How is electromagnetic wave variation distributed in space?

Question

Imagine an electromagnetic wave (a monochromatic one for example). The electric field amplitude and its variations travel in the propagation direction.

So, if there really exists a propagation direction, what happens in other directions? I mean, is there an oscillation in those other directions? Does something (electric or magnetic field) move in those directions? Does it's speed in those direction depends on frequency,or is this wave infinitely thin? Is it not covering any space in other directions?

I have seen many graphs showing that the field variation is not in the direction of propagation, that "means", based on polarization experiments that it's not a Longitudinal wave, It's a transverse wave.

If I accept that the field is not varying in the propagation direction (it just goes forward in that direction), then I could think it's varying in other directions, but in what direction?

Perhaps we can't say it's varying in any direction, it's not a spatial variable. I just can't imagine how the field variation is distributed in space.

EDIT

some options

1. Electromagnetic waves just move in one single spatial propagation direction.. then all spatial concepts that take account other directions are wrong, like "polarization" and "transverse wave" definition

2. Electromagnetic waves moves in others directions too, then speed of propagation in those other directions could depend on frequency, and thats weird, but who knows!

EDIT 2 (Add my comment as part of the question)

If there are infinite wave planes, then imagine a light that is turned OFF and then is turned ON, comparing the EM, first it is covering nothing, and then suddenly reach "infinity" (or at least very far..in fact any distance is enough), what is the speed of THAT propagation? because it is NOT a wave propagation, because is not in the direction of propagation, that infinite or very large field extends itself at infinite speed?

Answer 1

I am still not sure I understand your question but from what I gathered you are confused about the behavior of an EM wave (let's say monochromatic) so I'll discuss the properties of waves of increasing difficulty (eventually returning to EM).

EM waves are not easy to imagine because they are vectorial in nature. That is, in every point of the space there is an arrow describing the magnitude and direction of the $E$ field. Let's simplify this a imagine that at every point there is just a single value, say pressure, so we are be talking about sound waves. It's useful to consider the notion of a wave front. This is a collection of points that have the same phase (think of the circles that form if you throw a stone into the water). For a monochromatic wave $$ p = p_0 \exp(i\omega t - i{\mathbf k \cdot r})$$ (characterized by the wave-vector $\mathbf k$ and frequency $\omega$), it will be just a plane. You can imagine that plane moving at the speed of propagation $c = {\omega \over |{\mathbf k}|}$ into the direction given by $\mathbf k$. Let's emphasize that all points of that plane will oscillate in the same manner; which simplifies the picture by letting us consider just one point from that plane. We are left with the following image

Imagine that the direction of the propagation of the wave is the horizontal axis, the phase of the wave is carried out on the vertical axes and the plane is represented by any point of the curve (e.g. pick the top of the wave) that travels to the right.

Now, let's consider vector fields. These can be written as ${\mathbf E} = (E_x, E_y, E_z)$ and the previous discussion applies for every component of the field. So there are now three independent oscillations each with its own phase. Except that EM field is massless and so cannot oscillate in the direction of propagation (so called longitudinal polarization); this fact can be derived from Maxwell's equations. So this leaves us with two transversal polarizations $E_x$ and $E_y$ (taken as complex numbers because we also need to take account of their phases) by identifying the $\mathbf z$ axis with the direction of propagation $\mathbf k$. As in the previous case, for monochromatic wave the wavefront will be a plane. But now we have two independent polarization and instead of simple oscillation we will in general obtain an elliptical polarization

Again, single point on the curve represents a whole wavefront plane. Note that special cases include circular (when $E_y$ lags by $\pi /2$ behind $E_x$ but has equal magnitude) and linear polarizations (when $E_y$ and $E_x$ are in phase).

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Notes

1. We haven't discussed the magnetic part of EM wave $\mathbf B$ because it's determined by just knowing the wave vector $\mathbf k$ and the $\mathbf E$ field. The important point being that it will be perpendicular to both. So the above discussion about the travel of wavefront applies also to $\mathbf B$ field.

2. The monochromatic waves are not realistic as you might imagine. They are infinite in extent (whole wavefront plane oscillates and it sweeps entire universe during its travel). In reality the waves travel as concentrated wave-packets. But the above discussion is still useful because we can use Fourier analysis to pass to the frequency image of the wave and decompose the packet into monochromatic waves.

3. If we are not in the vacuum then the picture is much more complex. The $\mathbf E$ and $\mathbf B$ fields need not be perpendicular to the $\mathbf k$ vector (because the photons gain effective mass in the medium thanks to refractive index), the energy travels in different direction than the wave itself and so on.

Answer 2

I think you're trying to geometrise something that is not geometric. Your comment "It's no covering any space in other directions" is close to the truth.

A field is defined as a (scalar or vector) quantity with a value at every point in space. In the case of classical electromagnetism, this simply means that at each point along the propagation of an electromagnetic wave has an electric and magnetic field vector associated with it. The wave is called transverse because these vectors are both orthogonal to the direction of proportionate (phase) of the wave.

Answer 3

I'm not totally sure I'm parsing the question properly, but here's an attempt to answer part of it.

The derivation of electromagnetic waves from Maxwell's equations often presents infinite plane waves, which have definite momentum but are extended throughout all of space. An example would be:

$\vec{A}(x,y,z,t) = A_0 \hat{x} e^{i \omega(t - z/c)}$.

This wave propagates in the $z$ direction with frequency $\omega$, as you can see from the dependence on $t - z/c$. From ${\vec E} = -\partial_t {\vec A}$ you see that the electric field points in the $x$ direction, and from ${\vec B} = \nabla \times {\vec A}$ you see that the magnetic field points in the $y$ direction. But the field exists everywhere; no matter what point $(x,y,z)$ in space you look at, the field is oscillating there with amplitude $A_0$ and frequency $\omega$.

Now, as you know, in the real world electromagnetic waves don't look like this. When I turn on a lamp, EM waves come out of it, but they don't live everywhere in space; they propagate outward from the lamp and are absorbed or reflected by objects. So the infinite plane wave is an idealization. In the real world, a wave is a superposition of many plane waves. It might look like, for instance,

$A_\mu(t,{\bf x}) = \int \frac{d^3{\bf k}}{(2\pi)^3} A({\bf k}) \epsilon_\mu(k) e^{ik \cdot x}$.

(I've switched to a relativistic notation here out of laziness; if it isn't familiar to you, you might want to look at the simpler example at the Wikipedia page on wave packets.)

The key points are:

• Real world waves aren't really exactly monochromatic, hence the integral over wave vectors ${\bf k}$.
• By adding up plane waves of different wave vectors and frequencies, with a frequency and wave vector-dependent amplitude $A({\bf k})$, you can construct solutions to Maxwell's equations that have different types of dependence on space and time.
• There exist (though I haven't demonstrated it here) solutions that look approximately like plane waves over some region of space, and so are very nearly propagating in a definite direction with definite frequency within that region, but which fall off to zero outside that region.

So, the answer to one of your questions is that the infinite plane wave isn't "infinitely thin"; in fact, it's "infinitely thick" (spread out everywhere), but more realistic waves are localized over some finite region of space.

To give one more example: the beam that comes out of a laser is well-approximated as a Gaussian beam, with a field that decays like $e^{-kr^2/R(z)}$ as a function of transverse distance $r$. (Explaining what $R(z)$ means is beyond the scope of this answer, but you can read more at the link.)