Application
of
crystallography:
Fourier
transform
on
X-ray
1.0 Description of the problem .......................................................................... 2
2.0 Motivation ......................................................................................................... 2
2.1 Crystals: ........................................................................................................ 3
2.2 Bragg’s law: ................................................................................................. 5
3.0 Relation between Fourier transform and theory of diffraction: ......... 7
4.0 The structure factor equation: .................................................................... 8
5.0 Phase problem: ............................................................................................... 9
6.0 Solving the phase problem: ........................................................................ 9
6.1
Patterson map is also known as Patterson synthesis: .............. 9
6.2
The heavy atom method: .................................................................. 10
6.3
Isomorphous replacement: .............................................................. 10
6.4
Anomalous scattering: ...................................................................... 10
6.5
Direct methods: ................................................................................... 10
7.0 Collecting X-ray data: .................................................................................. 10
7.1 Crystal and unit cell:................................................................................ 11
7.2 Coordinate systems in crystallography: ............................................ 11
8.0 Appendix A..................................................................................................... 13
8.1 Mathematics: ............................................................................................. 13
8.2 Fourier series: ........................................................................................... 13
8.3 General Fourier transforms: .................................................................. 15
8.4 Fast Fourier transforms: ........................................................................ 16
9.0 Appendix B ..................................................................................................... 17
9.1Production of X-rays:................................................................................ 17
9.2 Detection of X-rays: ................................................................................. 18
10.0 References: .................................................................................................. 18
10.1 Books: ....................................................................................................... 18
10.2 Web sites: ................................................................................................. 18
1.0 Description of the problem
When the word microscopy comes in our mind we always think in
terms of lenses, which gather the light waves scattered by objects together,
and focuses them into an image. Similarly in our eyes, lenses make images
on retina by focusing light waves scattered by an object at a distance.
Moreover even magnifying lenses also magnifies an image by focusing the
light scattered by a minute object. [1a]
The above discussion is fine while trying to visualise small objects but
when we want to see something at atomic levels then light microscopes or
using visual light becomes useless. The reason being the object should be at
least half the wavelength of the electromagnetic wave used. [1b]
For instance an atom nucleus is about 10 -15 meters [2b] where as the
atomic diameter is about 10-11meters. Covalent bond length is about 0.96
Angstrom (one Angstrom (A)= 10-10 meters) and hydrogen bond length is
about 1.97 A.[3b] In contrast to all these visual light’s wavelength is much
bigger (7000 A).[4b]
So we have to use an electromagnetic wave with wavelength much
smaller then visual light, to be in the range of the atomic size or at least in the
inter-atomic distance. Hence we use X-rays (wavelength 0.30-30 A).
Theoretically microscopes using X-rays should be the best if we want
to see atomic level but the big disadvantage is that unlike the visual light Xrays cannot be focused to get an inverted image in the image plane at 2f
position. [1a]
2.0 Motivation
No matter what ever electromagnetic wave we use, every object makes
some pattern at f distance from the lens towards the image plane. This pattern
is also called diffraction pattern. The plane where these patterns are made is
