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1. Imaging through the turbulence

1.1. Ideal telescope

Airy function The image of a point source (star) in an ideal telescope without atmosphere is shaped by the diffraction and is described by an Airy function:


\begin{displaymath} P_0(\vec{\alpha}) = \frac{\pi D^2}{4 \lambda^2} \left[ \fra... .../\lambda)}{ \pi D \vert\vec{\alpha}\vert/\lambda)} \right] ^2, \end{displaymath} (1)

where:

The first dark ring is at an angular distance of $1.22 \lambda/D$ from the center. This is often taken as a measure of resolution in an ideal telescope.

Image $I(\vec{\alpha})$ of an astronomical object $O(\vec{\alpha})$ can be considered as a multitude of points, each point spread into an Airy function. This is written as a convolution:


\begin{displaymath} I(\vec{\alpha}) = \int {\rm d} \vec{\beta} \;\; O(\vec{\beta}) \;\; P_0(\vec{\alpha} - \vec{\beta}) = O \odot P_0. \end{displaymath} (2)

We call this imaging equation. The image, compared to object, is smoothed, its resolution is degraded. However, for a given telescope diameter $D$ this degradation is the least possible. We say that the image in this case is diffraction-limited. The example of an astronomical image (Galactic center) with different resolution is given below.

Galactic Center

Question: How the does diffraction-limited resolution depend on the wavelength?

Question: Compute the diffraction-limited resolution of human eye.

1.2. Point Spread Function

What happens if the telescope is not ideal? An image of a point source would not be as good as Airy function, the resolution would be degraded even more. But the imaging equation will still be valid! So, the Point Spread Function (PSF) $P(\vec{\alpha})$ is all we need to know to characterize the imaging. The width of PSF is a measure of resolution.

Note 1. Implicitly, we assumed in the above equations that $P(\vec{\alpha})$ is an image of a star of unit intensity, i.e. the integral of $P(\vec{\alpha})$ over $\alpha$ is equal to 1. Thus, the imaging equation preserves the total flux of an astronomical object, only distributes it differently between the pixels.

Note 2. We assumed that PSF has the same shape over the whole field of view. This condition is called isoplanatism. It is not always true in astronomical imaging, particularly with Adaptive Optics, because PSF slowly changes in the field. In this case imaging equation can be applied to parts of the field.


1.3. Resolution criteria

The shape of PSF may be irregular; what numeric measures of resolution are used in this case?

1. FWHM = Full Width at Half Maximum of the PSF.

2. Strehl ratio $ S = P(0)/P_0(0)$, i.e. the central intensity of PSF compared to the central intensity of the Airy function. The higher is Strehl ratio, the better is resolution. Diffraction-limited image is the best, hence $S \leq 1$ always.

3. Encircled energy. By definition, the integral of PSF is equal to 1. The PSF integral over the circle of radius $\beta$ is called the encircled energy. This characteristic is important for observations of faint objects, when one wants to concentrate the photons as much as possible.

The example of the PSF with compensation of turbulence is given in the picture below.

Question: Supposing that PSF is two times narrower, by how much the Strehl ratio would change?

Question: What would be the Strehl ratio if one half of the (ideal) telescope objective is given a phase delay of $\lambda/2$?


1.4. Optical Transfer Function

Another way to look at the imaging equation is to use the Fourier Transforms (FT; we denote it by tilde). The convolution becomes a product, so


\begin{displaymath} \tilde{I}(\vec{f}) = \tilde{O}(\vec{f}) \cdot \tilde{P}(\vec{f}). \end{displaymath} (3)

Here $\vec{f}$ is the spatial frequency (if $\alpha$ is measured in radians, $f$ is measured in inverse radians, or cycles per radian).

The $\tilde{P}(\vec{f})$ is called Optical transfer function (OTF). It describes the change of the modulus and phase of the object FT in the imaging process. The modulus of OTF is called Modulation transfer function (MTF). For astronomical (incoherent) imaging, $\vert\tilde{P}(\vec{f})\vert \leq 1$. Typically, MTF decreases with increasing frequencies, hence the small (high-frequency) details in the image are weakened and eventually lost.

