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Elastic Properties of Solids
Chapter 5
Elastic Properties of Solids Chapter Outline 5.1 5.2 5.3 5.4 5.5 5.6 5.7 5.8
Strain Tensor Dilation Stress Tensor Elastic Constants of Solids Elastic Energy Density Elastic Constants in Cubic Solids Elastic Energy Density in Cubic Solids Bulk Modulus in Cubic Solids
93 95 96 96 97 98 102 102
5.9 Elastic Waves in Cubic Solids 5.9.1 Elastic Waves in the [100] Direction 5.9.2 Elastic Waves in the [110] Direction 5.9.3 Elastic Waves in the [111] Direction 5.10 Isotropic Elasticity 5.11 Experimental Measurement of Elastic Constants Suggested Reading
103 105 106 108 110 111 113
The theoretical study of the various properties of crystalline solids requires a knowledge of the crystal potential, the reliable determination of which is difficult. One of the oldest and simplest methods is to view the crystal as an isotropic continuous elastic material with uniform density instead of as a discrete periodic array of atoms. The various properties of the solids, such as elastic constants, lattice vibrations, and thermal properties, can be studied using the continuum elasticity theory of solids. The study of elastic constants is of immense importance because they give information about the nature of binding forces in solids: a central physical quantity in studying various properties of solids. The continuum elasticity theory is valid for low-frequency waves, such as elastic waves with wavelength l > 106 cm or frequency n < 1011 Hz, as these waves are not able to see the atomic structure of a crystalline solid. The present chapter presents the calculation of elastic constants in the isotropic linear elasticity approximation in which Hooke’s law is valid.
5.1
STRAIN TENSOR
Consider a solid with Cartesian coordinate axes having unit vectors ^ia with a ¼ 1,2, and 3. The position vector of a point or a particle in the solid (Fig. 5.1) is defined as X ^i r (5.1) r¼ a a
a
where ra are the Cartesian components of the vector r. The application of a weak external force produces a small deformation in the solid that changes the orientation and magnitude of both the position vector and the unit vectors. The strain produced in the body can be studied in two approaches. In the first approach the unit vectors are kept unchanged and the orientation and magnitude of the Cartesian components of the position vector are changed. In the second approach the Cartesian components of the position vector are kept unchanged and the orientation and magnitude of the unit vectors are changed. It is more convenient to adopt the second approach as is done here. In a strained body, a particle at point r moves to r0 (Fig. 5.1) given by r0 ¼ r + uðrÞ where u(r) is the displacement produced. If ua is the Cartesian component of u(r) along the unit vector ^ia , then X uðrÞ ¼ ua^ia
(5.2)
(5.3)
a
Solid State Physics. https://doi.org/10.1016/B978-0-12-817103-5.00005-0 © 2019 Elsevier Inc. All rights reserved.
93
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Solid State Physics
FIG. 5.1 Coordinate axes in perfect and strained crystals. The position vector r goes to r0 after producing strain in the crystal.
Note that u(r) is not a constant quantity but is a function of the position vector. Therefore, u(r) varies continuously throughout the solid and forms a vector field usually called the strain field or displacement field. The components of 0 the unit vector after displacement, denoted as ^ia , are given as i Xh ^i0 ¼ ∂ ðr + uÞ ¼ ∂ rb^ib + ub^ib a ∂ra ∂ra b (5.4) X ^ ¼ dab + eab ib b
where eab ¼ ∂ub =∂ra
(5.5)
0
The position vector r0 can be written, in terms of ^ia , as r0 ¼
X
0
ra^ia
(5.6)
a
In the above equation the Cartesian components of r0 remain unchanged after deformation. To make the measurement of deformation free from the orientation of the coordinate axes, it is convenient to define a scalar product between the new unit vectors as X 0 0 eag ebg (5.7) gab ¼ ^ia ^ib ¼ dab + eab + eba + g
From the above equation the angle between the two deformed axes is h 0 0 i yab ¼ cos 1 gab = j ^ia k^ib j
(5.8)
Elastic Properties of Solids Chapter
In terms of gab the strain components eab are defined as 1 1 eab ¼ gab dab ¼ eab + eba 2" # 2 ∂u 1 ∂u b + a ¼ 2 ∂ra ∂rb
5
95
(5.9)
In defining the above equation, the second order terms in eab are neglected (the linear elasticity approximation). Eq. (5.9) $ shows that the strain tensor e is symmetric, that is, eab ¼ eba. The diagonal components of the strain tensor from Eq. (5.9) are given as eaa ¼
∂ua ¼ eaa ∂ra
(5.10)
When the direction of the axes is reversed, that is, a goes to a, then under the reversal of the coordinate axes one can write ra ¼ ra , ua ¼ ua From Eqs. (5.9), (5.10) one can write
∂ua ¼ eaa ∂ra " # 1 ∂ub ∂ua + ¼ eab eab ¼ ∂rb 2 ∂ra eaa ¼
(5.11)
(5.12)
(5.13)
Eqs. (5.12), (5.13) show that with the reversal of direction of one of the Cartesian coordinates, the nondiagonal components of the strain tensor change their sign while the diagonal components remain unchanged. Ideally speaking, two types of strain can exist in a solid. The first is hydrostatic pressure in which the volume of the solid $ changes without any change in its shape and is represented by the diagonal components of the strain tensor e . The second is pure shear in which the volume of the body remains unchanged by the deformation and only the shape changes. Pure shear is $ represented by the nondiagonal components of e . But in an actual strain, both the volume and shape of the solid may change. Therefore, the general strain may be represented as the linear combination of pure hydrostatic pressure and pure shear. To do so, the strain components are defined as " # 1 X 1 X egg + dab egg (5.14) eab ¼ eab dab 3 3 g g The first term on the right side is evidently a pure shear as the sum of its diagonal terms is zero, while the second term is a pure hydrostatic pressure. Hydrostatic pressure acts perpendicular to the surface of the solid, while pure shear acts along the surface. Therefore, these two scalar components are independent of each other and the square of the strain component eab can be obtained by the addition of the squares of these two components, that is, " #2 " #2 h i2 1 X 1 X egg + e (5.15) eab ¼ eab dab 3 3 g gg g The strain field, from Eqs. (5.1), (5.2), (5.4), and (5.6), can be written as i Xh ra eab ^ib uðrÞ ¼ r0 r ¼ a, b From Eqs. (5.3), (5.16) the components of the strain field are given by X ub ¼ ra eab
(5.16)
(5.17)
a
5.2
DILATION
Dilation is defined as the fractional change in volume of a solid due to the deformation produced by the application of an external force. It can be expressed in terms of the strain field components eab. Consider a solid in the form of a cube with unit edges ^i1 , ^i2 , and ^i3 . The volume of the solid V is given by V ¼ ^i1 ^i2 ^i3 ¼ 1
(5.18)
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Solid State Physics
0
0
0
After deformation the cube becomes a parallelepiped with edges ^i1 , ^i2 , and ^i3 , having volume V0 given by X 0 0 0 V0 ¼ ^i1 ^i2 ^i3 ¼ 1 + eaa
(5.19)
a
in the linear elasticity approximation. The dilation produced by the deformation becomes dD ¼
V0 V DV X eaa ¼ ¼ V V a
(5.20)
The symmetric strain tensor has six independent elements. For simplicity of notation, it is very convenient to use the Voigt notation for the strain components, according to which the subscripts are defined as 1 11 ¼ xx; 2 22 ¼ yy; 3 33 ¼ zz 4 23 ¼ yz; 5 31 ¼ zx; 6 12 ¼ xy
(5.21)
In the Voigt notation eab can be written as em where m takes the values 1, 2, 3, 4, 5, and 6.
5.3
STRESS TENSOR
There are two types of forces acting on a solid body. The first type is called the body force, which acts throughout the body and exerts influence on the whole of the mass distribution. The inertial and gravitational forces are examples of the body force. The body force is expressed per unit mass or per unit volume. If f(r) is the body force per unit volume, it can be written as X f a^ia (5.22) f ðrÞ ¼ a
Here fa is the a-component of the body force density. The second type is the surface force, which acts on the surface of the body. A surface force is expressed in units of force per unit area and is called stress. The stress is further divided into two categories: stress acting normal to the surface (normal stress) and stress acting along the surface (shear stress). As the normal stress and the shear stress act perpendicularly to each other, therefore, they are orthogonal stresses. Stress is repre$ sented by a tensor s with components sab where the first subscript indicates the direction of the stress and the second subscript the direction of the normal to the surface on which the stress is acting. Fig. 5.2 shows the different components of stress acting on a cubic solid. It can be easily shown that the stress tensor is symmetric, that is, sab ¼ sba
(5.23) $
which is a consequence of the fact that shear stress does not cause angular rotation. The symmetric stress tensor s has six independent elements, which, in the Voigt notation, can be written as sm where m can take values 1, 2, 3, 4, 5, and 6. If a body under stress is in the equilibrium state, its equation of motion is X ∂sab b
∂rb
+ fa ¼ 0
(5.24)
The first term gives the body force per unit volume applied externally on the body and the second term is the internal body force per unit volume.
5.4
ELASTIC CONSTANTS OF SOLIDS
In the linear elasticity approximation, valid for very small deformations only, the strain is linearly proportional to stress and vice versa. In mathematical language, we can write X sab ¼ Cabmn emn (5.25a) m, n X eab ¼ Csabmn smn (5.25b) m, n
Elastic Properties of Solids Chapter
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97
FIG. 5.2 Components sab of the stress tensor in a cubic crystal.
where Cabmn and Csabmn are called the elastic stiffness constants and elastic compliance constants. The quantity Cabmn has dimensions of energy/volume, while Csabmn has dimensions of volume/energy. In the Voigt notation Eqs. (5.25a), (5.25b) become X sm ¼ Cmn en (5.26a) n
em ¼
X
Csmn sn
(5.26b)
n
where m and n have integral values from 1 to 6. The constants Cmn are usually called the elastic constants of the crystalline solid.
5.5
ELASTIC ENERGY DENSITY
If U is the elastic energy density per unit volume, the stress components in terms of it are given by sab ¼
∂U ∂eab
(5.27a)
or dU ¼ sab deab
(5.27b)
The minus sign indicates that U is the work done on the system. In Voigt notation one can write (magnitude) dU ¼ sn den
(5.28)
From the above equation the stress components can be written in terms of energy density as sn ¼
dU den
(5.29)
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The total change in energy density due to all the stress components becomes X dU ¼ sn den
(5.30)
n
Substituting for sn from Eq. (5.26a), one can write dU ¼
X n, m
Therefore, the total energy density becomes
ð U ¼ dU ¼
Cnm em den
(5.31)
1X C e e 2 n, m nm m n
(5.32)
From Eqs. (5.29), (5.32) one can write s1 ¼
6 dU 1X ¼ C11 e1 + ðC + Cn1 Þen de1 2 n¼2 1n
(5.33)
Hence, in the stress-strain relations, the combination 1/2(Cmn + Cnm) appears in all the stress components. It follows that the elastic stiffness constants Cmn are symmetrical, that is, Cmn ¼ Cnm
(5.34)
Eq. (5.34) reduces the independent elastic constants from 36 to 21.
