# Photon Displacement in EM Waves (Amplitude)

1. Jun 25, 2015

### Steve13579

I am curious on what is meant by amplitude in an EM wave. From all the searching I have done the amplitude has something to do with the density of photons in the EM wave, and the frequency with energy of the photons. The part I am confused about is the fact that the wave is oscillating. I believe it's not correct to say there is displacement of the photons in such a way that you can characterize it by a distance measurement? I think of the EM wave as this. To me that makes much more sense.

I don't want to add any more since any answer to this may change my questions or eliminate them.

2. Jun 25, 2015

### norlesh

The way I like to think of it (undoubtedly a physicist would correct me) is at every point in space there are the electric (E) and magnetic (B) fields. In the absence of EM radiation E and B together will have approximately zero energy, but when an 'EM wave' passes through a point there is a rise in either the E or B field and at the same time a fall in the other to compensate so the net effect is even though the wave energy is passing through the point overall it still has zero total energy...

If the E and B fields at the point are plotted as axis on a graph then the changing energy at that point as it translated between the two fields would be a circle, the radius would be the wave amplitude, and the rate they were changing the frequency of the wave.

Last edited: Jun 25, 2015
3. Jun 25, 2015

### Staff: Mentor

The 'oscillation' in an EM wave is an oscillation of the electric and magnetic field vectors. The field vectors tell us the direction and magnitude of the electric and magnetic forces an EM wave would exert on a charged particle. I believe the amplitude of an EM wave is the magnitude of these vectors.

In the following picture, the blue arrows represent the electric field vectors and the red represents the magnetic field vectors. The best way to think of this picture is to think of it as graphing the changing strength and direction of the forces exerted by a single point of the wavefront over time.

You are correct that photons are NOT moving up and down or oscillating. A photon should not be thought of as a little particle moving along through space. Instead, think of it as just a little packet of energy that the EM wave gives up when it interacts with something. I'm sure that's not quite right in Quantum Electrodynamics, but for our purposes I think it's accurate enough.

The pictures in your link are representations of different aspects of antennas and EM waves. The one you directly linked to represents the gain of the antenna, which can be thought of as the relative amount of energy given off from, or captured by, an antenna plotted against direction. In other words, if you look at a dipole antenna, it actually emits EM waves in all directions, even directly up and down (because it isn't a 'perfect' dipole), but the relative strength of the wave will be stronger in directions that pass through the large, red portions of the picture. The part of the wave that is emitted close to vertical will be much, much smaller.

4. Jun 25, 2015

### Steve13579

Thanks that really cleared it up for me Drakkith! One last clarification I may need if you don't mind was on what norlesh had said:
Zero total energy when an EM wave passes a point in space?

5. Jun 25, 2015

### Staff: Mentor

I don't think that last part is correct. The electric and magnetic components of an EM wave are in phase with one another. When one increases, so does the other. But I'm not yet formally educated in classical EM theory, so I'm not certain. I also don't see how the energy remains zero when I know for a fact that EM waves transport energy. Perhaps I'm misunderstanding what norlesh meant?

6. Jun 25, 2015

### sophiecentaur

Not sure where you came across this - or how you have distilled it from various inputs. It really isn't a useful way of approaching things. The term"photon displacement" makes no sense at all. The nature of photons is very specific and really can't be lumped in with classical ideas of waves and transmitting antennae etc.. If you try to talk in terms of where a photon is or what it is doing at any time then you are in the realms of fantasy.
Take a region of space with a vast number of different EM waves passing through it. It is not like a busy traffic junction with loads of vehicles (photons) rushing through it and managing to avoid colliding with each other, however friendly such a model must seem. In pretty well all respects the photon has only any reality at the point where it involves the launching of an EM wave into space and at the point where it is detected. At any other time/place it is totally random and could be anywhere.
There are occasions where a photon approach is useful - for instance, in the photoelectric effect, where the existence of discrete quanta of em explains what is measured. Try to apply the same idea in many other situations and you can get things horribly twisted up.

7. Jun 25, 2015

### Steve13579

So could you say though that the amplitude is due to the rate of photons being emitted? Not trying saying the wave is photons though in that case.

8. Jun 25, 2015

### sophiecentaur

That sounds very reasonable. Energy per second (for a given frequency) is proportional to the number of Photons [per second - Edit].

Last edited: Jun 25, 2015
9. Jun 25, 2015

### Steve13579

Thanks for clarifying sophiecentaur!