6. Gegenschein

If you shine a light into a suspension of fine particles, the particles will scatter the light. This is easy enough to show with a little bit of flour stirred into a glass of water, or with a dilute solution of milk in water, in which case the particles are small globules of fat. You can see a beam of light passing through such a medium because of the scattering, which is known as the Tyndall effect.

We can model the interaction of light with the scattering particles using Mie scattering theory, named after German physicist Gustav Mie. This is essentially a set of solutions of Maxwell’s equations for the propagation of electromagnetic radiation (in this case, light) in the presence of the scattering objects. If you solve these equations for diffuse particles a bit bigger than the wavelength of light, you can derive the angular distribution of the scattered light. The scattering is far from uniform in all directions. Rather, it has two distinct lobes. Most of the light is scattered by very small angles, emerging close to the direction of the original incoming beam of light. As the scattering angle increases, less and less light is scattered in those directions. Until you reach a point somewhere around 90°, where the scattering is a minimum, and then the intensity of scattered light starts climbing up again as the angle continues to increase. It reaches its second maximum at 180°, where light is reflected directly back towards the source.

This bright spot of reflected light back towards the source is called backscatter. It can be seen when shining light into smoke or fog. It’s the reason why some cars have special fog lights, angled down to illuminate the road, rather than shine straight into the fog and reflect back into the driver’s eyes. Backscatter is also the reason for the bright spot you might have noticed on clouds around the shadow of a plane that you’re flying in (at the centre of the related optical phenomenon of glories).

Another place where there is a collection of smoke-sized particles is in interplanetary space. In the plane of the planets’ orbits around the sun, there is a considerable amount of left over material of sizes around 10 to 100 micrometres, constantly being replenished by asteroid collisions and outgassing from comets. This material is called the interplanetary dust cloud, or the zodiacal dust cloud, because it is densest in the ecliptic—the plane of the planets—which runs through the zodiac constellations. This dust has been sampled directly by several deep space probes: Pioneers 10 and 11, Ulysses, Galileo, Cassini, and New Horizons.

The brightest source of light in the solar system is the sun. As it shines through this interplanetary dust cloud, some of the light is scattered. Most of the scattered light is deflected only by small angles, in accordance with Mie theory. But some is backscattered, and in the backscatter direction there is a peak in brightness of the scattered light directly back towards the sun. S. S. Hong published a paper in 1985, with calculations of the scattering angles of light by the interplanetary dust cloud [1]. Here’s the pertinent plot from the paper:

Scattering intensity v. angle for interplanetary dust

Scattering intensity versus scattering angle for interplanetary dust. Figure reproduced from [1].

The different curves correspond to different choices of a power law to model the size distribution of the dust particles. In each case you can see that most of the scattering occurs at small angles, there is a minimum of scattering intensity around 90°, and the scattering increases again to a second maximum at 180°, the backscattering angle.

As an aside, this backscattering also occurs in interstellar dust, and here’s a figure from a paper by B. T. Draine showing scattering intensity versus angle for the measured dust distributions of the Small Magellanic Cloud, Large Magellanic Cloud, and Milky Way galaxy, plotted for several wavelengths of light [2]. The wavelengths are shown in Angstroms, and in these units visible light occurs between 4000 and 7000 Å (lower being ultraviolet and higher infrared). In these cases the models show minima in scattering around 130°, with the backscattering again being maximal at 180°.

Scattering intensity v. angle for interstellar dust

Scattering intensity versus scattering angle for interstellar dust. Figure reproduced from [2].

We’re not concerned with interstellar dust here, but it shows the general principle that there is a peak in scattered light directly back towards the light source, from fog, smoke, and space dust.

We are concerned with backscatter from the interplanetary dust cloud. Given that this phenomenon occurs, it implies that if we could look into space in the direction exactly opposite the direction of the sun, then we should see backscatter from the interplanetary dust.

If the Earth is spherical, then night time corresponds to the sun being behind the planet. You should see, in the night sky, the point exactly opposite the direction of the sun. You should be able to see, in that direction, the backscattered light of the sun from the interplanetary dust cloud.

