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I'm having trouble understanding the diagrams of elctromagnetic waves.

I have no problem with any concept in classical mechanics, and I think this can be answered without any relativity (which I don't know yet). I hope for an answer which is as intuitive as possible but any answer will be good.

So in the common diagram of an electromagnetic wave:

Diagram of EM wave

What exactly do the vectors represent? The wave is propagating though an eletromagnetic field. Are the lines just the field created by a particle?

We learned in physics that a charged particle creates a field in all of space, Is this a picture of changes in the field when it moves? And if so, how does it move? does it move up and down with the amplitude of this wave?

Lastly, It's not clear to me why the E-field part is not in the direction of the motion of the wave, because usually an electric field is depicted as so:

Electric field of a point charge

i.e it is in a direction "outwards" from the source of the charge.

This is really running through my mind for some time, so thanx for reading this!

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The axes aren't labelled in the first diagram. –  joseph f. johnson Feb 4 '12 at 22:54
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4 Answers 4

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Diagram of EM wave

What exactly do the vectors represent? The wave is propagating though an eletromagnetic field. Are the lines just the field created by a particle?

Each vector represents the value (magnitude and direction) of the electric (blue) or magnetic (field) at the point where the tail of the vector lies. Keep in mind that the electric and magnetic fields each have a value at every single point in space, but of course it's impossible to actually show all those values. Only the values at a selection of points are shown, enough to give you an idea of how the field behaves.

What may be confusing you is that this is not the field created by a particle. Electric and magnetic fields can be created in many different ways. One way is to have a charged particle sitting in space, or moving around in space, but a totally separate way is to have electric and magnetic fields propagating through the vacuum. In this latter case, one could say that the electric and magnetic fields create each other. (That's maybe a bit misleading when you get into the technical details of it, but it should be good enough as a high-level description.)

We learned in physics that a charged particle creates a field in all of space, Is this a picture of changes in the field when it moves? And if so, how does it move? does it move up and down with the amplitude of this wave?

If you had an electric dipole (a positive and negative charge very close to each other), and the two charges were oscillating up and down sinusoidally so that they switched positions with each half-oscillation, and if you looked only along a ray perpendicular to the axis along which the particles were oscillating, then you'd see something like what is shown in that picture. The particles would start the EM wave going, but then it would move away from them without involving any more particles.

Lastly, It's not clear to me why the E-field part is not in the direction of the motion of the wave, because usually an electric field is depicted as so:

Electric field of a point charge

Again, that is a totally different way of creating an electric field which has almost nothing to do with the electromagnetic wave shown at the top. When an electric charge creates an electric field, then yes, the field points away from the charge, but when a changing magnetic field "creates" an electric field, the electric field points perpendicular to the magnetic field (and also perpendicular to the direction in which the magnetic field is propagating).

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Great answer. How is the original illustration only like what you would expect from a direct-axis view of charge oscillation? Certainly the period is correct, since the speed of light is constant. Perhaps the magnitude would diminish as 1/r^2, but maybe this is also ignoring the wave-particle duality? Oh dear, have I asked too much? –  AlanSE Apr 5 '12 at 5:44
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I think I was probably alluding to the diminishing amplitude with that wording. (It would actually be more like $\frac{1}{r^3}$, unless you're very close to the dipole.) If you want to be precise, the EM wave shown in the picture should be considered a plot of the fields with respect to time, at a single point in space. –  David Z Apr 5 '12 at 6:12
    
That is, assuming none of the axes are spatial. If so, then at least the 1/r problem is irrelevant, since this diagram would be only trying to describe the field at one spatial point as time varies. And the way the axes are labelled in this diagram suggests that the z and y axes are not spatial. But it lables the x-axis with k, which is the wave vector, which is spatial, so this suggests that the x-axis is not time, but is the direction of propagation of a plane wave.....which is not produced by a dipole, as far as I know.... –  joseph f. johnson May 5 '12 at 4:24
    
What may be confusing you is that this is not the field created by a particle. Electric and magnetic fields can be created in many different ways. One way is to have a charged particle sitting in space, or moving around in space, but a totally separate way is to have electric and magnetic fields propagating through the vacuum. Yes, that's confusing, at least to me... I understand that em fields propagate through a vacuum, but aren't they initially created by charged particles? I guess this could be a separate question... –  monkut Jan 27 at 3:33
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So while the diagram might help it only shows you

  • one single propagation direction
  • only a static picture of the moving EM-wave
  • has constant amplitudes in k-direction
  • is only correct in the far-field zone

When you keep all this in mind it is actually a helpful picture but might not be enough as a visualization what is really going on. I have not looked deeply for a video but this diagram from wikipedia might help a bit:

enter image description here

What the author tried to visualize here is the moving EM-wave that has no E nor B component along the axis of the dipole but radiates outwards in circular symmetry.

The E-field lines at the beginning of one wave cycle start from the top of the dipole and end at the bottom of the dipole, exactly as you would have drawn the E-field of two opposite charges along the axis. Now these charges are not fixed but oscillate along the z-direction, this current creates a magnetic field perpendicular to the current direction, the red circles in the x-y-plane.

