- 173 Views
- Uploaded on
- Presentation posted in: General

Adaptive Optics in the VLT and ELT era Atmospheric Turbulence

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Adaptive Optics in the VLT and ELT era Atmospheric Turbulence

François Wildi

Observatoire de Genève

Credit for most slides : Claire Max (UC Santa Cruz)

- We, in astronomy, are essentially interested in the effect of the turbulence on the images that we take from the sky. This effect is to mix air masses of different index of refraction in a random fashion
- The dominant locations for index of refraction fluctuations that affect astronomers are the atmospheric boundary layer , the tropopause and for most sites a layer in between (3-8km) where a shearing between layers occurs.
- Atmospheric turbulence (mostly) obeys Kolmogorov statistics
- Kolmogorov turbulence is derived from dimensional analysis (heat flux in = heat flux in turbulence)
- Structure functions derived from Kolmogorov turbulence are r2/3

- Refractivity of air
where P = pressure in millibars, T = temp. in K, in microns

n = index of refraction. Note VERY weak dependence on

- Temperature fluctuations index fluctuations
(pressure is constant, because velocities are highly sub-sonic. Pressure differences are rapidly smoothed out by sound wave propagation)

tropopause

10-12 km

wind flow around dome

boundary layer

~ 1 km

stratosphere

Heat sources w/in dome

When a mirror is warmer than dome air, convective equilibrium is reached.

Remedies: Cool mirror itself, or blow air over it, improve mount

credit: M. Sarazin

credit: M. Sarazin

convective cells are bad

To control mirror temperature: dome air conditioning (day), blow air on back (night)

Cartoon (M. Sarazin): wind is from left, strongest turbulence on right side of dome

Computational fluid dynamics simulation (D. de Young) reproduces features of cartoon

- Wind speed must be zero at ground, must equal vwind several hundred meters up (in the “free” atmosphere)
- Boundary layer is where the adjustment takes place, where the atmosphere feels strong influence of surface
- Quite different between day and night
- Daytime: boundary layer is thick (up to a km), dominated by convective plumes
- Night-time: boundary layer collapses to a few hundred meters, is stably stratified. Perturbed if winds are high.

- Night-time: Less total turbulence, but still the single largest contribution to “seeing”

- Colors show intensity of radar return signal.
- Radio waves are backscattered by the turbulence.

solar

Outer scale L0

Inner scalel0

h

Wind shear

convection

h

ground

- Assume energy is added to system at largest scales - “outer scale” L0
- Then energy cascades from larger to smaller scales (turbulent eddies “break down” into smaller and smaller structures).
- Size scales where this takes place: “Inertial range”.
- Finally, eddy size becomes so small that it is subject to dissipation from viscosity. “Inner scale” l0
- L0 ranges from 10’s to 100’s of meters; l0 is a few mm

Questionable

- Medium is incompressible
- External energy is input on largest scales (only), dissipated on smallest scales (only)
- Smooth cascade

- Valid only in inertial range L0
- Turbulence is
- Homogeneous
- Isotropic

- In practice, Kolmogorov model works surprisingly well!

- What do you think really determines the outer scale in the boundary layer? At the tropopause?
- Hints:

- F. Martin et al. , Astron. Astrophys. Supp. v.144, p.39, June 2000
- http://www-astro.unice.fr/GSM/Missions.html

A structure function is measure of intensity of fluctuations of a random variable f (t) over a scale t :

Df(t) = < [ f (t + t) - f ( t) ]2 >

With the assumption that temperature fluctuations are carried around passively by the velocity field (for incompressible fluids), T and N have structure functions like

- DT ( r ) = < [ T (x ) - v ( T + r ) ]2 > = CT2 r 2/3
- DN ( r ) = < [ N (x ) - N ( x + r ) ]2 > = CN2 r 2/3
CN2 is a “constant” that characterizes the strength of the variability of N. It varies with time and location. In particular, for a static location (i.e. a telescope) CN2 will vary with time and altitude

10-14

- Index of refraction structure function
DN ( r ) = < [ N (x ) - N ( x + r ) ]2 > = CN2 r 2/3

- Night-time boundary layer: CN2 ~ 10-13 - 10-15 m-2/3

Paranal, Chile, VLT

Eight minute time period (C. Dainty, Imperial College)

Starfire Optical Range, Albuquerque NM

Siding Spring, Australia

- The dominant locations for index of refraction fluctuations that affect astronomers are the atmospheric boundary layer and the tropopause
- Atmospheric turbulence (mostly) obeys Kolmogorov statistics
- Kolmogorov turbulence is derived from dimensional analysis (heat flux in = heat flux in turbulence)
- Structure functions derived from Kolmogorov turbulence are r2/3
- All else will follow from these points!