5.6
ELASTIC CONSTANTS IN CUBIC SOLIDS
In the linear elasticity approximation, the stress-strain relation given by Eq. (5.26a) allows us to write the different components of stress as sxx ¼ C11 exx + C12 eyy + C13 ezz + C14 eyz + C15 ezx + C16 exy syy ¼ C21 exx + C22 eyy + C23 ezz + C24 eyz + C25 ezx + C26 exy szz ¼ C31 exx + C32 eyy + C33 ezz + C34 eyz + C35 ezx + C36 exy syz ¼ C41 exx + C42 eyy + C43 ezz + C44 eyz + C45 ezx + C46 exy
(5.35)
szx ¼ C51 exx + C52 eyy + C53 ezz + C54 eyz + C55 ezx + C56 exy sxy ¼ C61 exx + C62 eyy + C63 ezz + C64 eyz + C65 ezx + C66 exy The above equation can be written in matrix form as 1 0 0 sxx C11 C12 Bs C BC C B yy C B 21 22 C B B B szz C B C31 C32 C B B Bs C ¼ BC C B yz C B 41 42 C B B @ szx A @ C51 C52 sxy C61 C62
C13 C14 C15 C16
10
exx
1
B C C23 C24 C25 C26 C C B eyy C CB C C33 C34 C35 C36 C B ezz C CB C B C C43 C44 C45 C46 C C B eyz C CB C C53 C54 C55 C56 A @ ezx A C63 C64 C65 C66
(5.36)
exy
Consider an isotropic solid in which the physical properties do not alter under the symmetry operations of a cubic solid. The number of elastic constants can be reduced by considering the different symmetry relations of the cubic solid. Following are the important symmetry operations in a cubic solid. Fourfold (2p=4) rotation about one of the edges of a cube (Fig. 5.3), which changes the sign of the coordinates as follows (anticlockwise rotation) About z axis : x ! y, y ! x About y axis : z ! x, x ! z About x axis : y ! z, z ! y
(5.37)
Elastic Properties of Solids Chapter
5
99
FIG. 5.3 Twofold and fourfold rotations about the edges of a cube.
Twofold (2p/2) rotation about one of the edges of a cube, which changes the axes as follows (anticlockwise rotation): About z axis : x ! x, y ! y About y axis : z ! z, x ! x About x axis : y ! y, z ! z
(5.38)
Threefold (2 p/3) rotation about the diagonals of a cube (Fig. 5.4). There are four diagonals about which the transformation of the axes is as follows: x!y!z!x x ! z ! y ! x x ! z ! y ! x x ! y ! z ! x
(5.39)
We see from the above symmetry relations that the rotations either interchange the axes or reverse their sign. Let us apply the operation of a twofold rotation to the cubic solids in which the signs of the axes change. If the sign of the y-axis is reversed, then the left-hand side of the first expression of Eq. (5.35) is given as sxx ¼ C11 exx + C12 eyy + C13 ezz + C14 eyz + C15 ezx + C16 exy ¼ C11 exx + C12 eyy + C13 ezz C14 eyz + C15 ezx C16 exy
(5.40)
In the above equation we have used the properties (5.12), (5.13) of eab and we see that all the terms remain unchanged except the 4th and 6th terms of Eq. (5.40). Therefore, sxx is invariant under the transformation y ! y only if C14 ¼ C16 ¼ 0
(5.41)
The symmetry property of the elastic constants yields C41 ¼ C61 ¼ 0
(5.42)
Similarly, if we make the transformation z ! z in the first expression of Eq. (5.35), we get C14 ¼ C41 ¼ C15 ¼ C51 ¼ 0
(5.43)
Applying the transformation x ! x in the second expression of Eq. (5.35), we find syy ¼ C21 exx + C22 eyy + C23 ezz + C24 eyz C25 ezx C26 exy
(5.44)
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Solid State Physics
FIG. 5.4 Threefold rotation about the diagonal of a cubic structure.
and the transformation z ! z in the second expression of Eq. (5.35) yields syy ¼ C21 exx + C22 eyy + C23 ezz C24 eyz C25 ezx + C26 exy
(5.45)
The property of invariance yields, from Eqs. (5.44), (5.45), C24 ¼ C25 ¼ C26 ¼ C42 ¼ C52 ¼ C62 ¼ 0
(5.46)
The transformations x ! x and y ! y, when applied to the third expression of Eq. (5.35), reduce the following elastic constants to zero C34 ¼ C35 ¼ C36 ¼ C43 ¼ C53 ¼ C63 ¼ 0
(5.47)
In the above description we have considered the invariance of only the normal stress components sxx, syy, and szz in cubic crystals. Using Eqs. (5.41)–(5.43), (5.46), and (5.47) in Eq. (5.35), one can write sxx ¼ C11 exx + C12 eyy + C13 ezz syy ¼ C21 exx + C22 eyy + C23 ezz szz ¼ C31 exx + C32 eyy + C33 ezz syz ¼ C44 eyz + C45 ezx + C46 exy
(5.48)
szx ¼ C54 eyz + C55 ezx + C56 exy sxy ¼ C64 eyz + C65 ezx + C66 exy Let us now consider the shear stress defined by the last three expressions of Eq. (5.48). If we apply the transformation x ! x to the fourth expression of Eq. (5.48), the invariance property yields C45 ¼ C46 ¼ C54 ¼ C64 ¼ 0
(5.49)
We have used the fact that syz does not change under the transformation x ! x as it does not involve the x-coordinate. Similarly, the application of transformation y ! y to the fifth expression and of z ! z to the sixth expression of Eq. (5.48) give C54 ¼ C56 ¼ C45 ¼ C65 ¼ 0
(5.50)
C64 ¼ C65 ¼ C46 ¼ C56 ¼ 0
(5.51)
Elastic Properties of Solids Chapter
5
101
Substituting Eqs. (5.49)–(5.51) into Eq. (5.48), one gets sxx ¼ C11 exx + C12 eyy + C13 ezz syy ¼ C21 exx + C22 eyy + C23 ezz szz ¼ C31 exx + C32 eyy + C33 ezz syz ¼ C44 eyz szx ¼ C55 ezx sxy ¼ C66 exy
(5.52)
In the above discussion we have applied a twofold rotation about one of the edges of the cubic structure. Let us apply a fourfold rotation about one of the edges of the cube. In a fourfold rotation either the transformation x ! y, y ! z, z ! x or x ! y, y ! z, z ! x can take place. In both of these transformations, invariance is required. When fourfold rotation is applied about the x-axis, then the allowed transformations are y ! z and z ! y and in these transformations the first expression of Eq. (5.52) changes to (5.53) sxx ¼ C11 exx + C12 ezz + C13 eyy The first expression of Eqs. (5.52), (5.53) must be the same (invariance property) under the transformation, which gives C12 ¼ C13
(5.54)
Similarly, when fourfold rotation is applied to the second expression of Eq. (5.52) about the y-axis and to the third expression of Eq. (5.52) about the z-axis, one gets C21 ¼ C23 ¼ C31 ¼ C32
(5.55)
C12 ¼ C21 ¼ C31 ¼ C32 ¼ C13 ¼ C23
(5.56)
sxx ¼ C11 exx + C12 eyy + ezz syy ¼ C22 eyy + C12 ðezz + exx Þ szz ¼ C33 ezz + C12 exx + eyy syz ¼ C44 eyz szx ¼ C55 ezx sxy ¼ C66 exy
(5.57)
Eqs. (5.54), (5.55) collectively can be written as
Use of Eq. (5.56) in Eq. (5.52) yields
Let us apply a threefold rotation to Eq. (5.57) in which x ! y ! z ! x. The application of the transformation x ! y to the first expression of Eq. (5.57) yields syy ¼ C11 eyy + C12 ðexx + ezz Þ
(5.58)
Comparing Eq. (5.58) with the second expression of Eq. (5.57), the invariance of syy demands C11 ¼ C22
(5.59)
Similarly, when the transformations y ! z and z ! x are applied to the second and third terms, respectively, of Eq. (5.57) these give C22 ¼ C33 , C33 ¼ C11
(5.60)
The application of the transformation x ! y ! z ! x to the last three expressions of Eq. (5.57) yields C44 ¼ C55 ¼ C66
(5.61)
sxx ¼ C11 exx + C12 eyy + ezz syy ¼ C11 eyy + C12 ðezz + exx Þ szz ¼ C11 ezz + C12 exx + eyy syz ¼ C44 eyz szx ¼ C44 ezx sxy ¼ C44 exy
(5.62)
Using Eqs. (5.59)–(5.61) in Eq. (5.57) one gets
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Solid State Physics
Eq. (5.62) can be written in matrix form as 0 1 0 10 1 sxx exx C11 C12 C12 0 0 0 B syy C B C12 C11 C12 0 C B eyy C 0 0 B C B CB C B szz C B C12 C12 C11 0 B C 0 0 C B C¼B C B ezz C B syz C B 0 C C 0 CB 0 0 C44 0 B C B B eyz C @ szx A @ 0 A @ ezx A 0 0 0 C44 0 sxy exy 0 0 0 0 0 C44
(5.63)
Eq. (5.62) or Eq. (5.63) shows that in a solid with a cubic structure there are three independent elastic constants, C11, C12, and C44, the determination of which explains all of the elastic properties of these solids. From Eq. (5.63) one can derive the relation between the elastic stiffness and the elastic compliance constants for a cubic crystal. Table 5.1 presents the elastic stiffness constants for Al, Cu, Ag, and Au metals.
5.7
ELASTIC ENERGY DENSITY IN CUBIC SOLIDS
The elastic energy density, given by Eq. (5.32), is a quadratic function of the elastic strain. In cubic crystals there are only three nonzero elastic constants, C11, C12, and C44. Therefore, the nonzero terms in Eq. (5.32), with the help of Eq. (5.63), can be written as 1 1 U ¼ C11 e21 + e22 + e23 + C44 e24 + e25 + e26 2 2 + C12 ðe1 e2 + e1 e3 + e2 e3 Þ
(5.64)
The above equation can be written in terms of the x, y, and z axes as 1 1 U ¼ C11 e2xx + e2yy + e2zz + C44 e2yz + e2zx + e2xy 2 2 + C12 exx eyy + eyy ezz + ezz exx
5.8
(5.65)
BULK MODULUS IN CUBIC SOLIDS
The bulk modulus BM is defined, in terms of the energy density U, as 1 U ¼ BM d2D 2
(5.66)
where dD is the dilation. From Eq. (5.65), U can be calculated in terms of dD. For uniform dilation 1 exx ¼ eyy ¼ ezz ¼ dD 3
(5.67)
eyz ¼ ezx ¼ exy ¼ 0
(5.68)
and
TABLE 5.1 Adiabatic Elastic Stiffness Constants (1012 dyne/cm2) at Room Temperature (300 K) for Al, Cu, Ag, and Aua Metal
C11
C12
C44
Aas
Al
1.068
0.607
0.282
0.611
Cu
1.684
1.214
0.754
1.604
Ag
1.240
0.937
0.461
1.521
Au
1.923
1.631
0.420
1.438
a
Kittel, C. (1971). Introduction to solid state physics (4th ed.). New York: J. Wiley & Sons.
Elastic Properties of Solids Chapter
5
103
Substituting Eqs. (5.67), (5.68) into Eq. (5.65), one can write 1 U ¼ ðC11 + 2C12 Þd2D 6
(5.69)
Comparing Eqs.(5.66), (5.69), the bulk modulus becomes 1 BM ¼ ðC11 + 2C12 Þ 3
(5.70)
and the compressibility KC can be written as KC ¼
1 3 ¼ BM C11 + 2C12
(5.71)
With the knowledge of the elastic constants, the bulk modulus and compressibility of a cubic solid can be estimated.
