$$ 

 

 

Chapter 3

 

Mathematics Review

 

 

 

3.1. Vectors

 

3.1.1 Definition

 

For the purposes of this text, a vector is an object which has magnitude and direction.  Examples include forces, electric fields, and the normal to a surface.  A vector is often represented pictorially as an arrow and symbolically by an underlined letter $\underset{\_}{a}$ or using bold type .  Its magnitude is denoted  or .  There are two special cases of vectors: the unit vector $n$ has $\left|n\right|=1$; and the null vector  has $\left|0\right|=0$.

 

3.1.2 Vector Operations

 

 Addition

 

Let a and b be vectors.  Then $c=a+b$ is also a vector.  The vector c may be shown diagramatically by placing arrows representing a and b head to tail, as shown in the figure.

 

 Multiplication

 

1.      Multiplication by a scalar. Let a be a vector, and  a scalar.  Then $b=\alpha a$ is a vector.  The direction of $b$ is parallel to $a$ and its magnitude is given by $\left|b\right|=\alpha \left|a\right|$.

Note that you can form a unit vector n which is parallel to a by setting $n=\frac{a}{\left|a\right|}$.

2.      Dot Product (also called the scalar product). Let a and b be two vectors.  The dot product of a and b is a scalar denoted by $\alpha =a\cdot b$, and is defined by

$a\cdot b=\left|a\right|\left|b\right|\cos \theta (a,b)$,

where $\theta (a,b)$ is the angle subtended by a and b. Note that $a\cdot b=b\cdot a$, and $a\cdot a={\left|a\right|}^2$.  If $\left|a\right|\ne 0$ and $\left|b\right|\ne 0$ then $a\cdot b=0$ if and only if $\cos \theta (a,b)=0$; i.e. a and b are perpendicular.

3.      Cross Product (also called the vector product).  Let a and b be two vectors.  The cross product of a and b is a vector denoted by $c=a\times b$.  The direction of c is perpendicular to a and b, and is chosen so that (a,b,c) form a right handed triad, Fig. 3.  The magnitude of c is given by

$\left|c\right|=\left|a\times b\right|=\left|a\right|\left|b\right|\sin \theta (a,b)$

Note that $a\times b=-b\times a$ and $a\cdot (a\times b)=b\cdot (a\times b)=0$.

 

 Some useful vector identities

$\begin{array}{l}a\cdot (b\times c)=b\cdot (c\times a)=c\cdot (a\times b)\\ a\times (b\times c)=(a\cdot c)b-(a\cdot b)c\\ (a\times b)\cdot (c\times d)=(a\cdot c)(b\cdot d)-(b\cdot c)(a\cdot d)\end{array}$

 

3.1.3 Cartesian components of vectors

 

Let $(e_1,e_2,e_3)$ be three mutually perpendicular unit vectors which form a right handed triad, Fig. 4.  Then $\left\{e_1,e_2,e_3\right\}$ are said to form and orthonormal basis. The vectors satisfy

$\left|e_1\right|=\left|e_2\right|=\left|e_3\right|=1e_1\times e_2=e_3,e_1\times e_3=-e_2{e}_2\times e_3=e_1$

We may express any vector a as a suitable combination of the unit vectors $e_1$, $e_2$ and $e_3$.  For example, we may write

$a=a_1{e}_1+a_2{e}_2+a_3{e}_3={\displaystyle \sum _{i=1}^3{a}_i{e}_i}$

where  are scalars, called the components of a in the basis $\left\{e_1,e_2,e_3\right\}$.   The components of a have a simple physical interpretation.  For example, if we evaluate the dot product $a\cdot e_1$ we find that

$a\cdot e_1=(a_1{e}_1+a_2{e}_2+a_3{e}_3)\cdot e_1=a_1$

in view of the properties of the three vectors $e_1$, $e_2$ and $e_3$.  Recall that

$a\cdot e_1=\left|a\right|\left|e_1\right|\cos \theta (a,e_1)$

 

Then, noting that $\left|e_1\right|=1$, we have

$a_1=a\cdot e_1=\left|a\right|\cos \theta (a,e_1)$

 

Thus,  represents the projected length of the vector a  in the direction of $e_1$, as illustrated in the figure.  Similarly, $a_2$ and $a_3$ may be shown to represent the projection of $a$ in the directions $e_2$ and , respectively.

 

The advantage of representing vectors in a Cartesian basis is that vector addition and multiplication can be expressed as simple operations on the components of the vectors.  For example, let a, b and c be vectors, with components $(a_1,a_2,a_3)$, $(b_1,b_2,b_3)$ and $(c_1,c_2,c_3)$, respectively.  Then, it is straightforward to show that

$\begin{array}{c}c=a+b\iff c_1=a_1+b_1;c_2=a_2+b_2;c_3=a_3+b_3\\ a\cdot b={\displaystyle \sum _{i=1}^3{a}_i{b}_i}\\ c=a\times b\iff c_1=a_2{b}_3-a_3{b}_2;c_2=a_3{b}_1-a_1{b}_3;c_3=a_1{b}_2-a_2{b}_1\end{array}$

 

 

 

 

 

 

3.1.4 Change of basis

 

Let a be a vector, and let $\left\{e_1,e_2,e_3\right\}$ be a Cartesian basis.  Suppose that the components of a in the basis $\left\{e_1,e_2,e_3\right\}$ are known to be $(a_1,a_2,a_3)$.  Now, suppose that we wish to compute the components of a in a second Cartesian basis, $\left\{m_1,m_2,m_3\right\}$.  This means we wish to find components $({\alpha }_1,{\alpha }_2,{\alpha }_3)$, such that

$a={\alpha }_1{m}_1+{\alpha }_2{m}_2+{\alpha }_3{m}_3$

To do so, note that

$\begin{array}{l}{\alpha }_1=a\cdot m_1=a_1{e}_1\cdot m_1+a_2{e}_2\cdot m_1+a_3{e}_3\cdot m_1\\ {\alpha }_2=a\cdot m_2=a_1{e}_1\cdot m_2+a_2{e}_2\cdot m_2+a_3{e}_3\cdot m_2\\ {\alpha }_3=a\cdot m_3=a_1{e}_1\cdot m_3+a_2{e}_2\cdot m_3+a_3{e}_3\cdot m_3\end{array}$

This transformation is conveniently written as a matrix operation

$\left[\alpha \right]=\left[Q\right]\left[a\right]$,

where $\left[\alpha \right]$ is a matrix consisting of the components of a in the basis $\left\{m_1,m_2,m_3\right\}$, $\left[a\right]$ is a matrix consisting of the components of a in the basis $\left\{e_1,e_2,e_3\right\}$, and  is a `rotation matrix’ as follows

$\left[\alpha \right]=\left[\begin{array}{l}{\alpha }_1\\ {\alpha }_2\\ {\alpha }_3\end{array}\right]\left[a\right]=\left[\begin{array}{l}{a}_1\\ a_2\\ a_3\end{array}\right]\left[Q\right]=\left[\begin{array}{l}{m}_1\cdot e_1{m}_1\cdot e_2{m}_1\cdot e_3\\ m_2\cdot e_1{m}_2\cdot e_2{m}_2\cdot e_3\\ m_3\cdot e_1{m}_3\cdot e_2{m}_3\cdot e_3\end{array}\right]$

Note that the elements of  have a simple physical interpretation.  For example, $m_1\cdot e_1=\cos \theta (m_1,e_1)$, where $\theta (m_1,e_1)$ is the angle between the $m_1$ and  axes.  Similarly $m_1\cdot e_2=\cos \theta (m_1,e_2)$ where $\theta (m_1,e_2)$ is the angle between the $m_1$ and  axes.  In practice, we usually know the angles between the axes that make up the two bases, so it is simplest to assemble the elements of  by putting the cosines of the known angles in the appropriate places.

 

Index notation provides another convenient way to write this transformation:

${\alpha }_i=Q_{ij}{a}_j,Q_{ij}=e_i\cdot m_j$

You don’t need to know index notation in detail to understand this $–$ all you need to know is that

$Q_{ij}{a}_j\equiv {\displaystyle \sum _{j=1}^3{Q}_{ij}{a}_j}$

 

The same approach may be used to find an expression for $a_i$ in terms of ${\alpha }_i$.  If you work through the details, you will find that

$\left[\begin{array}{l}{a}_1\\ a_2\\ a_3\end{array}\right]=\left[\begin{array}{ccc}{m}_1\cdot e_1& m_2\cdot e_1& m_3\cdot e_1\\ m_1\cdot e_2& m_2\cdot e_2& m_3\cdot e_2\\ m_1\cdot e_3& m_2\cdot e_3& m_3\cdot e_3\end{array}\right]\left[\begin{array}{l}{\alpha }_1\\ {\alpha }_2\\ {\alpha }_3\end{array}\right]$

Comparing this result with the formula for ${\alpha }_i$ in terms of $a_i$, we see that

$\left[a\right]={\left[Q\right]}^T\left[\alpha \right]$

where the superscript T denotes the transpose (rows and columns interchanged). The transformation matrix  is therefore orthogonal, and satisfies

${\left[Q\right]}^{-1}={\left[Q\right]}^T\left[Q\right]{\left[Q\right]}^T={\left[Q\right]}^T\left[Q\right]=\left[I\right]$

where [I] is the identity matrix.

 

3.1.5 Useful vector operations

 Calculating areas

The area of a triangle bounded by vectors a, b¸and b-a is

$A=\frac{1}{2}|a\times b|$

The area of the parallelogram shown in the picture is 2A.

 

 Calculating angles

The angle between two vectors a and b is

$\theta ={\cos }^{-1}\left(a\cdot b/\left|a\right|\left|b\right|\right)$

 

 Calculating the normal to a surface.

If two vectors a and b can be found which are known to lie in the surface, then the unit normal to the surface is

$n=\pm \frac{a\times b}{\left|a\times b\right|}$

If the surface is specified by a parametric equation of the form $r=r(s,t)$, where s and t are two parameters and r is the position vector of a point on the surface, then two vectors which lie in the plane may be computed from

$a=\frac{\partial r}{\partial s},b=\frac{\partial r}{\partial t}$

 

 Calculating Volumes

The volume of the parallelopiped defined by three vectors a, b, c is

$V=|c\cdot \left(a\times b\right)|$

The volume of the tetrahedron shown outlined in red is V/6.

 

 

 

 

3.2 Vector Fields and Vector Calculus

 

 

3.2.1. Scalar field.

 

Let  $\left\{e_1,e_2,e_3\right\}$ be a Cartesian basis with origin O in three dimensional space.  Let

$r=x_1{e}_1+x_2{e}_2+x_3{e}_3$

denote the position vector of a point in space.  A scalar field is a scalar valued function of position in space.  A scalar field is a function of the components of the position vector, and so may be expressed as $\varphi (x_1,x_2,x_3)$. The value of  at a particular point in space must be independent of the choice of basis vectors.  A scalar field may be a function of time (and possibly other parameters) as well as position in space.

 

 

 

3.2.2. Vector field

 

Let  $\left\{e_1,e_2,e_3\right\}$ be a Cartesian basis with origin O in three dimensional space.  Let

$r=x_1{e}_1+x_2{e}_2+x_3{e}_3$

denote the position vector of a point in space.  A vector field is a vector valued function of position in space.  A vector field is a function of the components of the position vector, and so may be expressed as $v(x_1,x_2,x_3)$.  The vector may also be expressed as components in the basis $\left\{e_1,e_2,e_3\right\}$

$v(x_1,x_2,x_3)=v_1(x_1,x_2,x_3)e_1+v_2(x_1,x_2,x_3)e_2+v_3(x_1,x_2,x_3)e_3$

The magnitude and direction of  at a particular point in space is independent of the choice of basis vectors.   A vector field may be a function of time (and possibly other parameters) as well as position in space.

 

 

3.2.3. Change of basis for scalar fields.

Let $\left\{e_1,e_2,e_3\right\}$ be a Cartesian basis with origin O in three dimensional space. Express the position vector of a point relative to O in $\left\{e_1,e_2,e_3\right\}$ as

$r=x_1{e}_1+x_2{e}_2+x_3{e}_3$

and let $\varphi (x_1,x_2,x_3)$ be a scalar field.

Let $\left\{m_1,m_2,m_3\right\}$ be a second Cartesian basis, with origin P.  Let $c\equiv \overrightarrow{OP}$ denote the position vector of  P relative to O. Express the position vector of a point relative to P in $\left\{m_1,m_2,m_3\right\}$ as

$p={\xi }_1{m}_1+{\xi }_2{m}_2+{\xi }_3{m}_3$

 

To find $\varphi ({\xi }_1,{\xi }_2,{\xi }_3)$, use the following procedure.  First, express  p as components in the basis $\left\{e_1,e_2,e_3\right\}$, using the procedure outlined in Section 1.4:

$p=p_1{e}_1+p_2{e}_2+p_3{e}_3$

where

$\begin{array}{l}{p}_1={\xi }_1{e}_1\cdot m_1+{\xi }_2{e}_2\cdot m_1+{\xi }_3{e}_3\cdot m_1\\ p_2={\xi }_1{e}_1\cdot m_2+{\xi }_2{e}_2\cdot m_2+{\xi }_3{e}_3\cdot m_2\\ p_3={\xi }_1{e}_1\cdot m_3+{\xi }_2{e}_2\cdot m_3+{\xi }_3{e}_3\cdot m_3\end{array}$

or, using index notation

$p_i=Q_{ij}{\xi }_j$

where the transformation matrix $Q_{ij}$ is defined in Sect 1.4.

