Vector and tensor calculus - Strength and Structure

In this chapter, attention is given to the definition and manipulation of vector and tensor fields. This mathematical description is used in many physical disciplines, and mechanics in particular. Vectors and tensors enable a compact notation and manipulation, without the necessity to resort to individual components with respect to a particular basis. The latter is needed only at the point that quantitative calculations need to be done, which is generally postponed to the last step. Special attention is given to Cartesian tensor calculus, which provides the basis for the elaboration of coordinate-free equations. Within this specific framework, no distinction is made between contravariant and covariant tensors while the metric tensor is not introduced neither. The quantification into components measured with respect to a specific basis, and the related matrix calculus is also elaborated.

A.1Scalars¶

A scalar $a$ is a quantity which has only a measure but no direction. It is the most trivial tensor, the tensor of order zero. The value of a scalar does not depend on the choice of a basis. Well-known examples are the temperature $T$ or the density $\rho$ in a material point.

A.2Vectors¶

A.2.1Definition¶

A vector is a physical quantity possessing both a magnitude (or length) and a direction.

Magnitude and direction of a vector \V{a}. — Figure A.1:Magnitude and direction of a vector $\V{a}$ .

In case of a vector $\V{a}$ :

\V{a} \;=\; \| \V{a} \| \V{e}

(A.1)

the length is given by $\|\V{a}\|$ , and the direction is given by a unit vector $\V{e}$ , having a unit length. A vector is a first-order tensor. The zero vector $\V{0}$ is a special vector having zero length. The so-called index notation of a vector $\V{a}$ is given by

\V{a} \;\Longrightarrow\; a_i

(A.2)

A vector is thus indicated with a single index (i.c. $i$ ), the value of which ranges between 1 and the dimension of the Euclidian space it is used in (commonly this dimension equals 3). The order of this tensor (i.e. 1) thus equals its number of indices. The index notation obtains a specific meaning once a basis has been chosen, since the components of a vector are trivially identified with this notation, see Section A.2.3.

A.2.2Vector operations¶

Multiplication of a vector $\V{a} = \|\V{a}\| \V{e}$ by a scalar $\alpha$ yields a vector $\V{b}$ with length $\alpha \|\V{a}\|$ in the direction $\V{e}$ :

\V{b} \;=\; \alpha \V{a}

(A.3)

Multiplication of a vector: \V{b} = \alpha\V{a}. — Figure A.2:Multiplication of a vector: $\V{b} = \alpha\V{a}$ .

The sum of two vectors $\V{a}$ and $\V{b}$ is a new vector $\V{c}$ , equal to the diagonal of the parallelogram spanned by $\V{a}$ and $\V{b}$ .

\V{c} \;=\; \V{a} + \V{b}

(A.4)

The sum of two vectors: \V{c}=\V{a}+\V{b}. — Figure A.3:The sum of two vectors: $\V{c}=\V{a}+\V{b}$ .

scalar product, inner product or dot product^[1] of two vectors is a scalar quantity, defined as:

\V{a} \inp \V{b} \;=\; \| \V{a} \|\; \| \V{b} \| \cos(\phi)

(A.5)

where $\phi$ is the smallest angle between $\V{a}$ and $\V{b}$ . Note that the dot product of two nonzero mutually perpendicular vectors equals zero.

The inner product of two vectors: \V{a}\inp\V{b}. — Figure A.4:The inner product of two vectors: $\V{a}\inp\V{b}$ .

The product is commutative, i.e.:

\V{a} \inp \V{b} \;=\; \V{b} \inp \V{a}

(A.6)

and linear

\V{a} \inp (\V{b} + \V{c}) \;=\; \V{a}\inp\V{b} + \V{a}\inp\V{c}

(A.7)

Using index notations, Einstein’s summation convention has to be introduced. This convention states, that a summation is made over all repeated indices, where the range of the summation equals the dimension $n$ of the problem. For the inner product, this is written as

\V{a}\cdot\V{b} \;=\; a_ib_i \;=\; \sum_{i=1}^3 a_i b_i

(A.8)

The summation symbol is generally not written, but here simply added to clarify the convention.

The scalar product of a vector with itself yields the square of its length:

\V{a} \inp \V{a} \;=\; \| \V{a} \|^2 \geq 0

(A.9)

The vector product or cross product of two vectors is a vector quantity:

\V{a} \cross \V{b} \;=\; \V{c}

(A.10)

The length of the vector $\V{c}$ is given by:

\| \V{c} \| \;=\; \| \V{a} \|\; \| \V{b} \| \sin(\phi)

(A.11)

where $\phi$ is the smallest angle between $\V{a}$ and $\V{b}$ . Its value equals the area of the parallelogram spanned by the vectors $\V{a}$ and $\V{b}$ . The direction of $\V{c}$ is perpendicular to both $\V{a}$ and $\V{b}$ , such that the vectors $\V{a}$ , $\V{b}$ and $\V{c}$ form a righthanded system.

The cross product of two vectors: \V{c}=\V{a}\cross\V{b}. — Figure A.5:The cross product of two vectors: $\V{c}=\V{a}\cross\V{b}$ .

The linearity of the vector product is expressed through

\V{a} \cross (\V{b} + \V{c}) \;=\; \V{a} \cross \V{b} + \V{a} \cross \V{c}

(A.12)

The vector product is not commutative:

\V{b} \cross \V{a} \;=\; - \V{a} \cross \V{b}

(A.13)

which can be concluded from the definition. It can be shown that:

\V{a} \cross (\V{b} \cross \V{c}) \;=\; (\V{a} \inp \V{c}) \V{b} - (\V{a} \inp \V{b}) \V{c}

(A.14)

The triple product^[2] of three vectors is a scalar quantity, defined as:

\begin{aligned} \V{a} \cross \V{b} \inp \V{c} \;&\;= \| \V{a} \|\; \| \V{b} \| \sin(\phi) (\V{n} \inp \V{c}) \\ \;&=\; \| \V{a} \|\; \| \V{b} \| \sin(\phi) \| \V{c} \| \cos(\psi) \end{aligned}

(A.15)

where $\V{n}$ is the unit vector in the direction $\V{a} \cross \V{b}$ , $\phi$ is the smallest angle between $\V{a}$ and $\V{b}$ and $\psi$ is the smallest angle between $\V{c}$ and $\V{n}$ .

The triple product of three vectors \V{a}\cross\V{b}\inp\V{c}. — Figure A.6:The triple product of three vectors $\V{a}\cross\V{b}\inp\V{c}$ .

The absolute value of the triple product equals the volume of the parallelepiped spanned by the three vectors. If $\V{a} \cross \V{b} \inp \V{c} \neq 0$ , the vectors are not lying within one plane, and are called mutually independent.

The tensor product^[3] or open product of two vectors $\V{a}\V{b}$ is a linear vector transformation called a dyad. Application of the dyad $\V{a}\V{b}$ to a vector $\V{p}$ yields a vector $\V{a}$ , multiplied by the scalar $(\V{b} \inp \V{p})$ :

\V{a}\V{b} \inp \V{p} \;=\; \V{a} (\V{b} \inp \V{p})

(A.16)

or in index notation

(a_ib_j) p_j \;=\; a_i (b_jp_j)

(A.17)

The tensor product: \V{a} \V{b} \inp \V{p}. — Figure A.7:The tensor product: $\V{a} \V{b} \inp \V{p}$ .

The linearity of the transformation is expressed by:

\V{a}\V{b} \inp ( \alpha \V{p} + \beta \V{q} ) \;=\; \alpha \V{a} (\V{b} \inp \V{p}) + \beta \V{a} (\V{b} \inp \V{q})

(A.18)

The transpose of a dyad (also called conjugate dyad or adjoint dyad)^[4] is defined as:

( \V{a}\V{b} )^T \;=\; \V{b}\V{a} \neq \V{a}\V{b}

(A.19)

If $( \V{a}\V{b} )^T = \V{a}\V{b}$ , the dyad $\V{a}\V{b}$ is called symmetric.

A.2.3Vector basis¶

In three-dimensional space, a set of three vectors $\V{c}_1$ , $\V{c}_2$ and $\V{c}_3$ is called a basis, denoted by $\{ \V{c}_1, \V{c}_2, \V{c}_3 \}$ , if the vectors are independent, i.e. if the triple product of the vectors is nonzero:

\{ \V{c}_1, \V{c}_2, \V{c}_3 \} \quad \Rightarrow \quad \V{c}_1 \cross \V{c}_2 \inp \V{c}_3 \neq 0

(A.20)

The three vectors $\V{c}_1$ , $\V{c}_2$ and $\V{c}_3$ are called base vectors. If the basis consists of mutually perpendicular vectors, the basis is called an orthogonal basis.

If the basis consists of mutually perpendicular unit vectors, the basis is called an orthonormal basis. In the latter case the base vectors are usually denoted by $\V{e}_1$ , $\V{e}_2$ and $\V{e}_3$ . Thus it holds that:

\basis \quad \Rightarrow \quad \V{e}_i \inp \V{e}_j \;=\; \delta_{ij} \;=\; \left\{ \begin{array}{lcl} 0 & \mbox{if} & \mbox{$i \neq j$} \\ 1 & \mbox{if} & \mbox{$i = j$} \end{array} \right.

(A.21)

where $\delta_{ij}$ is called the Kronecker delta. A basis $\basis$ is called righthanded^[5] if the order of $\V{e}_1$ , $\V{e}_2$ and $\V{e}_3$ is such that:

\begin{aligned} \V{e}_1 \cross \V{e}_2 \;&=\; \V{e}_3 \\ \V{e}_2 \cross \V{e}_3 \;&=\; \V{e}_1 \\ \V{e}_3 \cross \V{e}_1 \;&=\; \V{e}_2 \end{aligned}

(A.22)

A Cartesian basis is an orthonormal basis in which the base vectors are independent of the location in three-dimensional space. It is usually denoted by $\basis$ or $\xyzbasis$ .

A Cartesian orthonormal basis \basis. — Figure A.8:A Cartesian orthonormal basis $\basis$ .

A cylindrical basis is an orthonormal basis as well. It is usually denoted by $\cylbasis$ . The bases $\basis$ and $\cylbasis$ are mutually related by:

\begin{aligned} \V{e}_r(\theta) \;&=\; \cos(\theta) \V{e}_1 + \sin(\theta) \V{e}_2 \\ \V{e}_t(\theta) \;&=\; - \sin(\theta) \V{e}_1 + cos(\theta) \V{e}_2 \\ \V{e}_z \;&=\; \V{e}_3 \end{aligned}

(A.23)

A cylindrical orthonormal basis \cylbasis. — Figure A.9:A cylindrical orthonormal basis $\cylbasis$ .