5.9
ELASTIC WAVES IN CUBIC SOLIDS
The propagation of elastic waves in a solid can be understood by considering the mechanical system shown in Fig. 5.5. Here P, Q, and R are wooden blocks joined together with springs A and B along the x-direction. Elastic waves are produced by stretching the springs along the x-direction. If the springs A and B are stretched equally in opposite directions, the force acting on the block Q is zero. Further, if uniform stress sxx is applied, the force acting on the blocks is again zero. If spring B is stretched more than spring A, nonuniform stress is produced in the system and the net stress acting on the block Q becomes Dsxx ¼ sxx ðBÞ sxx ðAÞ
(5.72)
Dsxx will make the block Q move in the x-direction, thereby producing an elastic wave. Let us now consider a small cubic solid with sides Dx, Dy, and Dz (Fig. 5.6). The application of a nonuniform stress sxx(x) in the x-direction produces a nonuniform strain in the cube. Let sxx(x + Dx) and sxx(x) be the stress applied to the opposite faces of the cube (see Fig. 5.6), then sxx ðx + DxÞ ¼ sxx ðxÞ +
∂sxx Dx ∂x
(5.73)
The net stress acting on the face ABCD is given by Dsxx ðDxÞ ¼ sxx ðx + DxÞ sxx ðxÞ ¼
∂sxx Dx ∂x
(5.74)
Therefore, the force DF(Dx) acting on the face ABCD becomes DFðDxÞ ¼ Dsxx DyDz ¼
∂sxx DxDyDz ∂x
(5.75)
The application of stress Dsxx(Dx) makes the cube move along the x-direction, thereby producing an elastic wave in the same direction. The stress Dsxx will also produce strain (deformation) in the y- (in the faces BCGF and ADHE) and z- (in the faces DCGH and ABFE) directions. The forces acting along the y- and z-directions are given by DFðDyÞ ¼ Fðy + DyÞ FðyÞ ¼
∂sxy DxDyDz ∂y
(5.76)
DFðDzÞ ¼ Fðz + DzÞ FðzÞ ¼
∂sxz DxDyDz ∂z
(5.77)
FIG. 5.5 A mechanical system of three wooden blocks P, Q, and R joined together with springs A and B along the x-direction.
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Solid State Physics
FIG. 5.6 A small cubic solid with sides Dx, Dy, and Dz along the three Cartesian directions. A nonuniform stress sxx(x) is applied along the x-direction.
Hence the total force acting on the cube due to the stress applied in the x-direction becomes DF ¼ DFðDxÞ + DFðDyÞ + DFðDzÞ ∂sxx ∂sxy ∂sxz + + DxDyDz ¼ ∂x ∂y ∂z
(5.78)
Let rm be the mass density of the homogeneous isotropic material, then according to Newton’s second law ðrm DxDyDzÞ
d2 u1 ¼ DF dt2
(5.79)
where u1 is the displacement in the x-direction. From Eqs. (5.78), (5.79) one can write rm
d2 u1 ∂sxx ∂sxy ∂sxz + + ¼ ∂x ∂y ∂z dt2
(5.80)
The solution of Eq. (5.80) requires knowledge of the elements of the stress tensor of a cubic solid. Differentiating the expressions for sxx, sxy, and sxz given by Eq. (5.62), and using Eqs. (5.9), (5.10) for the elements of the strain tensor, one gets 2 ∂eyy ∂ezz ∂sxx ∂e ∂2 u ∂ u2 ∂ 2 u3 ¼ C11 xx + C12 + ¼ C11 21 + C12 + (5.81) ∂x ∂x ∂x ∂x ∂x ∂x∂y ∂x∂z 2 ∂sxy ∂exy 1 ∂ u1 ∂2 u2 ¼ C44 ¼ C44 (5.82) + ∂y ∂y 2 ∂y2 ∂x∂y 2 ∂sxz ∂exz 1 ∂ u1 ∂ 2 u3 ¼ C44 ¼ C (5.83) + ∂z ∂z 2 44 ∂z2 ∂x∂z
Elastic Properties of Solids Chapter
Substituting Eqs. (5.81)–(5.83) into Eq. (5.80), one can write 2 2 ∂2 u1 ∂ 2 u1 1 ∂ u1 ∂2 u1 1 ∂ u2 ∂2 u3 + + 2 + C12 + C44 rm 2 ¼ C11 2 + C44 ∂t ∂x ∂y2 ∂z ∂x∂y ∂z∂x 2 2
5
105
(5.84)
Eq. (5.84) gives the equation of motion of the elastic wave produced in a cubic solid when stress is applied externally in the x-direction. In exactly the same manner, one can apply the nonhomogeneous stress in the y- and z-directions and can obtain the corresponding equations of motion for u2 and u3, which are given by 2 2 ∂ 2 u2 ∂ 2 u2 1 ∂ u2 ∂ 2 u2 1 ∂ u1 ∂ 2 u3 + (5.85) + 2 + C12 + C44 r 2 ¼ C11 2 + C44 ∂t ∂y ∂x2 ∂z ∂x∂y ∂y∂z 2 2 2 2 ∂2 u ∂2 u 1 ∂ u3 ∂ 2 u3 1 ∂ u1 ∂ 2 u2 + (5.86) rm 23 ¼ C11 23 + C44 + C C + + 12 ∂t ∂z ∂x2 ∂y2 2 2 44 ∂x∂z ∂y∂z Eqs. (5.84)–(5.86) can be written in a general form as rm
X 2 X ∂2 u ∂ ub ∂2 ua ∂ 2 ua 1 1 a C C ¼ C + + C + 11 44 12 44 2 2 2 ∂t ∂ra 2 ∂r ∂r 2 ∂rb b ðb6¼aÞ b ðb6¼aÞ a b
(5.87)
To examine the actual elastic waves produced in a cubic solid, let us apply Eq. (5.87) in different directions.
5.9.1
Elastic Waves in the [100] Direction
The elastic waves along the [100] direction may be longitudinal or transverse in nature. The displacement of longitudinal (L) elastic waves in the x-direction u1L is given by u1L ¼ u0L exp ½i ðKr1 oL tÞ ¼ u0L exp ½i ðKx oL tÞ
(5.88)
Here u0L and K are the amplitude and propagation wave vector along the x-direction and oL is the frequency of the L elastic wave. Substituting Eq. (5.88) into Eq. (5.84), the frequency of the elastic wave is given by sffiffiffiffiffiffiffi C11 K (5.89) oL ¼ rm Eq. (5.89) is called the dispersion relation because it relates the frequency to the wave vector. It immediately gives the linear velocity as sffiffiffiffiffiffiffi oL C11 ¼ (5.90) vL ¼ K rm Here the group velocity and the phase velocity are the same. There are two transverse (T) elastic waves with displacements along the y- and z-directions but with K along the xdirection and they are defined as h i (5.91) u2T1 ¼ u02 exp i Kr1 oT1 t h i (5.92) u3T2 ¼ u03 exp i Kr1 oT2 t where oT1 and oT2 are the frequencies of the T elastic waves. Substituting Eq. (5.91) into Eq. (5.85) and Eq. (5.92) into Eq. (5.86), one can solve for the velocities of the waves vT1 and vT2 to get sffiffiffiffiffiffiffiffiffi oT1 C44 ¼ vT 1 ¼ (5.93a) K 2rm sffiffiffiffiffiffiffiffiffi oT2 C44 ¼ vT 2 ¼ (5.93b) K 2rm Eqs. (5.93a), (5.93b) show that the velocities of the two T elastic waves are the same, but differ from that of the L elastic wave. The experimental measurement of the velocities of the L and T elastic waves in a cubic crystal allows us to find the
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Solid State Physics
FIG. 5.7 Longitudinal and transverse polarizations of the elastic wave traveling along the x-direction.
elastic constants C11 and C44 using Eqs. (5.90), (5.93a), (5.93b). Fig. 5.7 shows the L and T polarizations of the elastic wave with propagation wave vector K along the x-direction.
5.9.2
Elastic Waves in the [110] Direction
The propagation of elastic waves in the [100] direction gives information only about the elastic constants C11 and C44, but the elastic constant C12 remains undetermined. Let us consider the propagation of elastic waves along the [110] symmetry direction as shown in Fig. 5.8. The propagation wavevector K in the [110] direction is given by K K ¼ ^i1 K1 + ^i2 K2 ¼ pffiffiffi ^i1 + ^i2 (5.94) 2
FIG. 5.8 Longitudinal and transverse polarizations of an elastic wave traveling along the [110] direction.
Elastic Properties of Solids Chapter
5
107
The components K1 ¼ Kx and K2 ¼ Ky have the same magnitude. Fig. 5.8 shows the different polarizations (modes) of the elastic wave. It is evident from the figure that one of the transverse elastic waves T1 propagates in the xy-plane with particle displacement along the z-direction defined as K (5.95) u3T1 ¼ u03 exp i pffiffiffi ðx + yÞ oT1 t 2 Substituting Eq. (5.95) into Eq. (5.86), one gets
sffiffiffiffiffiffiffiffiffi C44 K oT1 ¼ 2rm
(5.96)
Therefore, the velocity of the T1 elastic wave becomes
sffiffiffiffiffiffiffiffiffi C44 vT 1 ¼ 2rm
(5.97)
The second transverse elastic wave T2 and the L elastic wave propagate in the xy-plane with particle displacement also in the xy-plane and they are defined as K (5.98) u1 ¼ u01 exp i pffiffiffi ðx + yÞ ot 2 K (5.99) u2 ¼ u02 exp i pffiffiffi ðx + yÞ ot 2 Substituting u1 and u2 into Eqs. (5.84), (5.85), one gets 2 2 1 K 1 K u1 + C12 + C44 u o2 rm u1 ¼ C11 + C44 2 2 2 2 2 2 2 1 K 1 K u2 + C12 + C44 u o2 rm u2 ¼ C11 + C44 2 2 1 2 2
(5.100) (5.101)
Eqs. (5.100), (5.101) have nontrivial solutions only if the determinant of the coefficients of u1 and u2 is zero, that is, 2 2
1 K
o2 rm + C11 + 1 C44 K
C + C 12 44
2 2
2 2
¼0 (5.102)
K2 1 K2
C +1C 2 o rm + C11 + C44
12
2 2 44 2 2 The above determinant gives a quadratic equation in o2 with the following solutions 1 o21 rm ¼ ðC11 + C12 + C44 Þ K2 2 1 o22 rm ¼ ðC11 C12 Þ K2 2
(5.103) (5.104)
So, the velocities of these elastic waves are given by
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi C11 + C12 + C44 v1 ¼ 2rm sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi C11 C12 v2 ¼ 2rm
(5.105)
(5.106)
Let us examine the nature of the two waves described by Eqs. (5.103), (5.104). Substituting Eq. (5.103) into Eq. (5.100) for the frequency o ¼ o1, one can prove that u1 ¼ u 2
(5.107)
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Solid State Physics
Thus, corresponding to the frequency o1, the displacement of the particle is in the [110] direction, which gives the L elastic wave. Similarly, substituting Eq. (5.104) into Eq. (5.100) for the frequency o ¼ o2, one can get u1 ¼ u2
(5.108)
According to Eq. (5.108) the displacement of the particle is along the 11 0 direction, which is perpendicular to the direction of propagation [1 1 0]. Hence Eq. (5.104) gives the frequency of the T2 elastic wave. Now, the velocities of the L and T2 elastic waves are given by sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi C11 + C12 + C44 vL ¼ (5.109) 2rm sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi C11 C12 vT 2 ¼ (5.110) 2rm Hence, from the experimental measurements of the velocities of the L and T elastic waves, given by Eqs. (5.97), (5.109), and (5.110), one can determine all of the elastic constants C11, C12 and C44 of a cubic crystal.
