Effects of the earth's atmosphere

The earth's atmosphere has several different effects: it emits light, it absorbs light, it shifts the apparent direction of incoming light, and it degrades the coherence of incoming light, leading to degradation of image quality when collecting light over a large aperture.

Night sky emission

We first consider emission from the sky, which comes from several sources:

• air glow

• zodiacal light

• sunlight

• moonlight

• aurorae

• light pollution

• unresolved stars and galaxies

• thermal emission: sky, telescope and dome!

Many of these source are emission line sources, not contiunuum sources, as shown in this plot How bright are these sources?

• air glow : strong in lines

• zodiacal light : m_V $\sim$ 22.2 - 23.5 depending on ecliptic latitude (and to a lesser degree on ecliptic longitude)

• sunlight : small away from twilight

• moonlight : variable, can be very bright! $\sim$ 10 times brighter at full moon.

  lunar age
  (days)	U	B	V	R	I (mag/square arcsec)
  0	22.0	22.7	21.8	20.9	19.9
  3	21.5	22.4	21.7	20.8	19.9
  7	19.9	21.6	21.4	20.6	19.7
  10	18.5	20.7	20.7	20.3	19.5
  14	17.0	19.5	20.0	19.9	19.2

  The significance of moonlight, especially in the optical, gives rise to splitting nights into dark, grey, and bright time, where dark time means no moon above horizon, grey time something like less than 50% moon above horizon, otherwise bright time.

• light pollution : strong in distinct lines

Optical:

For broadband work, for example in the V band, m_sky $\sim$ 22 mag/arcsec at good site so we switch over to background limited around m=22 for good image quality, switch over around m=20 for poorer image quality. Consequently, image quality matters for faint objects!! Moonlight is very significant, hence faint optical imaging requires dark time.

Optical spectroscopy: sky emission generally not much of a problem except around lines so long as moon is down (or work on bright objects); low dispersion observations can be background-limited for long exposures, but at higher dispersion or shorter exposures, spectroscopy is often readout-noise limited.

IR: For broadband work in H band, m $\sim$ 13.5 mag/arcsec; in K band, m $\sim$ 12.5 mag/arcsec. So for all except bright objects, we're background limited. This leads to some fundamental differences in data acquisition and analysis between the near-IR and the optical. For infrared spectra, it's hard to estimate S/N : depends on where your feature is located. Moonlight is not very significant, hence much IR work is done in bright time.

Sky brightness from most sources varies with time and position in the sky in an irregular fashion. Consqeuently, it's essentially impossible to estimate the sky a priori: sky must be determined from your observations, and if your observations don't distinguish object from sky, you'd better measure sky close by in location and in time: especially critical in the IR.

Transmission of atmosphere

Earth's atmosphere doesn't transmit 100% of light. Various things contribute to the absorption of light:

• scattering, e.g., Rayleigh scattering off molecules

• aerosols: scattering off larger particles

• variety of molecules:
  • ozone

  • H₂O

  • O₂

  • CO₂

  • N₂O

  • CH₄

All are functions of wavelength, time to some extent, and position in sky.

Sources of extinction

In the optical part of the spectrum, extinction is a roughly smooth function of wavelength and arises from a combination of ozone, Rayleigh scattering, and aerosols, as shown in this plot. The optical extinction can vary from night to night or season to season, as shown in this plot. Because of this variation, you must determine the amount of extinction on each night separately if you want accuracy better than a few percent. Generally, the shape of the extinction curve as a function of wavelength probably varies less than the amplitude at any given wavelength. Because of this, one commonly uses mean extinction coefficients when doing spectroscopy where one often only cares about relative fluxes.

In the infrared, the extinction does not vary so smoothly with wavelength because of the effect of molecular absorption. In fact, significant absorption bands define the so-called infrared windows (JHKLM), as shown in the near IR in this plot. At longer wavelengths, the broad absoprtion band behavior continues, as shown in this plot. In this figure, transmission = f (b_$\lambda$l ) where l is path length (units of airmass):

b$\lambda$l	f
-3	1
-2	0.97
-1	0.83
0	0.5
1	0.111
2	0.000

The L band is at 3.5$\mu$ , M band at 5$\mu$ .

