# What is the difference between far and near electromagnetic field?

1. Nov 10, 2014

### k9b4

I have read this wikipedia article, but I still don't understand: http://en.wikipedia.org/wiki/Near_and_far_field

Why can EM radiation only be transverse waves? Why can't it travel as a longitudinal wave?

If I have two charges in a vacuum, with the straight line between them called 'x direction', and I oscillate one of the charges in the x direction, the other charge will be caused to oscillate in the x direction.

Why is this situation not called EM radiation?

2. Nov 10, 2014

### Staff: Mentor

Far field terms fall off as 1/r. Near field terms fall off as 1/r^2 or 1/r^3 or faster.

The near field terms are not called radiation because the energy does not radiate off to infinity. It stays localized.

3. Nov 10, 2014

### k9b4

Then what is the difference between near and far field? Does it have to do with the magnetic field?

I learned in high school that a changing electric field creates a changing magnetic field, and a changing magnetic field creates a changing electric field. How does that work?

4. Nov 10, 2014

### Staff: Mentor

How can you ask me that? This is exactly the question I just answered.

5. Nov 10, 2014

### k9b4

Sorry, allow me to rephrase:

I understand that near field has a shorter range of effect than far field, as you stated.

Why does near field have a shorter range of effect than far field?

6. Nov 10, 2014

### nsaspook

The near field depends on the source characteristics of the antenna impedance mismatch to free space. Under one wavelength you have a boundary region of space between the usually lower impedance mainly magnetic H field of the antenna and the impedance (377 ohms) of free space that far field waves propagate in that generate standing-wave reactive fields near the antenna in that region close to the source antenna.

7. Nov 10, 2014

### Staff: Mentor

Because as r gets larger 1/r^2 gets much smaller than 1/r.

8. Nov 10, 2014

### k9b4

Yes. I know. Thanks. Why?

9. Nov 10, 2014

### Staff: Mentor

Just plot it out for yourself. At r=4 you get 1/r=1/4 and 1/r^2=1/16, and further out it gets even smaller faster. Every time r doubles 1/r^2 gets halved compared to 1/r.

You should be able to sketch these functions for yourself in a couple of minutes.

10. Nov 10, 2014

### k9b4

I don't need to plot it out. I believe you. I know that as the value of r is increased, the value of 1/r^2 is decreased faster than the value of 1/r is decreased.

I am asking for a physical explanation. An 'intuitive' explanation, if you will.

11. Nov 10, 2014

### Staff: Mentor

You want a physical explanation why one mathematical term decays faster than another? That doesn't make sense.

12. Nov 11, 2014

### Staff: Mentor

You take any arbitrary field. There is just the field. Now you want to understand your field so you expand it in term of a multiple expansion. Some terms have a 1/r term and some have a 1/r^2 term. We call the first kind far field and the next terms near field.

13. Nov 11, 2014

### k9b4

Perhaps I am misunderstanding what far and near field means.

Situation 1:
If I have two charges in a vacuum, with the straight line between them called 'x direction', and I oscillate one of the charges in x direction, the other charge will be caused to oscillate in x direction.

Situation 2:
If I have two charges in a vacuum, with the straight line between them called 'x direction', and the line perpendicular to x direction called 'y direction', and I oscillate one of the charges in y direction, the other charge will be caused to oscillate in y direction.

Why is the second situation called EM radiation and the first situation is not?

Or perhaps a better question is: Why does the second situation create EM radiation and the first situation does not?

Last edited: Nov 11, 2014
14. Nov 11, 2014

### mikeph

They are just defined that way. Oscillating the charge in situation 1 will produce near and far field components.

15. Nov 11, 2014

### Staff: Mentor

The EM field given by a point particle in arbitrary motion is given by the Lienard Wiechert potentials.

http://en.wikipedia.org/wiki/Liénar...onding_values_of_electric_and_magnetic_fields

Note that there are two terms. One term is proportional to 1/r^2 and the other is proportional to 1/r. The 1/r^2 term is called the near field and the 1/r term is called the far field. The energy in the 1/r^2 term stays near the charge so it is not said to radiate. The energy in the 1/r term escapes out to infinity so it is said to radiate.

Both of your situations produce EM radiation, it is just that the EM radiation goes in a different direction than where you placed the charge in the first situation. That is because the far field term is proportional to $(n-\beta) \times \dot{\beta}$ which is 0 in that orientation.

16. Nov 11, 2014

### k9b4

What is the difference in physical structure between the 1/r term and the 1/r^2 term which causes the 1/r term to radiate, and the 1/r^2 term to not radiate?

Does it have something to do with the magnetic field?

17. Nov 11, 2014

### my2cts

There are three zones.
a) Near field: non-propagating waves contributing very close to a source of the field. Besides a host of propagating waves may or may not also contribute.
b) Complicated mess zone. Non-propagating fields are negligible. If you understand this part you have mastered the theory. Congrats !
c) Far field: further than at least 3 wavelengths and far enough so that all source points can be considered equidistant. Huygens principle and Fraunhofer theory apply.

Longitudinal waves are inconsistent with charge conservation.

