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## (a) What happens in the general case if we ignore the phase and try to recombine with just the intensities?

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 a in Fig. 1 would correspond to vectors such as 6-4, 1-3, 7-2 or 9-8 and a peak such as b would correspond to 7-6, 1-5, 2-4 and 8-2. Even though this is quite a simple arrangement the interpretation of the Patterson map without some knowledge of the molecule would be very difficult if not impossible.

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