Transcript Document
Adaptive Optics in the VLT and ELT era
Atmospheric Turbulence
François Wildi
Observatoire de Genève
Credit for most slides : Claire Max (UC Santa Cruz)
Page 1
Atmospheric Turbulence Essentials
• 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
Fluctuations in index of refraction are
due to temperature fluctuations
• Refractivity of air
3
P
7.52
10
6
N (n 1) 10 77.6 1
2
T
where P = pressure in millibars, T = temp. in K, in microns
n = index of refraction. Note VERY weak dependence on
• Temperature fluctuations index fluctuations
N 77.6 (P / T )T
2
(pressure is constant, because velocities are highly sub-sonic. Pressure
differences are rapidly smoothed out by sound wave propagation)
Turbulence arises in several places
stratosphere
tropopause
10-12 km
wind flow around dome
boundary layer
~ 1 km
Heat sources w/in dome
Within dome: “mirror seeing”
• When a mirror is warmer than
dome air, convective
equilibrium is reached.
credit: M. Sarazin
• Remedies: Cool mirror itself,
or blow air over it, improve
mount
credit: M. Sarazin
convective
cells are bad
To control mirror temperature: dome air conditioning (day), blow air on
back (night)
Local “Seeing” Flow pattern around a telescope dome
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
Boundary layers: day and night
• 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”
Real shear generated turbulence (aka KelvinHelmholtz instability) measured by radar
• Colors show intensity of radar return signal.
• Radio waves are backscattered by the turbulence.
Kolmogorov turbulence, cartoon
solar
Outer scale L0
Inner scale l0
h
Wind shear
convection
h
ground
Kolmogorov turbulence, in words
• 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
Assumptions of Kolmogorov turbulence
theory
• 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
Questionable
• In practice, Kolmogorov model works surprisingly well!
Concept Question
• What do you think really determines the outer scale in
the boundary layer? At the tropopause?
• Hints:
Outer Scale ~ 15 - 30 m, from Generalized
Seeing Monitor measurements
• F. Martin et al. , Astron. Astrophys. Supp. v.144, p.39, June 2000
• http://www-astro.unice.fr/GSM/Missions.html
Atmospheric structure functions
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
Typical values of CN2
• 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
10-14
Paranal, Chile, VLT
Turbulence profiles from SCIDAR
Eight minute time period (C. Dainty, Imperial College)
Siding Spring, Australia
Starfire Optical Range,
Albuquerque NM
Atmospheric Turbulence: Main Points
• 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!
Phase structure function,
spatial coherence and r0
Definitions - Structure Function and
Correlation Function
• Structure function: Mean square difference
D (r ) (x ) (x r )
2
dx
(x
)
(x
r
)
• Covariance function: Spatial correlation of a
random variable with itself
B (r ) (x r ) (x )
dx (x r ) (x )
2
Relation between structure function and
covariance function
D (r ) 2B (0) B (r )
To derive this relationship, expand the product in the
definition of D ( r ) and assume homogeneity to take
the averages
Definitions - Spatial Coherence Function
For light wave exp[i(x)], phase is(x) kz t
• Spatial coherence function of field is defined as
r
r * r r
C (r ) ( x) ( x r )
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
•
Now evaluate spatial coherence
function C (r)
• For a Gaussian random variable with zero mean,
exp i exp 2 / 2
•
• So
r
r
r r
C (r ) expi[ ( x) ( x r )]
r
r r 2
r
exp ( x) ( x r ) / 2 exp D (r ) / 2
• So finding spatial coherence function C (r) amounts to evaluating
the structure function for phase D ( r ) !
Next solve for D ( r ) in terms of the
turbulence strength CN2
• We want to evaluate
C (r ) exp D (r )/2
• Remember that
r
r
D (r ) 2 B (0) B (r )
Solve for D ( r ) in terms of the
turbulence strength CN2, continued
• But
r
( x) k
h h
v
dz n( x, z)
for a wave propagating
h
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/ [radian/m]) x the Optical path
Here n(x, z) is the index of refraction.
