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Jun 7, 2017 at 0:20 comment added Phil Frost - W8II You can edit your answer instead of using comments.
Jun 7, 2017 at 0:09 comment added sm5bsz Comments are limited to 600 chars.... The essential thing is that waves in free space couples E and H fields over regions of space size wl/2. Maxwell equations. It is similar in all wave equations - corresponds to the uncertainty principle in quantum mechanics. Local fields that occupy regions of space much smaller than wl/2 do not radiate unless fields are VERY strong, but the radiated field always has "normal" size.
Jun 6, 2017 at 23:58 comment added sm5bsz Look at the antenna as a rx antenna. The incoming wave will generate a current in a matched load. The current creates a wave that partly cancels the incoming wave. If the feed-point is shorted, twice as much current flows, and the antenna reflects the energy from an area twice as large as the capture area. Any antenna that absorbs power generates a re-radiated field with a size proportional to wl/2 or bigger. Free fields can not exist with a smaller size. From that follows that absorbed fields have a size proportional to wl/2. (The above is no proof, maybe inspiration for thinking...)
Jun 6, 2017 at 12:38 comment added Phil Frost - W8II I know that. It's in the formula at the top of the question. Why?
Jun 6, 2017 at 3:14 comment added sm5bsz When the frequency decreases, the power you can extract from that tiny dipole becomes bigger (in proportion to wavelength squared.)
May 31, 2017 at 12:51 comment added Phil Frost - W8II So when I build that tiny dipole with unobtanium which is 100% and has similar gain to a dipole, why does the aperture increase as the frequency decreases?
May 30, 2017 at 22:11 review Late answers
May 31, 2017 at 2:23
May 30, 2017 at 21:53 history answered sm5bsz CC BY-SA 3.0