called focal or Fourier transform plane (figure 2.1). This is because the object
has a mathematical relationship with its diffraction pattern which is described
exactly by Fourier transform. X-rays cannot be focused to give an image but
they do give diffraction pattern (figure 2.2) at the focal plane which can be
mathematically interpreted into an image of the object. [1a]
Figure 2.1 shows Fourier transform (in the imaginary plane) of an object by a lens, f is the focal and  is the angle
with the plane of the scattering object. Figure taken from: - Introduction to macromolecular crystallographyAlexander Mcpherson.
Figure 2.2: A typical diffraction pattern obtain at f position in the reciprocal space. Figure taken from: - Physical
biochemistry –K.E. Van Holde 3rd edition
2.1 Crystals:
Pure substances in solid state can occur as amorphous or crystalline.
Crystals are solids made up of repeated units called unit cell. The difference
between a crystal and amorphous solid of the same chemical composition is
that the former is more ordered. Here ordered mean regular, symmetric, and
repeating. The application of crystals in structure determination using X-ray
diffraction is very important [2a]. It can be demonstrated as follows:
Figure 2.3: (a) is the object and (b) is the diffraction pattern at the Fourier transform plane. Figure taken from: Introduction to macromolecular crystallography- Alexander Mcpherson.
(The first image is a real image or in real space and the second image is in
imaginary or reciprocal space.) As shown in the figure (2.3) the diffraction
pattern formed at Focal plane by an object is continuous. Although the
diffraction pattern is a Fourier transform of the original object but to calculate it
mathematically back into its image is difficult.
Figure 2.4 (a) is the object in the real plane and (b) is the diffraction pattern in the reciprocal plane. Figure taken from:
- Introduction to macromolecular crystallography- Alexander Mcpherson.
In contrast to the previous example, figure (2.4) has an object (a) which
is more ordered (regular, symmetric, and repeating) and the diffraction pattern
in Figure (2.4 b) can now be easily converted back into the original object
mathematically.
Figure 2.5: first figure is an object and second figure is a diffraction pattern of the same object. Figure taken from: Introduction to macromolecular crystallography- Alexander Mcpherson.
Figure 2.5 is another example where the object (in the first figure) in
not ordered hence the diffraction pattern is not discrete as in figure (2.4).
Figure 2.6: First figure is an object and the second figure is again a diffraction pattern. Figure taken from: Introduction to macromolecular crystallography- Alexander Mcpherson.
Considering another example figure (2.6), here the object is ordered
(regular, symmetric, and repeating).The diffraction pattern at the focal point is
discrete points. Here in this diffraction pattern we can find the coordinates (or
indices). Moreover intensity data can also be used and hence mathematically
it is possible to get back the original object.
To conclude discrete diffraction pattern obtained from a crystal can
also be mathematically converted back into original object [1a].
2.2 Bragg’s law:
Discussion on X-ray diffraction is incomplete without mentioning the
Bragg’s law. It helps us in understanding the geometrical condition of
diffraction. It is expressed as
2d sin   n
(Figure 2.7)Where‘d’ is the spacing between the adjacent planes in the
parallel set,  is also called Bragg’s angle and  is the wavelength of X-ray
used. [8a]
Figure 2.7 diagrammatic view describing Bragg’s Law- Figure taken from Physical biochemistry –K.E. Van Holde 3rd
edition
Bragg’s law is actually for reflection and it is not an accurate representation of
the diffraction process. In the words of K.E. Van Holde [2a]”Atoms scatter Xrays, not because they are shiny spheres but because they are oscillating
dipoles. Still, the data observed during an X-ray diffraction experiment are
often called reflections because Bragg’s law correctly describes the conditions
of constructive interference in the X-ray diffraction.” So we can say that the
phenomenon of diffraction is not same as reflection but still Bragg’s equation
can be use.
A better explanation was given by a German scientist Max Von Laue.
Von Laue is believed to be the discoverer of the phenomenon of X-ray
diffraction and for this work he was awarded the Nobel Prize for physics in
1914. [7b]
Figure 2.8 diagrammatic representation of Van Laue’s explanation. Figure taken from Physical biochemistry –K.E.
Van Holde 3rd edition