If you want to calculate the direction of the energy flux of the wave you need the poynting vector $$\vec{S} = \vec{E}\times\vec{H}$$ In this example $\vec{S}$ points radially outwards or perpendicular to the dipole axis.

Additionally you can understand with this picture how an antenna works. If you reverse the time the waves arrive at your dipole instead of being sent out. The arriving waves induce a current only if the incoming field is not propagating along the z-axis of your dipole and the best efficiency is achieved for an wave perpendicular to the axis of the dipole.

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Two things: 1. Do I understand correctly - this is an electric dipole that is moving up and down constantly? 2. I don't understand what causes this shape of the electric field lines, can you give me some direction? it kinda looks like they are supposed to form a circular path, but I'd expect that they'd always head towards the bottom dipole –  fiftyeight Feb 1 '12 at 0:59
    
@fiftyeight: 1) No, not the dipole moves but the positive charges move down, negative up and then positive up, negative down. I am sorry that I don't have time to draw a better picture. 2) This is connected with 1) as they form a semicircle at one point in time, now the charges move and the next semicircle will have it's field lines point in the opposite direction to the first one. –  Alexander Feb 1 '12 at 9:20
    
Okay one more thing I don't understand is why this creates semi circles. Charges are moving up and down, the circular magnetic field makes since, but electric fields are usually depicted as just moving outwards from the charge. Is this the same effect as this picture in 2d? upload.wikimedia.org/wikipedia/commons/thumb/e/ed/… If so, why is this created since it needs two opposite charges –  fiftyeight Feb 1 '12 at 12:24
    
@fiftyeight: The effect is exactly same as in your linked picture. The two charges in a dipole antenna are electrons (-q) and missing electrons at the other end (+q). –  Alexander Feb 1 '12 at 13:09
    
OK, with the antenna it makes since, but if I just had one electron for example and I moved it up and down it wouldn't create such a wave since there is no positive charge at the other end? –  fiftyeight Feb 1 '12 at 17:06
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The electric and magnetic fields determine the force on a probe charge, according to the Lorentz formula:$$\vec{F}=q\vec{E}+\frac{q}{c}\vec{v}\times \vec{B}$$ The first picture you mention is about a plane propagating wave, i.e., it is sufficiently far from the source. You may consider the electric force by a mechanical analogy with a waving rope: the rope displacement is perpendicular to the propagation direction.

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But according to the Lorentz equation wouldn't there also be a component of the electric field in the direction "outwards" from the source of the electric field? Also, I still don't really understand in the first diagram you mentioned, how the particle is moving in order to create this wave –  fiftyeight Feb 1 '12 at 17:05
    
The particle moves along the vertical axes, it creates a longitudinal electric field too, but the latter decreases with distance as $1/R^2$ whereas the radiated electric field decays as $1/R$, so at large distances only the propagating field survives. –  Vladimir Kalitvianski Feb 1 '12 at 17:32
    
Why does the vertical component decrease at $1/R$? Coulomb's law says $1/R^2$, doesn't that gold for all components of the field? –  fiftyeight Feb 1 '12 at 23:06
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No, $1/R$ is only for a static or a quasi-static (near) field. The Maxwell equations include $\partial ^2 /\partial t^2$ that gives a wave equation with oscillating (propagating) solutions that have a $1/R$ dependence. –  Vladimir Kalitvianski Feb 1 '12 at 23:18
    
I think the Lorentz force is irrelevant to the OP. The picture seems to show an oscillating dipole, so the force is an external constraint, the charges are not free to move in response to the back-force they experience from the Lorentz law. –  joseph f. johnson May 4 '12 at 22:01
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Each vector represents the strength and direction of the field. Red for the electric field, blue for the other one. Only the base of the arrow counts for the physical location: that is, the field at the point in space where the base of the arrow is (which in this diagram is always some point on the $x$-axis) has direction the direction of the arrow and strength the length of the arrow.

Ummm...your diagram is not labelled with any axes. I assume it is the usual $x, y, z$. As already pointed out, this is all for one time-value only, changes in time are not being illustrated, you'd need a movie for that (or a good imagination, imagine yourself travelling in the opposite direction to the motion of the wave, at the speed of light).

It is also only good for that one line, but they could've done the same thing for all other points in space, the obstacle is merely practical: the diagram would be a mass of red and blue ink and you couldn't see anything...

B) the wave does its own propagatin, and there is no medium.

C) The wave is a separate issue from any particle. Ummm, what particle? You didn't mention any particle at first. A wave can propagate all by itself, there doesn't have to be any particle.

D) Your second diagram uses a different system. That system is usually reserved for a static field. That is the only way you can have an electric field without a magnetic field, since if the electric field changes, it produces a magnetic field. That is not the only difference. Those are lines of force, not vectors. (They are not vectors with infinite length, ja ja ja..) This system of representing a force field shows you the direction of the force field vector at any point: just pick your point you want to ask about, see which line it is on (not all the lines are drawn but you can interpolate visually in your imagination, of course, just as you had to with the other diagram anyway), and the directin the line goes in tells you the direction of the electric force field vector at that point. But not its length. The length (or strength) of the electric field is given by the density of the lines in that immediate region. Now you see what a different system this is, but you also see it has its advantages.

