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DoITPoMS Teaching & Learning Packages Tensors in Materials Science Tensors in Materials Science (all content)

# Tensors in Materials Science (all content)

Note: DoITPoMS Teaching and Learning Packages are intended to be used interactively at a computer! This print-friendly version of the TLP is provided for convenience, but does not display all the content of the TLP. For example, any video clips and answers to questions are missing. The formatting (page breaks, etc) of the printed version is unpredictable and highly dependent on your browser.

## Aims

On completion of this TLP you should:

• Understand what a tensor is
• Be familiar with some of the applications of tensors to Material Science
• Understand the significance of the representation surface of a tensor
• Be able to find the principal values and axes of a tensor
• Be able to transform tensors from one frame to another
• Be able to use the representation surface for a second rank matter tensor to describe and calculate material properties as a function of direction

## Before you start - The basics

This TLP requires a basic knowledge of vector and matrix algebra, including scalar products, matrix multiplication, 3x3 determinants and suffix notation. The TLP includes a section which revises these concepts; you can skip this section if you do not need to cover this material again.

If you are unfamiliar with the concept of anisotropy in materials, you should follow the TLP on Anisotropy before starting this one.

Similarly, you may need to read through the Crystallography TLP to gain the understanding of crystal systems and symmetry elements which will be needed in order to follow the section on the effects of symmetry on tensors.

## Introduction

Many physical phenomena of interest in materials science are naturally described by tensors including thermal, mechanical, electrical and magnetic properties. In isotropic materials, many properties (e.g. electrical conductivity) can be described by a single number, a scalar. However in a general crystalline solid, these properties can vary with the direction in which they are measured - and tensors are needed to describe them fully.

This TLP offers an introduction to the mathematics of tensors rather than the intricacies of their applications. Its aims are to familiarise the learner with tensor notation, how they can be constructed and how they can be manipulated to give numerical answers to problems.

## Scalars, Vectors and Matrices

Before we can move on to tensors, we must first be familiar with scalars, vectors and matrices. If you are comfortable with these concepts, you can move on to the next page.

## Scalars

• These are direction independent quantities that can be fully described by a single number, and are unaffected by rotations or changes in co-ordinate system. Examples of physical properties that are scalars: Energy, Temperature, Mass.
• For this TLP scalars will be written in italics.

## Vectors

• These are objects that possess a magnitude and a direction, and are referenced to a particular set of axes known as a basis. A basis is a set of unit vectors (vectors with a magnitude of 1) from which any other vector can be constructed by multiplication and addition.
• The vector is referenced to the basis by its components. If possible, the maths is simplified by using an orthonormal base with orthogonal (mutually perpendicular) unit vectors. Examples of physical properties that are described by vectors: Mechanical force, Heat flow, Electric field.
• Vectors will be written in bold and components of a vector, say x, will be written as xi

## Matrices

• A matrix is a mathematical object that contains a rectangular array of numbers that can be added and multiplied (according to matrix multiplication rules). They are very useful in many applications, for example in reducing a set of linear equations into a single equation, storing the coefficients of linear transformations (e.g. rotations), and as we shall see, in describing tensors.
• The components of matrix A are written aij where i refers to the row element and j refers to the column element.

## Scalar products

• For two vectors: a = (a1, a2, a3) and b = (b1, b2, b3) The scalar product (also known as the dot product) is defined as: a.b = a1b1 + a2b2 + a3b3
and so, for example, the vectors (1, 4, −3) and (2, 5, 1) have a scalar product of 1×2 + 4×5 − 3×1 = 19.
• The scalar product is related to θ, the angle between the two vectors, and can equivalently be written as: a.b = |a||b|cosθ.
• For vectors of unit length, we can see that the scalar product is equal to the cosine of the angle between them.

## Matrix multiplication

If we have two matrices,

A =

 a11 a12 a13 a21 a22 a23 a31 a32 a33

and  B =

 b11 b12 b13 b21 b22 b23 b31 b32 b33

Then the product C = AB is found by 3k=1 aikbkj where i, j and k are indices that represent the position of the element in the matrix.

