In order to answer this question we need to think back once more to the object
consisting of two points and its diffraction pattern which is a cosinusoidally
varying fringe pattern. Suppose now that the object is translated in its own
plane (which is assumed perpendicular to the light beam). It is well known that
the fringes will *not* move laterally. (This can be demonstrated easily if
a distant street lamp is viewed through a piece of fabric such as a handkerchief
or an umbrella: the diffraction pattern does not move on the retina when the
fabric is translated.) It is clear however that *some* representation of
the translation must be encoded in the diffraction pattern, since, if we allowed
the fringes to fall on a lens and be recombined to form an image, the
translation of the object would immediately become apparent. It can be shown
that the relative phases of the light arriving at various points of the pattern
change but since we cannot see or record phase we are not aware of the change.
It follows logically then that if we re-combine the *intensities* of the
diffraction pattern ignoring the phases, the information about *lateral*
position would be missing. The pairs of points which make up the object will
all be reproduced with the right separation from each other and in the right
orientation but they will all be symmetrically disposed about the centre of the
pattern instead of being properly distributed. In other words the resulting
distribution will contain information about all the vector distances that are
present between the various pairs of scattering points in the object but all
will be translated so that one end of every vector is the origin.

Since it is perfectly possible to record all the intensities in an X-ray photograph and to perform this recombination mathematically using a digital computer it should not be surprising to find that such a reconstruction is one of the standard methods of trying to decipher X-ray patterns. The process is known as the calculation of a Patterson function, after A. L. Patterson of Pennsylvania, U. S. A. who first suggested the technique. The problem is how to interpret the resulting map.

Figure 1 shows a simple object consisting of three points and Fig. 2 shows an
Idealised Patterson map showing how the six possible vectors **ab,
ba, bc, cb, ac and ca ** appear. The
relationship between the map and the object is not difficult to see. The
difficulty increases very rapidly, however, if the numbers of scattering points
increases. 10 scattering points would give rise to 90 peaks (though many would
overlap) and 100 would give 9900 peaks. To see how the complexity increases
even with simple structures we will consider an actual example. Figure 3 is a
typical map and Fig. 4 is one of the symmetrically related molecules in the
crystal giving rise to this distribution. A peak marked

**Copyright © 1997 International Union of Crystallography**