It is known that for any optical system $\vert\tilde{P}(\vec{f})\vert =0$ for $\vert\vec{f}\vert \ge f_c$ , where $f_c = D/\lambda$ is called cutoff frequency, and $D$ is the maximal size of aperture. It means that the information at the spatial frequencies above $f_c$ is irrevocably lost. We need larger telescopes to look at smaller objects!

The relation between PSF and OTF is Fourier transform, so if you know one you know another, they are the same thing expressed differently. From the FT properties it follows that $\tilde{P}(0) =1$ (PSF normalization), and that the Strehl ratio is proportional to the integral of OTF over frequencies.

Question: What is the minimum size of a telescope needed to resolve a fence with 10 cm bar spacing from a distance of 5 km?


1.5. Imaging through the atmosphere: long exposures

OTF plots Atmospheric turbulence may be regarded as random phase aberrations added to the telescope. These aberrations constantly change in time, so does the PSF. Here we consider the average PSF, corresponding to long exposure times. The theory leads to the expression


\begin{displaymath} \tilde{P}_{\rm LE}(\vec{f}) = \tilde{P}_0(\vec{f}) \tilde{P}_a(\vec{f}). \end{displaymath} (4)

Here $\tilde{P}_0(\vec{f})$ is the OTF of the telescope (see above), and $\tilde{P}_a(\vec{f})$ is the atmospheric transfer function. For large telescopes with good optical quality the resolution is entirely defined by the atmosphere, so we neglect the first term and $ \tilde{P}_{\rm LE} \approx \tilde{P}_a$. Of course, the atmospheric PSF $P_{\rm LE}$ is obtained by Fourier transform from $\tilde{P}_{\rm LE}$.

The atmospheric OTF is related to the statistics of atmospheric phase aberrations, the so-called phase structure function $D_{\phi}(\vec{r})$ (see the next section):


\begin{displaymath} \tilde{P}_a(\vec{f}) = \exp [-0.5 D_{\phi}(\lambda \vec{f}) ]. \end{displaymath} (5)

Note: In this formula, we pass from spatial coordinates in the plane of wave-front to spatial frequencies in the image plane multiplied by the wavelength. This relation follows from the wave optics: each Fourier component of an image is produced by the interference of light waves separated by a certain distance. This principle is used in radio and optical interferometers.

Question: Suppose that the shape of the atmospheric PSF is Gaussian. What is the corresponding shape of the structure function?

Question: Is the shape of the atmospheric PSF sensitive to the structure function in the regions where $D_{\phi} \ll 1$ and $D_{\phi} \gg 1$?


1.6. Turbulence statistics

The atmospherically distorted wave-front can be visualized as a smashed paper sheet. The wave arriving from a star is plane before entering the atmosphere. Then some parts of it go through the hotter-than-average air (less refractive index) and are advanced, other parts are retarded, and the plane wave-front is deformed. It is the purpose of Adaptive Optics to compensate these distortions. But first, we need to describe them in a statistical sense.

Air is slightly dispersive, but this is usually neglected, so that the perturbations of the optical path length $l(\vec{x})$ are considered as achromatic. However, the phase of the optical wave $\phi(\vec{x}) = \frac{2\pi}{\lambda}l(\vec{x})$ strongly depends on the wavelength $\lambda$! Speaking of perturbations, we assume that their average values are zero, $\langle \phi \rangle = 0$ (angular brackets denote statistical averaging).

Although random processes like $\phi(\vec{x})$ are typically described by correlation functions or covariances, in the atmospheric science the structure functions are preferred. Structure function is the average difference between the two values of a random process:


Distorted wave-front
\begin{displaymath} D_{\phi}(\vec{r}) = \langle [ \phi(\vec{x}+\vec{r}) - \phi(\vec{x})]^2 \rangle. \end{displaymath} (6)

Question: What is the relation between the structure function and the covariance function $B(\vec{r}) = \langle \phi(\vec{x}+\vec{r}) \phi(\vec{x}) \rangle$?

Question: How does the atmospheric $D_{\phi}(\vec{r})$ depend on the wavelength $\lambda$?

The Kolmogorov model of the turbulence distortions prescribes the specific form of the phase structure function, namely


\begin{displaymath} D_{\phi}(\vec{r}) = 6.88 \left( \frac{\vert\vec{r}\vert}{r_0} \right) ^{5/3}. \end{displaymath} (7)

This formula contains only one parameter, $r_0$, which is called atmospheric coherence radius, or Fried parameter. From the fact that the path length is achromatic, we get immediately that $r_0 \propto \lambda^{6/5}$. When specifying $r_0$, always indicate the corresponding wavelength!

