GEOL0026

Earth & Planetary Materials

Part 2a

A: The Physical Properties of

Anisotropic Materials

B: The Dynamics of Atoms in Crystals

C: Equations of State

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GEOL0026: Session A; The Physical Properties of Anisotropic Materials

Books

There is a truly great, in all senses of the word, book on this subject; The Physical

Properties of Crystals Their Representation by Tensors and Matrices, by J F Nye, pub.

Oxford Science. No scientist working in this area should be without a copy.

1/. Introduction to Isotropic and Anisotropic Materials

Even quite advanced solid-state physics courses tend to assume that materials are

isotropic – whereas, in real life, they are not. What does this word mean in this context?

Isotropic materials

Suppose we apply an electric field, E, of so many Volts per metre, to our sample; if it is

made of an isotropic material (e.g. copper), we will make a current density, j, of so

many Amps per metre2 flow in a direction parallel to E. Provided that E is not too large,

the magnitude of j will be proportional to the magnitude of E and so we may write

j = σE

where σ is the electrical conductivity of the material.

Diagrammatically, we can picture this as shown

on the right

Anisotropic materials

Now consider the case where we apply our electric field to a crystalline substance such

as graphite. Graphite is anisotropic; the atoms are arranged in planar layers and there is

no reason for us to suppose that the electrical conductivity parallel to the layers will be

the same as that perpendicular to the layers; indeed, there is every reason for us to

suppose that it will not be the same. Dutta (Phys. Rev. B, 90, 187-192, 1953) measured

the conductivity of graphite and found that the conductivity in a direction perpendicular

to the hexagonal axis is about 104 times greater than that along the hexagonal axis.

The consequence of this is that if we apply our field, E, to a graphite crystal in a general

direction, j will tend to be directed towards the “direction of easy flow of current”, i.e.

towards the planar layers. Diagrammatically, we usually now have j no longer parallel

to E. Clearly the constant of proportionality, σ, between j and E will have to vary also.

Note: (a) this figure is only schematic not quantitative; (b) I have greatly diminished the

amount of anisotropy actually present in this material.

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We find that for some directions (those perpendicular to and parallel to the layers) j and

E are parallel, but with a very different constant of proportionality (σ) between them. In

all other directions, j and E are no longer parallel, and the value of σ varies with

orientation, from a minimum value (in this case when E is perpendicular to the layers)

to a maximum value (in this case when E is parallel to the layers).

2/. Scalars, Vectors and Tensors

How can we describe behaviour such as that shown above by graphite quantitatively?

We shall continue to use our example of electrical conductivity (σ) to illustrate how this

may be done.

For an isotropic substance, σ is a scalar quantity; it has magnitude but no directional

dependence.

E and j are both vectors (you may have noticed that I have been writing them in bold

type) – and so they have both magnitude and direction; note that when referring to just

the magnitude of the vector I shall use normal italic type, i.e. E, j.

Now suppose that we choose three orthogonal reference axes in our crystal (it doesn’t

matter how we orient these axes at this stage – we just suppose that they exist);

following the notation in Nye’s book, we will refer to these axes as Ox1, Ox2, Ox3

(note that I shall sometimes abbreviate this to x1, x2, x3 when there is no danger of

confusing the axes with the coordinates of a point in space).

We can then fully describe both E and j,

in the usual way, in terms of their

components resolved onto these

three axes. Thus

E = [E1, E2, E3] and j = [j1, j2, j3]

with E = (E1 2 + E2 2 + E3 2)1/2 etc.

As our axes are, as yet, in no way constrained to be oriented with respect to the

crystallographic axes of the material we are studying (we shall see later that it is usually

highly convenient if they are!), there is every reason to suppose that the components of

our electric field, E1 etc., will each produce a current density in a general orientation.

Thus, in general, we must write

j1 = σ11E1 + σ12E2 + σ13E3

j2 = σ21E1 + σ22E2 + σ23E3

j3 = σ31E1 + σ32E2 + σ33E3

i.e., each component of E contributes to each component of j. The nine coefficients, σij,

now needed to describe our electrical conductivity may be written in the form of a

matrix

          333231 232221 131211   

Note that in this notation, the first suffix denotes the row and the second the column

and are called a second-rank tensor.

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It is important that you have a clear understanding at this stage of what is meant by, e.g.

the term σ21; this term tells us what the current density flowing in direction Ox2 will be

when we apply an electric field in direction Ox1, i.e. it relates j2 to E1.

Exercise 1 should help you to understand this.

The Hierarchy of Tensors

All of the quantities that we have encountered so far – scalar, vector and second-rank

tensor - are actually part of a continuing hierarchy of tensors.

A scalar is a zero-rank tensor – no sets of axes are involved in specifying this quantity

and it is, therefore, specifiable by a single number, i.e. it has 0 subscripts.

A vector is a first-rank tensor – one set of axes is involved in specifying this quantity,

as it requires its components on the 3 axes to be specified, i.e. it requires 1 subscript.

A second-rank tensor – has two sets of axes involved in its specification; it therefore

requires 9 numbers to specify it, each associated with a pair of axes, i.e. it requires 2

subscripts.

In general, if a property, T, relates a vector, p to a vector q, via

p1 = T11q1 + T12q2 + T13q3

p2 = T21q1 + T22q2 + T23q3

p3 = T31q1 + T32q2 + T33q3

the constants Tij are said to form a second-rank tensor.

Some examples of second rank tensors are given in the table below (from Nye).

We may extend these ideas to tensors of still higher rank. Thus, for example:

the piezoelectric coefficients, dijk, are a third-rank tensor relating the resulting

electrical polarization to the applied stress producing it

the elastic stiffnesses, cijkl , are a fourth-rank tensor, giving the stress on a crystal for a

given strain.

Note that the rank of the tensor = the number of subscripts required to specify it.

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3/. Principal Axes and Representation Surfaces of Second-Rank Tensors

As with the reciprocal lattice construction in Part 1 of this course, I think that at this

point it is as well to put aside the algebraic formulation of tensors that we used in

Section 2, in favour of a graphical approach. Note that the physics is exactly the same,

no additional approximations are made; what we shall aim to do, however, is to gain a

better understanding of the way in which we can use tensors.

We shall state without proof that the second-rank tensors that we are considering here

are all symmetric, i.e. that T21 = T12, etc., or, in general Tij = Tji. (see Nye for further

details).

