Transcript Symmetry_of_Properties - IITK - Indian Institute of Technology

Symmetry of Properties
 Various properties:
 Mechanical, Electrical, Magnetic, Thermal, etc.
 Tensorial nature of properties.
 Symmetry imposed restraints.
Part of
MATERIALS SCIENCE
& A Learner’s Guide
ENGINEERING
AN INTRODUCTORY E-BOOK
Anandh Subramaniam & Kantesh Balani
Materials Science and Engineering (MSE)
Indian Institute of Technology, Kanpur- 208016
Email: [email protected], URL: home.iitk.ac.in/~anandh
http://home.iitk.ac.in/~anandh/E-book.htm
Advanced Reading
 Physical Properties of Crystals: Their Representation by Tensors and Matrices
J.F. Nye
Clarendon Press, Oxford, 1985.
 Properties of Materials: Anisotropy, Symmetry, Structure
Robert E Newnham
Oxford University Press, Oxford, 2005.
Overview of Properties
For now we concern ourselves with properties of crystals
 For every stimulus there is a natural response. E.g.:
 when electric field is applied to a dielectric material it can lead to electric polarization,
 when stresses develop inside a material, it can lead to strains,
 application of magnetic field can lead to magnetization.
 In addition there can be cross-coupling terms, i.e. one kind of stimulus can lead to properties,
which are not naturally associated with it. E.g.:
 application of pressure can lead to electric polarization (piezoelectricity),
 presence of temperature difference can cause electrical potential difference
(pyroelectricity),
 the electrical resistance is altered in the presence of a magnetic field (magnetoresistance).
 A stimulus can be connected to a response via a material property. E.g.:
 M = H,
  = E,
 P = pi T.
The material property has to be measured via experiments (or computed from more
fundamental properties).
 In reality the stimulus and response are tensors (of various orders) and hence the material
property is also a tensor of some rank (i.e. many values may be required to describe the property). The equations
(as above) in tensorial form become:  Mi = ijHj,  ij = Eijklkl,  Pi = pi T.
Classification of material properties
Polar tensors (rank 0-4)
Axial tensors (rank 0-4)
Material Properties
Transport properties
 Relate flow to a gradient
Properties with hysterisis  Involve domain wall motion
Examples of 1st to 4th ranked tensor properties The number of subscripts denotes the order of the tensor
Phenomena
Pyroelectricity
Thermal Expansion
Piezoelectricity
Elasticity
Cause
Effect
Pi
T
(Temperature (Electric polarization)
difference)
T
ij
(Strain)
Pi
ij
(Electric polarization)
(Stress)
kl
ij
(strain)
(Stress)
Coefficient
pi
(1st rank tensor)
Equation
Pi = pi T
αij
(2nd rank tensor)
dijk
rd
(3 rank tensor)
ij = αijT
Eijkl
(4th rank tensor)
Pi = dijkjk
ij = Eijklkl
Scalar, Vector and Tensor Quantities
 To describe a property at a point inside the material we may require to specify just a number
(magnitude of that property like temperature or density), a magnitude and direction (i.e. 3
numbers in 3D like for electric field or pyroelectric coefficient) or even more numbers. The
number of values required is given by 3n in 3D (2n in 2D); where ‘n’ is called the rank of the
tensor.
 To understand why tensors are required let us consider a force ‘F’ (with a F) applied along
the x-axis and ask the question*: “what will happen to the body?”.
 Clearly the information available is insufficient to answer unequivocally. If the forces are
applied as in Fig.1b the body will elongate, while if applied as in Fig.1c the body will shear.
 Hence, we need to specify the plane on which the force is acting. In case in Fig.1b the +F
force is acting on the +x-plane along the +x direction this is written as Fxx. In Fig.1c the
force +F is acting along the +y-plane along the +x direction written as Fyx.
 This implies (in this example of mechanical deformation) two directions are required to specify the force
(and hence determine its effect):
(i) the direction of the plane normal and (ii) the direction of the force.
 A combination of these forces Fxx & Fxy could also be acting on the body.
Fig.1a
F
F
F
Fig.1b
F
Fxx
Note. For tension/compression: +F on +x
face is positive and similarly, F force on x
plane is also positive. For shear: shear
causing clockwise rotation is positive.
* For now we will assume that other forces are present to give us force and moment balance (i.e. equilibrium condition).
F
Fig.1c
F
Fyx
Tensors of various ranks
 Tensors can be used to describe: (i) fields or (ii) properties.
 These tensors can belong to various ranks: zeroth rank, first rank, second rank, etc.
  E.g. Temperature field is a scalar field where each point in space is described by one
number the ‘T’ at that point (T(x,y,z)). Scalar fields are tensor fields of rank-0. On the other
hand some fields require more numbers to be specified at each point in space.
 Electric field (polar vector) and Magnetic field (axial vector) require three numbers (in 3D)
to be specified at each point. These 3 numbers are the components along the coordinate axes
and give the direction and magnitude of the vector. Such a field is a vector field. Vectors are
tensors of rank-1.
 Some other fields require more numbers to be specified. E.g. to describe the state of stress
at a point, we need 9 numbers (in 3D) in general (stress being a symmetric tensor we actually
need only 6 numbers). Stress is a tensor of rank-2.
 The number of values required to be specified can be thought of as the number of directions
involved. This is specified by adding subscripts to the parameter.
 A scalar involves zero directions and does not have a subscript (e.g. T)*.This implies a
scalar quantity is the same in all coordinate axes.
 A vector involves one direction and has a single subscript (Ei , with i= 1,2,3 in 3D, i.e.
(E1, E2, E3) or (Ex, Ey, Ez)).
 A tensor (rank-2) has two directions involved and has two subscripts (ij , with i,j = 1,2,3
in 3D).
* Note: only one number needed at each point. If there is a to point to point variation of temperature a set of T have to be specified for each
(x,y,z) which gives rise to the temperature field.
Rank
n (2D)
n (3D)
Field Tensors
0
1
1
Temperature
1
2
3
2
4
9
Electric field, Magnetic
field
Stress field, Strain field
3
8
4
16
Property tensors
Polar
Axial
Specific heat
Rotatory power
Pyroelectricity
Pyromagnetism
Thermal expansion
Magnetoelectricity
27
Piezoelectricity
Piezomagnetism
81
Elastic Stiffness
Piezogyrotropy
n- How many numbers need to be specified
Vectors and their transformations
 The direction and magnitude of a vector (vector is a first rank tensor) is same in any
coordinate system. Let us consider a vector P in (x,y) coordinate axes with intercepts (Px,
Py). Let us consider rotation of axes by an angle  (using z-axis) as in the figure below.
 The new intercepts along the coordinate axes (Px’ ,Py’) can be related to the old ones (x,y) by
a transformation matrix (aij). Needless to say the magnitude and direction of the vector
remains unchanged during the rotation of the coordinate system only the intercepts (x,y)
along the coordinate axes change. The aij are nothing but the direction cosines between the
new and the old axes.
 For a certain specific orientation (=0) the component along y-axis (or for that matter the x-axis) will
be zero.
 Px '   cos sin   Px 
P  
 P 

