Appendix B
Introduction to Tensors and their properties

 

 

 

B.1. Basic properties of tensors

 

 

B.1.1 Examples of Tensors

 

The gradient of a vector field is a good example of a tensor.  Visualize a vector field: at every point in space, the field has a vector value $u(x_1,x_2,x_3)$.  Let $G=u\nabla$ represent the gradient of u.  By definition, G enables you to calculate the change in u when you move from a point x in space to a nearby point at $x+dx$:

$du=Gdx$

G is a second order tensor.  From this example, we see that when you multiply a vector by a tensor, the result is another vector. 

 

This is a general property of all second order tensors.  A tensor is a linear mapping of a vector onto another vector.  Two examples, together with the vectors they operate on, are:

 

· The stress tensor

$t=n\sigma$

where n is a unit vector normal to a surface, $\sigma$ is the stress tensor and t is the traction vector acting on the surface.

 

· The deformation gradient tensor

$dw=Fdx$

where dx is an infinitesimal line element in an undeformed solid, and dw is the vector representing the deformed line element.

 

 

 

B.1.2 Matrix representation of a tensor

 

To evaluate and manipulate tensors, we express them as components in a basis, just as for vectors.  We can use the displacement gradient to illustrate how this is done.  Let $u(x_1,x_2,x_3)$ be a vector field, and let $G=u\nabla$ represent the gradient of u.  Recall the definition of G

$du=Gdx$

Now, let $\left\{e_1,e_2,e_3\right\}$ be a Cartesian basis, and express both du and dx as components.  Then, calculate the components of du in terms of dx using the usual rules of calculus

$\begin{array}{l}d{u}_1={\displaystyle \frac{\partial u_1}{\partial x_1}d{x}_1+{\displaystyle \frac{\partial u_1}{\partial x_2}d{x}_2+{\displaystyle \frac{\partial u_1}{\partial x_3}d{x}_3}}}\\ d{u}_2={\displaystyle \frac{\partial u_2}{\partial x_1}d{x}_1+{\displaystyle \frac{\partial u_2}{\partial x_2}d{x}_2+{\displaystyle \frac{\partial u_2}{\partial x_3}d{x}_3}}}\\ d{u}_3={\displaystyle \frac{\partial u_3}{\partial x_1}d{x}_1+{\displaystyle \frac{\partial u_3}{\partial x_2}d{x}_2+{\displaystyle \frac{\partial u_3}{\partial x_3}d{x}_3}}}\end{array}$

We could represent this as a matrix product

$\left[\begin{array}{c}d{u}_1\\ d{u}_2\\ d{u}_3\end{array}\right]=\left[\begin{array}{ccc}{\displaystyle \frac{\partial u_1}{\partial x_1}}& {\displaystyle \frac{\partial u_1}{\partial x_2}}& {\displaystyle \frac{\partial u_1}{\partial x_3}}\\ {\displaystyle \frac{\partial u_2}{\partial x_1}}& {\displaystyle \frac{\partial u_2}{\partial x_2}}& {\displaystyle \frac{\partial u_2}{\partial x_3}}\\ {\displaystyle \frac{\partial u_3}{\partial x_1}}& {\displaystyle \frac{\partial u_3}{\partial x_2}}& {\displaystyle \frac{\partial u_3}{\partial x_3}}\end{array}\right]\left[\begin{array}{c}d{x}_1\\ d{x}_2\\ d{x}_3\end{array}\right]$

From this we see that G can be represented as a $3\times 3$ matrix.  The elements of the matrix are known as the components of G in the basis $\left\{e_1,e_2,e_3\right\}$.  All second order tensors can be represented in this form.  For example, a general second order tensor S could be written as

$S\equiv \left[\begin{array}{ccc}{S}_{11}& S_{12}& S_{13}\\ S_{21}& S_{22}& S_{23}\\ S_{31}& S_{32}& S_{33}\end{array}\right]$

You have probably already seen the matrix representation of stress and strain components in introductory courses.

 

Since S can be represented as a matrix, all operations that can be performed on a $3\times 3$ matrix can also be performed on S.  Examples include sums and products, the transpose, inverse, and determinant.  One can also compute eigenvalues and eigenvectors for tensors, and thus define the log of a tensor, the square root of a tensor, etc.  These tensor operations are summarized below.

 

Note that the numbers $S_{11}$, $S_{12}$, … $S_{33}$ depend on the basis $\left\{e_1,e_2,e_3\right\}$, just as the components of a vector depend on the basis used to represent the vector.  However, just as the magnitude and direction of a vector are independent of the basis, so the properties of a tensor are independent of the basis.  That is to say, if S is a tensor and u is a vector, then the vector

$v=Su$

has the same magnitude and direction, irrespective of the basis used to represent u, v, and S.