Now, express c as components in $\left\{e_1,e_2,e_3\right\}$, and note that

$\begin{array}{l}r=p+c\\ \Rightarrow x_1{e}_1+x_2{e}_2+x_3{e}_3=p_1{e}_1+p_2{e}_2+p_3{e}_3+c_1{e}_1+c_2{e}_2+c_3{e}_3\\ \Rightarrow x_1=p_1+c_1,x_2=p_2+c_2,x_3=p_3+c_3\\ \Rightarrow x_i=Q_{ij}{\xi }_j+c_i\end{array}$

so that

$\begin{array}{l}\varphi (x_1,x_2,x_3)=\varphi (p_1+c_1,p_2+c_2,p_3+c_3)\\ =\varphi (Q_{1j}{\xi }_j+c_1,Q_{2j}{\xi }_j+c_2,Q_{3j}{\xi }_j+c_3)\end{array}$

 

3.2.4. Change of basis for vector fields.

Let $\left\{e_1,e_2,e_3\right\}$ be a Cartesian basis with origin O in three dimensional space. Express the position vector of a point relative to O in $\left\{e_1,e_2,e_3\right\}$ as

$r=x_1{e}_1+x_2{e}_2+x_3{e}_3$

and let $v(x_1,x_2,x_3)$ be a vector  field, with components

$v(x_1,x_2,x_3)=v_1(x_1,x_2,x_3)e_1+v_2(x_1,x_2,x_3)e_2+v_3(x_1,x_2,x_3)e_3$

Let $\left\{m_1,m_2,m_3\right\}$ be a second Cartesian basis, with origin P.  Let $c\equiv \overrightarrow{OP}$ denote the position vector of  P relative to O. Express the position vector of a point relative to P in $\left\{m_1,m_2,m_3\right\}$ as

$p={\xi }_1{m}_1+{\xi }_2{m}_2+{\xi }_3{m}_3$

 

To express the vector field as components in $\left\{m_1,m_2,m_3\right\}$ and as a function of the components of p, use the following procedure.  First, express $(v_1,v_2,v_3)$ in terms of $({\xi }_1,{\xi }_2,{\xi }_3)$ using the procedure outlined for scalar fields in the preceding section

$\begin{array}{l}{v}_k(x_1,x_2,x_3)=v_k(p_1+c_1,p_2+c_2,p_3+c_3)\\ =v_k(Q_{1j}{\xi }_j+c_1,Q_{2j}{\xi }_j+c_2,Q_{3j}{\xi }_j+c_3)\end{array}$

for k=1,2,3.  Now, find the components  of v in $\left\{m_1,m_2,m_3\right\}$ using the procedure outlined in Section 1.4.  Using index notation, the result is

$\begin{array}{l}v=Q_{1i}{v}_i(Q_{1j}{\xi }_j+c_1,Q_{2j}{\xi }_j+c_2,Q_{3j}{\xi }_j+c_3)e_1\\ +Q_{2i}{v}_i(Q_{1j}{\xi }_j+c_1,Q_{2j}{\xi }_j+c_2,Q_{3j}{\xi }_j+c_3)e_2\\ +Q_{2i}{v}_i(Q_{1j}{\xi }_j+c_1,Q_{2j}{\xi }_j+c_2,Q_{3j}{\xi }_j+c_3)e_3\end{array}$

 

 

3.2.5. Time derivatives of vectors

 

Let a(t) be a vector whose magnitude and direction vary with time, t.  Suppose that $\left\{i,j,k\right\}$ is a fixed basis, i.e. independent of time.  We may express a(t) in terms of components $(a_x,a_y,a_z)$ in the basis $\left\{i,j,k\right\}$ as

.

The time derivative of a is defined using the usual rules of calculus

$\dot{a}(t)=\frac{d}{dt}a(t)=\underset{\in \to 0}{\lim }\frac{a(t+\in )-a(t)}{\in }$,

or in component form as

$\dot{a}(t)={\dot{a}}_{x}i+{\dot{a}}_{y}j+{\dot{a}}_{z}k$

The definition of the time derivative of a vector may be used to show the following rules

$\begin{array}{l}\frac{d}{dt}\left[\alpha (t)a(t)\right]=\dot{\alpha }(t)a(t)+\alpha (t)\dot{a}(t)\\ \frac{d}{dt}\left[a(t)\cdot b(t)\right]=\dot{a}(t)\cdot b(t)+a(t)\cdot \dot{b}(t)\\ \frac{d}{dt}\left[a(t)\times b(t)\right]=\dot{a}(t)\times b(t)+a(t)\times \dot{b}(t)\end{array}$

 

 

3.2.6. Using a rotating basis

 

It is often convenient to express position vectors as components in a basis which rotates with time.  To write equations of motion one must evaluate time derivatives of rotating vectors.

 

Let $\left\{e_1,e_2,e_3\right\}$ be a basis which rotates with instantaneous angular velocity $\Omega$.  Then,

$\frac{d{e}_1}{dt}=\Omega \times e_1,\frac{d{e}_2}{dt}=\Omega \times e_2,\frac{d{e}_3}{dt}=\Omega \times e_3$

 

 

3.2.7. Gradient of a scalar field.

 

Let  be a scalar field in three dimensional space.  The gradient of  is a vector field denoted by  or $\varphi \nabla$, and is defined so that

$(\varphi \nabla )\cdot a=\underset{\in \to 0}{\lim }\frac{\varphi (r+\in a)-\varphi (r)}{\in }$

for every position r in space and for every vector a.

 

Let   be a Cartesian basis with origin O in three dimensional space.  Let

$r=x_1{e}_1+x_2{e}_2+x_3{e}_3$

denote the position vector of a point in space.  Express  as a function of the components of r $\varphi =\varphi (x_1,x_2,x_3)$.  The gradient of   in this basis is then given by

$\varphi \nabla =\frac{\partial \varphi }{\partial x_1}{e}_1+\frac{\partial \varphi }{\partial x_2}{e}_2+\frac{\partial \varphi }{\partial x_3}{e}_3$

 

3.2.8. Gradient of a vector field

 

Let v be a vector field in three dimensional space.  The gradient of v is a tensor field denoted by $\text{grad}(v)$ or $\nabla v$, and is defined so that

$(\nabla v)\cdot a=\underset{\in \to 0}{\lim }\frac{v(r+\in a)-v(r)}{\in }$

for every position r in space and for every vector a.

 

Let  $\left\{e_1,e_2,e_3\right\}$ be a Cartesian basis with origin O in three dimensional space.  Let

$r=x_1{e}_1+x_2{e}_2+x_3{e}_3$

denote the position vector of a point in space.  Express v as a function of the components of r, so that $v=v(x_1,x_2,x_3)$.  The gradient of  v in this basis is then given by

$\nabla v=\left[\begin{array}{ccc}\frac{\partial v_1}{\partial x_1}& \frac{\partial v_1}{\partial x_2}& \frac{\partial v_1}{\partial x_3}\\ \frac{\partial v_2}{\partial x_1}& \frac{\partial v_2}{\partial x_2}& \frac{\partial v_2}{\partial x_3}\\ \frac{\partial v_3}{\partial x_1}& \frac{\partial v_3}{\partial x_2}& \frac{\partial v_3}{\partial x_3}\end{array}\right]$

Alternatively, in index notation

${\left[\nabla v\right]}_{ij}\equiv \frac{\partial v_i}{\partial x_j}$

 

3.2.9. Divergence of a vector field

 

Let v be a vector field in three dimensional space.  The divergence of v is a scalar field denoted by $\text{div}(v)$ or $\nabla \cdot v$.  Formally, it is defined as $\text{trace(grad(}v\text{))}$ (the trace of a tensor is the sum of its diagonal terms). 

 

Let   be a Cartesian basis with origin O in three dimensional space.  Let

$r=x_1{e}_1+x_2{e}_2+x_3{e}_3$

denote the position vector of a point in space.  Express v as a function of the components of r: $v=v(x_1,x_2,x_3)$. The divergence of v is then

 

$\text{div(}v\text{)=}\frac{\partial v_1}{\partial x_1}+\frac{\partial v_2}{\partial x_2}+\frac{\partial v_3}{\partial x_3}$

 

 

3.2.10. Curl of a vector field.

 

Let v be a vector field in three dimensional space.  The curl of  v  is a vector field denoted by $\text{curl}(v)$ or $\nabla \times v$.  It is best defined in terms of its components in a given basis, although its magnitude and direction are not dependent on the choice of basis.

 

Let  $\left\{e_1,e_2,e_3\right\}$ be a Cartesian basis with origin O in three dimensional space.  Let

$r=x_1{e}_1+x_2{e}_2+x_3{e}_3$

denote the position vector of a point in space.  Express v as a function of the components of r  $v=v(x_1,x_2,x_3)$. The curl of  v in this basis is then given by

$\nabla \times v=\left|\begin{array}{ccc}{e}_1& e_2& e_3\\ \frac{\partial }{\partial x_1}& \frac{\partial }{\partial x_2}& \frac{\partial }{\partial x_3}\\ v_1& v_2& v_3\end{array}\right|=\left(\frac{\partial v_3}{\partial x_2}-\frac{\partial v_2}{\partial x_3}\right)e_1+\left(\frac{\partial v_1}{\partial x_3}-\frac{\partial v_3}{\partial x_1}\right)e_2+\left(\frac{\partial v_2}{\partial x_1}-\frac{\partial v_1}{\partial x_2}\right)e_3$

Using index notation, this may be expressed as

${\left[\nabla \times v\right]}_i={\in }_{ijk}\frac{\partial v_j}{\partial x_k}$

 

 

3.2.11 The Divergence Theorem.

 

Let V be a closed region in three dimensional space, bounded by an orientable surface S. Let n  denote the unit vector normal to S, taken so that n points out of V. Let u be a vector field which is continuous and has continuous first partial derivatives in some domain containing T.  Then

${\displaystyle \underset{V}{\int }\text{div(}u\text{)}}dV={\displaystyle \underset{S}{\int }u\cdot n}dA$

alternatively, expressed in index notation

${\displaystyle \underset{V}{\int }\frac{\partial u_i}{\partial x_i}dV={\displaystyle \underset{S}{\int }{u}_i{n}_{i}dA}}$

For a proof of this extremely useful theorem consult e.g. Kreyzig, Advanced Engineering Mathematics, Wiley, New York, (1998).

 

 

 

3.3 MATRICES

 

3.3.1 Definition

 

An $(n\times m)$ matrix $[A]$ is a set of numbers, arranged in m rows and n columns

$\left[A\right]=\left[\begin{array}{l}{a}_{11}{a}_{12}{a}_{13}\cdots a_{1n}\\ a_{21}{a}_{22}{a}_{23}\cdots a_{2n}\\ ⋮⋮⋮\ddots ⋮\\ a_{m1}{a}_{m2}{a}_{m3}\cdots a_{mn}\end{array}\right]$

 

 A square matrix has equal numbers of rows and columns

 A diagonal matrix is a square matrix with elements such that  for

 The identity matrix $\left[I\right]$ is a diagonal matrix for which all diagonal elements $a_{ii}=1$

 A symmetric matrix is a square matrix with elements such that $a_{ij}=a_{ji}$

 A skew symmetric matrix is a square matrix with elements such that $a_{ij}=-a_{ji}$

 

 

3.3.2 Matrix operations

 

 Addition  Let  $\left[A\right]$ and $\left[B\right]$ be two matrices of order $(m\times n)$ with elements $a_{ij}$ and $b_{ij}$.  Then

$\left[C\right]=\left[A\right]+\left[B\right]\iff c_{ij}=a_{ij}+b_{ij}$

 

 

 Multiplication by a scalar.  Let $\left[A\right]$ be a matrix with elements $a_{ij}$, and let k be a scalar.  Then

$\left[B\right]=k\left[A\right]\iff b_{ij}=k{a}_{ij}$

 

 

 Multiplication by a matrix. Let $\left[A\right]$ be a matrix of order $(m\times n)$ with elements $a_{ij}$, and let $\left[B\right]$ be a matrix of order $(p\times q)$ with elements .  The product $\left[C\right]=\left[A\right]\left[B\right]$ is defined only if n=p, and is an $(m\times q)$ matrix such that

$\left[C\right]=\left[A\right]\left[B\right]\iff c_{ij}={\displaystyle \sum _{k=1}^n{a}_{ik}{b}_{kj}}$

Note that multiplication is distributive and associative, but not commutative, i.e.

$\left[A\right]\left(\left[B\right]+\left[C\right]\right)=\left[A\right]\left[B\right]+\left[A\right]\left[C\right]\left[A\right]\left(\left[B\right]\left[C\right]\right)=\left(\left[A\right]\left[B\right]\right)\left[C\right]\left[A\right]\left[B\right]\ne \left[B\right]\left[A\right]$

The multiplication of a vector by a matrix is a particularly important operation.  Let b and c be two vectors with n components, which we think of as $(1\times n)$ matrices.  Let $\left[A\right]$ be an $(m\times n)$ matrix.  Thus

$b=\left[\begin{array}{l}{b}_1\\ b_2\\ b_3\\ ⋮\\ b_n\end{array}\right]c=\left[\begin{array}{l}{c}_1\\ c_2\\ c_3\\ ⋮\\ c_n\end{array}\right]\left[A\right]=\left[\begin{array}{l}{a}_{11}{a}_{12}{a}_{13}\cdots a_{1n}\\ a_{21}{a}_{22}{a}_{23}\cdots a_{2n}\\ ⋮⋮⋮\ddots ⋮\\ a_{m1}{a}_{m2}{a}_{m3}\cdots a_{mn}\end{array}\right]$

Now,

$c=\left[A\right]b\iff c_i={\displaystyle \sum _{j=1}^n{a}_{ij}{b}_j}$

i.e.