C = AB =

 a11 a12 a13 a21 a22 a23 a31 a32 a33

 b11 b12 b13 b21 b22 b23 b31 b32 b33

=

 a11b11 + a12b21 + a13b31 a11b12 + a12b22 + a13b32 a11b13 + a12b23 + a13b33 a21b11 + a22b21 + a23b31 a21b12 + a22b22 + a23b32 a21b13 + a22b23 + a23b33 a31b11 + a32b21 + a33b31 a31b12 + a32b22 + a33b32 a31b13 + a32b23 + a33b33

Don't bother trying to remember the above result, remember the rule:

ROW × COLUMN = RC = "Race Car" or "Really Cool!" or make up your own acronym to remember it. This is also useful to remember the conventional order of suffices, where the first suffix indicates the row and the second indicates the column.

You can use the following activity to practice more matrix multiplication.

## 3×3 determinants

The determinant of a 3×3 matrix can be calculated along any row by 'expanding by minors'. For the matrix

A =

 a11 a12 a13 a21 a22 a23 a31 a32 a33

The determinant is:

|A| =
 a11 a12 a13 a21 a22 a23 a31 a32 a33
=  a11
 a22 a23 a32 a33
- a12
 a21 a23 a31 a33
+ a13
 a21 a22 a31 a32

=  a11(a22a33 - a23a32) - a12(a21a33 - a23a31) + a13(a21a32 - a22a31)

The minor of a11 is the 2×2 determinant made up from the elements not in its row or column. The minus sign appears because expanding along the rows or columns follows the cofactor pattern:

 + - + - + - + - +

For more of a reminder about determinants see here.

## What is a Tensor?

Tensors are simply mathematical objects that can be used to describe physical properties, just like scalars and vectors. In fact tensors are merely a generalisation of scalars and vectors; a scalar is a zero rank tensor, and a vector is a first rank tensor.

The rank (or order) of a tensor is defined by the number of directions (and hence the dimensionality of the array) required to describe it. For example, properties that require one direction (first rank) can be fully described by a 3×1 column vector, and properties that require two directions (second rank tensors), can be described by 9 numbers, as a 3×3 matrix. As such, in general an nth rank tensor can be described by 3n coefficients.

The need for second rank tensors comes when we need to consider more than one direction to describe one of these physical properties. A good example of this is if we need to describe the electrical conductivity of a general, anisotropic crystal. We know that in general for isotropic conductors that obey Ohm's law:

j = σE

Which means that the current density j is parallel to the applied electric field, E and that each component of j is linearly proportional to each component of E. (e.g. j1 = σE1).

However in an anisotropic material, the current density induced will not necessarily be parallel to the applied electric field due to preferred directions of current flow within the crystal (a good example of this is in graphite). This means that in general each component of the current density vector can depend on all the components of the electric field:

j1 = σ11E1 + σ12E2 + σ13E3
j2 = σ21E1 + σ22E2 + σ23E3
j3 = σ31E1 + σ32E2 + σ33E3

So in general, electrical conductivity is a second rank tensor and can be specified by 9 independent coefficients, which can be represented in a 3×3 matrix as shown below:

σ =

 σ11 σ12 σ13 σ21 σ22 σ23 σ31 σ32 σ33

Other examples of second rank tensors include electric susceptibility, thermal conductivity, stress and strain. They typically relate a vector to another vector, or another second rank tensor to a scalar. Tensors of higher rank are required to fully describe properties that relate two second rank tensors (e.g. Stiffness (4th rank): stress and strain) or a second rank tensor and a vector (e.g. Piezoelectricity (3rd rank): stress and polarisation).

To view these and more examples, and to investigate how changing the components of the tensors affect these properties, go through the flash program below.

## Tensor Usage

Lots of physical quantities of interest can be described by tensors, and a small subset of the common ones is shown in the flash animation below. The tensors in the circles are those that can be applied and measured in any orientation with respect to the crystal (e.g. stress, electric field) and are known as field tensors. The tensors that link these properties are those that are intrinsic properties of the crystal and must conform to its symmetry (e.g. thermal conductivity), and are known as matter tensors.