5.9.3
Elastic Waves in the [111] Direction
Another high-symmetry direction of interest in cubic crystals is the [111] direction. Consider an elastic wave propagating in the [111] direction with wave vector K (Fig. 5.9) defined as K ¼ ^i1 K1 + ^i2 K2 + ^i3 K3 ¼ ^i1 Kx + ^i2 Ky + ^i3 Kz K ¼ pffiffiffi ^i1 + ^i2 + ^i3 3
(5.111)
Here we have used the fact that, in the [111] direction, the magnitude of all the Cartesian components of K is equal. Therefore, the Cartesian components of displacement are given by K (5.112) u1 ðrÞ ¼ u01 exp i pffiffiffi ðx + y + zÞ ot 3
FIG. 5.9 Longitudinal and transverse polarizations of an elastic wave traveling along the [111] direction.
Elastic Properties of Solids Chapter
K u2 ðrÞ ¼ u02 exp i pffiffiffi ðx + y + zÞ ot 3 K u3 ðrÞ ¼ u03 exp i pffiffiffi ðx + y + zÞ ot 3
5
109
The total displacement u(r) becomes K uðrÞ ¼ u01^i1 + u02^i2 + u03^i3 exp i pffiffiffi ðx + y + zÞ ot 3
(5.113) (5.114)
(5.115)
One is interested in finding the velocities of the different modes of the elastic wave propagating in the [111] direction. Substituting Eqs. (5.112)–(5.114) into Eq. (5.84) and simplifying, we obtain 1 1 ðC11 + C44 3LÞu1 + C12 + C44 u2 + C12 + C44 u3 ¼ 0 (5.116) 2 2 where L ¼ rm
o2 K2
Similarly, substituting Eqs. (5.112)–(5.114) into Eqs. (5.85), (5.86), one can write 1 1 C12 + C44 u1 + ðC11 + C44 3LÞu2 + C12 + C44 u3 ¼ 0 2 2 1 1 C12 + C44 u1 + C12 + C44 u2 + ðC11 + C44 3LÞu3 ¼ 0 2 2
(5.117)
(5.118) (5.119)
Eqs. (5.116), (5.118), and (5.119) have nontrivial solutions only if the determinant of the coefficient of u1, u2, and u3 is zero, that is,
A B B
B A B ¼ 0 (5.120)
B B A where A ¼ C11 + C44 3L
(5.121)
and B ¼ C12 +
1 C 2 44
(5.122)
Expanding the determinant of Eq. (5.120), we get ðA BÞ2 ðA + 2BÞ ¼ 0
(5.123)
The above equation gives three solutions (out of which one solution is doubly degenerate) given by A¼B
(5.124)
A ¼ 2B
(5.125)
Eqs. (5.124), (5.125), with the help of Eqs. (5.121), (5.122), yield vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u 1 u uC11 C12 + C44 t 2 K o1 ¼ 3rm sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi C11 + 2C12 + 2C44 K o2 ¼ 3rm
(5.126)
(5.127)
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Solid State Physics
Let us determine the nature of the polarization of the elastic waves with frequencies o1 and o2. Eq. (5.125) can be written as 3L2 ¼ C11 + 2C12 + 2C44
(5.128)
Substituting Eq. (5.128) into Eq. (5.116), one gets 2u1 ¼ u2 + u3
(5.129)
Similarly, substituting Eq. (5.128) into Eqs. (5.118), (5.119), we obtain 2u2 ¼ u3 + u1
(5.130)
2u3 ¼ u1 + u2
(5.131)
From Eqs. (5.129)–(5.131) one can immediately write u1 ¼ u 2 ¼ u 3
(5.132)
Hence, for the eigenvalue L2, the three components of displacement are equal, which yields the displacement u along the [111] direction. Therefore, the elastic wave corresponding to the eigenvalue L2 or o2 represents the L elastic wave. If oL represents the frequency of the L wave, then from Eq. (5.127) its velocity is given by o vL ¼ L ¼ K
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi C11 + 2C12 + 2C44 3rm
(5.133)
Let us now consider the wave with frequency o1, which from Eqs. (5.124), (5.121), and (5.122) gives 3L1 ¼ C11 C12 +
1 C 2 44
(5.134)
Substituting Eq. (5.134) into Eq. (5.116), one immediately gets u1 + u2 + u3 ¼ 0
(5.135)
If we substitute Eq. (5.134) into Eqs. (5.118), (5.119) we obtain the same expression as given by Eq. (5.135). Therefore, Eq. (5.134) represents two degenerate T modes of the elastic wave propagating in the [111] direction and these must be orthogonal to each other and also to the L elastic wave. Let us take 110 as the direction of displacement of one of the transverse waves (say T1). Then the third displacement vector X3, representing the second transverse wave (say T2), must be perpendicular to both X1 ¼ [111] and X2 ¼ 110 , that is, X3 ¼ X1 X2 ¼ ^i1 + ^i2 + ^i3 ^i1 ^i2 + 0^i3
(5.136) ¼ ^i1 + ^i2 2^i3 Hence the displacement vector X3 of the T2 elastic wave is in the 112 direction. The polarizations of three elastic waves are shown in Fig. 5.9. From Eq. (5.126) the velocities of both the T elastic waves are the same and are given by vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u 1 u C C12 + C44 oT u t 11 2 ¼ vT 1 ¼ vT 2 ¼ vT ¼ K 3rm
(5.137)
An alternate method for determining the eigenvectors of the elastic wave is given in Appendix C. The above discussion yields only two velocity equations, that is, Eqs. (5.133), (5.137), but there are three elastic constants. Therefore, one cannot determine all the elastic constants of a cubic solid in this case. It is worthwhile to note that, in general, the elastic waves in an isotropic medium are mixtures of both the L and T polarizations, depending on the direction of propagation of the wave: only in high-symmetry directions do these possess pure L or T polarization. Further, the two T elastic waves may not, in general, have the same velocity.
5.10
ISOTROPIC ELASTICITY
The substitution C11 C12 ¼ C44
(5.138)
Elastic Properties of Solids Chapter
5
111
into Eqs. (5.90), (5.109), and (5.133) gives the same velocity of the L elastic waves in the different symmetry directions [100], [110], and [111]. Further, the velocities of the T elastic waves, given by Eqs. (5.93a), (5.93b), (5.97), (5.110), and (5.137) along the different directions [100], [110], and [111], also become the same, although different from that of the L elastic waves. In other words, regardless of the direction of propagation, the velocity of an elastic wave with a particular polarization becomes the same, subject to the condition (5.138). Therefore, this is known as the elastic isotropy condition. The anisotropy factor Aas in cubic crystals is defined as the square of the ratio of the velocities of the T elastic waves propagating in the [100] and [110] directions, that is, Aas ¼
C44 C11 C12
(5.139)
Aas is also equal to the ratio of the squares of the velocities of the two T elastic waves propagating in the [110] direction. Aas is unity for elastic isotropy. The departure of the values of Aas from unity is a measure of the anisotropy in cubic crystals (see Table 5.1).
5.11 EXPERIMENTAL MEASUREMENT OF ELASTIC CONSTANTS Ultrasonic waves are elastic waves whose velocity in solids can be measured experimentally. The elastic constants can be evaluated from the experimentally measured velocities of ultrasonic waves with different polarizations propagating in different symmetry directions in cubic solids. One of the most commonly used methods to measure the velocities is the ultrasonic pulse method, a schematic setup of which is shown in Fig. 5.10. In this method, ultrasonic pulses at regular intervals are produced by a quartz crystal fixed at one end of the specimen crystal and these are allowed to travel through it, as shown. The pulses are reflected back at the opposite end of the specimen, which ultimately reach the quartz crystal again. The time t taken by an ultrasonic pulse to travel the forward and backward journey in the crystal is measured experimentally. If d is the length of the specimen, then the velocity of the ultrasonic waves is given by v¼
2d t
(5.140)
Actually, the ultrasonic waves are allowed to travel along one of the symmetry directions, say [110], and the velocity of the waves with different polarizations is measured. Then the elastic constants of the cubic solid are evaluated using Eqs. (5.97), (5.109), and (5.110). In the above discussion we have defined the second-order elastic constants assuming Hooke’s law to be valid. In this approximation, the elastic energy density is a quadratic function of the strain. But for large stresses and strains, the Hooke’s law is not valid and one has to consider higher-order terms in strain and stress-strain relations (Eqs. 5.26a, 5.26b) and the energy density expression (Eq. 5.32). The elastic energy density involving cubic terms of the strain elements should be considered and these are manifestations of nonlinear effects, such as the interaction of phonons and thermal expansion. Therefore, one can define the third-order elastic constants from the energy density involving cubic terms of the strain elements.
FIG. 5.10 Schematic diagram for the ultrasonic pulse method used for determining velocities of ultrasonic waves in solids.
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Solid State Physics
Problem 5.1
If the factor of one-half is neglected in Eq. (5.9) and the strain components are defined as eab ¼
∂ub ∂ua ∂u + ,e ¼ a ∂ra ∂rb aa ∂ra
(5.141)
prove that the equation of motion for any component ua of the displacement field is given by rm
X ∂2 ub X ∂2 u ∂ 2 ua ∂2 ua a ¼ C + C + ð C + C Þ 11 44 12 44 ∂t2 ∂r2a ∂r ∂r ∂r2b b ðb6¼aÞ b ðb6¼aÞ a b
(5.142)
In some books, the equation of motion given by Eq. (5.142) is used.
Problem 5.2
Assume the running wave-like solution for the displacement u(r) defined as uðrÞ ¼ u0 ei ðK ro tÞ
(5.143)
where K is the propagation wave vector for the elastic wave. (a) If the elastic wave is traveling in the [100] direction, then from Eq. (5.142) prove that the velocities for the longitudinal and transverse waves are given by
vL ¼
sffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffi C11 C44 , vT1 ¼ vT2 ¼ rm rm
(5.144)
(b) If the elastic wave is traveling in the [110] direction, then from Eq. (5.142) prove that the velocities for the longitudinal and transverse waves are given by sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi C11 + C12 + 2C44 vL ¼ 2rm sffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi C44 C11 C12 ,v ¼ vT1 ¼ rm T2 2rm
(5.145)
(5.146)
(c) If the elastic wave is traveling in the [111] direction, then from Eq. (5.142) prove that the velocities for the longitudinal and transverse waves are given by sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi C11 + 2C12 + 4C44 vL ¼ 3rm sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi C11 C12 + C44 vT1 ¼ vT2 ¼ 3rm
(5.147)
(5.148)
Elastic Properties of Solids Chapter
5
113
Problem 5.3
Using the Newton second law of force in equation fa ¼
X ∂sab ∂rb b
prove that for a wave-like solution, the equation of motion is given by i X h Cabmn Kb Kn o2 rm dam um0 ¼ 0 m, n, b
(5.149)
(5.150)
Here u0 is the amplitude of the elastic wave and the other symbols have their usual meanings. Eq. (5.150) is called the Christoffel equation.
Problem 5.4
The free energy of a deformed body is defined as F ¼ F0 +
" #2 X 1 lL eaa + mL e2ab 2 a
(5.151)
where F0 is a constant quantity and lL and mL are called the Lame’s coefficients. Express the free energy as a sum of the pure shear strain and pure hydrostatic compression. Further, find the stress components from the free energy.
SUGGESTED READING Eshelby, J. D. (1956). The continuum theory of lattice defects. F. Seitz, & D. Turnbull (Eds.), Solid state physics (pp. 79–144). Vol. 3. New York: Academic Press. Huntington, H. B. (1958). The elastic constants of crystals. F. Seitz, & D. Turnbull (Eds.), Solid state physics (pp. 213–351). Vol. 7. New York: Academic Press. Landau, L. D., & Lifshitz, E. M. (1970). Theory of elasticity. London: Pergamon Press. Timoshenko, S., & Doodier, J. N. (1970). Theory of elasticity (3rd ed.). New York: McGraw-Hill Book Co.