Okay, by now I hope you see that in the first diagram, the lines are not the field created by a particle, although they are in your second diagram. I also hope you now see that it is false to say « The wave is propagating though an electromagnetic field.» The field is given by that diagram, it propagates in empty space, or in filled-up space, makes no difference. Whether it is a wave or not is partly a matter of taste: if it looks, over time, as if it was bobbing up and down like an ocean wave, you can call it a wave for that reason. Or if, more abstractly, you notice the potential functions satisfy the wave equation, you can call it a wave for that abstract reason. The wave is not something different from the field as it changes over time. The field is the wave, and vice versa.

E) Yes, a charged particle creates a field in all of space, but not that field, not the one you drew, unless the particle is moving or did some moving (it could then stop).

Now since we cannot in fact see the motion in time of the field from your diagram, which is only one time-slice, so to speak, I cannot answer very precisely, nor can you. The relation between the motion of a charge and the propagation of a field can be rather complicated and indirect, it depends on the acceleration of the charge more than anything else. So the very direct connection you ask about is sometimes, but not usually, true.

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Beginning to answer your further questions. I am convinced there is something slightly wrong with the diagram. The two charges labelled q form what is called a dipole. A moving dipole radiates, but it produces a spherical wave, with a very complicated structure. But if you consider only the x-axis in space, then this diagram is only slightly wrong in that the amplitude of the field should decay the further out it gets.

This looks like a slight adaptation of the common diagram of a plane wave. The reality or accuracy of the notion of a plane wave was addressed in another answer to one of your other questions. It could be produced by an infinite number of oscillating charges and is a very useful approximation. It is also simpler to understand than more realistic waves and that is why it appears so often in beginning texts. From a theoretical point of view, it is of fundamental importance because every possible wave can be analysed as a linear combination of plane waves. A spherical wave is next in importance, and the analysis of a spherical wave as a linear combination of plane waves is a standard exercise.

But if the x-axis in this diagram is time and not spatial, then the diagram shows the oscillation of a wave at a point. So you cannot tell from one point whether it is a plane wave, or a spherical wave, has constant amplitude in space, or decays as it should from a dipole. So if this is the meaning of the diagram, it is completely accurate: it shows a sinusoidal wave, and the oscillating dipole produces a sinusoidal spherical wave if the charges oscillate sinusoidally. But even if this is the intention of the artist of the diagram, the label `k' is then incorrect, since that means the wave vector which indicates the direction of propagation of a plane wave (usually, it is not usually used for spherical waves although it could be). More later.

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For a plane wave, the concept of direction of propagation is easy, and usually the diagram you have is for a plane wave. But there seems to be a dipole here, and for dipole radiation there is no one « direction » in which the wave propagates, nor is the electric field vector always in the same direction. But along the axis that is here labelled with `k', perpendicular to the axis of the dipole, the E vector is always parallel to the vector connecting the two elements of the dipole, as shown in the diagram. But in other locations in space, the E vector does not have to have that direction.

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This diagram is an incorrect mixture of the idea of a plane wave and the idea of a dipole radiating a wave. If the x-axis is spatial and the dipole is oscillating sinusoidally then the diagram only needs two slight corrections: the amplitude of the wave should be decreasing roughly speaking as $1/x$ (there are also terms which decrease faster), and the $k$ should be omitted from the labels, since for dipole radiation there is basically spherical propagation, not planar propagation with one direction.

IN particular, the wave also propagates « up » in the z direction the axis along which the dipole is oscillating up and down. Its amplitude is weaker, it falls off as $1/r^2$, but the E field here is in the z direction, parallel to the dipole axis, and in the direction of propagation, which is what you were asking about. So it can happen, just not for plane waves. The wave produced by a dipole is more like a spherical wave than it is like a plane wave, and this diagram hides that fact.

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You explain really well, thank you. The only thing I still don't understand is why the electromagnetic "wave" is always illustrated in this manner if it depends so much on the motion of the particle, and can you give an example of a simple system that will create "wave" like in my first drawing and explain why the field lines go up and down like this and have no component in the direction of the motion –  fiftyeight Feb 6 '12 at 19:36
    
and idea about these questions? –  fiftyeight Feb 13 '12 at 20:25
    
Let's see it's spacial axes, E field being in the x direction, B field in y direction and z in the direction of propagation –  fiftyeight Mar 1 '12 at 18:36
    
Well, suppose the three axes are x,y,z. So time is fixed. The field is created, at first, near the moving charges: they are accelerating, which produces a magnetic field. They are charged, which produces an electric field. These field then propagates through space away from the charges in all directions, at the speed of light. In any volume free from charges, the E vector is always perpendicular to the B vector. (That is a law of Nature.) So that part seems right. But as several people have mentioned, the amplitude of E and B should be decreasing in space as 1/R, so that seems wrong. –  joseph f. johnson May 5 '12 at 3:19
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