sin

cos

 y 
 y'  
Px '  Cos .Px  Sin Py  a11Px  a12 Py
Py '   Sin .Px  Cos Py  a21Px  a22 Py
Cos(90   ) 
 a11 a12   Cos
aij  


a
a
Cos
(90


)
Cos

 21 22  

These equations can be written as:
P
 Px '   a11 a12   Px 
P  
 P 
a
a
 y '   21 22   y 
Using
Einstein’s
summation
convention
Pi '  aij Pj
Pi '  aij Pj
In this equation ‘i’ is free index and ‘j’ is the dummy index (over which the sum is performed)
i=1,2 & j = 1,2 (here the ‘=‘ means ‘takes values of’ ). ‘x’, ‘y’ can also b replaced with indices ‘1’, ‘2’.
In 3D the aij will be a 3×3 matrix and the equation using direction cosines (as in table below) can be written as:
 Px '   a11 a12
P   a
 y '   21 a22
 P  a
 z '   31 a32
a13   Px  The condensed equation (in 3D)
a23   Py  still remains the same except
 that i = 1,2,3 & j = 1,2,3 Pi  aij Pj
a33 
P
 z 
Direction cosines (3D)
x1'
x2'
x3'
x1
a11
a12
a13
x2
a21
a22
a23
Sometimes we need to go from the old to the new coordinate system*. x3
In this case the equations turn out to be:
a31
a32
a33
The reverse transformation
P1  a11P1'  a21P2'  a31P3'
P2  a12 P1'  a22 P2'  a32 P3'
Pi  a ji Pj'
P3  a13P1'  a23P2'  a33P3'
* Hey! Can’t I just call the new old and the old new?
Note the position of the dummy index now (in aji)
Transformation of Tensors
 As mentioned before there are two kinds of tensors (here we are keeping our focus on 2nd and higher rank
tensors): (i) field tensors and (ii) property tensors. Many material/physical properties have a
tensorial nature (i.e. are tensors).
 The effect of symmetry on physical properties can be determined by how the tensor
transforms under a symmetry operation.
 The magnitude of a property in any arbitrary direction can be evaluated by transforming the
tensor.
 As expected, under the action of a symmetry operator of a crystal the Tensor properties do
not change.
 Scalar such as Temperature is same in all coordinates. So, if the temperature is T at a point in
space, it is still same whatever coordinates we choose.
 We have already seen how vectors transform; now let us see how 2nd rank tensors transform
(due to coordinate transformation). Let us consider a vector (pi) related to another vector (qj)
via a tensor (Tij). [p & q could be fields as we shall see later]. Let p,q be polar vectors for now.
p1  T11q1  T12q2  T13q3
p2  T21q1  T22q2  T23q3
p3  T31q1  T32q2  T33q3
In matrix
form
 p1  T11 T12 T13  q1 
 p   T T T   q 
 2   21 22 23   2 
 p  T T T   q 
 3   31 32 33  3 
Using
Einstein
summation
convention
pi  Tij q j
‘i’ is free index and ‘j’ is the dummy index (over which the sum is performed)
Continued…
We have
pi  Tij q j
(1)
The vector pi (or pk) transforms to the new coordinate system as:
In terms of the alternate indices (1) can be written as:
pi'  aik pk
pk  Tkl ql
(2)
(3)
The vector qi (or qk) transforms to the old coordinate system from new as:
ql  a jl q'j
(4)
Putting together equations (2,3,4):
p i'  aik  pk   aik Tkl ql   aik Tkl  ql   aikTkl  a jl q'j    aik a jlTkl  q'j
That is:
p i'   aik a jlTkl  q'j
Now this is nothing but the relation between p and q in the new coordinate system (the ‘’’
system). This implies that the quantity in the brackets must be Tij’  The transformation matrix
in the primed coordinate system.
Tij   aik a jlTkl 
‘i,j’ are free indices and ‘k,l’ are dummy indices (over which the sum is performed)
Alternate Matrix method for Tensor Transformation
We have noted that we can transform 2nd rank tensors using the relation
Tij   aik a jlTkl 
A alternate matrix form of this equation is also very useful, which can be derived as below.
The relations we have already seen are:
Vectors from old to
new coordinate system
 p   a  p 
1

q

a
q
      q    a   q
One vector to another in old
coordinate system
One vector to another in new
coordinate system
 p   T  q 
 For a orthogonal matrix the transpose is the
inverse: (a)T = (a)1
 p  T  q
Next we substitute one by one these relations to identify [T’]
 p   a  p    a T  q    a T  a   q   a T  a   q  T  q
1
T    a T  a 
T
T
Noting that:
 a T  a 
T
 

  a  T  a 
T
We get:
T    a  T   a 
T
Notes on transformation matrices
 The new set of axes is related to the old set by three direction cosines (3 nos.). This implies that
not all the 9 terms in the transformation matrices (aij) are independent and only 3 are
independent. This implies that there must be 6 relations between the (aij).
 As each row in the (aij) matrix represents a direction of a vector with respect to 3 orthogonal
axes (i.e as direction cosines), we have three relations of the type: Cos2α+ Cos2β+ Cos2 = 1.
a112  a122  a132  1
2
2
2
2
2
2
 a32
 a33
1
a21
 a22
 a23
 1 a31
 As any row represents one of the orthogonal directions, product of any two rows is zero.
a11a21  a12a22  a13a23  0 a11a31  a12a32  a13a33  0
a21a31  a22a32  a23a33  0
Determinant of the transformation matrices
a11 a12
aij  a21 a22
a31 a32
a13
a23
a33
For a transformation that leaves the handedness of
+1 an axis unchanged (e.g. a rotation)
1 For a transformation that changes the handedness
of an axis (e.g. a inversion or mirror)
 The determinant of any orthogonal matrix is either +1 or −1.
 All the matrices associated with symmetry operations (rotations, inversion, mirror) are
orthogonal.
Transformation of Tensors of all ranks (polar)
 We can generalize the transformation law to any rank tensor as in the table below.
 These transformation laws serve as the very definition of these tensors.
 If these tensors represent physical quantities, it becomes clear that the components of the
tensor change (from one coordinate system to another) by the physical quantity itself
remains unchanged. E.g. the electric field vector may be represented by a sent of 3 different
numbers in two different coordinate systems, however (as obvious) the electric field strength
itself remains the same.
 Confusion may arise in the case of 2nd rank tensors in that they resemble a transformation
matrix (both are specified by 9 numbers and can be written as a 3×3 matrix). However, they
represent different quantities. Tensors (2nd rank) can be transformed to different axes using
these transformation matrices; but (clearly) the transformation matrices themselves cannot be
transformed (from one axes set to another).
Table-T
Rank
0 (scalar)
1 (vector)
Values required
2D
3D
Transformation law
Old  New
New  Old
' = 
 = '
1
1
2
3
pi   aij  p j
pi   a ji  pj
2 (*)
4
9
Tij   aik a jl  Tkl
Tij   aki alj  Tkl
3
8
27
Tijk   ail a jm akn  Tlmn