It is important to note that the base vectors $\V{e}_r$ and $\V{e}_t$ depend on the circumferential coordinate $\theta$ . The derivatives of the vectors with respect to $\theta$ are given by:

\begin{aligned} \frac{d\V{e}_r(\theta)}{d\theta} \;&=\; \V{e}_t(\theta)\\ \frac{d\V{e}_t(\theta)}{d\theta} \;&=\; - \V{e}_r(\theta) \end{aligned}

(A.24)

A.2.4Vector components; matrix notation of a vector¶

Let $\{ \V{c}_1, \V{c}_2, \V{c}_3 \}$ be a basis in three-dimensional space. Then any vector $\V{a}$ can be expressed as a unique linear combination of the base vectors. In case of an orthonormal basis $\basis$ we may write:

\V{a} \;=\; a_1 \V{e}_1 + a_2 \V{e}_2 + a_3 \V{e}_3 = a_i\V{e}_i,

(A.25)

where we again used the Einstein summation convention, i.e. we sum over the repeated index $i$ .

Components a_i of a vector \V{a} with respect to a Cartesian
basis \basis. — Figure A.10:Components $a_i$ of a vector $\V{a}$ with respect to a Cartesian basis $\basis$ .

The coefficients $a_i$ are called the components of the vector $\V{a}$ with respect to the basis. Note that, since the basis is orthonormal, a component $a_i$ of a vector $\V{a}$ with respect to the base vector $\V{e}_i$ can be found by:

a_i \;=\; \V{a}\inp\V{e}_i

(A.26)

The component $a_i$ represents the length of the projection of the vector $\V{a}$ on the line with direction $\V{e}_i$ . For convenience we can use a matrix notation in which we store base vectors and components in a column:

\begin{aligned} \C{a} \;&=\; \begin{bmatrix} a_1 \\ a_2 \\ a_3 \end{bmatrix} \\ \C{\V{e}} \;&=\; \begin{bmatrix} \V{e}_1 \\ \V{e}_2 \\ \V{e}_3 \end{bmatrix} \end{aligned}

(A.27)

Thus we may write:

\V{a} \;=\; \begin{bmatrix} a_1 & a_2 & a_3 \end{bmatrix} \begin{bmatrix} \V{e}_1 \\ \V{e}_2 \\ \V{e}_3 \end{bmatrix} \;=\; \C{a}^T \C{\V{e}}

(A.28)

or, equivalently:

\V{a} \;=\; \begin{bmatrix} \V{e}_1 & \V{e}_2 & \V{e}_3 \end{bmatrix} \begin{bmatrix} a_1 \\ a_2 \\ a_3 \end{bmatrix} = \C{\V{e}}^T \C{a}

(A.29)

The relations $a_i = \V{a}\inp\V{e}_i$ can now be written as:

\C{a} \;=\; \V{a}\inp\C{\V{e}}

(A.30)

A.3Second order tensors¶

A.3.1Definition¶

A second order tensor is a linear transformation of a vector into another vector:

\T{A} \inp \V{p} \;=\; \V{q}

(A.31)

or in index notation

A_{ij}p_j \;=\; q_i

(A.32)

Note that we limit ourselves here to so-called Cartesian tensors, which allow orthogonal linear coordinate transformations only. In a three-dimensional space, a second-order tensor is fully defined by the image of three independent vectors, see Figure A.11.

A second-order tensor \T{A} in a three-dimensional space. — Figure A.11:A second-order tensor $\T{A}$ in a three-dimensional space.

Indeed, any set of three independent vectors constitutes a vector basis by means of which any other vector can be expressed. The image of an arbitrary vector can be found trivially through the images of three such vectors. This property is a direct consequence of the linearity of the transformation, which is expressed by:

\T{A} \inp ( \alpha \V{m} + \beta \V{n} ) \;=\; \alpha \T{A} \inp \V{m} + \beta \T{A} \inp \V{n}

(A.33)

For an arbitrary vector $\V{a}=a_1\V{e}_1 + a_2\V{e}_2 + a_3\V{e}_3$ the image vector thus equals

\T{A}\inp\V{a} \;=\; a_1(\T{A}\inp\V{e}_1) + a_2(\T{A}\inp\V{e}_2) + a_3(\T{A}\inp\V{e}_3)

(A.34)

A.3.2Some properties¶

Any second order tensor may be represented by a finite, nonunique, sum of dyads:

\T{A} \;=\; \alpha_1 \V{a}_1 \V{b}_1 + \alpha_2 \V{a}_2 \V{b}_2 + \alpha_3 \V{a}_3 \V{b}_3 + \dots

(A.35)

Two special tensors are the zero tensor $\T{O}$ and the unit tensor^[6] $\T{I}$ . If applied to vector $\V{p}$ , these tensors yield the following:

\T{O} \inp \V{p} \;=\; \V{0}

(A.36)

\T{I} \inp \V{p} \;=\; \V{p}

(A.37)

The index notations for $\T{O}$ and $\T{I}$ are respectively

\begin{aligned} \T{0} \;&\Leftrightarrow\; 0_{ij}\\ \T{I} \;&\Leftrightarrow\; \delta_{ij} \end{aligned}

(A.38)

where the Kronecker delta $\delta_{ij}$ is used for the unit tensor.

The transpose or conjugate $\T{A}^T$ of a tensor $\T{A}$ is defined by the following relation:

\T{A}^T \inp \V{p} \;=\; \V{p} \inp \T{A}

(A.39)

In index notation it holds that

\T{A}\;\Leftrightarrow\;A_{ij} \quad\Longrightarrow\quad \T{A}^T\;\Leftrightarrow\;A_{ji}

(A.40)

For the special case $\T{A} = \V{a}\V{b}$ , we find:

\begin{aligned} (\V{a}\V{b})^T \inp \V{p} \;&=\; \V{p} \inp (\V{a}\V{b}) \\ \;&=\; (\V{p} \inp \V{a}) \V{b} \\ \;&=\; \V{b} (\V{p} \inp \V{a}) \\ \;&=\; \V{b} (\V{a} \inp \V{p}) \\ \;&=\; (\V{b} \V{a}) \inp \V{p} \end{aligned}

(A.41)

showing that:

(\V{a}\V{b})^T \;=\; \V{b} \V{a}

(A.42)

If $\T{A}$ is written as a sum of dyads, the conjugate $\T{A}^T$ is obtained by reversing the two vectors in each dyad. Tensors can be multiplied by a scalar, summed with another tensor, or multiplied by another tensor:

\begin{aligned} (\alpha \T{A}) \inp \V{v} \;&=\; \T{A} \inp (\alpha \V{v}) ,\\ (\T{A} + \T{B}) \inp \V{v} \;&=\; \T{A} \inp \V{v} + \T{B} \inp \V{v} ,\\ (\T{A}\inp\T{B})\inp\V{v} \;&=\; \T{A}\inp(\T{B}\inp\V{v}) \end{aligned}

(A.43)

Note that (in general):

\T{A}\inp\T{B} \;\neq\; \T{B}\inp\T{A}

(A.44)

The inner product of a tensor with itself is called the square of a tensor, i.e.

\T{A}^2 \;=\; \T{A}\inp\T{A}

(A.45)

Likewise, higher-order exponents are defined through repeated inner products, e.g.,

\T{A}^3 \;=\; \T{A}\inp\T{A}\inp\T{A}

(A.46)

A.3.3Tensor components; matrix notation of a tensor¶

Consider a three-dimensional space with a basis $\basis$ . Then, according to (A.25), all vectors $\V{a}_i$ and $\V{b}_i$ in relation (A.35) can be written with respect to this basis. Substitution of these expressions in (A.35) results into:

\begin{aligned} \T{A} \;=\; & A_{11}\V{e}_1\V{e}_1 + A_{12}\V{e}_1\V{e}_2 + A_{13}\V{e}_1\V{e}_3 + \\ & A_{21}\V{e}_2\V{e}_1 + A_{22}\V{e}_2\V{e}_2 + A_{23}\V{e}_2\V{e}_3 + \\ & A_{31}\V{e}_3\V{e}_1 + A_{32}\V{e}_3\V{e}_2 + A_{33}\V{e}_3\V{e}_3 \end{aligned}

(A.47)

where the coefficients $A_{ij}$ are functions of the coefficients $\alpha_i$ and the components of the vectors $\V{a}$ and $\V{b}$ . Note that the coefficients now coincide with the index notation. Thus the tensor $\T{A}$ can be expressed as a finite sum of 9 dyads of the base vectors. The coefficients $A_{ij}$ are called the components of the tensor $\T{A}$ with respect to the basis $\basis$ . In general, a component $A_{ij}$ of a tensor $\T{A}$ can be obtained by:

A_{ij} \;=\; \V{e}_i \inp \T{A} \inp \V{e}_j

(A.48)

as can be verified from substitution of (A.47) in this equation. For convenience, we can store the components $A_{ij}$ in a matrix $\M{A}$ :

\begin{aligned} \M{A} \;&=\; \begin{bmatrix} A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} & A_{32} & A_{33} \end{bmatrix} \\ \;&=\; \begin{bmatrix} (\V{e}_1 \inp \T{A} \inp \V{e}_1) & (\V{e}_1 \inp \T{A} \inp \V{e}_2) & (\V{e}_1 \inp \T{A} \inp \V{e}_3) \\ (\V{e}_2 \inp \T{A} \inp \V{e}_1) & (\V{e}_2 \inp \T{A} \inp \V{e}_2) & (\V{e}_2 \inp \T{A} \inp \V{e}_3) \\ (\V{e}_3 \inp \T{A} \inp \V{e}_1) & (\V{e}_3 \inp \T{A} \inp \V{e}_2) & (\V{e}_3 \inp \T{A} \inp \V{e}_3) \end{bmatrix} \\ \;&=\; \begin{bmatrix} \V{e}_1 \\ \V{e}_2 \\ \V{e}_3 \end{bmatrix} \inp \T{A} \inp \begin{bmatrix} \V{e}_1 & \V{e}_2 & \V{e}_3 \end{bmatrix} \\ \;&=\; \C{\V{e}} \inp \T{A} \inp \C{\V{e}}^T \end{aligned}

(A.49)

Using the matrix $\M{A}$ , relation (A.47) can be rewritten as:

\begin{aligned} \T{A} \;&=\; \begin{bmatrix} \V{e}_1 & \V{e}_2 & \V{e}_3 \end{bmatrix} \begin{bmatrix} A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} & A_{32} & A_{33} \end{bmatrix} \begin{bmatrix} \V{e}_1 \\ \V{e}_2 \\ \V{e}_3 \end{bmatrix} \\ \;&=\; \C{\V{e}}^T \M{A} \; \C{\V{e}} \end{aligned}

(A.50)

Together, equations (A.49) and (A.50) describe the transformation from tensor to matrix and back.