- Structure function: Mean square difference
- Covariance function: Spatial correlation of a random variable with itself

To derive this relationship, expand the product in the definition of D ( r ) and assume homogeneity to take the averages

- Spatial coherence function of field is defined as
Covariance for complex fn’s

C (r) measures how “related” the field is at one position x to its values at neighboring positions x + r .

Do not confuse the complex field with its phase f

- For a Gaussian random variable with zero mean,
- So
- So finding spatial coherence function C (r) amounts to evaluating the structure function for phase D ( r ) !

- We want to evaluate
- Remember that

- But for a wave propagating
vertically (in z direction) from height h to height h + h. This means that the phase is the product of the wave vector k (k=2p/l [radian/m]) x the Optical path

Here n(x, z) is the index of refraction.

- Hence

- Change variables:
- Then

Algebra…

- Now we can evaluate D ( r )

Algebra…

- But

For a slant path you can add factor( sec )5/3 to account for dependence on zenith angle

Concept Question: Note the scaling of the coherence function with separation, wavelength, turbulence strength. Think of a physical reason for each.

- Define optical transfer functions of telescope, atmosphere
- Definer0 as the telescope diameter where the two optical transfer functions are equal
- Calculate expression for r0

- Imaging in the presence of imperfect optics (or aberrations in atmosphere): in intensity units
Image = Object Point Spread Function

I = O PSF dx O(r - x) PSF( x )

- Take Fourier Transform: F( I ) = F(O )F( PSF )
- Optical Transfer Function is Fourier Transform of PSF:
OTF = F( PSF )

convolved with

Seeing limited OTF

Seeing limited PSF

Intensity

-1

l / D

l / r0

r0 / l

D / l

Diffraction limited PSF

Diffraction limited OTF

Intensity

-1

l / r0

l / D

D / l

r0 / l

- OTF for the whole imaging system (telescope plus atmosphere)
S ( f ) = B ( f ) T ( f )

Here B ( f ) is the optical transfer fn. of the atmosphere and T ( f) is the optical transfer fn. of the telescope (units of f are cycles per meter).

f is often normalized to cycles per diffraction-limit angle (l / D).

- Measure the resolving power of the imaging system by
R = dfS ( f ) = dfB ( f ) T ( f )

- R of a perfect telescope with a purely circular aperture of (small) diameter d is
R= df T ( f ) =( p / 4 ) ( d / l )2

(uses solution for diffraction from a circular aperture)

- Define a circular aperture r0 such that the R of the telescope (without any turbulence) is equal to the R of the atmosphere alone:
dfB ( f ) = dfT ( f ) ( p / 4 ) ( r0/ l )2

- Now we have to evaluate the contribution of the atmosphere’s OTF: dfB ( f )
- B ( f ) = C ( l f ) (to go from cycles per meter to cycles per wavelength)

Algebra…

(6p / 5) G(6/5) K-6/5

- Now we need to do the integral in order to solve for r0 :
( p / 4 ) ( r0/ l )2 =dfB ( f ) = df exp (- K f 5/3)

- Now solve for K:
K = 3.44 (r0/ l )-5/3

B ( f ) = exp - 3.44 ( l f / r0 )5/3 = exp - 3.44 ( / r0 )5/3

Algebra…

Replace by r

- r0is size of subaperture, sets scale of all AO correction
- r0gets smaller when turbulence is strong (CN2 large)
- r0gets bigger at longer wavelengths: AO is easier in the IR than with visible light
- r0gets smaller quickly as telescope looks toward the horizon (larger zenith angles )

- Usually r0 is given at a 0.5 micron wavelength for reference purposes. It’s up to you to scale it by -1.2 to evaluate r0 at your favorite wavelength.
- At excellent sites such as Paranal, r0 at 0.5 micron is 10 - 30 cm. But there is a big range from night to night, and at times also within a night.
- r0 changes its value with a typical time constant of 5-10 minutes

- Using the Kolmogorov turbulence hypothesis, the atmospheric phase PSD can be derived and is

- This expression canbeused to compute the amount of phase error over an uncorrectedpupil

Tip-tilt is single biggest contributor

Focus, astigmatism, coma also big

High-order terms go on and on….

Units: Radians of phase / (D / r0)5/6

Reference: Noll76