I don't have a scientific education, yet I'm scientifically curious. Among other things, I'm struggling to understand the nature of electromagnetic waves.

What I have recently realized is that the depiction of a EM wave as wobbly rope is inherently wrong.

Until recently I was imagining radio waves to be invisible wobbly lines piercing the space in all directions. I bet that the absolute majority of common people visualizes radio waves like that.

em wave graph

But the wobbly rope image is not a depiction of a radio wave itself. It's a graph where X axis is distance (in the direction of the wave propagation) and Y axis is the "strength" (or "tension") of the field in each point of that distance. And a graph is not a depiction of an object. It's a documentation of a phenomenon. And the form of the actual wave (and even the concept of the wave itself!) has nothing to do with the wobbly rope.

After investigating this matter further, reading answers on this site (specifically from dmckee and CuriousOne, thanks!) and a couple of sleepless nights, I've figured out that:

  • In almost every graph, one of the axes is time. But the wobbly rope graph is an exception, it doesn't demonstrate time. To demonstrate time, those graphs are animated. This is the main reason of the wobbly rope discrepancy: a picture associated with a moving wave actually looks like a moving wave, but that's merely an unfortunate coincidence! :-O
  • Sometimes, Z axis is added. Y and Z both mean the same thing: "strength", but one is for the magnetic field and the other one is for the electric field.
  • The fields don't wobble. It's just every point of the field changes its state. Each point of the field is a vector. This means that each point's state consists of two values: a "strength" and a direction. Each point of a field is always stationary, but each is associated with a direction and "strength". When an electric or a magnetic wave propagates through a field, the points of the field change their directions and "strengths" but they themselves never move.
  • The direction of each point of the field is always perpendicular to the direction of the wave that propagates through that region of the field.
  • Electric and magnetic waves always travel together, and their directions in every point are perpendicular to each other. On the wobbly graph, the Y and Z axes are also perpendicular, but that's just a coincidence. Y and Z demonstrate amplitude, the "strength" of the field in every point, but the X axis demonstrates distance. Distance and "strength" are different things and don't correspond with each other! That's why those images are graphs, not illustrations of waves themselves.
  • Electromagnetic waves are not rays and neither they are composed of rays! (Mind: blown.) EM waves travel in fronts, i.e., they spread spatially like sound waves or water ripples. The graph represents not a spatial wave but a single longitudinal line copy-pasted from a spatial wave. That's merely a thought experiment.
  • Though it's technically possible to emit exactly one photon (a photon is a single unit of a wave), a photon is a quantum particle. It conforms to quantum laws. Human-world laws do not apply to photons. This means that it is not correct to speak about such thing as a "trajectory" of a single photon. Photons are kinda probabilities, not moving objects. (I can't get my head around this statement, please help me find a mental model for it! SOS!) This means that the wobbly rope illustration does not demonstrate a "single" wave. It merely demonstrates an 1D fragment of a 3D wave.
  • When an electromagnetic wave travels as a front, the photons interact with each other, they kinda form a single "object" -- a spatial wave. This wave would, roughly speaking, bounce off a metal mesh (a Faraday cage), even though a single photon could easily come between wires of the mesh.


  1. Are the above conclusions correct?
  2. Say a spatial EM wave propagates through the EM fields. We "take a snapshot" of the field and from that snapshot we crop an 1D line perpendicular to the direction of the wave. How will vectors of each point on that field line relate to each other? Will they all be pointed at the same direction? Or will they form a sine-like wobbly pattern? If yes, is that pattern's period equal to (or derived from) the wavelength? See two variants below. The second variant would explain for me the Faraday cage: how the size of the mesh step (a transverse parameter) can depend on the wavelength (a longitudinal parameter). The second variant might also explain circularly polarized waves.

Imgur Imgur


Here's a fresh one:

Light: Crash Course Astronomy #24

And it's a poster of an educational video about the nature of light!


2 Answers 2


Many of the things you write sound OK. But I wouldn't say that the other directions in the graph are mere strengths, they indicate the actual value (strength/magnitude and direction as well) of the electric and magnetic fields.

First note that technically the electric field is a vector and it is a field so it should have a vector (possibly zero at every point). Each vector has a head and a tail. You can think of the location of the tail as telling you the place where the field has a particular value. Then you can think if the difference between the head and the tail as telling you both the magnitude and the direction of the electric field at that point, in some specified unit system. So imagine a bunch of arrows all the same color the location of each is telling you where it is telling you the field and how the arrow points from there tells you the value.

Then draw the magnetic fields in a different color. And in both cases you can't draw a vector at every single point because it would just be too much to see them all.