The above model, although it may seem primitive, is the basis of the whole theory of imaging through the turbulence, including Adaptive Optics. Of course, at large distances (more than few meters) and small distances (less than 1 cm) the model is not good, but this turns out to be not very important.

Question: What is the r.m.s. (root-mean-square) atmospheric phase difference at the baseline of $r=r_0$ in radians and wavelengths?

Question: If $r_0 =15$ cm at 0.5 micron wavelength, what is $r_0$ at 2.2 micron wavelength?

Now we put this model into long-exposure atmospheric OTF, and get it in the form:


\begin{displaymath} \tilde{P}_a(\vec{f}) = \exp [-3.44 (\lambda \vert\vec{f}\vert/r_0 )^{5/3} ]. \end{displaymath} (8)

The atmospheric long-exposure PSF is obtained by taking the Fourier transform of this equation. Is it a Gaussian curve? The numerical calculations give a relation between the FWHM of the atmospheric PSF (called $\beta_{0.5}$, or seeing) and $r_0$:


\begin{displaymath} \beta_{0.5} = 0.98 \lambda / r_0. \end{displaymath} (9)

At the wavelength of 0.5 micron, 1 arcsecond seeing corresponds to $r_0$=10.1 cm.

The Strehl number of the atmospheric PSF is exactly the same as in an ideal telescope of diameter $r_0$ (this is the reason why a strange coefficient 6.88 appears). So, for a large telescope, $D \ll r_0$, the Strehl ratio is simply $S=(r_0/D)^2$.

Question: What is the Strehl ratio of long-exposure images at the 4 m telescope under 1 arcsecond seeing at the wavelengths of 0.5 and 2.2 microns?

The Fried radius $r_0$ is sometimes identified with the characteristic scale of atmospheric perturbations. This is not quite true: we see that Kolmogorov law does not have any characteristic scale. However, only the perturbations of the size of the order of $r_0$ are relevant for long-exposure imaging. At smaller scales, the distortions are much less than $\lambda$, at larger scales the $D_{\phi}$ becomes so large that the atmospheric OTF is zero.

Locally, the strength of turbulent fluctuations of refractive index in the air are described by a refractive index structure constant $C_n^2$ which is measured in strange units, m$^{-2/3}$. The dependence of $C_n^2$ on altitude is called the turbulence profile. Seeing depends on the integrated effect of all atmospheric layers:


\begin{displaymath} r_0^{-5/3} = 0.423 \frac{2 \pi}{\lambda} \sec z \int_0^{H_max} C_n^2(h) {\rm d}h, \end{displaymath} (10)

where $h$ is the altitude, $z$ is the zenith angle, and the integration is done from telescope up to the maximum altitude of turbulence (something like 20 km).

Turbulence profile.

This figure shows an example of the turbulence profile at Cerro Paranal, plotted in relative units (full line). The fraction of the turbulent energy up to a given altitude is overplotted in dashed line. Although in this instance a significant part of turbulence was concentrated in only two laters, still some 1/3 of the total energy is distributed continuously over all altitudes.

Question: From this relation, find how $r_0$ and seeing depend on the zenith angle $z$.


1.7. Atmospheric time constant

Atmospheric time constant Turbulence can often be modeled as fixed phase screens which are driven by the wind in front of the telescope. Knowing the spatial properties of the phase screens (structure function) and the wind velocity, we know also the temporal behavior of the perturbations. The atmospheric time constant $\tau_0$ is defined as


\begin{displaymath} \tau_0 = 0.31 \frac{r_0}{\bar{V}}, \end{displaymath} (11)

where $\bar{V}$ is the wind velocity averaged over the altitude. The parameter $\tau_0$ defines how fast an Adaptive Optics system need to be.

Question: Taking the typical value of $\bar{V}$=20 m/s, what would be the atmospheric time constant at 0.5 and 2.2 micron wavelengths under 1 arcsecond seeing?

The images of astronomical objects taken with exposure time $\tau_0$ or shorter are called short-exposure images. They correspond to fixed (frozen) atmospheric aberrations. At longer exposure times, the aberrations are averaged, and for exposures much longer than $\tau_0$ the long-exposure PSF is obtained.