 

 

 

B.1.3 The difference between a matrix and a tensor

 

If a tensor is a matrix, why is a matrix not the same thing as a tensor?  Well, although you can multiply the three components of a vector u by any $3\times 3$ matrix,

$\left[\begin{array}{c}{b}_1\\ b_2\\ b_3\end{array}\right]=\left[\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33}\end{array}\right]\left[\begin{array}{c}{u}_1\\ u_2\\ u_3\end{array}\right]$

the resulting three numbers $(b_1,b_2,b_3)$ may or may not represent the components of a vector.  If they are the components of a vector, then the matrix represents the components of a tensor A, if not, then the matrix is just an ordinary old matrix.

 

 To check whether $(b_1,b_2,b_3)$ are the components of a vector, you need to check how $(b_1,b_2,b_3)$ change due to a change of basis.  That is to say, choose a new basis, calculate the new components of u in this basis, and calculate the new matrix in this basis (the new elements of the matrix will depend on how the matrix was defined.  The elements may or may not change $–$ if they don’t, then the matrix cannot be the components of a tensor).  Then, evaluate the matrix product to find a new left hand side, say $\left({\beta }_1,{\beta }_2,{\beta }_3\right)$.  If  $\left({\beta }_1,{\beta }_2,{\beta }_3\right)$ are related to $(b_1,b_2,b_3)$ by the same transformation that was used to calculate the new components of u, then $(b_1,b_2,b_3)$ are the components of a vector, and, therefore, the matrix represents the components of a tensor.

 

 

 

B.1.4 Formal definition of a tensor

 

Tensors are rather more general objects than the preceding discussion suggests.   There are various ways to define a tensor formally.  One way is the following:

 

A tensor is a linear vector valued function defined on the set of all vectors

 

More specifically, let $S(v)$ denote a tensor operating on a vector.  Linearity then requires that, for all vectors $v,w$ and scalars $\alpha$

 

· $S(v+w)=S(v)+S(w)$

 

· $S(\alpha v)=\alpha S(v)$

 

Alternatively, one can define tensors as sets of numbers that transform in a particular way under a change of coordinate system.  In this case, we suppose that n dimensional space can be parameterized by a set of n  real numbers $x_i$.   We could change coordinate system by introducing a second set of real numbers ${x^{\prime }}_i(x_k)$ which are invertible functions of $x_i$.   Tensors can then be defined as sets of real numbers that transform in a particular way under this change in coordinate system.  For example

 

· A tensor of zeroth rank is a scalar that is independent of the coordinate system.

 

· A covariant tensor of rank 1 is a vector that transforms as ${v^{\prime }}_i={\displaystyle \frac{\partial x_j}{\partial {x^{\prime }}_i}{v}_j}$ 

· A contravariant tensor of rank 1 is a vector that transforms as ${v^{\prime }}^i={\displaystyle \frac{\partial {x^{\prime }}_i}{\partial x_j}{v}^j}$

 

· A covariant tensor of rank 2 transforms as ${S^{\prime }}_{ij}={\displaystyle \frac{\partial x_i}{\partial {x^{\prime }}_k}{\displaystyle \frac{\partial x_j}{\partial {x^{\prime }}_l}{S}_{kl}}}$

 

· A contravariant tensor of rank 2 transforms as ${S^{\prime }}^{kl}={\displaystyle \frac{\partial {x^{\prime }}_i}{\partial x_k}{\displaystyle \frac{\partial {x^{\prime }}_i}{\partial x_l}{S}^{ij}}}$

 

· A mixed tensor of rank 2 transforms as ${S^{\prime }}_j^i={\displaystyle \frac{\partial {x^{\prime }}_k}{\partial x_i}{\displaystyle \frac{\partial x_j}{\partial {x^{\prime }}_l}{S}_l^k}}$

 

 

Higher rank tensors can be defined in similar ways.  In solid mechanics we nearly always use Cartesian tensors, (i.e. we work with the components of tensors in a Cartesian coordinate system) and this level of generality is not needed (and is rather mysterious).  We might occasionally use a curvilinear coordinate system, in which we do express tensors in terms of covariant or contravariant components (see Chapter 10, for example) $–$ this gives some sense of what these quantities mean.   But since solid mechanics idealizes the world as a Euclidean space we don’t see some of the subtleties that arise, e.g. in the theory of general relativity.

 

 

 

B.1.5 Creating a tensor using a dyadic product of two vectors.

 

Let a and b be two vectors.  The dyadic product of a and b  is a second order tensor S denoted by

$S=a\otimes b$.

with the property

$Su=\left(a\otimes b\right)u=a(b\cdot u)$

for all vectors u.  (Clearly, this maps u onto a vector parallel to a with magnitude $\left|a\right|\left(b\cdot u\right)$ )

 

The components of $a\otimes b$ in a basis $\left\{e_1,e_2,e_3\right\}$ are

$\left[\begin{array}{ccc}{a}_1{b}_1& a_1{b}_2& a_1{b}_3\\ a_2{b}_1& a_2{b}_2& a_2{b}_3\\ a_3{b}_1& a_3{b}_2& a_3{b}_3\end{array}\right]$

 

Note that not all tensors can be constructed using a dyadic product of only two vectors (this is because $\left(a\otimes b\right)u$ always has to be parallel to a, and therefore the representation cannot map a vector onto an arbitrary vector).  However, if a, b, and c are three independent vectors (i.e. no two of them are parallel) then all tensors can be constructed as a sum of scalar multiples of the nine possible dyadic products of these vectors. 

