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For atmospheric refraction of visible light, Wikipedia gives the order of 1 arc minute at 45° altitude above the horizon, and 5.3 arc minutes at 10°. This is caused by the dielectric polarizability of all of the bound electrons in all the atoms of the atmosphere.

At the much lower HF frequencies of radio, the free electrons and ions will contribute, and some forms of radio communication have relied on refraction at large incident angles to deflect terrestrial signals back to the Earth at a distant ground station.

So I expect that at the lower frequencies used in radio astronomy, corrections to the observed location of radio sources due to ionospheric refraction could be much larger than those at visible wavelengths, but I am not sure.

How large does can this effect ever get? At what frequency? Are there ever corrections as large as 1 degree?


I started thinking about this after asking How many stations could one hear with an AM/FM radio in front of the ISS' cupola window? which includes the image below.

below: from the Radio Jove Project's exercise The Effects of Earth's Upper Atmosphere on Radio Signals.

enter image description here

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  • $\begingroup$ If you track down tables of n (index of refraction) at radio wavelengths, your problem is solved. Your diagrams appear to show purely reflective effects. $\endgroup$ – Carl Witthoft Aug 21 '17 at 13:11
  • $\begingroup$ @CarlWitthoft The diagram is obviously a cartoon, not to scale, and not to be taken too literally. However the process it illustrates is indeed refraction, and not reflection. Refraction by the plasma of the ionosphere is a complex problem and not something you can just look up in tables. In fact the index is literally a complex number n+jk. $\endgroup$ – uhoh Aug 21 '17 at 15:30
  • $\begingroup$ Nothing wrong with complex n = nx + i*ny $\endgroup$ – Carl Witthoft Aug 22 '17 at 12:29
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    $\begingroup$ Do a search on "LOFAR position refraction", I think you will find useful material. $\endgroup$ – Rob Jeffries Sep 5 '17 at 6:33
  • $\begingroup$ @RobJeffries omg what a nightmare, no wonder there is (at least some) interest in putting a low frequency array in space. On slide 29 here I see > 1 degree for 20-30 MHz. At some point, do you think you might augment your current answer with a mention of the the more exotic LF regime and ionosphere? There really is no maximum or answer for "how large" but the transition from arcseconds in multi-GHz to degrees at tens of MHz is pretty amazing. $\endgroup$ – uhoh Sep 5 '17 at 6:49
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For plane parallel refraction an approximation for the deviation you are talking about is $$\Delta \theta \simeq (n-1) \cot \theta,$$ where $\theta$ is the observed elevation, $\Delta \theta$ is the change in elevation from its true value due to refraction and $n$ is the refractive index averaged over airmass.

According to this source from the Green bank radio telescope, they use something like this, with an added model for how $n$ varies with height, scaled by the atmospheric pressure. The largest value of $n$ quoted is 1.00031 at ground level. This is basically the same as the refractive index of air at visible wavelengths.

So, to my surprise, the effects of refraction on radio telescope pointing are similar to those for optical telescopes. It simply turns out that the real part of the refractive index (that controls the phase velocity of light and hence refraction) is just as close to 1 for radiowaves as it is for visible light.

Here is another source that gives some algorithms to calculate the effective (small) real refractive index for radio waves.

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  • $\begingroup$ For a slab (goes in and then goes out again, which is the case for the ionosphere), the deviation has to be zero. Is that equation for one interface only, like from space to air for visible wavelengths? The ionospheric layers are more like shells, with no charged particles both above and below, so I don't think that equation applies - if it is for planar surfaces. $\endgroup$ – uhoh Aug 30 '17 at 9:27
  • $\begingroup$ @uhoh The equation is the approximation for the deviation of a radio source from its expected position due to the atmosphere above the observer. $\endgroup$ – Rob Jeffries Aug 30 '17 at 10:08
  • $\begingroup$ I see, this is for any EM wave and a simple dielectric, not the interaction with the plasma in the ionosphere. OK I'll take a look, thanks! $\endgroup$ – uhoh Aug 30 '17 at 11:42
  • $\begingroup$ @uhoh The radio waves have to pass through everything above the Earth's surface. Only the real part of the refractive index changes the phase velocity and causes refraction. This appears to be very small. $\endgroup$ – Rob Jeffries Aug 30 '17 at 12:12
  • $\begingroup$ The refraction can be so strong that lower frequencies transmitted from Earth are bent back to the surface; "skip". I am pretty sure that at those frequencies, incoming waves from space would be reflected back into space - there are no one-way mirrors. I think your "very small" only applies to higher frequencies, and the lowest frequencies really do experience substantial effects. I may have to read up and answer my own question. I wish there was a way to "escalate to an expert" in this case a low-frequency observational radio astronomer. $\endgroup$ – uhoh Sep 1 '17 at 11:45
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I found some interesting information in this vulgarization paper by Ian Poole.

A first point is electron density in ionosphere changes between day and night, so the resulting bend will be different: enter image description here

This very interesting site explains notably that there is a

cut-off frequency for the ionosphere beyond which it loses its capacity to reflect shortwaves. Depending the latitude, the season and the solar activity mainly, during the day this frequency is around 3-10 MHz and goes down to about 2-6 MHz during the night

The article includes an illustration of the transition from angular deflection to complete reflection as a function of angle (click for full size):

enter image description here

caption: Space wave, ground wave, and ionospheric waves. Above a critical angle, waves escape in free space while waves emitted under a low incidence angle can reach very far countries. This is valid between approx. 1-500 MHz.

But the best site I found on the subject is that one. It states that

Ionospheric reflection (not absorption) prevents photons with wavelengths > 30 m (f< 10 MHz) to reach the ground [...]

Total internal reflection in the ionosphere at longer wavelengths makes the Earth look like a silvery ball from space, like the glass face of an underwater wristwatch viewed obliquely.

It goes on saying the atmosphere is not perfectly transparent at any radio frequency. And moreover it adds noise. It explains why the best sites for radio observation at higher frequencies are exceptionally high and dry.

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  • $\begingroup$ Those are some great illustrations! Figure 5 is noteworthy, because it shows that for transmitted rays, they return approximately to their original direction once they leave the ionized regions. This is different than for refraction by air, because once the ray enters air, it says in air. But the ionized layers are shells. There will be an effect because the shells are curved and not slabs, but now I realize it will be a much smaller effect than I thought. OK I'll take a closer look at your links. Thanks! $\endgroup$ – uhoh Aug 29 '17 at 14:36
  • $\begingroup$ Refraction by a slab with parallel faces does not change the direction of a ray: i.stack.imgur.com/k0QK4.png and also i.stack.imgur.com/ZMrgu.png $\endgroup$ – uhoh Aug 29 '17 at 14:37
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    $\begingroup$ This is a helpful answer. HF waves experience much stronger effects than VHF or UHF. There is plenty of short-wave radio and amateur radio experience here, but most radio astronomy is done at much higher frequency where the ionospheric deflection is much smaller. This is due in part to the science available at those wavelengths, and in part due to the better angular resolution a given instrument has as the wavelength decreases. I added an image from one of your links, it's only an illustration but it helps illustrate that the effects are larger at lower frequency. $\endgroup$ – uhoh Sep 3 '17 at 3:54
  • $\begingroup$ I won't accept an answer yet, still holding off for an answer that has addresses the angular deflection quantitatively. But here's the bounty. I may add another later if necessary. $\endgroup$ – uhoh Sep 3 '17 at 3:57

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