Tijk   ali amj ank  Tlmn
4
16
81
   aima jn ako alp  Tmnop
Tijkl

Tijkl   ami anj aok a pl  Tmnop
* Causally called a tensor (i.e.
when the word tensor is used
without specifying the rank,
this implies a rank-2 tensor)
Polar and axial tensor properties of rank 0, 1, 2, 3 and 4
Symmetry of property & symmetry of a crystal
 In a amorphous material there is no symmetry at the atomic level. This implies that there is
no preferred direction in a glass  i.e. glasses are isotropic* (any property measured along
any direction will have the same value).
 On the other hand crystals can be anisotropic (i.e. properties can be direction dependent).
 It is perhaps obvious that the symmetry of a property should have$ the symmetry of the
crystal ($as we shall see soon this is actually ‘at least the symmetry of the crystal’). E.g. for a crystal with 4mm
symmetry a property measured along +x will have the same value as that along +y. E.g. if a
field is applied along +x and the current is measured along +x (Jx) this will be equal to the
current measured along +y (if field is applied along +y).
 We expect from 4-fold symmetry and inversion centre (mirrors) present:
J+x = J+y = Jx = Jy
Ey
Jy
Jx
Ex
* It is important to note that glasses with no symmetry at the atomic level have the highest
symmetry with respect to properties!
Neumann principle
 We have already noted that the symmetry of a property should be equal to that of the crystal.
 However, a property may have higher symmetry. So the correct statement is:
the symmetry of a property can be equal to that or greater than that of the crystal.
Other statements of the Neumann principle.
 Symmetry elements of a physical property of a crystal must include all the symmetry
elements of its point group (all its rotational axes, mirror planes, etc.).
Examples of a property having a higher symmetry (than the point group symmetry). We will
understand the details soon.
 All properties that can be represented by tensors of rank up to 2 are isotropic for cubic
crystals. Electrical conductivity of cubic crystals is isotropic.
How to use Symmetry & Neumann principle to determine the number of independent constants
in a property tensor?
1) Start with the property tensor (say in matrix form).
2) Reduce the number of independent components based on any constrains out the symmetry of
the crystal (e.g. symmetry of the stimulus, response, energy considerations, etc.). List out the
independent components at the end of all these considerations.
3) Use the appropriate transformation law based on the rank of the tensor (Table-T). Use
alternate matrix based transformation rules where applicable (and easier).
4) Apply the appropriate transformation matrices corresponding to the symmetry operators of
the crystal (only the bare minimum symmetry operators need to be used (the generator matrices)). In some cases where
direct application of transformation principles are possible, the same can be used (e.g. a 3fold along [111] in cubic crystals takes xy, yz & zx).
5) After the application of each transformation, apply Neumann’s principle. I.e. the transformed
components must be equal to the original components. Based on this determined the ‘equal’
components of the property tensor and the zero components.
These steps will become clear once we take up some actual solved examples in this chapter.
How to understand anisotropy?
 Let us start with a toy model to ‘visualize’ anisotropy in crystals. If the glass ball is pulled as
shown then it will not move along the direction of pull, but will move along the direction as
marked. I.e. even if the stimulus is along ‘P’, the response is along ‘A’.
P
Tension
Shear
A
 We can think of two alternate ways to understand anisotropy:
 The direction of the response is different from the direction of the stimulus (if electric
field is applied along ‘S’ direction of a crystal the current may flow along ‘R’).
 The response may involve other terms than the ‘natural’ expectation due to the stimulus
(e.g. if a crystal is stretched along ‘F’, it may shear in addition to being elongated).
Funda Check
What can happen if we pull, bend or twist an anisotropic crystal?
 If we pull the crystal it may elongate and shear.
 If we bend the crystal using pure bending moments, it may bend and twist.
 If we twist the crystal using pure torsional moments, it may twist and bend.
 The words in red are not the natural responses to the applied load.
 In an isotropic material only the green responses will exist.
How is that a block of Copper is isotropic (say with respect to its elastic properties)?
Origin of isotropy at the level of the microstructure and partial anisotropy in crystalline materials
 At the level of the crystal structure (in a single crystal), a given property may be isotropic
due to Neumann’s principle (e.g. cubic crystals like NaCl). However, isotropy may also arise
due to spatial averaging of properties at the level of the microstructure.
 A single crystal of Cu will be anisotropic with respect to its elastic properties (we will see
later that 3 independent elastic constants are required: E11, E12, E44).
 However, a block of Cu is polycrystalline with multiple grains oriented randomly: thus there
is no preferred direction for the Cu block and its elastic properties are isotropic* (we require
only two independent moduli: E (Young’s modulus) and  (Poisson’s ratio) to describe the its
elastic response).
 In a wire of Cu, the grains may not be oriented randomly and such a material is said to
possess crystallographic texture. I.e. in a textured material there is (some degree of)
preferential orientation of grains. Texture can develop during rolling, extrusion, wiredrawing, thin film growth, etc.
 In textured materials, the anisotropy present at the level of the single crystal is partially
recovered.
Atomic structure level
Single crystals
Origin of Isotropy
Due to Neumann’s principle
* Assuming that the block has many many grains
and hence all orientations are represented.
Microstructure level
Also for textured materials
Stimulus and Response