 

 

 

B.2. Operations on Second Order Tensors

 

 

B2.1 Tensor components. 

 

Let $\left\{e_1,e_2,e_3\right\}$ be a Cartesian basis, and let S be a second order tensor.  The components of S in $\left\{e_1,e_2,e_3\right\}$ may be represented as a matrix

$\left[\begin{array}{ccc}{S}_{11}& S_{12}& S_{13}\\ S_{21}& S_{22}& S_{23}\\ S_{31}& S_{32}& S_{33}\end{array}\right]$

where

$\begin{array}{l}{S}_{11}=e_1\cdot \left(S{e}_1\right),S_{12}=e_1\cdot \left(S{e}_2\right),S_{13}=e_1\cdot \left(S{e}_3\right),\\ S_{21}=e_2\cdot \left(S{e}_1\right),S_{22}=e_2\cdot \left(S{e}_2\right),S_{23}=e_2\cdot \left(S{e}_3\right),\\ S_{31}=e_3\cdot \left(S{e}_1\right),S_{32}=e_3\cdot \left(S{e}_2\right),S_{33}=e_3\cdot \left(S{e}_3\right),\end{array}$

 

The representation of a tensor in terms of its components can also be expressed in dyadic form as

$S={\displaystyle \sum _{j=1}^3{\displaystyle \sum _{i=1}^3{S}_{ij}{e}_i\otimes e_j}}$

This representation is particularly convenient when using polar coordinates, as described in Appendix E.

 

 

 

B2.2 Addition

Let S and T be two tensors.  Then $U=S+T$ is also a tensor. Denote the Cartesian components of U, S and T by matrices as defined above.  The components of U are then related to the components of S and T by

$\left[\begin{array}{ccc}{U}_{11}& U_{12}& U_{13}\\ U_{21}& U_{22}& U_{23}\\ U_{31}& U_{32}& U_{33}\end{array}\right]=\left[\begin{array}{ccc}{S}_{11}+T_{11}& S_{12}+T_{12}& S_{13}+T_{13}\\ S_{21}+T_{21}& S_{22}+T_{22}& S_{23}+T_{23}\\ S_{31}+T_{31}& S_{32}+T_{32}& S_{33}+T_{33}\end{array}\right]$

 

 

 

 

B2.3 Product of a tensor and a vector

 

Let u be a vector and S a second order tensor.  Then

$v=Su$

is a vector.  Let $\left(u_1,u_2,u_3\right)$ and $\left(v_1,v_2,v_3\right)$ denote the components of vectors u and v in a Cartesian basis $\left\{e_1,e_2,e_3\right\}$, and denote the Cartesian components of S as described above.  Then

$\left[\begin{array}{c}{v}_1\\ v_2\\ v_3\end{array}\right]=\left[\begin{array}{ccc}{S}_{11}& S_{12}& S_{13}\\ S_{21}& S_{22}& S_{23}\\ S_{31}& S_{32}& S_{33}\end{array}\right]\left[\begin{array}{c}{u}_1\\ u_2\\ u_3\end{array}\right]=\left[\begin{array}{c}{S}_{11}{u}_1+S_{12}{u}_2+S_{13}{u}_3\\ S_{21}{u}_1+S_{22}{u}_2+S_{23}{u}_3\\ S_{31}{u}_1+S_{32}{u}_2+S_{33}{u}_3\end{array}\right]$

 

The product

$v=uS$

is also a vector.  In component form

$\left[\begin{array}{ccc}{v}_1& v_2& v_3\end{array}\right]=\left[\begin{array}{ccc}{u}_1& u_2& u_3\end{array}\right]\left[\begin{array}{ccc}{S}_{11}& S_{12}& S_{13}\\ S_{21}& S_{22}& S_{23}\\ S_{31}& S_{32}& S_{33}\end{array}\right]=\left[\begin{array}{c}{u}_1{S}_{11}+u_2{S}_{21}+u_3{S}_{31}\\ u_1{S}_{12}+u_2{S}_{22}+u_3{S}_{32}\\ u_1{S}_{13}+u_2{S}_{23}+u_3{S}_{33}\end{array}\right]$

Observe that $uS\ne Su$ (unless S is symmetric).

 

 

 

B2.4 Product of two tensors

 

Let T and S be two second order tensors.  Then $U=TS$ is also a tensor.

