Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I just started learning about optics, and in the book I'm reading they explain how the electrical field caused by a single charged particle could be described by a series of field lines, and compare them to ropes, to provide an intuition of the concept.

Then they say that and that if we wiggle the particle up and down, that would produce transversal waves in the horizontal field lines, but no waves in the vertical lines. I know that the physical analogy is not to be taken literally, but I don't understand why wouldn't that cause compression waves in the vertical lines.

I mean, even though the direction of the field in the points directly above and below the particle doesn't change, the intensity does. And I assume it wouldn't instantly. So what am I missing?

share|cite|improve this question
It follows from Maxwell's equations in vacuum, that the $\vec{E}$ and $\vec{B}$ fields are divergenceless, i.e., incompressible. By a Fourier transform one sees that $\vec{E}$ and $\vec{B}$ must propagate as transversal degrees of freedom. – Qmechanic Jul 5 '11 at 21:15
To add to what Qmechanic wrote: the divergenceless property of the electric and magnetic fields are general properties of electromagnetic theories. The same expression also holds for non-linear theories of electromagnetism, as well as propagation in homogeneous (but not necessarily isotropic) media (for example, crystal optics). And in those cases you also have only transversal degrees of freedom. – Willie Wong Jul 5 '11 at 21:21
Thanks, but as I said, I am (1) just starting to learn about the subject (so I haven't yet reached Maxwell's equations), and (2) looking for an intuitive way to understand why this happens -- i.e., I'm not doubting that it does. – waldyrious Jul 5 '11 at 21:26
up vote 8 down vote accepted

If you are going to pursue this physical analogy, which can be useful at times, then you must consider the electric field lines to be of constant tension. That is, the tension of these lines is a constant no matter how much you stretch them. This is different from ordinary ropes or strings or whatever, where the more you stretch them, the higher the tension.

More technically, if you examine the Maxwell stress tensor for a pure electric field, you will find a tension term along the direction of the field and a pressure term transverse to the field. So you in the static case, you can think of the electric field lines as being in balance between tension along field lines and a pressure pushing different lines apart.

For an ordinary stretched string or something like that, if you move the end of the string longitudinally then the stretching or compression changes the tension and the difference in tension will propagate along the string, producing a longitudinal wave. In the case of electric field lines, there is no change in tension to propagate along the line, since the tension is a fixed constant. I hope this helps with your intuition.

share|cite|improve this answer
Thank you. I am afraid I am not yet prepared to understand all the details of your answer, but it gives me a general idea about how this differs from day-to-day physics and why the analogy fails in that particular aspect. – waldyrious Jul 6 '11 at 21:53

It depends on what you mean by "compression wave". When we typically think of compression wave, we think of sound waves, where the air (the medium) has a pressure differential between the peak and trough of the wave.

In Electromagnetism, the wave is not a change in the medium^, it is a change in the electromagnetic field.* Because of this, we have to ask, what "compresses" in the compression wave? One possible answer is that the "EM Field" gets more dense, or more strongly positive, at which point we are back where we started: the analogy gets us nowhere, it is neither wrong, nor more insightful. We also find that it starts to break down (what about "strongly negative" E-Field, this doesn't really work in a pressure analogy).

So, the E-Field doesn't have compression waves because it doesn't modify the medium in which it is traveling.

^In this case, medium is understood to be the vacuum, or space-time, not the macroscopic medium (or dielectric). In a dielectric, it is kind of possible for EM waves to PRODUCE compression waves (waves of varying density of the medium), but they cannot fundamentally BE compression waves.

*For a long time, this wasn't well understood, which is why (pre-Einstein), the dominant belief in physics was in a "luminiferous aether" as the medium in which EM waves traveled. Michelson and Morley actually "disproved" this in 1887 with their seminal experiment (though I believe Michelson spent the rest of his career trying to improve upon his initial measurement and find the aether). Combined with their null result, and Einstein's Theory of Special Relativity which came out 34 years later, the idea of a "medium" in which EM waves propagate is largely considered false.

share|cite|improve this answer
You're right, "compression" was the incorrect word to use -- what I meant is a longitudinal wave. But since I am aware that the whole analogy wasn't literal, I didn't think too hard about using the scientifically correct words, but rather about posing the problem in a way I could visualize it more easily. In this case, I assumed that moving the charged particle up would mean that a point directly above it would be under a stronger influence of its field, since it's closer, and that this increase would not happen instantaneously, but propagate at the speed of light -- hence a longitudinal wave. – waldyrious Jul 6 '11 at 21:48
Something to keep in mind is that longitudinal waves are actually possible in EM, however they require a medium in which to propagate (you can't do this in a vacuum). Typically they show up in waveguides in a dielectric. – Andrew Spott Jul 6 '11 at 23:09