Now let’s imagine the Earth is flat. The sun shines on some part of the Earth at all times, so therefore it must be above the plane of the Earth at all times. (How some parts of that plane are in the dark of night is a question for another time. Some Flat Earth models propose a sort of cosmic lampshade for the sun, which makes it more like a spotlight.) At any rate, it should never be possible to look into the sky in the exact opposite direction to the sun. So there should be no point in the night sky with a peak of backscattered sunlight.

Now that we’ve made the predictions from our models, what do we actually see? It turns out that the backscattered sunlight is a visible phenomenon, and it can be seen exactly where predicted by the spherical Earth model. It’s a faint glow in the night sky, centred on the point in the exact opposite direction to the sun. The earliest recorded description of it comes from 1730, by the French astronomer Esprit Pézenas. The German explorer Alexander von Humboldt wrote about it around 1800 on a trip to South America, and gave it the name Gegenschein (German for “counter-shine”).

Unfortunately, in our modern industrial society light pollution is so bad that it’s almost impossible to see the gegenschein anywhere near where people live. You need to go somewhere remote and far away from any settlements, where it is truly dark at night. If you do that, you can see something like this:


Gegenschein, as seen from the Very Large Telescope site, Cerro Paranal, Chile. (Photo: Creative Commons Attribution 4.0 International by the European Southern Observatory.)

The gegenschein is the glow in the sky just above the centre of the image. Heck, it’s so beautiful, here’s another one:


Gegenschein, as seen from the Very Large Telescope site, Cerro Paranal, Chile. (Photo: Creative Commons Attribution 4.0 International by the European Southern Observatory.)

This is a fisheye image, with the band of the Milky Way and the horizon wrapped around the edge of the circle. Here the gegenschein is the broad glow centred around a third of the way from the centre, at the 1 o’clock angle.

The visibility of the gegenschein shows that, in places where it is night time, the sun is actually behind the Earth. On a flat Earth, the sun can never be behind the Earth, so the gegenschein would never be visible. And so the optical effect of backscatter provides evidence that the Earth is a globe.


[1] Hong, S. S. “Henyey-Greenstein representation of the mean volume scattering phase function for zodiacal dust”. Astronomy and Astrophysics, 146, p. 67-75, 1985. http://adsabs.harvard.edu/abs/1985A%26A…146…67H

[2] Draine, B. T. “Scattering by Interstellar Dust Grains. I. Optical and Ultraviolet”. The Astrophysical Journal, 598, p. 1017-1025, 2003. https://doi.org/10.1086/379118

9 thoughts on “6. Gegenschein”

  1. Typo, “The gegesnschein is the glow in the sky just above the centre of the image.”

    Obviously the gegenschein is caused by the counter-sun, which is on the far side of the flat Earth and moves in a symmetrical manner to the our-side main sun.

  2. I think this would be more convincing if you first proved that there is always a part of the earth in sunlight at all times rather than taking it as a given. Otherwise the argument is that the gegenschein is what happens when the sun goes on the underside of the earth. You allude to this being one of your 100 proofs (“a question for another time”), so I guess my suggestion is that it covering it before gegenschein is better than after.

    1. I don’t think I’ve ever heard of flat-earther’s disputing that one though, because it’s so easy to prove, you just call up your buddy who lives a few timezones away. Or travel across timezones, and don’t reset your watch. You’ll see that the sun rises and sets at a different time than where you started, and the effect reverses when you return. So therefore there are some places where the sun is set at the same time that it is not at others. I think this is why they came up with the whole “spotlight sun” idea. Because a flat earth that the sun went around would normally experience almost universal sunset with no timezones.

      1. Not only that, but the sun would rise and set in different directions in such a flat earth model. It would be impossible for the sun to consistently rise in the east and set in the west.

        1. Strictly, you just need a cylindrical Earth to explain all of that; nothing here speaks to latitudinal curvature. :p

  3. “The visibility of the gegenschein shows that at in places where it is night time, …”

    I think the “at” is not supposed to be there, and you are missing a comma after the “that”

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