Elastic Properties of Solids Chapter Outline 5.1 5.2 5.3 5.4 5.5 5.6 5.7 5.8
Strain Tensor Dilation Stress Tensor Elastic Constants of Solids Elastic Energy Density Elastic Constants in Cubic Solids Elastic Energy Density in Cubic Solids Bulk Modulus in Cubic Solids
93 95 96 96 97 98 102 102
5.9 Elastic Waves in Cubic Solids 5.9.1 Elastic Waves in the [100] Direction 5.9.2 Elastic Waves in the [110] Direction 5.9.3 Elastic Waves in the [111] Direction 5.10 Isotropic Elasticity 5.11 Experimental Measurement of Elastic Constants Suggested Reading
103 105 106 108 110 111 113
The theoretical study of the various properties of crystalline solids requires a knowledge of the crystal potential, the reliable determination of which is difficult. One of the oldest and simplest methods is to view the crystal as an isotropic continuous elastic material with uniform density instead of as a discrete periodic array of atoms. The various properties of the solids, such as elastic constants, lattice vibrations, and thermal properties, can be studied using the continuum elasticity theory of solids. The study of elastic constants is of immense importance because they give information about the nature of binding forces in solids: a central physical quantity in studying various properties of solids. The continuum elasticity theory is valid for low-frequency waves, such as elastic waves with wavelength l > 106 cm or frequency n < 1011 Hz, as these waves are not able to see the atomic structure of a crystalline solid. The present chapter presents the calculation of elastic constants in the isotropic linear elasticity approximation in which Hooke’s law is valid.
5.1
STRAIN TENSOR
Consider a solid with Cartesian coordinate axes having unit vectors ^ia with a ¼ 1,2, and 3. The position vector of a point or a particle in the solid (Fig. 5.1) is defined as X ^i r (5.1) r¼ a a
a
where ra are the Cartesian components of the vector r. The application of a weak external force produces a small deformation in the solid that changes the orientation and magnitude of both the position vector and the unit vectors. The strain produced in the body can be studied in two approaches. In the first approach the unit vectors are kept unchanged and the orientation and magnitude of the Cartesian components of the position vector are changed. In the second approach the Cartesian components of the position vector are kept unchanged and the orientation and magnitude of the unit vectors are changed. It is more convenient to adopt the second approach as is done here. In a strained body, a particle at point r moves to r0 (Fig. 5.1) given by r0 ¼ r + uðrÞ where u(r) is the displacement produced. If ua is the Cartesian component of u(r) along the unit vector ^ia , then X uðrÞ ¼ ua^ia
(5.2)
(5.3)
a
Solid State Physics. https://doi.org/10.1016/B978-0-12-817103-5.00005-0 © 2019 Elsevier Inc. All rights reserved.
93
94
Solid State Physics
FIG. 5.1 Coordinate axes in perfect and strained crystals. The position vector r goes to r0 after producing strain in the crystal.
Note that u(r) is not a constant quantity but is a function of the position vector. Therefore, u(r) varies continuously throughout the solid and forms a vector field usually called the strain field or displacement field. The components of 0 the unit vector after displacement, denoted as ^ia , are given as i Xh ^i0 ¼ ∂ ðr + uÞ ¼ ∂ rb^ib + ub^ib a ∂ra ∂ra b (5.4) X ^ ¼ dab + eab ib b
where eab ¼ ∂ub =∂ra
(5.5)
0
The position vector r0 can be written, in terms of ^ia , as r0 ¼
X
0
ra^ia
(5.6)
a
In the above equation the Cartesian components of r0 remain unchanged after deformation. To make the measurement of deformation free from the orientation of the coordinate axes, it is convenient to define a scalar product between the new unit vectors as X 0 0 eag ebg (5.7) gab ¼ ^ia ^ib ¼ dab + eab + eba + g
From the above equation the angle between the two deformed axes is h 0 0 i yab ¼ cos 1 gab = j ^ia k^ib j
(5.8)
Elastic Properties of Solids Chapter
In terms of gab the strain components eab are defined as 1 1 eab ¼ gab dab ¼ eab + eba 2" # 2 ∂u 1 ∂u b + a ¼ 2 ∂ra ∂rb
5
95
(5.9)
In defining the above equation, the second order terms in eab are neglected (the linear elasticity approximation). Eq. (5.9) $ shows that the strain tensor e is symmetric, that is, eab ¼ eba. The diagonal components of the strain tensor from Eq. (5.9) are given as eaa ¼
∂ua ¼ eaa ∂ra
(5.10)
When the direction of the axes is reversed, that is, a goes to a, then under the reversal of the coordinate axes one can write ra ¼ ra , ua ¼ ua From Eqs. (5.9), (5.10) one can write
∂ua ¼ eaa ∂ra " # 1 ∂ub ∂ua + ¼ eab eab ¼ ∂rb 2 ∂ra eaa ¼
(5.11)
(5.12)
(5.13)
Eqs. (5.12), (5.13) show that with the reversal of direction of one of the Cartesian coordinates, the nondiagonal components of the strain tensor change their sign while the diagonal components remain unchanged. Ideally speaking, two types of strain can exist in a solid. The first is hydrostatic pressure in which the volume of the solid $ changes without any change in its shape and is represented by the diagonal components of the strain tensor e . The second is pure shear in which the volume of the body remains unchanged by the deformation and only the shape changes. Pure shear is $ represented by the nondiagonal components of e . But in an actual strain, both the volume and shape of the solid may change. Therefore, the general strain may be represented as the linear combination of pure hydrostatic pressure and pure shear. To do so, the strain components are defined as " # 1 X 1 X egg + dab egg (5.14) eab ¼ eab dab 3 3 g g The first term on the right side is evidently a pure shear as the sum of its diagonal terms is zero, while the second term is a pure hydrostatic pressure. Hydrostatic pressure acts perpendicular to the surface of the solid, while pure shear acts along the surface. Therefore, these two scalar components are independent of each other and the square of the strain component eab can be obtained by the addition of the squares of these two components, that is, " #2 " #2 h i2 1 X 1 X egg + e (5.15) eab ¼ eab dab 3 3 g gg g The strain field, from Eqs. (5.1), (5.2), (5.4), and (5.6), can be written as i Xh ra eab ^ib uðrÞ ¼ r0 r ¼ a, b From Eqs. (5.3), (5.16) the components of the strain field are given by X ub ¼ ra eab
(5.16)
(5.17)
a
5.2
DILATION
Dilation is defined as the fractional change in volume of a solid due to the deformation produced by the application of an external force. It can be expressed in terms of the strain field components eab. Consider a solid in the form of a cube with unit edges ^i1 , ^i2 , and ^i3 . The volume of the solid V is given by V ¼ ^i1 ^i2 ^i3 ¼ 1
(5.18)
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Solid State Physics
0
0
0
After deformation the cube becomes a parallelepiped with edges ^i1 , ^i2 , and ^i3 , having volume V0 given by X 0 0 0 V0 ¼ ^i1 ^i2 ^i3 ¼ 1 + eaa
(5.19)
a
in the linear elasticity approximation. The dilation produced by the deformation becomes dD ¼
V0 V DV X eaa ¼ ¼ V V a
(5.20)
The symmetric strain tensor has six independent elements. For simplicity of notation, it is very convenient to use the Voigt notation for the strain components, according to which the subscripts are defined as 1 11 ¼ xx; 2 22 ¼ yy; 3 33 ¼ zz 4 23 ¼ yz; 5 31 ¼ zx; 6 12 ¼ xy
(5.21)
In the Voigt notation eab can be written as em where m takes the values 1, 2, 3, 4, 5, and 6.
5.3
STRESS TENSOR
There are two types of forces acting on a solid body. The first type is called the body force, which acts throughout the body and exerts influence on the whole of the mass distribution. The inertial and gravitational forces are examples of the body force. The body force is expressed per unit mass or per unit volume. If f(r) is the body force per unit volume, it can be written as X f a^ia (5.22) f ðrÞ ¼ a
Here fa is the a-component of the body force density. The second type is the surface force, which acts on the surface of the body. A surface force is expressed in units of force per unit area and is called stress. The stress is further divided into two categories: stress acting normal to the surface (normal stress) and stress acting along the surface (shear stress). As the normal stress and the shear stress act perpendicularly to each other, therefore, they are orthogonal stresses. Stress is repre$ sented by a tensor s with components sab where the first subscript indicates the direction of the stress and the second subscript the direction of the normal to the surface on which the stress is acting. Fig. 5.2 shows the different components of stress acting on a cubic solid. It can be easily shown that the stress tensor is symmetric, that is, sab ¼ sba
(5.23) $
which is a consequence of the fact that shear stress does not cause angular rotation. The symmetric stress tensor s has six independent elements, which, in the Voigt notation, can be written as sm where m can take values 1, 2, 3, 4, 5, and 6. If a body under stress is in the equilibrium state, its equation of motion is X ∂sab b
∂rb
+ fa ¼ 0
(5.24)
The first term gives the body force per unit volume applied externally on the body and the second term is the internal body force per unit volume.
5.4
ELASTIC CONSTANTS OF SOLIDS
In the linear elasticity approximation, valid for very small deformations only, the strain is linearly proportional to stress and vice versa. In mathematical language, we can write X sab ¼ Cabmn emn (5.25a) m, n X eab ¼ Csabmn smn (5.25b) m, n
Elastic Properties of Solids Chapter
5
97
FIG. 5.2 Components sab of the stress tensor in a cubic crystal.
where Cabmn and Csabmn are called the elastic stiffness constants and elastic compliance constants. The quantity Cabmn has dimensions of energy/volume, while Csabmn has dimensions of volume/energy. In the Voigt notation Eqs. (5.25a), (5.25b) become X sm ¼ Cmn en (5.26a) n
em ¼
X
Csmn sn
(5.26b)
n
where m and n have integral values from 1 to 6. The constants Cmn are usually called the elastic constants of the crystalline solid.
5.5
ELASTIC ENERGY DENSITY
If U is the elastic energy density per unit volume, the stress components in terms of it are given by sab ¼
∂U ∂eab
(5.27a)
or dU ¼ sab deab
(5.27b)
The minus sign indicates that U is the work done on the system. In Voigt notation one can write (magnitude) dU ¼ sn den
(5.28)
From the above equation the stress components can be written in terms of energy density as sn ¼
dU den
(5.29)
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Solid State Physics
The total change in energy density due to all the stress components becomes X dU ¼ sn den
(5.30)
n
Substituting for sn from Eq. (5.26a), one can write dU ¼
X n, m
Therefore, the total energy density becomes
ð U ¼ dU ¼
Cnm em den
(5.31)
1X C e e 2 n, m nm m n
(5.32)
From Eqs. (5.29), (5.32) one can write s1 ¼
6 dU 1X ¼ C11 e1 + ðC + Cn1 Þen de1 2 n¼2 1n
(5.33)
Hence, in the stress-strain relations, the combination 1/2(Cmn + Cnm) appears in all the stress components. It follows that the elastic stiffness constants Cmn are symmetrical, that is, Cmn ¼ Cnm
(5.34)
Eq. (5.34) reduces the independent elastic constants from 36 to 21.
