Electromagnetic wave attenuation

[petermer]

Hi,
I am just curious; do EM waves attenuate in a vacuum? If yes, how does this happen? Also, how do they faint through a medium?

 

[fantispug]

EM waves do not attenuate at all in a vacuum for a simple reason - conservation of momentum and energy. EM waves carry energy and if they attenuated this energy would have to be transferred to something else, but in a vacuum there is nothing else!

I assume by faint you mean dissipate. EM waves interact with the charged particles in the medium (usually electrons and protons) exchanging momentum and energy. The majority of the energy will go to random motion of the particles in the medium; that is be dissipated as heat.

 

[petermer]

But what about http://en.wikipedia.org/wiki/Free-space_path_loss" ? Isn't the spreading out of an E/M Wave considered to be a type of attenuation?

 

[burgman]

Propagation Constant = Attenuation constant(Real) + Phase Constant(Imaginary).
For the above mention, attenuation is min if not zero for vacuum.

Power loss due to spreading is not consider an attenuation as conservation of energy still holds for a given solid angle.

 

[petermer]

So that makes the equation $$Propagation Constant = Phase Contant*i$$. Does this explain a point's energy loss as the E/M wave spreads out?

 

[burgman]

No ... that equation ensures that the wave propagates.

a sphere have a surfacce area of 4*pi*r*r . As the wave spreads r increase, thus the surface area increase. The initial point source energy is now integrated over a larger area.

 

[petermer]

Ok, got it now. But, speaking for a path, as the wave propagates, it does lose energy, doesn't it?

 

[Born2bwire]

Ok, got it now. But, speaking for a path, as the wave propagates, it does lose energy, doesn't it?

Not in a lossless medium. In a lossless medium, the wave's energy gets spread out as the wavefront expands out in space, which is the free space loss factor. Only in a lossy medium will the wave actually be attenuated as it propagates.

 

[petermer]

Yes, I can understand that. However, what happens when we measure the energy of points that lie on a line which crosses the E/M wave's source? Won't we spot a reduction of energy as we move further away the source?

 

[Bob S]

Hello petermer-
EM waves do not "lose energy" in a vacuum, but they do disperse, due to the initial beam divergence. Consider a laser beam with a divergence of 1 minute of arc trying to illuminate a spot on the moon. By the time it gets there, the laser beam is 70 miles wide. Also it is hard to communicate with spacecraft leaving the solar system, because the radio cummunication beam has diverged significantly and the power density (watts per square meter) is extermely low.
Bob S

 

[sophiecentaur]

It's interesting that losses due to absorption relate exonentially to the path distance whilst 'spreading losses' (inverse square) are less severe for long paths.

I.e. The absorptive loss in dB is proportional to the dB loss per metre but the spreading loss only decreases by 6dB every time you double the distance.

The loss through cable or even optical fibre will always, eventually, be worse than the loss through 'free space' as distances get bigger and bigger. Very lucky for space exploration.

 

[Born2bwire]

Yes, I can understand that. However, what happens when we measure the energy of points that lie on a line which crosses the E/M wave's source? Won't we spot a reduction of energy as we move further away the source?

Any finite source will eventually succumb to space loss, so depending on how you plot your line of measurements, you will measure successively decreasing energy levels.

 

[petermer]

Ok, you all covered my question, thanks. I've got another relevant question though: Do parabolic antennae (in vacuum) have zero spreading, thus zero energy loss on a path that crosses the source point?

 

[sophiecentaur]

No. You may produce a nominally parallel beam by putting the feed at the focus of the dish but it must diverge due to diffraction. In the end, the inverse square law will always kick in because the source will behave like a point when you are far enough away.

 

[petermer]

No. You may produce a nominally parallel beam by putting the feed at the focus of the dish but it must diverge due to diffraction. In the end, the inverse square law will always kick in because the source will behave like a point when you are far enough away.

But how can the inverse square law hold when talking about a line?

 

[sophiecentaur]

What I mean is that the 'parallel' beam will, in fact, end up diverging as if from a point some (possibly large) distance behind the actual source. The inverse square spreading will act as if the source were at this point. At a large enough receiving distance, the difference in actual distance and 'virtual' distance becomes negligable.

 

[Born2bwire]

But how can the inverse square law hold when talking about a line?

Any finite source will "look" like a point source from far enough away. Eventually, no matter how highly directional the original beam was, it will always be hit with space loss factor. No finite source can give you a perfectly focused beam of radiation.

 

[sophiecentaur]

It is not necessary for the radiation to be spreading out in a sphere - just a cone will do for the inverse square law to apply. After all, parts of a spherical surface don't 'know' what the others are doing.