Observation
Airmass & Atmospheric Extinction
Why stars dim and redden near the horizon
Airmass is the relative length of the path starlight travels through Earth's atmosphere — normalized to 1.0 straight overhead — and atmospheric extinction is the wavelength-dependent dimming and reddening that path imposes. To first order airmass equals sec(z), where z is the angle from the zenith: 1.0 overhead, about 2 at 60° from the zenith, and roughly 38 right at the horizon. Extinction follows the magnitude relation m = m₀ + k·X, where k is the extinction coefficient (~0.15 mag per airmass in the visual band at a good site). Because Rayleigh scattering scales as 1/λ⁴, blue light is stripped away far faster than red — which is why low stars, and the setting Sun, glow orange.
- Airmass at zenithX = sec(z) = 1.0
- Airmass at 60° from zenithX = 2.0
- Airmass at the horizonX ≈ 38 (curved-atmosphere)
- Visual extinction coefficientk_V ≈ 0.10–0.20 mag/airmass
- Scattering wavelength lawRayleigh ∝ 1/λ⁴
- Horizon vs. zenith dimming≈ 5.6 mag (×170) in V
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What airmass actually measures
Every photon that reaches a ground-based telescope has just finished a swim through the atmosphere. Airmass is the bookkeeping for how long that swim was. We define it as the column of air the light traversed relative to the shortest possible column — the one straight up. Look at the zenith (directly overhead) and the light crossed exactly one airmass; that is the definition, X = 1. Tip your telescope toward the horizon and the same atmosphere is now a long slanted slab, so the light crosses many airmasses.
For a flat, uniform, plane-parallel atmosphere the geometry is pure trigonometry. If z is the zenith angle (0° overhead, 90° at the horizon), the path length is the vertical thickness divided by cos(z):
X = sec(z) = 1 / cos(z)
That single formula carries the whole intuition: at the zenith cos(0°) = 1 so X = 1; at 60° from the zenith cos(60°) = 0.5 so X = 2; at 70°, X ≈ 2.92. But sec(z) diverges to infinity as z → 90°, which is physically wrong — the real atmosphere is a thin curved shell wrapped on a round planet, so the slant path stays finite. The widely used Kasten & Young (1989) fit replaces the bare secant with a curved-atmosphere correction and yields a maximum of about X ≈ 38 at the true horizon. In practice the divergence matters only below ~10° altitude; above 70° altitude the plain secant is accurate to better than a percent.
Extinction: where the photons go
Crossing more atmosphere means losing more light, and that loss is atmospheric extinction. It is the sum of several physical channels:
- Rayleigh scattering — light bounces off air molecules (mostly N₂ and O₂). The cross-section scales as 1/λ⁴, so it dominates in the blue and ultraviolet.
- Aerosol (Mie) scattering — dust, sea salt, smoke, and volcanic haze. The wavelength dependence is gentle (roughly 1/λ to 1/λ¹·³) and highly variable night to night.
- Molecular absorption — ozone removes a broad chunk of the green-yellow (the Chappuis band) and nearly all of the ultraviolet; water vapor and O₂ carve absorption bands in the red and near-infrared.
Because extinction acts multiplicatively on flux — each layer transmits a fixed fraction — it adds linearly in magnitudes, which are logarithmic. That gives astronomy's workhorse relation:
mobs = m₀ + k · X
Here mobs is the measured magnitude (fainter = larger number), m₀ is the magnitude you would measure with no atmosphere at all (the "above-the-atmosphere" value), and k is the extinction coefficient in magnitudes per airmass. The relation is a straight line in the plane of magnitude versus airmass; observe a star at several altitudes through the night, fit the line, and the slope is k while the intercept at X = 0 is the corrected magnitude. This is the heart of all-sky photometric calibration.
Reddening: why low stars turn orange
The 1/λ⁴ steepness of Rayleigh scattering means k is not one number — it depends on wavelength. Blue light is scattered out of the beam far more aggressively than red. At a good dark site the visual (V, ~550 nm) coefficient is around 0.15 mag/airmass, the blue (B, ~440 nm) is closer to 0.25–0.30, and the red (R, ~640 nm) drops to ~0.10. So as airmass grows, a star's blue light is depleted faster than its red light, and the transmitted color shifts redward — this is atmospheric reddening.
It is exactly the everyday physics of sunset. At the zenith the noon Sun crosses one airmass and looks white; at the horizon it crosses ~38 airmasses, the blue is almost entirely scattered into the surrounding sky (which is why the daytime sky is blue), and what remains is the familiar orange-red disk. The same effect makes bright low stars like Sirius "twinkle" through a rainbow of colors: turbulence steers the already-reddened, dispersed beam, and your eye samples a flickering spectrum.
Putting numbers on it
The table below shows airmass and the corresponding extinction in the V band for a representative coefficient kV = 0.15 mag/airmass. The brightness ratio is 100.4·Δm relative to the zenith.
| Altitude | Zenith angle z | Airmass X | Extra extinction (k·X, V) | Brightness vs. zenith |
|---|---|---|---|---|
| 90° (zenith) | 0° | 1.00 | 0.15 mag | 1.00× |
| 60° | 30° | 1.15 | 0.17 mag | 0.98× |
| 45° | 45° | 1.41 | 0.21 mag | 0.95× |
| 30° | 60° | 2.00 | 0.30 mag | 0.87× |
| 20° | 70° | 2.90 | 0.44 mag | 0.76× |
| 10° | 80° | 5.6 | 0.84 mag | 0.53× |
| ~0° (horizon) | ~90° | ~38 | ~5.6 mag | ~0.006× (×1/170) |
The lesson observers internalize: keep targets high. Beyond an airmass of ~2 (altitude 30°) the extinction, the differential color term, and the atmospheric turbulence all rise steeply, and the limiting magnitude of a survey collapses. This is why telescope queues prioritize objects near transit and why "airmass" is a column in every observing log.
