# 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

These are my personal answers.
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.

Then your second question:
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.

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