The explanation can be shown as follows: l=c cos 
Where l is an integer,  is wavelength, c is spacing between the points and 
is the angle between the direction of the scattered radiation and the row of
scatterers.
Von Laue further extended this to three dimensions and the relation it
has with Braggs law can be shown as follows:
(h2/a2+k2/b2+l2/c2)1/2 = 2 sin / = n/d
Where 2= and n and d are analogous to l and c respectively. Here hkl could
be described as coordinates of reciprocal space and abc as coordinates for
real space [2a]
3.0 Relation between Fourier transform and theory of diffraction:
An excellent description of such a correlation is given by
Ramachandran and Srinivasan [3a], assuming P to be either an electron or an
atom (figure 3.1). Considering a monochromatic parallel beam of X-rays such
that it makes P as scattering centre Let Si and Sr be unit vector along the
direction of incident and scattered radiation.
Figure 3.1 (a) shows scattering of x-rays from a point P and the origin O (b) shows the vector product of Sr and Si.
Figure taken from: Fourier Methods in Crystalloraphy- Ramachandran and Srinivasan.
First we have to calculate the resultant scattered radiation which
makes an angle 2with the incident direction. The phase of wave scattered
by the point P in the direction sr, is ahead of the phase of the wave scattered
by a point at the origin O, by a quantity , which is given by
  2 /  (AO-PB)= 2 /  (r.( sr  si ))
Where r is the position vector of P with O as origin
The wave vectors Si and Sr can be defined as
Si= (1 /  ) si
and Sr= (1 /  ) sr
Hence the phase difference can be written as
  2 .r.( Sr  Si )  2 .r.r *
Where r* = Sr-Si
Considering “a” as the amplitude of the wave scattered by p where as the
amplitude of the incident wave is unity. Hence the resultant scattered radiation
from P can be written as
a. exp( 2 .i.r * .r )
Ramachandran further extended this equation by taking in account a
set of n scattering points at positions rj (where j=1 to N). So the total wave
scattered is obtained by summing all the scattered waves given by F(r*).
F(r*) =  j 1 aj. exp( 2 .i.r * .rj )
N
Where vector r* represents the scattering phenomenon and relates
wavelength of the radiation with direction of incident and the scattered waves.
According to Ramachandran [3a] on further modifying this equation to
continuous medium the amplitude of the wave scattered by the matter in a
volume element dv at r is (r)dv. Hence the total scattering in the scattering
direction is given by
F (r )    (r ) exp( 2 .i.r * .r )dv
“Thus the wave scattered by a continuous medium can be represented as
Fourier transform of the scattering density, and it also follows, from the theory
of Fourier transforms, that the scattering density is, in turn, given by the
inverse Fourier transform namely”[3a]
 (r )   F (r*) exp( 2 .i.r * .r )dv
The mathematics of Fourier transform is described in Appendix A
4.0 The structure factor equation:
In equation (9 from appendix A) electron density is related to sum of all
scattered waves and dv which is the volume element. Here we introduce
another factor called structure factor. It depends on crystal structure and
hence the name structure factor. It can be mathematically described as:
N
F (h, k , l )   f ( j ) exp 2 .i(hx( j )  ky( j )  lz ( j )) …………………….(10)
j 1
So the electron density can be described in terms of structural factor as:
F (h, k , l )    ( x, y, z ) exp( 2 .i(hx  ky  lz )).dx.dy.dz …………………(11)
hkl
or
F (h, k , l )    ( x, y, z ) exp( 2 .i(hx  ky  lz )).dV ………………………..(12)
V
Structure factor is also among many factors which affect the intensity of X-ray
in the diffraction pattern. Hence it has a relationship with the intensity which
will be discussed in the next section. [2a]
5.0 Phase problem:
From the above discussion it is clear that if we have a way by which we
can calculate structure factor then solving electron density equation will be
very easy and thus we can solve the structure of the molecule in the crystal.
In NMR and infra red spectroscopy we can detect both amplitude and
phase for light of long wavelengths, but unfortunately we can’t do this thing for
short wavelength like x-rays, but we can measure the total energy by
measuring the intensity of the diffraction pattern. Since intensity is
proportional to the square of the amplitude so we have the amplitude of
structure factor but not the phase angle [9a]. So the equation (10) of structure
factor can be rewritten as:
F (hkl)  F (hkl) exp( i (hkl)) ………………………………………………(13)
Structural factor obtained in equation (13) can be inserted in electron density
equation (12) as follows:
 ( xyz)  1 / V