Many of these quantities are described by symmetrical tensors (e.g. stress, electrical susceptibility), for which the off diagonal components Tij and Tji are equal (i.e. T12 = T21). Taking electrical susceptibility as an example, this means that applying a field in the 1 direction, produces a polarisation in the 2 direction, and this is equal in magnitude to the polarisation produced in the 1 direction if the field is applied in the 2 direction. Whilst this seems intuitively reasonable, the explanation for it is not immediately obvious and the mathematical proof is in fact quite complex. Readers who would like to follow the detailed argument can refer to the textbook by Nye (Physical Properties of Crystals: see the Going further page)

## Suffix notation

Suffices are used to represent components of tensors and vectors. For example in the case of a vector x = (x1 x2 x3) we can then refer to its jth component as xj. We can also refer to x as the vector xj where we know that j can take the values 1, 2 and 3 ( j is then known as a free suffix).

It is important to note that tensors are defined with respect to a basis just like with vectors, and that the individual components of the tensor change when the basis is changed, while the magnitude and physical meaning stay the same. Note that there are different conventions for the order of the suffices. In this TLP we use the tensor component Tij to represent the effect on the i axis due to action on the j axis.

## Einstein summation convention

Let us consider the equation: x.y = x1y1 + x2y2 + x3y3. This can be written as x.y = 3i=0 xiyi.

Using the Einstein summation convention, we can drop the sigma and just write this as x.y = xiyi, remembering to sum over all the indices. Another example of this is the equation y = (a.b)x, which can be written using the summation convention as yi = ajbjxi where j is summed over (known as a dummy suffix) and the value of i can be 1, 2 or 3 (i.e. is a free suffix). Note that in effect this represents 3 separate equations, one for each vector component.

If a suffix appears twice in a term it is a dummy suffix and is summed over, whereas free suffices appear once in every term.

A more complex example is: (|a|2 - c.a)x + |b|2y = zφ can be rewritten as (ajaj  clal)xi + bkbkyi = ziφ.

Second rank tensors have components in two directions. This leads to the components of the tensor A being written aij such that a tensor operating on a vector to give another vector y = Ax can be written yi = aijxj where we see that the suffix j is summed over. This also applies for tensor multiplication, for which C = AB becomes cij = aikbkj where k is summed over. Making use of this convention is a useful simplifying technique in proving tensor and vector properties.

## Voigt Notation

As we have seen, many physical quantities are described by symmetric tensors. Voigt notation (also known as matrix notation) is an alternative way of representing and simplifying these tensors. An example using a symmetrical second rank tensor (e.g. stress) is shown below:

 T11 T12 T13 T12 T22 T23 T13 T23 T33

=

 T1 T6 T5 . T2 T4 . . T3

=

 T1 T2 T3 T4 T5 T6

 Tensor Notation Voigt Notation 11 22 33 23,32 13,31 12,21 1 2 3 4 5 6

These substitutions allow us to represent a symmetric second rank tensor as a 6-component vector. Likewise a third rank tensor can be represented as a 3×6 matrix (keeping the first suffix e.g. T123 = T14), and a fourth rank tensor as a 6×6 matrix (doing the operation on the first two and then the last two suffices e.g. T1322 = T52). This is very useful as we can display every tensor up to 4th rank as a single two-dimensional matrix, simplifying the maths and making them easier to visualise. It is particularly useful for the equations of elasticity where σij = Cijklεkl can be converted to σi = Cijεj:

 σ1 σ2 σ3 σ4 σ5 σ6

=

 C11 C12 C13 C14 C15 C16 C21 C22 C23 C24 C25 C26 C31 C32 C33 C34 C35 C36 C41 C42 C43 C44 C45 C46 C51 C52 C53 C54 C55 C56 C61 C62 C63 C64 C65 C66

 ε1 ε2 ε3 ε4 ε5 ε6

It should be noted that for convenience some scaling factors are often introduced when converting tensors into Voigt notation. For example, by convention the off-diagonal (shear) components of the strain tensor ε are converted such that in Voigt notation they are equal to the engineering shear strain:

 ε11 ε12 ε13 ε21 ε22 ε23 ε31 ε32 ε33

=

 ε1 ½γ12 ½γ13 ½γ21 ε2 ½γ23 ½γ31 ½γ32 ε3

=

 ε1 ½ε6 ½ε5 . ε2 ½ε4 . . ε3

=

 ε1 ε2 ε3 ½ε4 ½ε5 ½ε6

As such, care must be taken when looking up numerical values and converting between notations to check that consistent definitions are used.

