Does EM Radiation Dissipate at a 1/r^5 Ratio for Circularly Polarized Waves?

[counterphit]

inverse square vs cube = ?

If the electric field falls off at the inverse square ratio, and magnetic at inverse cube, does EM radiation dissipate at 1/r^5 ratio for circularly polarized waves?

 

[Drakkith]

No, the inverse cube applies to the magnetic strength of a magnet, not the magnetic field of an EM wave. Remember that an EM wave is a disturbance in the EM field that carries energy, it is not the force from a magnet. The magnetic field of the EM wave really isn't a "field" like you would describe around a magnet, but an oscillating field vector. What I'm getting at is that if you let out a quick "pulse" of light and followed right behind the wavefront you will not feel a force from either the electric or magnetic fields of the EM wave because they are not sources of charge like a particle or magnet is, but a change in the EM field. I hope that makes sense.

 

[counterphit]

Thanks so much for the reply! Let me be more specific. Say I have a sphere, free space capacitance. Radius is .75. Around the equator you have an inductance (disk, close approximation for modeling cube ratio at close distance) periphery 1.5.

 

[counterphit]

Got cut off... Radius 1.5, say the say the wave length is such that the quarter (90 degrees) periphery falls at radius 6. At r=3 both electric at in inverse sqare and magnetic at inverse cube, are equal both in terms of intensity (0.707) in terms of phase relation and square vs cube relationship. I why is it not a 1/r^5 at this point. Near field.

 

[Menaus]

but a change in the EM field.

What EM field? The EM field emitted by the antenna?

Oh, and light does induce some sort of pressure when it hits objects.

http://en.wikipedia.org/wiki/Radiation_pressure

 

[sophiecentaur]

@counterphit
The inverse square law doesn't apply in the near field of a radiating dipole (so what you say is reasonable). In the near field, the magnetic and electric fields have different ratios but settle down to a constant ratio at a distance. Not only do the amplitudes change but also the phases. They are more or less in quadrature right next to the radiator but, once the fields are purely radiative, they are in phase and the ISL kicks in (for the Power flux density).

 

[Drakkith]

What EM field? The EM field emitted by the antenna?

The EM field in space. The change is caused by accelerating charges in an antenna. (Among other ways)

Oh, and light does induce some sort of pressure when it hits objects.

http://en.wikipedia.org/wiki/Radiation_pressure

Of course it does, it carries energy and momentum with it.

 

[sophiecentaur]

If the electric field falls off at the inverse square ratio, and magnetic at inverse cube, does EM radiation dissipate at 1/r^5 ratio for circularly polarized waves?

I just spotted the bit about Circular Polarisation. What do you think would be so special about CP, compared with plane polarisation?

 

[counterphit]

I knew there was a misunderstanding on this topic because I was on my phone and could not present my case well, let me gather some things and Ill post again a complete answer to what I found for my question.

 

[counterphit]

In this formula, GRX and GTX are the receive and transmit antenna gains (respectively), d is the distance between the antennas, λ is wavelength, and k = 2π/λ is the wave number. The reason for writing Friis’s law in a non-standard way (using wave number) will become clear momentarily. The upshot of Friis’s Law is that the far-field power rolls off as the inverse square of the distance (1/d^2). Near-field links do not obey this relationship. Near field power rolls off as powers higher than inverse square, typically inverse fourth (1/d^4) or higher.

This is one example of what I found. Reprinted From: 2005 Antennas and Propagation Society International Symposium, Washington, DC, USA; Volume 3A, 3-8 July 2005, Pages: 237 - 240.