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 ifV1 = 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 formulationwhere 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 symmetrici.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(R)/(R)• 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) whenR=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(s – baseline) 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=.3Identical 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…

Alf eri, Ellipticity observed using VINCI On the 140 meter baseline: On the 66 meter baseline:

Apparent diameter vs. position angle for an ellipse, should be a sine wave.

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( -42 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 diskUse linear limb darkening law in

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 ).

Limb darkened disk with k=.5

Dotted line = UD visibility function for comparison

Limb darkened disk with k=.5Detail 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 R=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!)