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This is perhaps a total newbie question, and I will try to formulate it the best I can, so here it goes. How does an electromagnetic wave travel through for example, the vacuum of space?

I usually see that waves are explained using analogies with water, pieces of rope, the strings of a guitar, etc, but it seems to me that all those waves need a medium to propagate. In fact, from my point of view, in those examples the wave as a "thing" does not exist, it's just the medium that moves (involuntary reference to The Matrix, sorry).

But in space there is no medium, so how does a wave travel? Are there free particles of some sort in this "vacuum" or something? I believe the existence of "ether" was discarded by Michelson and Morley, so supposedly there isn't a medium for the wave to travel through.

Moreover, I've seen other answers that describe light as a perturbation of the electromagnetic field, but isn't the existence of the field, potential until disturbed? How can it travel through something it does not exist until it's disturbed by the traveling light in the first place? (this last sentence is probably a big misconception by me).

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The particles associated with the electromagnetic waves, described by Maxwell's equations, are the photons. Photons are massless gauge bosons, the so called "force-particles" of QED (quantum electrodynamics).

While sound or the waves in water are just fluctuations (or differences) in the densities of the medium (air, solid material, water, ...), the photons are actual particles, i.e. excitations of a quantum field. So the "medium" where photons propagate is just space-time which is still there, even in most abandoned places in the universe.

The analogies you mentioned are still not that bad. Since we cannot visualize the propagation of electromagnetic waves, we have to come up with something we can, which is unsurprisingly another form of a wave, e.g. water or strings.

As PotonicBoom already mentioned, the photon field exists everywhere in space-time. However, only the excitation of the ground state (the vacuum state) is what we mean by the particle called photon.

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  • $\begingroup$ Nicely said +1 from me! It is interesting what you said about spacetime being some sort of a 'medium'. The metric does not appear in the QED Lagrangian, so how is the claim above justified? Just curious, I'm probably missing something obvious! $\endgroup$ – Constandinos Damalas Jan 4 '15 at 23:23
  • $\begingroup$ The metric does not appear explicitly in the Lagrangian, i.e. the Lagrangian density $\mathcal{L}$. It does however in scalar products, i.e. $F^{\mu\nu}=g^{\mu\rho}g^{\nu\sigma}F_{\rho\sigma}$, and in the action, i.e. the integral over $\mathcal{L}$. Usually a quantum field theory is defined in Minkowski space-time with the common $d^4x$. $\endgroup$ – Clever Jan 5 '15 at 8:20
  • $\begingroup$ But you could just as well define it in some curved space-time with metric $g_{\mu\nu}$ and measure $\sqrt{g}d^4x$, with $g=\det g_{\mu \nu}$. Note that in this case, all scalar products and the lowering and raising of indices have to be done with the metric $g_{\mu\nu}$, which is not as simple as in the Minkowskian case. $\endgroup$ – Clever Jan 5 '15 at 8:20
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If we simplify to the classical electromagnetism then the electromagnetic field is a vector field that exists in all of space. A time-dependent electromagnetic field has an electric field part and a magnetic field part associated with it and both are changing with time. They are described by Maxwell's equations.

This website has an nice animated gif which shows how the two vector fields propagates in 3D space. Notice that the electric and magnetic fields oscillate (change values) perpendicular to each other at all times. What we call then radiation is just the travelling disturbance of energy.

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  • $\begingroup$ Very cool resources, thanks! I think I need to look at that hypnotizing gif for a while :) $\endgroup$ – jotadepicas Jan 4 '15 at 21:48
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    $\begingroup$ It's not what was asked. $\endgroup$ – Sofia Jan 4 '15 at 21:48
  • $\begingroup$ @Sofia Which question did it not address? This is how the field propagates. $\endgroup$ – Constandinos Damalas Jan 4 '15 at 21:51
  • $\begingroup$ @PhotonicBoom your answer makes me think that I cannot grasp the classical idea of a vector field that exists in all of space. Is a vector field a concept, a mathematical "description" if you will, that only comes to "reality" when it is disturbed by something else? (an experiment, a particle, etc) and it presents itself as a force, current, etc. Or is it a physical thing? (perhaps there is no answer to this question without going into the Quantum-Physics realm?). Thanks again! $\endgroup$ – jotadepicas Jan 4 '15 at 21:57
  • $\begingroup$ @jotadepicas You are dwelling into philosophy here so be careful. The concept of a field is a mathematical one, but it has been shown that this description works, i.e it describes all observed phenomena. Now if the field exists or not is not really a valid physics question. What we know is that if you treat radiation like that, you need no medium to describe it. $\endgroup$ – Constandinos Damalas Jan 4 '15 at 21:59
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Since, electro magnetic waves have electric and magnetic vector. Due to this EM waves show electric and magnetic field. An electric and magnetic field have no need a medium to show thier effect. Hence in the presence of electric and magnetic field vector which vibrate perpendeculer to each other and get pertervation EM waves travels in vacuum.

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Electromagnetic waves are only an observed phenomenon. There is no travel.

As an example we can choose a photon which is "traveling" from Sun to Earth (for simplicity we do not care about gravity issues here). That means that Sun is emitting a bit of energy (a momentum) which is received by Earth. So far everything is OK. But what is happening between emission and absorption? The spacetime interval is zero, because the worldline of the photon is lightlike. An empty zero interval cannot be travelled, the transmission of the momentum is happening directly.

The same phenomenon in a reduced form may be observed in a thought experiment where an astronaut makes a space roundtrip near speed of light. He may travel 2000 light years in 2001 years which are reduced for him by time dilation and length contraction to about 2 light years in about 2 years. Due to the high velocity, the spacetime interval (and his proper time) has shrunken to ca. 2 years. When he is coming back to Earth after 2001 years, only 2 years have passed for him (the so-called twin paradox).

The spacetime interval of photons is reduced to zero, that means that they do not travel through spacetime, their momentum is transmitted directly from one place to another. The only observable trace they are leaving in spacetime is an electromagnetic wave moving at light speed c. The observed electromagnetic wave is telling us that at its place spacetime reduced to zero in order to transfer a momentum. This fact is complying with the second postulate of Einstein's relativity which does not say that light is moving at c, but that light is observed by any observer as moving at c.

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  • $\begingroup$ Photons don't have a rest frame, according to the answers to my question $\endgroup$ – Physiks lover Mar 25 '17 at 22:09
  • $\begingroup$ I'm not sure that you read well my answer. I am not talking about a rest frame of a photon. $\endgroup$ – Moonraker Mar 26 '17 at 12:13
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Photons are released in packets. So,when the source is generating the electromagnetic wave it's actually being released in packet forms. These packets are self entities that travel on it's own and doesn't need any medium to sustain it.

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protected by ACuriousMind Mar 25 '17 at 21:52

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