1. Understanding interferometric visibility functions J. Meisner

2. 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!)

3. The very underresolved case:

4. Principle of superposition of visibility functions Must use the ACTUAL visibility V(f), not V 2 ! 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 V 1 = v 1 /p 1 and V 2 = v 2 /p 2 Then: V 3 = (v 1 + v 2 ) / (p 1 + p 2 ) Can also express as weighted average of V 1 and V 2 : V 3 = (k V 1 + (1-k) V 2 ) <<<< Simplest formulation where k = p 1 / (p 1 + p 2 )

5. Two important examples of real visibilities adding 1) Object F 1 (x) whose visibility = V 1 (f) with k of flux, PLUS extended emission over large x (overresolved) whose V=0 with (1-k) of flux. V net = k * V 1 (f) + 0 2) Object F 1 (x) whose visibility = V 1 (r) with k of flux, PLUS point source (unresolved) at center (x=0) whose V=1 with (1-k) of flux. V net = k * V 1 (f) + (1-k)

6. Visibility of a square star Obtain Sinc(R) Envelope falls off as R -1

7. Two approaches for obtaining the visibility function of a circularly symmetric object The image is circularly symmetric: F(x,y) = F(r) where r 2 = x 2 + y 2 ; The visibility function will ALSO be circularly symmetric i.e. V(u,v) = V(w) where w 2 = u 2 + v 2 ; 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!)

8. Uniform Disk case Diameter = D Call R = f s * D Take fourier transform of collapsed density (or 2-D FT of disk, or Hankel transform of radial density function), get: V= 2 J 1 (  R)/(  R) Similar in appearance to Sinc function. At large R, envelope falls off ~ R -3/2 (?).

10. Definition of “Resolvability” R The resolvability R for an observation at a particular spatial frequency f s 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 = f s * 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).

11. 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 * r mean where r mean is the 1 st moment of radial position averaged over the disk. (For a uniform disk, r mean = 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 = f s (dot) S where f s is the spatial frequency vector. 4.For a gaussian “disk”, D= the full width 1/e intensity diameter.

12. Uniform disk, D=1

13. Uniform disk, but D=.3 Identical visibility curve with respect to (normalized) Resolvability R

14. Another example of adding real visibilities Disk of diameter D out MINUS a removed inner disk (hole) of diameter D in (i.e. a ring). So amount of light removed = (D in /D out ) 2 = A Then V net (F s ) = (V UD (F s D out ) – A V UD (F s D in )) / (1-A) Limiting case: J 0.

15. Uniform disk D=1, minus hole D=.2

16. Uniform disk D=1, minus hole D=.5

17. 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…

18. 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.

19. 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( -4  2 R 2 ) Note: The “Full Width Half Maximum” (often quoted) for a gaussian is then given by: FWHM =.832 D

20. Gaussian “disk”, D=1 Dotted line = UD visibility function for comparison

21. Gaussian disk with a hole removed

22. Another example of adding real visibilities Gaussian disk MINUS a removed inner disk (hole) of diameter D in 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.

23. Gaussian disk with a hole removed

24. Limb darkened disk Use linear limb darkening law in  For a Limb Darkened disk, we call D = 3 * r mean where r mean is the 1 st moment of radial position averaged over the disk. (For a uniform disk, r mean = 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  ).

25. Limb darkened disk with k=.5 Dotted line = UD visibility function for comparison

26. Limb darkened disk with k=.5 Detail at high spatial frequencies Dotted line = UD visibility function for comparison

27. “Fully-darkened disk”, K=1 Dotted line = UD visibility function for comparison

28. Example: psi phe beyond the first null, from VINCI Limb darkening model used:

29. Visibility of a binary star. Define D = separation. Then visibility magnitude is oscillatory with period  R=1

30. Visibility of binary star, equal brightness

31. Visibility of binary, brightness =.7,.3 Magnitude Real Imaginary

32. Another example where PHASE is the most sensitive quantity: planet transiting a star Non-zero imaginary part

33. Central Hotspot

34. Central Hotspot – high spatial freq.

35. The End (?) (not really!)