   F (hkl) cos2 (hx  ky  lz )   (hkl) …………………(14)
h   k   l  
6.0 Solving the phase problem:
Mcpherson [1a] has described many methods which are been
developed to solve this problem.
1.
2.
3.
4.
5.
6.
De-convolution of the Patterson map.
The Heavy atom method
Isomorphous replacement
Anomalous scattering
Molecular replacement
Direct methods
6.1 Patterson map is also known as Patterson synthesis:
This method can solve a structure on its own provided the structure
contains very few atoms (maximum up to twelve). Except the direct
method it works in conjunction with nearly all the above mentioned
methods. [1a].
The biggest draw back of this method is that it does Fourier
calculation using the phase less quantity F 2 as a coefficient.
Patterson equation is written as:
P(u, v, w)  1 /V
F (hkl)
2
 F (hkl)
hkl
2
cos 2 (hu  kv  lw)
2
is simply the measured intensity I(hkl)
The P(u,v,w) calculated though this way is then plotted on graph.[6a]
6.2 The heavy atom method:
Here a heavy atom is been introduced in the sample containing lighter
atoms. Heavy atom has greater electron density and can be
recognised easily. Once the heavy atom coordinates are known then it
is relatively easy to find the positions of lighter atoms. [1a]
6.3 Isomorphous replacement:
In this method phases from structure factors of a known molecule is
used as initial estimates of phases for a new molecule. [4a] The phasing
model and the new molecule should be isomorphous i.e. unit cells should
be nearly identical in shape and size. The replacement part of the method
indicates that a heavy atom is added in the unit cell [2a].
6.4 Anomalous scattering:
Sometimes it is also called resonance scattering because it is produce
by the electrons producing Thompson scattering of X-rays [1a]. When xrays frequency is close to electron resonance frequency then the electrons
start absorbing the X-ray. The electron then undergoes a quantum
transition and then reemits the energy as a photon of the same
wavelength. The emitted X-rays have a phase changed by 90 degrees [4a].
6.5 Direct methods:
In 1985 two scientist Jerome karle and Herbert Hauptman were
awarded the Nobel prize in chemistry for the achievements in developing
Direct methods for X-ray crystallography. It is due to direct methods that Xray crystallography became so famous. One of the direct methods relies
on the existence of mathematical relationships among certain combination
of phases. Another method work such that it tries out random
arrangements of atoms, predicts the diffraction patterns they would
produce and compares the predicted patterns with those obtained from the
crystals. [4a]
7.0 Collecting X-ray data:
The modern day X-ray instruments measures the position and the
intensity of each reflection by the help of an optical scanner. This information
is transmitted in digital form to a computer for analysis.
 We get the intensity value by measuring the optical absorbance of the
spot
 Position helps us in obtaining the direction of diffraction.
Figure 7.1: diagrammatic view of x-ray diffraction. Figure taken from Crystallography made crystal clear- Gale Rodes
7.1 Crystal and unit cell:
Crystals are three dimensional structure made up of identical unit cells (in
other words, a three dimensional arrangement of identical, repetitive unit
cells).
Figure 7.2 The figure shown is of a simple cubic unit cell. Figure taken from http://cst-www.nrl.navy.mil/lattice/
7.2 Coordinate systems in crystallography:
Since the crystals are made up of three dimension unit cells so each reflection
can be assigned three coordinates or indices in the reciprocal or imaginary
space.
Figure 7.3: diagrammatic representation of 3 dimension view of a diffraction pattern. Figure taken from
Crystallography made crystal clear- Gale Rodes
In X-ray crystallography h, k, l are designated to the positions in 3 dimension
scale. The central spot or the round solid central reflection is given the
coordinates (h, k, l) = (0, 0, 0)
Another way to give coordinates is by measuring the actual distance as the
dimensions of reciprocal space are inverse of dimensions of the real space of
the crystals.
Figure 7.4: diagrammatic view of object in the real space (the first part of the figure) and the diffraction pattern in the
reciprocal space. Figure taken from Crystallography made crystal clear- Gale Rodes
8.0 Appendix A
8.1 Mathematics:
The Fourier transform is named after Jean Baptiste Joseph. In short “it
is an integral transform that re-expresses a function in terms of sinusoidal
based functions.” [6b]
Fourier transforms have variety of applications in many fields e.g.
In physics- signal processing, cryptography, acoustics, optics.
In maths-number theory, probability theory, statistics, geometry.
In biology-oceanography
In chemistry-x-ray crystallography [6b]
8.2 Fourier series:
For periodic functions, no matter how much complicated they are, they
can be described as the sum of simple waves. This sum is called Fourier
series [4a].
e.g. [4a] F ( x)  Fo. cos 2 (0 x   0)
+
F1. cos 2 (1x  1)
+
F 2. cos 2 (2 x   2)
+
.
.
.
Fn. cos 2 (n   .n)
Or equivalently
N
F ( x   Fh. cos 2 (hx   .h))
h 0
e.g. [5a] F () is a function such that F()=F(). They are said to be
periodic with period 2