Airmass and zenith distance dependence

Clearly, if the light has to pass through a larger path in the Earth's atmosphere, more light will be scattered/absorbed; hence one expects the least amount of absorption directly overhead (zenith), increasing as one looks down towards the horizon.

Definition of airmass: path length that light takes through atmosphere relative to length at zenith: X $\equiv$ 1 vertically (at z = 0 ). Given the zenith distance, z , which can be computed from:

sec z = (sin$$\phi$$sin$$\delta$$ + cos$$\phi$$cos$$\delta$$cos h)^-1

where $\phi$ is the latitude, $\delta$ is the declination, and h is the hour angle ( = localsiderealtime - $\alpha$ , where $\alpha$ is the right ascension), we have

X $$\sim$$ sec z

which is exactly true in the case of a plane parallel atmosphere. Since the earth's atmosphere is not a plane, the plane parallel approximation breaks down for larger airmasses. For X > 2 , a more precise formula is needed, the following gives a higher order approximation:

X = sec z - 0.0018167(sec z - 1) - 0.002875(sec z - 1)² -0.0008083(sec z - 1)³

How much light is lost going through the atmosphere?

Consider a thin sheet of atmosphere, with incident flux F , and outcoming flux F + dF . Let the thin sheet have opacity $\kappa$ = N$\sigma$ , where N is the number density of absorbers/scatterers, and $\sigma$ is the cross-section/absorber-scatterer.

dF = - $$\kappa$$Fdx

F = F_tope^{-$\int$$\kappa$dx} $$\equiv$$ F_tope^-$\tau$

where $\tau$ is the optical depth of atmosphere, which parameterizes how much light is lost.

If the optical depth through the atmosphere is just proportial to the physical path length (true if same atmospheric structure is sampled in different directions), then

$$\tau$$(X) $$\sim$$ $$\tau$$(z = 0)X

F = F_tope^-$\tau_{0}$X

Expressing things in magnitudes, we have:

m = m₀ +1.086$$\tau_{0}^{}$$X

We can define the extinction coefficient k_$\lambda$ :

m₀ = m + k_$\lambda$X

k_$\lambda$ $$\equiv$$ -1.086$$\tau$$(z = 0)

so the amount of light lost in magnitudes can be specified by a set of extinction coefficients. Note by this definition, the extinction coefficient will be negative; others may use the opposite sign convention (e.g. defining m₀ = m - k_$\lambda$X ). Of course, use of the scaling of $\tau$ or k with airmass assumes photometric weather!!

We will talk later about some details of determining extinction coefficient, but the basic idea is that you can determine the extinction by monitoring the brightness of a star (or a set of stars of known brightness) at a range of different airmasses.

Atmospheric refraction

The direction of light as it passes through the atmosphere is also changed because of refraction since the index of refraction changes through the atmosphere. The amount of change is characterized by Snell's law:

$$\mu_{1}^{}$$sin$$\theta_{1}^{}$$ = $$\mu_{2}^{}$$sin$$\theta_{2}^{}$$

Let z₀ be the true zenith distance, z be the observed zenith distance, z_n be the observed zenith distance at layer n in the atmosphere, $\mu$ be the index of refraction at the surface, and $\mu_{n}^{}$ be the index of refraction at layer n. At the top of the atmosphere:

$${\sin z_0\over \sin z_N}$$ = $${\mu_N \over 1}$$

At each infinitessimal layer:

$${\sin z_n \over \sin z_{n-1}}$$ = $${\mu_{n-1}\over \mu_n}$$

as so on for each layer down to the lowest layer:

$${\sin z_1 \over \sin z}$$ = $${\mu\over \mu_1}$$

Multiply these to get:

sin z₀ = $$\mu$$sin z

from which we can see that the refraction depends only the index of refraction near the earth's surface.