We can apply various kinds of stimuli on a material (like Force, Electric Field,
Magnetic Field, Temperature difference, etc.). These fields may give rise to natural
responses (like elongation, electric current or polarization, magnetization, heat flow, etc.).
 However, we may also get responses which are usually associated with other kinds of
stimuli. E.g. the application of pressure may lead to the polarization of the crystal (Piezoelectric effect) or the application of a magnetic field may lead to strain in the material
(magnetostriction).
 A material/physical property connects the stimulus to the response.
 The stimulus response plot may be linear, non-linear or may even show hysteresis.
(As stated before)
Crystallographic axes vs Property Axes
 The crystallographic axis (a,b,c) may or may NOT coincide with the property axis (Z1, Z2,
Z3). The property axes are always orthogonal, while the crystallographic axes may or may
not be orthogonal. The standard settings are as in the table below.
Property axis
Z1
Z2
Z3
Cubic
a
b
c
Hexagonal
a
 (Z1, Z3)
c (6-fold)
Tetragonal
a
b
c (4-fold)
Trigonal
(hexagonal setting)
a
 (Z1, Z3)
c
(3-fold, [001])
Orthorhombic
(c < a < b)
a
b
c
Monoclinic
 (Z2, Z3)
b [010]
2-fold or  m
c
Triclinic
 (Z2, Z3)
On (010)
 (010)
c
Electrical Properties
 The response of an electric field on a material can be: (i) motion of charges (electrons or
ions), (ii) polarization. In metals electrons flow on the application of an electric field, while
in materials like solid electrolytes ions carry the current. Materials which lack free charges
are called di-electrics (should have been called dia-electrics, which would have made it
consistent with dia-magnetics). In dielectrics, the internal field weakly opposes the external
field (like in diamagnetic materials the internal magnetic field weakly opposes the external
field).
 In some crystals the centre of mass of the positive charges does not coincide with the centre
of mass of negative charges, leading to a net electric dipole. These crystals are spontaneously
polarized in the absence of an external field  polar materials. Such crystals lack a centre of
symmetry (inversion centre).
 Polar materials come in two types: (i) those in which the reversal in field direction can
reverse the direction of spontaneous polarization (called Ferro-electrics) and (ii) those in
which this does not happen. Ferroelectric materials show a hysteresis in the electric fieldpolarization plot. Materials which show hyteresis in stimulus-response plots are called
Ferroics (in general).
 In certain crystals pressure (mechanical stress) can lead polarization (or change in
polarization if the crystal already has spontaneous polarization). These materials are termed
as Piezoelectric. It is also possible that electric field can lead to mechanical strain in the
material this is termed as Inverse Piezoelectric effect.
Symmetry and Pyroelectric, Piezoelectric and Ferroelectric Effects
32 point groups
Crystal Symmetry Point Groups
21 point groups
Noncentrosymmetric Groups
1 point group 432
Possessing
Other Symmetry
Elements (Nonpolar)
Do not show Pyro-, Piezoor Ferroelectric effects
11 point groups
Centrosymmetric Groups
20 point groups
Piezoelectric Effect
(Polarized under
Mechanical Stress)
1, 2, m, 222, mm2, 4, 4, 422,
4mm, 42m, 3, 32, 3m, 6, 6,
622, 6mm, 62m, 23, 43m
Out of the 20 PG 10 show pyroelectricity
Polarized in the absence
of an electric field
Polar crystals
10 point groups
Pyroelectric Effect
(Spontaneously Polarized)
1, 2, m, mm2, 3, 3m, 4, 4mm, 6, 6mm
Out of the 10 PG a subset show Ferroelectric effect
Polar crystals
Subgroup of the point groups
Ferroelectric Effect
(Spontaneously Polarized
With Reversible Polarization)
Pyroelectricity
 On heating/cooling in some materials (like tourmaline*, gallium nitride (GaN) based materials,
caesium nitrate (CsNO3), Lithium tantalate (LiTaO3, LiNbO3)) voltage develops, this phenomenon
is called pyroelectricity. On heating/cooling charge develops on the opposite faces of crystals
lacking centre of inversion.
 Heating leads to shift in atomic positions so as to lead to a voltage.
 All pyroelectric materials are also piezoelectric, but the vice-versa is not true.
 The temperature change (T) is related to the polarization (P) by:
Pi  pi T
 Pi → polarization vector [C/m2]
 pi → pyro-electric coefficient (polar first rank tensor/vector property)
 On heating/cooling in some materials (like tourmaline, gallium nitride (GaN), caesium nitrate
(CsNO3), Lithium tantalate (LiTaO3)) voltage develops, this phenomenon is called
pyroelectricity. On heating/cooling charge develops on the opposite faces of crystals lacking
centre of inversion.
 As temperature change (stimulus) does NOT have a directionality, the response
(polarization) must correspond to the symmetry of the crystal.
 All pyroelectric materials are also piezoelectric, but the vice-versa is not true.
 The temperature change (T) is related to the polarization (P) by:
'
 Pi is a polar vector, which transforms as: Pi  aij Pj
* Common black tourmaline has a nominal formula NaFe2+3Al6Si6O18(BO3)3(OH)4
0

0

1 0 
0 1
The coefficients after the application of the inversion operator is:
 p1'   1 0 0   p1    p1 
 ' 
  

 p 2    0 1 0   p 2     p 2 
 p3'   0 0 1 p3    p3 



 

 p1'    p1 
 ' 

 p2     p2 
 p3'    p3 

 
 If a component of a tensor changes sign after a
transformation of coordinates, the component must
vanish (= 0) for that crystal in accordance with
Neumann’s principle.
By Neumann’s principle:  p1'   p1    p1 
 '   

 p2    p2     p2 
 p3'   p3    p3 


 

 p1   0 
 p   0
 2  
 p   0
 3  
Point groups with inversion symmetry not showing pyroelectricity
 Crystals with centre of symmetry do not show pyroelectric effect. Let the property axes be
the orthogonal set (Z1, Z2, Z3) and let the components of the pyroelectric coefficient be: (p1,
p2, p3). The matrix operator for centre of inversion is:  1 0 0 
Crystal
System
Group with
Centre of
symmetry (i)
Cubic
(2)
4 2
3
m m
2
3
m
Hexagonal
(2)
6 2 2
m m m
6
m
Trigonal
(2)
3
2
m
3
Tetragonal
(2)
4 2 2
m m m
4
m
Orthorhombic
(1)
2 2 2
m m m
Monoclinic
(1)
2
m
Triclinic (1)
1
This implies that in all crystals with centre of symmetry pyroelectricity is absent (i.e. if we heat
these type of crystals no charge/potential will develop):
(i) 11 of the 32 point groups &
(ii) 2 out of 7 Curie groups (/m m, m).
 We have noted that not all crystals which lack an inversion centre are pyroelectric. This
implies that not all piezo-electric crystals are pyroelectric. Quartz is a well known
piezoelectric crystal, which is not pyroelectric. Quartz has a point group symmetry 32 (2 ||
Z1 & 3 || Z3). This implies that the pyroelectric coefficients will remain the same under these
transformations. Note: all cubic crystals have 2-fold || <001> and 3-fold || <111>.
3 2 0   p1    p1 2  3 p2 2 
 p1'   1 2
  

 ' 
 3 p1
p2
3-fold along Z3
2 
2
 p2     3 2  1 2 0   p 2   

 p3'   0
0
1   p3  
p3

  



