@FC-in-the-UK said in #39:
> at Thalassokrator if the wave length of the electric current was smaller than the distance between the feet of the bird, would the bird be electrocuted?
Sorry that I didn't get around to answering you sooner. I hadn't checked on Lichess this past week, I hope my restarting of this thread will not inconvenience anyone.
I actually have to say that I'm initially not quite sure (although perhaps I ought to know). I'm not an electrical engineer. From my rudimentary understanding of electrodynamics and avian physiology I can say that whenever there's a nonzero electric potential difference, i.e. a voltage, the macroscopic approximation of Ohm's law should apply.
Therefore current should flow through a conducting material connecting the two points over which the potential difference is measured in proportion to the ratio between voltage (electrical potential difference) and electrical resistance of said material.
It should be noted that even air (usually a very bad conductor) can become ionised and conduct electricity across the spark gap at very high electric potential difference (=high voltage), so connecting doesn't necessarily mean touching. Never climb the roof of an electric tram or subway and NEVER EVER move your hand close to the overhead wire in any way! You'll die before you touch it.
Now, with a bird sitting on a electric power transmission line, it's my impression that unless the bird were to come into contact with the ground (having one foot somehow touching the ground (or sufficiently close to it as to ionise the air), the other still touching the power transmission line) no electric potential difference could be measured between the bird's feet. Electric power transmission lines use three-phase electric power, a form of alternating current:
en.wikipedia.org/wiki/Alternating_current#Mathematics_of_AC_voltagesWhat I'm referring to in the above wiki article might be too simplistic to be applicable to the real world (I don't know, as I said I'm not an EE), but assuming that it is (even ignoring the three-phase part) I understand it to mean that AC voltages change only in time, not in space. Therefore I take it that an AC power line carrying an alternating current (and equally alternating although phase-shifted voltage) at a frequency of say 60 Hz will be at the same voltage (relative to the ground) everywhere along the wire at a particular moment in time. The AC doesn't seem to have a spatial "wavelength". Only a temporal one if you will. Which would be a sixtieth of a second in this case if I'm not mistaken (it's less weird to think in terms of frequency instead of wavelength here, I think).
The voltage between power line and ground changes 60 times a second, but is constant along the length of the wire at any moment in time (it's the same at every point along the wire). Therefore the frequency of the AC voltage and the distance between a bird's feet should not matter. The bird cannot be electrocuted while solely standing on a wire that's at the same voltage relative to the ground everywhere along its length. Both feet are at the same potential difference to the ground at any moment and therefore there's never any electric potential difference between the feet themselves and never any reason for current to flow through the bird.
At least that's my (possibly flawed) understanding.