We also state without proof (at this stage) that there must exist an orientation within the

crystal for our reference axes (Ox1, Ox2, Ox3), such that only the diagonal terms of Tij

are then non-zero, i.e. such that the tensor becomes

          33 22 11 00 00 00 S S S

We shall use the symbol Sii to denote that we are now referring to the tensor described

in terms of these special axes, called its principal axes. The values S11, S22 and S33 are

called the principal components of the tensor (note that Nye usually abbreviates these

further to S1, S2 and S3).

(Some of you may recall from earlier mathematics courses that we can express Tij in this way by

determining the eigenvalues, i.e. the values of Sii, and the eigenvectors, i.e. the orientation of the principal

axes, of the general 3x3 matrix describing the tensor)

Using for convenience our principal axes, we may now define a representation surface

for our tensor (for a second-rank tensor, this will be a representation quadric) by the

equation

12333 2 222 2 111 =++ xSxSxS

If S11, S22 and S33 are all positive and all different from each other, this surface is a

triaxial ellipsoid, as shown below – a shape that might well be familiar to some of you

from the “Earth Materials” course in the first year.

By putting x2 = x3 = 0, and solving the resulting equation, it is very easy to show that the

ellipsoid cuts the Ox1 principal axis at the point x1 = 1/√S11; similarly it must cut the Ox2

and Ox3 principal axes at x2 = 1/√S22 and x3 = 1/√S33 as shown.

Note that in this figure,

which is taken from Nye’s

book, he has used the condensed

notation S1, S2 and S3 to denote the

principal components.

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In some cases, one or more of the principal components, Sii, may be negative.

If only one principal component is negative, then the representation surface becomes a

hyperboloid of one sheet, as shown on the left below; if two are negative it is a

hyperboloid of two sheets as shown on the right.

A hyperboloid of one sheet

A hyperboloid of two sheets

If all three principal components are negative, the representation surface becomes an

imaginary triaxial ellipsoid.

For GEOL0026, we shall confine ourselves to situations where all three Sii are positive!

4/. Using the Representation Surface

These surfaces are all very well, but what do we do with them?

Let us return to our example of electrical conductivity. We wish to be able to determine

the magnitude of this property in a given direction; however, the concept of magnitude

needs some thought here as, in general, j and E are not parallel.

Suppose we have a very wide, thin slab of material across which we apply an electric

field, E; if we measure the current density j in this experiment, what we shall be

measuring is the flow of current perpendicular to the slab, i.e., the component of j that is

parallel to the applied field, E.

We can define, therefore,

the magnitude of the conductivity, σ, in

the direction of E, as the component

of j parallel to E, divided by E, i.e. σ = j||/E

How can we determine the magnitude of σ in any direction from our representation

surface?

We do this as shown on the right

Having drawn the representation surface, we

mark on it a line parallel to the direction of E.

this line cuts through the ellipsoid at P, thereby

defining a radius OP (r); this radius, r, is equal to 1/√σ

In general terms, for our tensor, S, therefore, the length of any radius vector, r, of the

representation quadric is equal to the reciprocal of the square root of the magnitude,

S, of the property in that direction.

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Thus if vectors p and q are related by the tensor, S, and we find that a line on the quadric

parallel to q, has radius r, then the component of p parallel to q is given by p|| = Sq,

where S = 1/r2

This property is very useful, but it is not the only thing that we can determine from the

representation quadric. Considering our example above, it turns out, also, that if we

draw a line perpendicular to the representation surface at the point P, this line will be

pointing in the direction of total current flow, j, for our applied field, E.

This is difficult to draw in 3-dimensions

but is reasonably clear from the picture

on the right, which shows the central

section through the quadric – the tangent

plane to the representation surface at P

lies perpendicular to the plane of the paper.

The diagram is multi-purpose, showing both

how to determine the direction of j for a given E

and also, more generally, the direction of p for a given q.

Thus, the direction of p for a given q may be found by first drawing, parallel to q, a

radius vector, OP, of the representation quadric, and then taking the normal to the

quadric at P.

Note that only if q lies along a principal axis will q and p be parallel to each other.

Time for another example, I think. Try Exercise 2 now.

4a/. Finding the magnitude of a property in a given direction numerically

Instead of using the representation surface, we may use a numerical method to find the

magnitude, in a given direction, of a property described by a second-rank tensor.

Consider again the case of the electric field and electric current density that we

discussed above. We can define the direction of the electric field, E, in terms of its

magnitude, E, and the three direction cosines, li, of a unit vector (as shown in the

diagram) parallel to the direction of E. Thus, the three components of E along any

general set of perpendicular reference axes (making no assumption about their

orientation) are given by

E1 = El1

E2 = El2

E3 = El3

and so

j1 = σ11El1 + σ12El2 + σ13El3

j2 = σ21El1 + σ22El2 + σ23El3

j3 = σ31El1 + σ32El2 + σ33El3

We now need to find the component of j, j||, that is parallel

to E. To do this we resolve j1, j2 and j3 along this direction, i.e.

j|| = j1l1 + j2l2 + j3l3

=

(σ11El1 + σ12El2 + σ13El3)l1 +

(σ21El1 + σ22El2 + σ23El3)l2 +

(σ31El1 + σ32El2 + σ33El3)l3

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Remembering that the magnitude of the conductivity, σ, in the direction of E is the

component of j parallel to E, divided by E, i.e. σ = j||/E we finally obtain

σ = (σ11l1 + σ12l2 + σ13l3)l1 + (σ21l1 + σ22l2 + σ23l3)l2 + (σ31l1 + σ32l2 + σ33l3)l3

i.e.

σ = σ11l1l1 + σ12l1l2 + σ13l1l3 + σ21l2l1 + σ22l2l2 + σ23l2l3 +σ31l3l1 + σ32l3l2 + σ33l3l3

As we shall see in Section 6, this equation may be written very compactly as

σ = σijlilj

This is a very useful equation as it allows you, for example, to find the magnitude of the

thermal expansion coefficient of any crystal in any general direction.

5/. The Effects of Crystal Symmetry on Second-Rank Tensors

The use of our representation quadric enables us to see the effects of crystal symmetry

very easily. Neumann’s principle states that

The symmetry elements of any physical property of a crystal must include the

symmetry elements of the point group of the crystal.

You should notice that Neumann carefully uses the word “include”; there is nothing to

prevent the physical properties having higher symmetry, provided that this higher

symmetry includes all the symmetry elements (rotation axes, mirror planes etc.) of the

crystal’s point group – what they cannot ever have is lower symmetry.

We can now think of Neumann’s principle being applied to our representation surface,

within the various crystal classes (or point groups).

We note, firstly, that all properties describable by our second rank tensor must be

centrosymmetric, as the representation surface must have a centre of symmetry – if a

vector r is a radius vector of the quadric, then so must be a vector –r.

