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Next: What does seeing cause Up: Effects of the earth's Previous: Atmospheric refraction

Seeing: theory and practice

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

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:

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:

\begin{displaymath}D_N(x) == < \vert N(r+x) - N(r)\vert^2 >\end{displaymath}

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

Kolmogorov turbulence gives:

\begin{displaymath}D_n(x) = C_n^2 x^2/3\end{displaymath}

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

\begin{displaymath}D_\phi(x) = 6.88 {x\over r_o}^{5/3}\end{displaymath}

where the coherence length $r_0$ (also known as the Fried parameter) is:

\begin{displaymath}r_0 = 0.185 \lambda^{6/5} \cos^{3/5} z[ \int(C_n^2 dh) ]^{-3/5}\end{displaymath}

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

Physically, $r_0$ is (roughly) inversely proportional to the image size from seeing:

\begin{displaymath}d \sim \lambda/r_0\end{displaymath}

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

\begin{displaymath}d \sim \lambda / D.\end{displaymath}

Seeing dominates when $r_0 < D$; a larger $r_0$ means better seeing.

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

The meat of $r_0$ is in $\int(C_n^2 dh)$; 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 site probably has $r_0 \sim 10$ 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.

\begin{displaymath}\theta \sim 0.314 r_0 / H\end{displaymath}

where $H$ is the average distance of the seeing layer:

\begin{displaymath}H = \sec z [ \int(C_n^2 h^{5/3} dh) / \int (C_n^2 dh) ] ^{3/5}\end{displaymath}

For $r_0 = 10$ cm, $H \sim 5000$ m , $\theta \sim 1.3$ arcsec.

In the infrared $r_0 \sim 70$ cm, $H \sim 5000$ m, $\theta \sim 9$ arcsec.

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


next up previous
Next: What does seeing cause Up: Effects of the earth's Previous: Atmospheric refraction
Rene Walterbos 2003-04-14