# EM Waves, how are they created?

1. May 13, 2013

### mesa

A moving charge creates a Bf so is an EM wave created (for example) by an electron oscillating in a conductor producing a changing Bf which then induces an Ef with both perpendicular to the direction of propagation? If so I am presuming the point where the Bf crosses the axis represents the point where the electron had changed direction, is this correct?

2. May 13, 2013

### ZombieFeynman

Light waves are produced by accelerating charges. When a charge accelerates, it creates a kink in the fields that emanate from it. This kink, starting from the location of the charge, is seen by an observer to move away from the charges in a direction perpendicular to the directions of the fields in the kink. Whats happening is that the fields nearby the charge are replaced by what the charge has just recently done. Farther away from the charge we observe what the charge has done a while longer ago.

3. May 14, 2013

### mesa

Propagation of the Bf due to a moving charge seems straightforward. The field strength is changing based on the velocity of the charge at a given time (depending on F.o.R.) so when it crosses the 'axis' that is the point when the charge changes direction, very good.

4. May 14, 2013

### ZombieFeynman

It is crucial, I think, to note that EM radiation is from accelerating, not just moving charges.

Im also not sure what axis you are talking about.

5. May 14, 2013

### mesa

Along x in the image below. So I am assuming the point where the Bf crosses is where the direction of velocity of our oscillating electron (at distance traveled at velocity 'c') is changing so had no velocity and therefore no Bf, is this correct?

So you are saying I should be more specific as it is the acceleration that causes the change in magnitude of the Bf and resulting induced Ef as there could be no 'wave' with constant velocity?

6. May 14, 2013

### toneboy1

I didn't really understand your last reply mesa but to weigh in:
I thought the electron emitting a photon and absorbing was more to do with concervation of energy than B being induced then E and over and over. As I mean, when you heat something up it emitts photons. You know, like the whole band gap thing, electron moves down an energy level emitts a photon.

Though I have a question which the diagram helps, so the electron jossels about and you're saying where the acceleration changes is like the origin of the diagram, well I remeber hearing that each electron only emitts one photon, well which way does it propagate? In the x direction or -x direction?

7. May 15, 2013

### mesa

I am looking at an example from the standpoint of an oscillating electron in a conductor and the Bf created by this movement which produces an EM wave and does the point where the Bf and Ef cross over the axis represent the change in direction of the electron? Maybe I am missing something here but I thought EM waves could be created by oscillating charges?

8. May 15, 2013

### toneboy1

obviously what your sayings right concidering radio and telly but I think the electron has to move down an energy level or more to create a photon; I suspect a radio or lightbulb does this in some way. If anyone has anything to complete this picture for us that'd be good.

9. May 15, 2013

### mesa

Agreed, in the meantime here is how I see it:

All three figures show the Bf created by a moving electron oscillating in the negative 'y' from the origin with an EM wave moving out along 'x' at 'c'. The magnitude (arbitrary scale of course) at each point is shown by the arrows with the point of maximum velocity being at the center of oscillation. Obviously this doesn't include the induced Ef but that is a basic relationship of cross product of the changing Bf through space so can be omitted since this is understood.

Figures 1 and 2 represent 2λ and fig(3) 1/2λ. So we know there will be a magnetic field creating an EM wave propagating through space but due to the oscillation of the electron the points of highest magnitude/velocity are not typically at their highest magnitude over the majority of any axis which is drawn (although through symmetry it can be drawn the way we typically see EM waves as in the previous image)

As such the magnitude will be stronger outside of that axis at any point except (assuming the 'new' axis is drawn bisecting the oscillation) 1,3,5,7,... will be accurate representations and since points 0,2,4,6... being the points of no velocity of the charge moving back to the oscillation at 'c'

An issue I see with these figures is the magnitude of the Bf dropping off at 1/r^2 from the oscillation.

Any thoughts?

10. May 17, 2013

### toneboy1

Not to add anything to what you've said, maybe to answer my own question of which way along the axis it travels; I should probably of kept in mind that when we talk about a photon as a wavefunction it's all with respect to probability. As in, we don't actually know which way it's traveling due to H.U.P just probabilities of what it's 'expecation value' would be.
with all these different models of light for different circumstances it's easy to get confused.