We define astronomical refraction, r , to be the angular amount that the object is displaced by the refraction of the Earth's atmosphere:

sin(z + r) = $$\mu$$sin z

In cases where r is small (pretty much always):

sin z + r cos z = $$\mu$$sin z

r = ($$\mu$$ -1)tan z

$$\equiv$$ R tan z

where we have defined R , known as the ``constant of refraction''.

A typical value of the index of refraction is $\mu$ $\sim$ 1.00029 , which gives R = 60 arcsec (red light).

The direction of refraction is that a star apparently moves towards the zenith. Consequently in most cases, star moves in both RA and DEC:

r_$\alpha$ = r sin q

r_$\delta$ = r cos q

where q is the parallactic angle, the angle between N and the zenith:

sin q = cos$$\phi$$$${\sin h\over \sin z}$$

Note that the expression for r is only accurate for small zenith distances (z < 45 ). At larger z , can't use plane parallel approximation. Observers have empirically found:

r = A tan z + B tan³z

A = ($$\mu$$ - 1) + B

B $$\sim$$ -0.07"

but these vary with time, so for precise measurements, you'd have to determine A and B on your specific night of observations.

Of course, the index of refraction varies with wavelength, so consequently does the astronomical refraction:

$\lambda$	R
3000	63.4
4000	61.4
5000	60.6
6000	60.2
7000	59.9
10000	59.6
40000	59.3

This gives rise to the phenomenon of atmospheric dispersion, or differential refraction. Because of the variation of index of refraction with wavelenth, every object actually appears as a little spectrum with the blue end towards the zenith. The spread in object position is proportional to tan z .

Note the importance of this effect for spectroscopy, and the consequent importance of the relation between a slit orientation and the parallactic angle.

Seeing: theory and practice

References: Coulson, ARAA 23,19; Beckers, ARAA 31, 13; Schroeder 16.II.

Generally, a perfect astronomical optical system will make a perfect (diffraction-limited) image for an incoming plane wavefront of light. The Earth's atmosphere is turbulent and variations in the index of refraction cause the plane wavefront from distant objects to be distorted. This distortion introduces amplitude variations, positional shifts and also image degradation.

This causes two astronomical effects:

• scintillation, which is amplitude variations, which typically vary over scales of cm: generally very small for larger apertures, and
• seeing: positional changes and image quality changes. The effect of seeing depends on aperture size: for small apertures, one sees a diffraction pattern moving around, while for large apertures, one sees a set of diffraction patterns (speckles) moving around on scale of $\sim$ 1 arcsec. These observations imply:
  • local wavefront curvatures flat on scales of small apertures, and
  • instantaneous slopes vary by $\sim$ an arcsec.

The time variation scales are several milliseconds and up.

The effect of seeing can be derived from theories of atmospheric turbulence, worked out originally by Kolmogorov, Tatarski, Fried. Here, I'll quote some pertinent results, without derivation.

A turbulent field can be described statistically by a structure function:

D_N(x) = = < | N(r + x) - N(r)|² >

where x is separation of points, N is any variable (e.g. tempereature, index of refraction, etc), r is position.

Kolmogorov turbulence gives:

D_n(x) = C_n²x^2/3

where C_n is the refractive index structure constant. From this, one can derive the phase structure function at the telescope aperture:

D_$\phi$(x) = 6.88$${\left(x\over r_o\right)}^{{5/3}}_{}$$

where the coherence length r₀ (also known as the Fried parameter) is:

r₀ = 0.185$$\left(\vphantom{\lambda^{6/5}}\right.$$$$\lambda^{{6/5}}_{}$$$$\left.\vphantom{\lambda^{6/5}}\right)$$$$\left(\vphantom{\cos^{3/5} z}\right.$$cos^3/5z$$\left.\vphantom{\cos^{3/5} z}\right)$$$$\left[\vphantom{ \int(C_n^2 dh) }\right.$$$$\int$$(C_n²dh)$$\left.\vphantom{ \int(C_n^2 dh) }\right]^{{-3/5}}_{}$$

where z is zenith angle, $\lambda$ is wavelength. Using optics theory, one can convert D_$\phi$ into an image shape.