5.6
ELASTIC CONSTANTS IN CUBIC SOLIDS
In the linear elasticity approximation, the stress-strain relation given by Eq. (5.26a) allows us to write the different components of stress as sxx ¼ C11 exx + C12 eyy + C13 ezz + C14 eyz + C15 ezx + C16 exy syy ¼ C21 exx + C22 eyy + C23 ezz + C24 eyz + C25 ezx + C26 exy szz ¼ C31 exx + C32 eyy + C33 ezz + C34 eyz + C35 ezx + C36 exy syz ¼ C41 exx + C42 eyy + C43 ezz + C44 eyz + C45 ezx + C46 exy
(5.35)
szx ¼ C51 exx + C52 eyy + C53 ezz + C54 eyz + C55 ezx + C56 exy sxy ¼ C61 exx + C62 eyy + C63 ezz + C64 eyz + C65 ezx + C66 exy The above equation can be written in matrix form as 1 0 0 sxx C11 C12 Bs C BC C B yy C B 21 22 C B B B szz C B C31 C32 C B B Bs C ¼ BC C B yz C B 41 42 C B B @ szx A @ C51 C52 sxy C61 C62
C13 C14 C15 C16
10
exx
1
B C C23 C24 C25 C26 C C B eyy C CB C C33 C34 C35 C36 C B ezz C CB C B C C43 C44 C45 C46 C C B eyz C CB C C53 C54 C55 C56 A @ ezx A C63 C64 C65 C66
(5.36)
exy
Consider an isotropic solid in which the physical properties do not alter under the symmetry operations of a cubic solid. The number of elastic constants can be reduced by considering the different symmetry relations of the cubic solid. Following are the important symmetry operations in a cubic solid. Fourfold (2p=4) rotation about one of the edges of a cube (Fig. 5.3), which changes the sign of the coordinates as follows (anticlockwise rotation) About z axis : x ! y, y ! x About y axis : z ! x, x ! z About x axis : y ! z, z ! y
(5.37)
Elastic Properties of Solids Chapter
5
99
FIG. 5.3 Twofold and fourfold rotations about the edges of a cube.
Twofold (2p/2) rotation about one of the edges of a cube, which changes the axes as follows (anticlockwise rotation): About z axis : x ! x, y ! y About y axis : z ! z, x ! x About x axis : y ! y, z ! z
(5.38)
Threefold (2 p/3) rotation about the diagonals of a cube (Fig. 5.4). There are four diagonals about which the transformation of the axes is as follows: x!y!z!x x ! z ! y ! x x ! z ! y ! x x ! y ! z ! x
(5.39)
We see from the above symmetry relations that the rotations either interchange the axes or reverse their sign. Let us apply the operation of a twofold rotation to the cubic solids in which the signs of the axes change. If the sign of the y-axis is reversed, then the left-hand side of the first expression of Eq. (5.35) is given as sxx ¼ C11 exx + C12 eyy + C13 ezz + C14 eyz + C15 ezx + C16 exy ¼ C11 exx + C12 eyy + C13 ezz C14 eyz + C15 ezx C16 exy
(5.40)
In the above equation we have used the properties (5.12), (5.13) of eab and we see that all the terms remain unchanged except the 4th and 6th terms of Eq. (5.40). Therefore, sxx is invariant under the transformation y ! y only if C14 ¼ C16 ¼ 0
(5.41)
The symmetry property of the elastic constants yields C41 ¼ C61 ¼ 0
(5.42)
Similarly, if we make the transformation z ! z in the first expression of Eq. (5.35), we get C14 ¼ C41 ¼ C15 ¼ C51 ¼ 0
(5.43)
Applying the transformation x ! x in the second expression of Eq. (5.35), we find syy ¼ C21 exx + C22 eyy + C23 ezz + C24 eyz C25 ezx C26 exy
(5.44)
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Solid State Physics
FIG. 5.4 Threefold rotation about the diagonal of a cubic structure.
and the transformation z ! z in the second expression of Eq. (5.35) yields syy ¼ C21 exx + C22 eyy + C23 ezz C24 eyz C25 ezx + C26 exy
(5.45)
The property of invariance yields, from Eqs. (5.44), (5.45), C24 ¼ C25 ¼ C26 ¼ C42 ¼ C52 ¼ C62 ¼ 0
(5.46)
The transformations x ! x and y ! y, when applied to the third expression of Eq. (5.35), reduce the following elastic constants to zero C34 ¼ C35 ¼ C36 ¼ C43 ¼ C53 ¼ C63 ¼ 0
(5.47)
In the above description we have considered the invariance of only the normal stress components sxx, syy, and szz in cubic crystals. Using Eqs. (5.41)–(5.43), (5.46), and (5.47) in Eq. (5.35), one can write sxx ¼ C11 exx + C12 eyy + C13 ezz syy ¼ C21 exx + C22 eyy + C23 ezz szz ¼ C31 exx + C32 eyy + C33 ezz syz ¼ C44 eyz + C45 ezx + C46 exy
(5.48)
szx ¼ C54 eyz + C55 ezx + C56 exy sxy ¼ C64 eyz + C65 ezx + C66 exy Let us now consider the shear stress defined by the last three expressions of Eq. (5.48). If we apply the transformation x ! x to the fourth expression of Eq. (5.48), the invariance property yields C45 ¼ C46 ¼ C54 ¼ C64 ¼ 0
(5.49)
We have used the fact that syz does not change under the transformation x ! x as it does not involve the x-coordinate. Similarly, the application of transformation y ! y to the fifth expression and of z ! z to the sixth expression of Eq. (5.48) give C54 ¼ C56 ¼ C45 ¼ C65 ¼ 0
(5.50)
C64 ¼ C65 ¼ C46 ¼ C56 ¼ 0
(5.51)
Elastic Properties of Solids Chapter
5
101
Substituting Eqs. (5.49)–(5.51) into Eq. (5.48), one gets sxx ¼ C11 exx + C12 eyy + C13 ezz syy ¼ C21 exx + C22 eyy + C23 ezz szz ¼ C31 exx + C32 eyy + C33 ezz syz ¼ C44 eyz szx ¼ C55 ezx sxy ¼ C66 exy
(5.52)
In the above discussion we have applied a twofold rotation about one of the edges of the cubic structure. Let us apply a fourfold rotation about one of the edges of the cube. In a fourfold rotation either the transformation x ! y, y ! z, z ! x or x ! y, y ! z, z ! x can take place. In both of these transformations, invariance is required. When fourfold rotation is applied about the x-axis, then the allowed transformations are y ! z and z ! y and in these transformations the first expression of Eq. (5.52) changes to (5.53) sxx ¼ C11 exx + C12 ezz + C13 eyy The first expression of Eqs. (5.52), (5.53) must be the same (invariance property) under the transformation, which gives C12 ¼ C13
(5.54)
Similarly, when fourfold rotation is applied to the second expression of Eq. (5.52) about the y-axis and to the third expression of Eq. (5.52) about the z-axis, one gets C21 ¼ C23 ¼ C31 ¼ C32
(5.55)
C12 ¼ C21 ¼ C31 ¼ C32 ¼ C13 ¼ C23
(5.56)
sxx ¼ C11 exx + C12 eyy + ezz syy ¼ C22 eyy + C12 ðezz + exx Þ szz ¼ C33 ezz + C12 exx + eyy syz ¼ C44 eyz szx ¼ C55 ezx sxy ¼ C66 exy
(5.57)
Eqs. (5.54), (5.55) collectively can be written as
Use of Eq. (5.56) in Eq. (5.52) yields
Let us apply a threefold rotation to Eq. (5.57) in which x ! y ! z ! x. The application of the transformation x ! y to the first expression of Eq. (5.57) yields syy ¼ C11 eyy + C12 ðexx + ezz Þ
(5.58)
Comparing Eq. (5.58) with the second expression of Eq. (5.57), the invariance of syy demands C11 ¼ C22
(5.59)
Similarly, when the transformations y ! z and z ! x are applied to the second and third terms, respectively, of Eq. (5.57) these give C22 ¼ C33 , C33 ¼ C11
(5.60)
The application of the transformation x ! y ! z ! x to the last three expressions of Eq. (5.57) yields C44 ¼ C55 ¼ C66
(5.61)
sxx ¼ C11 exx + C12 eyy + ezz syy ¼ C11 eyy + C12 ðezz + exx Þ szz ¼ C11 ezz + C12 exx + eyy syz ¼ C44 eyz szx ¼ C44 ezx sxy ¼ C44 exy
(5.62)
Using Eqs. (5.59)–(5.61) in Eq. (5.57) one gets
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Solid State Physics
Eq. (5.62) can be written in matrix form as 0 1 0 10 1 sxx exx C11 C12 C12 0 0 0 B syy C B C12 C11 C12 0 C B eyy C 0 0 B C B CB C B szz C B C12 C12 C11 0 B C 0 0 C B C¼B C B ezz C B syz C B 0 C C 0 CB 0 0 C44 0 B C B B eyz C @ szx A @ 0 A @ ezx A 0 0 0 C44 0 sxy exy 0 0 0 0 0 C44
(5.63)
Eq. (5.62) or Eq. (5.63) shows that in a solid with a cubic structure there are three independent elastic constants, C11, C12, and C44, the determination of which explains all of the elastic properties of these solids. From Eq. (5.63) one can derive the relation between the elastic stiffness and the elastic compliance constants for a cubic crystal. Table 5.1 presents the elastic stiffness constants for Al, Cu, Ag, and Au metals.
5.7
ELASTIC ENERGY DENSITY IN CUBIC SOLIDS
The elastic energy density, given by Eq. (5.32), is a quadratic function of the elastic strain. In cubic crystals there are only three nonzero elastic constants, C11, C12, and C44. Therefore, the nonzero terms in Eq. (5.32), with the help of Eq. (5.63), can be written as 1 1 U ¼ C11 e21 + e22 + e23 + C44 e24 + e25 + e26 2 2 + C12 ðe1 e2 + e1 e3 + e2 e3 Þ
(5.64)
The above equation can be written in terms of the x, y, and z axes as 1 1 U ¼ C11 e2xx + e2yy + e2zz + C44 e2yz + e2zx + e2xy 2 2 + C12 exx eyy + eyy ezz + ezz exx
5.8
(5.65)
BULK MODULUS IN CUBIC SOLIDS
The bulk modulus BM is defined, in terms of the energy density U, as 1 U ¼ BM d2D 2
(5.66)
where dD is the dilation. From Eq. (5.65), U can be calculated in terms of dD. For uniform dilation 1 exx ¼ eyy ¼ ezz ¼ dD 3
(5.67)
eyz ¼ ezx ¼ exy ¼ 0
(5.68)
and
TABLE 5.1 Adiabatic Elastic Stiffness Constants (1012 dyne/cm2) at Room Temperature (300 K) for Al, Cu, Ag, and Aua Metal
C11
C12
C44
Aas
Al
1.068
0.607
0.282
0.611
Cu
1.684
1.214
0.754
1.604
Ag
1.240
0.937
0.461
1.521
Au
1.923
1.631
0.420
1.438
a
Kittel, C. (1971). Introduction to solid state physics (4th ed.). New York: J. Wiley & Sons.
Elastic Properties of Solids Chapter
5
103
Substituting Eqs. (5.67), (5.68) into Eq. (5.65), one can write 1 U ¼ ðC11 + 2C12 Þd2D 6
(5.69)
Comparing Eqs.(5.66), (5.69), the bulk modulus becomes 1 BM ¼ ðC11 + 2C12 Þ 3
(5.70)
and the compressibility KC can be written as KC ¼
1 3 ¼ BM C11 + 2C12
(5.71)
With the knowledge of the elastic constants, the bulk modulus and compressibility of a cubic solid can be estimated.