In (A.37), the unit tensor $\T{I}$ was defined. After writing $\T{I}$ in the general form (A.47) and applying definition (A.37), it can be derived that the unit tensor can be written as:

\T{I} \;=\; \V{e}_1\V{e}_1 + \V{e}_2\V{e}_2 + \V{e}_3\V{e}_3

(A.51)

Thus, for the unit matrix $\M{I}$ belonging to the unit tensor $\T{I}$ it holds:

\M{I} \;=\; \C{\V{e}}\inp\C{\V{e}}^T = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}

(A.52)

A.3.4Scalar functions of a tensor¶

A.3.4.1Euclidean norm¶

The Euclidean norm^[7] $m$ of a tensor, a measure of its magnitude, is defined by:

m \;=\; \| \T{A} \| = \max_{\V{e}} \| \T{A} \inp \V{e} \| \quad \forall\; \V{e} \text{ with } \| \V{e} \| = 1 .

(A.53)

The norm is an invariant quantity^[8], i.e. it is independent of the choice of the basis. With respect to the norm of a tensor, the following properties hold:

\begin{aligned} \| \T{A} \| \;&\geq\; 0 \\ \| \alpha \T{A} \| \;&=\; |\alpha|\; \| \T{A} \| \\ \| \T{A} \inp \T{B} \| \;&\leq\; \| \T{A} \|\; \| \T{B} \| \\ \| \T{A} + \T{B} \| \;&\leq\; \| \T{A} \| + \| \T{B} \| \end{aligned}

(A.54)

A.3.4.2The first invariant¶

The first invariant of a tensor $\T{A}$ , also called the trace^[9] of $\T{A}$ , is defined as:

\begin{alignedat}{2} J_1(\T{A}) \;&=\; \tr(\T{A}) && \\ &\;=\; \tfrac{1}{\V{c}_1 \cross \V{c}_2 \inp \V{c}_3} [&& \V{c}_1 \inp \T{A} \inp ( \V{c}_2 \cross \V{c}_3 ) + \\ & && \V{c}_2 \inp \T{A} \inp ( \V{c}_3 \cross \V{c}_1 ) + \V{c}_3 \inp \T{A} \inp ( \V{c}_1 \cross \V{c}_2 ) ] \end{alignedat}

(A.55)

where $\{ \V{c}_1, \V{c}_2, \V{c}_3 \}$ represents a vector basis. Since $J_1(\T{A})$ is an invariant quantity, it can also be calculated using an orthonormal basis $\basis$ :

\begin{aligned} J_1(\T{A}) \;&=\; \V{e}_1 \inp \T{A} \inp \V{e}_1 + \V{e}_2 \inp \T{A} \inp \V{e}_2 + \V{e}_3 \inp \T{A} \inp \V{e}_3 \\ \;&=\; A_{11} + A_{22} + A_{33}\\ \;&=\; \tr{\M{A}} \end{aligned}

(A.56)

The following properties hold for the first invariant:

\begin{aligned} J_1(\T{A}) \;&=\; J_1(\T{A}^T) \\ J_1(\T{I}) \;&=\; 3 \\ J_1(\alpha \T{A}) \;&=\; \alpha J_1(\T{A}) \\ J_1(\T{A} + \T{B}) \;&=\; J_1(\T{A}) + J_1(\T{B}) \end{aligned}

(A.57)

A.3.4.3The second invariant¶

The second invariant of a tensor $\T{A}$ is defined as:

\begin{alignedat}{2} J_2(\T{A}) \;&=\; \half \{ \tr^2(\T{A}) - &&\tr(\T{A}^2) \} \\ \;&=\; \tfrac{1}{\V{c}_1 \cross \V{c}_2 \inp \V{c}_3} [&& \V{c}_1 \inp (\T{A} \inp \V{c}_2) \cross (\T{A} \inp \V{c}_3) + \\ &&& \V{c}_2 \inp (\T{A} \inp \V{c}_3) \cross (\T{A} \inp \V{c}_1) + \\ &&& \V{c}_3 \inp (\T{A} \inp \V{c}_1) \cross (\T{A} \inp \V{c}_2) ] \end{alignedat}

(A.58)

The following properties hold for the second invariant:

\begin{aligned} J_2(\T{A}) \;&=\; J_2(\T{A}^T) \\ J_2(\T{I}) \;&=\; 3 \\ J_2(\alpha \T{A}) \;&=\; \alpha^2 J_2(\T{A}) \end{aligned}

(A.59)

A.3.4.4The third invariant¶

The third invariant of a tensor $\T{A}$ , also called the determinant of $\T{A}$ , is defined as:

J_3(\T{A}) \;=\; \det(\T{A}) \;=\; \tfrac{1}{\V{c}_1 \cross \V{c}_2 \inp \V{c}_3} (\T{A} \inp \V{c}_1) \inp (\T{A} \inp \V{c}_2) \cross (\T{A} \inp \V{c}_3)

(A.60)

For the third invariant, the following properties hold:

\begin{aligned} J_3(\T{A}) \;&=\; J_3(\T{A}^T)\\ J_3(\T{I}) \;&=\; 1 \\ J_3(\alpha \T{A}) \;&=\; \alpha^3 J_3(\T{A}) \\ J_3(\T{A} \inp \T{B}) \;&=\; J_3(\T{A}) J_3(\T{B}) \end{aligned}

(A.61)

The determinant of a tensor equals the determinant of its matrix with respect to an orthonormal basis.

\det(\T{A}) \;=\; \det(\M{A})

(A.62)

If $\det(\T{A})$ is nonzero, i.e. the triple product of the transformed base vectors is nonzero, the tensor is called a regular tensor^[10]. If $\det(\T{A})$ is zero, the tensor $\T{A}$ is called a singular tensor^[11]. In the latter case all vectors in three-dimensional space are transformed into vectors lying in a plane (or on a line). Vectors perpendicular to that plane (or that line), are transformed into the zero vector.

A.3.4.5Scalar product¶

The scalar product, double dot product or double inner product or convolution^[12] of two tensors $\T{A}$ and $\T{B}$ is the contraction of these two tensors, denoted by $\T{A}\dinp\T{B}$ , is a scalar, defined as:

\T{A}\dinp\T{B} \;=\; \tr (\T{A}\inp\T{B} )

(A.63)

In index notation, this product is the full contraction of indices

\T{A}\dinp\T{B} \;=\; A_{ij} B_{ji}

(A.64)

Some properties are:

\begin{aligned} \T{A}\dinp\T{I} \;&=\; \tr(\T{A})\\ \T{A}\dinp\T{B} \;&=\; \T{B}\dinp\T{A} \\ \;&=\; \T{A}^T\dinp\T{B}^T\\ \;&=\; \T{B}^T\dinp\T{A}^T \end{aligned}

(A.65)

A.3.4.6Some properties of the invariants of a tensor¶

The three principal invariants of a tensor are related through the Caley-Hamilton theorem:

\T{A}^3 - J_1(\T{A})\T{A}^2 + J_2(\T{A})\T{A} - J_3(\T{A})\T{I} \;=\; \T{0}

(A.66)

The conjugated tensor $\T{A}^T$ satisfies the same equation.

For any non-singular tensor it holds that:

\begin{aligned} J_1(\T{A}^{-1}) \;&=\; \frac{J_2(\T{A})}{J_3(\T{A})}\\ J_2(\T{A}^{-1}) \;&=\; \frac{J_1(\T{A})}{J_3(\T{A})}\\ J_3(\T{A}^{-1}) \;&=\; \frac{1}{J_3(\T{A})} \end{aligned}

(A.67)

Any function of the principal invariants of a tensor $\T{A}$ is an invariant as well.

A.3.5Eigenvalues and eigenvectors of a tensor¶

The eigenvalues^[13] $\lambda$ and the eigenvectors^[14] $\V{n}$ of a tensor $\T{A}$ are defined as those scalars and vectors that satisfy the following equation:

\T{A} \inp \V{n} \;=\; \lambda \V{n} \quad \text{ with } \quad \V{n} \neq \V{0}

(A.68)

To compute the eigenvalues and eigenvectors of a tensor, we proceed as follows. First equation (A.68) is rewritten as:

\begin{aligned} \T{A} \inp \V{n} \;&=\; \lambda \V{n} \quad \Rightarrow \\ \T{A} \inp \V{n} - \lambda \V{n} \;&=\; \V{0} \quad \Rightarrow \\ \T{A} \inp \V{n} - \lambda \T{I} \inp \V{n} \;&=\; \V{0} \quad \Rightarrow \\ (\T{A} - \lambda \T{I}) \inp \V{n} \;&=\; \V{0} \end{aligned}

(A.69)

Since $\V{n} \neq \V{0}$ , the tensor $\T{A} - \lambda \T{I}$ must be singular, meaning that:

\det(\T{A} - \lambda \T{I}) \;=\; 0 \quad \Rightarrow \quad \det(\M{A} - \lambda \M{I}) \;=\; 0

(A.70)

Elaboration of the latter equation leads to the characteristic equation:

\lambda^3 - J_1(\T{A}) \lambda^2 + J_2(\T{A}) \lambda - J_3(\T{A}) \;=\; 0

(A.71)

This equation has three roots $\lambda_1$ , $\lambda_2$ and $\lambda_3$ . The eigenvector $\V{n}_i$ corresponding with the eigenvalue $\lambda_i$ is found from the following set of linear equations:

(\T{A} - \lambda_i \T{I}) \inp \V{n}_i \;=\; \V{0} \quad \Rightarrow \quad (\M{A} - \lambda_i \M{I}) \C{n}_i \;=\; \C{0}

(A.72)

Since this is not an independent set of equations, it yields only the ratio $n_{i1} \dinp n_{i2} \dinp n_{i3}$ of the components of the eigenvector. A unique solution is obtained by the additional requirement that the eigenvector has unit length:

\| \V{n}_i \| \;=\; 1 \quad \Rightarrow \quad n_{i1}^2 + n_{i2}^2 + n_{i3}^2 = 1

(A.73)

The principal invariants of a tensor are then equal to:

\begin{aligned} J_1(\T{A}) \;&=\; \lambda_1 + \lambda_2 + \lambda_3 \\ J_2(\T{A}) \;&=\; \lambda_1\lambda_2 + \lambda_1\lambda_3 + \lambda_2\lambda_3\\ J_3(\T{A}) \;&=\; \lambda_1 \lambda_2 \lambda_3 \end{aligned}

(A.74)

A.3.6Some special tensors¶

A.3.6.1Inverse tensor¶

The inverse tensor $\T{A}^{-1}$ of a tensor $\T{A}$ is defined as the tensor that satisfies the following equation:

\T{A}^{-1} \inp \T{A} \;=\; \T{I}

(A.75)

$\T{A}^{-1}$ only exists if $\T{A}$ is regular. The components of the inverse tensor are given by the inverse matrix $\M{A}^{-1}$

[A^{-1}]_{ij} \;=\; \frac{1}{\det(\M{A})}\left[ (-1)^{i+j} M_{ij}\right]^T

(A.76)

where $M_{ij}$ is the minor of the component $A_{ij}$ , i.e. the determinant of the matrix that is obtained by deleting row $i$ and column $j$ .

