page_title: Van Cittert–Zernike theorem
section_title: Proof of the theorem
hierarchical_section_title: Van Cittert–Zernike theorem / Proof of the theorem
context_page_description: The Van Cittert–Zernike theorem, named after physicists Pieter Hendrik van Cittert and Frits Zernike, is a formula in coherence theory that states that under certain conditions the Fourier transform of the intensity distribution function of a distant, incoherent source is equal to its complex visibility. This implies that the wavefront from an incoherent source will appear mostly coherent at large distances. Intuitively, this can be understood by considering the wavefronts created by two incoherent sources. If we measure the wavefront immediately in front of one of the sources, our measurement will be dominated by the nearby source. If we make the same measurement far from the sources, our measurement will no longer be dominated by a single source; both sources will contribute almost equally to the wavefront at large distances.
  This reasoning can be easily visualized by dropping two stones in the center of a calm pond. Near the center of the pond, the disturbance created by the two stones will be very complicated. As the disturbance propagates towards the edge of the pond, however, the waves will smooth out and will appear to be nearly circular.
context_section_description: Let  and  be respectively the cartesian coordinates of the source plane and the observation plane.  Suppose the electric field due to some point from the source in the source plane is measured at two points,  and , in the observation plane.  The position of a point in the source may be referred to by its direction cosines .  (Since the source is distant, its direction should be the same at  as at .)  The electric field measured at  can then be written using phasors:
  where  is the distance from the source to ,  is the angular frequency of the light, and  is the complex amplitude of the electric field.  Similarly, the electric field measured at  can be written as
  Let us now calculate the time-averaged cross-correlation between the electric field at  and :
  Because the quantity in the angle brackets is time-averaged an arbitrary offset to the temporal term of the amplitudes may be added as long as the same offset is added to both.  Let us now add  to the temporal term of both amplitudes.  The time-averaged cross-correlation of the electric field at the two points therefore simplifies to
  But if the source is in the far field then the difference between  and  will be small compared to the distance light travels in time .  ( is on the same order as the inverse bandwidth.)  This small correction can therefore be neglected, further simplifying our expression for the cross-correlation of the electric field at  and  to
  Now,  is simply the intensity of the source at a particular point, .  So our expression for the cross-correlation simplifies further to
  To calculate the mutual coherence function from this expression, simply integrate over the entire source.
  Note that cross terms of the form  are not included due to the assumption that the source is incoherent.  The time-averaged correlation between two different points from the source will therefore be zero.
  Next rewrite the  term using  and .  To do this, let  and .  This gives
  where  is the distance between the center of the plane of observation and the center of the source.  The difference between  and  thus becomes
  But because  and  are all much less than , the square roots may be Taylor expanded, yielding, to first order,
  which, after some algebraic manipulation, simplifies to
  Now,  is the midpoint along the -axis between  and , so  gives us , one of the direction cosines to the sources.  Similarly, .  Moreover, recall that  was defined to be the number of wavelengths along the -axis between  and .  So
  Similarly,  is the number of wavelengths between  and  along the -axis, so
  Hence
  Because  and  are all much less than , .  The differential area element, , may then be written as a differential element of solid angle of .  Our expression for the mutual coherence function becomes
  Which reduces to
  But the limits of these two integrals can be extended to cover the entire plane of the source as long as the source's intensity function is set to be zero over these regions.  Hence,
  which is the two-dimensional Fourier transform of the intensity function.  This completes the proof.
caption: The source is in the XY-plane, shown at the top of the figure, and the detector is in the xy-plane, shown at the bottom of the figure. Consider the electric field at two points, 
    
      
        
          
            P
            
              1
            
          
        
      
      {\displaystyle P_{1}}
    
   and 
    
      
        
          
            P
            
              2
            
          
        
      
      {\displaystyle P_{2}}
    
  , in the detection plane due to some point in the source whose coordinates are given by the direction cosines 
    
      
        
          l
        
      
      {\displaystyle l}
    
   and 
    
      
        
          m
        
      
      {\displaystyle m}
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