If we now consider the crystal systems in turn, we find that:

For cubic crystals – the representation surface must be a sphere, as the only way our

quadric can have four 3-fold axes is if S11 = S22 = S33.

For hexagonal, trigonal and tetragonal crystals - the representation surface must be an

ellipsoid of revolution, as this is the only way in which our quadric can have a 6-fold

axis, a 3-fold axis or a 4-fold axis. If we adopt the usual convention of choosing this

high-symmetry axis of the crystal to lie along Ox3, then S11 = S22 ≠

S33

(note that the usual crystallographic caveat about the meaning of the “≠ ” sign applies).

For orthorhombic crystals, there are no restrictions on the magnitudes of S11, S22 and S33

and the representation surface is thus a triaxial ellipsoid, but to conform to the crystal

symmetry, the principal axes of the representation quadric must lie along the

crystallographic axes (i.e. parallel to the 2-fold axes and perpendicular to the mirror

planes).

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For monoclinic crystals, the representation surface is again triaxial, but one principal

axis must lie along the 2-fold axis, and/or perpendicular to the mirror plane of the

monoclinic crystal.

For triclinic crystals, whose maximum symmetry is a centre of symmetry, there are no

restrictions on the size or the orientation of the triaxial representation surface.

These properties are summarised in the table below (from Nye).

In some cases (e.g. hexagonal and trigonal) it is necessary to state the relationship

between the orthogonal axes, Ox1, Ox2, Ox3, used to specify our tensor, and the

crystallographic axes, a, b, c ( or x, y, z in the Table above). The usual convention is:

Cubic, tetragonal and orthorhombic, Ox1, Ox2, Ox3 parallel, respectively, to a, b, c.

Trigonal and hexagonal crystals, Ox3 parallel to c, Ox1 parallel to a, Ox2 parallel to

[120] (since γ = 120o).

Monoclinic, Ox2 parallel to b (the other axes are not fixed by the crystal symmetry)

Triclinic, no restrictions.

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6. Some More Mathematics

a/. The dummy suffix notation

We shall now go back to our original statement of a second-rank tensor, T, relating two

vectors, p and q, via

p1 = T11q1 + T12q2 + T13q3

p2 = T21q1 + T22q2 + T23q3

p3 = T31q1 + T32q2 + T33q3

This is a rather cumbersome formula to write. We could shorten it to

j j j j qTp  = = = 3 1 11

j j j j qTp  = = = 3 1 22

j j j j qTp  = = = 3 1 33

or even more compactly to

)3,2,1( 3 1 == = = iqTp j j j iji

Even this is a bit of a handful to write so, instead, we use the Einstein summation

convention and write

pi = Tij qj

where the repeated suffix, j, implies summation over j (the range of j, here from j = 1 to j

= 3, is understood from the context). The other suffix, i, is a free suffix.

Because the repeated (“dummy”) suffix “disappears” when the adding up is done, we

can change its label to whichever letter we choose; thus, pi = Tij qj =

Tik qk. = Til ql etc.

b/. Changing Axes

We can learn more about the properties of tensors if we consider what happens when we

change the orientation of our orthogonal reference axes. Clearly, the physical property

must remain invariant – if we apply an electric field to a crystal we cannot change the

current that flows just by drawing our reference axes in a different way.

We specify the relationship between our “new” axes ( 1xO  , 2xO  , 3xO  ) and our “old”

axes (Ox1, Ox2, Ox3) in terms of direction cosines; these are simply the cosines of the

angles between one axis and another, as shown in the matrix and diagram below.

Old

Ox1 Ox2

Ox3

1xO 

a11 a12 a13

New 2xO 

a21 a22 a23

3xO 

a31 a32 a33

11

(Note that only 3 aij can be independent of each other – it is more convenient at present,

however, to treat all nine values as if they were)

c/. Transformation of a Vector

Suppose we have a vector, p, specified on our “old” axes and we wish to transform this

so that it is specified on our “new” axes (we then call it

p' ) – clearly, the vector itself

must not change magnitude or direction as a result of this operation.

To do this, we:

i/. Write down the components of the vector on the old axes [p1, p2, p3]

ii/.

find the component of the vector on the first of the new axes, 1xO  , by taking

[p1

x (the projection of Ox1 on to 1xO  ] + [p2

x (the projection of Ox2 on to 1xO  ] +

[p3

x (the projection of Ox3 on to 1xO  ]

Thus

1p

= p1 a11 + p2 a12 + p3 a13

=

a1j pj

(using the summation convention)

iii/. and iv/. by repeating this procedure for 2xO 

and 3xO 

we obtain

2p

= p1 a21 + p2 a22 + p3 a23

=

a2j pj

and

3p

= p1 a31 + p2 a32 + p3 a33

=

a3j pj

i.e.

jiji pap =

(6.c.1)

Suppose that we want to go the other way, i.e. to turn

p' into p. By following exactly

the same method, you can show that

jjii pap =

(6.c.2)

Now try doing this yourselves: Exercises 3 & 4 give you some practice in this.

d/. Transformation of a Second-rank Tensor

We know that our tensor, T, relates vectors p and q (measured on the “old” axes) via

pi = Tij qj

(6.d.1)

We now wish to know the coefficients of the tensor T ', that will relate the transformed

vectors p' and q', to each other via

jiji qTp =

(6.d.2)

Clearly, T and T ' must be related, as the physical relationship between the vectors

cannot be changed by the use of different axes.

To find the relations ship between T and T ', we first find p' in terms of p. From

Equation (6.c.1), we have

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kiki pap =

(Note that I have changed one of the suffixes from j to k; this does not matter as it is a

“dummy suffix” over which we are summing)

We now replace pk in the above using equation (6.d.1). Thus,

lkliki qTap =

(again, I have changed the label of the dummy suffix of T from j to l).

Next, we replace ql by expressing it in terms of q'. The result from Exercise 4 gives us

jjlkliki qaTap =

This may be rearranged as

jkljliki qTaap =

Comparing this with Equation (6.d.2) gives us finally

kljlikij TaaT =

(6.d.3)

which is the transformation law for our second-rank tensor.

Let us write out one of the components, e.g. 12T  , so that we can see what this really

means. We have

12T 

=

a11a21T11 + a11a22T12 + a11a23T13

+ a12a21T21 + a12a22T22 + a12a23T23

+ a13a21T31 + a13a22T32 + a13a23T33

Equation (6.d.3) is incredibly compact. Each of the 9 components of T' is related to all 9

components of T, so it actually represents nine equations, each with nine terms in them.