 

Denote the components of U, S and T by $3\times 3$ matrices.  Then,

$\begin{array}{l}\left[\begin{array}{ccc}{U}_{11}& U_{12}& U_{13}\\ U_{21}& U_{22}& U_{23}\\ U_{31}& U_{32}& U_{33}\end{array}\right]=\left[\begin{array}{ccc}{T}_{11}& T_{12}& T_{13}\\ T_{21}& T_{22}& T_{23}\\ T_{31}& T_{32}& T_{33}\end{array}\right]\left[\begin{array}{ccc}{S}_{11}& S_{12}& S_{13}\\ S_{21}& S_{22}& S_{23}\\ S_{31}& S_{32}& S_{33}\end{array}\right]\\ =\left[\begin{array}{ccc}{T}_{11}{S}_{11}+T_{12}{S}_{21}+T_{13}{S}_{31}& T_{11}{S}_{12}+T_{12}{S}_{22}+T_{13}{S}_{32}& T_{11}{S}_{13}+T_{12}{S}_{23}+T_{13}{S}_{33}\\ T_{21}{S}_{11}+T_{22}{S}_{21}+T_{23}{S}_{31}& T_{21}{S}_{12}+T_{22}{S}_{22}+T_{23}{S}_{32}& T_{21}{S}_{12}+T_{22}{S}_{22}+T_{23}{S}_{32}\\ T_{31}{S}_{11}+T_{32}{S}_{21}+T_{33}{S}_{31}& T_{31}{S}_{12}+T_{32}{S}_{22}+T_{33}{S}_{32}& T_{31}{S}_{13}+T_{32}{S}_{23}+T_{33}{S}_{33}\end{array}\right]\end{array}$

Note that tensor products, like matrix products, are not commutative; i.e. $TS\ne ST$

 

 

 

B2.5 Transpose

 

Let S be a tensor.  The transpose of S is denoted by $S^T$ and is defined so that

$u{S}^T=Su$

 

Denote the components of S by a 3x3 matrix.  The components of  $S^T$ are then

$S^T\equiv \left[\begin{array}{ccc}{S}_{11}& S_{21}& S_{31}\\ S_{12}& S_{22}& S_{32}\\ S_{13}& S_{23}& S_{33}\end{array}\right]$

i.e. the rows and columns of the matrix are switched.

Note that, if A and B are two tensors, then

${\left(AB\right)}^T=B^T{A}^T$

 

 

B2.6 Trace

 

Let S be a tensor, and denote the components of S by a $3\times 3$ matrix.  The trace of S is denoted by tr(S) or trace(S), and can be computed by summing the diagonals of the matrix of components

$\text{trace}\left(S\right)=S_{11}+S_{22}+S_{33}$

More formally, let $\left\{e_1,e_2,e_3\right\}$ be any Cartesian basis.  Then

$\text{trace}\left(S\right)=e_1\cdot \left(S{e}_1\right)+e_2\cdot \left(S{e}_2\right)+e_3\cdot \left(S{e}_3\right)$

The trace of a tensor is an example of an invariant of the tensor $–$ you get the same value for trace(S) whatever basis you use to define the matrix of components of S.

 

 

 

B2.7 Contraction.

 

Inner Product: Let S and T be two second order tensors.  The inner product of S and T is a scalar, denoted by $S:T$.  Represent S and T by their components in a basis.  Then

$\begin{array}{l}S:T=S_{11}{T}_{11}+S_{12}{T}_{12}+S_{13}{T}_{13}\\ +S_{21}{T}_{21}+S_{22}{T}_{22}+S_{23}{T}_{23}\\ +S_{31}{T}_{31}+S_{32}{T}_{32}+S_{33}{T}_{33}\end{array}$

Observe that $S:T=T:S$, and also that $S:I=\text{trace(}S\text{)}$, where I is the identity tensor.

Outer product: Let S and T be two second order tensors.  The outer product of S and T is a scalar, denoted by $S\cdot \cdot T$.  Represent S and T by their components in a basis.  Then

$\begin{array}{l}S\cdot \cdot T=S_{11}{T}_{11}+S_{21}{T}_{12}+S_{31}{T}_{13}\\ +S_{12}{T}_{21}+S_{22}{T}_{22}+S_{32}{T}_{23}\\ +S_{13}{T}_{31}+S_{23}{T}_{32}+S_{33}{T}_{33}\end{array}$

Observe that $S\cdot \cdot T=S^T:T$

 

 

 

B2.8 Determinant

 

The determinant of a tensor is defined as the determinant of the matrix of its components in a basis.  For a second order tensor

$\begin{array}{l}\det S=\det \left[\begin{array}{l}{S}_{11}{S}_{12}{S}_{13}\\ S_{21}{S}_{22}{S}_{23}\\ S_{31}{S}_{32}{S}_{33}\end{array}\right]\\ =S_{11}(S_{22}{S}_{33}-S_{23}{S}_{32})+S_{12}(S_{23}{S}_{31}-S_{21}{S}_{33})+S_{13}(S_{21}{S}_{32}-S_{31}{S}_{22})\end{array}$

In index notation this would read

$\det (S)={\displaystyle \frac{1}{6}{\in }_{ijk}{\in }_{lmn}{S}_{li}{S}_{mj}{S}_{nk}}$

 

Note that if S and T are two tensors, then

$\det (S)=\det \left(S^T\right)\det (ST)=\det (S)\det (T)$

In solid mechanics we often have to find the derivative of the determinant of a tensor with respect to the tensor.   The following formula is helpful

${\displaystyle \frac{\partial \det (A)}{\partial A}=\det (A)A^{-T}{\displaystyle \frac{\partial \det (A_{ij})}{\partial A_{ij}}=\det (A_{ij})A_{ji}^{-1}}}$

 

 

 

B2.9 Inverse

 

Let S be a second order tensor.  The inverse of S exists if and only if $\det (S)\ne 0$, and is defined by

$S^{-1}S=I$

where $S^{-1}$ denotes the inverse of S and I is the identity tensor.

