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I'm recently picking up interest about some aspects of physics, I never really studied physics apart from some basics and I'm having trouble finding a good course explaining the theorical approach without diving into maths (too much).

I'm interested in understanding how does electromagnetic waves work and was wondering what makes them propagate. I'm so used to mechanical waves needing a material to propagate that I'm having a hard time grasping that it's doable in a void.

From what I understood, for an electromagnetic wave to appear we need some energy exciting the electronic cloud of an atom to make its electric field "flickers" and thus creating a magnetic field, which will "disturb" the electric field, which will "disturb" the magnetic field and back and forth...

Do I have this right so far ? And if yes are those disturbances what we call electromagnetic waves and are traveling at speed of light ? I just can't grasp how a field can propagate so fast and so far

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    $\begingroup$ Light and other electromagnetic waves do not require a medium. It carries the energy itself. Change in electric field, creates a magnetic field. Change in magnetic field, creates an electric field. It keeps rippling forward. $\endgroup$
    – Mast
    May 21 at 19:15
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    $\begingroup$ @JosB: the "amount of fields" is only as gigantic as it was in the first instant, because there is always energy conservation. So, when a spherical wave spreads from say an antenna about 0.1m wide (microwaves) to say 100m, its field amplitude will have shrunk to 0.1/100=0.001 to the power of two, i.e. 1/millionth. You can imagine how small the amplitudes get when the EM fields travel for one second (gigantic 2.99 x 10^9 m). $\endgroup$
    – oliver
    May 21 at 19:28
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    $\begingroup$ Here is a math-free classical explanation - In what medium are non-mechanical waves a disturbance? The aether?. And here is a more quantum mechanical one. How can a red light photon be different from a blue light photon? $\endgroup$
    – mmesser314
    May 21 at 19:35
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    $\begingroup$ The field doesn't shrink much since there is conservation of energy. Light travels nearly without losses throughout vacuum, 8 light minutes is a relatively short distance. $\endgroup$
    – Mast
    May 22 at 6:36
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    $\begingroup$ @JosB: no, the frequency does not change. If it's red it stays red, if it's blue it stays blue, and if it's infrared, it stays so. The intensity is what decreases. The fact that we can see light from the sun although the sun is 159 million kilometers away is because the sun at its surface shines so blindingly bright that after travelling 8 lightminutes there is still enough intensity so we can see it. If the sun was as bright as a candle, we wouldn't see it over that distance. The intensity fall-off is proportional to the inverse area of the sphere the light is currently at, i.e. prop to 1/r^2 $\endgroup$
    – oliver
    May 22 at 8:47
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Many concepts in modern physics boil down to looking at something that worked in classical physics, and realizing that apparently crucial features are not really as important as they appeared at first glance.

Your description of the propagation of electromagnetic fields is spot on. While it's very counterintuitive for us to imagine a wave propagating without a medium, in fact it turns out that there is no need for particles to move for a wave to exist. The crucial feature of a wave (in our current understanding) is that energy is transported from one region of spacetime to another. Maxwell's equations allow propagating disturbances in the electromagnetic field to carry energy.

The trick is that these disturbances do not imply the movement of any particle and there is no rest frame for a medium in which the waves propagate. But it turns out this is just extra baggage that our common experience causes us to expect, but which is not really necessary. If you look at the equations (Maxwell's equations) defining the laws of motion for the electric and magnetic fields, you can show that the electric and magnetic fields obey a wave equation (ie -- the same equation obeyed by small ripples on a pond, or waves on a string). This equation has exactly the same form in any reference frame. The modern point of view is to take these governing equations as the definition of a wave -- even though this is an abstraction from our common sense intuition that waves should involve a disturbance in a medium.

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    $\begingroup$ there is no need for particles to move for a wave to exist : isn't it a contradiction with the wave-particle duality concept? $\endgroup$
    – manu190466
    May 21 at 20:59
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    $\begingroup$ @manu190466 I don't think so. A wave traveling in a medium is by definition a disturbance of the positions of (classical) particles -- this feature of waves is not important, even classically, as shown in Maxwell's equations. Quantum mechanically, it is true that we can think of a classical wave as being a coherent state consisting of a superposition of many quanta (photons in this case). But the existence of photons doesn't change the fact that there is no medium in which the wave (or superposition of photons) is traveling. $\endgroup$
    – Andrew
    May 21 at 21:04
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    $\begingroup$ In equations what I mean is that in the wave equation obeyed by the vector potential, $\square A_\mu = 0$, it is not necessary that $A_\mu$ represents displacements of particles from equilibrium (unlike what you would expect based on an analysis of waves in an elastic medium). $\endgroup$
    – Andrew
    May 21 at 21:07

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