Exercise 5 gives you some practice in doing these transformations.

Mathematically, it is actually the transformation laws, (6.c.1) and (6.d.3) that allow us

to define p and T, respectively, as first and second-rank tensors. The laws for tensors

of higher rank are given below (Nye); mercifully, I do not think it necessary for us to

prove them.

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7. Crystal Symmetry and Second-Rank Tensors: a Mathematical Approach

We can also use our transformation laws for second-rank tensors to deduce which of

their coefficients are forced by the crystal symmetry to be equal to each other or equal

to zero.

We will take as an example the tetragonal point group 422.

Firstly, we note that the effects of symmetry may be viewed in two equivalent ways, as

shown in the diagrams below. The object shown in (a) has four-fold symmetry. We can

see this either: (b) by rotating the object by 90o, keeping the axes fixed and noting it

looks the same after the transformation or (c) by rotating the axes by 90o, keeping the

object fixed and noting it looks the same after the transformation. It does not matter

which method we use, they are entirely equivalent.

(a)

(b)

(c)

We can apply the second method to our second-rank tensors; if our change in axes

corresponds to a symmetry operation of the crystal, then the coefficients of the tensor

must be the same before and after the transformation of axes (i.e. the tensor must “look”

exactly the same on the new set of axes); thus, the coefficients must transform such that

Tij = ijT  .

Let us firstly consider the effect of the 4-fold axis in point group 422; this axis lies

parallel to Ox3. The matrix of direction cosines for a transformation of axes

corresponding to this symmetry element is then

Old

Ox1 Ox2

Ox3

1xO 

0

1

0

New 2xO 

-1

0

0

3xO 

0

0

1

and so

T11

= 11T 

= a11 a11T11

+

a11 a12T12

+ a11 a13T13

+ a12 a11T21

+

a12 a12T22

+ a12 a13T23

+ a13 a11T31

+

a13 a12T32

+ a13 a13T33

= 0T11

+

0T12

+

0T13

+ 0T21

+

1T22

+

0T23

+ 0T31

+

0T32

+ 0T33

= T22

Thus, the four-fold symmetry requires that T11

=

T22

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What about T33?

T33

= 33T 

= a31 a31T11

+

a31 a32T12

+ a31 a33T13

+ a32 a31T21

+

a32 a32T22

+ a32 a33T23

+ a33 a31T31

+

a33 a32T32

+ a33 a33T33

= 0T11

+

0T12

+

0T13

+ 0T21

+

0T22

+

0T23

+ 0T31

+

0T32

+ 1T33

= T33

Thus, the four-fold symmetry places no restriction on T33

(as it is unchanged by the

transformation of axes)

Now let’s try T12

T12 = 12T 

= a11 a21T11

+

a11 a22T12

+ a11 a23T13

+ a12 a21T21

+

a12 a22T22

+ a12 a23T23

+ a13 a21T31

+

a13 a22T32

+ a13 a23T33

= 0T11

+

0T12

+

0T13

+ (-1)T21

+

0T22

+

0T23

+ 0T31

+

0T32

+ 0T33

= - T21

=

-T12

(since the tensor is symmetric; see Section 3/.)

The only way in which the relationship T12 = -T12 can be true is if T12 = 0.

Thus, the four-fold symmetry requires that T12 = 0

What about T13?

T13

= 13T 

= a11 a31T11

+

a11 a32T12

+ a11 a33T13

+ a12 a31T21

+

a12 a32T22

+ a12 a33T23

+ a13 a31T31

+

a13 a32T32

+ a13 a33T33

= 0T11

+

0T12

+

0T13

+ 0T21

+

0T22

+

1T23

+ 0T31

+

0T32

+ 0T33

= T23

Thus, the four-fold symmetry requires that T13

= T23

Finally, we consider the effect of one of the 2-fold axes in 422; we shall choose the 2- fold axis parallel to Ox1. The matrix of direction cosines for a transformation of axes

corresponding to this symmetry element is then

Old

Ox1 Ox2

Ox3

1xO 

1

0

0

New 2xO 

0

-1

0

3xO 

0

0

-1

15

Once again we consider the transformation of T13

T13

= 13T 

= a11 a31T11

+

a11 a32T12

+ a11 a33T13

+ a12 a31T21

+

a12 a32T22

+ a12 a33T23

+ a13 a31T31

+

a13 a32T32

+ a13 a33T33

= 0T11

+

0T12

+

(-1)T13

+ 0T21

+

0T22

+

0T23

+ 0T31

+

0T32

+ 0T33

= -T13

Thus, the two-fold symmetry requires that T13

= 0

(and so T23 = 0 also)

We have now shown that, when described with respect to the standard

crystallographic axes, any second-rank tensor in point group 422, must have the form

T11 0 0

0 T11 0

0 0 T33

and its representation surface is, therefore, an ellipsoid of revolution.

Now try another one for yourself in Exercise 6.

8/. Some examples of Second-Rank tensors

We have already encountered, in some detail, two second-rank tensors: electrical

conductivity and thermal conductivity. Some other examples include:

a/. Stress

Stress is defined as force per unit area. Suppose that we have a unit cube of material, cut

perpendicular to our reference axes Ox1, Ox2, Ox3, as shown below.

We may resolve any forces acting on each

face of the cube into their components

parallel to our reference axes.

We may then specify the stress by the nine coefficients, ij, where the first suffix, i,

denotes the direction in which the force acts and the second suffix, j, denotes the face on

which it acts (e.g. j = 1, denotes the face perpendicular to Ox1 etc.).

16

It is apparent from the figure that:

11, 22, and 33 are normal components of stress (i.e. purely tensile or compressive),

whereas,

12, 13, 23 etc. are the shear stresses.

It can be shown that the components of ij, thus defined, form a second-rank tensor.

By taking moments, it can also be shown that, if our cube is to remain stationary, the

tensor must be symmetrical, i.e. 23 = 32 etc.

Some special forms of the stress tensor are:

A Uniaxial stress

Hydrostatic Pressure

A Pure Shear

          000 000 00

          − − − p p p 00 00 00

          000 00 00  

Note that the stress tensor does not represent a crystal property, but instead is akin to a

force applied to the crystal; there is thus no requirement that it conform to any crystal

symmetry. It is therefore, termed a field tensor, rather than a matter tensor (such as

electrical conductivity).

b/. Strain

Strain is defined by “change in length/original length”. Consider the 1-dimensional

example shown below.