 

The inverse of a tensor may be computed by calculating the inverse of the matrix of its components.  Formally, the inverse of a second order tensor can be written in a simple form using index notation as

$S_{ji}^{-1}={\displaystyle \frac{1}{2\det (S)}{\in }_{ipq}{\in }_{jkl}{S}_{pk}{S}_{ql}}$

In practice it is usually faster to compute the inverse using methods such as Gaussian elimination.

 

 

 

B2.10 Eigenvalues and Eigenvectors (Principal values and direction)

 

Let S be a second order tensor.  The scalars $\lambda$ and unit vectors m which satisfy

$Sm=\lambda m$

are known as the eigenvalues and eigenvectors of S, or the principal values and principal directions of S. Note that $\lambda$ may be complex.  For a second order tensor in three dimensions, there are generally three values of $\lambda$ and three unique unit vectors m which satisfy this equation.  Occasionally, there may be only two or one value of $\lambda$.  If this is the case, there are infinitely many possible vectors m that satisfy the equation.  The eigenvalues of a tensor, and the components of the eigenvectors, may be computed by finding the eigenvalues and eigenvectors of the matrix of components.

 

The eigenvalues of a symmetric tensor are always real, and its eigenvectors are mutually perpendicular (these two results are important and are proved below).  The eigenvalues of a skew tensor are always pure imaginary or zero.

 

The eigenvalues of a second order tensor are computed using the condition $\det (S-\lambda I)=0$.  This yields a cubic equation, which can be expressed as

$\begin{array}{l}{\lambda }^3-I_1{\lambda }^2+I_2\lambda -I_3=0\\ I_1=trace(S)I_2=(I_1^2-S\cdot \cdot S)/2I_3=\det (S)\end{array}$

There are various ways to solve the resulting cubic equation explicitly $–$ a solution for symmetric S is given below, but the results for a general tensor are too messy to be given here.   The eigenvectors are then computed from the condition $(S-\lambda I)m=0$.

 

 

 

B2.11 Change of Basis.

 

Let S be a tensor, and let $\left\{e_1,e_2,e_3\right\}$ be a Cartesian basis.  Suppose that the components of S in the basis $\left\{e_1,e_2,e_3\right\}$ are known to be

$[S^{(e)}]=\left[\begin{array}{ccc}{S}_{11}^{(e)}& S_{12}^{(e)}& S_{13}^{(e)}\\ S_{21}^{(e)}& S_{22}^{(e)}& S_{23}^{(e)}\\ S_{31}^{(e)}& S_{32}^{(e)}& S_{33}^{(e)}\end{array}\right]$

 

Now, suppose that we wish to compute the components of  S in a second Cartesian basis, $\left\{m_1,m_2,m_3\right\}$.  Denote these components by

$[S^{(m)}]=\left[\begin{array}{ccc}{S}_{11}^{(m)}& S_{12}^{(m)}& S_{13}^{(m)}\\ S_{21}^{(m)}& S_{22}^{(m)}& S_{23}^{(m)}\\ S_{31}^{(m)}& S_{32}^{(m)}& S_{33}^{(m)}\end{array}\right]$

To do so, first compute the components of the transformation matrix [Q]

$\left[Q\right]=\left[\begin{array}{l}{m}_1\cdot e_1{m}_1\cdot e_2{m}_1\cdot e_3\\ m_2\cdot e_1{m}_2\cdot e_2{m}_2\cdot e_3\\ m_3\cdot e_1{m}_3\cdot e_2{m}_3\cdot e_3\end{array}\right]$

(this is the same matrix you would use to transform vector components from $\left\{e_1,e_2,e_3\right\}$ to $\left\{m_1,m_2,m_3\right\}$ ).  Then,

$[S^{(m)}]=[Q][S^{(e)}]{[Q]}^T$

or, written out in full

$\left[\begin{array}{ccc}{S}_{11}^{(m)}& S_{12}^{(m)}& S_{13}^{(m)}\\ S_{21}^{(m)}& S_{22}^{(m)}& S_{23}^{(m)}\\ S_{31}^{(m)}& S_{32}^{(m)}& S_{33}^{(m)}\end{array}\right]=\left[\begin{array}{l}{m}_1\cdot e_1{m}_1\cdot e_2{m}_1\cdot e_3\\ m_2\cdot e_1{m}_2\cdot e_2{m}_2\cdot e_3\\ m_3\cdot e_1{m}_3\cdot e_2{m}_3\cdot e_3\end{array}\right]\left[\begin{array}{ccc}{S}_{11}^{(e)}& S_{12}^{(e)}& S_{13}^{(e)}\\ S_{21}^{(e)}& S_{22}^{(e)}& S_{23}^{(e)}\\ S_{31}^{(e)}& S_{32}^{(e)}& S_{33}^{(e)}\end{array}\right]\left[\begin{array}{l}{m}_1\cdot e_1{m}_2\cdot e_1{m}_3\cdot e_1\\ m_1\cdot e_2{m}_2\cdot e_2{m}_3\cdot e_2\\ m_1\cdot e_3{m}_2\cdot e_3{m}_3\cdot e_3\end{array}\right]$