Third question.
The two charges will repel or attract each other, so either fly apart, or be retained by a potential well such as formed by a string, or perhaps rotate about each other in some Kepler orbit. You may want to specify which of these cases applies.
Anyway if one of the charges on top of this would also be accelerated, e.g. by oscillating, then there would be radiative coupling. If these are macroscopic charged objects (please specify ;-) ), then the wavelength of this radiation would probably be larger than the distance between these objects (please specify ;-) ). In this case it is not helpful to speak of radiation.

This is not a simple answer but I hope that it makes sense to you and that it shows how rich the subject of electromagnetic waves is.

Last edited: Nov 11, 2014
18. Nov 11, 2014

### my2cts

The 1/r part does not propagate because it does not solve the homogeneous wave equation, so that E#pc.

19. Nov 11, 2014

### Staff: Mentor

It is just a basic energy analysis. Energy density goes as the square of the field, so if the field goes as 1/r then the energy density goes as 1/r^2. Since the surface area goes as r^2 the energy at each radius is constant for the 1/r term only. For all higher order terms the energy decreases with r, meaning that it stays localized.

20. Nov 11, 2014

### DrZoidberg

You can also think about it in terms of particle physics.
The far field is just a synonym for real photons. While the near field consists of virtual photons.
And since photons are particles it's obvious that the far field has to fall of as 1/r because the number of particles and their energy can't just change for no reason.
Let's assume an antenna sends out a very short pulse of photons. There are x photons comprising that pulse. Those particles will now spread out in a sphere around the antenna. If the radius of that sphere doubles it's surface area will increase four fold. Which means the energy density will drop to 1/4th. And since energy density depends on the square of the field strength the field strength has to drop to 1/2.

21. Nov 12, 2014

### FrankJ777

Sorry, I'm not sure if I'm mistaken, but it seems that some people are saying that the far field falls off at 1/r and the near field falls off at 1/r2. It's my understanding that its the other way around.

For example the power density at a point away from an isotropic radiator is:
P=Pt/4πr2
where P is the power density and P2 is the power radiated by the isotropic radiator.
It's part of the Friis equation, which we use to to figure out how powerful a radio signal will be at a given distance, and in this case we are talking about the far field.

I think it's easy to understand how it's derived if you imagine that a transmitter radiates an EM signal equally out in all directions, essentially a sphere of EM energy. As the energy propagates away from transmitter the sphere increases in size, with the transmitter all ways at the centre. As the sphere increases in size, it maintains the same amount of power at its surface with the power distributed equally over its surface, but as it increases in size the power density at the surface decreases as it has to cover more area. So it seems that P=Pt/4πr2 describes the power of from the transmitter divided out over the surface area of a sphere.

22. Nov 12, 2014

### FrankJ777

Also I believe that's another difference.
The far field propagates energy away from the source pretty much like you said. The changing electric field produces a changing magnetic field, which are 90 degree angles from one another, and are proportional to the intrinsic impedance of the medium which they are propagating through.
The near field if believe is a local to the radiator.

Believe me, I've been trying to understand the near field better, like why it drops off at 1/r. It seems intuitively obvious why the far field would drop off at 1/r2 considering the the energy is distributed out over the surface of a ball that expands from a radiator or antenna. I'm having a harder time understanding what phenomenon causes the the 1/r drop off of the near field. I'm planning to jump back into my old EM textbook and seeing how its derived once i get the time.

23. Nov 12, 2014

### Staff: Mentor

The 2nd charge will not oscillate in the first situation. No radiation is emitted in the direction that you oscillate a charge. This is also the case when you have a simple dipole antenna. No signal is emitted along the antenna's axis. The intensity of the EM waves is maximum in the directions perpendicular to the direction of oscillation and falls off from there, eventually reaching zero.

There is no dividing line between the near field and far field effects. The closer you are to the oscillating charges, the more the near field comes into effect. We arbitrarily define the far field to be about 2 wavelengths away from the emitter because at this distance there is very little feedback from receiving sources on the emitter.The closer the receiver is to the source, the more the emitter "notices" the receiver and is influenced by it. We say that the near field is "coupled" to the emitter. In contrast, the far field is "de-coupled" from the emitter, since no matter what happens to the far field radiation, it has no effect on the emitter.

A good example is a transformer versus two antenna. We can consider an electrical transformer as an emitter with a receiver placed nearly in contact with it. Loads on the receiving circuit heavily influence how the emitter circuit functions. In contrast, an antenna consists of an emitter and receive placed a great distance apart. At this distance the receiver has practically no influence on the emitter. No matter what the load is, the emitter behaves nearly identically.

The amplitude of the changes in the EM field fall off at 1/r in the far field, which means that the energy falls off at 1/r2. Near to a source there are other effects that behave differently and have effects that fall off in amplitude faster.
http://en.wikipedia.org/wiki/Near_and_far_field#Classical_EM_modelling

24. Nov 12, 2014

### Staff: Mentor

You are definitely mistaken. The near field clearly must fall off faster than the far field. So there is no way it could ever be the other way around.

Yes. And power is the square of the field. So since the far field power falls off as 1/r^2 that implies that the far field itself falls off as 1/r.

Last edited: Nov 12, 2014
25. Nov 12, 2014