## Transformation of axes

As with a vector, every tensor is described with respect to a basis, and if we choose a different basis or different orientation from which to look at the problem, the physical meaning is the same but the components of the tensor will change.

Some orientations are easier to work in than others, due to the geometry of the problem or properties of the physical situation. We must learn how to move our problem from one frame into another. The transformation matrices which we require are pure rotations and are therefore given the symbol R.

## Transforming the basis

Let us consider the 2-dimensional simplification first:

.

We are rotating from the basis with basis vectors x and y into a new basis with basis vectors x' and y'. The new basis can be written in terms of the old basis by resolving the vectors: x' = xcosθ + ysinθ and y' = −xsinθ + ycosθ, which can be written as a matrix as:

 x' y'

=

 cosθ sinθ -sinθ cosθ

 x y

The components of this rotation matrix, R, are the cosines of the angles involved (known as direction cosines).

The component rij is the cosine of the angle between xj (old basis) and xi' (new basis) i.e. the component of xj resolved along xi'. For example, in general we can say:

x1' = r11x1 + r12x2 + r13x3

Since for unit vectors the scalar product is just the cosine of the angle between the two vectors, we can write: rij = xi'.xj

Written in full we get:

R =

 x1'.x1 x1'.x2 x1'.x3 x2'.x1 x2'.x2 x2'.x3 x3'.x1 x3'.x2 x3'.x3

This is the transformation matrix to go from old to new basis. To go from new to old basis, it is easily seen that the matrix is the transpose of the one above. So rotating and then rotating back gives: RRT = I i.e. the original result.

Therefore the inverse of the rotation matrix is its transpose.

## Transforming a vector

Consider the vector: a = a1x1 +  a2x2 +  a3x3 =  a1'x1' +  a2'x2' +  a3'x3'

We know that the component of a resolved onto the new 1 axis is: a1' = a.x1' = a1r11 + a2r12 + a3r13
The other components can be similarly resolved, giving the above result.

Therefore, to rotate a vector we use the following equation: a' Ra

## Transforming a second rank tensor

To derive the transformation law for a second rank tensors, let us consider the general tensor equation: p Tq (in the old basis), and p' T'q' (in the new basis). To transform the vectors we use: p' Rp and q' Rq

The above knowledge allows us to make some simple substitutions to see that: p' Rp RTq RTR-1q' RTRTq', and also that p' T'q'

Hence: T' = RTRT

In suffix notation, this can be written as the final transformation law: Tij' = rimrjnTmn, (or conversely) Tij = rmirnjTmn'

Note again that these both represent 9 equations, one for each component of the tensor.

## Transforming an nth rank tensor

For an nth rank tensor the transformation law is as follows Tijk...'   = rimrjnrko...Tmno... where there are n transformation matrices. The transformation laws are useful as we can then give a mathematical definition of a tensor as 'an object whose coefficients transform according to the rules above.'

This results in an object that retains its physical meaning whatever basis is used to describe it. This is an important concept, because transforming to a well-known basis usually simplifies the mathematics of a problem, as we will see in the next section.

## Principal axes

As we have seen, a general second rank tensor has the form:

 T11 T12 T13 T21 T22 T23 T31 T32 T33

However, in a particular basis, this takes a simpler form:

 T11' 0 0 0 T22' 0 0 0 T33'

i.e. All off diagonal elements are zero. These basis vectors are known as the principal axes (or directions) and the non-zero tensor components as the principal values. Note that in general the principal axes for a given property will not necessarily coincide with the crystal axes.

Using principal axes simplifies the mathematics and highlights the symmetry of the situation. Considering once again the case of electrical conductivity, when working in an arbitrary basis the equations take the form:

j1 = σ11E1 + σ12E2 + σ13E3
j2 = σ21E1 + σ22E2 + σ23E3
j3 = σ31E1 + σ32E2 + σ33E3

In the principal basis they take the form:

j1 = σ11E1
j2 = σ22E2
j3 = σ33E3

i.e. The effect of an action along a principal axis is also directed along that axis (the conductivities along each principal axis are of course different from each other).

## Finding the principal axes

As we have seen, in the principal basis the component equations become Tx = λx where λ is a constant of proportionality. This represents 3 different linear equations where λ has 3 possible values (the principal values).

