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If someone were to transmit short wave "Skywave" from the equator in the UTC+0 timezone (Greenwich Time) with vertical polarization, and the receiver is at the equator ~10000km away in the UTC-6 timezone (Galapagos Time), given enough transmission power, what would be the polarization of the received signal?

Assuming that the magnetic and gravitational field strength are consistent in all locations along the path of travel and that the reflections off of the surface of the earth do not affect the polarization. Also assuming that the transmitter and receiver are both at sea level, and that the reflections off of earths surface are all at sea level on a perfectly smooth surface. The time is also midnight half-way between the receiver and transmitter, so "Skywave" conditions are optimal.

EDIT: Knowing that the ionospheric radio reflection may or may not randomize the polarization of a reflection, lets simply assume all reflections act as if they are reflecting off of an idealized radio reflector.

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  • \$\begingroup\$ Explain what you mean by this: "Disregarding the polarization effects due to collision with matter." If you remove earth from the picture, then there are two radio stations floating in space and the concept of "vertical" and "horizontal" do not mean anything.\ \$\endgroup\$
    – mkeith
    Apr 13, 2021 at 19:57
  • \$\begingroup\$ is this a school test question? \$\endgroup\$
    – jsotola
    Apr 13, 2021 at 19:59
  • \$\begingroup\$ And when you say "sky bounce" do you mean sky wave? \$\endgroup\$
    – mkeith
    Apr 13, 2021 at 19:59
  • \$\begingroup\$ @mkeith if you remove collisions with earth from the picture, there are still concepts of vertical and horizontal. \$\endgroup\$
    – user253751
    Apr 13, 2021 at 20:05
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    \$\begingroup\$ @user8079 Draw a diagram showing the path the wave takes and the polarization, and you should see. \$\endgroup\$
    – user253751
    Apr 13, 2021 at 20:05

1 Answer 1

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There's no way to know. The polarization of the signal is rotated as it passes through the ionosphere, with the amount of rotation depending on the frequency, the path length through the ionosphere, and the electron density at every point along the way. These are subject to change, sometimes rapidly, and I believe that over typical skywave paths at HF frequencies, the total rotation is quite a bit more than 360°, so that even fractionally small shifts in the path length (caused by changes in the effective reflection height) or electron profile can alter the received polarization by 90°.

For this reason, every resource I know of treats the polarization of any signal received via skywave as effectively random.

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  • \$\begingroup\$ @user8079 not meaningfully. Things do change, but you still don't know what rotational sense you will get, how squashed the ellipse will be, or what the tilt angle will be. The stuff that jonk referred to (particularly regarding the "ordinary" and "extraordinary" waves) is relevant. \$\endgroup\$
    – hobbs
    Apr 15, 2021 at 1:52
  • \$\begingroup\$ Right, I am starting to sort of get it now, so essentially gravity has no noticeable effect on the polarization of radio, nor does really any other external field. However the properties of the ionized particles in the atmosphere cause multi-path effects and refraction, which gives essentially random polarization on the receiver. However, a horribly simplistic model states that vertical polarization should be received. So hobbs and @jonk , you are both helpful and have answered my question. \$\endgroup\$
    – user8079
    Apr 15, 2021 at 5:20
  • \$\begingroup\$ Yeah, if the antennas were installed on a sphere having the mass of earth, but without atmosphere and transparent to radio waves, then the received signal would be vertically polarized (VPol). But that scenario is so divorced from reality that I think we all kind of struggled with it. \$\endgroup\$
    – mkeith
    Apr 15, 2021 at 5:37

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