By Neumann’s principle:  p1    p1 2  3 p2 2 
3 p1 = p2 Both these conditions
 p     3 p1  p2  
are satisfied only
2
2
 2 
p
=
3
p
1
2
if p1 = p2 = 0

 p  
p3

 3 

Now we apply the other symmetry operation (2-fold along Z1 (after the 3-fold operation)
2-fold along Z1
 p1'   1 0 0   0   0 
 ' 
  

 p2    0  1 0   0    0 
 p3'   0 0 1 p3    p3 



 
 p1   0 
p  0 

By Neumann’s principle:  2  
 p  p 
 3  3


 p1   0 
p  0 
 2  
p  p 
 3  3
 p1   0 
 p   0
 2  
 p   0
 3  
 It is to be noted that though we are deriving the effect of symmetry on the pyroelectric
coefficients, it is applicable to all 1st order tensor (vector) properties (as there is nothing specific to
pyroelectricity in the matrix operations).
 Now let us look at the tourmaline crystal, which addition to being pyroelectric is
piezoelectric. Tourmaline has a point group symmetry 3m (m  Z1 & 3 || Z3). This implies
that the pyroelectric coefficients will remain the same under these transformations.
m  Z1
 p1'   1 0 0   p1    p1 
 ' 
  

 p2    0 1 0   p2    p2 
 p3'   0 0 1  p3   p3 



 
By Neumann’s principle:  p1    p1 
 p2    p 2 
p   p 
 3  3 

 p1   0 
p p 
 2  2
p  p 
 3  3
3 2 0   0   3 p2 2 
 p1'    1 2
  

 ' 
p2
p


3
2

1
2
0
p




3-fold along Z3
2 
 2
 2 


'
 p3   0
0
1   p3   p3 
  

 p1   0 
  p2    0 
Since p3 survives tourmaline is pyroelectric
p  p 
 3  3

3 p2/2 = 0
p2 =  p2/2
 p2 = 0
Polar axis
 The direction of polarization (polar axis) is determined by the symmetry of the crystal. If the
polarization can be visualized as an arrow mark (), the ends of the arrow cannot be related by
any symmetry operator of the crystal. The disawllowed operations are as below.
 As expected pyroelectricity always occurs along a polar axis, this does
not imply that all polar axis will show pyroelectricity. E.g. Z1 axis of
quartz (32 symmetry) is a polar axis (along ‘a’ direction in hexagonal
setting of the trigonal crystal), however, pyroelectric charges do not
accumulate along this direction. This happens due to the presence of the
3-fold orthogonal to the Z1 (which implies that there will be three
equivalent directions 120 apart which will lead to zero net
polarization).
How to locate the Polar axis?
 If vectors are drawn from the origin to each
of the equivalent points and the resultant
obtained, this will be along the polar
direction. Examples are shown for point
groups m and 2.
 If the vectoral sum leads to a zero
resultant then there will be no net
polarization and hence no polar
direction. E.g. for the point group 2/m.
Pyroelectricity in cubic materials
 All cubic crystals have 2-fold || <001> and 3-fold || <111>. 3-fold along <111> will take
xy, yz & zx.
3-fold along [111]
 p1'   0 1 0   p1   p2 
 ' 
   
 p2    0 0 1   p2    p3 
 p3'   1 0 0  p3   p1 



 
By Neumann’s principle:  p1   p2 
 p2    p3 
p  p 
 3  1

 p1   p1 
p p 
 2   1
p  p 
 3  1
Now we apply the other symmetry operation (2-fold along Z1 (after the 3-fold operation)
2-fold along Z1
 p1'   1 0 0   p1    p1 
 ' 
  

 p2    0 1 0   p1     p1 
 p3'   0 0 1  p1   p1 



 
By Neumann’s principle:  p1   0 
p  0
 2  
 p   0
 3  
 It is to be noted that though we are deriving the effect of symmetry on the pyroelectric
coefficients, it is applicable to all 1st order tensor (vector) properties (as there is nothing specific to
pyroelectricity in the matrix operations). In cubic crystals first rank tensor properties vanish.
Pyroelectricity in textured materials
 Suppose we are not considering a single crystal but a material with texture with m
symmetry (a strong electric field along Z3, which aligns domains parallel to Z3). Will
pyrolectricity survive in the textured material?
 || Z3
 p1'   Cos
 ' 
 p2     Sin
 p3'   0
 
Sin
Cos
0
0   p1   p1 Cos  p2 Sin 
0   p2     p1 Sin  p2 Cos 
 

1 
p3
  p3  

 p Cos  p2 Sin   p1  If equality in row-1 and
By Neumann’s principle:  1
p 
p
Sin


p
Cos

1
2
  2  row-2 are to be applicable


  p  for all  values
p3

  3
m  Z1


 p1'   1 0 0   0   0 
 ' 
 0    0 
p

0
1
0
2
  
   
 p3'   0 0 1  p3   p3 



 
 p1   0 
p  0 
 2  
p  p 
 3  3
Since p3 survives the textured material with m symmetry is pyroelectric
 p1   0 
p  0 
 2  
p  p 
 3  3
Pyroelectric crystal classes
 We have noted that there are 10 points groups, which show pyroelectricity. Two Curie groups
also show pyroelectricity. These are listed below. The number of independent measurements
required is the number of non-zero coefficients in each case.
Point groups
Curie Groups
Pyroelectric coefficients
1
 p1 
p 
 2
p 
 3
(3 Nos.)
2
0
p 
 2
0
 
(1 No)
m
 p1 
0
 
p 
 3
(2 Nos.)
mm2 (OR),
3, 3m (Trigonal),
4, 4mm (Tetragonal)
6, 6mm (Hexagonal)
0
0
 