T11x1 + T12x2 + T13x3 = λx1
T21x2 + T22x2 + T23x3 = λx2
T31x3 + T32x2 + T33x3 = λx3

There is a useful solution for this when |T − λI| =  0, i.e. when:

 T11 − λ T12 T13 T21 T22 − λ T23 T31 T32 T33 − λ
= 0

This gives a cubic equation in λ called the secular equation. To find the principal values we must solve this equation for λ. For each of the three solution for λ we find the vector x that solves the equation above. Each of theses solutions for x is a vector parallel to one of the principal axes. This vector can be of any length so long as it points along the principal axis, so generally we scale the vector so that it is of unit length, giving us an orthonormal basis.

It is worth noting that the principal values are called the eigenvalues of the matrix representing T, and the unit vectors along the principal axes are its eigenvectors. The general operation of finding these is not only useful when simplifying tensors, but is used throughout physics and chemistry for example in studying modes of vibration, and calculating energies in quantum mechanics. The activity below shows you how to find the secular equation and principal values of a symmetric second rank tensor.

If we are working with a tensor where the one of the principal values is given, i.e. a tensor of the form:

T =

 T11 T12 0 T21 T22 0 0 0 T33

Then we can use the Mohr's circle construct to geometrically find the two unknown principal values. This is demonstrated further in the Theory of Metal Forming TLP. Transforming our tensor into the principal basis Using what we know about transformation matrices, i.e. that rij = xi'.xj, we can see that the transformation matrix to rotate from the old into the principal basis is simply the matrix of normalised eigenvectors (e1, e2 and e3).

R =

 | | | e1 e2 e3 | | |

Below are two more flash programs to show you another example of finding the principal values and principal axes.

## The representation surface

The representation surface (or representation quadric) is a geometrical representation of a second rank tensor and is useful for giving us a visual image of the tensor as well as being useful for example in calculating magnitudes of material properties described by second rank tensors.

Let us consider the equation Tijxixj = 1. Here Tij represents a second rank tensor and xi and xj are coordinates.

This can be written in full as:

T11x12 + T12x1x2 + T13x1x3 + T21x2x1 + T22x22 + T23x2x3 + T31x3x1 + T32x3x2 + T33x32 = 1

which can be plotted to obtain a 3-dimensional graph. This graph is in fact a surface that is a complete description of T.

If we want to transform the surface to a new basis, making the substitutions xi = rkixk' and xj = rljxl', we can write the equation of the representation surface as Tijrkixk'rljxl' = 1, or equivalently as Tkl'xk'xl' = 1.
This means that: Tkl' = rklrijTij

Therefore any transformations of the tensor result in identical transformations of the 3D plot.

For symmetric tensors, the quadric equation for the representation surface simplifies to:

T11x12 + T22x22 + T33x32 + 2T12x1x2 + 2T13x1x3 + 2T23x2x3 = 1, which is the equation of an ellipsoid.

For non-symmetric tensors and those with negative principal values, the representation surfaces are more complex. As with the tensor itself, the representation surface has its simplest form when referred to the principal axes (when the basis vectors are in line with the radii of the ellipsoid), where the equation becomes: T1x12 + T2x22 + T3x32 = 1 (where T1, T2 and T3 are the principal values of the tensor).

## Magnitude of a property in a given direction

We will often want to talk about the magnitude of a property in a particular direction. For example if we apply an electric field to graphite in the [143] direction and measure the current density in the same direction, we would like to be able to describe the result as a measurement of the conductivity of graphite in the [143] direction.

Formally, the applied electric field E is described by a vector, as is the resultant current j. However conductivity requires a second rank matter tensor, and this means that these two vectors (E and j) will not in general be parallel to each other, and we will only measure the component of j which is parallel to E. For practical reasons it is therefore sensible to define the conductivity in a particular direction as the component of j that is parallel to E divided by the magnitude of E, i.e. j||/E.

We can apply this definition to a general second rank matter tensor T. If a field q = q(l1 l2 l3) is applied then the response of the material will be p = Tq.