The line element, x, at a point x on the line, is altered in length to (x + u) by the

deformation. Thus the “change in length/original length” = u/x. The strain e is

defined by the value of u/x as x tends to zero, i.e.

dx du e =

In 3-dimensions, the point P will have coordinates [x1, x2, x3], and a general line element

at this point will be [x1, x2, x3]. Each of the three components of this line element

can have components of displacement parallel to each of the three axes. Thus, to specify

the strain, in the limit of the line element tending to zero, we need the 9 coefficients

j i ij x u e   =

which tell us the displacements in directions “i” of the line elements that were parallel

to directions “j”.

The strain tensor, ij, is then defined as the symmetrical part of eij; the antisymmetric

part of eij (Nye calls this ij ) corresponds to a rotation of the body.

17

It is probably easiest to visualise the meaning of the strains by considering the diagrams

below;

11, 22 and 33 correspond to tensile strains, e.g. 11 is the extension, parallel to Ox1 of a

line of unit length originally (i.e. before deformation) parallel to Ox1.

12, 13 and 23 correspond to shear strains, e.g. 12 is the displacement, parallel to Ox1, of

a line of unit length originally parallel to Ox2. Since the displacements are relative to

lines of unit length, the tangents of the angles marked are equal to 12 and 21. As the

displacements are assumed to be small, the tangent ≈

the angle measured in radians and

so (12 +21) = 212 is the total change in angle between line elements originally parallel

to Ox1 and Ox2

The tensile strain 11

The shear strains 12 and 21

Exercise 7 gives you some experience in defining stresses and evaluating strains.

c/. Thermal Expansion

A small temperature change, T, will produce a strain in a crystal

Tijij = 

Since T is a scalar, the coefficients of thermal expansion (like the strains they produce)

must constitute a second-rank tensor.

Note, however, that the expansion coefficients are a matter tensor and their values must

be in accordance with the crystal symmetry via Neumann’s principle.

Exercise 8 lets you derive the principal coefficients and the directions of the

principal axes of a section of the thermal expansion tensor of the mineral afwillite.

d/. Relative Dielectric Impermeability, and “The Optical Indicatrix”

We shall consider, for the moment, an isotropic material.

When an electric field, E, is applied to this substance it induces an electrical

polarisation, P, which is given by

P = χ εo E

Where χ is the electric susceptibility of the material and εo is a scalar constant, called the

permittivity of free space (εo = 8.854 x 10-12 Farad m-1).

The electric displacement vector, D, is then defined as

D = oE + P = o(1 + χ) E = or E

18

where r = (1 + χ), is called the relative permittivity (in older books this may still be

called the dielectric constant).

NB: in Section 8.4 no longer denotes a strain – I have not changed the symbol

to something else as  is almost universally used in electricity and magnetism

textbooks.

For materials that are not strongly magnetic, electromagnetic wave theory (derived via

Maxwell’s equations) then tells us that the refractive index, n = √r

We may define another quantity, B, the relative dielectric impermeability, as the

reciprocal of the relative permittivity; thus, B = 1/r = (1/n)2 and n = 1/√B

In an anisotropic material, r and B will become second-rank tensors. The equation of

the representation surface for B (referred to the principal axes) is then

12333 2 222 2 111 =++ xBxBxB

This surface will be an ellipsoid, cutting the principal axes at points 1/√B11, 1/√B22,

1/√B33; i.e. at n1, n2, n3, the values of the three principal refractive indices.

The representation surface for Bij is, therefore, exactly the same surface as that of the

optical indicatrix.

You should note that the refractive indices of a crystal do not, themselves, form a

tensor, but their variation with direction is governed by the behaviour of the dielectric

properties of the material, which are tensors. Thus, we find that

The optical indicatrix is the representation surface (quadric) of the relative dielectric

impermeability tensor (when I taught optical mineralogy to the first-year students, I

used to not tell them this, lest it frighten them too much!).

As well as the information in Nye’s book, there is an excellent discussion of crystal

optics written by Peter Gay, from the “physics of crystals” viewpoint, rather than simply

describing “how to use optics to identify minerals”; it is called An Introduction to

Crystal Optics (currently out of print but second-hand copies are easily obtainable).

The effect of the anisotropy of a crystal on its optical properties is far more profound

than most geologists realise. When dealing with these materials, we must now

distinguish between the ray direction of our light beam (this is the direction in which

the energy of the light travels, i.e. the direction that we “see”) and the wave normal

direction (the line at right-angles to the wave crests of the light). In free space and in

isotropic materials, these directions are the same.

However, in anisotropic materials, this may no longer be the case; for example, in

uniaxial crystals, the ray direction and wave normal direction of the ordinary ray are

the same (i.e. the two lines are parallel), but for the extraordinary ray they are, in

general, not the same (i.e. they are no longer parallel).

We may use the optical indicatrix to find not only the refractive indices, but also the ray

direction of the light, for any specified wave normal direction. The figures below show

how to do this, for the case of a uniaxial negative mineral such as calcite.

19

The figures above assume that on entering the crystal, the wave normal directions of the

extraordinary and ordinary rays are the same (this will be the case if the light enters the

crystal at right angles to its surface, i.e. “at normal incidence”, as shown in the figure

below).

The figure on the left, above, gives a 3-dimensional view of the uniaxial indicatrix; the

one on the right gives the principal cross section containing the optic axis and the wave

normal.

The refractive indices (no and ne') and the polarisation directions for the ordinary and

extraordinary ray are found, exactly as you learnt in “Earth Materials”, by taking a slice

through the indicatrix at right angles to the wave normal direction. For the ordinary ray,

the wave normal and the ray directions are always the same (i.e. direction ON in the

figure above on the right) and the ray is polarised perpendicular to the paper. For the

extraordinary ray, the ray direction is found by finding the point, R, where the tangent

to the surface of the indicatrix is perpendicular to the wave normal and the polarisation

direction is parallel to this tangent. Inspection of the figures above readily shows that

the only orientations for which the extraordinary ray direction and wave normal

direction are parallel occur when the extraordinary ray is travelling perpendicular to

the optic axis.

The effect of the different ray directions is shown in the diagram below

When light hits a block of calcite at normal incidence, the ordinary ray travels straight

through (this wave is polarised at right angles to the plane of the paper and its

polarisation direction is shown as a set of dots). The extraordinary ray (polarised in the

plane of the paper, as indicated by “arrows”) travels like a crab through the crystal, with

20

the ray having a component of “sideways” motion. This is why the two beams emerge

from a parallel plate of calcite displaced from each other.

As thin sections of rocks are typically only 30 μm thick this “sideways offset” of the

light does not present a problem, and so it is commonly “glossed over” by those

lecturing on this subject!