How extinction varies with site, band, and night
The extinction coefficient is not a universal constant — it is a property of the air above a particular telescope on a particular night. The table contrasts typical values.
| Condition / band | k (mag/airmass) | Dominant cause |
|---|---|---|
| U band (~365 nm), good site | 0.45–0.60 | Rayleigh + ozone, steep in UV |
| B band (~440 nm), good site | 0.25–0.30 | Rayleigh scattering |
| V band (~550 nm), good site | 0.10–0.20 | Rayleigh + aerosols + ozone |
| R band (~640 nm), good site | 0.07–0.12 | Aerosols |
| I band (~800 nm), good site | 0.04–0.08 | Aerosols + water bands |
| High-altitude site (Mauna Kea, 4200 m) | lower across all bands | less air, less aerosol/water above |
| After a major volcanic eruption | +0.1 to +0.3 in V | stratospheric aerosol layer |
High, dry sites (Mauna Kea at 4200 m, the Atacama at 2400–5000 m) win twice: there is simply less air column overhead, and that column is poorer in the water vapor and aerosols that drive the most variable part of extinction. Stratospheric aerosols injected by eruptions such as Pinatubo (1991) raised measured V extinction at observatories worldwide for years.
The second-order color term
For precise photometry the simple m = m₀ + k·X is not quite enough, because extinction varies across a filter's bandpass. A blue star and a red star seen through the same V filter at the same airmass are not dimmed by exactly the same amount — the blue star's light is weighted toward the steeper part of the extinction curve. Astronomers add a second-order color term:
mobs = m₀ + k′·X + k″·(color)·X
where k′ is the first-order (gray) coefficient and k″ couples airmass to the star's color index (e.g. B−V). The term is small — a few hundredths of a magnitude per airmass per magnitude of color — but it is the difference between 1% and 3% photometry, and it matters enormously for surveys like LSST that chase millimagnitude calibration across the whole sky.
Common misconceptions
- Airmass is a mass. No — it is a dimensionless path-length ratio, normalized to 1 at the zenith. The name is historical.
- sec(z) works everywhere. It is excellent above ~70° altitude but diverges at the horizon; use Kasten & Young near the ground.
- Extinction dims all colors equally. It is strongly wavelength-dependent (∝1/λ⁴ for Rayleigh) — that is the whole reason for reddening.
- The extinction coefficient is fixed. It changes with site, altitude, season, humidity, dust, and volcanic activity, and must be re-measured each night for precise work.
- Reddening here equals interstellar reddening. Both redden, but atmospheric reddening is removable by observing higher; interstellar reddening from dust between stars is intrinsic to the line of sight.
- A star at the horizon is only a little fainter. At the horizon it is dimmed by ~5.6 magnitudes in V — a factor of ~170 — which is why deep imaging at low altitude is futile.
Frequently asked questions
What is airmass?
Airmass is the relative length of the path that starlight travels through the atmosphere, normalized so that looking straight up (the zenith) equals 1.0. For a plane-parallel atmosphere it is simply X = sec(z), where z is the angle from the zenith. At 60° from the zenith X = 2; at 70° X ≈ 2.9; near the horizon the simple formula blows up, but a curved-atmosphere model gives a finite X ≈ 38.
What is atmospheric extinction?
Atmospheric extinction is the dimming of starlight as it passes through the air, caused by Rayleigh scattering off molecules, scattering and absorption by aerosols (dust, smoke, haze), and absorption by ozone and water vapor. It is quantified by an extinction coefficient k in magnitudes per airmass. At a good dark site k is about 0.15 mag/airmass in V; in the blue (B) it is larger, around 0.25–0.30 mag/airmass.
Why do stars look redder near the horizon?
Rayleigh scattering removes light in proportion to 1/λ⁴, so blue photons (short λ) are scattered out of the beam far more efficiently than red ones. Near the horizon the airmass is large, so much more blue is stripped away, leaving the transmitted light reddened. The same physics turns the setting Sun orange-red and is why low stars like Sirius can flash red and green.
How do astronomers correct photometry for extinction?
They observe standard stars across a range of airmasses through the night and fit the relation m_obs = m_0 + k·X, where m_0 is the magnitude that would be measured above the atmosphere. The slope k is the extinction coefficient and the intercept m_0 is the corrected value. A second-order color term, k″·(color)·X, accounts for extinction varying across a filter's bandpass.
Why is the simple sec(z) formula wrong near the horizon?
sec(z) assumes a flat, plane-parallel atmosphere, which diverges to infinity as z → 90°. The real atmosphere is a thin curved shell on a round Earth, so the path length stays finite. Empirical fits such as Kasten & Young (1989) replace sec(z) with a curved-atmosphere expression that gives a maximum airmass near 38 at the true horizon.
How much fainter is a star at the horizon than at the zenith?
With k ≈ 0.15 mag/airmass in V, the extra extinction at the horizon (X ≈ 38) versus the zenith (X = 1) is about 0.15 × 37 ≈ 5.6 magnitudes — a factor of roughly 170 in brightness. That is why limiting magnitude collapses near the horizon and why observers avoid targets below about 30° altitude (airmass 2).