11. May 17, 2013

### mesa

Well then let us start with the simplest idea and break it down from there before getting into Heisenberg. How about an oscillating electron inducing a Bf? ;)

The Figures are a 2 dimensional representation of a plane of the Bf (at some point in time) created by our moving charge. There is no 'probability' to this behavior so makes drawing a representative wave, it's direction, and amplitude straightforward.

The accompanying figures should be an satisfactory 2D representation of that wave propagating outward from the oscillation at 'c' unless you see (or someone else for that matter) something I missed. I understand that the Ef is not drawn but as mentioned earlier since this can be shown as nothing more than a simple representation of cross product it can be omitted as this is understood.

It is interesting that an EM wave drawn along a straight axis is not a true representation for the maximum amplitude of our magnitude of the wave along most points of our axis although since the drift speed of electrons is negligible (and therefore only a small distance is traveled) compared to the propagation speed of our wave at 'c' I suppose to some extent it can be disregarded for now.

What do you think?

12. May 22, 2013

### toneboy1

I think that your drawing is wrong in the same way that showing a yr 9 student an electron revolving around a nucleus is wrong (orbital theory etc)
Extrapolating on your point that drift velocity is negligable I might submit that perhaps the electron doesn't move at all for like kHz just hovers oscilating inpercievably small distances, and instead of there being some sort of physical acceleration for your zero point, that it is to do with something like molecular spin.
As Feynman said, we don't know that photons actually travel in straight lines.

Other than that I'm out of ideas, maybe the lack of response indicates that science just doesn't know yet.

13. May 22, 2013

### mesa

Is the way I understand how a Bf ind is created from a current in a wire wrong? We could dig deeper and look at individual electrons and the such but this is just an over view of magnetic fields created by an oscillating electron in a conductor and how the magnitude of that field permeates space resulting in an EM wave. The Ef is simply a cross product of the magnetic field and so is left out for simplification. What am I missing?

14. May 22, 2013

### toneboy1

I wouldnt say that because I don't know, I can understand your frustration, what source did you read that indicated to you that B was created from the current?
What did you say B was crossed with to produce E ? I thought you meant that the photon was = |B| x |E|.

15. May 23, 2013

### mesa

Bf is induced by current, or in other words any time there is a charge in motion to your F.o.R. You can find out more in any Physics texts that covers electricity and magnetism.

As far as the Ef is concerned it is nothing more than the result of the Bf but perpendicular to it and to that of the direction of motion. I feel like I was able to answer my original question and am happy with the results shown here unless someone on the forum has a reasonable objection.

The only thing I would add is it would be better to view the EM wave from a standpoint far away from the oscillating electron as the waves amplitude decreases by 1/r^2 and therefore would be more similar to a conventional representation (like the first diagram above) when viewed at far distances from this oscillation. Anyway, it's been fun, see you around the forum.

16. May 24, 2013

### toneboy1

I would disagree with your assertion that the B created from the electricity is the same B present in the propagation of a photon, that's more conventional classical physics, photons are a different Quantum relm; but if you're happy than good.

17. May 24, 2013

### nsaspook

Explain your disagreement please.

18. May 25, 2013

### toneboy1

I'm tempted to ask what you don't understand about the distinction already made. Where to start; the B created in an electric circuit is vertual photon field due to a flow, not oscillation, it is not comprised of an E component nor is it's magnitude related to the frequency of electron oscillation. Again the diagram posted doesn't illustrate any usable concept of B (or E) because photons are described more by de broglie waves. *insert reiteration of HUP*...look up Biot–Savart.

19. May 25, 2013

### ZapperZ

Staff Emeritus
This is incorrect. In a synchrotron light source, free electrons are made to oscillate back and forth as they pass through insertion devices such as an undulator or a wiggler. This can be accurately described via straightforward classical E&M without invoking any "energy level".

As far as I know, the description of antenna and EM wave being generated by transmission sources are done using classical E&M, not via quantum mechanical "energy levels".

Zz.

20. May 25, 2013

### toneboy1

correct me if I'm mistaken but I think that's misleading as the synchrotron must create a photon and positron if it's creating EMR from free electrons. Which isn't a classical electromagnetism explanation.

my background on the matter is merely second year QM and semiconductor phys.

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