The ray directions for biaxial minerals are yet more complex; those of you who are

interested are referred to Peter Gay’s book.

Exercise 9 lets you use some of these additional properties of the optical indicatrix.

9/. Elasticity; a Fourth-Rank Tensor Property

a/. Stiffness and Compliance

Stresses and strains are related by elastic moduli. Since both stress and strain are

second-rank tensors, with 9 components each, we might naively expect to need 81

coefficients to link them. However, the fact that the stress and strain are symmetric etc.

reduces the maximum number of independent coefficients to 21 (for triclinic crystals).

The elastic stiffnesses, c, are defined by

σij =

cijkl εkl

and the elastic compliances, s, by

εij

= sijkl σkl

Remember that the Einstein summation convention applies to the right-hand side of

the equations.

Clearly, c = 1/s (though the inversion, which involves a 6x6 matrix – see below, is not

trivial; note, however, that spreadsheet programs such as Excel can now do this).

Although the equations above look rather horrid, they are easy enough to use; for

example,

σ11 =

c1111 ε11 + c1112 ε12 + c1113 ε13 + c1121 ε21 + c1122 ε22 + c1123 ε23 + c1131 ε31

+ c1132 ε32 + c1133 ε33

Thus, we find, for example, that c1111 is the coefficient that relates a tensile stress σ11 to

a tensile strain ε11.

In practice, for higher symmetry crystals many of the coefficients cijkl may be required

by symmetry to be either zero or to be equal to each other.

b/. Matrix Notation for Elasticity

The equations above are rather along-winded, especially as the tensors are all

symmetrical.

We may make them more compact by putting:

σ11 = σ1 σ22 = σ2 σ33 = σ3 σ23 = σ4 σ13 = σ5 σ12 = σ6

ε11 = ε1 ε22 = ε2 ε 33 = ε3 ε23 = ½ε4 ε13 = ½ε5 ε12 = ½ε6

i.e.

21

The stiffness tensor, cijkl, then becomes, cmn, where the pairs of suffixes ij and kl (i, j, k, l

= 1, 2, 3) are each replaced by suffixes m and n which take the values m, n = 1, 2, 3, 4,

5, 6.

However, in the case of the compliances, sijkl, it is also necessary to introduce some

factors of 2 and 4 when making this change, as follows (see Nye for details):

sijkl = smn

(when

m and n are 1, 2, 3),

e.g. s12

= s1122

2sijkl = smn (when either m or n are 4, 5, 6) e.g. s16 = 2s1112 (or s16 = s1112 + s1121)

4sijkl = smn (when both m and n are 4, 5, 6)

e.g. s46 = 4s2312 (or s46 =s2312+s3212+s2321+s3221)

In this new notation, our equations take the form

σi =

cij εj

εi = sij σj (where, in both cases, the suffixes, i and j, run from 1 to 6)

and the elastic constant matrices are written as

Note that, when written in this way, cij and sij have the appearance of second-rank

tensors – but they are not; if we wish to transform axes we must go back to the full four- suffix notation.

The figure at the end of this section shows the form of the elastic constants for all 32

crystal classes and also those for a truly isotropic material (e.g. a glass); information on

how to use the figure is given in the key. Note that for these fourth-rank tensor

properties, cubic crystals and truly isotropic materials behave differently.

c/. Some Examples

i/. The Incompressibility (Bulk Modulus) of a Cubic Crystal.

The incompressibility, K, relates a change in volume to an applied hydrostatic pressure,

by

K = -V(dP/dV)

If an applied pressure, ΔP, is hydrostatic, σ1 = σ2 = σ3 = -ΔP and σ4 = σ5 = σ6 = 0 (i.e.

there are no shear stresses). Also, we may assume (from the symmetry) that ε1 = ε2 = ε3

= ε and ε4 = ε5 = ε6 = 0.

Remembering the meaning of ε1 (see the diagram in Section 8b), we find that the

volume of a strained cube of crystal, originally with sides 1 unit in length, is

(1+ ε1) (1+ ε2) (1+ ε3) ≈

[1 + (ε1 + ε2 + ε3)]

(terms in ε2 and ε3 may be ignored as ε<<1)

Thus (ΔV/V) = (ε1 + ε2 + ε3) = 3ε

22

From our elasticity theory, we have

σ1 =

c11 ε1 + c12 ε2 + c13 ε3 + c14 ε4 + c15 ε5 + c16 ε6

which reduces, in our special case of hydrostatic pressure, to

-ΔP = c11 ε + c12 ε + c13 ε

Also, for cubic crystals, c12 = c13, and so

-ΔP = (c11+ 2c12)ε = (c11+ 2c12)(ΔV/3V)

i.e. the incompressibility,

3 )2( 1211 ccK + =

ii/. The Elastic Anisotropy of Cubic Crystals

We may define a “Young’s Modulus” , E, for any crystal in any direction by cutting a

bar of the crystal in this direction (which we may call 1xO  ) and then measuring the ratio

of the applied load per unit area to the change in length of the bar.

Thus, for a bar of a cubic crystal cut parallel to [100]

ε1 =

s11 σ1 and, therefore, E100 = 1/ s11

Suppose we now choose a direction parallel to [110]

We must now transform our compliance tensor, using the matrix below, to find s11'

Old

Ox1 Ox2

Ox3

1xO 

1/√2 1/√2

0

New 2xO 

-1/√2

1/√2

0

3xO 

0

0

1

To do this, we have to return to the four-suffix notation and then re-transform to two

suffixes; we then find that

s11' = (2s11 + 2 s12 + s44)/4 = 1/E110

Thus, even for a cubic crystal, the modulus in directions [110] and [100] is different

In general, if we consider deformation of a cubic crystal under uniaxial load in terms of

the change in d-spacing of planes of index (hkl), we may define a single-crystal, plane- specific, Young’s modulus, Ehkl, which is given by the equation

( )         ++ ++       −−−= 2222 222222 44 121111 2 21 lkh lklhkhs sss Ehkl

where the term in brackets has limiting values of 0 for (h00) planes and 1/3 for (hhh)

planes; for (110) planes this is easily reduced to the expression given above.

23

The condition for a material to be truly isotropic under uniaxial loading is, therefore,

that

0 2 44 1211 =      −− s ss

It took me a while to do the transformation above; I suggest that you have a go in

Exercise 10.

d/. Polycrystalline Aggregates

In general, Earth and Planetary scientists tend to require knowledge of the elastic

properties of assemblages of minerals, rather than those of single crystals.