$\begin{array}{l}{c}_1=a_{11}{b}_1+a_{12}{b}_2+a_{13}{b}_3+\cdots a_{1n}{b}_n\\ c_2=a_{21}{b}_1+a_{22}{b}_2+a_{23}{b}_3+\cdots a_{2n}{b}_n\\ ⋮\\ c_m=a_{m1}{b}_1+a_{m2}{b}_2+a_{m3}{b}_3+\cdots a_{mn}{b}_n\end{array}$

 

 Transpose. Let $\left[A\right]$ be a matrix of order $(m\times n)$ with elements $a_{ij}$.  The transpose of $\left[A\right]$ is denoted ${\left[A\right]}^T$.  If $\left[B\right]$ is an $(n\times m)$ matrix such that $\left[B\right]={\left[A\right]}^T$, then $b_{ij}=a_{ji}$, i.e.

${\left[A\right]}^T={\left[\begin{array}{l}{a}_{11}{a}_{12}{a}_{13}\cdots a_{1n}\\ a_{21}{a}_{22}{a}_{23}\cdots a_{2n}\\ ⋮⋮⋮\ddots ⋮\\ a_{m1}{a}_{m2}{a}_{m3}\cdots a_{mn}\end{array}\right]}^T=\left[\begin{array}{l}{a}_{11}{a}_{21}{a}_{31}\cdots a_{n1}\\ a_{12}{a}_{22}{a}_3{}_2\cdots a_{n2}\\ ⋮⋮⋮\ddots ⋮\\ a_{1m}{a}_{2m}{a}_{3m}\cdots a_{nm}\end{array}\right]$

Note that

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

 

 Determinant  The determinant is defined only for a square matrix.  Let $\left[A\right]$ be a $(2\times 2)$ matrix with components $a_{ij}$.  The determinant of  $\left[A\right]$ is denoted by $\det \left[A\right]$ or $\left|A\right|$ and is given by

$\left|A\right|=\left|\begin{array}{l}{a}_{11}{a}_{12}\\ a_{21}{a}_{22}\end{array}\right|=a_{11}{a}_{22}-a_{12}{a}_{21}$

Now, let $\left[A\right]$ be an $\left(n\times n\right)$ matrix.  Define the minors $M_{ij}$ of $\left[A\right]$ as the determinant formed by omitting the ith row and jth column of $\left[A\right]$.  For example, the minors $M_{11}$ and $M_{12}$ for a $\left(3\times 3\right)$ matrix are computed as follows.   Let

$\left[A\right]=\left[\begin{array}{l}{a}_{11}{a}_{12}{a}_{13}\\ a_{21}{a}_{22}{a}_{23}\\ a_{31}{a}_{32}{a}_{33}\end{array}\right]$

Then

$M_{11}=\left|\begin{array}{l}{a}_{22}{a}_{23}\\ a_{32}{a}_{33}\end{array}\right|=a_{22}{a}_{33}-a_{32}{a}_{23}{M}_{12}=\left|\begin{array}{l}{a}_{21}{a}_{23}\\ a_{31}{a}_{33}\end{array}\right|=a_{21}{a}_{33}-a_{31}{a}_{23}$

Define the cofactors $C_{ij}$ of $\left[A\right]$ as

$C_{ij}={\left(-1\right)}^{i+j}{M}_{ij}$

Then, the determinant of the $\left(n\times n\right)$ matrix $\left[A\right]$ is computed as follows

$\left|A\right|={\displaystyle \sum _{j=1}^n{a}_{ij}{C}_{ij}}$

The result is the same whichever row i is chosen for the expansion.  For the particular case of a $\left(3\times 3\right)$ matrix

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

The determinant may also be evaluated by summing over rows, i.e.

$\left|A\right|={\displaystyle \sum _{i=1}^n{a}_{ij}{C}_{ij}}$

and as before the result is the same for each choice of column j.  Finally, note that

 

 Inversion.  Let $\left[A\right]$ be an $\left(n\times n\right)$ matrix.  The inverse of $\left[A\right]$ is denoted by  and is defined such that

The inverse of $\left[A\right]$ exists if and only if $\det \left[A\right]\ne 0$.  A matrix which has no inverse is said to be singular.  The inverse of a matrix may be computed explicitly, by forming the cofactor matrix $\left[C\right]$ with components $c_{ij}$ as defined in the preceding section.  Then

${\left[A\right]}^{-1}=\frac{1}{\det \left[A\right]}{\left[C\right]}^T$

In practice, it is faster to compute the inverse of a matrix using methods such as Gaussian elimination. 

 

Note that

${\left(\left[A\right]\left[B\right]\right)}^{-1}={\left[B\right]}^{-1}{\left[A\right]}^{-1}$

For a diagonal matrix, the inverse is

$\left[A\right]=\left[\begin{array}{l}{a}_{11}00\cdots 0\\ 0a_{22}0\cdots 0\\ ⋮⋮⋮\ddots ⋮\\ 000\cdots a_{nn}\end{array}\right]=\left[\begin{array}{l}1/a_{11}00\cdots 0\\ 01/a_{22}0\cdots 0\\ ⋮⋮⋮\ddots ⋮\\ 000\cdots 1/a_{nn}\end{array}\right]$

For a $\left(2\times 2\right)$ matrix, the inverse is

$\left[\begin{array}{l}{a}_{11}{a}_{12}\\ a_{21}{a}_{22}\end{array}\right]=\frac{1}{a_{11}{a}_{22}-a_{12}{a}_{21}}\left[\begin{array}{l}{a}_{22}-a_{12}\\ -a_{21}{a}_{11}\end{array}\right]$

 

 Eigenvalues and eigenvectors. Let $\left[A\right]$ be an $\left(n\times n\right)$ matrix, with coefficients $a_{ij}$.  Consider the vector equation

$\left[A\right]x=\lambda x$                                                (1)

where x is a vector with n components, and $\lambda$ is a scalar (which may be complex).  The n nonzero vectors x and corresponding scalars $\lambda$ which satisfy this equation are the eigenvectors and eigenvalues of $\left[A\right]$.

 

Formally, eighenvalues and eigenvectors may be computed as follows.  Rearrange the preceding equation to

$\left(\left[A\right]-\lambda \left[I\right]\right)x=0$                                     (2)

This has nontrivial solutions for x only if the determinant of the matrix $\left(\left[A\right]-\lambda \left[I\right]\right)$ vanishes.  The equation

$\det \left(\left[A\right]-\lambda \left[I\right]\right)=0$

is an nth order polynomial which may be solved for $\lambda$.  In general the polynomial will have n roots, which may be complex.  The eigenvectors may then be computed using equation (2).  For example, a $\left(2\times 2\right)$ matrix generally has two eigenvectors, which satisfy

$\left|A-\lambda I\right|=\left|\begin{array}{l}{a}_{11}-\lambda a_{12}\\ a_{21}{a}_{22}-\lambda \end{array}\right|=(a_{11}-\lambda )(a_{22}-\lambda )-a_{12}{a}_{21}=0$

Solve the quadratic equation to see that

$\begin{array}{l}{\lambda }_1=\frac{1}{2}\left(a_{11}+a_{22}\right)-\frac{1}{2}{\left\{{\left(a_{11}+a_{22}\right)}^2-4\left(a_{11}{a}_{22}-a_{12}{a}_{21}\right)\right\}}^{1/2}\\ {\lambda }_2=\frac{1}{2}\left(a_{11}+a_{22}\right)+\frac{1}{2}{\left\{{\left(a_{11}+a_{22}\right)}^2-4\left(a_{11}{a}_{22}-a_{12}{a}_{21}\right)\right\}}^{1/2}\end{array}$

The two corresponding eigenvectors may be computed from (2), which shows that

$\left[\begin{array}{l}{a}_{11}-{\lambda }_i{a}_{12}\\ a_{21}{a}_{22}-{\lambda }_i\end{array}\right]\left[\begin{array}{l}{x}_1^{(i)}\\ x_2^{(i)}\end{array}\right]=0$

so that, multiplying out the first row of the matrix (you can use the second row too, if you wish $–$ since we chose $\left(2\times 2\right)$ to make the determinant of the matrix vanish, the two equations have the same solutions.  In fact, if $a_{12}=0$, you will need to do this, because the first equation will simply give 0=0 when trying to solve for one of the eigenvectors)

$\begin{array}{l}\left(\frac{1}{2}\left(a_{11}-a_{22}\right)+\frac{1}{2}{\left\{{\left(a_{11}+a_{22}\right)}^2-4\left(a_{11}{a}_{22}-a_{12}{a}_{21}\right)\right\}}^{1/2}\right)x_1^{(1)}+a_{12}{x}_2^{(1)}=0\\ \left(\frac{1}{2}\left(a_{11}-a_{22}\right)-\frac{1}{2}{\left\{{\left(a_{11}+a_{22}\right)}^2-4\left(a_{11}{a}_{22}-a_{12}{a}_{21}\right)\right\}}^{1/2}\right)x_1^{(2)}+a_{12}{x}_2^{(2)}=0\end{array}$

which are satisfied by any vector of the form

$\begin{array}{l}{x}^{(1)}=\left[\begin{array}{l}2a_{12}\\ \left(a_{11}-a_{22}\right)+{\left\{{\left(a_{11}+a_{22}\right)}^2-4\left(a_{11}{a}_{22}-a_{12}{a}_{21}\right)\right\}}^{1/2}\end{array}\right]p\\ x^{(2)}=\left[\begin{array}{l}2a_{12}\\ \left(a_{11}-a_{22}\right)-{\left\{{\left(a_{11}+a_{22}\right)}^2-4\left(a_{11}{a}_{22}-a_{12}{a}_{21}\right)\right\}}^{1/2}\end{array}\right]q\end{array}$

where p and q are arbitrary real numbers.

 

It is often convenient to normalize eigenvectors so that they have unit `length’.  For this purpose, choose p and q so that $x^{(i)}\cdot x^{(i)}=1$.  (For vectors of dimension n, the generalized dot product is defined such that $x\cdot x={\displaystyle {\sum }_{i=1}^n{x}_i{x}_i}$ )

 

One may calculate explicit expressions for eigenvalues and eigenvectors for any matrix up to order $\left(4\times 4\right)$, but the results are so cumbersome that, except for the $\left(2\times 2\right)$ results, they are virtually useless.  In practice, numerical values may be computed using several iterative techniques.  Packages like Mathematica, Maple or Matlab make calculations like this easy.

 

The eigenvalues of a real symmetric matrix are always real, and its eigenvectors are orthogonal, i.e. the ith and jth eigenvectors (with ) satisfy $x^{(i)}\cdot x^{(j)}=0$.

 

The eigenvalues of a skew symmetric matrix are pure imaginary.

 

 Spectral and singular value decomposition.  Let $\left[A\right]$ be a real symmetric  $\left(n\times n\right)$ matrix. Denote the n (real) eigenvalues of $\left[A\right]$ by ${\lambda }_i$, and let $w^{(i)}$ be the corresponding normalized eigenvectors, such that $w^{(i)}\cdot w^{(i)}=1$.  Then, for any arbitrary vector b,

$\left[A\right]b={\displaystyle \sum _{i=1}^n{\lambda }_i\left(w^{(i)}\cdot b\right)w^{(i)}}$

Let $\left[\Lambda \right]$ be a diagonal matrix which contains the n eigenvalues of $\left[A\right]$ as elements of the diagonal, and let $\left[Q\right]$ be a matrix consisting of the n eigenvectors as columns, i.e.

$\left[\Lambda \right]=\left[\begin{array}{l}{\lambda }_{1}00\cdots 0\\ 0{\lambda }_{2}0\cdots 0\\ ⋮⋮⋮\ddots ⋮\\ 000\cdots {\lambda }_n\end{array}\right]\left[Q\right]=\left[\begin{array}{l}{w}_1^{(1)}{w}_1^{(2)}{w}_1^{(3)}\cdots w_1^{(n)}\\ w_2^{(1)}{w}_2^{(2)}{w}_2^{(3)}\cdots w_2^{(n)}\\ ⋮⋮⋮\ddots ⋮\\ w_n^{(1)}{w}_n^{(2)}{w}_n^{(3)}\cdots w_n^{(n)}\end{array}\right]$

Then

$\left[A\right]=\left[Q\right]\left[\Lambda \right]{\left[Q\right]}^T{\left[Q\right]}^T\left[Q\right]=\left[Q\right]{\left[Q\right]}^T=\left[I\right]{\left[Q\right]}^T\left[A\right]\left[Q\right]=\left[\Lambda \right]$

Note that this gives another (generally quite useless) way to invert $\left[A\right]$

${\left[A\right]}^{-1}=\left[Q\right]{\left[\Lambda \right]}^{-1}{\left[Q\right]}^T$

where ${\left[\Lambda \right]}^{-1}$ is easy to compute since $\left[\Lambda \right]$ is diagonal.