• Hence
B (r ) k
2
h h
h h
h
h
dz dz n(x, z)n(x r, z)
Solve for D ( r ) in terms of the
turbulence strength CN2, continued
z z z
• Change variables:
• Then
B (r ) k 2
k2
h h
h h z
h
h z
h h
h h z
dz dz
dz dz B
N
(r ,z)
h z
h
B (r ) k 2h
n(x, z)n(x r , z z)
h h z
h z
2
dzB
(r
,z)
k
h dzBN (r,z)
N
Solve for D ( r ) in terms of the
turbulence strength CN2, continued
• Now we can evaluate D ( r )
D (r ) 2 B (0) B (r ) 2k 2h dz BN (0,z) BN (r ,z)
D (r ) 2k 2h dz BN (0,0) BN (r ,z) BN (0,0) BN (0,z)
D (r ) k 2h dz DN (r ,z) DN (0,z)
Solve for D ( r ) in terms of the
turbulence strength CN2, completed
• But
r
r
DN (r ) CN2 r
2/3
r
D (r ) k 2 hCN2
C
2
N
r
2
z
r 2 z2
dz
2 1/ 3
1/ 3
so
z2 / 3
2 (1/ 2)(1/ 6) 5 / 3
5 /3
r 2.914r
5
(2 / 3)
•
r
2 5/3 2
2 5/3
2
D (r ) 2.914 k r CN h 2.914 k r dhCN (h)
0
Finally we can evaluate the spatial
coherence function C (r)
1
r
r
2 5/3
2
C (r ) exp D (r ) / 2 exp 2.914 k r dhCN (h)
2
0
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.
Given the spatial coherence function, calculate
effect on telescope resolution
• Define optical transfer functions of telescope,
atmosphere
• Define r0 as the telescope diameter where the two
optical transfer functions are equal
• Calculate expression for r0
Define optical transfer function (OTF)
• Imaging in the presence of imperfect optics (or
aberrations in atmosphere): in intensity units
Image = Object Point Spread Function
convolved with
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 )
Examples of PSF’s and their
Optical Transfer Functions
Seeing limited OTF
Intensity
Seeing limited PSF
/D
/ r0
D/
Diffraction limited OTF
Intensity
Diffraction limited PSF
r0 /
-1
/D
/ r0
r0 /
D/
-1
Now describe optical transfer function of the
telescope in the presence of turbulence
• 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 ( / D).
• Measure the resolving power of the imaging system by
R = df S ( f ) = df B ( f ) T ( f )
Derivation of r0
• R
of a perfect telescope with a purely circular aperture of (small)
diameter d is
R
=
df T ( f ) = ( p / 4 ) ( d / )
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:
df B ( f ) = df T ( f ) ( p / 4 ) ( r0 / )2
Derivation of r0 , continued
• Now we have to evaluate the contribution of the
atmosphere’s OTF:
df B ( f )
• B ( f ) = C ( f ) (to go from cycles per meter to cycles
per wavelength)
1
r
2 5/3
2
C (r ) exp 2.914 k r dhCN (h)
2
0
B( f ) C (f ) exp Kf
5/ 3
Derivation of r0 , continued
• Now we need to do the integral in order to solve for r0 :
( p / 4 ) ( r0 / ) 2 =
df B ( f ) = df exp (- K f 5/3)
(6p / 5) (6/5) K-6/5
• Now solve for K:
K = 3.44 (r0 / )-5/3
B ( f ) = exp - 3.44 ( f / r0 )5/3 = exp - 3.44 ( / r0 )5/3
Replace by r
Derivation of r0 , concluded
r 5/ 3 1
2 5 /3
2
3.44 2.914k r sec dhCN (h)
2
r0
r0 0.423k sec
2
dhC (h)
2
N
5/ 3
D (r) 6.88(r / ro )
3 / 5
Scaling of r0
• r0 is size of subaperture, sets scale of all AO correction
2
2
r0 0.423k sec CN (z)dz
0
H
3 / 5
6 /5
sec
3 / 5
2
CN (z)dz
-3/5
• r0 gets smaller when turbulence is strong (CN2 large)
• r0 gets bigger at longer wavelengths: AO is easier in the IR than with visible
light
• r0 gets smaller quickly as telescope looks toward the horizon (larger zenith
angles )
Typical values of r0
• 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
Phase PSD, another important parameter
• Using the Kolmogorov turbulence hypothesis, the
atmospheric phase PSD can be derived and is
0.023 11/ 3
( k ) 5 / 3 k
r0
• This expression can be used to compute the amount of
phase error over an uncorrected pupil
D
1.03
r0
2
5/3
Units: Radians of phase / (D / r0)5/6
Tip-tilt is single biggest contributor
Focus, astigmatism,
coma also big
Reference: Noll76
High-order terms go
on and on….