 

To prove this result, let u and v be vectors satisfying

$v=S\cdot u$

Denote the components of u and v in the two bases by  $\underset{¯}{u^{(e)}},\underset{¯}{u^{(m)}}$ and $\underset{¯}{v^{(e)}},\underset{¯}{v^{(m)}}$, respectively.  Recall that the vector components are related by

$\begin{array}{l}\underset{¯}{u^{(m)}}=[Q]\underset{¯}{u^{(e)}}\underset{¯}{u^{(e)}}={\left[Q\right]}^T\underset{¯}{u^{(m)}}\\ \underset{¯}{v^{(m)}}=[Q]\underset{¯}{v^{(e)}}\underset{¯}{v^{(e)}}={\left[Q\right]}^T\underset{¯}{v^{(m)}}\end{array}$

Now, we could express the tensor-vector product in either basis

$\underset{¯}{v^{(m)}}=\left[S^{(m)}\right]\underset{¯}{u^{(m)}}\underset{¯}{v^{(e)}}=\left[S^{(e)}\right]\underset{¯}{u^{(e)}}$

Substitute for $\underset{¯}{u^{(e)}},\underset{¯}{v^{(e)}}$ from above into the second of these two relations, we see that

${\left[Q\right]}^T\underset{¯}{v^{(m)}}=\left[S^{(e)}\right]{\left[Q\right]}^T\underset{¯}{u^{(m)}}$

Recall that

$\left[Q\right]{\left[Q\right]}^T=\left[I\right]\left[I\right]\underset{¯}{v^{(m)}}=\underset{¯}{v^{(m)}}$

so multiplying both sides by [Q] shows that

$\underset{¯}{v^{(m)}}=\left[Q\right]\left[S^{(e)}\right]{\left[Q\right]}^T\underset{¯}{u^{(m)}}$

so, comparing with the first of equation (1)

$[S^{(m)}]=[Q][S^{(e)}]{[Q]}^T$

as stated.

 

 

 

B2.12 Invariants

 

Invariants of a tensor are functions of the tensor components which remain constant under a basis change.  That is to say, the invariant has the same value when computed in two arbitrary bases $\left\{e_1,e_2,e_3\right\}$ and $\left\{m_1,m_2,m_3\right\}$.  A symmetric second order tensor always has three independent invariants.

 

Examples of invariants are

 

1. The three eigenvalues

 

2. The determinant

 

3. The trace

 

4. The inner and outer products

 

These are not all independent $–$ for example any of 2-4 can be calculated in terms of 1.

 

In practice, the most commonly used invariants are:

$\begin{array}{l}{I}_1=\mathrm{trace}(S)=S_{kk}\\ I_2={\displaystyle \frac{1}{2}\left(\mathrm{trace}{\left(S\right)}^2-S\cdot \cdot S\right)={\displaystyle \frac{1}{2}(S_{ii}{S}_{jj}-S_{ij}{S}_{ji})}}\\ I_3=\det (S)={\displaystyle \frac{1}{6}{\in }_{ijk}{\in }_{pqr}{S}_{ip}{S}_{jq}{S}_{kr}={\in }_{ijk}{S}_{i1}{S}_{j2}{S}_{k3}}\end{array}$

 

 

 

B2.13 The Cayley-Hamilton Theorem

 

Let S be a second order tensor and let $I_1=trace(S),\text{}{I}_2=(I_1^2-S\cdot \cdot S)/2I_3=\det (S)$ be the three invariants.   Then

$S^3-I_1{S}^2+I_{2}S-I_{3}I=0$

(i.e. a tensor satisfies its characteristic equation).   There is an obscure trick to show this… Consider the tensor $S-\alpha I$ (where $\alpha$ is an arbitrary scalar), and let T be the adjoint of $S-\alpha I$, (the adjoint is just the inverse multiplied by the determinant) which satisfies

$T(S-\alpha I)=\det (S-\alpha I)I=(-{\alpha }^3+I_1{\alpha }^2-I_2\alpha +I_3)I$

Assume that T= ${\alpha }^2{T}_1+\alpha T_2+T_3$.   Substituting in the preceding equation shows that

$T_1=I{T}_{1}S-T_2=I_{1}I{T}_3-T_{2}S=I_{2}I{T}_{3}S=I_{3}I$

Use these to substitute for $I_1,I_2,I_3$ into

$S^3-I_1{S}^2+I_{2}S-I{I}_3=S^3-(S-T_2)S^2+(T_3-T_{2}S)S-T_{3}S=0$

 

 

 

B3. Special tensors

 

 

B3.1 Identity tensor 

 