p 
 3
(1 No)
, m
Number of
independent
values required
Ferroelectricity
With special reference to Perovskites
 BaTiO3 (a model material w.r.t to ferroelectricity) has a cubic structure with centre of
symmetry above 120C and hence has no permanent dipole moment. Below 120C the
crystal structure is tetragonal, with the ‘Ti’ shifted from the centroid of the octahedra formed
by ‘O’. This is an example of a Perovskite structure, which have a general formula ABO3
(other examples are: SrTiO3, KTaO3, SrSnO3, and many more).
 In fact BaTiO3 [(Ba+2Ti+4) (O2)3] can exist in hexagonal, cubic, tetragonal, orthorhombic,
and rhombohedral crystal structures. All of these phases exhibit the ferroelectric effect
except the cubic phase (with an inversion centre).
 Not all Perovskites are ferroelectric, SrTiO3 (at RT) with centre of symmetry is cubic (cP5,
Pm3m) and is not a ferroelectric material. In BaTiO3 the Ti+4 ion is slightly smaller than the
space given by the octahedral position and hence the ion slightly shifts along one of the
<001> directions; thus lowering the symmetry of the crystal to tetragonal. This further
introduces a permanent dipole to the structure. Note: SrTiO3 becomes tetragonal at low T (below 168C).
 Regions within the material where the electric dipoles are parallely oriented (say along
[001]) form a ferroelectric domain. In neighbouring domains the dipole may be oriented
along one of the other members of the <001> family (e.g. [010]).
 Different Perovskites can exhibit a variety of properties like they can be: dielectric (CaTiO3),
ferroelectric (BaTiO3), piezoelectric (Pb(Zr,Ti)O3), semiconducting ((Ba,La)TiO3),
superconducting, etc. (they can even show GMR effect). It is to be noted that some of these compounds have
alloying in one of the sublattices and some depend on off-stoichiometry for their properties.
BaTiO3
 The structure below 120 is tetragonal (next slide for figures and data) with the Ti+4 ion
slightly displaced towards one of the <001> directions.
BaTiO3
Ti not at the centre of
the cell (slightly
displaced up)
O
Ti
Ba
BaTiO3
Lattice parameter(s)
a = 3.99 Å, c = 4.033 Å
Space Group
P4mm (99)
Strukturbericht notation
Pearson symbol
Other examples with this
structure
tP5
Wyckoff
position
Site
Symmetry
x
y
z
Occ
Ba
1a
4mm
0
0
0
1
Ti
1b
4mm
½
½
0.515
1
O1
1b
4mm
½
½
0.025
1
O2
2c
2mm.
0
½
0.483
1
The displacement of Ti
gives BaTiO3 a net
electric dipole
SrTiO3
Ti not at the centre of the
cell (unlike in BaTiO3)
O
Ti
Sr
SrTiO3
Lattice parameter(s)
a = 3.90 Å
Space Group
Pm3m (221)
Strukturbericht notation
Pearson symbol
Other examples with this
structure
cP5
Wyckoff
position
Site
Symmetry
x
y
z
Occ
Ti
1a
m3m
0
0
0
1
Sr
1b
m3m
½
½
½
1
O
3d
4/mm.m
½
0
0
1
Piezoelectricity
Thermal Expansion
 When a material is heated it usually expands* (conversely a material will contract on cooling). The strains
introduced are stress free strains (i.e. the body will be stress free in the expanded state, in the
absence of any external constraints).
 Since the stimulus (T) has no directions involved, the response (thermal strains) must
correspond to the symmetry of the crystal. This implies that none of the symmetry elements
of a crystal can be destroyed during free thermal expansion.
 Thermal expansion (strains) can be related to the temperature change by:
2nd rank tensor (field)
 ij   ij T
Scalar
Thermal expansion coefficient 2nd rank tensor (material property)
Written in full
 11 12 13  11 12 13 
  

 .T




21
22
23
21
22
23
 




 31  32  33   31  32 33 
 As strain is a symmetric tensor**, so is the tensor representing the coefficient of thermal
expansion. Note that the strain components are scaled-up versions of the coefficients of thermal expansion (with appropriate change
in units).
* Examples to the contrary exist (in some materials coefficient of thermal expansion may be negative in some temperature regimes).
** In usual materials.
Thermal strains along principal directions
 The directions of principal strains will coincide with the principal directions of αij. Along the
principal directions the strains are:
1  1 T  2   2 T  3   3 T
 This implies that a sphere in the crystal will become an ellipsoid (the strain ellipsoid) on
heating/cooling. This is illustrated in the 2D example below, wherein the new lengths along x1
& x2 are: (1+α1T) & (1+α2T).
heat
 The coefficient of bulk expansion is: (α11 + α22 + α33) = (α1 + α2 + α3) is an invariant.
Isotropic versus anisotropic thermal expansion
 In a isotropic material: α11 = α22 = α33 (& α12 = α23 = α31 = 0)*. To understand the effect of
isotropic expansion (in 2D) let us consider a unit square which is randomly oriented w.r.t to the
principal axes (x1, x2). Before and after thermal expansion (as the green circle goes to the red
circle) the square (yellow) remains a square (pink).
Fig.1: Isotropic expansion
 In the case of an anisotropic crystal (in
2D), α11  α22 & hence 11  22. In such a
crystal on thermal expansion the green
circle becomes the red ellipse and a
square (yellow in Fig.2) becomes a
parallelogram (pink) after thermal
expansion.
* We will soon see “why?”.
Fig.2: Anisotropic crystal
Circular hole
remains circular
Isotropic
Anisotropic
Circular hole
becomes ellipse
Symmetry imposed restrictions on the number of independent coefficients
We have noted before that axes transformations can be carried out by:
T    a  T   a 
T
Let us consider a cubic crystal with (4/m 3 2/m) symmetry (highest available for cubic crystals)
11' 12' 13'   0 1 0  11 12 13   0 1 0 
 '
 1 0 0  
'
' 
  1 0 0 






21
22
23
21
22
23

 



'
'
'
 31  32  33   0 0 1   31  32  33   0 0 1 


4-fold
along Z3
Like in tetragonal crystals
By Neumann’s 11
principle:


 21
 31
 0 1 0   12 11 13    22
  1 0 0    22  21  23    12

 
 0 0 1   

  32 31 33    32
12 13    22  21  23 
 22  23    12 11 13 
32  33    32  31 33 

 21  23 
11 13 
31 33 
0 
11 12 13  11 0

 0 



0
22
23 
11
 21


 31  32  33   0
0  33 
 If a crystal had only a 4-fold axis then, two independent coefficients would survive: α11 &
α33. This implies that all tetragonal crystals have only two independent coefficients (out of a
possible 6).
 It is to be noted that though we are deriving the effect of symmetry on the thermal expansion
coefficients, it is applicable to all 2nd order tensor properties (as there is nothing specific to thermal expansion
coefficient in the matrix operations).
Now we apply the other symmetry operation m  Z3 (after the 4-fold operation)
m  Z3
11' 12' 13'   1 0 0  11 0
0 1 0 0 
 '
0 1 0  0 
'
' 
0 1 0 




0
21
22
23
11

 



'
'
'




 31  32  33   0 0 1  0

0

0
0

1
33





0  11
 1 0 0  11 0
  0 1 0   0 11
0  0

 
 0 0 1  0
0 33   0


11' 12' 13'   0
3-fold || [111]
 '
1
'
' 




21
22
23

 
Which is present in all
'
'
'
 31  32  33   0
cubic crystals


 0 0 1   0 11 0   33
  1 0 0   0
0 11    0

 
 0 1 0  
0   0

  33 0
0
11
0
0 
0 

33 
We got no reduction in the number of
coefficients based on the mirror plane.
Now we try the 3-fold along [111].
0 1  11 0
0 0 1 0
0 0   0 11 0   0 0 1 



1 0   0
0  33   1 0 0 
0
0  By Neumann’s principle:
0
11 0  11 12 13  11 0

 0 



0
22
23 
11
0 11   21


 31  32  33   0
0 11 
 This implies that for (4/m 3 2/m) crystals there is only one independent coefficient.
 We could have change the order of the operations (i.e. we could have operated the 3-fold
first) and this should not make any difference. Also, we could have operated the 3 instead
of the 3-fold (as this is available for the (4/m 3 2/m) crystals.
 Next, we ask the question: “how many independent coefficients will a second rank tensor
property have for cubic crystals”. This includes cubic crystals lower point group symmetry like ’23’.
All cubic crystals have a 3-fold || to <111> and 2-fold || <100>*
Now let us try the other option pointed out before; i.e. operate the 3-fold axis first.
3-fold || [111]
11' 12' 13'   0 0 1  11 12 13   0 1 0 
 '
 1 0 0  
'
' 
0 0 1