F ( )  1 / 2a0   (an. cos n  bn sin n ) ………….(1)
1
Here ½ a0 is the coefficient of the constant function 1= cos0and ½ is
incorporated for later convenience. There is no b0 as sin0=0.
Note : Sometimes it is better to work with complex components than to work
with trigonometric functions (e.g. cos)
 cos   exp( i )  exp( i ) / 2
sin   exp( i )  exp( i ) / 2i
exp( i )  cos   i sin 
By using the above formula we can say.
cos n  (exp( in )  exp( in ) / 2)
sin n  (exp( in )  exp( in ) / 2i )
 Equation (1) can be rewritten as.

F ( )   Cn exp( in  ) ………………(2)

Where
C 0  1 / 2a 0
Cn  1 / 2(an  ibn )
C  n  1 / 2(an  ibn )
a 0  2co
an  Cn  C (  n)
bn  i (Cn  C (  n))
n  1,2,3,........
We can also say
exp( in  )  cos n  i sin n
cos( n )  cos n
sin( n)   sin n
Finding the value of Cn, an and bn:
Multiplying both sides of equation (2) by exp(-ik)(where k is integer)
and integrating from –  to +






 F ( ) exp( ik ).d   Cn  exp( i(n  k ) ).d ………………(3)
Solving RHS:
(i) When n is not equal to k




  Cn  exp( i(n  k ) ).d  1 / i(n  k ) exp( i(n  k ) ) limits - to 
= (-1)n-k –(-1)n-k/i(n-k)=0
(ii) When n=k








  Cn  exp( i(n  k ) ).d   Cn  d  2
Hence the outcome of (i) and (ii) will be (ii) only. So the equation (3) can be
written as follows:
2Ck 

 F ( ) exp( ik ).d ……………………….(4)

Re-labeling the integer k as n

 Cn  1 / 2  F ( ) exp( n ).d


a0  2C 0  1 /   F ( ).d
(exp(0)=1)


an  Cn  C (n)  1 / 2  F ( )(exp( in )  exp( in )).d

By replacing cos n  exp( in )  exp( in ) / 2

an  Cn  C (n)  1 /   F ( ). cos n .d


bn  i(Cn  C (n))  i / 2  F (exp( in )  exp( in )).d

By replacing sin n  (exp( in )  exp( in )) / 2i

bn  i(Cn  C (n))  1 /   F ( ) sin n .d

an and bn are Fourier’s coefficients.
8.3 General Fourier transforms:
In the previous example we saw how sin and cosine function of a
Fourier series is converted to exponential form. Gale Rodes [4a] described the
above shown exponential form of Fourier series in a more general form. An
example of F(x) and F(h) was taken such that F(h) is a Fourier transform (FT)
of F(x).

F ( h) 
 F ( x) exp( 2 .i(hx)).dx ……………………….(5)

The basic property of Fourier transform that it is a linear invertible transform.
The inverse of the Fourier transform is the following

F ( x) 
 F (h) exp( 2 .i(hx)).dh ……………………..(6)

A perfect example to show this property can be when x is time in
seconds and h is reciprocal time or frequency (1/seconds).
“Taking the FT of time-dependent functions is a means of decomposing these
functions into their component frequencies and is sometimes referred to as
Fourier analysis.” [4a]. This way frequencies of many spectral lines can be
obtained simultaneously. A perfect example of this is spectroscopy (NMRnuclear magnetic resonance and infrared spectroscopy).
Extending this view Gale Rhodes further pointed out the application of
Fourier transform in 3 dimensions. For example taking F(h) in three
dimensions to F(h, k, l) and F(x) to F(x, y, z). Hence the equations can be
written as
F ( x, y, z )   F (h, k , l ) exp( 2 .i(hx  ky  lz )).dh.dk .dl ……………(7)
hkl
F (h, k , l )   F ( x, y, z ) exp( 2 .i (hx  ky  lz )).dx.dy.dz …………….(8)
xyz
Our electron density equation
 (r )   F (r*) exp( 2 .i.r * .r )dv
can also be written as
 ( x, y, z )  1 / V h k l Fhkl. exp( 2 .i(hx  ky  lz )) ………………..(9)
The reason is that diffraction is a regular pattern (as described in previous
section of Von Laue’ equation). Hence the electron density can be calculated
from a sum of F for all Miller indices (h, k, l) (coordinates of reciprocal
space).[2a]
8.4 Fast Fourier transforms:
Stout and Jensen [6a] describe a modification of Fourier transform which
is very commonly used these days. According to them from 1960s it was
possible to speed the calculation described in equation (9) by
1. Suitably separating the terms in the summation
2. Performing shorter and faster transforms on the separated elements
3. And then combining these transforms at the end
This can be demonstrated by considering one dimensional example