The identity tensor I satisfies

$\begin{array}{l}Iv=vI=v\\ SI=IS=S\end{array}$

for any tensor S or vector v.  In any Cartesian basis, the identity tensor has components

$\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

 

 

B3.2 Symmetric Tensors

 

 A symmetric tensor S has the property

$S=S^T$

The components of a symmetric tensor have the form

$\left[\begin{array}{ccc}{S}_{11}& S_{12}& S_{13}\\ S_{12}& S_{22}& S_{23}\\ S_{13}& S_{23}& S_{33}\end{array}\right]$

so that there are only six independent components of the tensor, instead of nine.   Symmetric tensors have some nice properties:

 

· The eigenvectors of a symmetric tensor with distinct eigenvalues are orthogonal.   To see this, let $u,v$ be two eigenvectors, with corresponding eigenvalues ${\lambda }_u,{\lambda }_v$. Then

$\begin{array}{l}v\cdot \left[Su\right]=u\cdot \left[S^{T}v\right]=u\cdot \left[Sv\right]\\ \Rightarrow v\cdot {\lambda }_{u}u=u\cdot {\lambda }_{v}v\\ \Rightarrow ({\lambda }_u-{\lambda }_v)u\cdot v=0\\ \Rightarrow u\cdot v=0\end{array}$.

 

· The eigenvalues of a symmetric tensor are real.  To see this, suppose that $\lambda ,u$ are a complex eigenvalue/eigenvector pair, and let $\overline{\lambda },\overline{u}$ denote their complex conjugates.   Then, by definition $S\cdot u=\lambda u\Rightarrow S\overline{u}=\overline{\lambda }\overline{u}$.  And hence $\overline{u}\cdot \left[Su\right]=\lambda \overline{u}\cdot u,u\cdot \left[S\overline{u}\right]=\overline{\lambda }u\cdot \overline{u}$.  But note that for a symmetric tensor $\overline{u}\cdot \left[Su\right]=u\cdot \left[S^T\overline{u}\right]=u\cdot \left[S\overline{u}\right]$.  Thus $\lambda \overline{u}\cdot u=\overline{\lambda }u\cdot \overline{u}\Rightarrow \lambda =\overline{\lambda }$.

 

The eigenvalues of a symmetric tensor can be computed as

$\begin{array}{l}{\lambda }_k={\displaystyle \frac{I_1}{3}+2\sqrt{{\displaystyle \frac{-p}{3}}}\cos \left\{{\displaystyle \frac{1}{3}{\cos }^{-1}\left({\displaystyle \frac{3q}{2p}\sqrt{{\displaystyle \frac{-3}{p}}}}\right)-{\displaystyle \frac{2(k-1)\pi }{3}}}\right\}k=1,2,3}\\ p=I_2-{\displaystyle \frac{1}{3}{I}_1^{2}q=-{\displaystyle \frac{2I_1^3-9I_1{I}_2+27I_3}{27}}}\\ I_1=trace(S)I_2={\displaystyle \frac{1}{2}\left(I_1^2-S:S\right)I_3=\det (S)}\end{array}$

The eigenvectors can then be found by back-substitution into $\left[S-\lambda I\right]m=0$. To do this, note that the matrix equation can be written as

$\left[\begin{array}{ccc}{S}_{11}-\lambda & S_{12}& S_{13}\\ S_{12}& S_{22}-\lambda & S_{23}\\ S_{13}& S_{23}& S_{33}-\lambda \end{array}\right]\left[\begin{array}{c}{m}_1\\ m_2\\ m_3\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$

Since the determinant of the matrix is zero, we can discard any row in the equation system and take any column over to the right hand side.   For example, if the tensor has at least one eigenvector with $m_3\ne 0$ then the values of $m_1,m_2$ for this eigenvector can be found by discarding the third row, and writing

$\left[\begin{array}{cc}{S}_{11}-\lambda & S_{12}\\ S_{12}& S_{22}-\lambda \end{array}\right]\left[\begin{array}{c}{m}_1\\ m_2\end{array}\right]=-m_3\left[\begin{array}{c}{S}_{13}\\ S_{23}\end{array}\right]$

 

· Spectral decomposition of a symmetric tensor  Let S be a symmetric second order tensor, and let $\{{\lambda }_i,e_i\}$ be the three eigenvalues and eigenvectors of S.  Then S can be expressed as

$S={\displaystyle \sum _{i=1}^3{\lambda }_i{e}_i\otimes e_i}$

To see this, note that S can always be expanded as a sum of 9 dyadic products of an orthogonal basis.  $S=S_{ij}{e}_i\otimes e_j$.  But since $e_i$ are eigenvectors it follows that

$e_m\cdot (S_{ij}{e}_i\otimes e_j)\cdot e_k=S_{mk}=\left\{\begin{array}{c}{\lambda }_{k}m=k\\ 0m\ne k\end{array}\right.$

 

 

B3.3 Skew Tensors 

 

A skew tensor W has the property

$W^T=-W$

The components of a skew tensor have the form

$\left[\begin{array}{ccc}0& W_{12}& W_{13}\\ -W_{12}& 0& W_{23}\\ -W_{13}& -W_{23}& 0\end{array}\right]$

 

Dual vector of a skew tensor:  Every skew tensor W has a dual vector $\omega$ that satisfies

$Wu=\omega \times u$

for all vectors u.  To see this, relate the components of W and $\omega$ as follows

$\omega =\left[\begin{array}{c}{\omega }_1\\ {\omega }_2\\ {\omega }_3\end{array}\right]W=\left[\begin{array}{ccc}0& -{\omega }_3& {\omega }_2\\ {\omega }_3& 0& -{\omega }_1\\ -{\omega }_2& {\omega }_1& 0\end{array}\right]$

Then evaluate the tensor-vector product and the cross product to see they are equivalent.