The inverse of the product of two tensors is given by:

(\T{A} \inp \T{B} )^{-1} \;=\; \T{B}^{-1} \inp \T{A}^{-1}

(A.77)

The proof is obtained by comparing the following expressions for $\V{a}$ :

\begin{aligned} (\T{A} \inp \T{B}) \inp \V{a} \;&=\; \V{b} \quad \Rightarrow \\ \V{a} \;&=\; (\T{A} \inp \T{B})^{-1} \inp \V{b} \end{aligned}

(A.78)

and

\begin{aligned} (\T{A} \inp \T{B}) \inp \V{a} \;&=\; \T{A} \inp (\T{B} \inp \V{a}) = \V{b} \quad \Rightarrow \\ \T{B} \inp \V{a} \;&=\; \T{A}^{-1} \inp \V{b} \qquad \Rightarrow \\ \V{a} \;&=\; \T{B}^{-1} \inp \T{A}^{-1} \inp \V{b} \end{aligned}

(A.79)

A.3.6.2Deviatoric part of a tensor¶

The deviatoric part^[15] $\T{A}^d$ of a tensor $\T{A}$ is defined as:

\T{A}^d \;=\; \T{A} - \tfrac{1}{3} \tr(\T{A}) \T{I}

(A.80)

where $\tfrac{1}{3} \tr(\T{A}) \T{I}$ is called the hydrostatic or spherical^[16] part of $\T{A}$ . For the deviatoric part, the following properties hold:

\begin{aligned} (\T{A} + \T{B})^d \;&=\; \T{A}^d + \T{B}^d \\ \tr(\T{A}^d) \;&=\; 0 \end{aligned}

(A.81)

A.3.6.3Symmetric tensor¶

A tensor $\T{A}$ is called symmetric if:

\T{A}^T \;=\; \T{A}

(A.82)

For the components $A_{ij}$ of a symmetric tensor $\T{A}$ with respect to an orthonormal basis it holds that:

A_{ij} \;=\; A_{ji}

(A.83)

i.e. the matrix $\M{A}$ corresponding to $\T{A}$ is symmetric too. The eigenvalues $\lambda_i$ and eigenvectors $\V{n}_i$ of a symmetric tensor have the following important properties:

the eigenvalues and eigenvectors are real
if two eigenvalues are different, the corresponding eigenvectors are mutually perpendicular
if two eigenvalues are equal, the corresponding eigenvectors can be chosen mutually perpendicular

Thus, the set of eigenvectors can be used as an orthonormal basis.

The orthonormal basis of eigenvectors \{ \V{n}_1, \V{n}_2, \V{n}_3 \}
of a symmetric tensor. — Figure A.12:The orthonormal basis of eigenvectors $\{ \V{n}_1, \V{n}_2, \V{n}_3 \}$ of a symmetric tensor.

If a tensor $\T{A}$ is considered with respect to the basis of its eigenvectors, we find the spectral form^[17] of the tensor:

\begin{aligned} \T{A} \;&=\; \T{A} \inp \T{I} \\ \;&=\; \T{A} \inp ( \V{n}_1\V{n}_1 + \V{n}_2\V{n}_2 + \V{n}_3\V{n}_3 ) \\ \;&=\; \lambda_1 \V{n}_1\V{n}_1 + \lambda_2 \V{n}_2\V{n}_2 + \lambda_3 \V{n}_3\V{n}_3 \end{aligned}

(A.84)

The matrix $\M{A}$ of $\T{A}$ with respect to this basis is a diagonal matrix:

\M{A} \;=\; \begin{bmatrix} \lambda_1 & 0 & 0 \\ 0 & \lambda_2 & 0 \\ 0 & 0 & \lambda_3 \end{bmatrix}

(A.85)

Using the spectral form, some functions of a tensor can easily be calculated. E.g. for the inverse, the square root, and the natural logarithm of a tensor $\T{A}$ it holds that:

\begin{aligned} \T{A}^{-1} \;&=\; \lambda_1^{-1} \V{n}_1\V{n}_1 + \lambda_2^{-1} \V{n}_2\V{n}_2 + \lambda_3^{-1} \V{n}_3\V{n}_3 \\ \T{A}^{1/2} \;&=\; \lambda_1^{1/2} \V{n}_1\V{n}_1 + \lambda_2^{1/2} \V{n}_2\V{n}_2 + \lambda_3^{1/2} \V{n}_3\V{n}_3 \\ \ln \T{A} \;&=\; \ln(\lambda_1) \V{n}_1\V{n}_1 + \ln(\lambda_2) \V{n}_2\V{n}_2 + \ln(\lambda_3) \V{n}_3\V{n}_3 \end{aligned}

(A.86)

A.3.6.4Skewsymmetric tensor¶

A tensor $\T{A}$ is called skewsymmetric^[18] or antisymmetric if:

\T{A}^T \;=\; - \T{A}

(A.87)

For the components $A_{ij}$ of a skewsymmetric tensor $\T{A}$ with respect to an orthonormal basis it holds that:

A_{ij} \;=\; - A_{ji} \quad \Rightarrow \quad A_{ii} \;=\; 0

(A.88)

If $\T{A}$ is skewsymmetric and $\T{B}$ is symmetric, then:

\T{A}:\T{B}=0 \quad {\rm especially} \quad \T{A}:\T{I} = \tr\T{A} = 0

(A.89)

The proof follows from:

\T{A} \dinp \T{B} \;=\; \T{A}^T \dinp \T{B}^T \;=\; - \T{A} \dinp \T{B}

(A.90)

If $\T{A}$ is skewsymmetric, then a unique axial vector^[19] $\V{\omega}$ exists, such that:

\T{A} \inp \V{q} \;=\; \V{\omega} \cross \V{q}

(A.91)

The proof of the existence of $\V{\omega}$ is as follows:

\begin{aligned} \T{A} \inp \V{q} \;&=\; \V{p} \quad \Rightarrow \\ \V{q} \inp \T{A} \inp \V{q} \;&=\; \V{q} \inp \T{A}^T \inp \V{q} \\ \;&=\; - \; \V{q} \inp \T{A} \inp \V{q} \quad \Rightarrow \\ \V{q} \inp \T{A} \inp \V{q} \;&=\; 0 \quad \Rightarrow \\ \V{q} \inp \V{p} \;&=\; 0 \quad \Rightarrow \\ \V{q} \;&\perp\; \V{p} \quad \Rightarrow \quad \exists\;\V{\omega} \; \text{ such that } \V{p} \;=\; \V{\omega} \cross \V{q} \end{aligned}

(A.92)

On the other hand, if:

\T{A} \inp \V{q} \;=\; \V{\omega} \cross \V{q} \quad \forall\ \V{q}

(A.93)

comparison of the following expressions:

\begin{aligned} \T{A} \inp \V{q} \;&=\; \begin{bmatrix} \V{e}_1 & \V{e}_2 & \V{e}_3 \end{bmatrix} \begin{bmatrix} A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} & A_{32} & A_{33} \end{bmatrix} \begin{bmatrix} q_1 \\ q_2 \\ q_3 \end{bmatrix} \\ \;&=\; \begin{bmatrix} \V{e}_1 & \V{e}_2 & \V{e}_3 \end{bmatrix} \begin{bmatrix} A_{11}q_1 + A_{12}q_2 + A_{13}q_3 \\ A_{21}q_1 + A_{22}q_2 + A_{23}q_3 \\ A_{31}q_1 + A_{32}q_2 + A_{33}q_3 \end{bmatrix} \end{aligned}

(A.94)

and

\begin{alignedat}{2} \V{\omega} \cross \V{q} \;&=\; &&( \omega_1 \V{e}_1 + \omega_2 \V{e}_2 + \omega_3 \V{e}_3 ) \cross ( q_1 \V{e}_1 + q_2 \V{e}_2 + q_3 \V{e}_3 ) \\ \;&=\; &&\omega_1 q_2 ( \V{e}_3) + \omega_1 q_3 (-\,\V{e}_2) + \omega_2 q_1 (-\,\V{e}_3) + \omega_2 q_3 ( \V{e}_1) + \\ &&& \omega_3 q_1 ( \V{e}_2) + \omega_3 q_2 (-\,\V{e}_1) \\ \;&=\; && \begin{bmatrix} \V{e}_1 & \V{e}_2 & \V{e}_3 \end{bmatrix} \begin{bmatrix} \omega_2 q_3 - \omega_3 q_2 \\ \omega_3 q_1 - \omega_1 q_3 \\ \omega_1 q_2 - \omega_2 q_1 \end{bmatrix} \end{alignedat}

(A.95)

shows that:

\M{A} \;=\; \begin{bmatrix} 0 & -\,\omega_3 & \omega_2 \\ \omega_3 & 0 & -\,\omega_1 \\ -\,\omega_2 & \omega_1 & 0 \end{bmatrix}

(A.96)

and thus that the tensor $\T{A}$ is skewsymmetric.