It would be better to answer this question in another way, rather than by trying to think about moving field lines, which themselves are a pure analogy to explain field strength. If one has a charged particle lets say an electron, which itself has mass and therefore will not move at light speed. If this electron is impressed with an oscillatory energy such that it is forced to move in a cyclic manner the field around it will also oscillate with the electron and as such also produce a quadrature phase component which is the magnetic field. This proves by observation that a moving electron charge has some form of relationship with free space. Also remember that free space passes through every molecular system and thus these characteristics are also available in all material bodies. It has been formulated by Maxwell and others that free space has two characteristics which can be measured by extrapolation. These are called Permittivity and Permeability. Permittivity is the characteristic of space which supports electric fields and permeability is the characteristic which supports magnetic fields. Both these characteristics take a finite time to set up and collapse. If one produces an instant velocity change to an electric field the associated magnetic field will take a short time to be set up in free space. If one gives enough energy to the electron by increasing its frequency there will come a time where the electric field which is produced by the moving electron cannot collapse fast enough in free space before the next oscillatory part of the field comes behind it. This then forces away the previous cycle with its quadrature magnetic component. This is the point where one has an EM oscillatory field travelling away from the source. Also each oscillatory wavelet of electric and magnetic fields can be considered a single photon which is a self contained energy source without mass. This then has to move away at light speed because this is the speed at which the changing electric field takes to set up the quadrature magnetic field. This is a free space characteristic and is why photons or oscillatory EM wavelets have to travel only at that speed. This was found well before Einstein used the concept and is the manner in which electromagnetic communication operates. Stan

share|cite|improve this answer
This answer would be more readable if it was structured into paragraphs. – Claudius Nov 15 '12 at 20:13

Imagine you standing some distance from me, and you move a charge back and forth along the line joining us. Waldir, you are quite right that the electric field I observe will fluctuate, and that these fluctuations will not reach me instantly - they will travel at the speed of light. However, this is not electromagnetic radiation. Why?-

The electric field falls off with the square of the distance ($1/R^2$), and in fact the the variation in the field due to a given movement of the charge falls off with the cube of the distance ($1/R^3$). On the other hand, the electric and magnetic fields in electromagnetic radiation emitted from an oscillating charge fall off only with the first power of the distance ($1/R$). This may seem surprising, but remember that the power carried by the wave is proportional to the product of the electric and magnetic fields - so it follows the standard inverse square law.

The fluctuations in the electric field described in the first paragraph are known as near field effects, and are negligible at great distances The important feature of electromagnetic waves is that they can be detected far from the charges that originally created them. They derive this character from the fact that as previous answers have explained, they are self-sustaining - the changing magnetic field creates the electric field and vice versa - unlike the near field effects that are directly caused by the moving charges.

share|cite|improve this answer
Very interesting! I can see myself getting an intuitive view of what's happening with this perspective. Would it be acceptable to relate it to @jartza's answer? For example: assuming an ideal space geometry, and that we have a laser pointer pointer aimed horizontally at a surface perpendicular to the beam, moving the pointer up and down would move the spot at the surface an equal amount (with some delay due to the speed of light), but the size of the spot would not change significantly (also with a delay) if the pointer is sufficiently far from the surface. Is this an acceptable analogy? – waldyrious Sep 4 '14 at 2:04

Let us consider a horizontally pointing laser pointer that is being wiggled up and down. At some distance away the laser beam makes a bright spot on a wall. The spot moves up and down, say 10 meters, when the pointer device is moving up and down two millimeters. This is caused by relativistic aberration:

Okay, now we wiggle the laser pointer horizontally and in the beam's direction. Now the bright spot on the wall does not move at all, just the size of the spot changes a tiny amount.

Now Quantum Electro Dynamics tells us that electric field consists of stream of virtual photons, but I'm not saying that aberration of virtual photons is an explanation of anything at all.

I'm just saying that transverse wiggle and longitudinal wiggle are very different, because of relativistic effects, that are different in transverse and longitudinal directions.

Oh yes I must add that a sensitive instrument can detect a movement of a charge from any direction, and the information travels at the speed of light.

share|cite|improve this answer

Well, is not 100% percent accurate to say that there aren't longitudinal EM waves. In a waveguide there are allowed propagation modes that have non-zero electric and magnetic components in the direction of propagation:

share|cite|improve this answer

Imagine an electron being a sphere with an infinite number of lines extending radially from it's surface. These lines extend an infinite length. As you move the sphere towards an observer electron, the number of lines interfering with the observer will not change. They miss the observer electron. There will be one possible line of interference but your movement is along that one line, resulting in no exchange of information. Moving the same electron perpendicular to the observer electron and the number of interfering lines becomes a function of distance between the two electrons and frequency of oscillation.

share|cite|improve this answer

protected by Qmechanic Feb 14 '15 at 22:55

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

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