However, even if the assemblage contains only one mineral, it is not straightforward,

because of the crystalline anisotropy, to calculate the elastic properties. These will lie

somewhere between those of the Reuss bound, assuming uniform stress (i.e. all

crystallites in the assemblage supporting an equal load), and the Voigt bound, assuming

uniform strain (i.e. all crystallites in the assemblage being equally deformed); in

practice an average of the two bounds, called the Voigt-Reuss-Hill average is

commonly used (see e.g. Poirier, Introduction to the Physics of the Earth’s Interior,

pub. Cambridge University Press; den Toonder et al. Modelling and Simulation in

Materials Science and Engineering 7, 909-928, 1999 for further discussion). Assuming

that the “rock” is a fully dense, single-phase aggregate, the incompressibility (K) of the

aggregate (which is isotropic) must have the same value as can be obtained by applying

the Reuss or Voigt bounds to the individual crystals – but, except for cubic crystals, the

Reuss and Voigt values for K will differ (in the case of cubic crystals the Voigt bound is

effectively always imposed by the symmetry when calculating K). Clearly for poly- mineralic rocks, or those which are not fully dense, the position is still more complex.

For an isotropic aggregate, we need only two independent elastic constants, and the

non-zero terms in the cij matrix are:

c11 = c22 = c33

c12 = c13 = c23

c44 = c55 = c66 = ½( c11 – c12)

Note that these cij refer to the values for the aggregate, not to those of the crystals from

which the aggregate is composed.

Just as we found for the example in Section 9.c.i above, the bulk modulus, K, of an

isotropic material is given by

3 )2( 1211 ccK + =

Also, for isotropic materials, c44 = ½( c11 – c12) is usually termed the shear modulus, μ.

Poisson’s ratio, ν, is another quantity commonly used for isotropic materials. Although

Poisson’s ratio applies generally, it is less useful for crystals as its value will then vary

with direction. For a material loaded uniaxially along Ox1 (i.e. σ1 ≠

0 but all other σi =

0), Poisson’s ratio is given by the ratio of the thinning to the elongation; thus,

24

1 3 1 2      −=−=

Exercise 11 lets you derive Poisson’s ratio in terms of both sij and cij.

e/. Seismic Waves in Isotropic Aggregates

The seismic P-wave and S-wave velocities, VP and VS, are given by the expressions

2/1 3 4             + =   K VP

and

2/1       =   SV

We can also write ν in terms of K and μ, obtaining

      + −      = 1 3 2 2 3    K K

From the expressions for the seismic velocities, we find that

3 4 2 +=       K V V S P

and so

3 4 2 −      = S P V VK 

By substituting this into the equation for ν, above, we finally obtain

        −      −      = 12 2 2 2 S P S P V V V V 

Therefore, Poisson’s ratio (a measurable property for materials in the laboratory) can be

obtained directly from PREM and other seismic models as a function of depth within

the Earth

Just Exercise 12 to try now; it gives you a chance to derive some of the equations

quoted above for yourselves. 25

Form of the Elastic Constant Matrices in the 32 Crystal Classes

(taken from Nye)

26

Exercise 1:

With respect to axes Ox1, Ox2, Ox3, the electrical conductivity, σij, of a crystal can be

represented by the matrix

          − − 77 77 7 10x1310x2.50 10x2.510x70 0010x25

Ohm-1 m-1

Calculate the components of the current density, ji, along each of the axes and the total

magnitude of the current density, j, when an electric field E of 104 V m-1 is applied in

the following directions; for (a), (b) and (c), calculate also the angle between j and Ox2.

(a) parallel to Ox1

(b) parallel to Ox2

(c) parallel to Ox3

(d) in the Ox1 - Ox2 plane, midway between the two axes.

(e) in the Ox1 – Ox3 plane, midway between the two axes.

(f) such that E makes equal angles with Ox1, Ox2, Ox3 and lies in the octant where all of

its components are positive,

Exercise 2:

Constructing and Using a Representation Surface for Thermal Conductivity

Thermal conductivity is governed by the equation (dQ/dt) = - ΚA(dT/dx), where (dQ/dt)

is the flow of heat in J s-1, A is the cross-sectional area (m2), (dT/dx) is the temperature

gradient (K m-1) and Κ is the thermal conductivity (J s-1 m-1 K-1).

The principal thermal conductivity coefficients of a certain tetragonal crystal are:

Κ11 = Κ22 = 100 J s-1 m-1 K-1

Κ33 = 400 J s-1 m-1 K-1

1/. Using a scale of 1 m = 1 [J s-1 m-1 K-1]-1/2 draw the Ox2 – Ox3 section of the

conductivity representation quadric.

Hint for drawing the ellipse. First calculate where it crosses the principal axes, Ox2 and

Ox3. Next use the equation of the ellipse, here 1 2 333 2 222 =+ xSxS

(since x1 = 0) to

calculate values of x3 for a set of suitably chosen values of x2; you can then sketch in the

ellipse by hand between these points.

2/. A line, OP, lies in the Ox2 – Ox3 plane, making angles of 30o with respect to Ox2

and 60o with respect to Ox3.

Draw OP on your diagram, measure the radius of the representation surface and

thus find the magnitude of the thermal conductivity in this direction.

3/. Draw a line perpendicular to the representation surface at point P.

Probably the easiest way to do this is to use a plastic ruler to draw a line parallel to the

surface and then draw a line at right-angles to this.

This line will show the direction of heat flow, dQ/dt, when a temperature gradient,

-dT/dx, is applied along the line OP.

Measure the angle between this heat flow direction and the axis Ox2.

Contd. on next page

27

(Exercise 2/. Contd.)

4/. We shall now use numerical methods to check that our construction has, indeed,

given the correct values for the thermal conductivity and the heat flow direction.

We suppose that a temperature gradient of 104 K m-1 is established along OP; note that

this is a vector quantity.

a/. Calculate the components of this vector parallel to Ox2 and Ox3

A little thought shows that, if the temperature gradient, -dT/dx, is directed along one of

the principal axes of the representation quadric, the corresponding heat flow, dQ/dt, will

be parallel to it.

b/. Calculate the components of the heat flow vector, dQ/dt, parallel to Ox2 and Ox3

(you may assume unit area).

c/. Draw these vector components of the heat flow to scale on your diagram and

hence find the net direction of heat flow. Calculate the angle that this vector makes

to the Ox2 and Ox3 axes. This vector should be parallel to the line that you obtained

in (3/.) – is it?

d/. Calculate the angle between the heat-flow vector and the line OP. Resolve the

heat flow vector so as to find the component of heat flow along OP, and hence find

the magnitude of the thermal conductivity in this direction.