 

 Square root of a matrix.   Let $\left[A\right]$ be a real symmetric  $\left(n\times n\right)$ matrix.  Denote the singular value decomposition of $\left[A\right]$ by $\left[A\right]=\left[Q\right]\left[\Lambda \right]{\left[Q\right]}^T$ as defined above.  Suppose that $\left[S\right]={\left[A\right]}^{1/2}$ denotes the square root of $\left[A\right]$, defined so that

$\left[S\right]\left[S\right]=\left[A\right]$

One way to compute $\left[S\right]$ is through the singular value decomposition of $\left[A\right]$

$\left[S\right]=\left[Q\right]{\left[\Lambda \right]}^{1/2}{\left[Q\right]}^T$

where

${\left[\Lambda \right]}^{1/2}=\left[\begin{array}{l}\sqrt{{\lambda }_1}00\cdots 0\\ 0\sqrt{{\lambda }_2}0\cdots 0\\ ⋮⋮⋮\ddots ⋮\\ 000\cdots \sqrt{{\lambda }_n}\end{array}\right]$

 

3.4 Brief Introduction to Tensors

 

3.4.1 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 $G=\nabla u$ 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 $x+dx$:

$du=G\cdot dx$

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

$t=n\cdot \sigma$

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

$dw=F\cdot dx$

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

 

 

3.4.2 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

$du=G\cdot dx$

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

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

We could represent this as a matrix product

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

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.  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

$v=S\cdot u$

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

 

 

 

3.4.3 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.

 

 

3.4.4 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

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

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 $\left(a\otimes b\right)\cdot u$ 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. 

 

 

 

 

 

 

3.4.5. Operations on Second Order Tensors

 

 Tensor components. 

 

Let  be a Cartesian basis, and let S be a second order tensor.  The components of S in  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

$\begin{array}{l}{S}_{11}=e_1\cdot \left(S\cdot e_1\right),S_{12}=e_1\cdot \left(S\cdot e_2\right),S_{11}=e_1\cdot \left(S\cdot e_3\right),\\ S_{21}=e_2\cdot \left(S\cdot e_1\right),S_{22}=e_2\cdot \left(S\cdot e_2\right),S_{21}=e_2\cdot \left(S\cdot e_3\right),\\ S_{31}=e_3\cdot \left(S\cdot e_1\right),S_{32}=e_3\cdot \left(S\cdot e_2\right),S_{31}=e_3\cdot \left(S\cdot e_3\right),\end{array}$

 

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.

 

 

 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

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

 

 

 Product of a tensor and a vector

 

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

$v=S\cdot u$

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

$v=u\cdot S$

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 $u\cdot S\ne S\cdot u$ (unless S is symmetric).

 

 

 Product of two tensors

 

Let T and S be two second order tensors.  Then $U=T\cdot S$ 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. $T\cdot S\ne S\cdot T$

 

 Transpose

 

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

$u\cdot S^T=S\cdot u$

 

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

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

 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

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

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.

 

 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 $S\cdot \cdot T=S^T:T$

 

 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

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

 

Note that if S and T are two tensors, then

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

 

 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

$S^{-1}\cdot S=I$

where  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.  The result cannot be expressed in a compact form for a general three dimensional second order tensor, and is best computed by methods such as Gaussian elimination.

 

 

 

 

 

 Eigenvalues and Eigenvectors (Principal values and direction)

 

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

$S\cdot m=\lambda m$

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 (see A.3.2)

 

The eigenvalues of a symmetric tensor are always real.  The eigenvalues of a skew tensor are always pure imaginary or zero.

 

 

 Change of Basis.

 

Let S be a tensor, and let  be a Cartesian basis.  Suppose that the components of S in the basis  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

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

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.

 

 

 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.

 

 

3.4.6

 

 Identity tensor  The identity tensor I is the tensor such that, for any tensor S or vector v

$\begin{array}{l}I\cdot v=v\cdot I=v\\ S\cdot I=I\cdot S=S\end{array}$

In any basis, the identity tensor has components

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

 

 Symmetric Tensor 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.

 

 Skew Tensor  A skew tensor S has the property

$S^T=-S$

The components of a skew tensor have the form

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

 

 Orthogonal Tensors An orthogonal tensor S has the property

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

An orthogonal tensor must have $\det (S)=\pm 1$; a tensor with $\det (S)=+1$ is known as a proper orthogonal tensor.

 

 

3.5 Vectors and Tensors in Spherical-Polar Coordinates

 

3.5.1 Specifying points in spherical-polar coordinates

To specify points in space using spherical-polar coordinates, we first choose two convenient, mutually perpendicular reference directions (i and k in the picture).  For example, to specify position on the Earth’s surface, we might choose k to point from the center of the earth towards the North Pole, and choose i to point from the center of the earth towards the intersection of the equator (which has zero degrees latitude) and the Greenwich Meridian (which has zero degrees longitude, by definition).

 

Then, each point P in space is identified by three numbers, $R,\theta ,\varphi$ shown in the picture above.  These are not components of a vector.

 

In words:

R is the distance of P from the origin

$\theta$ is the angle between the k direction and OP

$\varphi$ is the angle between the i direction and the projection of OP onto a plane through O normal to k

 

By convention, we choose  $R\ge 0$,  $0\le \theta \le {180}^o$ and $0\le \varphi \le {360}^o$

 

3.5.2 Converting between Cartesian and Spherical-Polar representations of points

 

When we use a Cartesian basis, we identify points in space by specifying the components of their position vector relative to the origin (x,y,z), such that $r=xi+yj+zk$  When we use a spherical-polar coordinate system, we locate points by specifying their spherical-polar coordinates $R,\theta ,\varphi$

 

The formulas below relate the two representations.  They are derived using basic trigonometry

$\begin{array}{l}x=R\sin \theta \cos \varphi \\ y=R\sin \theta \sin \varphi \\ z=R\cos \theta \end{array}$                    $\begin{array}{l}R=\sqrt{x^2+y^2+z^2}\\ \theta ={\cos }^{-1}z/R\\ \varphi ={\tan }^{-1}y/x\end{array}$

 

3.5.3 Spherical-Polar representation of vectors

 

When we work with vectors in spherical-polar coordinates, we abandon the {i,j,k} basis.  Instead, we specify vectors as components in the $\{e_R,e_{\theta },e_{\varphi }\}$ basis shown in the figure.  For example, an arbitrary vector a is written as $a=a_R{e}_R+a_{\theta }{e}_{\theta }+a_{\varphi }{e}_{\varphi }$, where $(a_R,a_{\theta },a_{\varphi })$ denote the components of a.

 

The basis is different for each point P.  In words

$e_R$ points along OP

$e_{\theta }$ is tangent to a line of constant longitude through P

$e_{\varphi }$ is tangent to a line of constant latitude through P.

 

For example if polar-coordinates are used to specify points on the Earth’s surface,  you can visualize the basis vectors like this.  Suppose you stand at a point P on the Earths surface.  Relative to you: $e_R$ points vertically upwards; $e_{\theta }$ points due South; and $e_{\varphi }$ points due East. Notice that the basis vectors depend on where you are standing.

 

You can also visualize the directions as follows.  To see the direction of $e_R$, keep $\theta$ and $\varphi$ fixed, and increase R. P is moving parallel to $e_R$.  To see the direction of $e_{\theta }$, keep R and $\varphi$ fixed, and increase $\theta$. P now moves parallel to  $e_{\theta }$.  To see the direction of $e_{\varphi }$, keep R and $\theta$ fixed, and increase $\varphi$.  P now moves parallel to $e_{\varphi }$.  Mathematically, this concept can be expressed as follows.  Let r be the position vector of P.  Then

$e_R=\frac{1}{\left|\frac{\partial r}{\partial R}\right|}\frac{\partial r}{\partial R}{e}_{\theta }=\frac{1}{\left|\frac{\partial r}{\partial \theta }\right|}\frac{\partial r}{\partial \theta }{e}_{\varphi }=\frac{1}{\left|\frac{\partial r}{\partial \varphi }\right|}\frac{\partial r}{\partial \varphi }$

By definition, the `natural basis’ for a coordinate system is the derivative of the position vector with respect to the three scalar coordinates that are used to characterize position in space (see Chapter 10 for a more detailed discussion).  The basis vectors for a polar coordinate system are parallel to the natural basis vectors, but are normalized to have unit length.  In addition, the natural basis for a polar coordinate system happens to be orthogonal. Consequently, $\{e_R,e_{\theta },e_{\varphi }\}$ is an orthonormal basis (basis vectors have unit length, are mutually perpendicular and form a right handed triad)

 

 

3.5.4 Converting vectors between Cartesian and Spherical-Polar bases

 

Let $a=a_R{e}_R+a_{\theta }{e}_{\theta }+a_{\varphi }{e}_{\varphi }$ be a vector, with components $(a_R,a_{\theta },a_{\varphi })$ in the spherical-polar basis $\{e_R,e_{\theta },e_{\varphi }\}$.  Let $a_x,a_y,a_z$ denote the components of a in the basis {i,j,k}.

 

The two sets of components are related by

$\left[\begin{array}{c}{a}_x\\ a_y\\ a_z\end{array}\right]=\left[\begin{array}{ccc}\sin \theta \cos \varphi & \cos \theta \cos \varphi & -\sin \varphi \\ \sin \theta \sin \varphi & \cos \theta \sin \varphi & \cos \varphi \\ \cos \theta & -\sin \theta & 0\end{array}\right]\left[\begin{array}{c}{a}_R\\ a_{\theta }\\ a_{\varphi }\end{array}\right]$

while the inverse relationship is

$\left[\begin{array}{c}{a}_R\\ a_{\theta }\\ a_{\varphi }\end{array}\right]=\left[\begin{array}{ccc}\sin \theta \cos \varphi & \sin \theta \sin \varphi & \cos \theta \\ \cos \theta \cos \varphi & \cos \theta \sin \varphi & -\sin \theta \\ -\sin \varphi & \cos \varphi & 0\end{array}\right]\left[\begin{array}{c}{a}_x\\ a_y\\ a_z\end{array}\right]$

 

Observe that the two 3x3 matrices involved in this transformation are transposes (and inverses) of one another.   The transformation matrix is therefore orthogonal, satisfying $[Q]{[Q]}^T=[I]$, where $[I]$ denotes the 3x3 identity matrix.

 

Derivation: It is easiest to do the transformation by expressing each basis vector $\{e_R,e_{\theta },e_{\varphi }\}$ as components in {i,j,k}, and then substituting.  To do this, recall that $r=xi+yj+zk$, recall also the conversion

$x=R\sin \theta \cos \varphi y=R\sin \theta \sin \varphi z=R\cos \theta$

and finally recall that by definition

$e_R=\frac{1}{\left|\frac{\partial r}{\partial R}\right|}\frac{\partial r}{\partial R}{e}_{\theta }=\frac{1}{\left|\frac{\partial r}{\partial \theta }\right|}\frac{\partial r}{\partial \theta }{e}_{\varphi }=\frac{1}{\left|\frac{\partial r}{\partial \varphi }\right|}\frac{\partial r}{\partial \varphi }$

Hence, substituting for x,y,z and differentiating

$\begin{array}{l}r=R\sin \theta \cos \varphi i+R\sin \theta \sin \varphi j+R\cos \theta k\\ \Rightarrow \frac{\partial r}{\partial R}=\sin \theta \cos \varphi i+\sin \theta \sin \varphi j+\cos \theta k\end{array}$

Conveniently we find that $\left|\frac{\partial r}{\partial R}\right|=1$. Therefore

$e_R=\sin \theta \cos \varphi i+\sin \theta \sin \varphi j+\cos \theta k$

Similarly

$\begin{array}{l}\frac{\partial r}{\partial \theta }=R\cos \theta \cos \varphi i+R\cos \theta \sin \varphi j-R\sin \theta k\\ \frac{\partial r}{\partial \varphi }=-R\sin \theta \sin \varphi i+R\sin \theta \cos \varphi j\end{array}$

while $\left|\frac{\partial r}{\partial \theta }\right|=R$, $\left|\frac{\partial r}{\partial \varphi }\right|=R\sin \theta$ so that

$e_{\theta }=\cos \theta \cos \varphi i+\cos \theta \sin \varphi j-\sin \theta k$     $e_{\varphi }=-\sin \varphi i+\cos \varphi j$

Finally, substituting

$\begin{array}{l}a=a_R[\sin \theta \cos \varphi i+\sin \theta \sin \varphi j+\cos \theta k]\\ \text{}\text{}\text{}\text{}+a_{\theta }[\cos \theta \cos \varphi i+\cos \theta \sin \varphi j-\sin \theta k]\\ \text{}\text{}\text{}\text{}\text{}+a_{\varphi }[-\sin \varphi i+\cos \varphi j]\end{array}$

Collecting terms in i, j and k, we see that

$\begin{array}{l}{a}_x=\sin \theta \cos \varphi a_R+\cos \theta \cos \varphi a_{\theta }-\sin \varphi a_{\varphi }\\ a_y=\sin \theta \sin \varphi a_R+\cos \theta \sin \varphi a_{\theta }+\cos \varphi a_{\varphi }\\ a_z=\cos \theta a_R-\sin \theta a_{\theta }\end{array}$

This is the result stated.

 

To show the inverse result, start by noting that

$\begin{array}{l}a=a_R{e}_R+a_{\theta }{e}_{\theta }+a_{\varphi }{e}_{\varphi }=a_{x}i+a_{y}j+a_{z}k\\ \Rightarrow a\cdot e_R=a_R=a_{x}i\cdot e_R+a_{y}j\cdot e_R+a_{z}k\cdot e_R\end{array}$

(where we have used $e_{\theta }\cdot e_R=e_{\varphi }\cdot e_R=0$ ).  Recall that

$\begin{array}{l}{e}_R=\sin \theta \cos \varphi i+\sin \theta \sin \varphi j+\cos \theta k\\ \Rightarrow i\cdot e_R=\sin \theta \cos \varphi j\cdot e_R=\sin \theta \sin \varphi k\cdot e_R=\cos \theta \end{array}$

Substituting, we get

$a_R=\sin \theta \cos \varphi a_x+\sin \theta \sin \varphi a_y+\cos \theta a_z$

Proceeding in exactly the same way for the other two components gives the remaining expressions

$\begin{array}{l}{a}_{\theta }=\cos \theta \cos \varphi a_x+\cos \theta \sin \varphi a_y-\sin \theta a_z\\ a_{\varphi }=-\sin \varphi a_x+\cos \varphi a_y\end{array}$

Re-writing the last three equations in matrix form gives the result stated.