 

 

 

B3.4 Orthogonal Tensors

 

An orthogonal tensor R has the property

$\begin{array}{l}R{R}^T=R^{T}R=I\\ R^{-1}=R^T\end{array}$

An orthogonal tensor must have $\det (R)=\pm 1$; a tensor with $\det (R)=+1$ is known as a proper orthogonal tensor.  Orthogonal tensors also have some interesting and useful properties:

 

· Orthogonal tensors map a vector onto another vector with the same length.  To see this, let u be an arbitrary vector.  Then, note that ${\left|Ru\right|}^2=\left[Ru\right]\cdot \left[Ru\right]=u\cdot R^{T}Ru=u\cdot u={\left|u\right|}^2$

 

· The eigenvalues of an orthogonal tensor are $1,e^{\pm i\theta }$ for some value of $\theta$.  To see this, let $u$ be an eigenvector, with corresponding eigenvalue $\lambda$.  By definition, $Ru=\lambda u$.  Hence

$\left[Ru\right]\cdot \left[Ru\right]=\lambda u\cdot \lambda u\Rightarrow u\cdot u={\lambda }^{2}u\cdot u\Rightarrow {\lambda }^2=1$.

Similarly, $\lambda \overline{\lambda }=1$.  Since the characteristic equation is cubic, there must be at most three eigenvalues, and at least one eigenvalue must be real.

 

 

Proper orthogonal tensors can be visualized physically as rotations.  A rotation can also be represented in several other forms besides a proper orthogonal tensor.   For example

 

1. The Rodriguez representation quantifies a rotation as an angle of rotation $\theta$ (in radians) about some axis n (specified by a unit vector).  Given R, there are various ways to compute n  and $\theta$.  For example, one way would be find the eigenvalues and the real eigenvector.  The real eigenvector (suitably normalized) must correspond to n; the complex eigenvalues give $e^{i\theta }$.  A faster method is to note that

$\text{trace}(R)=1+2\cos \theta 2\sin \theta n=\text{dual}(R-R^T)$

where $\omega =\text{dual(}W\text{)}$ denotes the dual vector of W.

 

2. Alternatively, given n and $\theta$, R can be computed from

$R=\cos \theta I+WW(1-\cos \theta )+W\sin \theta$

where W is the skew tensor that has n as its dual vector, i.e. $W_{ij}=-{\in }_{ijk}{n}_k$.  In index notation, this formula is

$R_{ij}=\cos \theta {\delta }_{ij}+n_i{n}_j(1-\cos \theta )-\sin \theta {\in }_{ijk}{n}_k$

 

 

Another useful result is the Polar Decomposition Theorem, which states that invertible second order tensors can be expressed as a product of a symmetric tensor with an orthogonal tensor: 

$A=RU=VRR{R}^T=IU=U^{T}V=V^T$

Moreover, the tensors $R,U,V$ are unique.  To see this, note that

 

1. $A^{T}A$ is symmetric and has positive eigenvalues (to see that it’s symmetric, simply take the transpose, and to see that the eigenvalues are positive, note that $dx\cdot (A^{T}A)dx>0$ for all vectors dx). 

2. Let  ${\lambda }_k^2$ and $m_k$ be the three eigenvalues and eigenvectors of $A^{T}A$.  Since the eigenvectors are orthogonal, we can write $A^{T}A={\displaystyle \sum _{k=1}^3{\lambda }_k^2{m}_k\otimes m_k}$. 

 

3. We can then set $U={\displaystyle \sum _{k=1}^3{\lambda }_k{m}_k\otimes m_k}$ and define $R=A{U}^{-1}$.  U is clearly symmetric, and also $U^2=A^{T}A$. To see that R is orthogonal note that

$R^{T}R=U^{-T}{A}^{T}A{U}^{-1}=U^{-T}{U}^2{U}^{-1}=I$.

 

 

4. Given that U and R exist we can write $RU=\left[RU{R}^T\right]R$ so if we define $V=RU{R}^T$ then $A=VR$.   It is easy to show that V is symmetric.

 

5. To see that the decomposition is unique, suppose that $A=\widehat{R}\widehat{U}$ for some other tensors $\widehat{R},\widehat{U}$.  Then $A^{T}A={\widehat{U}}^2$.  But $A^{T}A$ has a unique square root so $\widehat{U}=U$.  The uniqueness of R follows immediately.