Along the vector q the magnitude of T is given by:

$T = \frac{{{\bf{p}} \cdot {\bf{q}}}}{q} \times \frac{1}{q} = \frac{{{\bf{Tq}} \cdot {\bf{q}}}}{{{q^2}}} = \frac{{{T_{ij}}{q_i}{q_j}}}{{{q^2}}} = {T_{ij}}{l_i}{l_j}$

This can be simply related to the representation surface, as we know that the surface is decribed by the equation Tij xi xj = 1, where xi = r li and r is the radius. This gives us r2 Tij li lj = r2 T = 1

Hence the radius of the surface and the magnitude of the property it describes in a given direction are related by: = 1 / r2 or = 1 / √T

## The radius-normal property

The radius-normal property of a representation surface gives us a geometrical method of finding the effect of a second rank tensor for a given action, for example finding the current density for a given electric field.

The property states that for the tensor equation p = Tq, if we draw the representation surface of T and take q from the origin, the vector normal to the surface where q meets it, is parallel to the direction of p. The size of p is given by the previously explained 'magnitude in a given direction' formulae, i.e.

|p= |q| / r2

where r is the radius to the point on the representation surface.

The property will be demonstrated along the principal axes of T but it can be generalised to any basis.

Let vector q = q(l1 l2 l3) where q is the magnitude and l1, l2 and l3 are the direction cosines of q. The point Q is the point on the representation such that OQ is parallel to q, so that OQ = r(l1 l2 l3).

As the equation of the surface is T1 x12 + T2 x22 + T3 x32 = 1, the tangent plane at the point (a1 a2 a3) has equation T1 x1 a1 + T2 x2 a2 + T3 x3 a3 = 1.

Hence the normal is n = (T1 a1  T2 a2  T3 a3)  = r (T1 l1  T2 l2  T3 l3).

Now as q = q(l1 l2 l3) and we are working in the principal basis we find simply that p = q (T1 l1  T2 l2  T3 l3), showing us that p is parallel to n.

These properties of the representation surface give us a simple way of finding the magnitude of property in a certain direction.

## The effects of crystal symmetry

Matter tensors abide by a fundamental postulate of crystal physics known as Neumann's Principle. This 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' .

As we know, the physical properties of crystals are described by tensors and the point group of a crystal is the set of its macroscopic symmetry elements such as rotation axes, mirror planes, and centres of symmetry.

Taken with the 7 different crystal systems, the possible combinations of symmetry elements gives rise to the 32 crystal classes.

This postulate essentially puts conditions on the form of matter tensors depending on the crystal symmetry - the tensors describing the matter property must be invariant under its symmetry operations.

## Effects of symmetry on first rank matter tensors

• The vectors describing the matter property must be invariant under the symmetry operations. Straight away we see that any crystal with a centre of inversion cannot hold a first rank property since for a general vector p, ( p1, p2, p3 ) ≠ ( p1p2p3 ).
• We can also see with a little thought that if there is a rotation axis the vector property must lie along the rotation axis. An immediate consequence of this is that if the crystal structure has more than one rotation axis, then once again the crystal cannot possess the vector property since it cannot lie along two different rotation axes.
• If the crystal includes a mirror plane , the vector must lie within the plane. If there is more than one mirror plane, the vector must lie in the intersection.
• Finally if the crystal system contains a mirror plane and a rotation axes, the vector is non-zero only if the rotation axis is contained within the crystal plane.

The crystal classes that can possess a first rank matter tensor property, along with the number of independent components and the form of the vector are shown below.

 Crystal system Crystal class Number of independent components Form of the vector Triclinic 1 3 (p1, p2, p3) Monoclinic (diad axis parallel to x2) 2 1 (0, p, 0) Monoclinic m 2 (p1, 0, p3) Orthorhombic mm2 1 (0, 0, p) Tetragonal 4 1 (0, 0, p) Tetragonal 4mm 1 (0, 0, p) Triagonal 3 1 (0, 0, p) Triagonal 3m 1 (0, 0, p) Hexagonal 6 1 (0, 0, p) Hexagonal 6mm 1 (0, 0, p)

## Effects of symmetry on second rank tensors

Straight away we see that properties relating to a second rank tensor are centrosymmetric since by inverting the vectors in the equation pi = Tijqj, the same Tij satisfy the equation. So although the crystal may not have a centre of inversion, the tensor property does.

The best way to consider the conditions imposed by the crystal systems is to consider the representation surface and expressing its axes relative to the crystallographic axes. We shall consider rotations only as it can be demonstrated that mirror symmetries are covered by the rotation results.

