power cables - not in the signal path?

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

Geoff, no, I'm not referring to absorption and scattering. I'm referring simply to a matter of geometry. A radio antenna, a radar antenna, a flashlight, etc., will not radiate energy in a beam that remains at precisely/exactly/100.00000000% the same diameter is it travels out over an extended distance. The cross-sectional area of the beam will increase in size as the distance from the source increases. That will happen to the greatest degree, of course, if there is no beam to begin with, i.e., if the radiating antenna or other source is omnidirectional. The surface area of a sphere increases exponentially as its radius increases (or more precisely, in proportion to the square of the radius).

Therefore the fraction of the originally radiated energy which intersects a GIVEN cross-sectional area will decrease as the distance of that cross-sectional area from the source becomes greater.

That is the main reason why although a radio station's transmitter may send out many thousands of watts, the signal seen by the input circuit of a radio receiver may be only a few microvolts, corresponding as a rough order of magnitude to perhaps a trillionth of a watt.

Low level signals for cell phones work pretty well even though they are "spread out." RF is not necessarily line of sight.

True, but not relevant to my point.

Regards,
-- Al

"...not relevant..."

Hey a new advertising slogan for many MD products!!! %^)