 

 

 

3.5.5 Spherical-Polar representation of tensors

 

The triad of vectors $\{e_R,e_{\theta },e_{\varphi }\}$ is an orthonormal basis (i.e. the three basis vectors have unit length, and are mutually perpendicular).  Consequently, tensors can be represented as components in this basis in exactly the same way as for a fixed Cartesian basis $\{e_1,e_2,e_3\}$.  In particular, a general second order tensor S can be represented as a 3x3 matrix

$S\equiv \left[\begin{array}{ccc}{S}_{RR}& S_{R\theta }& S_{R\varphi }\\ S_{\theta R}& S_{\theta \theta }& S_{\theta \varphi }\\ S_{\varphi R}& S_{\varphi \theta }& S_{\varphi \varphi }\end{array}\right]$

You can think of $S_{RR}$ as being equivalent to $S_{11}$, $S_{R\theta }$ as $S_{12}$, and so on. All tensor operations such as addition, multiplication by a vector, tensor products, etc can be expressed in terms of the corresponding operations on this matrix, as discussed in Section B2 of Appendix B.

 

 

The component representation of a tensor can also be expressed in dyadic form as

$\begin{array}{l}S=S_{RR}{e}_R\otimes e_R+S_{R\theta }{e}_R\otimes e_{\theta }+S_{R\varphi }{e}_R\otimes e_{\varphi }\\ +S_{\theta R}{e}_{\theta }\otimes e_R+S_{\theta \theta }{e}_{\theta }\otimes e_{\theta }+S_{\theta \varphi }{e}_{\theta }\otimes e_{\varphi }\\ +S_{\varphi R}{e}_{\varphi }\otimes e_R+S_{\varphi \theta }{e}_{\varphi }\otimes e_{\theta }+S_{\varphi \varphi }{e}_{\varphi }\otimes e_{\varphi }\end{array}$

 

Furthermore, the physical significance of the components can be interpreted in exactly the same way as for tensor components in a Cartesian basis.  For example, the spherical-polar coordinate representation for the Cauchy stress tensor has the form

$\sigma \equiv \left[\begin{array}{ccc}{\sigma }_{RR}& {\sigma }_{R\theta }& {\sigma }_{R\varphi }\\ {\sigma }_{\theta R}& {\sigma }_{\theta \theta }& {\sigma }_{\theta \varphi }\\ {\sigma }_{\varphi R}& {\sigma }_{\varphi \theta }& {\sigma }_{\varphi \varphi }\end{array}\right]$

The component ${\sigma }_{\theta R}$ represents the traction component in direction $e_R$ acting on an internal material plane with normal $e_{\theta }$, and so on.  Of course, the Cauchy stress tensor is symmetric, with ${\sigma }_{\theta R}={\sigma }_{R\theta }$

 

 

 

3.5.6 Constitutive equations in spherical-polar coordinates

 

The constitutive equations listed in Chapter 3 all relate some measure of stress in the solid (expressed as a tensor) to some measure of local internal deformation (deformation gradient, Eulerian strain, rate of deformation tensor, etc), also expressed as a tensor.  The constitutive equations can be used without modification in spherical-polar coordinates, as long as the matrices of Cartesian components of the various tensors are replaced by their equivalent matrices in spherical-polar coordinates.

 

For example, the stress-strain relations for an isotropic, linear elastic material in spherical-polar coordinates read

$\left[\begin{array}{c}{\epsilon }_{RR}\\ {\epsilon }_{\theta \theta }\\ {\epsilon }_{\varphi \varphi }\\ 2{\epsilon }_{\theta \varphi }\\ 2{\epsilon }_{R\varphi }\\ 2{\epsilon }_{R\theta }\end{array}\right]=\frac{1}{E}\left[\begin{array}{cccccc}1& -\nu & -\nu & 0& 0& 0\\ -\nu & 1& -\nu & 0& 0& 0\\ -\nu & -\nu & 1& 0& 0& 0\\ 0& 0& 0& 2\left(1+\nu \right)& 0& 0\\ 0& 0& 0& 0& 2\left(1+\nu \right)& 0\\ 0& 0& 0& 0& 0& 2\left(1+\nu \right)\end{array}\right]\left[\begin{array}{c}{\sigma }_{RR}\\ {\sigma }_{\theta \theta }\\ {\sigma }_{\varphi \varphi }\\ {\sigma }_{\theta \varphi }\\ {\sigma }_{R\varphi }\\ {\sigma }_{R\theta }\end{array}\right]+\alpha \Delta T\left[\begin{array}{c}1\\ 1\\ 1\\ 0\\ 0\\ 0\end{array}\right]$

 

HEALTH WARNING: If you are solving a problem involving anisotropic materials using spherical-polar coordinates, it is important to remember that the orientation of the basis vectors $\{e_R,e_{\theta },e_{\varphi }\}$ vary with position.   For example, for an anisotropic, linear elastic solid you could write the constitutive equation as

$\begin{array}{l}\sigma =C(\epsilon -\alpha \Delta T)\\ \sigma =\left[\begin{array}{c}{\sigma }_{RR}\\ {\sigma }_{\theta \theta }\\ {\sigma }_{\varphi \varphi }\\ {\sigma }_{\theta \varphi }\\ {\sigma }_{R\varphi }\\ {\sigma }_{R\theta }\end{array}\right]C=\left[\begin{array}{cccccc}{c}_{11}& c_{12}& c_{13}& c_{14}& c_{15}& c_{16}\\ c_{12}& c_{22}& c_{23}& c_{24}& c_{25}& c_{26}\\ c_{13}& c_{23}& c_{33}& c_{34}& c_{35}& c_{36}\\ c_{14}& c_{24}& c_{34}& c_{44}& c_{45}& c_{46}\\ c_{15}& c_{25}& c_{35}& c_{45}& c_{55}& c_{56}\\ c_{16}& c_{26}& c_{36}& c_{46}& c_{56}& c_{66}\end{array}\right]\epsilon =\left[\begin{array}{c}{\epsilon }_{RR}\\ {\epsilon }_{\theta \theta }\\ {\epsilon }_{\varphi \varphi }\\ 2{\epsilon }_{\theta \varphi }\\ 2{\epsilon }_{R\varphi }\\ 2{\epsilon }_{R\theta }\end{array}\right]\alpha =\left[\begin{array}{c}{\alpha }_{RR}\\ {\alpha }_{\theta \theta }\\ {\alpha }_{\varphi \varphi }\\ 2{\alpha }_{\theta \varphi }\\ 2{\alpha }_{R\varphi }\\ 2{\alpha }_{R\theta }\end{array}\right]\end{array}$

however, the elastic constants $c_{11},c_{12},\mathrm{...}$ would need to be represent the material properties in the basis $\{e_R,e_{\theta },e_{\varphi }\}$, and would therefore be functions of position (you would have to calculate them using the lengthy basis change formulas listed in Section 3.2.11).  In practice the results are so complicated that there would be very little advantage in working with a spherical-polar coordinate system in this situation.

 

 

3.5.7 Converting tensors between Cartesian and Spherical-Polar bases

 

Let S be a tensor, with components

$S\equiv \left[\begin{array}{ccc}{S}_{RR}& S_{R\theta }& S_{R\varphi }\\ S_{\theta R}& S_{\theta \theta }& S_{\theta \varphi }\\ S_{\varphi R}& S_{\varphi \theta }& S_{\varphi \varphi }\end{array}\right]\left[\begin{array}{ccc}{S}_{xx}& S_{xy}& S_{xz}\\ S_{yx}& S_{yy}& S_{yz}\\ S_{zx}& S_{xy}& S_{zz}\end{array}\right]$

in the spherical-polar basis $\{e_R,e_{\theta },e_{\varphi }\}$ and the Cartesian basis {i,j,k}, respectively.  The two sets of components are related by

 

 

 

 

$\left[\begin{array}{ccc}{S}_{xx}& S_{xy}& S_{xz}\\ S_{yx}& S_{yy}& S_{yz}\\ S_{zx}& S_{xy}& S_{zz}\end{array}\right]=\left[\begin{array}{ccc}\sin \theta \cos \varphi & \cos \theta \cos \varphi & -\sin \varphi \\ \sin \theta \sin \varphi & \cos \theta \sin \varphi & \cos \varphi \\ \cos \theta & -\sin \theta & 0\end{array}\right]\left[\begin{array}{ccc}{S}_{RR}& S_{R\theta }& S_{R\varphi }\\ S_{\theta R}& S_{\theta \theta }& S_{\theta \varphi }\\ S_{\varphi R}& S_{\varphi \theta }& S_{\varphi \varphi }\end{array}\right]\left[\begin{array}{ccc}\sin \theta \cos \varphi & \sin \theta \sin \varphi & \cos \theta \\ \cos \theta \cos \varphi & \cos \theta \sin \varphi & -\sin \theta \\ -\sin \varphi & \cos \varphi & 0\end{array}\right]$

$\left[\begin{array}{ccc}{S}_{RR}& S_{R\theta }& S_{R\varphi }\\ S_{\theta R}& S_{\theta \theta }& S_{\theta \varphi }\\ S_{\varphi R}& S_{\varphi \theta }& S_{\varphi \varphi }\end{array}\right]=\left[\begin{array}{ccc}\sin \theta \cos \varphi & \sin \theta \sin \varphi & \cos \theta \\ \cos \theta \cos \varphi & \cos \theta \sin \varphi & -\sin \theta \\ -\sin \varphi & \cos \varphi & 0\end{array}\right]\left[\begin{array}{ccc}{S}_{xx}& S_{xy}& S_{xz}\\ S_{yx}& S_{yy}& S_{yz}\\ S_{zx}& S_{xy}& S_{zz}\end{array}\right]\left[\begin{array}{ccc}\sin \theta \cos \varphi & \cos \theta \cos \varphi & -\sin \varphi \\ \sin \theta \sin \varphi & \cos \theta \sin \varphi & \cos \varphi \\ \cos \theta & -\sin \theta & 0\end{array}\right]$

 

These results follow immediately from the general basis change formulas for tensors given in Appendix B.

 

 

3.5.8 Vector Calculus using Spherical-Polar Coordinates

 

Calculating derivatives of scalar, vector and tensor functions of position in spherical-polar coordinates is complicated by the fact that the basis vectors are functions of position.  The results can be expressed in a compact form by defining the gradient operator, which, in spherical-polar coordinates, has the representation

$\nabla \equiv \left(e_R\frac{\partial }{\partial R}+e_{\theta }\frac{1}{R}\frac{\partial }{\partial \theta }+e_{\varphi }\frac{1}{R\sin \theta }\frac{\partial }{\partial \varphi }\right)$

In addition, the derivatives of the basis vectors are

$\begin{array}{l}\frac{\partial e_R}{\partial R}=\frac{\partial e_{\theta }}{\partial R}=\frac{\partial e_{\varphi }}{\partial R}=0\frac{\partial e_R}{\partial \theta }=e_{\theta }\frac{\partial e_{\theta }}{\partial \theta }=-e_R\frac{\partial e_{\varphi }}{\partial \theta }=0\\ \frac{\partial e_R}{\partial \varphi }=\sin \theta e_{\varphi }\frac{\partial e_{\theta }}{\partial \varphi }=\cos \theta e_{\varphi }\frac{\partial e_{\varphi }}{\partial \varphi }=-\sin \theta e_R-\cos \theta e_{\theta }\end{array}$

You can derive these formulas by differentiating the expressions for the basis vectors in terms of {i,j,k}

$e_R=\sin \theta \cos \varphi i+\sin \theta \sin \varphi j+\cos \theta k$   $e_{\theta }=\cos \theta \cos \varphi i+\cos \theta \sin \varphi j-\sin \theta k$     $e_{\varphi }=-\sin \varphi i+\cos \varphi j$

and evaluating the various derivatives. When differentiating, note that {i,j,k} are fixed, so their derivatives are zero.  The details are left as an exercise.

 

The various derivatives of scalars, vectors and tensors can be expressed using operator notation as follows. 

