7 Physical properties
7.4 Third-rank tensors
7.4.1 Piezoelectricity
The definition of the piezoelectric tensor
Arguably one of the most important third-rank tensors for 21st-century applications is thepiezoelectric tensor, dij k. This links an induced polarization to an applied stress (thedirect piezoelectric effect), or an induced strain to an applied electric field (theindirect piezoelectric effect):
Pi =
j k
dij kσj k (7.38)
j k =
k
dij kEi (7.39)
The equivalence of these forms can be obtained by a thermodynamic argument.
We write the internal energy as
dU =
j k
σj kd j k+
i
EidPi (7.40)
The free energy function, which we denote as , can be written as
= U −
j k
σj k j k−
i
EiPi (7.41)
This can be written in differential form as d= −
j k
j kdσj k−
i
PidEi (7.42)
Thus we have the following two differentials:
∂
∂σj k
E
= − j k (7.43)
∂
∂Ei
σ
= −Pi (7.44)
From the second differential we have
− ∂2
∂Ei∂σj k = ∂ j k
∂Ei = ∂Pi
∂σj k = dij k (7.45) Both first differentials relate to the piezoelectric effect, as the indirect and direct effect respectively. This demonstrates that the same coefficient applies to both effects.
Symmetry and piezoelectricity: example of the tetragonal ferroelectric phase of perovskite
Symmetry plays an important role in determining the piezoelectric properties of a material. Consider the direct piezoelectric effect. The simplest example is the role of a centre of symmetry. An applied stress will not remove the centre (stress always has a symmetry corresponding to a centrosymmetric point group), and so application of stress by itself cannot create an electrical polarization in a centrosymmetric crystal. Therefore the first symmetry condition that must be met for a piezoelectric material is that it will not have a centre of symmetry.
There are further symmetry conditions on the type of stress that can induce an electrical polarization, and on the direction of that polarization.
The role of symmetry can be illustrated by the example of the piezoelectric effect given by the ferroelectric tetragonal phase of a perovskite material such as BaTiO3or PbTiO3, which is shown schematically in Fig. 7.9. The symmetry of the crystal is point group 4mm, that is, there is a 4-fold rotation axis along[001], and mirror planes perpendicular to the equivalent100 and 110 directions.
Because of the symmetry of the ferroelectric phase there is already an electrical polarization along[001] in the unstressed crystal, together with an elongation of the crystal in this direction. Application of the stress σ3, that is a force along [001] applied on the (001) face, will reduce the extent of the elongation along the [001] direction. This will squash the TiO6 octahedra and force the Ti4+ cations back towards the centre of the octahedra. Similarly, there will be a compression of the oxygen atoms around the Ba/Pb sites, forcing the cation to move towards the centre of the 12 oxygens. The effects of these induced cation displacements will be to reduce the overall electrical polarization along x3= [001].
Fig. 7.9 Schematic representation of the piezoelectric effect in a ferroelectric per-ovskite of point group symmetry 4mm. [001]
(x3) is vertical, and [100] (x1) is horizontal.
σ1
σ4 σ3
Now consider application of the stress σ1(because of the 4-fold rotational symmetry the same arguments here will also apply to σ2). There is a mirror plane normal to the face on which the force is applied, and the stress is not able to change this symmetry. Therefore there will be no induced electrical polarization along the[100] direction. However, by squashing together the oxygen atoms in the TiO6octahedra, the Ti4+cation will be further displaced along[001], and so an additional contribution to the electrical polarization along[001] will be induced.
We now consider the effects of an applied shear stress. First we consider σ6, which will cause a shear of the a–b plane (not illustrated in Fig. 7.9). This will have the effect of moving two of the oxygens in the TiO6octahedra closer together, and of moving two others further apart. This shear does not destroy the orthogonal mirror planes on the{110} planes, so there can be no polarization in the a–b plane. To first order, the changes in the distances between the two pairs of oxygen atoms in the a–b plane along the110 directions will be of equal size but opposite sign, so there will no net effect on the position of the Ti4+ cation to first order (although there may be a second-order effect that would be represented by a tensor of higher order). Thus σ6is not able to produce an induced electrical polarization. On the other hand σ5, which causes a shear on the a–c plane, removes all mirror planes other than the (010) mirror plane, and therefore a polarization is allowed on a general direction in the a–c plane.
Writing everything out full, we can state the results of this discussion in matrix form:
Piezoelectricity of non-polar systems: example of a tetrahedron
The example of the tetragonal ferroelectric phase of perovskite is a case where the existing polarization of the crystal is changed by the application of stress.
Piezoelectricity is also found in crystals without a centre of symmetry even when there is no polarization in the stress-free crystal. We explore this point by the example of a single tetrahedral group of ions, as illustrated in Fig. 7.10.
In this case we assume that the vertices of the tetrahedra are anions, of charge
−Q, and the centre of the cluster is a cation, of charge +4Q. Note that the point symmetry of the tetrahedral cluster is 43m – there is no centre of symmetry.
We will assume for the sake of the discussion that the anion–cation bond is strong, and remains of constant length when a small stress is applied. This can be achieved if the anion–cation–anion bonds can flex with the applied stress to give an angle different to the ideal tetrahedral angle of 109.47◦. In Fig. 7.10 the axes are defined along the 4 axes, with the central cation at the origin.