A.3.6.5Positive-definite tensor¶

A tensor $\T{A}$ is called positive-definite^[20] if:

\V{a} \inp \T{A} \inp \V{a} \;>\; 0 \quad \forall\;\V{a} \neq \V{0}

(A.97)

A positive-definite tensor $\T{A}$ cannot be skewsymmetric. This follows from:

\begin{aligned} \V{a} \inp \T{A} \inp \V{a} \;&=\; \V{a} \inp \T{A}^T \inp \V{a} \\ \;&>\; 0 \quad \forall\; \V{a} \neq \V{0} \quad \Rightarrow \\ \V{a} \inp (\T{A} - \T{A}^T) \inp \V{a} \;&=\; 0 \quad \wedge \quad \V{a} \neq \V{0} \end{aligned}

(A.98)

If $\T{A}$ would be skewsymmetric, the latter equation would yield zero, since then $\T{A}^T = -\T{A}$ .

A.3.6.6Orthogonal tensor¶

A tensor $\T{A}$ is called orthogonal^[21] if:

(\T{A} \inp \V{a}) \inp (\T{A} \inp \V{b}) \;=\; \V{a} \inp \V{b} \quad \forall\ \V{a},\V{b}

(A.99)

For the special case $\V{b} = \V{a}$ we get:

(\T{A} \inp \V{a}) \inp (\T{A} \inp \V{a}) \;=\; \V{a} \inp \V{a} \quad \Rightarrow \quad \| \T{A} \inp \V{a} \| \;=\; \| \V{a} \|

(A.100)

Thus, an orthogonal transformation preserves lengths of vectors and angles between vectors. The inverse of an orthogonal tensor is identical to its conjugate:

\T{A}^T \;=\; \T{A}^{-1}

(A.101)

as follows from:

\begin{aligned} (\T{A} \inp \V{a}) \inp (\T{A} \inp \V{b}) \;&=\; \V{a} \inp \T{A}^T \inp \T{A} \inp \V{b} \\ \;&=\; \V{a} \inp \V{b} \quad \Rightarrow \\ \T{A} \inp \T{A}^T \;&=\; \T{I} \quad \Rightarrow \\ \T{A}^T \;&=\; \T{A}^{-1} \end{aligned}

(A.102)

Moreover, this equation shows that:

\begin{aligned} \det(\T{A} \inp \T{A}^T) \;&=\; [\det(\T{A})]^2 = \det(\T{I}) = 1 \quad \Rightarrow \\ \det(\T{A}) \;&=\; \pm 1 \end{aligned}

(A.103)

Thus, an orthogonal tensor is a regular tensor. If $\det(\T{A}) = 1$ , $\T{A}$ can be shown to describe a rotation. If $\det(\T{A}) = -1$ , $\T{A}$ can be shown to describe a reflection.

A.4Spatial derivatives of scalar and vector fields¶

A.4.1Derivatives of functions of one or more variables¶

A function f of one variable x and its derivative. — Figure A.13:A function $f$ of one variable $x$ and its derivative.

Consider a scalar function $f(x)$ of one variable $x$ . Using a Taylor series, the change $df$ of the value of $f$ due to a change $dx$ in the parameter $x$ can be expressed as:

\begin{aligned} df \;&=\; f(x+dx) - f(x) \\ \;&=\; f(x) + \left.\tfrac{df}{dx}\right|_x dx + \half \left.\tfrac{d^2f}{dx^2}\right|_x dx^2 + \dots - f(x) \\ \;&\approx\; \left.\tfrac{df}{dx}\right|_x dx \end{aligned}

(A.104)

Extension to a function of two variables $f(x,y)$ yields:

\begin{aligned} df \;&=\; f(x+dx,y+dy) - f(x,y) \\ \;&=\; f(x,y) + \left.\parder{f}{x}\right|_{x,y} dx + \left.\parder{f}{y}\right|_{x,y} dy + \dots - f(x,y) \\ \;&\approx\; \left.\parder{f}{x}\right|_{x,y} dx + \left.\parder{f}{y}\right|_{x,y} dy \end{aligned}

(A.105)

And for a function $a(x_1, x_2, x_3)$ of three variables, the variation $da$ of $a$ , due to (infinitesimal) variations $dx_1$ , $dx_2$ and $dx_3$ can be written as:

da \;=\; dx_1 \parder{a}{x_1} + dx_2 \parder{a}{x_2} + dx_3 \parder{a}{x_3}

(A.106)

The index notation for differentiation with respect to one of the variables (i.c. $x_i$ ) is commonly denoted with a comma between or in front of the indices. The equation (A.106) for $da$ with this convention reads

da \;=\; dx_i a_{,i}

(A.107)

A.4.2Components of the position vector¶

In three-dimensional space, we may interpret the variables $x_1$ , $x_2$ and $x_3$ in the function $a(x_1, x_2, x_3)$ as the components of the position vector^[22] $\V{x}$ with respect to a Cartesian basis $\basis$ . The position vector $\V{x}$ of a point with respect to a fixed point in space, the origin $O$ , can be expressed as:

\V{x} \;=\; x_1 \V{e}_1 + x_2 \V{e}_2 + x_3 \V{e}_3

(A.108)

Components x_i of the position vector \V{x} with respect to a Cartesian
basis \basis. — Figure A.14:Components $x_i$ of the position vector $\V{x}$ with respect to a Cartesian basis $\basis$ .

For the position vector $\V{x} + d\V{x}$ of a neighbouring point we can write:

\V{x} + d\V{x} \;=\; (x_1 + dx_1) \V{e}_1 + (x_2 + dx_2) \V{e}_2 + (x_3 + dx_3) \V{e}_3

(A.109)

and thus the vector $d\V{x}$ can be expressed as:

d\V{x} \;=\; dx_1 \V{e}_1 + dx_2 \V{e}_2 + dx_3 \V{e}_3

(A.110)

From the latter relation it is obvious that the changes $dx_1$ , $dx_2$ and $dx_3$ can be related to the vector $d\V{x}$ according to:

\begin{aligned} dx_1 \;&=\; d\V{x} \inp \V{e}_1 \\ dx_2 \;&=\; d\V{x} \inp \V{e}_2 \\ dx_3 \;&=\; d\V{x} \inp \V{e}_3 \end{aligned}

(A.111)

We could also express the function $a(\V{x})$ in terms of the components $r$ , $\theta$ , $z$ of the position vector $\V{x}$ with respect to a cylindrical basis $\cylbasis$ . Then the position vector equals:

\V{x} \;=\; r \V{e}_r(\theta) + z \V{e}_z

(A.112)

Components r, \theta, z of the position vector \V{x} with respect to a
cylindrical basis \cylbasis. — Figure A.15:Components $r$ , $\theta$ , $z$ of the position vector $\V{x}$ with respect to a cylindrical basis $\cylbasis$ .

Within the cylindrical basis, the position $\V{x} + d\V{x}$ of a neighbouring point equals:

\begin{alignedat}{2} \V{x} + d\V{x} \;&=\; &&(r + dr) \V{e}_r(\theta + d\theta) + (z + dz) \V{e}_z \\ \;&=\; &&(r + dr) \{ \V{e}_r(\theta) + \tfrac{d\V{e}_r}{d\theta} d\theta \} + (z + dz) \V{e}_z \\ \;&=\; &&r \V{e}_r(\theta) + z \V{e}_z + \\ & &&r \V{e}_t(\theta) d\theta + dr \V{e}_r(\theta) + \V{e}_t(\theta) dr d\theta + dz \V{e}_z \end{alignedat}

(A.113)

from which it can be concluded that, in a linear approximation, the vector $d\V{x}$ equals:

d\V{x} \;\approx\; dr \V{e}_r(\theta) + r d\theta \V{e}_t(\theta) + dz \V{e}_z

(A.114)

The changes $dr$ , $d\theta$ and $dz$ are related to the vector $d\V{x}$ according to:

\begin{aligned} dr \;&=\; d\V{x} \inp \V{e}_r \\ d\theta \;&=\; \tfrac{1}{r} d\V{x} \inp \V{e}_t\\ dz \;&=\; d\V{x} \inp \V{e}_z \end{aligned}

(A.115)

A.4.3The gradient operator¶

Considering $a$ as a function of the position vector $\V{x}$ , we can investigate the variation $da$ due to a variation $d\V{x}$ :

\begin{aligned} da \;&=\; dx_1 \parder{a}{x_1} + dx_2 \parder{a}{x_2} + dx_3 \parder{a}{x_3} \\ \;&=\; (d\V{x} \inp \V{e}_1) \parder{a}{x_1} + (d\V{x} \inp \V{e}_2) \parder{a}{x_2} + (d\V{x} \inp \V{e}_3) \parder{a}{x_3} \\ \;&=\; d\V{x} \inp \left[ \V{e}_1 \parder{a}{x_1} + \V{e}_2 \parder{a}{x_2} + \V{e}_3 \parder{a}{x_3} \right] \\ \;&=\; d\V{x} \inp (\grad a) \end{aligned}

(A.116)

The symbol $\grad$ is called the gradient operator^[23]:

\begin{aligned} \grad \;&=\; \left[ \V{e}_1 \parder{}{x_1} + \V{e}_2 \parder{}{x_2} + \V{e}_3 \parder{}{x_3} \right] \\ \;&=\; \C{\V{e}}^T \C{\nabla} \end{aligned}

(A.117)

and the expression $\grad a$ is called the gradient of $a$ . The gradient of a vector function $\V{a}$ is obtained similarly:

\begin{aligned} d\V{a} \;&=\; dx_1 \parder{\V{a}}{x_1} + dx_2 \parder{\V{a}}{x_2} + dx_3 \parder{\V{a}}{x_3} \\ \;&=\; (d\V{x} \inp \V{e}_1) \parder{\V{a}}{x_1} + (d\V{x} \inp \V{e}_2) \parder{\V{a}}{x_2} + (d\V{x} \inp \V{e}_3) \parder{\V{a}}{x_3} \\ \;&=\; d\V{x} \inp \left[ \V{e}_1 \parder{\V{a}}{x_1} + \V{e}_2 \parder{\V{a}}{x_2} + \V{e}_3 \parder{\V{a}}{x_3} \right] \\ \;&=\; d\V{x} \inp (\grad \V{a}) \end{aligned}

(A.118)

or in index notation

da_j \;=\; dx_i a_{j,i}

(A.119)

For the special case $\V{a} = \V{x}$ we find:

\grad \V{x} \;=\; \T{I}

(A.120)

The gradient operator $\grad$ can be applied to a vector $\V{a}$ in several manners, the result of which is called the gradient of $\V{a}$ (grad $\V{a}$ ), the divergence^[24] of $\V{a}$ (div $\V{a}$ ), or the rotation^[25] or curl of $\V{a}$ (rot $\V{a}$ ):

\begin{aligned} {\rm grad\ }\V{a} \;&=\; \grad \V{a} = \V{e}_1 \parder{\V{a}}{x_1} + \V{e}_2 \parder{\V{a}}{x_2} + \V{e}_3 \parder{\V{a}}{x_3} ,\\ {\rm div\ }\V{a} \;&=\; \grad \inp \V{a} = \V{e}_1 \inp \parder{\V{a}}{x_1} + \V{e}_2 \inp \parder{\V{a}}{x_2} + \V{e}_3 \inp \parder{\V{a}}{x_3} ,\\ {\rm rot\ }\V{a} \;&=\; \grad \cross \V{a} = \V{e}_1 \cross \parder{\V{a}}{x_1} + \V{e}_2 \cross \parder{\V{a}}{x_2} + \V{e}_3 \cross \parder{\V{a}}{x_3} \end{aligned}

(A.121)

Note that div $\V{a}$ yields a scalar, rot $\V{a}$ yields a vector, and grad $\V{a}$ yields a second order tensor. Applied to tensors in general, the gradient increases the order by 1, the divergence diminishes the order by 1, while the order remains unaffected by the curl.