 

Gradient of a scalar function: Let $f(R,\theta ,\varphi )$ denote a scalar function of position.  The gradient of f is denoted by

$\nabla f=f\left(e_R\frac{\partial }{\partial R}+e_{\theta }\frac{1}{R}\frac{\partial }{\partial \theta }+e_{\varphi }\frac{1}{R\sin \theta }\frac{\partial }{\partial \varphi }\right)=e_R\frac{\partial f}{\partial R}+e_{\theta }\frac{1}{R}\frac{\partial f}{\partial \theta }+e_{\varphi }\frac{1}{R\sin \theta }\frac{\partial f}{\partial \varphi }$

Alternatively, in matrix form

$\nabla f={\left[\frac{\partial f}{\partial R},\frac{1}{R}\frac{\partial f}{\partial \theta },\frac{1}{R\sin \theta }\frac{\partial f}{\partial \varphi }\right]}^T$

 

Gradient of a vector function Let $v=v_R{e}_R+v_{\theta }{e}_{\theta }+v_{\varphi }{e}_{\varphi }$ be a vector function of position. The gradient of v is a tensor, which can be represented as a dyadic product of the vector with the gradient operator as

$v\otimes \nabla =\left(v_R{e}_R+v_{\theta }{e}_{\theta }+v_{\varphi }{e}_{\varphi }\right)\otimes \left(e_R\frac{\partial }{\partial R}+e_{\theta }\frac{1}{R}\frac{\partial }{\partial \theta }+e_{\varphi }\frac{1}{R\sin \theta }\frac{\partial }{\partial \varphi }\right)$

The dyadic product can be expanded $–$ but when evaluating the derivatives it is important to recall that the basis vectors are functions of the coordinates $(R,\theta ,\varphi )$ and consequently their derivatives do not vanish.  For example

$\frac{1}{R}\frac{\partial }{\partial \theta }\left(v_R{e}_R\right)\otimes e_{\theta }=\frac{1}{R}\frac{\partial v_R}{\partial \theta }{e}_R\otimes e_{\theta }+\frac{v_R}{R}\frac{\partial e_R}{\partial \theta }\otimes e_{\theta }=\frac{1}{R}\frac{\partial v_R}{\partial \theta }{e}_R\otimes e_{\theta }+\frac{v_R}{R}{e}_{\theta }\otimes e_{\theta }$

Verify for yourself that the matrix representing the components of the gradient of a vector is

$v\otimes \nabla \equiv \left[\begin{array}{ccc}\frac{\partial v_R}{\partial R}& \frac{1}{R}\frac{\partial v_R}{\partial \theta }-\frac{v_{\theta }}{R}& \frac{1}{R\sin \theta }\frac{\partial v_R}{\partial \varphi }-\frac{v_{\varphi }}{R}\\ \frac{\partial v_{\theta }}{\partial R}& \frac{1}{R}\frac{\partial v_{\theta }}{\partial \theta }+\frac{v_R}{R}& \frac{1}{R\sin \theta }\frac{\partial v_{\theta }}{\partial \varphi }-\cot \theta \frac{v_{\varphi }}{R}\\ \frac{\partial v_{\varphi }}{\partial R}& \frac{1}{R}\frac{\partial v_{\varphi }}{\partial \theta }& \frac{1}{R\sin \theta }\frac{\partial v_{\varphi }}{\partial \varphi }+\cot \theta \frac{v_{\theta }}{R}+\frac{v_R}{R}\end{array}\right]$

 

Divergence of a vector function Let $v=v_R{e}_R+v_{\theta }{e}_{\theta }+v_{\varphi }{e}_{\varphi }$ be a vector function of position. The divergence of v is a scalar, which can be represented as a dot product of the vector with the gradient operator as

$\nabla \cdot v=\left(e_R\frac{\partial }{\partial R}+e_{\theta }\frac{1}{R}\frac{\partial }{\partial \theta }+e_{\varphi }\frac{1}{R\sin \theta }\frac{\partial }{\partial \varphi }\right)\cdot \left(v_R{e}_R+v_{\theta }{e}_{\theta }+v_{\varphi }{e}_{\varphi }\right)$

Again, when expanding the dot product, it is important to remember to differentiate the basis vectors. Alternatively, the divergence can be expressed as $\text{trace}(v\otimes \nabla )$, which immediately gives

$\nabla \cdot v\equiv \frac{\partial v_R}{\partial R}+2\frac{v_R}{R}+\frac{1}{R}\frac{\partial v_{\theta }}{\partial \theta }+\frac{1}{R\sin \theta }\frac{\partial v_{\varphi }}{\partial \varphi }+\cot \theta \frac{v_{\theta }}{R}$

 

Curl of a vector function Let $v=v_R{e}_R+v_{\theta }{e}_{\theta }+v_{\varphi }{e}_{\varphi }$ be a vector function of position. The curl of v is a vector, which can be represented as a cross product of the vector with the gradient operator as

$\nabla \times v=\left(e_R\frac{\partial }{\partial R}+e_{\theta }\frac{1}{R}\frac{\partial }{\partial \theta }+e_{\varphi }\frac{1}{R\sin \theta }\frac{\partial }{\partial \varphi }\right)\times \left(v_R{e}_R+v_{\theta }{e}_{\theta }+v_{\varphi }{e}_{\varphi }\right)$

The curl rarely appears in solid mechanics so the components will not be expanded in full

 

 

Divergence of a tensor function.   Let $S$ be a tensor, with dyadic representation

$\begin{array}{l}S=S_{RR}{e}_R\otimes e_R+S_{R\theta }{e}_R\otimes e_{\theta }+S_{R\varphi }{e}_R\otimes e_{\varphi }\\ +S_{\theta R}{e}_{\theta }\otimes e_R+S_{\theta \theta }{e}_{\theta }\otimes e_{\theta }+S_{\theta \varphi }{e}_{\theta }\otimes e_{\varphi }\\ +S_{\varphi R}{e}_{\varphi }\otimes e_R+S_{\varphi \theta }{e}_{\varphi }\otimes e_{\theta }+S_{\varphi \varphi }{e}_{\varphi }\otimes e_{\varphi }\end{array}$

The divergence of S is a vector, which can be represented as

$\nabla \cdot S=\left(e_R\frac{\partial }{\partial R}+e_{\theta }\frac{1}{R}\frac{\partial }{\partial \theta }+e_{\varphi }\frac{1}{R\sin \theta }\frac{\partial }{\partial \varphi }\right)\cdot \left(\begin{array}{c}{S}_{RR}{e}_R\otimes e_R+S_{R\theta }{e}_R\otimes e_{\theta }+S_{R\varphi }{e}_R\otimes e_{\varphi }\\ +S_{\theta R}{e}_{\theta }\otimes e_R+S_{\theta \theta }{e}_{\theta }\otimes e_{\theta }+S_{\theta \varphi }{e}_{\theta }\otimes e_{\varphi }\\ +S_{\varphi R}{e}_{\varphi }\otimes e_R+S_{\varphi \theta }{e}_{\varphi }\otimes e_{\theta }+S_{\varphi \varphi }{e}_{\varphi }\otimes e_{\varphi }\end{array}\right)$

Evaluating the components of the divergence is an extremely tedious operation, because each of the basis vectors in the dyadic representation of S must be differentiated, in addition to the components themselves.  The final result (expressed as a column vector) is

$\nabla \cdot S\equiv \left[\begin{array}{c}\frac{\partial S_{RR}}{\partial R}+2\frac{S_{RR}}{R}+\frac{1}{R}\frac{\partial S_{\theta R}}{\partial \theta }+\cot \theta \frac{S_{\theta R}}{R}+\frac{1}{R\sin \theta }\frac{\partial S_{\varphi R}}{\partial \varphi }-\frac{1}{R}\left(S_{\theta \theta }+S_{\varphi \varphi }\right)\\ \frac{\partial S_{R\theta }}{\partial R}+2\frac{S_{R\theta }}{R}+\frac{1}{R}\frac{\partial S_{\theta \theta }}{\partial \theta }+\cot \theta \frac{S_{\theta \theta }}{R}+\frac{1}{R\sin \theta }\frac{\partial S_{\varphi \theta }}{\partial \varphi }+\frac{S_{\theta R}}{R}-\cot \theta \frac{S_{\varphi \varphi }}{R}\\ \frac{\partial S_{R\varphi }}{\partial R}+2\frac{S_{R\varphi }}{R}+\frac{\sin \theta }{R}\frac{\partial S_{\theta \varphi }}{\partial \theta }+\cos \theta \frac{S_{\theta \varphi }}{R}+\frac{1}{R\sin \theta }\frac{\partial S_{\varphi \varphi }}{\partial \varphi }+\frac{1}{R}\left(S_{\varphi R}+S_{\varphi \theta }\right)\end{array}\right]$

 

 

 

 

 

3.6 Vectors and Tensors in Cylindrical-Polar Coordintes

3.6.1 Specifying points in space using in cylindrical-polar coordinates

 

To specify the location of a point in cylindrical-polar coordinates, we choose an origin at some point on the axis of the cylinder, select a unit vector k to be parallel to the axis of the cylinder, and choose a convenient direction for the basis vector i, as shown in the picture.  We then use the three numbers $r,\theta ,z$ to locate a point inside the cylinder, as shown in the picture.  These are not components of a vector.

 

In words

r is the radial distance of P from the axis of the cylinder

 is the angle between the i direction and the  projection of OP onto the i,j plane

z is the length of the projection of OP on the axis of the cylinder.

By convention r>0 and $0\le \theta \le {360}^o$

 

3.6.2 Converting between cylindrical polar and rectangular cartesian coordinates

 

When we use a Cartesian basis, we identify points in space by specifying the components of their position vector relative to the origin (x,y,z), such that $r=xi+yj+zk$  When we use a spherical-polar coordinate system, we locate points by specifying their spherical-polar coordinates $r,\theta ,z$

 

The formulas below relate the two representations.  They are derived using basic trigonometry

$\begin{array}{l}x=r\cos \theta \\ y=r\sin \theta \\ z=z\end{array}$                  $\begin{array}{l}r=\sqrt{x^2+y^2}\\ \theta ={\tan }^{-1}y/x\\ z=z\end{array}$

 

 

 

 

 

 

 

3.6.3 Cylindrical-polar representation of vectors

When we work with vectors in spherical-polar coordinates, we specify vectors as components in the .. basis shown in the figure.  For example, an arbitrary vector a is written as $a=a_r{e}_r+a_{\theta }{e}_{\theta }+a_z{e}_z$, where $(a_r,a_{\theta },a_z)$ denote the components of a.

 

The basis vectors are selected as follows

$e_r$ is a unit vector normal to the cylinder at P

$e_{\theta }$ is a unit vector circumferential to the cylinder at P, chosen to make $\{e_r,e_{\theta },e_z\}$ a right handed triad

$e_z$ is parallel to the k vector.

 

You will see that the position vector of point P would be expressed as

$r=r{e}_r+z{e}_z=r\cos \theta i+r\sin \theta j+zk$

 

Note also that the basis vectors are intentionally chosen to satisfy

$e_r=\frac{1}{\left|\frac{\partial r}{\partial r}\right|}\frac{\partial r}{\partial r}{e}_{\varphi }=\frac{1}{\left|\frac{\partial r}{\partial \theta }\right|}\frac{\partial r}{\partial \theta }{e}_z=\frac{1}{\left|\frac{\partial r}{\partial z}\right|}\frac{\partial r}{\partial z}$

The basis vectors have unit length, are mutually perpendicular, and form a right handed triad and therefore $\{e_r,e_{\theta },e_z\}$ is an orthonormal basis.  The basis vectors are parallel to (but not equivalent to) the natural basis vectors for a cylindrical polar coordinate system (see Chapter 10 for a more detailed discussion).

 

 

3.6.4 Converting vectors between Cylindrical and Cartesian bases

Let $a=a_r{e}_r+a_{\theta }{e}_{\theta }+a_z{e}_z$ be a vector, with components $(a_r,a_{\theta },a_z)$ in the spherical-polar basis $\{e_R,e_{\theta },e_{\varphi }\}$.  Let $a_x,a_y,a_z$ denote the components of a in the basis {i,j,k}.

 

The two sets of components are related by

$\left[\begin{array}{c}{a}_x\\ a_y\\ a_z\end{array}\right]=\left[\begin{array}{ccc}\cos \theta & -\sin \theta & 0\\ \sin \theta & \cos \theta & 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{c}{a}_r\\ a_{\theta }\\ a_z\end{array}\right]$     $\left[\begin{array}{c}{a}_r\\ a_{\theta }\\ a_z\end{array}\right]=\left[\begin{array}{ccc}\cos \theta & \sin \theta & 0\\ -\sin \theta & \cos \theta & 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{c}{a}_x\\ a_y\\ a_z\end{array}\right]$

 

Observe that the two 3x3 matrices involved in this transformation are transposes (and inverses) of one another.   The transformation matrix is therefore orthogonal, satisfying $[Q]{[Q]}^T=[I]$, where $[I]$ denotes the 3x3 identity matrix.

 

The derivation of these results follows the procedure outlined in E.1.4 exactly, and is left as an exercise.

 

 

 

 

3.6.5 Cylindrical-Polar representation of tensors

 

The triad of vectors $\{e_r,e_{\theta },e_z\}$ is an orthonormal basis (i.e. the three basis vectors have unit length, and are mutually perpendicular).  Consequently, tensors can be represented as components in this basis in exactly the same way as for a fixed Cartesian basis $\{e_1,e_2,e_3\}$.  In particular, a general second order tensor S can be represented as a 3x3 matrix

$S\equiv \left[\begin{array}{ccc}{S}_{rr}& S_{r\theta }& S_{rz}\\ S_{\theta r}& S_{\theta \theta }& S_{\theta z}\\ S_{zr}& S_{z\theta }& S_{zz}\end{array}\right]$

You can think of $S_{rr}$ as being equivalent to $S_{11}$, $S_{r\theta }$ as $S_{12}$, and so on. All tensor operations such as addition, multiplication by a vector, tensor products, etc can be expressed in terms of the corresponding operations on this matrix, as discussed in Section B2 of Appendix B.

