum mechanical level, it is the probabilistic interaction of the charged particle with the field that is accelerating it.

The lower photon is off mass shell giving the energy, from the accelerating field, for the radiation of the real photon (upper).

If a charged particle is then hit by the radiation emitted by the second animation, would it just feel a force in a single direction, rather than an oscillatory force?

It depends on the size of the charged particle. At the quantum level it is only energies and spins that are exchanged. The single photon does not oscillate, only its wave function whose complex conjugate squared will give the probability density for the interaction to occur.

In terms of the actual photons, it would seem the first animation emits only photons with the same frequency as the particle's oscillation.

The dipole emits classical EM light with zillions of photons of energy = $h \ \nu$, the frequency, because it needs the ~1023 atoms of the dipole antenna. A single quantum mechanical particle is not a dipole (it can radiate as a dipole in an imposed field which has alternating frequency), it has to be imaged by Feynman diagrams with individual photon probabilistic radiation.

so would it would emit the exact same photons for that period?

The emission will follow the probability distribution for the solution of the above feynman diagram.

As your question hints, frequency domain analysis is a convenience which often obscures an intuition what the fields are actually doing at any given time. It takes some practice to learn to mentally jump back and forth between the frequency and time domains, but for inquiries like this, it certainly pays dividends.

The recipe is simple:

1. Pick a location in space and time.

2. From that point, begin to measure the electric field as a function of time.

3. Fourier transform the resulting waveform to obtain the frequency domain representation.

What you'll find for the second animation is that your temporal E-field waveform will be a short pulse with some duration (the pulse duration follows the duration of the acceleration). Since time and frequency are conjugate variables, the corresponding frequency will be: all of them, from close to 0 Hz up to approximately the inverse duration of the pulse (as an aside, the spectral phase will be flat, meaning that all the frequencies are in phase with each other). That is to say, a short pulse in time is a broadband pulse in frequency. There is no single frequency describing this event, unlike the oscillating charge of the first animation, which must oscillate forever to achieve a truly single-frequency ($\delta$ function) emission.

• Interesting. So when the second animation starts to undergo its acceleration, how does it "know" it needs to emit a broadband range of frequencies rather than "wait" to see if its sinusoidal acceleration will be completed to form a single definite frequency (sorry if this sounds stupid)? – jaydnul Aug 25 '17 at 5:44
• It doesn't know, the frequency emitted are a consequence of the movement (acceleration) and the shape of the consequent field. To have one single frequency you would need a charge oscillating sinusoidally from the infinite past to the infinite future. That is what a single frequency is. Any shape other that a pure sinusoid is composed of more than one frequency. So as Gilbert says, Fourier transform theory is essential to analyse those phenomenon. – EigenDavid Aug 25 '17 at 6:39
• Hmm, so is any charged particle following a sinusoidal oscillation emitting a broadband of frequencies? What I mean is, moment to moment, the particle only "knows" its under an acceleration, but it doesn't know that acceleration is achieving an oscillatory pattern in space. If it is emitting a broadband, how do those other frequencies wash out to where we detect something "near" the fundamental oscillation frequency emitting from the particle? – jaydnul Aug 25 '17 at 7:10
• There is no need for the particle to know its own movement before and after because it does not really emit photon of one given frequency at each instant (anyway thinking of photon is really not the best approach here). The corollary to the fact that one frequency is a infinite signal is that if you want to measure one frequency, you also have to do it for an infinite time. But again if you really want to understand you need Fourier analysis.Look at physics.stackexchange.com/questions/267474/… and maybe ask a question on this specifically – EigenDavid Aug 25 '17 at 9:46
• I found an interesting post here: math.stackexchange.com/questions/749065/… So maybe my questions is if particle starts oscillating at a constant frequency, based the its Fourier transform, it too is emitting a wide range of frequencies (since it hasn't been oscillating forever)? – jaydnul Aug 25 '17 at 17:05