power cables - not in the signal path?

06-11-14: Cbozdog
... EM radiation decays rapidly with distance...

06-11-14: Geoffkait
RF waves! to take an example do not attenuate much over distance, traveling at the speed of light or close to it.

You're both right, sort of. I believe that the main reason RF attenuates as distance increases is that it "spreads out," and therefore while the total amount of energy that is present at a given distance does not decrease a great deal as distance increases (assuming propagation through a medium that is conductive to it, such as air or a vacuum), the amount of energy reaching a given cross-sectional area at which it may be received will decrease considerably as distance increases.

The degree to which that happens will of course vary greatly depending on the directionality of how the energy is launched, e.g., omni-directionally, or as a beam that is focused with some particular degree of sharpness.

Regards,
-- Al

Geoffkait and Almarg - agree with both, with the small exception that attenuation of radiation takes place in any medium that is not vacuum to various degrees depending on medium and frequency (with the most striking example being metals - which are very effective in absorbing said radiation for practically all frequencies relevant here). The reason is (again grossly simplifying) that metals have those free carriers ready to move when they get pushed around by EM, so the EM energy reaching them is converted to electron motion.

The EM discussion touches on another topic (which I just realized). Kinda fun actually. Why do the "magic stones" work? (not that I have ever tried one, but I think I might).

The EM radiation at these frequencies is fairly delocalized - according to Heisenberg - hehe (the wavelengths can be on the order of yards to miles, compared to the wavelength of visible light which is on the order of fractions of micron). As such - any given quantum of radiation can be anywhere within that radius (with a certain probability). However, as soon as the wavefunction of said radiation overlaps with a strong absorber - there is a probability that it will get absorbed into it - and thus disappear for all other possible absorbers (such as unsuspecting cables lying around).

With that in mind - having such "magic stones" placed around the source of EM (but not too close so that their action as secondary emitters would not have least impact) would effectively suck RF from the environment.

Ha! If true then here goes the "magic stone" business out of business (now anyone can dump their spare iron and wrenches and such around their electronics and get crystal clear audio).

(incidentally - delocalization of wavefunction can be viewed as a reason for diminished RF influence with distance - the probability that the wave/quantum will be absorbed does decrease with distance - the more localized the wf, the quicker the probability that it will affect nearby cables diminishes).

06-12-14: Cbozdog
Geoffkait and Almarg - agree with both, with the small exception that attenuation of radiation takes place in any medium that is not vacuum to various degrees depending on medium and frequency (with the most striking example being metals - which are very effective in absorbing said radiation for practically all frequencies relevant here). The reason is (again grossly simplifying) that metals have those free carriers ready to move when they get pushed around by EM, so the EM energy reaching them is converted to electron motion.

Cbozdog, agreed. As I indicated, I was "assuming propagation through a medium that is conductive to it, such as air or a vacuum," air being a bit less conductive than a vacuum, consistent with your comment.

It should perhaps also be stated that the earlier comments pertain to propagation of electromagnetic energy through free space, as opposed to through coaxial cables, waveguides, fiberoptic cables, etc., for which the factors affecting attenuation as a function of distance are of course completely different than the "spreading out" effect I referred to.

Regards,
-- Al

Well, not sure I agree with your detective work, unless by "spreads out" you mean absorption and scattering. Low level signals for cell phones work pretty well even though they are "spread out." RF is not necessarily line of sight. Cheers, Geoff