The expression for the gradient operator in cylindrical coordinates can be found following a similar derivation as in the case of a Cartesian coordinate system:

\begin{aligned} da \;&=\; dr \parder{a}{r} + d\theta \parder{a}{\theta} + dz \parder{a}{z} \\ \;&=\; (d\V{x} \inp \V{e}_r) \parder{a}{r} + (\tfrac{1}{r} d\V{x} \inp \V{e}_t) \parder{a}{\theta} + (d\V{x} \inp \V{e}_z) \parder{a}{z} \\ \;&=\; d\V{x} \inp \left[ \V{e}_r \parder{a}{r} + \tfrac{1}{r} \V{e}_t \parder{a}{\theta} + \V{e}_z \parder{a}{z} \right] \\ \;&=\; d\V{x} \inp (\grad a) \end{aligned}

(A.122)

Thus the expression for the gradient operator in a cylindrical coordinate system equals:

\grad \;=\; \V{e}_r \parder{}{r} + \V{e}_t \frac{1}{r} \parder{}{\theta} + \V{e}_z \parder{}{z}

(A.123)

A.5Tensor formula involving first and second-order tensors¶

The following relations on differentiable scalar functions $\phi$ , vectors $\V{a},\V{b}$ , tensors $\T{A},\T{B},\T{C}$ , inner/outer products and their derivatives can be proven:

\begin{aligned} \T{A} \dinp (\T{B}\inp\T{C}) \;&=\; (\T{A}\inp\T{B}) \dinp \T{C}\\ \grad(\phi\V{a}) \;&=\; (\grad\phi)\V{a} + \phi\grad\V{a}\\ \grad \inp (\phi \V{a}) \;&=\; (\grad\phi) \inp \V{a} + \phi \grad \inp \V{a}\\ \grad(\V{a}\inp\V{b}) \;&=\; (\grad\V{a})\inp\V{b} + (\grad\V{b})\inp\V{a}\\ \grad \inp (\V{a}\V{b}) \;&=\; (\grad \inp \V{a}) \V{b} + \V{a} \inp \grad\V{b}\\ \grad \inp (\phi \T{A}) \;&=\; (\grad\phi) \inp \T{A} + \phi \grad \inp \T{A}\\ \grad \inp (\T{A} \inp \V{a}) \;&=\; (\grad \inp \T{A}) \inp \V{a} + \T{A} :(\grad\V{a})^T\\ \grad \inp \V{a} \;&=\; (\grad\V{a}) \dinp \T{I} = \tr(\grad\V{a}) \end{aligned}

(A.124)

A.6Gauss’ divergence theorem¶

Gauss’ divergence theorem relates volume integrals and surface integrals. Let’s consider a closed volume $V$ with and external closed surface $A$ , which has an outward normal $\vec{n}$ in each point on that surface. Let $\vec{a}$ be a continuous vector field that is defined in the entire volume $V$ . The divergence theorem then states that the volume integral over $V$ of the divergence of this vector field $\vec{p}$ equals the surface integral on $A$ of the normal projection of this vector field $\vec{p}$ on the external surface.

\int\int\int_V \: \grad \inp \vec{p} \: dV \;=\; \int\int_A \: \vec{n} \inp \vec{p} \: dA \;=\; \int\int_A \: \vec{p}\inp\vec{n} \: dA

(A.125)

The divergence theorem is often used in deriving balance equations in continua. Using the divergence theorem, it is often possible to convert volume integrals into surface integrals, if a vector field can be found of which the divergence equals the integrand in the volume integral.

The generalised form of Gauss’ divergence theorem applies to a tensor field $\T{P}$ , for which an equal results hold (even though the integrand in the surface integral is not longer a scalar):

\int\int\int_V \: \grad \inp \T{P} \: dV \;=\; \int\int_A \: \vec{n} \inp \T{P} \: dA

(A.126)

A.7Higher-order and fourth-order tensors¶

A.7.1Definition¶

Some problems invoke the transformation of tensors. Such transformations are carried out by means of higher-order tensors. A general definition of a tensor could be: an invariant object with an arbitrary number of indices. The number of indices is the order (or rank) of the tensor. The transformation which is most relevant in the mechanics of materials is the transformation of a second-order tensor into another second-order tensor by means of a fourth-order tensor.

A fourth-order tensor \TQ{C} that maps a (linear) second-order tensor \T{A}
onto a second-order tensor \T{B}. — Figure A.16:A fourth-order tensor $\TQ{C}$ that maps a (linear) second-order tensor $\T{A}$ onto a second-order tensor $\T{B}$ .

An arbitrary tensor $\T{A}$ (which is a linear transformation of vectors) is mapped onto a different tensor $\T{B}$ (which is another linear transformation of vectors) by means of the double inner product with a fourth-order tensor $\TQ{C}$ , see Figure A.16.

\T{B} \;=\; \TQ{C} \dinp \T{A}

(A.127)

Using index tensor notations, the tensor $A_{lk}$ transforms to $B_{ij}$ according to

B_{ij} \;=\; C_{ijkl}\,A_{lk}

(A.128)

The tensor $C_{ijkl}$ which was used to perform this transformation has 4 indices.

A.7.2Properties¶

Since a fourth-order tensor is built on quadrades rather than dyades, some new calculus rules apply. Each fourth-order tensor can be written as a finite sum of quadrades. With respect to a two-dimensional Cartesian basis, a tensor $\TQ{C}$ would be given by $sixteen$ terms, i.e.

\begin{split} \TQ{C} \;=\; & C_{1111}\V{e}_1\V{e}_1\V{e}_1\V{e}_1 + C_{1112}\V{e}_1\V{e}_1\V{e}_1\V{e}_2 + C_{1121}\V{e}_1\V{e}_1\V{e}_2\V{e}_1 + C_{1122}\V{e}_1\V{e}_1\V{e}_2\V{e}_2\\ & C_{1211}\V{e}_1\V{e}_2\V{e}_1\V{e}_1 + C_{1212}\V{e}_1\V{e}_2\V{e}_1\V{e}_2 + C_{1221}\V{e}_1\V{e}_2\V{e}_2\V{e}_1 + C_{1222}\V{e}_1\V{e}_2\V{e}_2\V{e}_2\\ & C_{2111}\V{e}_2\V{e}_1\V{e}_1\V{e}_1 + C_{2112}\V{e}_2\V{e}_1\V{e}_1\V{e}_2 + C_{2121}\V{e}_2\V{e}_1\V{e}_2\V{e}_1 + C_{2122}\V{e}_2\V{e}_1\V{e}_2\V{e}_2\\ & C_{2211}\V{e}_2\V{e}_2\V{e}_1\V{e}_1 + C_{2212}\V{e}_2\V{e}_2\V{e}_1\V{e}_2 + C_{2221}\V{e}_2\V{e}_2\V{e}_2\V{e}_1 + C_{2222}\V{e}_2\V{e}_2\V{e}_2\V{e}_2 \end{split}

(A.129)

In a three-dimensional space a tensor $\TQ{C}$ has $3 \times\ 3 \times 3 \times 3 = 81$ components. The correct full matrix representation of a higher-order tensor can only be achieved with multi-dimensional matrices (which exist in modern computing languages). A fourth-order 3D tensor is then represented by a $[3 \times 3 \times 3 \times 3]$ matrix.

Different types of transposed tensors are defined: the total transpose $\TQ{C}^{T}$ , the left transpose $\TQ{C}^{LT}$ or the right transpose $\TQ{C}^{RT}$ . If the tensor $\TQ{C}$ is denoted $C_{ijkl}$ in index notation, each of these transposes is given by

\begin{aligned} \TQ{C}^{T} \;&\Leftrightarrow\; C_{lkji} \\ \TQ{C}^{LT} \;&\Leftrightarrow\; C_{jikl} \\ \TQ{C}^{RT} \;&\Leftrightarrow\; C_{ijlk} \end{aligned}

(A.130)

Different types of minor symmetries can be defined for a fourth-order tensor. A left symmetric fourth-order tensor, is a tensor for which

\TQ{C} \;=\;\TQ{C}^{LT} \quad\Longrightarrow\quad C_{ijkl}\;=\;C_{jikl}

(A.131)

In a 3D space, it has $81 - 1 \times 3 \times 3 \times 3 = 54$ distinct components. A right symmetric fourth-order tensor is given by

\TQ{C} \;=\;\TQ{C}^{RT} \quad\Longrightarrow\quad C_{ijkl}\;=\;C_{ijlk}

(A.132)

In a 3D space, it has $81 - 3 \times 3 \times 3 \times 1 = 54$ distinct components. A tensor which is left and right symmetric, has only $54 - 1 \times 3 \times 3 \times 2 = 36$ distinct components. A middle symmetric fourth-order tensor complies with

\TQ{C} \;=\;\TQ{C}^{T} \quad\Longrightarrow\quad C_{ijkl}\;=\;C_{lkji}

(A.133)

In a 3D space, it has $81 - 2 \times 3 \times 3 \times 2 = 45$ distinct components. A total symmetric fourth-order tensor equals is left, right and central symmetric, i.e.

\TQ{C} \;=\; \TQ{C}^{LT} \;=\; \TQ{C}^{RT} \;=\; \TQ{C}^{T}

(A.134)

In a 3D space, it has $36 - \half 6 \times 5 = 21$ distinct components.