5.9
ELASTIC WAVES IN CUBIC SOLIDS
The propagation of elastic waves in a solid can be understood by considering the mechanical system shown in Fig. 5.5. Here P, Q, and R are wooden blocks joined together with springs A and B along the x-direction. Elastic waves are produced by stretching the springs along the x-direction. If the springs A and B are stretched equally in opposite directions, the force acting on the block Q is zero. Further, if uniform stress sxx is applied, the force acting on the blocks is again zero. If spring B is stretched more than spring A, nonuniform stress is produced in the system and the net stress acting on the block Q becomes Dsxx ¼ sxx ðBÞ sxx ðAÞ
(5.72)
Dsxx will make the block Q move in the x-direction, thereby producing an elastic wave. Let us now consider a small cubic solid with sides Dx, Dy, and Dz (Fig. 5.6). The application of a nonuniform stress sxx(x) in the x-direction produces a nonuniform strain in the cube. Let sxx(x + Dx) and sxx(x) be the stress applied to the opposite faces of the cube (see Fig. 5.6), then sxx ðx + DxÞ ¼ sxx ðxÞ +
∂sxx Dx ∂x
(5.73)
The net stress acting on the face ABCD is given by Dsxx ðDxÞ ¼ sxx ðx + DxÞ sxx ðxÞ ¼
∂sxx Dx ∂x
(5.74)
Therefore, the force DF(Dx) acting on the face ABCD becomes DFðDxÞ ¼ Dsxx DyDz ¼
∂sxx DxDyDz ∂x
(5.75)
The application of stress Dsxx(Dx) makes the cube move along the x-direction, thereby producing an elastic wave in the same direction. The stress Dsxx will also produce strain (deformation) in the y- (in the faces BCGF and ADHE) and z- (in the faces DCGH and ABFE) directions. The forces acting along the y- and z-directions are given by DFðDyÞ ¼ Fðy + DyÞ FðyÞ ¼
∂sxy DxDyDz ∂y
(5.76)
DFðDzÞ ¼ Fðz + DzÞ FðzÞ ¼
∂sxz DxDyDz ∂z
(5.77)
FIG. 5.5 A mechanical system of three wooden blocks P, Q, and R joined together with springs A and B along the x-direction.
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Solid State Physics
FIG. 5.6 A small cubic solid with sides Dx, Dy, and Dz along the three Cartesian directions. A nonuniform stress sxx(x) is applied along the x-direction.
Hence the total force acting on the cube due to the stress applied in the x-direction becomes DF ¼ DFðDxÞ + DFðDyÞ + DFðDzÞ ∂sxx ∂sxy ∂sxz + + DxDyDz ¼ ∂x ∂y ∂z
(5.78)
Let rm be the mass density of the homogeneous isotropic material, then according to Newton’s second law ðrm DxDyDzÞ
d2 u1 ¼ DF dt2
(5.79)
where u1 is the displacement in the x-direction. From Eqs. (5.78), (5.79) one can write rm
d2 u1 ∂sxx ∂sxy ∂sxz + + ¼ ∂x ∂y ∂z dt2
(5.80)
The solution of Eq. (5.80) requires knowledge of the elements of the stress tensor of a cubic solid. Differentiating the expressions for sxx, sxy, and sxz given by Eq. (5.62), and using Eqs. (5.9), (5.10) for the elements of the strain tensor, one gets 2 ∂eyy ∂ezz ∂sxx ∂e ∂2 u ∂ u2 ∂ 2 u3 ¼ C11 xx + C12 + ¼ C11 21 + C12 + (5.81) ∂x ∂x ∂x ∂x ∂x ∂x∂y ∂x∂z 2 ∂sxy ∂exy 1 ∂ u1 ∂2 u2 ¼ C44 ¼ C44 (5.82) + ∂y ∂y 2 ∂y2 ∂x∂y 2 ∂sxz ∂exz 1 ∂ u1 ∂ 2 u3 ¼ C44 ¼ C (5.83) + ∂z ∂z 2 44 ∂z2 ∂x∂z
Elastic Properties of Solids Chapter
Substituting Eqs. (5.81)–(5.83) into Eq. (5.80), one can write 2 2 ∂2 u1 ∂ 2 u1 1 ∂ u1 ∂2 u1 1 ∂ u2 ∂2 u3 + + 2 + C12 + C44 rm 2 ¼ C11 2 + C44 ∂t ∂x ∂y2 ∂z ∂x∂y ∂z∂x 2 2
5
105
(5.84)
Eq. (5.84) gives the equation of motion of the elastic wave produced in a cubic solid when stress is applied externally in the x-direction. In exactly the same manner, one can apply the nonhomogeneous stress in the y- and z-directions and can obtain the corresponding equations of motion for u2 and u3, which are given by 2 2 ∂ 2 u2 ∂ 2 u2 1 ∂ u2 ∂ 2 u2 1 ∂ u1 ∂ 2 u3 + (5.85) + 2 + C12 + C44 r 2 ¼ C11 2 + C44 ∂t ∂y ∂x2 ∂z ∂x∂y ∂y∂z 2 2 2 2 ∂2 u ∂2 u 1 ∂ u3 ∂ 2 u3 1 ∂ u1 ∂ 2 u2 + (5.86) rm 23 ¼ C11 23 + C44 + C C + + 12 ∂t ∂z ∂x2 ∂y2 2 2 44 ∂x∂z ∂y∂z Eqs. (5.84)–(5.86) can be written in a general form as rm
X 2 X ∂2 u ∂ ub ∂2 ua ∂ 2 ua 1 1 a C C ¼ C + + C + 11 44 12 44 2 2 2 ∂t ∂ra 2 ∂r ∂r 2 ∂rb b ðb6¼aÞ b ðb6¼aÞ a b
(5.87)
To examine the actual elastic waves produced in a cubic solid, let us apply Eq. (5.87) in different directions.
5.9.1
Elastic Waves in the [100] Direction
The elastic waves along the [100] direction may be longitudinal or transverse in nature. The displacement of longitudinal (L) elastic waves in the x-direction u1L is given by u1L ¼ u0L exp ½i ðKr1 oL tÞ ¼ u0L exp ½i ðKx oL tÞ
(5.88)
Here u0L and K are the amplitude and propagation wave vector along the x-direction and oL is the frequency of the L elastic wave. Substituting Eq. (5.88) into Eq. (5.84), the frequency of the elastic wave is given by sffiffiffiffiffiffiffi C11 K (5.89) oL ¼ rm Eq. (5.89) is called the dispersion relation because it relates the frequency to the wave vector. It immediately gives the linear velocity as sffiffiffiffiffiffiffi oL C11 ¼ (5.90) vL ¼ K rm Here the group velocity and the phase velocity are the same. There are two transverse (T) elastic waves with displacements along the y- and z-directions but with K along the xdirection and they are defined as h i (5.91) u2T1 ¼ u02 exp i Kr1 oT1 t h i (5.92) u3T2 ¼ u03 exp i Kr1 oT2 t where oT1 and oT2 are the frequencies of the T elastic waves. Substituting Eq. (5.91) into Eq. (5.85) and Eq. (5.92) into Eq. (5.86), one can solve for the velocities of the waves vT1 and vT2 to get sffiffiffiffiffiffiffiffiffi oT1 C44 ¼ vT 1 ¼ (5.93a) K 2rm sffiffiffiffiffiffiffiffiffi oT2 C44 ¼ vT 2 ¼ (5.93b) K 2rm Eqs. (5.93a), (5.93b) show that the velocities of the two T elastic waves are the same, but differ from that of the L elastic wave. The experimental measurement of the velocities of the L and T elastic waves in a cubic crystal allows us to find the
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Solid State Physics
FIG. 5.7 Longitudinal and transverse polarizations of the elastic wave traveling along the x-direction.
elastic constants C11 and C44 using Eqs. (5.90), (5.93a), (5.93b). Fig. 5.7 shows the L and T polarizations of the elastic wave with propagation wave vector K along the x-direction.
5.9.2
Elastic Waves in the [110] Direction
The propagation of elastic waves in the [100] direction gives information only about the elastic constants C11 and C44, but the elastic constant C12 remains undetermined. Let us consider the propagation of elastic waves along the [110] symmetry direction as shown in Fig. 5.8. The propagation wavevector K in the [110] direction is given by K K ¼ ^i1 K1 + ^i2 K2 ¼ pffiffiffi ^i1 + ^i2 (5.94) 2
FIG. 5.8 Longitudinal and transverse polarizations of an elastic wave traveling along the [110] direction.
Elastic Properties of Solids Chapter
5
107
The components K1 ¼ Kx and K2 ¼ Ky have the same magnitude. Fig. 5.8 shows the different polarizations (modes) of the elastic wave. It is evident from the figure that one of the transverse elastic waves T1 propagates in the xy-plane with particle displacement along the z-direction defined as K (5.95) u3T1 ¼ u03 exp i pffiffiffi ðx + yÞ oT1 t 2 Substituting Eq. (5.95) into Eq. (5.86), one gets
sffiffiffiffiffiffiffiffiffi C44 K oT1 ¼ 2rm
(5.96)
Therefore, the velocity of the T1 elastic wave becomes
sffiffiffiffiffiffiffiffiffi C44 vT 1 ¼ 2rm
(5.97)
The second transverse elastic wave T2 and the L elastic wave propagate in the xy-plane with particle displacement also in the xy-plane and they are defined as K (5.98) u1 ¼ u01 exp i pffiffiffi ðx + yÞ ot 2 K (5.99) u2 ¼ u02 exp i pffiffiffi ðx + yÞ ot 2 Substituting u1 and u2 into Eqs. (5.84), (5.85), one gets 2 2 1 K 1 K u1 + C12 + C44 u o2 rm u1 ¼ C11 + C44 2 2 2 2 2 2 2 1 K 1 K u2 + C12 + C44 u o2 rm u2 ¼ C11 + C44 2 2 1 2 2
(5.100) (5.101)
Eqs. (5.100), (5.101) have nontrivial solutions only if the determinant of the coefficients of u1 and u2 is zero, that is, 2 2
1 K
o2 rm + C11 + 1 C44 K
C + C 12 44
2 2
2 2
¼0 (5.102)
K2 1 K2
C +1C 2 o rm + C11 + C44
12
2 2 44 2 2 The above determinant gives a quadratic equation in o2 with the following solutions 1 o21 rm ¼ ðC11 + C12 + C44 Þ K2 2 1 o22 rm ¼ ðC11 C12 Þ K2 2
(5.103) (5.104)
So, the velocities of these elastic waves are given by
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi C11 + C12 + C44 v1 ¼ 2rm sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi C11 C12 v2 ¼ 2rm
(5.105)
(5.106)
Let us examine the nature of the two waves described by Eqs. (5.103), (5.104). Substituting Eq. (5.103) into Eq. (5.100) for the frequency o ¼ o1, one can prove that u1 ¼ u 2
(5.107)
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Solid State Physics
Thus, corresponding to the frequency o1, the displacement of the particle is in the [110] direction, which gives the L elastic wave. Similarly, substituting Eq. (5.104) into Eq. (5.100) for the frequency o ¼ o2, one can get u1 ¼ u2
(5.108)
According to Eq. (5.108) the displacement of the particle is along the 11 0 direction, which is perpendicular to the direction of propagation [1 1 0]. Hence Eq. (5.104) gives the frequency of the T2 elastic wave. Now, the velocities of the L and T2 elastic waves are given by sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi C11 + C12 + C44 vL ¼ (5.109) 2rm sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi C11 C12 vT 2 ¼ (5.110) 2rm Hence, from the experimental measurements of the velocities of the L and T elastic waves, given by Eqs. (5.97), (5.109), and (5.110), one can determine all of the elastic constants C11, C12 and C44 of a cubic crystal.