1.8. Isoplanatic angle

Isoplanatic angle The long-exposure atmospheric PSF is independent of the viewing direction (isoplanatic), because the turbulence and its structure function are statistically the same everywhere in the field. But the instantaneous atmospheric phase aberrations do depend on the direction: the telescope beam as projected on the atmospheric layer at 10 km shifts by 0.5 m for an angular offset of 10 arcseconds.

The standard definition of the atmospheric isoplanatic angle $\theta_0$ is

\begin{displaymath} \theta_0 = 0.31 \frac{r_0}{\bar{h}}, \end{displaymath} (12)

where $\bar{h}$ is some characteristic average turbulence altitude. The averaging is done by weighting the $C_n^2(h)$ profile with $h^{5/3}$, as a result a relatively high $\bar{h} \approx 10$ km is obtained for typical conditions.

Question: What is the isoplanatic angle $\theta_0$ at wavelengths of 0.5 and 2.2 micron under 1 arcsecond seeing?

This phenomenon is very troublesome for Adaptive Optics, because it limits the distance between guide star and the scientific objects. It turns out that for most objects there is no suitable (bright and close-by) guide star, hence artificial laser guide stars are required. Alternatively, a 3-dimensional correction of turbulence ( Multi-Conjugate Adaptive Optics, MCAO) may be attempted to increase the corrected field.


1.9. Zernike modes

In optics, the aberrations are often represented as a sum of special polynomials, called Zernike polynomials. Atmospheric random aberrations can be considered in the same way; however, the coefficients of these aberrations (defocus, astigmatism, etc.) are now random functions changing in time.

A Zernike polynomial $Z_n^m(r,\theta)$ is defined in polar coordinates $(r,\theta)$ on a circle of unit radius $(r<1)$. They are characterized by a radial order $n$ and an azimuthal order $m$ (for a given $n$, $m$ takes the values from 0 to $n$). Frequently, instead of two indices $n$ and $m$ a continuous numeration with single index $j$ is used. For a given radial order $N$, there are a total of $(N+1)(N+2)/2$ Zernike polynomials.

The first Zernike modes are named as common aberrations, and have simple meaning (see the Table of the first 15 Zernike modes).


The power of Zernike modes comes from the fact that they are orthonormal, that is the scalar product $(Z_i, Z_j)$ is equal to 1 if $i=j$ and zero otherwise. The scalar product is defined as integral over telescope aperture:


\begin{displaymath} (Z_i, Z_j) = \pi ^{-1} \int_{\vert\vec{r}\vert<1 } {\rm d} \vec{r} Z_i(\vec{r}) Z_j(\vec{r}). \end{displaymath} (13)

Now, any phase aberration $\phi (\vec{r})$ inside the telescope pupil can be represented as an infinite sum of Zernike polynomials


\begin{displaymath} \phi(\vec{r}) = \sum_{j=1}^{\infty} a_j Z_j(\vec{r}), \end{displaymath} (14)

and the coefficients are found as scalar products:


\begin{displaymath} a_j = (\phi, z_j). \end{displaymath} (15)

Often a limited number of Zernike modes gives already a good enough representation of atmospheric aberrations. If these modes are corrected by Adaptive Optics, an almost diffraction-limited image quality is obtained.

Piston mode corresponds to a constant phase, which does not influence image. Usually the piston mode is ignored.

Question: A 4 m telescope, $f/15$, is defocussed by 1 mm. Compute the resulting aberration $a_4$ for wavelengths of 0.5 and 2.2 microns.

Question: Suppose that atmospheric aberration contains only random tips and tilts with equal amplitudes $\langle a_2 \rangle = \langle a_3 \rangle = A$. Write the corresponding phase structure function.

The orthonormality of Zernike modes gives an easy way to compute the phase variance integrated over the pupil. For one mode, it is $a_j^2$. The variance for all modes is equal to the sum of squared coefficients starting from 2 (piston excluded).