Physically, r₀ is (roughly) inversely proportional to the image size from seeing:

d $$\sim$$ $$\lambda$$/r₀

as compared with the image size from diffraction-limited images:

d $$\sim$$ $$\lambda$$/D.

Seeing dominates when r₀ < D ; a larger r₀ means better seeing.

Seeing is more important than diffraction at shorter wavelengths (and for larger apertures), diffraction more important at longer wavelengths (and for smaller apertures); the effects of diffraction and seeing cross over in the IR for most astronomical-sized telescopes ($\sim$ 5 microns for 4m); the crossover falls at a shorter wavelength for smaller telescope or better seeing.

The meat of r₀ is in $\int$(C_n²dh) ; as you might expect, this varies from site to site and also in time. At most sites, there seems to be three regimes of ``surface layer" (wind-surface interactions and manmade seeing), ``planetary boundary layer" ( influenced by diurnal heating), and ``free atmosphere" (10 km is tropopause: high wind shears), as seen in this plot. A typical astronomical site has r₀ $\sim$ 10 cm at 5000Å.

We also have to consider the coherence of the same turbulence pattern over the sky: coherence angle call the isoplanatic angle, and region over which the turbulence pattern is the same is called the isoplanatic patch.

$$\theta$$ $$\sim$$ 0.314r₀/H

where H is the average distance of the seeing layer:

H = sec z[$$\int$$(C_n²h^5/3dh)/$$\int$$(C_n²dh)]^3/5

For r₀ = 10 cm, H $\sim$ 5000 m , $\theta$ $\sim$ 1.3 arcsec.

In the infrared r₀ $\sim$ 70 cm, H $\sim$ 5000 m, $\theta$ $\sim$ 9 arcsec.

Note however, that the ``isoplanatic patch for image motion" (not wavefront) is $\sim$ 0.3D/H . For D = 4 m, H $\sim$ 5000 m, $\theta_{{kin}}^{}$ = 50 arcsec.

Other sources of seeing

Dome seeing.

Mirror seeing.

What does seeing cause the image to look like?

The ``quality'' of an image can be described in many different ways. The overall shape of the distribution of light from a point source is specified by the point spread function. Diffraction gives a basic limit to the quality of the PSF, but any aberrations or image motion add to structure/broadening of the PSF.

Another way of describing the quality of an image is to specify it's modulation transfer function (MTF). The MTF and PSF are a Fourier transform pair. Turbulence theory gives:

MTF($$\nu$$) = exp[- 3.44($$\lambda$$$$\nu$$/r_o)^5/3]

where $\nu$ is the spatial frequency.

Note that a gaussian goes as $\nu^{2}_{}$ , so this is close to a gaussian. The shape of seeing-limited images is roughly Gaussian in core but has more extended wings This is relevant because the seeing is often described by fitting a Gaussian to a stellar profile. A potentially better empirical fitting function is a Moffat function :

I = p₁(1 + (x - p₂)²/p₄² + (y - p₃)²/p₅²)^-p₆

Measuring the Seeing

Probably the most common way of describing the seeing is by specifying the full-width-half-maximum (FWHM) of the image, which may be estimated either by direct inspection or by fitting a function (usually a Gaussian); note the correspondence of FWHM to $\sigma$ of a gaussian: FWHM = 2.355$\sigma$ . However, beware of effects of sampling of the PSF: you're really getting the PSF integrated over pixels, not the PSF! Remember that the FWHM doesn't fully specify a PSF, and one should always consider how applicable the quantity is.

Another way of characterizing the PSF is by giving the encircled energy as a function of radius, or at some specified radius. This is often used for specifying optical tolerances.

A final way of characterizing the image quality, more commonly used in adaptive optics applications, is the Strehl ratio. The Strehl ratio is the ratio between the peak amplitude of the PSF and the peak amplitude expected in the presence of diffraction only. With normal atmospheric seeing, the Strehl ratio is very low. However, the Strehl ratio is often used when discussing the performance of adaptive optics systems (more later).