Airglow

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The airglow is the very weak emission of light by the earth's atmosphere, which means that the night sky is never completely dark. It was first noticed in 1868 by Anders Ångström. It is caused by various processes in the upper atmosphere, such as the recombination of ions which were photoionised by the sun during the day, luminescence caused by cosmic rays striking the upper atmosphere, and chemiluminescence caused mainly by oxygen and nitrogen reacting with hydroxyl ions at heights of a few hundred kilometres. It is not noticeable during the daytime because of the scattered light from the Sun.

Even at the best ground-based observatories, airglow limits the sensitivity of telescopes at visible wavelengths. Partly for this reason, space-based telescopes such as the Hubble Space Telescope can observe much fainter objects than current ground-based telescopes at visible wavelengths.

The airglow at night may be bright enough to be noticed by an observer. Although airglow emission is fairly uniform across the atmosphere, to an observer on the ground it appears brightest at about 10 degrees above the horizon, because the lower one looks the greater the depth of atmosphere one is looking through. Very low down, however, atmospheric extinction reduces the apparent brightness of the airglow.

The sky brightness is typically quoted in units of astronomical magnitudes per square arcsecond of sky.

1 How to calculate the effects of airglow
2 See also
3 References
4 External links

[edit] How to calculate the effects of airglow

We need first to convert apparent magnitudes into fluxes of photons; this clearly depends on the spectrum of the source, but we will ignore that initially. At visible wavelengths we need the parameter S₀(V), the power per square centimetre of aperture and per micron of wavelength produced by a zeroth-magnitude star, to convert apparent magnitudes into fluxes -- Reference 1 tells us this is $S_{0}(V)=4.0\times 10^{-12}$ W cm^-2 µm^-1. If we take the example of a V=28 star observed through a normal V band filter ( $B = 0.2$ µm bandpass, frequency $\nu \sim 6\times10^{14}$ Hz), the number of photons we receive per square meter of telescope aperture per second from the source is $N s$ :

$N_{s}=10^{-28/2.5}\times\frac{S_{0}(V) \times B}{h\nu}$

(where $h$ is Planck's constant; $h ν$ is the energy of a single photon of frequency $ν$ ).

At V band, the emission from airglow is V = 22 per square arcsecond at a high-altitude observatory on a moonless night; in excellent seeing conditions, the image of a star will be about 0.7 arc-seconds across with an area of 0.4 square arc-seconds, and so the emission from airglow over the area of the image corresponds to about V = 23. This gives the number of photons from airglow, $N a$ :

$N_{a}=10^{-23/2.5}\times\frac{S_{0}(V) \times B}{h\nu}$

The signal-to-noise for an ideal groundbased observation with a telescope of area $A$ (ignoring losses and detector noise), arising from Poisson statistics, is just:

$S/N = \sqrt{A}\times\frac{N_{s}}{\sqrt{N_{s}+N_{a}}}$

Now we can do a quick calculation for a 10 m diameter ideal ground-based telescope and an unresolved star: every second, over a patch the size of the seeing-enlarged image of the star, 35 photons arrive from the star and 3500 from air-glow. So, over an hour, roughly $1.3\times 10^7 \pm 3500$ photons arrive from the air-glow, and approximately $1.3 \times 10^5$ arrive from the source; so the S/N ratio is about 35.

We can compare this with "real" answers from exposure time calculators. For an 8 m VLT telescope, according the FORS exposure time calculator you need 40 hours of observing time to reach V = 28, while the 2.4m Hubble only takes 4 hours according to the ACS exposure time calculator. A hypothetical 8m Hubble would take nearer 30 minutes.

It should be clear from this calculation that reducing the size of the seeing disc can make much fainter objects detectable against the air-glow; unfortunately, adaptive optics techniques that reduce the diameter of the seeing disc of an Earth-based telescope by an order of magnitude only as yet work in the infra-red, where the sky is in any case much brighter. Space telescopes don't have to worry about seeing discs.