We first consider the application of the stress σ1. This has the effect of squashing the tetrahedral cluster along the x1direction. This causes the 3-fold
Fig. 7.10 Schematic representation of the piezoelectric effect in a tetrahedron of initial point group symmetry 43m. The x1axis is hor-izontal, and the x2axis is vertical. The lightly-shaded anions (large spheres) are above the plane of the diagram, and the darkly-shaded anions are below the plane of the diagram; the central cation (small sphere) is in the plane of the diagram.
σ2 σ4
rotational symmetry to be lost along each of the axes, and the 4 rotoinversion axes along both the x2and x3axes are reduced to 2-fold rotation axes along these directions. However, the 4 rotoinversion axis along x1is retained. With the presence of these three orthogonal rotation and rotoinversion axes, there can be no polarization.
Now we consider the application of a shear stress, σ6. This has the effect of moving the top two anions away from each other, and the lower two anions apart. Since the system tries to keep the cation–anion distances constant, the cation is moved upwards. It can be shown (Problem 7.4) that the displacement of the cation along the x3direction, u+(z), is related to the displacements of the anions in the x1− x2(x− y) direction, ±u−(xy), by
|u+(z)| = |u−(xy)|/√
2 (7.47)
where we use the modulus because the different directions of motion all cause the same displacement. The induced polarization is simply equal to 4Qu+(z).
The final result can be summarized by the matrix equation:
Given that a perfect tetrahedral arrangement of cations and anions has a non-zero piezoelectric tensor, we might expect crystals that contain tetrahedral arrangements of atoms to also have non-zero piezoelectric tensor. This applies to the semiconducting cubic ZnS and related materials, where each cation is in a tetrahedral coordination with neighbouring anions (and vice versa) and to systems such as quartz that consist of networks of linked SiO4tetrahedra.
Of course, the arrangements of the tetrahedral groups may give additional symmetry that changes the piezoelectric properties – for example, the crystal structure of the high-temperature cristobalite phase of silica has a centre of symmetry – but the point is that a crystal without a centre of symmetry can have
a piezoelectric effect even without there being a polarization in the stress-free crystal.
7.4.2 Use of Voigt notation for third-rank tensors
We have made use of the Voigt notation in the example above, but to perform operations such as the rotation of axes we need to convert to the long-handed three-index mode. In long hand, we can write the equation for the direct piezoelectric effect as
P1= d111σ11+ d112σ12+ d113σ13 + d121σ21+ d122σ22+ d123σ23
+ d131σ31+ d132σ32+ d133σ33
and similarly for P2and P3. When we take account of the symmetry of the σ tensor by using Voigt notation for this tensor, i.e. σ12= σ21= σ6, etc., we can rewrite this equation as
P1= d111σ1+ (d112+ d121)σ6+ (d113+ d131)σ5
+ d122σ2+ (d123+ d132)σ4+ d133σ3
By comparison with the full Voigt expansion:
P1= d11σ1+ d12σ2+ d13σ3+ d14σ4+ d15σ5+ d16σ6 (7.49) we see that d11 = d111, d12 = d122, and d13 = d133, whereas d14 = 2d123, d15 = 2d113, and d16 = 2d112. Similar results follow from an analysis of P2
and P3. The new factor of 2 is important!
7.4.3 Transformations of third-rank tensors
Transformation equations
Earlier we used physical arguments to obtain the non-zero components of the piezoelectric tensor for a tetragonal crystal and a tetrahedron. The symmetry properties can be obtained by using symmetry transformations, but for this we need to see how to transform a third-rank tensor by a change in the definitions of the axes. This will follow the approach taken in transformation of second-rank tensors. We define a first-second-rank tensor P, a second-second-rank tensor Q, and a third-rank tensor T. These are linked by P = T · Q. Similarly we can define a set of transformed tensors: P= T· Q. Following the earlier procedure, we obtain
P= a · P = a · (T · Q) P= T· Q= T· (a · Q · aT)
⇒ a · T · Q = T· a · Q · aT= T· a · a · Q
⇒ T= a · T · aT· aT= a · a · a · T
(7.50)
where we have made use of the fact that QT= Q and TT= T, for example in These equations can be readily extended for higher-rank tensor transformations.
Effect of symmetry
We now use the transformation equation to demonstrate how symmetry can determine which elements of the piezoelectric tensor are zero. We will consider the effect of 4-fold rotation axis along x3. The transformation matrix for a 90◦ rotation is
First we consider d11and d22. The transformation equations give d111 = a12a12a12d222= d222
We now consider the pairs involving the tensile stresses, using the fact that we expect pairs to be equal by symmetry:
d122 = a12a21a21d211= d211
Now we consider the piezoelectric coefficients involving the shear stresses:
d123 = a12a21a33d213= −d213
This gives the piezoelectric tensor for a material with point group 4 of
d=
0 0 0 d14 d15 0
0 0 0 d15 −d14 0
d31 d31 d33 0 0 0
(7.55)
As a result of symmetry, the number of non-zero coefficients of the piezoelectric coefficient will be reduced to a relatively small number in most crystal classes.