The general representation surface has 3 mutually perpendicular diads, three planes of symmetry perpendicular to the diad axes and is centrosymmetric.

• Triclinic - Since there are no symmetry elements not possessed by the general representation surface, there are no restrictions on its components and so stays at 6 independent components. These components contain information on the magnitude of the principal values and the 3 angles required to define the orientation of the quadric axes to the crystallographic axes.
• Monoclinic - A diad of the representation surface must be aligned with the diad of the crystal system. Apart from this the surface is free to take any orientation, and so its independent components contain information about the three principal values and the one angle required to orientate the 2 free axes to the crystallographic axes.
• Orthorhombic - The crystal system contains 3 mutually perpendicular diads. On aligning the surface with the crystallographic axes we find only the principal value information is required. This also holds true for the mmm class.
• Uniaxial systems (tetragonal, triagonal and hexagonal) - The only way for the representation surface to possess 3-, 4- or 6-fold rotation symmetry is to align a diad along the crystallographic direction and revolve around it. This results in only 2 independent components since 2 of the principal values must be equal.
• Cubic - The four triad axes of the cubic system force the surface to become a sphere and so only a single component is required to define it.
 Crystal system Number of independent components Form of the tensor Triclinic 6 $\left( {\begin{array}{*{20}{c}} {{T_{11}}}&{{T_{12}}}&{{T_{13}}}\\ {{T_{21}}}&{{T_{22}}}&{{T_{23}}}\\ {{T_{31}}}&{{T_{32}}}&{{T_{33}}} \end{array}} \right)$ Monoclinic (diad axis parallel to x2) 4 $\left( {\begin{array}{*{20}{c}} {{T_{11}}}&0&{{T_{13}}}\\ 0&{{T_2}}&0\\ {{T_{31}}}&{{T_0}}&{{T_{33}}} \end{array}} \right)$ Orthorhombic 3 $\left( {\begin{array}{*{20}{c}} {{T_1}}&0&0\\ 0&{{T_2}}&0\\ 0&0&{{T_3}} \end{array}} \right)$ Tetragonal, Triagonal, Hexagonal 2 $\left( {\begin{array}{*{20}{c}} {{T_1}}&0&0\\ 0&{{T_1}}&0\\ 0&0&{{T_3}} \end{array}} \right)$ Cubic 1 $\left( {\begin{array}{*{20}{c}} T&0&0\\ 0&T&0\\ 0&0&T \end{array}} \right)$

## Summary

• Tensors can be used in a wide variety of Material Science fields, including but not limited to stress and strain, temperature and entropy, electricity and magnetism.
• A tensor is a set of coefficients which transform from one basis to another according to the transformation law: Tijk...'   = rimrjnrko...Tmno...
• The representation surface of a second rank symmetric tensor is an ellipsoid constructed from the equation: Tijxixj = 1.
• The principal values, λ, of a tensor can be found by solving the equation |T − λI| =  0 and the principal axes by finding the vector such that (T − λIx= 0.
• We can impose conditions on the components of matter tensors as they must adhere to the symmetry of the crystal class.

## Questions

### Quick questions

You should be able to answer these questions without too much difficulty after studying this TLP. If not, then you should go through it again!

1. Let a = (2,3,5) and b = (-1, 2, -4).
Calculate (the scalar product)

2. Let a = (2,3,5) and b = (-1, 2, -4).
Calculate the tensor product).

3. Let,  and . Find the matrix products AB, BA, AC, CAB and the determinants |A|, |AB| and |BA|.

4. What is the transformation matrix for a rotation through angle θ about Ox2?

5. A vector is first rotated through angle θ about Ox2 and then through φ about Ox1. What is the combined transform matrix?

6. Show that a pure shear stress field  can be represented as a pure normal stress field by rotating through 45° about the vertical axis.

7. The conductivity tensor of a crystal is found to be . Show that the crystal does not conduct in one direction and find this direction relative to the lab basis.

8. For the stress state with normal stresses σxx = 50 GPa, σyy = -70 GPa, σzz = 20 GPa and shear stresses σxy = 30 GPa, σxz = 45 GPa and σyz = 60 GPa, find the normal stress on the (263) plane in a cubic system.