1.10. Statistics of atmospheric Zernike coefficients

Given the Kolmogorov turbulence model, we can obtain the statistical properties of the coefficients $a_j$ corresponding to the atmospheric phase aberrations. The mathematical manipulations lead to a simple formula:


\begin{displaymath} \langle a_i a_j \rangle = c_{ij} \left( \frac{D}{r_0} \right) ^{5/3}, \end{displaymath} (16)

where the coefficients $c_{ij}$ are the elements of the so-called Noll matrix. The low-order coefficients (up to radial order 3) are given below.


i \ j 2 3 4 5 6 7 8 9 10
2 0.449 0 0 0 0 0 0.0142 0 0
3 0 0.449 0 0 0 0.0142 0 0 0
4 0 0 0.0232 0 0 0 0 0 0
5 0 0 0 0.0232 0 0 0 0 0
6 0 0 0 0 0.0232 0 0 0 0
7 0 0.0142 0 0 0 0.00619 0 0 0
8 0.0142 0 0 0 0 0 0.00619 0 0
9 0 0 0 0 0 0 0 0.00619 0
10 0 0 0 0 0 0 0 0 0.00619


As you may see, the Noll matrix is almost diagonal (less so for higher orders, however). Why the coefficient $c_{11}$ is missing? For Kolmogorov turbulence it is infinite! However, the first (piston) mode is of no importance to imaging.

Question: For 4m telescope and 1 arcsecond seeing, compute the r.m.s. amplitude of tilt in radians (for 0.5 micron wavelength). Convert it to the r.m.s. amplitude of stellar image motion. Does this amplitude depend on the wavelength?

What happens if we correct the first modes with Adaptive Optics? The corresponding coefficients become zero, and the total phase variance is reduced. Denoting the pupil-averaged phase variance as $\langle \epsilon ^2 \rangle$, we write


\begin{displaymath} \langle \epsilon ^2 \rangle = \left( \frac{D}{r_0} \right) ^... ...{\infty} c_{jj} = \left( \frac{D}{r_0} \right) ^{5/3} \Delta_J \end{displaymath} (17)

when the first $J$ Zernike modes are corrected.

The full, uncorrected atmospheric phase variance (all modes except piston) corresponds to $\Delta_1 = 1.0299$. In other words, in a telescope with aperture diameter of $r_0$ the atmospheric phase variance is about 1 square radians. If tip and tilt are corrected, then $\Delta_3 = 0.134$. It means that tip and tilt contribute 87% of the total phase variance. Correcting the modes up to radial order 2 leaves us with $\Delta_6 = 0.0648$, the radial order 3 leaves uncorrected variance of $\Delta_{10} = 0.0401$. As you see, further reduction of phase variance demands the correction of larger and larger number of Zernike modes.

For a large number of corrected modes $J$ ($J>10$, as happens in real systems), the assymptotic formula of Noll is very useful:


\begin{displaymath} \Delta_J \approx 0.2944 J^{-\sqrt{3}/2} \end{displaymath} (18)

Question: Using the values of $\Delta_J$ and $c_{ij}$ given above, compute $\Delta_4$.

How many modes do we need to correct? Opticians know that when the residual phase is less than 1 radian, the image quality approaches the diffraction limit. You have by now all the tools to predict the required number of modes as a function of telescope diameter, seeing, and wavelength! It is sufficient to write $\langle \epsilon ^2 \rangle =1$ and to work back all the formulas ( try to do it!). The result is


\begin{displaymath} J \approx 0.24 \left( \frac{D}{r_0} \right) ^{1.92}. \end{displaymath} (19)

Question: How many Zernike modes must be corrected at the 4 m telescope for imaging at 0.5 and 2.2 microns under 1 arcsecond seeing?

Is it necessary to correct the turbulence using Zernike modes? Of course not, the phase aberrations can be measured and corrected with any other set of basis functions, or without any modes at all, operating directly on the wave-fronts. It turns out that Zernike modes are the second best choice (the best set of modes is called Karunen-Loeve modes). The choice depends on the number of controlled parameters (modes) needed to achieve a given degree of correction; for Zernike modes it is less than for local wave-front control.

Summary. In this chapter, the fundamentals of imaging in ideal and aberrated telescopes were reminded (PSF, OTF, diffraction limit, Strehl ratio). Then the basic atmospheric parameters relevant for Adaptive Optics (phase structure function, seeing, $r_0$, time constant and isoplanatic angle) were introduced. The de-composition of the random phase aberrations on Zernike modes was studied. Now we are able to predict the number of modes that need to be corrected.


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