Transcript Visibility functions and image parameter estimation

Understanding interferometric
visibility functions
J. Meisner
Computing the visibility function V(u,v)
from the image function F(x,y)
Possible approaches:
1) Do the 2-D fourier transform (not so intuitive)
2) In a circularly symmetric case, perform the
Hankel transform (not so intuitive)
3) In either case, decide on a baseline direction,
then take the image and collapse it onto an
axis in that direction. THEN take the 1-D
fourier transform! (more intuitive!)
The very underresolved case:
Principle of superposition of visibility functions
• Must use the ACTUAL visibility V(f), not V2 !
• Visibilities add as COMPLEX numbers in general
• Special case: symmetric images. Visibilities are purely
real.
• Resulting normalized visibility is the sum of the
unnormalized visibilities divided by the TOTAL light.
Thus if
V1 = v1 /p1 and V2 = v2 /p2 Then:
V3 = (v1 + v2 ) / (p1 + p2)
• Can also express as weighted average of V1 and V2:
V3 = (k V1 + (1-k) V2 )
<<<< Simplest formulation
where k = p1 / (p1 + p2 )
Two important examples of real
visibilities adding
1) Object F1(x) whose visibility = V1(f) with k of flux,
PLUS extended emission over large x (overresolved)
whose V=0 with (1-k) of flux.
Vnet = k * V1(f) + 0
2) Object F1(x) whose visibility = V1(r) with k of flux,
PLUS point source (unresolved) at center (x=0)
whose V=1 with (1-k) of flux.
Vnet = k * V1(f) + (1-k)
Visibility of a square star 
Obtain Sinc(R)
Envelope falls off
as R-1
Two approaches for obtaining the visibility
function of a circularly symmetric object
The image is circularly symmetric:
F(x,y) = F(r) where r2 = x2 + y2 ;
The visibility function will ALSO be circularly symmetric
i.e. V(u,v) = V(w) where w2 = u2 + v2 ; Then:
To find V(w) either:
1) Find the Hankel Transform of F(r) which is then V(w)
2) Collapse the object along one (any!) position angle,
and take the fourier transform of that.
(Of course these are mathematically equivalent!)
Uniform Disk case
• Diameter = D
• Call R = fs * D
• Take fourier transform of collapsed
density (or 2-D FT of disk, or Hankel
transform of radial density function), get:
• V= 2 J1(pR)/(pR)
• Similar in appearance to Sinc function.
At large R, envelope falls off ~ R-3/2 (?).
Definition of “Resolvability” R
•
The resolvability R for an observation at a
particular spatial frequency fs is a dimensionless
number which describes an observation of some
class without respect to the specific image size or
the specific baseline, but is a proportional measure
of the baseline length with respect to the image
size.
•
R = fs * D where D is some full-width-like
measurement specific to a class of observations
(and therefore subject to arbitrary definition). The
definition is chosen to be convenient and
appropriate (commensurate with the uniform disk
case where D = the diameter).
Definition of “Resolvability” R:
Normalized “diameter” metrics
My definitions:
1. For a Uniform Disk: D= the diameter.
Thus the disk is resolved (at the first visiblity null) when
R=1.22
2. For a Limb Darkened disk, D = 3 * rmean where rmean is the 1st
moment of radial position averaged over the disk. (For a
uniform disk, rmean = 2/3 * Radius)
3. For a Binary Star: D = the separation collapsed into the position
angle of the baseline, or:
D= |S| cos(fs – fbaseline) where S is the separation vector.
Thus R = fs (dot) S where fs is the spatial frequency vector.
4. For a gaussian “disk”, D= the full width 1/e intensity diameter.
Uniform disk, D=1
Uniform disk, but D=.3
Identical visibility curve with respect to (normalized) Resolvability R
Another example of adding real visibilities
Disk of diameter Dout MINUS a removed inner disk
(hole) of diameter Din (i.e. a ring).
So amount of light removed = (Din/Dout)2 = A
Then Vnet(Fs) = (VUD(FsDout) – A VUD(FsDin)) / (1-A)
Limiting case: J0 .
Uniform disk D=1, minus hole D=.2
Uniform disk D=1, minus hole D=.5
Solving for an ELLIPSE:
Visibility curve identical
to UD.
To solve for parameters
of ellipse, need to
measure visibilities at 3
(or more) distinct position
angles…
Apparent diameter vs. position angle for an ellipse, should be a sine wave.
Alf eri, Ellipticity observed using VINCI
On the 140 meter baseline:
On the 66 meter baseline:
Special case: Gaussian nebulosity
The 2D -> 1D collapse of a gaussian is a gaussian
The fourier transform of a gaussian is a gaussian
Note that the visibility therefore NEVER goes negative:
V(R) > 0.
Define D in this case as the full diameter between 1/e points.
Then find the visibility is given by:
V(R) = exp( -4p2 R2)
Note: The “Full Width Half Maximum” (often quoted) for a
gaussian is then given by:
FWHM = .832 D
Gaussian “disk”, D=1
Dotted line = UD visibility
function for comparison
Gaussian disk with a hole removed
Another example of adding real visibilities
Gaussian disk MINUS a removed inner disk (hole) of
diameter Din
In this case, the gaussian has almost no visibility
contributions at larger spatial frequencies. What we
see then is the visibility function of the HOLE which
looks like a uniform disk (but with NEGATIVE visibility).
Note: What is important is NOT that the original image
is a gaussian, but only that it is very overresolved at
the spatial frequencies produced by the (negative)
inner disk.
Gaussian disk with a hole removed
Limb darkened disk
Use linear limb darkening law in m
For a Limb Darkened disk, we call D = 3 * rmean where
rmean is the 1st moment of radial position averaged
over the disk.
(For a uniform disk, rmean = 2/3 * Radius)
This causes the V(R) at small R to match that of a
uniform disk with the same D.
Significant difference only observable beyond first null
(at R=1.22) where the amplitudes of the sidelobes
are decreased relative to a uniform disk.
Details of the limb darkening are even more difficult to
observe! Therefore just use a standard/simple limb
darkening law (i.e. linear in m).
Limb darkened disk with k=.5
Dotted line = UD visibility
function for comparison
Limb darkened disk with k=.5
Detail at high spatial frequencies
Dotted line = UD visibility
function for comparison
“Fully-darkened disk”, K=1
Dotted line = UD visibility
function for comparison
Example: psi phe beyond the first null,
from VINCI
Limb darkening model used:
Visibility of a
binary star.
Define
D = separation.
Then visibility
magnitude is
oscillatory
with period
DR=1
Visibility of binary star, equal brightness
Visibility of binary, brightness = .7, .3
Magnitude
Real
Imaginary
Another example where PHASE is the most
sensitive quantity: planet transiting a star
Non-zero
imaginary
part
Central Hotspot
Central Hotspot – high spatial freq.
The End (?)
(not really!)