For a plane wave traveling through vacuum there is a great deal of regularity. If the wave travels in the x direction then the electric and magnetic fields all point in the y-z plane. And their values only depend on time and on x, so you can draw just one longitudinal line and it will tell you about all of the values at all the locations. And since the electric fields and magnetic fields all point in the y-z plane you can reimagine the y-z plane as like the independent axis of a graph.

And for a classical plane wave travelling through a vacuum your picture should have the electric and magnetic fields both be strong together at one plane of a fixed value in the longitudinal direction and then at a different plane corresponding to a different fixed value of the longitudinal direction. Strong together and weak together and then strong again but pointing in the opposite direction.

If so, you get a picture much like the first one you drew (though that is just for one the electric and magnetic fields, a wave has both). That is all fine for a classical wave travelling through a vacuum.

For your other questions I think you should look at existing questions about the different topics.

From a quantum perspective it is much more complicated. From a quantum view there isn't an electric or a magnetic field, there is a photon field. When you have a large number of photons all in phase with each other it can look or act like an electromagnetic wave, but it is still different, and if you have a small number or they aren't in phase then it just truly different.

The electric and magnetic fields classically are stand-ins for saying how charges interact (though they do have their own energy and momentum, pressure and stress, etc.) And in quantum mechanics different interactions are possible and so the photon field is a stand in for saying how those different interactions go.

Since in quantum mechanics the charges don't have a location and momentum saying that their momentum changes based on the field where the charge is located just isn't going to be possible because you have none of those things. And if you can't verify the field's values at any location it becomes difficult to say it is there in that way.

We make the objects we need to predict results, for quantum mechanics we need different objects because we are predicting different interactions. Only in some limits (that only hold sometimes) do we expect quantum mechanical effects to start to approximately look like classical effects.

As for a transverse spacing in a Faraday cage stopping a wave that varies only in the longitudinal direction remember that you were drawing only the simplest case, the case of a plane wave.

In a plane wave the entire plane parallel to the y-z plane has the exact same electromagnetic field. This very simple (to describe) solution allows the wave to travel entirely in the x direction, over time each part of the wave simply slides over, the new value at distance $x=x_0+\Delta x$ over is the old value at $x_0$ a time $\Delta x/c$ earlier.

But no real wave is ever that perfect. They might expand in a spherical front instead of a plane front. They might have a central region where it is strong and it gets weaker farther out (like a Gaussian beam) some people even make beams where the energy and momentum travels along a cone so there is a central strong region but if you place a small object there the region beyond it still gets light because the light from farther out came from a base of the cone that started out farther from the central region.

Sort of like if you had some racers all lined up for a race but instead if running in their own tracks they ran towards the same point that is in front on the center runner. The center runner gets there first, but if parts of the center of the track are damaged the runners that left later will eventually close in and it won't matter that that the runners originally closer to the center got affected.

So a real wave has a more complicated wave front. And in fact if you are trying to avoid the wires of the Faraday cage you need your beam to be focused to not extend out too far in the direction transverse to the direction of propagation. And this isn't restricting the amplitude, this is saying that the actual wave shouldn't have field values spread out in the yz plane, it should be focused to a small portion of the yz plane.

Imagine you shot narrow beams all towards the same hole in the Faraday cage, you need very precision aiming each point has to has "its x" aimed very precisely to get in that same hole.

In general if you aim your initial propagation directions with accuracy on the order of the wavelength then you gave changed the wave.

Thus is because the direction of propagation isn't a magic thing you can freely associate with a point. It depends on how the electromagnetic field varied in space. When you field didn't vary in the y and z directions then the field travelled in the x direction. Now you are trying to get each region to aim slightly differently so you try to adjust the electric and magnetic field here and there to have each part aim just right.

Waves tend to spread, so if you aim some at that one whole, lots will be lost in other directions and what aimed well for one whole is nor going to aim for the other holes, so the Faraday cage succeeds at blocking almost all fields coming at it assuming the cage is big enough to have many many holes.

I think many of these question can be (and have been) asked as separate questions.


What Maxwell derived was from radio waves. Radio waves are modulated photon radiation. Electrons in an antenna rod get accelerated at once and emit photons. The density of this photons distributes in space. Detecting this radiation with a receiver one get a sinus wave form. The frequency of this wave has to do with the frequency of the antenna generator and has nothing to do with the frequency of the single photons in this wave.

But all this photons form the electric and the magnetic component of the radio wave. Hence each photon has an electric and a magnetic component. Photons once emitted are indivisible units. Imagine photons as bubbles, inflated at one moment to the left and right as a magnetic dipole and in the next moment inflated to the top and bottom as an electric dipole. Both inflations ar e perpendicular to the propagation of the photon. More details see in my paper.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.