22
23 
22
23  
 21
  21


'
'
'




 31  32  33   0 1 0   31  32  33   1 0 0 


 0 0 1  11 12 13  0 1 0   0 0 1  13 11 12   33  31  32 
  1 0 0   21  22  23   0 0 1    1 0 0   23  21  22   13 11 12 



 

 0 1 0  




 1 0 0   0 1 0   33  31  32   23  21  22 

  31 32  33 
By Neumann’s 11 12

principle:
 22
21

 31  32

13   33 31 32 
 23   13 11 12 
33   23  21  22 

α11 = α22 = α33
α12 = α13 = α23
11 12 13  11 12 12 

  





22
23 
11
12 
 21
 12
 31  32  33  12 12 11 
* Note: not all cubic crystals have 4-fold || <100>. In addition, cubic crystal may have 2-fold || <110>
Next we operate the 2-fold || [001]
2-fold || [001]
11' 12' 13'   1 0 0  11 12 12   1 0 0 
 '
 0 1 0  
'
' 
  0 1 0 






21
22
23
12
11
12

 



'
'
'




 31  32  33   0 0 1  12 12 11   0 0 1 


 1 0 0   11 12 12   11
  0 1 0  12 11 12    12

 
 0 0 1  

  12 12 11  12
By Neumann’s 11 12

principle:


12 12
12

11
12 12 
11 12 
12 11 
12   11 12 12 
12    12 11 12 
11   12 12 11 

α12 = 0
0
11 12 13  11 0

 0 


0
21
22
23
11

 

 31  32  33   0
0 11 
 All cubic crystals are isotropic with respect to second rank tensor properties
(require only one independent constant).
Centrosymmetry in 2nd rank tensor properties
 Let us consider a vector stimulus qj connected to a vector response pi via a tensor property
Tij. Let us reverse the direction of the stimulus, this will reverse the direction of the response
without affecting Tij.
 This implies that all second rank tensor properties are centrosymmetrical.
pi  Tij q j
 Tij   q j   Tij  q j     pi 
 p1  T11 T12 T13  q1 
 p   T T
q 
T
2
21
22
23
  
  2  Reversing stimulus
 p  T T T   q 
 3   31 32 33  3 
  p1  T11 T12 T13  q1 
  p   T T
  q 
T
 2   21 22 23   2 
  p  T T T   q 
 3   31 32 33  3 
The effect of crystal symmetry on properties represented by symmetrical second-rank tensors
Optical
classifica
tion
System
Characteristic
symmetry[1]
Nature of
representation quadric
and its orientation
Number of
independent
coefficients
Cubic
Four 3-fold axes
Sphere
1
S
0

 0
0
S
0
0
0

S 
Tetragonal
One 4-fold axis
2
Hexagonal
One 6-fold axis
Trigonal
One 3-fold axis
Quadric of revolution
about the principal
symmetry axis  x3  z 
 S1
0

 0
0
S1
0
0
0 
S3 
Orthorhombic
Three mutually
perpendicular 2-fold
axes; no axes of
higher order
General quadric with axes
 x1, x2 , x3  to the diad
3
 S1
0

 0
0
S2
0
0
0 
S3 
One 2-fold axis
General quadric with one
axis  x2  to the diad
4
 S11
0

 S31
0
S2
0
S31 
0 
S33 
General quadric. No fixed
relation to
crystallographic axes
6
 S11
S
 12
 S31
S12
S22
S23
S31 
S23 
S33 
Isotropic
(anaxial)
Uniaxial
Tensor[2]
Monoclinic
axes  x, y, z 
axis  y 
Biaxial
Triclinic
A centre of symmetry
or no symmetry
[1] Axes of symmetry may be rotation axes or inversion axes.
[2] The setting of the reference axes
x1 , x2 , x3 in column 6 in relation to the crystallographic axes x, y, z and to the symmetry
elements is that shown in column 4.
4th Order Tensor Properties (mechanics)
 The stresses are related to strains by Hooke’s law via the 4th order tensor properties: Stiffness
(Eijkl) and Compliance (Sijkl). Note: Usually ‘C’ is the symbol for stiffness and the symbol for compliance is not ‘C’ and
confusing with ‘Stiffness’!
 ij  Eijkl  kl  ij  Sijkl  kl
 Being 4th order tensors there are 3×3×3×3 (= 81) components (in 3D). Not all these
components are independent. The number of independent components reduces 81 to 36
given that stress and strain are symmetric tensors. For diagonal terms:
11  S111212  S1121 21
 is a symmetric tensor:
 12   21 11  ( S1112  S1121 )12
 S1112 & S1121 always occur together and cannot be experimentally separated and keeping this
in view S1112 is assumed to be equal to S1121. For off-diagonal terms:
12  ( S1233 ) 33  21  ( S2133 ) 33

Eij kl
9×9
If all components are
independent
As 12 = 21
As ‘ij’ & ‘kl’ refer to strain &
stress tensors which have 6
independent components (3D)
S2133  S1233
Eij kl
6×6
Thus not all components are
independent
The single index notation for stress & strain Voigt’s notation
 Sometimes the indices are condensed (from 2 to 1) according to the notation as below. The
two index notation is the matrix notation and has to be converted back to tensor notation so
that transformations can be carried out (as usual for tensors).
 11  12  13   1  6  5 







22
23 
2
4



 33  
 3 
Tensor
Matrix
11 12 13   1




22
23 



 33  
1 6
1 6 6
dW   i d  i   Eij j d  i
2 i 1
2 j 1 i 1
5 

1

2 4
 3 
1
2
Only one symmetric
half of the matrix is written
Matrix
Tensor
 1   E11
  
This will lead to condensed notation  2   E21
version of the stress-strain relation  3   E31
 
(matrix form)
 4   E41
 5   E51
E.g. E43  E2333
  
 6   E61
6
2
1
2
E12
E22
E32
E42
E52
E62
E13
E23
E33
E43
E53
E63
E14
E24
E34
E44
E54
E64
E15
E25
E35
E45
E55
E65
E16   1 
E26   2 
 
E36   3 
 
E46   4 
E56   5 
 
E66   6 
 Further reduction in the number of independent moduli is possible using the concept of
strain energy of a hyper-elastic material. If ‘W’ is the strain energy per unit volume:
1 3 3
1 3 3
dW   ij d  ij   Eijkl  kl d  ij
2 i 1 i 1
2 i 1 i 1
 Let us consider the specific case where the only stress applied is 11 (= 1 in condensed
index notation) and the body is in plane strain condition (i.e. 13, 23 & 33 are zero). The
strain energy is given by (Taylor series expansion for small strain in a hyperelastic material):
 
   2



1

W
1

W
W ( ij )  W (0)   
  ij    
  ij kl   ...