5.9.3
Elastic Waves in the [111] Direction
Another high-symmetry direction of interest in cubic crystals is the [111] direction. Consider an elastic wave propagating in the [111] direction with wave vector K (Fig. 5.9) defined as K ¼ ^i1 K1 + ^i2 K2 + ^i3 K3 ¼ ^i1 Kx + ^i2 Ky + ^i3 Kz K ¼ pffiffiffi ^i1 + ^i2 + ^i3 3
(5.111)
Here we have used the fact that, in the [111] direction, the magnitude of all the Cartesian components of K is equal. Therefore, the Cartesian components of displacement are given by K (5.112) u1 ðrÞ ¼ u01 exp i pffiffiffi ðx + y + zÞ ot 3
FIG. 5.9 Longitudinal and transverse polarizations of an elastic wave traveling along the [111] direction.
Elastic Properties of Solids Chapter
K u2 ðrÞ ¼ u02 exp i pffiffiffi ðx + y + zÞ ot 3 K u3 ðrÞ ¼ u03 exp i pffiffiffi ðx + y + zÞ ot 3
5
109
The total displacement u(r) becomes K uðrÞ ¼ u01^i1 + u02^i2 + u03^i3 exp i pffiffiffi ðx + y + zÞ ot 3
(5.113) (5.114)
(5.115)
One is interested in finding the velocities of the different modes of the elastic wave propagating in the [111] direction. Substituting Eqs. (5.112)–(5.114) into Eq. (5.84) and simplifying, we obtain 1 1 ðC11 + C44 3LÞu1 + C12 + C44 u2 + C12 + C44 u3 ¼ 0 (5.116) 2 2 where L ¼ rm
o2 K2
Similarly, substituting Eqs. (5.112)–(5.114) into Eqs. (5.85), (5.86), one can write 1 1 C12 + C44 u1 + ðC11 + C44 3LÞu2 + C12 + C44 u3 ¼ 0 2 2 1 1 C12 + C44 u1 + C12 + C44 u2 + ðC11 + C44 3LÞu3 ¼ 0 2 2
(5.117)
(5.118) (5.119)
Eqs. (5.116), (5.118), and (5.119) have nontrivial solutions only if the determinant of the coefficient of u1, u2, and u3 is zero, that is,
A B B
B A B ¼ 0 (5.120)
B B A where A ¼ C11 + C44 3L
(5.121)
and B ¼ C12 +
1 C 2 44
(5.122)
Expanding the determinant of Eq. (5.120), we get ðA BÞ2 ðA + 2BÞ ¼ 0
(5.123)
The above equation gives three solutions (out of which one solution is doubly degenerate) given by A¼B
(5.124)
A ¼ 2B
(5.125)
Eqs. (5.124), (5.125), with the help of Eqs. (5.121), (5.122), yield vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u 1 u uC11 C12 + C44 t 2 K o1 ¼ 3rm sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi C11 + 2C12 + 2C44 K o2 ¼ 3rm
(5.126)
(5.127)
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Solid State Physics
Let us determine the nature of the polarization of the elastic waves with frequencies o1 and o2. Eq. (5.125) can be written as 3L2 ¼ C11 + 2C12 + 2C44
(5.128)
Substituting Eq. (5.128) into Eq. (5.116), one gets 2u1 ¼ u2 + u3
(5.129)
Similarly, substituting Eq. (5.128) into Eqs. (5.118), (5.119), we obtain 2u2 ¼ u3 + u1
(5.130)
2u3 ¼ u1 + u2
(5.131)
From Eqs. (5.129)–(5.131) one can immediately write u1 ¼ u 2 ¼ u 3
(5.132)
Hence, for the eigenvalue L2, the three components of displacement are equal, which yields the displacement u along the [111] direction. Therefore, the elastic wave corresponding to the eigenvalue L2 or o2 represents the L elastic wave. If oL represents the frequency of the L wave, then from Eq. (5.127) its velocity is given by o vL ¼ L ¼ K
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi C11 + 2C12 + 2C44 3rm
(5.133)
Let us now consider the wave with frequency o1, which from Eqs. (5.124), (5.121), and (5.122) gives 3L1 ¼ C11 C12 +
1 C 2 44
(5.134)
Substituting Eq. (5.134) into Eq. (5.116), one immediately gets u1 + u2 + u3 ¼ 0
(5.135)
If we substitute Eq. (5.134) into Eqs. (5.118), (5.119) we obtain the same expression as given by Eq. (5.135). Therefore, Eq. (5.134) represents two degenerate T modes of the elastic wave propagating in the [111] direction and these must be orthogonal to each other and also to the L elastic wave. Let us take 110 as the direction of displacement of one of the transverse waves (say T1). Then the third displacement vector X3, representing the second transverse wave (say T2), must be perpendicular to both X1 ¼ [111] and X2 ¼ 110 , that is, X3 ¼ X1 X2 ¼ ^i1 + ^i2 + ^i3 ^i1 ^i2 + 0^i3
(5.136) ¼ ^i1 + ^i2 2^i3 Hence the displacement vector X3 of the T2 elastic wave is in the 112 direction. The polarizations of three elastic waves are shown in Fig. 5.9. From Eq. (5.126) the velocities of both the T elastic waves are the same and are given by vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u 1 u C C12 + C44 oT u t 11 2 ¼ vT 1 ¼ vT 2 ¼ vT ¼ K 3rm
(5.137)
An alternate method for determining the eigenvectors of the elastic wave is given in Appendix C. The above discussion yields only two velocity equations, that is, Eqs. (5.133), (5.137), but there are three elastic constants. Therefore, one cannot determine all the elastic constants of a cubic solid in this case. It is worthwhile to note that, in general, the elastic waves in an isotropic medium are mixtures of both the L and T polarizations, depending on the direction of propagation of the wave: only in high-symmetry directions do these possess pure L or T polarization. Further, the two T elastic waves may not, in general, have the same velocity.
5.10
ISOTROPIC ELASTICITY
The substitution C11 C12 ¼ C44
(5.138)
Elastic Properties of Solids Chapter
5
111
into Eqs. (5.90), (5.109), and (5.133) gives the same velocity of the L elastic waves in the different symmetry directions [100], [110], and [111]. Further, the velocities of the T elastic waves, given by Eqs. (5.93a), (5.93b), (5.97), (5.110), and (5.137) along the different directions [100], [110], and [111], also become the same, although different from that of the L elastic waves. In other words, regardless of the direction of propagation, the velocity of an elastic wave with a particular polarization becomes the same, subject to the condition (5.138). Therefore, this is known as the elastic isotropy condition. The anisotropy factor Aas in cubic crystals is defined as the square of the ratio of the velocities of the T elastic waves propagating in the [100] and [110] directions, that is, Aas ¼
C44 C11 C12
(5.139)
Aas is also equal to the ratio of the squares of the velocities of the two T elastic waves propagating in the [110] direction. Aas is unity for elastic isotropy. The departure of the values of Aas from unity is a measure of the anisotropy in cubic crystals (see Table 5.1).
5.11 EXPERIMENTAL MEASUREMENT OF ELASTIC CONSTANTS Ultrasonic waves are elastic waves whose velocity in solids can be measured experimentally. The elastic constants can be evaluated from the experimentally measured velocities of ultrasonic waves with different polarizations propagating in different symmetry directions in cubic solids. One of the most commonly used methods to measure the velocities is the ultrasonic pulse method, a schematic setup of which is shown in Fig. 5.10. In this method, ultrasonic pulses at regular intervals are produced by a quartz crystal fixed at one end of the specimen crystal and these are allowed to travel through it, as shown. The pulses are reflected back at the opposite end of the specimen, which ultimately reach the quartz crystal again. The time t taken by an ultrasonic pulse to travel the forward and backward journey in the crystal is measured experimentally. If d is the length of the specimen, then the velocity of the ultrasonic waves is given by v¼
2d t
(5.140)
Actually, the ultrasonic waves are allowed to travel along one of the symmetry directions, say [110], and the velocity of the waves with different polarizations is measured. Then the elastic constants of the cubic solid are evaluated using Eqs. (5.97), (5.109), and (5.110). In the above discussion we have defined the second-order elastic constants assuming Hooke’s law to be valid. In this approximation, the elastic energy density is a quadratic function of the strain. But for large stresses and strains, the Hooke’s law is not valid and one has to consider higher-order terms in strain and stress-strain relations (Eqs. 5.26a, 5.26b) and the energy density expression (Eq. 5.32). The elastic energy density involving cubic terms of the strain elements should be considered and these are manifestations of nonlinear effects, such as the interaction of phonons and thermal expansion. Therefore, one can define the third-order elastic constants from the energy density involving cubic terms of the strain elements.
FIG. 5.10 Schematic diagram for the ultrasonic pulse method used for determining velocities of ultrasonic waves in solids.
112
Solid State Physics
Problem 5.1
If the factor of one-half is neglected in Eq. (5.9) and the strain components are defined as eab ¼
∂ub ∂ua ∂u + ,e ¼ a ∂ra ∂rb aa ∂ra
(5.141)
prove that the equation of motion for any component ua of the displacement field is given by rm
X ∂2 ub X ∂2 u ∂ 2 ua ∂2 ua a ¼ C + C + ð C + C Þ 11 44 12 44 ∂t2 ∂r2a ∂r ∂r ∂r2b b ðb6¼aÞ b ðb6¼aÞ a b
(5.142)
In some books, the equation of motion given by Eq. (5.142) is used.
Problem 5.2
Assume the running wave-like solution for the displacement u(r) defined as uðrÞ ¼ u0 ei ðK ro tÞ
(5.143)
where K is the propagation wave vector for the elastic wave. (a) If the elastic wave is traveling in the [100] direction, then from Eq. (5.142) prove that the velocities for the longitudinal and transverse waves are given by
vL ¼
sffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffi C11 C44 , vT1 ¼ vT2 ¼ rm rm
(5.144)
(b) If the elastic wave is traveling in the [110] direction, then from Eq. (5.142) prove that the velocities for the longitudinal and transverse waves are given by sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi C11 + C12 + 2C44 vL ¼ 2rm sffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi C44 C11 C12 ,v ¼ vT1 ¼ rm T2 2rm
(5.145)
(5.146)
(c) If the elastic wave is traveling in the [111] direction, then from Eq. (5.142) prove that the velocities for the longitudinal and transverse waves are given by sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi C11 + 2C12 + 4C44 vL ¼ 3rm sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi C11 C12 + C44 vT1 ¼ vT2 ¼ 3rm
(5.147)
(5.148)
Elastic Properties of Solids Chapter
5
113
Problem 5.3
Using the Newton second law of force in equation fa ¼
X ∂sab ∂rb b
prove that for a wave-like solution, the equation of motion is given by i X h Cabmn Kb Kn o2 rm dam um0 ¼ 0 m, n, b
(5.149)
(5.150)
Here u0 is the amplitude of the elastic wave and the other symbols have their usual meanings. Eq. (5.150) is called the Christoffel equation.
Problem 5.4
The free energy of a deformed body is defined as F ¼ F0 +
" #2 X 1 lL eaa + mL e2ab 2 a
(5.151)
where F0 is a constant quantity and lL and mL are called the Lame’s coefficients. Express the free energy as a sum of the pure shear strain and pure hydrostatic compression. Further, find the stress components from the free energy.
SUGGESTED READING Eshelby, J. D. (1956). The continuum theory of lattice defects. F. Seitz, & D. Turnbull (Eds.), Solid state physics (pp. 79–144). Vol. 3. New York: Academic Press. Huntington, H. B. (1958). The elastic constants of crystals. F. Seitz, & D. Turnbull (Eds.), Solid state physics (pp. 213–351). Vol. 7. New York: Academic Press. Landau, L. D., & Lifshitz, E. M. (1970). Theory of elasticity. London: Pergamon Press. Timoshenko, S., & Doodier, J. N. (1970). Theory of elasticity (3rd ed.). New York: McGraw-Hill Book Co.