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Article #267Re: Cut-projection method
From: Dale E. Tronrud
Date: Tue, 28 Nov 1995 00:00
Date: Tue, 28 Nov 1995 00:00
55 lines
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3113 bytes
In article <Pine.SGI.3.91.951127153848.8482A-100000@helios.physics> Daniel Cote <dcote@Physics.UToronto.CA> writes: >From: Daniel Cote <dcote@Physics.UToronto.CA> >Subject: Cut-projection method >Date: Mon, 27 Nov 1995 20:44:02 GMT >This is about the cut-projection method (and some other related stuff). >I know this is kind of basic, but I don't understand how an intersection=20 >in real space becomes a projection in Fourier space. I looked at many,=20 >many books, and they all state "We know that...". But I don't. >So my question is extremely simple: >How do you show (mathematically) that an intersection in real space=20 >becomes a projection in fourier space? >I really need to understand the basic idea behind that technique, and I=20 >can't see it. >Thank you, >------------- >Daniel C=F4t=E9 "If it moves, it's Biology. >U of Toronto Physics Dept. If it stinks, it's Chemistry. >dcote@physics.utoronto.ca If it doesn't work, it's Physics= >." A single section of electron density can be represented as the product of the full 3-d map and a "plane". The plane would be zero everywhere except at the location of the section where it would have a value of one. By the convolution thereom the Fourier transform of the section would be the convolution of the transform of the 3-d map and the transform of the plane. The transform of the map is your structure factors. The Fourier transform of the plane is more difficult to justify. A simple way to explore the function is to look at its variability in different directions. Assume the plane is z=5. For any x and y, as you vary z to cross the plane you will find a delta function. The Fourier transform of a delta function is a constant for all resolutions. For any z, as you vary x or y you will never have any variation in the function. The Fourier transform of a constant function is zero for all resolutions (except for the 000 term). While this is not a mathematical proof it indicates that the Fourier transform of a plane in a line, perpendicular to the plane and passing through the origin. Now all you need to do is to convolute the structure factors with the line to get the Fourier transform of the section. You should be able to work out that this is a projection operator. Of course this argument works both directions. This is why a precession photo gives you a projection in real space. Dale Tronrud University of Oregon
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X-Google-Language: ENGLISH,ASCII-7-bit X-Google-Thread: fcf0b,f2ec2fc127edf6aa,start X-Google-Attributes: gidfcf0b,public X-Google-Thread: f804b,f2ec2fc127edf6aa,start X-Google-Attributes: gidf804b,public From: dale@uoxray.uoregon.edu (Dale E. Tronrud) Subject: Re: Cut-projection method Date: 1995/11/28 Message-ID: <dale.7.30BB62F5@uoxray.uoregon.edu> X-Deja-AN: 120383293 references: <Pine.SGI.3.91.951127153848.8482A-100000@helios.physics> organization: University of Oregon newsgroups: sci.techniques.xtallography,bionet.xtallography