Does your result agree with the value that you found in (2/.)?

Exercise 3: Changing Axes (1)

(i) Write down the direction-cosine matrices for the three pairs of axes ( 1xO  , 2xO  , 3xO  )

and (Ox1, Ox2, Ox3), where ( 1xO  , 2xO  , 3xO  ) are obtained as follows:

a/. by rotation through 90o about Ox3 (as shown on the first page of Section 7.)

b/. by rotation through 180o about Ox2

c/. by rotation through 45o about Ox3 (rotate the axes in the same direction as in a/.)

(ii) A vector, q, has components [1, 1, 0]

What are its components on the three new sets of axes (a/., b/., c/.) above?

Confirm in each case that the magnitude of the vector is unchanged by the

transformation of axes

Draw diagrams to confirm that your transformed coordinates are correct.

Exercise 4: Changing Axes (2)

Using the method given in the lecture notes, prove the general relationship jjll qaq =

where q is a vector and both suffixes, l and j, run from 1 to 3.

Exercise 5: Changing Axes (3)

For the three transformations of axes given in Exercise 3 above, find the new values of

the following components of the second-rank tensor Tij

11T 

33T 

12T 

13T 

28

Exercise 6: Symmetry and Second-Rank Tensors

A monoclinic crystal, of point group 2, has its 2-fold axis parallel to Ox2.

(i) Write down the direction cosine matrix to transform a set of orthogonal axes by this

symmetry operation.

(ii) By transforming T11, T22, T33, T12, T13 and T23, determine the restrictions placed by

this 2-fold symmetry upon the coefficients of the tensor.

Exercise 7: Stresses and Strains

i/. Express in words the meaning of σ11, σ12 and σ21.

ii/. Using a scale of 10 cm = 1 unit, draw diagrams to show the meaning of:

(a) ε33, when ε33 = 0.05.

(b) e13 and e31 (use one diagram) with e13 = 0.04 and e31 = 0.12

(c) Now evaluate the symmetric (ε13 and ε31) and antisymmetric ( 13 and 31 ) parts of

e13 and e31, using the expressions εij = ½(eij + eji) and ij = ½(eij - eji). Plot the shear

strains ε13 and ε31 on the diagram you used for (b). You should find (allowing for the

fact that the strains are not infinitesimally small) that the unequal shear strains of Part

(b) have been resolved into a symmetric strain + a rotation.

Exercise 8: The Thermal Expansion Tensor

The monoclinic crystal afwillite, Ca3(SiO3OH)2.2H2O has cell dimensions

a = 16.21 Å b = 5.63 Å c = 13.23 Å 

= 134.8o

X-ray measurements of the thermal expansion perpendicular to the following Bragg

planes h0l gave the following results

h

0

l



h

0

l



4

0

14 13.7 x 10-6 K-1

20

0

8

24.3 x 10-6 K-1

0

0

12 17.1 ”

16

0

12

11.8 ”

6

0

8 21.4 ”

8

0

12

8.2 ”

18

0

4

28.2 ”

Using a sheet of A3 graph paper and scale of 1 cm = 1 Å, draw the cross section of the

unit cell perpendicular to the b-axis and mark on it the directions of the normals to the

planes listed above (note that when making the drawing it is sometimes easier to

consider a parallel plane to the one listed, e.g. use 4

0

3̅ instead of 16

0

12).

Along these normals, plot points at distances 1/ from the origin (a suitable scale is

“distance in cm = 0.025*1/”).

By connecting these points you will be able to draw the principal section of the thermal

expansion tensor perpendicular to [010].

By measuring your diagram, find the values of the principal coefficients and the

directions of the principal axes (after measuring the diagram, don’t forget to take out the

scale factor when calculating the values).

29

Exercise 9: Ray Directions in Uniaxial Crystals

A uniaxial crystal has principal refractive indices ne = 1.3 and no = 2.0

(Note that it is extremely unlikely that any real substance could be this birefringent!)

(a) What is its optic sign?

(b) By drawing a cross-section of the indicatrix, determine the two refractive indices, ne'

and no for light with a wave normal direction inclined at 30 o to the optic axis.

(10 cm = 2 units of refractive index, should give a suitable scale)

(c) Using the construction given in the lecture notes, find the difference in angle

between the ray directions for the ordinary and extraordinary rays.

Exercise 10: Elasticity

(i) Explain what is meant by ε1, ε6, c12 and s66.

(ii)

Using the diagram from Nye (given at the end of the lecture notes on page 25)

showing the form of the sij and cij matrices, write down all of the non-zero components

of sij for a cubic crystal, grouping together the terms that are equal. Now write out the

corresponding sets of coefficients on the full, four-suffix notation, sijkl, once again

grouping together terms that are equal; you should find that there are 21 non-zero

values.

(iii) A bar is cut from a cubic crystal parallel to [111].

Choose a new set of axes with 1xO 

parallel to [111] and write down the three direction

cosines, a11, a12 and a13 (you will not require any other values of aij).

Now transform the sijkl values to obtain 1111s

on the new axes. Having transformed your

tensor, return the sijkl involved in the expression for 1111s

to two-suffix notation

(remembering to include, where necessary, the factors of 2 and 4 – see lecture notes,

page 21).

Congratulations (!), you have now evaluated 1/E[111], where E[111] is the Young’s

modulus (E = [load per unit area]/[strain produced in direction of load]) of the crystal

in the [111] direction.

Exercise 11: Elastic properties of Isotropic Substances

A uniaxial stress, σ1, is applied to an isotropic material (i.e. all other σi = 0).

i/. Use the equations, εi = sij σj

to determine the values of ε1, ε2 and ε3, and hence the

value of Poisson’s ratio, ν, in terms of sij.

ii/. Now use the equations, εi = sij σj

to show that, for this uniaxial load, ε4, ε5 and ε6 are

all zero.

iii/. Using your knowledge of the non-zero strains from parts (i) and (ii) and the

equations σi = cijεj, write down the expression for σ2. We know that this stress is zero;

use the resulting equation to determine ν in terms of cij.

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Exercise 12: Applications to Seismology

i/. By using the values of K and μ in terms of cij for an isotropic material, show that the

expression

      + −      = 1 3 2 2 3    K K

reduces to the expression for ν that you found in Exercise 11, Part (iii.).

ii/. Starting with the expressions for VP and VS given in the lecture notes (page 24) and

the expression for ν given above, prove that

        −      −      = 12 2 2 2 S P S P V V V V 

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