9. Graphite has a layered hexagonal structure with cell dimensions a = 0.246 nm and c = 0.679 nm and has electrical conductivities parallel and perpendicular to the planes of σparallel = 1.02 × 105 S m-1 and σperpendicular = 0.24 × 105 S m-1 respectively.
A sample is mounted such that an electric field is set up along the [112] direction. The current density is measured parallel to the electric field.
By first constructing the conductivity tensor and the direction cosine vector, find the expected current density in this direction for a 100 V m-1 electric field.

## Going further

### Books

Physical Properties of Crystals by J. F. Nye, OUP - N.B. latest edition ISBN: 0198511655
Fundamental basis for the description of material properties using tensors. A complete reference for the major tensor uses.

Tensor Properties of Crystals by D. R. Lovett, IoP - N.B. latest edition ISBN: 0750306262
A much more concise treatment that gives a good overview of tensors in Material Science.

## Tangent plane proof

(811, #45) Show that the equation of the tangent plane to the ellipsoid

Consider the general ellipsoid

$\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} + \frac{{z{}^2}}{{{c^2}}} = 1$

At the point $$\left( {x_0},{y_0},{z_0} \right)$$ the tangent plane can be written as

$\frac{{x{x_0}}}{{{a^2}}} + \frac{{y{y_0}}}{{{b^2}}} + \frac{{z{z_0}}}{{{c^2}}} = 1$

## Proof

Remember that the normal vector to a surface is proportional to the gradient vector at that point. Also remember the equation of a plane written in vector notation is

${\bf{r}} \cdot {\bf{n}} = d$

where r is the general position vector, n is the unit normal to the plane and d is the distance from origin to plane.

Now the gradient at point  $$\left( {x_0},{y_0},{z_0} \right)$$is given by

$$\nabla F\left( {\begin{array}{*{20}{c}} {{x_0},}&{{y_0},}&{{z_0}} \end{array}} \right) =$$$$\left( {\begin{array}{*{20}{c}} {\frac{{2{x_0}}}{{{a^2}}},}&{\frac{{2{y_0}}}{{{b^2}}},}&{\frac{{2{z_0}}}{{{c^2}}}} \end{array}} \right)$$ = $$\lambda {\bf{n}}$$

for some value λ.

Now  $${\bf{r}} \cdot {\bf{n}} = d$$ can be written as

$$\left( {\begin{array}{*{20}{c}} {x,}&{y,}&z \end{array}} \right) \cdot$$$$\left( {\begin{array}{*{20}{c}} {\frac{{2{x_0}}}{{\lambda {a^2}}},}&{\frac{{2{y_0}}}{{\lambda {b^2}}},}&{\frac{{2{z_0}}}{{\lambda {c^2}}}} \end{array}} \right)$$ = $$\left( {\begin{array}{*{20}{c}} {{x_0},}&{{y_0},}&{{z_0}} \end{array}} \right) \cdot$$$$\left( {\begin{array}{*{20}{c}} {\frac{{2{x_0}}}{{\lambda {a^2}}},}&{\frac{{2{y_0}}}{{\lambda {b^2}}},}&{\frac{{2{z_0}}}{{\lambda {c^2}}}} \end{array}} \right)$$

and upon expanding we get

$\frac{{2x{x_0}}}{{{a^2}}} + \frac{{2y{y_0}}}{{{b^2}}} + \frac{{2z{z_0}}}{{{c^2}}} = \frac{{2x_0^2}}{{{a^2}}} + \frac{{2y_0^2}}{{{b^2}}} + \frac{{2z_0^2}}{{{c^2}}}$

and by cancelling the 2s and realising the RHS satisfies the ellipsoid equation we arrive at

$\frac{{x{x_0}}}{{{a^2}}} + \frac{{y{y_0}}}{{{b^2}}} + \frac{{z{z_0}}}{{{c^2}}} = 1$

## Einstein summation convention

Academic consultant: Paul Midgley (University of Cambridge), Erica Bithell (University of Cambridge)
Content development: Iain Rist, James McGinnigle
Web development: Lianne Sallows and David Brook

DoITPoMS is funded by the UK Centre for Materials Education.

Additional support for the development of this TLP came from the Worshipful Company of Armourers and Brasiers'