 1   ij 
  2   ij  kl 



0


0
i
j
i
j

 

W ( ij )  W (0)   ij   ij   T 3  ... Ignoring higher order terms and observing that the
quantity in blue font is Eijkl
Zero if no
residual strain
Zero if no
residual stress
Noting that the order of differentiation does not matter
  2

  2



1

W
1
1

W
1



W ( ij )   



E








 ijkl  ij kl  2     ij kl  2  Eklij   ij kl
ij kl
 2   ij  kl 
 2
 ij  0


  kl ij ij 0


E   E 
ijkl
klij
Eijkl is a symmetric tensor
Effect of symmetry elements on Eijkl
 We have noted that even for crystals without any symmetry, only 21 of the 81 possible
elastic constants are independent. For crystals with symmetry this number is further reduced.
Let us consider a tetragonal crystal with 4 symmetry.
 The 4-fold will take: (a)(b), (b)(a), (c)(c). Matrix of surviving coefficients is shaded
blue only 7 independent terms survive: E1111, E3333, E1122, E1133, E1313, E1212, E1112.
 1111 1122 1133 1123 1131 1112 

2222 2233 2223 2231 2212 


3333 3323 3331 3312 


2323 2331 2312 


3131 3112 


1212 

 2222 2211 2233 2213 2232

1111 1133 1113 1132


3333 3313 3332

1313 1332


3232


1131=1113=2223=2232=1113
 all these coefficients are zero
2213=1123=1132=2231=2213
 all these coefficients are zero
3331=3332=3323=3313=3331
 all these coefficients are zero
2312=3221=1321=3112=3221
 all these coefficients are zero
2221
1121
3321

1321 
3221

2121 
0
0
1112 
 1111 1122 1133


1111
1133
0
0

111
2




3333
0
0
0
4  fold
Eijkl


1313
0
0



1313
0 


1212 

3312=3321=3312
 all these coefficients are zero
2331=1332=3213=2331
 all these coefficients are zero
Components of Eijkl for cubic crystals
Cubic point groups  23, 43m,
2
4 2
3, 432,
3
m
m m
 Cubic crystals belong to one of the 5 point groups as above. All the point groups have at
least a 3-fold symmetry (along <111>) along with at least a 2-fold along <100>.
  The 2-fold (along <100>) will take: (a)(a), (b)(b), (c)(c).
 The 3-fold (along <111>) will take: (a)(b), (b)(c), (c)(a).
 In the end we have only 3 independent terms: E1111 , E1122 , E2323. In 2-index notation: E11, E12, E44.
 1111 1122 1133 1123 1131 1112 

2222 2233 2223 2231 2212 


3333 3323 3331 3312 


2323 2331 2312 


3131 3112 


1212 

0
0
 1111 1122 1133

2222 2233
0
0


3333
0
0

2323 2331


3131



Action of 2-fold
1112 
2212 
3312 

0 
0 

1212 
 1111 1122 1133 1123 1131 1112 

2222 2233 2223 2231 2212 


3333 3323 3331 3312 


2323 2331 2312 


3131 3112 


1212 

Action of 3-fold
0
0
2223 
 2222 2233 2211

3333 2233
0
0
3323 


1111
0
0
1123 


3131 3112
0 


1212
0 


2323 

We see that some of the terms are equal to one another. We
have already noted that some of these terms are zero.
Hence, the surviving terms (in the symmetric half) are :
Note: the total number of non-zero terms is 12 out of
possible 36 (24 zero terms). But, only 3 are independent.
0
0
0 
 1111 1122 1122

1111 1122
0
0
0 


1111
0
0
0 


2
3
23
0
0



2323
0 


2323 

Pathway for reduction of number of independent Eijkl
81
Due to symmetry of stress
and strain tensors
36
Due to strain energy
considerations
21
Due to symmetry present in
cubic crystals
3
Curie Principle
Click here for introduction to Curie groups
 When certain stimuli lead to certain responses, the symmetry elements of the stimuli should
be seen in the response.
 A crystal subjected to certain stimulus will display only those symmetry elements, which are
common to the crystal without the stimulus and the stimulus without the crystal.
 In terms of groups this can be written as:
R
crystal with stimulus
  G
cystal symmetry
  S
Stimulus

 The symmetry of the crystal in the presence of the stimulus is the intersection
group of the symmetry of the crystal (in the absence of the stimulus) with the
symmetry of the stimulus (in the absence of the crystal).
 R, G, S are point groups (and not space groups).
 Let us consider a cubic crystal with 4/m 3 2/m symmetry, which is loaded along [111]
direction. The symmetry of the stimulus is m (cylindrical symmetry). The symmetry of the
strained crystal is 3m (the common symmetry to both the crystal and the stimulus).
 To illustrate the Curie principle, let us consider the example of Ammonium Dihydrogen
Phosphate (ADP, 42m) subjected to electric filed along the 4 direction of crystal (along
[001], with a pure rotation axis of 2-fold). The stimulus (field) has a symmetry of m. The
resultant symmetry of the crystal in the presence of the electric field (stimulus) is 2mm.
 This can be understood as follows: 2-fold is a subgroup of 4. This is common to that
between 4 and  and survives. The horizontal 2-fold present in 42m is lost as this is not
present in m. The surviving symmetry is 2mm as shown and belongs to the orthorhombic
class (can also be written as mm2).
m
42m
2mm


4
[001]crystal
 Let us now impose the electric field along other directions and see it effect.
42m
[100]crystal
2



[uvw]crystal
42m
 m along a
m || 2
random direction
[110]crystal
42m

m || m


m
1
1
W   1111
2
2W  11  11  ( E11kl kl ) 11  ( E111111  E111212  E1121 21  E1122 22 ) 11
W
 ( E1122 ) 11
Taking partial derivatives (1st wrt to 22 then wrt 11) 2
 22
 2W
2
 E1122
11 22
Now reversing the order of the partial derivatives (1st wrt to 11 then wrt 22 )
2
W
 (2 E111111  E111212  E1121 21  E1122 22 )
11
 2W
2
 E1122
 2211
2
Since the order of differentiation should not matter as far as the energy density (W) is
concerned, (1) = (2). This implies E1122 = E2211.
1
Now we apply m  Z3 (after the 3-fold operation)
m  Z3
11' 12' 13'   1 0 0  11 12 12   1 0 0 
 '
 0 1 0  
'
' 
0 1 0 






22
23 
11
12  
 21
  12


'
'
'



 31  32  33   0 0 1 12 12 11   0 0 1


1 0 0  11 12 12  1 0 0   1 0 0  11 12 12 
0 1 0  12 11 12   0 1 0    0 1 0  12 11 12 









 0 0 1  0 0 1 12 12 11 
0 0 1 12 12 11 
incomplete
m  Z3
 11' 12' 13'   1 0 0   11 12 12   1 0 0 
 '
 0 1 0 
 0 1 0  
'
' 






22
23 
11
12  
 21
  12


'
'
'




  31  32  33   0 0 1  12 12 11  0 0 1