 

The component representation of a tensor can also be expressed in dyadic form as

$\begin{array}{l}S=S_{rr}{e}_r\otimes e_r+S_{r\theta }{e}_r\otimes e_{\theta }+S_{rz}{e}_r\otimes e_z\\ +S_{\theta r}{e}_{\theta }\otimes e_r+S_{\theta \theta }{e}_{\theta }\otimes e_{\theta }+S_{\theta z}{e}_{\theta }\otimes e_z\\ +S_{zr}{e}_z\otimes e_r+S_{z\theta }{e}_z\otimes e_{\theta }+S_{zz}{e}_z\otimes e_z\end{array}$

 

 

3.6.6 Constitutive equations in cylindrical-polar coordinates

 

The constitutive equations listed in Chapter 3 all relate some measure of stress in the solid (expressed as a tensor) to some measure of local internal deformation (deformation gradient, Eulerian strain, rate of deformation tensor, etc), also expressed as a tensor.  The constitutive equations can be used without modification in cylindrical-polar coordinates, as long as the matrices of Cartesian components of the various tensors are replaced by their equivalent matrices in spherical-polar coordinates.

 

For example, the stress-strain relations for an isotropic, linear elastic material in cylindrical-polar coordinates read

$\left[\begin{array}{c}{\epsilon }_{rr}\\ {\epsilon }_{\theta \theta }\\ {\epsilon }_{zz}\\ 2{\epsilon }_{\theta z}\\ 2{\epsilon }_{rz}\\ 2{\epsilon }_{r\theta }\end{array}\right]=\frac{1}{E}\left[\begin{array}{cccccc}1& -\nu & -\nu & 0& 0& 0\\ -\nu & 1& -\nu & 0& 0& 0\\ -\nu & -\nu & 1& 0& 0& 0\\ 0& 0& 0& 2\left(1+\nu \right)& 0& 0\\ 0& 0& 0& 0& 2\left(1+\nu \right)& 0\\ 0& 0& 0& 0& 0& 2\left(1+\nu \right)\end{array}\right]\left[\begin{array}{c}{\sigma }_{rr}\\ {\sigma }_{\theta \theta }\\ {\sigma }_{zz}\\ {\sigma }_{\theta z}\\ {\sigma }_{rz}\\ {\sigma }_{r\theta }\end{array}\right]+\alpha \Delta T\left[\begin{array}{c}1\\ 1\\ 1\\ 0\\ 0\\ 0\end{array}\right]$

The cautionary remarks regarding anisotropic materials in E.1.6 also applies to cylindrical-polar coordinate systems.

 

 

 

 

 

 

 

 

 

 

3.6.7 Converting tensors between Cartesian and Spherical-Polar bases

                                               

Let S be a tensor, with components

$S\equiv \left[\begin{array}{ccc}{S}_{rr}& S_{r\theta }& S_{rz}\\ S_{\theta r}& S_{\theta \theta }& S_{\theta z}\\ S_{zr}& S_{z\theta }& S_{zz}\end{array}\right]\equiv \left[\begin{array}{ccc}{S}_{xx}& S_{xy}& S_{xz}\\ S_{yx}& S_{yy}& S_{yz}\\ S_{zx}& S_{xy}& S_{zz}\end{array}\right]$

in the cylindrical-polar basis $\{e_r,e_{\theta },e_z\}$ and the Cartesian basis {i,j,k}, respectively.  The two sets of components are related by

 

 

 

 

$$ 

$\left[\begin{array}{ccc}{S}_{xx}& S_{xy}& S_{xz}\\ S_{yx}& S_{yy}& S_{yz}\\ S_{zx}& S_{xy}& S_{zz}\end{array}\right]=\left[\begin{array}{ccc}\cos \theta & -\sin \theta & 0\\ \sin \theta & \cos \theta & 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{ccc}{S}_{rr}& S_{r\theta }& S_{rz}\\ S_{\theta r}& S_{\theta \theta }& S_{\theta z}\\ S_{zr}& S_{z\theta }& S_{zz}\end{array}\right]\left[\begin{array}{ccc}\cos \theta & \sin \theta & 0\\ -\sin \theta & \cos \theta & 0\\ 0& 0& 1\end{array}\right]$

$\left[\begin{array}{ccc}{S}_{rr}& S_{r\theta }& S_{rz}\\ S_{\theta r}& S_{\theta \theta }& S_{\theta z}\\ S_{zr}& S_{z\theta }& S_{zz}\end{array}\right]=\left[\begin{array}{ccc}\cos \theta & \sin \theta & 0\\ -\sin \theta & \cos \theta & 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{ccc}{S}_{xx}& S_{xy}& S_{xz}\\ S_{yx}& S_{yy}& S_{yz}\\ S_{zx}& S_{xy}& S_{zz}\end{array}\right]\left[\begin{array}{ccc}\cos \theta & -\sin \theta & 0\\ \sin \theta & \cos \theta & 0\\ 0& 0& 1\end{array}\right]$

 

 

 

3.6.8 Vector Calculus using Cylindrical-Polar Coordinates

 

Calculating derivatives of scalar, vector and tensor functions of position in cylindrical-polar coordinates is complicated by the fact that the basis vectors are functions of position.  The results can be expressed in a compact form by defining the gradient operator, which, in spherical-polar coordinates, has the representation

$\nabla \equiv \left(e_r\frac{\partial }{\partial r}+e_{\theta }\frac{1}{r}\frac{\partial }{\partial \theta }+e_z\frac{\partial }{\partial z}\right)$

In addition, the nonzero derivatives of the basis vectors are

$\frac{\partial e_r}{\partial \theta }=e_{\theta }\frac{\partial e_{\theta }}{\partial \theta }=-e_r(\text{all other derivatives are zero)}$

 

The various derivatives of scalars, vectors and tensors can be expressed using operator notation as follows. 

 

Gradient of a scalar function: Let $f(r,\theta ,z)$ denote a scalar function of position.  The gradient of f is denoted by

$\nabla f=\left(e_r\frac{\partial }{\partial r}+e_{\theta }\frac{1}{r}\frac{\partial }{\partial \theta }+e_z\frac{\partial }{\partial z}\right)f=e_r\frac{\partial f}{\partial r}+e_{\theta }\frac{1}{r}\frac{\partial f}{\partial \theta }+e_z\frac{\partial f}{\partial z}$

Alternatively, in matrix form

$\nabla f={\left[\frac{\partial f}{\partial r},\frac{1}{r}\frac{\partial f}{\partial \theta },\frac{\partial f}{\partial z}\right]}^T$

 

Gradient of a vector function Let $v=v_r{e}_r+v_{\theta }{e}_{\theta }+v_z{e}_z$ be a vector function of position. The gradient of v is a tensor, which can be represented as a dyadic product of the vector with the gradient operator as

$v\otimes \nabla =\left(v_r{e}_r+v_{\theta }{e}_{\theta }+v_z{e}_z\right)\otimes \left(e_r\frac{\partial }{\partial r}+e_{\theta }\frac{1}{r}\frac{\partial }{\partial \theta }+e_z\frac{\partial }{\partial z}\right)$

The dyadic product can be expanded $–$ but when evaluating the derivatives it is important to recall that the basis vectors are functions of the coordinate $\theta$ and consequently their derivatives may not vanish.  For example

$\frac{1}{r}\frac{\partial }{\partial \theta }\left(v_r{e}_r\right)\otimes e_{\theta }=\frac{1}{r}\frac{\partial v_r}{\partial \theta }{e}_r\otimes e_{\theta }+\frac{v_r}{r}\frac{\partial e_r}{\partial \theta }\otimes e_{\theta }=\frac{1}{r}\frac{\partial v_r}{\partial \theta }{e}_r\otimes e_{\theta }+\frac{v_r}{r}{e}_{\theta }\otimes e_{\theta }$

Verify for yourself that the matrix representing the components of the gradient of a vector is

$v\otimes \nabla \equiv \left[\begin{array}{ccc}\frac{\partial v_r}{\partial r}& \frac{1}{r}\frac{\partial v_r}{\partial \theta }-\frac{v_{\theta }}{r}& \frac{\partial v_r}{\partial z}\\ \frac{\partial v_{\theta }}{\partial r}& \frac{1}{r}\frac{\partial v_{\theta }}{\partial \theta }+\frac{v_r}{r}& \frac{\partial v_{\theta }}{\partial z}\\ \frac{\partial v_z}{\partial r}& \frac{1}{r}\frac{\partial v_z}{\partial \theta }& \frac{\partial v_z}{\partial z}\end{array}\right]$

 

Divergence of a vector function Let $v=v_r{e}_r+v_{\theta }{e}_{\theta }+v_z{e}_z$ be a vector function of position. The divergence of v is a scalar, which can be represented as a dot product of the vector with the gradient operator as

$\nabla \cdot v=\left(e_r\frac{\partial }{\partial r}+e_{\theta }\frac{1}{r}\frac{\partial }{\partial \theta }+e_z\frac{\partial }{\partial z}\right)\cdot \left(v_r{e}_r+v_{\theta }{e}_{\theta }+v_z{e}_z\right)$

Again, when expanding the dot product, it is important to remember to differentiate the basis vectors. Alternatively, the divergence can be expressed as $\text{trace}(v\otimes \nabla )$, which immediately gives

$\nabla \cdot v\equiv \frac{\partial v_r}{\partial r}+\frac{v_r}{r}+\frac{1}{r}\frac{\partial v_{\theta }}{\partial \theta }+\frac{\partial v_z}{\partial z}$

 

Curl of a vector function Let $v=v_R{e}_R+v_{\theta }{e}_{\theta }+v_z{e}_z$ be a vector function of position. The curl of v is a vector, which can be represented as a cross product of the vector with the gradient operator as

$\nabla \times v=\left(e_r\frac{\partial }{\partial r}+e_{\theta }\frac{1}{r}\frac{\partial }{\partial \theta }+e_z\frac{\partial }{\partial z}\right)\times \left(v_r{e}_r+v_{\theta }{e}_{\theta }+v_z{e}_z\right)$

The curl rarely appears in solid mechanics so the components will not be expanded in full

 

 

Divergence of a tensor function.   Let $S$ be a tensor, with dyadic representation

$\begin{array}{l}S=S_{rr}{e}_r\otimes e_r+S_{r\theta }{e}_r\otimes e_{\theta }+S_{rz}{e}_r\otimes e_z\\ +S_{\theta r}{e}_{\theta }\otimes e_r+S_{\theta \theta }{e}_{\theta }\otimes e_{\theta }+S_{\theta z}{e}_{\theta }\otimes e_z\\ +S_{zr}{e}_z\otimes e_r+S_{z\theta }{e}_z\otimes e_{\theta }+S_{zz}{e}_z\otimes e_z\end{array}$

The divergence of S is a vector, which can be represented as

$\nabla \cdot S=\left(e_r\frac{\partial }{\partial r}+e_{\theta }\frac{1}{r}\frac{\partial }{\partial \theta }+e_z\frac{\partial }{\partial z}\right)\cdot \left(\begin{array}{l}{S}_{rr}{e}_r\otimes e_r+S_{r\theta }{e}_r\otimes e_{\theta }+S_{rz}{e}_r\otimes e_z\\ +S_{\theta r}{e}_{\theta }\otimes e_r+S_{\theta \theta }{e}_{\theta }\otimes e_{\theta }+S_{\theta z}{e}_{\theta }\otimes e_z\\ +S_{zr}{e}_z\otimes e_r+S_{z\theta }{e}_z\otimes e_{\theta }+S_{zz}{e}_z\otimes e_z\end{array}\right)$

Evaluating the components of the divergence is an extremely tedious operation, because each of the basis vectors in the dyadic representation of S must be differentiated, in addition to the components themselves.  The final result (expressed as a column vector) is

$\nabla \cdot S\equiv \left[\begin{array}{c}\frac{\partial S_{rr}}{\partial r}+\frac{S_{rr}}{r}+\frac{1}{r}\frac{\partial S_{\theta r}}{\partial \theta }+\frac{\partial S_{zR}}{\partial z}-\frac{S_{\theta \theta }}{r}\\ \frac{1}{r}\frac{\partial S_{\theta \theta }}{\partial \theta }+\frac{\partial S_{r\theta }}{\partial r}+\frac{S_{r\theta }}{r}+\frac{S_{\theta r}}{r}+\frac{\partial S_{z\theta }}{\partial z}\\ \frac{\partial S_{zz}}{\partial z}+\frac{\partial S_{rz}}{\partial r}+\frac{S_{rz}}{r}+\frac{1}{r}\frac{\partial S_{\theta z}}{\partial \theta }\end{array}\right]$

 

 

 

 

 

3.7 Index Notation for Vector and Tensor Operations

 

 

Operations on Cartesian components of vectors and tensors may be expressed very efficiently and clearly using index notation.