H /2
H /2
0
0
 F (2h) exp( 2 .i(2h) x)   F (2h  1) exp( 2 .i(2h  1) x)
This equation can be modified as

H /2
H /2
0
0
 F (2h) exp( 2 .i(2h) x)  exp( 2 .ix )  F (2h  1) exp( 2 .i(2h) x)
Hence with same exponential term on both the sides will make calculation
easier as it will be computed once and used twice. The idea is also expressed
as a constant common phase is taken out, because the functions are periodic
in nature and so it becomes faster than the ordinary Fourier transform.
9.0 Appendix B
Rontgen in 1895 was the first one to discover X-rays. He realized that the
nature of these rays was unknown so he used letter X. Since then these rays
are called X-rays.
Some of the important properties which he saw were:
 These rays were invisible (just opposite to ordinary light)
 They travelled in straight line (similar to ordinary light)
 They were more penetrating than the visible light. They could even
pass human body, metal, wood and other opaque objects where visible
light could not.
By 1912 many properties of x-rays were explored to study the nature of
these rays (including x-ray diffraction of crystals). [7a]
9.1Production of X-rays:
In an atom the nucleus is surrounded by electrons, which orbit around
the nucleus in their own shells. The shell nearest to the nucleus is called k
shell and so the electrons in that shell are called k shell electrons. In short Xrays are produced when any electrically charged particle of sufficient kinetic
energy,
knocks
off
or
removes
these
k
shell
electrons.
Figure 9.1: diagrammatic representation of a nucleus and its shells. Figure taken from: - Crystal Structure AnalysisW.Clegg.
Another electron enters the k shell. In k shell the electron’s energy should be
lowest possible, hence this new electron librates energy in the form of
radiation (one of which will be X-rays). As shown in the diagram if the electron
comes form L shell the radiation is called k and if the electron comes from M
shell then the radiation is called K [7a].
The assembly for x-ray production usually consist of
(a) a source of electrons (b) a high accelerating voltage and (c) a metal target
An x-ray tube contains two electrodes and when a high voltage is applied
across these electrodes then electrons move towards the anode or the target.
Electrons thus strike the target at a very high speed and so x-rays are
produced and radiated in all directions. Some of metals used as targets are
copper, cobalt, iron, and chromium.
Some other examples of x-ray sources are as follows Gas tubes,
Filament tubes, rotating anode tubes, micro-focus tubes, pulsed (or flashed)
tubes, miniature tubes, high voltage tubes and linear accelerators and
Synchrotron radiation. Synchrotron radiations are currently the most popular
source of X-rays. The principle is based on the fact that acceleration of
charged particles produces electromagnetic radiations [7a].
9.2 Detection of X-rays:
Previous methods were fluorescent screens, photographic film, and
electronic detectors. Currently the most popular one image (storage) plates
[7a].
10.0 References:
10.1 Books:
[1a] Introduction to macromolecular crystallography- Alexander Mcpherson.
[2a] Physical biochemistry –K.E. Van Holde 3rd edition
[3a] Fourier Methods in Crystalloraphy- Ramachandran and Srinivasan
[4a] Crystallography made crystal clear- Gale Rodes
[5a] Fourier analysis and its application- Gerald B Folland
[6a] X-ray structure determination –George. H. Stout and Lyle. H. Jensen
[7a] Elements of x-ray Diffraction- B.D.Cullity and S.R.Stock
[8a] Crystal Structure Analysis- W.Clegg.
[9a] An introduction to x-ray x-raystallography- Michael M Woolfson
10.2 Web sites:
[1b] http://www.stolaf.edu/people/hansonr/mo/x-ray.html
[2b] http://www.elmhurst.edu/~chm/vchembook/161Ahydrogenbond.html
[3b] http://www.triumf.ca/safety/rpt/rpt_1/node7.html
[4b] http://imagers.gsfc.nasa.gov/ems/visible.html
[5b] http://cst-www.nrl.navy.mil/lattice/
[6b] http://en.wikipedia.org/wiki/Fourier_transform
[7b] http://www.norskfysikk.no/nfs/epsbiografer/VONLAUE.PDF