The fourth-order null tensor $\TQ{0}$ is defined by

\TQ{0}\dinp\T{A} \;=\; \T{0}

(A.135)

The fourth-order unit tensor $\TQ{I}$ maps a second-order tensor $\T{A}$ on itself

\TQ{I}\dinp\T{A} \;=\; \T{A}

(A.136)

This tensor is given in index notation by

\TQ{I} \;\Longrightarrow\; I_{ijkl} \;=\; \delta_{il}\delta_{jk}

(A.137)

or with respect to a 3D Cartesian basis

\begin{split} \TQ{I} \;=\; &\V{e}_1\V{e}_1\V{e}_1\V{e}_1 + \V{e}_1\V{e}_2\V{e}_2\V{e}_1 + \V{e}_1\V{e}_3\V{e}_3\V{e}_1 \\ &\V{e}_2\V{e}_1\V{e}_1\V{e}_2 + \V{e}_2\V{e}_2\V{e}_2\V{e}_2 + \V{e}_2\V{e}_3\V{e}_3\V{e}_2 \\ &\V{e}_3\V{e}_1\V{e}_1\V{e}_3 + \V{e}_3\V{e}_2\V{e}_2\V{e}_3 + \V{e}_3\V{e}_3\V{e}_3\V{e}_3 \end{split}

(A.138)

The symmetric fourth-order unit tensor $\TQ{I}^s$ maps a second-order tensor $\T{A}$ on the symmetric part of $\T{A}$ , denoted $\T{A}^s$ :

\TQ{I}^s\dinp\T{A} \;=\; \T{A}^s \;=\; \tfrac{1}{2}(\T{A} + \T{A}^T)

(A.139)

This tensor is given in index notation by

\TQ{I}^s \;\Leftrightarrow\; I_{ijkl}^s \;=\; \tfrac{1}{2}\left(\delta_{il}\delta_{jk} + \delta_{ik}\delta_{jl}\right)

(A.140)

This tensor is a total symmetric tensor, i.e. left symmetric, right symmetric and central symmetric. Note that this tensor is also given by

\TQ{I}^s \;=\; \tfrac{1}{2}(\TQ{I}+\TQ{I}^{RT})

(A.141)

or with respect to a Cartesian basis by

\begin{split} \TQ{I}^s \;=\; &\V{e}_1\V{e}_1\V{e}_1\V{e}_1 + \tfrac{1}{2}\V{e}_1\V{e}_2\V{e}_2\V{e}_1 + \tfrac{1}{2}\V{e}_1\V{e}_2\V{e}_1\V{e}_2 + \tfrac{1}{2}\V{e}_1\V{e}_3\V{e}_3\V{e}_1 + \tfrac{1}{2}\V{e}_1\V{e}_3\V{e}_1\V{e}_3 \\ & \tfrac{1}{2}\V{e}_2\V{e}_1\V{e}_1\V{e}_2 + \tfrac{1}{2}\V{e}_2\V{e}_1\V{e}_2\V{e}_1 + \V{e}_2\V{e}_2\V{e}_2\V{e}_2 + \tfrac{1}{2}\V{e}_2\V{e}_3\V{e}_3\V{e}_2 + \tfrac{1}{2}\V{e}_2\V{e}_3\V{e}_2\V{e}_3 \\ & \tfrac{1}{2}\V{e}_3\V{e}_1\V{e}_1\V{e}_3 + \tfrac{1}{2}\V{e}_3\V{e}_1\V{e}_3\V{e}_1 + \tfrac{1}{2}\V{e}_3\V{e}_2\V{e}_2\V{e}_3 + \tfrac{1}{2}\V{e}_3\V{e}_2\V{e}_3\V{e}_2 + \V{e}_3\V{e}_3\V{e}_3\V{e}_3 \end{split}

(A.142)

Fourth-order tensors can be partially modified through inner products with second-order tensors or double inner products with other fourth-order tensors. The inner product between a fourth-order tensor $\TQ{A}$ and a second-order tensor $\T{B}$ yields a fourth-order tensor given by

\begin{aligned} \TQ{C} \;&=\; \TQ{A} \inp \T{B} \\ C_{ijkl} \;&=\; A_{ijkm}B_{ml} \end{aligned}

(A.143)

The double inner product between a fourth-order tensor $\TQ{A}$ and another fourth-order tensor $\TQ{B}$ again leads to a fourth-order tensor $\TQ{C}$ according to

\begin{aligned} \TQ{C} \;&=\; \TQ{A}\dinp\TQ{B} \\ C_{ijkl} \;&=\; A_{ijmn}B_{nmkl} \end{aligned}

(A.144)

Note that, in general, the double inner product used in combination with fourth-order tensors is no longer commutative

\begin{aligned} \TQ{A}\dinp\T{B} \;&\neq\; \T{B}\dinp\TQ{A}\\ \TQ{A}\dinp\TQ{B} \;&\neq\; \TQ{B}\dinp\TQ{A} \end{aligned}

(A.145)

A.7.3Tensor rules¶

\begin{aligned} \TQ{A}\dinp(\T{B}\inp\T{C}) \;&=\; (\TQ{A}\inp\T{B})\dinp\T{C} \\ \T{A}\inp\T{B} + \T{B}^T\inp\T{A}^T \;&=\; \TQ{I}^s \dinp (\T{A}\inp\T{B}) \\ \;&=\; (\TQ{I}^s \inp \T{A})\dinp\T{B} \end{aligned}

(A.146)

A.7.4Matrix condensation of the double inner product with a second-order tensor¶

In many numerical implementations, tensor products are condensed in a two-dimensional matrix format once a Cartesian basis has been chosen. Making use of special properties of the tensors involved then permits to omit a considerable amount of computations for a double inner product. Components of a second-order tensor $\T{B}$ are normally represented as a matrix

\M{B} \;=\; \begin{bmatrix} B_{11} & B_{12} & B_{13} \\ B_{21} & B_{22} & B_{32} \\ B_{31} & B_{32} & B_{33} \end{bmatrix}

(A.147)

For the computation of a double inner product with a fourth-order tensor, it is often more convenient to store the 9 components in a column $\C{B}$ Taking into account the symmetry of a second-order tensor $\T{B}$ permits to store the 6 distinct components in a column rather than in a matrix, e.g.,

\C{B} \;=\; \begin{bmatrix} B_{11} \\ B_{12} \\ B_{13} \\ B_{21} \\ B_{22} \\ B_{23} \\ B_{31} \\ B_{32} \\ B_{33} \end{bmatrix}

(A.148)

On the basis of this notation, it is possible to obtain the components of second-order tensor $\T{C}$ as a similarly structured column, which is the double inner product between a fourth-order tensor $\TQ{A}$ and a second-order tensor $\T{B}$ , as a matrix-column product, i.e. the components of

\T{C} \;=\; \TQ{A} \dinp \T{B}

(A.149)

are obtained through

\begin{aligned} \C{C} \;&=\; \MQ{A}\, \C{B} \\ \left[\begin{smallmatrix} C_{11} \\ C_{12} \\ C_{13} \\ C_{21} \\ C_{22} \\ C_{23} \\ C_{31} \\ C_{32} \\ C_{33} \end{smallmatrix} \right] \;&=\; \left[\begin{smallmatrix} A_{1111} & A_{1121} & A_{1131} & A_{1112} & A_{1122} & A_{1132} & A_{1113} & A_{1123} & A_{1133} \\ A_{1211} & A_{1221} & A_{1231} & A_{1212} & A_{1222} & A_{1232} & A_{1213} & A_{1223} & A_{1233} \\ A_{1311} & A_{1321} & A_{1331} & A_{1312} & A_{1322} & A_{1332} & A_{1313} & A_{1323} & A_{1333} \\ A_{2111} & A_{2121} & A_{2131} & A_{2112} & A_{2122} & A_{2132} & A_{2113} & A_{2123} & A_{2133} \\ A_{2211} & A_{2221} & A_{2231} & A_{2212} & A_{2222} & A_{2232} & A_{2213} & A_{2223} & A_{2233} \\ A_{2311} & A_{2321} & A_{2331} & A_{2312} & A_{2322} & A_{2332} & A_{2313} & A_{2323} & A_{2333} \\ A_{3111} & A_{3121} & A_{3131} & A_{3112} & A_{3122} & A_{3132} & A_{3113} & A_{3123} & A_{3133} \\ A_{3211} & A_{3221} & A_{3231} & A_{3212} & A_{3222} & A_{3232} & A_{3213} & A_{3223} & A_{3233} \\ A_{3311} & A_{3321} & A_{3331} & A_{3312} & A_{3322} & A_{3332} & A_{3313} & A_{3323} & A_{3333} \\ \end{smallmatrix} \right] \left[\begin{smallmatrix} B_{11} \\ B_{12} \\ B_{13} \\ B_{21} \\ B_{22} \\ B_{23} \\ B_{31} \\ B_{32} \\ B_{33} \end{smallmatrix}\right] \end{aligned}

(A.150)

Note that the matrix $\MQ{A}$ has indeed 81 components (all the components of $\TQ{A}$ ), although this matrix is only valid in conjunction with a double inner product with a second-order tensor.