 

3.7.1. Vector and tensor components.

 

Let x be a (three dimensional) vector and let S be a second order tensor.   Let $\left\{e_1,e_2,e_3\right\}$ be a Cartesian basis. Denote the components of x in this basis by $\left(x_1,x_2,x_3\right)$, and denote the components of S by

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

Using index notation, we would express x and S as

$x\equiv x_{i}S\equiv S_{ij}$

 

3.7.2. Conventions and special symbols for index notation

 

 Range Convention: Lower case Latin subscripts (i, j, k…) have the range $\left(1,2,3\right)$.  The symbol $x_i$ denotes three components of a vector $x_1,x_2$ and $x_3$.  The symbol $S_{ij}$ denotes nine components of a second order tensor, $S_{11},S_{12},S_{13},S_{21}\dots S_{33}$

 

 Summation convention (Einstein convention): If an index is repeated in a product of vectors or tensors, summation is implied over the repeated index.  Thus

$\begin{array}{l}\lambda =a_i{b}_i\equiv \lambda ={\displaystyle \sum _{i=1}^3{a}_i{b}_i\equiv \lambda =a_1{b}_1+a_2{b}_2+a_3{b}_3}\\ c_i=S_{ik}{x}_k\equiv c_i={\displaystyle \sum _{k=1}^3{S}_{ik}{x}_k}\equiv \{\begin{array}{l}{c}_1=S_{11}{x}_1+S_{12}{x}_2+S_{13}{x}_3\\ c_2=S_{21}{x}_1+S_{22}{x}_2+S_{23}{x}_3\\ c_3=S_{31}{x}_1+S_{32}{x}_2+S_{33}{x}_3\end{array}\\ \lambda =S_{ij}{S}_{ij}\equiv \lambda ={\displaystyle \sum _{i=1}^3{\displaystyle \sum _{j=1}^3{S}_{ij}{S}_{ij}}}\equiv \lambda =S_{11}{S}_{11}+S_{12}{S}_{12}+\dots +S_{31}{S}_{31}+S_{32}{S}_{32}+S_{33}{S}_{33}\\ C_{ij}=A_{ik}{B}_{kj}\equiv C_{ij}={\displaystyle \sum _{k=1}^3{A}_{ik}{B}_{kj}}\equiv \left[C\right]=\left[A\right]\left[B\right]\\ C_{ij}=A_{ki}{B}_{kj}\equiv C_{ij}={\displaystyle \sum _{k=1}^3{A}_{ki}{B}_{kj}}\equiv \left[C\right]={\left[A\right]}^T\left[B\right]\end{array}$

 

In the last two equations, $\left[A\right]$, $\left[B\right]$ and $\left[C\right]$ denote the $\left(3\times 3\right)$ component matrices of A, B and C.

 

 

 The Kronecker Delta:  The symbol ${\delta }_{ij}$ is known as the Kronecker delta, and has the properties

${\delta }_{ij}=\{\begin{array}{l}1i=j\\ 0i\ne j\end{array}$

thus

${\delta }_{11}={\delta }_{22}={\delta }_{33}=1{\delta }_{12}={\delta }_{21}={\delta }_{13}={\delta }_{31}={\delta }_{32}={\delta }_{32}=0$

You can also think of ${\delta }_{ij}$ as the components of the identity tensor, or a $\left(3\times 3\right)$ identity matrix.  Observe the following useful results

$\begin{array}{l}{\delta }_{ij}={\delta }_{ji}\\ {\delta }_{kk}=3\\ a_i={\delta }_{ik}{a}_k\\ A_{ij}={\delta }_{ik}{A}_{kj}\end{array}$

 

 The Permutation Symbol: The symbol ${\in }_{ijk}$ has properties

${\in }_{ijk}=\{\begin{array}{l}1i,j,k=1,2,3;2,3,1\text{or}3,1,2\\ -1i,j,k=3,2,1;2,1,3\text{or }\text{1,3,2}\\ \text{0}\text{otherwise}\end{array}$

thus

$\begin{array}{l}{\in }_{123}={\in }_{231}={\in }_{312}=1\\ {\in }_{321}={\in }_{213}={\in }_{132}=-1\\ {\in }_{111}={\in }_{112}={\in }_{113}={\in }_{121}={\in }_{122}={\in }_{131}={\in }_{133}=0\\ {\in }_{211}={\in }_{212}={\in }_{221}={\in }_{222}={\in }_{223}={\in }_{232}={\in }_{233}=0\\ {\in }_{311}={\in }_{313}={\in }_{322}={\in }_{323}={\in }_{321}={\in }_{332}={\in }_{333}=0\end{array}$

Note that

$\begin{array}{l}{\in }_{ijk}={\in }_{kij}={\in }_{jki}=-{\in }_{jik}=-{\in }_{kji}=-{\in }_{kji}\\ {\in }_{kki}=0\\ {\in }_{ijk}{\in }_{imn}={\delta }_{jm}{\delta }_{kn}-{\delta }_{jn}{\delta }_{mk}\\ {\in }_{ijk}{\in }_{lmn}={\delta }_{il}\left({\delta }_{jm}{\delta }_{kn}-{\delta }_{jn}{\delta }_{km}\right)-{\delta }_{im}\left({\delta }_{jl}{\delta }_{kn}-{\delta }_{jn}{\delta }_{kl}\right)+{\delta }_{in}\left({\delta }_{jl}{\delta }_{km}-{\delta }_{jm}{\delta }_{kl}\right)\end{array}$

 

3.7.3. Rules of index notation

 

1. The same index (subscript) may not appear more than twice in a product of two (or more) vectors or tensors.  Thus

are valid, but

are meaningless

 

 

 

 

 

2. Free indices on each term of an equation must agree.  Thus

are valid, but

are meaningless.

 

3.  Free and dummy indices may be changed without altering the meaning of an expression, provided that rules 1 and 2 are not violated. Thus

 

3.7.4. Vector operations expressed using index notation

 

 Addition.   $c=a+b\equiv c_i=a_i+b_i$

 

 Dot Product  $\lambda =a\cdot b\equiv \lambda =a_i{b}_i$

 

 Vector Product $c=a\times b\equiv c_i={\in }_{ijk}{a}_j{b}_k$

 

 Dyadic Product  

 

 Change of Basis.  Let a be a vector. Let  be a Cartesian basis, and denote the components of a in this basis by .  Let  be a second basis, and denote the components of a in this basis by ${\alpha }_i$.  Then, define

$Q_{ij}=m_i\cdot e_j=\cos \theta (m_i,e_j)$

where  denotes the angle between the unit vectors   and .  Then

${\alpha }_i=Q_{ij}{a}_j$

 

3.7.5. Tensor operations expressed using index notation.

 

 Addition.   $C=A+B\equiv C_{ij}=A_{ij}+B_{ij}$

 

 Transpose  $A=B^T\equiv A_{ij}=B_{ji}$

 

 Scalar Products

$\begin{array}{l}\lambda =A:B\equiv \lambda =A_{ij}{B}_{ij}\\ \lambda =A\cdot \cdot B\equiv \lambda =A_{ji}{B}_{ij}\end{array}$

 

 Product of a tensor and a vector

$\begin{array}{l}c=Ab\equiv c_i=A_{ij}{b}_j\\ c=A^{T}b\equiv c_i=A_{ji}{b}_j\end{array}$

 

 Product of two tensors

$\begin{array}{l}C=AB\equiv C_{ij}=A_{ik}{B}_{kj}\\ C=A^{T}B\equiv C_{ij}=A_{ki}{B}_{kj}\end{array}$

 Determinant

$\begin{array}{l}\lambda =\det A\equiv \lambda =\frac{1}{6}{\in }_{ijk}{\in }_{lmn}{A}_{li}{A}_{mj}{A}_{nk}\\ \iff {\in }_{lmn}\lambda ={\in }_{ijk}{A}_{li}{A}_{mj}{A}_{nk}\end{array}$

 Change of Basis.  Let A be a second order tensor. Let  be a Cartesian basis, and denote the components of A in this basis by $A_{ij}$.  Let $\left\{m_1,m_2,m_3\right\}$ be a second basis, and denote the components of A in this basis by .  Then, define

where  denotes the angle between the unit vectors   and .  Then

 

3.7.6. Calculus using index notation

 

The derivative $\partial x_i/\partial x_j$ can be deduced by noting that $\partial x_i/\partial x_j=1i=j$ and  $\partial x_i/\partial x_j=0i\ne j$.  Therefore

                                                                  $\frac{\partial x_i}{\partial x_j}={\delta }_{ij}$

The same argument can be used for higher order tensors

                                                               $\frac{\partial A_{ij}}{\partial A_{kl}}={\delta }_{ik}{\delta }_{jl}$

 

3.7.7. Examples of algebraic manipulations using index notation

 

1. Let a, b, c, d be vectors.  Prove that

$\left(a\times b\right)\cdot \left(c\times d\right)=\left(a\cdot c\right)\left(b\cdot d\right)-\left(b\cdot c\right)\left(a\cdot d\right)$

 

Express the left hand side of the equation using index notation (check the rules for cross products and dot products of vectors to see how this is done)

$\left(a\times b\right)\cdot \left(c\times d\right)\equiv {\in }_{ijk}{a}_j{b}_k{\in }_{imn}{c}_m{d}_n$

Recall the identity

${\in }_{ijk}{\in }_{imn}={\delta }_{jm}{\delta }_{kn}-{\delta }_{jn}{\delta }_{mk}$

so

${\in }_{ijk}{a}_j{b}_k{\in }_{imn}{c}_m{d}_n=\left({\delta }_{jm}{\delta }_{kn}-{\delta }_{jn}{\delta }_{mk}\right)a_j{b}_k{c}_m{d}_n$

Multiply out, and note that

${\delta }_{jm}{a}_j=a_m{\delta }_{kn}{b}_k=b_n$

(multiplying by a Kronecker delta has the effect of switching indices…) so

$\left({\delta }_{jm}{\delta }_{kn}-{\delta }_{jn}{\delta }_{mk}\right)a_j{b}_k{c}_m{d}_n=a_m{b}_n{c}_m{d}_n-a_n{b}_m{c}_m{d}_n$

Finally, note that

$a_m{c}_m\equiv a\cdot c$

and similarly for other products with the same index, so that

$a_m{b}_n{c}_m{d}_n-a_n{b}_m{c}_m{d}_n=a_m{c}_m{b}_n{d}_n-b_m{c}_m{a}_n{d}_n\equiv \left(a\cdot c\right)\left(b\cdot d\right)-\left(b\cdot c\right)\left(a\cdot d\right)$

 

2. The stress$—$strain relation for linear elasticity may be expressed as

${\sigma }_{ij}=\frac{E}{1+\nu }\left({\epsilon }_{ij}+\frac{\nu }{1-2\nu }{\epsilon }_{kk}{\delta }_{ij}\right)$

where ${\sigma }_{ij}$ and ${\epsilon }_{ij}$ are the components of the stress and strain tensor, and $E$ and $\nu$ denote Young’s modulus and Poisson’s ratio.  Find an expression for strain in terms of stress.

 

Set i=j to see that

${\sigma }_{ii}=\frac{E}{1+\nu }\left({\epsilon }_{ii}+\frac{\nu }{1-2\nu }{\epsilon }_{kk}{\delta }_{ii}\right)$

Recall that ${\delta }_{ii}=3$, and notice that we can replace the remaining ii by kk

$\begin{array}{l}{\sigma }_{kk}=\frac{E}{1+\nu }\left({\epsilon }_{kk}+\frac{\nu }{1-2\nu }3{\epsilon }_{kk}\right)=\frac{E}{1-2\nu }{\epsilon }_{kk}\\ \iff {\epsilon }_{kk}=\frac{1-2\nu }{E}{\sigma }_{kk}\end{array}$

Now, substitute for ${\epsilon }_{kk}$ in the given stress$—$strain relation

$\begin{array}{l}{\sigma }_{ij}=\frac{E}{1+\nu }\left({\epsilon }_{ij}+\frac{\nu }{E}{\sigma }_{kk}{\delta }_{ij}\right)\\ \iff {\epsilon }_{ij}=\frac{1+\nu }{E}\left({\sigma }_{ij}-\frac{\nu }{1+\nu }{\sigma }_{kk}{\delta }_{ij}\right)\end{array}$

 

3. Solve the equation

$\mu \left\{{\delta }_{kj}{a}_i{a}_i+\frac{1}{1-2\nu }{a}_k{a}_j\right\}{U}_k=P_j$

for $U_k$ in terms of $P_i$ and $a_i$

 

Multiply both sides by  to see that

$\begin{array}{l}\mu \left\{a_j{\delta }_{kj}{a}_i{a}_i+\frac{1}{1-2\nu }{a}_k{a}_j{a}_j\right\}{U}_k=P_j{a}_j\\ \iff \mu \left\{a_k{a}_i{a}_i+\frac{1}{1-2\nu }{a}_k{a}_j{a}_j\right\}{U}_k=P_j{a}_j\\ \iff \mu U_k{a}_k\frac{2\left(1-\nu \right)}{1-2\nu }{a}_i{a}_i=P_j{a}_j\iff U_k{a}_k=\frac{(1-2\nu )P_j{a}_j}{2\mu \left(1-\nu \right)a_i{a}_i}\end{array}$

Substitute back into the equation given for $U_k{a}_k$ to see that

$\mu U_j{a}_i{a}_i+\frac{P_k{a}_k}{2(1-\nu )a_i{a}_i}{a}_j=P_j\Rightarrow U_j=\frac{1}{\mu a_i{a}_i}\left(P_j-\frac{P_k{a}_k}{2(1-\nu )a_n{a}_n}{a}_j\right)$

 

4. Let $r=\sqrt{x_k{x}_k}$.  Calculate $\frac{\partial r}{\partial x_i}$

 

We can just apply the usual chain and product rules of differentiation

$\frac{\partial r}{\partial x_i}=\frac{1}{2}\frac{1}{\sqrt{x_k{x}_k}}\left(x_k\frac{\partial x_k}{\partial x_i}+\frac{\partial x_k}{\partial x_i}{x}_k\right)=\frac{1}{\sqrt{x_k{x}_k}}{x}_k{\delta }_{ik}=\frac{x_i}{\sqrt{x_k{x}_k}}=\frac{x_i}{r}$

 

5. Let $\lambda =A_{ij}{A}_{ij}$.  Calculate $\partial \lambda /\partial A_{kl}$

 

Using the product rule

$\frac{\partial \lambda }{\partial A_{kl}}=A_{ij}{\delta }_{ik}{\delta }_{jl}+{\delta }_{ik}{\delta }_{jl}{A}_{ij}=2A_{kl}$