In the particular case of mechanics, symmetric second-order tensors play an important role. Taking into account the symmetry of a second-order tensor $\T{B}$ permits to store the 6 distinct components in a smaller column with a somewhat different structure

\C{B} \;=\; \begin{bmatrix} B_{11} \\ B_{22} \\ B_{33} \\ B_{12} \\ B_{13} \\ B_{23} \end{bmatrix}

(A.151)

If the resulting tensor $\T{C}$ is not symmetric, it is represented by a column with 9 components. The matrix $\MQ{A}$ is restructured and condensed to a $[9\times 6]$ matrix by adding up the appropriate columns in equation (A.150), i.e. if $\C{B}$ has the format (A.151) and $\C{C}$ maintains the format (A.148) then $\T{C}=\TQ{A} \dinp \T{B}$ is represented in component form by

\begin{aligned} \C{C} \;&=\; \MQ{A}\, \C{B} \\ \left[\begin{smallmatrix} C_{11} \\ C_{12} \\ C_{13} \\ C_{21} \\ C_{22} \\ C_{23} \\ C_{31} \\ C_{32} \\ C_{33} \end{smallmatrix}\right] &= \left[\begin{smallmatrix} A_{1111} & A_{1122} & A_{1133} & (A_{1112}+A_{1121}) & (A_{1113}+A_{1131}) & (A_{1123}+A_{1132}) \\ A_{1211} & A_{1222} & A_{1233} & (A_{1212}+A_{1221}) & (A_{1213}+A_{1231}) & (A_{1223}+A_{1232}) \\ A_{1311} & A_{1322} & A_{1333} & (A_{1312}+A_{1321}) & (A_{1313}+A_{1331}) & (A_{1323}+A_{1332}) \\ A_{2111} & A_{2122} & A_{2133} & (A_{2112}+A_{2121}) & (A_{2113}+A_{2131}) & (A_{2123}+A_{2132}) \\ A_{2211} & A_{2222} & A_{2233} & (A_{2212}+A_{2221}) & (A_{2213}+A_{2231}) & (A_{2223}+A_{2232}) \\ A_{2311} & A_{2322} & A_{2333} & (A_{2312}+A_{2321}) & (A_{2313}+A_{2331}) & (A_{2323}+A_{2332}) \\ A_{3111} & A_{3122} & A_{3133} & (A_{3112}+A_{3121}) & (A_{3113}+A_{3131}) & (A_{3123}+A_{3132}) \\ A_{3211} & A_{3222} & A_{3233} & (A_{3212}+A_{3221}) & (A_{3213}+A_{3231}) & (A_{3223}+A_{3232}) \\ A_{3311} & A_{3322} & A_{3333} & (A_{3312}+A_{3321}) & (A_{3313}+A_{3331}) & (A_{3323}+A_{3332}) \end{smallmatrix}\right] \left[\begin{smallmatrix} B_{11} \\ B_{22} \\ B_{33} \\ B_{12} \\ B_{13} \\ B_{23} \end{smallmatrix}\right] \end{aligned}

(A.152)

If the fourth-order tensor presents a symmetry, the equations can be simplified considerably. A right symmetric tensor has only 54 distinct components (3 columns equal 3 other columns in the matrix $\MQ{A}$ ). A left symmetric tensor has also only 54 distinct components (3 lines equal 3 other lines in the matrix $\MQ{A}$ ). A tensor which is left and right symmetric has only 36 independent components (only 6 lines and columns remain in the matrix $\MQ{A}$ ). The case where the tensor $\TQ{A}$ is left symmetric results in a tensor $\T{C}$ which is symmetric. Applying $\T{C}=\TQ{A} \dinp \T{B}$ again to a symmetric $\T{B}$ can then be done with columns $\C{B}$ and $\C{C}$ of the form (A.151). The result in this particular case is then condensed as

\begin{aligned} \C{C} \;&=\; \MQ{A}\, \C{B} \\ \left[\begin{smallmatrix} C_{11} \\ C_{22} \\ C_{33} \\ C_{12} \\ C_{13} \\ C_{23} \end{smallmatrix} \right] \;&=\; \left[\begin{smallmatrix} A_{1111} & A_{1122} & A_{1133} & A_{1112}+A_{1121} & A_{1113}+A_{1131} & A_{1123}+A_{1132} \\ A_{2211} & A_{2222} & A_{2233} & A_{2212}+A_{2221} & A_{2213}+A_{2231} & A_{2223}+A_{2232} \\ A_{3311} & A_{3322} & A_{3333} & A_{3312}+A_{3321} & A_{3313}+A_{3331} & A_{3323}+A_{3332} \\ A_{1211} & A_{1222} & A_{1233} & A_{1212}+A_{1221} & A_{1213}+A_{1231} & A_{1223}+A_{1232} \\ A_{1311} & A_{1322} & A_{1333} & A_{1312}+A_{1321} & A_{1313}+A_{1331} & A_{1323}+A_{1332} \\ A_{2311} & A_{2322} & A_{2333} & A_{2312}+A_{2321} & A_{2313}+A_{2331} & A_{2323}+A_{2332} \end{smallmatrix} \right] \left[\begin{smallmatrix} B_{11} \\ B_{22} \\ B_{33} \\ B_{12} \\ B_{13} \\ B_{23} \end{smallmatrix}\right] \end{aligned}

(A.153)

Other simplifications are easy to make and left to the reader.

A.8Summary¶

Vector operations


Multiplication	$\V{b} = \alpha \V{a}$
Sum	$\V{c} = \V{a} + \V{b}$
Inner product	$\V{a} \inp \V{b} = \lVert \V{a} \rVert\; \lVert \V{b} \rVert\cos(\phi)$
	$\V{a} \inp \V{b} = \V{b} \inp \V{a}$
Cross product	$\V{a} \cross \V{b} = \V{c} = \lVert \V{a} \rVert \; \lVert \V{b} \rVert \sin(\phi)$
	$\V{b} \cross \V{a} = - \V{a} \cross \V{b}$
Triple product	$\V{a} \cross \V{b} \inp \V{c} = \lVert \V{a} \rVert \; \lVert \V{b} \rVert \sin(\phi) \lVert \V{c} \rVert \cos(\psi)$
Tensor product (dyad)	$\V{a} \V{b}$
	$\V{a} \V{b} \inp \V{p} = \V{a} (\V{b} \inp \V{p})$
	$(\V{a} \V{b})^T = \V{b} \V{a}$

Gradient operators


Cartesian	$\grad = \V{e}_1 \parder{}{x_1} + \V{e}_2 \parder{}{x_2} + \V{e}_3 \parder{}{x_3}$
$\text{grad\ }\V{a}$	$\text{grad\ }\V{a} = \grad \V{a} = \V{e}_1 \parder{\V{a}}{x_1} + \V{e}_2 \parder{\V{a}}{x_2} + \V{e}_3 \parder{\V{a}}{x_3}$
	$\grad \V{a} = a_{j,i} \V{e}_i\V{e}_j$
$\text{div\ }\V{a}$	$\text{div\ }\V{a} = \grad \inp \V{a} = \V{e}_1 \inp \parder{\V{a}}{x_1} + \V{e}_2 \inp \parder{\V{a}}{x_2} + \V{e}_3 \inp\parder{\V{a}}{x_3}$
$\text{rot\ }\V{a}$	$\text{rot\ }\V{a} = \grad \cross \V{a} = \V{e}_1 \cross \parder{\V{a}}{x_1} + \V{e}_2 \cross \parder{\V{a}}{x_2} + \V{e}_3 \cross \parder{\V{a}}{x_3}$
Cylindrical	$\grad = \V{e}_r \parder{}{r} + \V{e}_t \frac{1}{r} \parder{}{\theta} + \V{e}_z \parder{}{z}$

Second-order Tensors


Second order tensor	$\T{A} \inp \V{p} = \V{q}$
	$\T{A} \inp (\alpha \V{m} + \beta \V{n}) = \alpha \T{A} \inp \V{m} + \beta \T{A} \inp \V{n}$
	$(\alpha \T{A}) \inp \V{v} = \T{A} \inp (\alpha \V{v})$
	$(\T{A} + \T{B}) \inp \V{v} = \T{A} \inp \V{v} + \T{B} \inp \V{v}$
	$(\T{A}\inp\T{B})\inp\V{v} = \T{A}\inp(\T{B}\inp\V{v})$
Conjugate of tensor	$\T{A}^T \inp \V{p} = \V{p} \inp \T{A}$
Unit tensor	$\T{I} = \V{e}_1\V{e}_1 + \V{e}_2\V{e}_2 + \V{e}_3\V{e}_3$
First invariant	$J_1(\T{A}) = \tr(\T{A})$
Second invariant	$J_2(\T{A}) = \half \{ \tr^2(\T{A}) - \tr(\T{A}^2) \}$
Third invariant	$J_3(\T{A}) = \det(\T{A})$
Double dot product	$\T{A}\dinp\T{B} = \tr (\T{A}\inp\T{B} )$
	$\T{A} \dinp \T{I} = \tr(\T{A})$
	$\T{A} \dinp \T{B} = \T{A}^T \dinp \T{B}^T$
Inverse tensor	$\T{A}^{-1} \inp \T{A} = \T{I}$
	$(\T{A} \inp \T{B} )^{-1} = \T{B}^{-1} \inp \T{A}^{-1}$
Deviatoric part	$\T{A}^d = \T{A} - \tfrac{1}{3} \tr(\T{A}) \T{I}$
Symmetric tensor	$\T{A}^T = \T{A}$
Skew-symmetric tensor	$\T{A}^T = -\T{A}$
Positive definite tensor	$\V{a} \inp \T{A} \inp \V{a} > 0 \quad \forall\;\V{a} \neq \V{0}$
Orthogonal tensor	$(\T{A} \inp \V{a}) \inp (\T{A} \inp \V{b}) = \V{a} \inp \V{b} \quad \forall\ \V{a},\V{b}$

Fourth-order Tensors


Transformation	$\T{B} = \TQ{C} \dinp \T{A}$
Components	$[3 \times 3 \times 3 \times 3]$ matrix
Total transpose	$\TQ{C}^{T} = [C_{ijkl}]^{T} = C_{lkji}$
Left transpose	$\TQ{C}^{LT} = [C_{ijkl}]^{LT} = C_{jikl}$
Right transpose	$\TQ{C}^{RT} = [C_{ijkl}]^{RT} = C_{ijlk}$
Left symmetric	$\TQ{C} = \TQ{C}^{LT}$
Right symmetric	$\TQ{C} = \TQ{C}^{RT}$
Middle symmetric	$\TQ{C} = \TQ{C}^{T}$
Total symmetric	$\TQ{C} = \TQ{C}^{LT} = \TQ{C}^{RT} = \TQ{C}^{T}$
Fourth-order unit tensor	$\TQ{I}\dinp\T{A} = \T{A}$
Symmetric fourth-order unit	$\TQ{I}^s\dinp\T{A} = \T{A}^s = \tfrac{1}{2}(\T{A} + \T{A}^T)$
Products	$\TQ{C} = \TQ{A} \inp \T{B}$
	$\TQ{C} = \TQ{A}\dinp\TQ{B}$
Matrix condensation	$\T{B} = \TQ{C} \dinp \T{A} \Longrightarrow \C{B} = \MQ{C}\, \C{A}$

Footnotes¶

inwendig produkt
↩
tripel produkt
↩
dyadisch produkt
↩
geconjugeerde dyade
↩
rechtshandige basis
↩
tweede-orde eenheidstensor
↩
Euclidische norm
↩
invariant
↩
spoor
↩
reguliere tensor
↩
singulier
↩
dubbel inwendig produkt
↩
eigenwaarden
↩
eigenvectoren
↩
deviatorisch deel
↩
hydrostatisch of sferisch deel
↩
spectrale vorm
↩
scheefsymmetrisch
↩
axiaalvector
↩
positief definiet
↩
orthogonale tensor
↩
positievector
↩
gradiënt
↩
divergentie
↩
rotatie
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