0.1 Cartesian vectors and tensors: their algebra  (Page 2/5)

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Unit vectors

The unit vectors in the direction of a set of mutually orthogonal coordinate axis are defined as follows.

${\mathbf{e}}_{\left(1\right)}=\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right],\text{}\text{}{\mathbf{e}}_{\left(2\right)}=\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right],\text{}\text{}{\mathbf{e}}_{\left(3\right)}=\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]$

The suffixes to e are enclosed in parentheses to show that they do not denote components. A vector, a , can be expressed in terms of its components, ( a 1 , a 2 , a 3 ) and the unit vectors.

$\mathbf{a}={a}_{1}{\mathbf{e}}_{\left(1\right)}+{a}_{2}{\mathbf{e}}_{\left(2\right)}+{a}_{3}{\mathbf{e}}_{\left(3\right)}$

This equation can be multiplied and divided by the magnitude of a to express the vector in terms of its magnitude and direction.

$\begin{array}{c}\mathbf{a}=|\mathbf{a}|\left(\frac{{a}_{1}}{|\mathbf{a}|}{\mathbf{e}}_{\left(1\right)}+\frac{{a}_{2}}{|\mathbf{a}|}{\mathbf{e}}_{\left(2\right)}+\frac{{a}_{3}}{|\mathbf{a}|}{\mathbf{e}}_{\left(3\right)}\right)\\ =|\mathbf{a}|\left({\lambda }_{1}{\mathbf{e}}_{\left(1\right)}+{\lambda }_{2}{\mathbf{e}}_{\left(2\right)}+{\lambda }_{3}{\mathbf{e}}_{\left(3\right)}\right)\end{array}$

where λ i are the directional cosines of a .

A special unit vector we will use often is the normal vector to a surface, n . The components of the normal vector are the directional cosines of the normal direction to the surface.

Scalar product – orthogonality

The scalar product (or dot product ) of two vectors, a and b is defined as

$\mathbf{a}•\mathbf{b}=|\mathbf{a}||\mathbf{b}|\mathrm{cos}\theta$

where θ is the angle between the two vectors. If the two vectors are perpendicular to each other, i.e., they are orthogonal , then the scalar product is zero. The unit vectors along the Cartesian coordinate axis are orthogonal and their scalar product is equal to the Kronecker delta.

$\begin{array}{ccc}{\mathbf{e}}_{\left(i\right)}•{\mathbf{e}}_{\left(j\right)}& =& {\delta }_{ij}\\ & =& \left\{\begin{array}{c}1,i=j\\ 0,i\ne j\end{array}\end{array}$

The scalar product is commutative and distributive. The cosine of the angle measured from a to b is the same as measured from b to a . Thus the scalar product can be expressed in terms of the components of the vectors.

$\begin{array}{ccc}\mathbf{a}•\mathbf{b}& =& \left({a}_{1}{\mathbf{e}}_{\left(1\right)}+{a}_{2}{\mathbf{e}}_{\left(2\right)}+{a}_{3}{\mathbf{e}}_{\left(3\right)}\right)•\left({b}_{1}{\mathbf{e}}_{\left(1\right)}+{b}_{2}{\mathbf{e}}_{\left(2\right)}+{b}_{3}{\mathbf{e}}_{\left(3\right)}\right)\hfill \\ & =& {a}_{i}{b}_{j}{\delta }_{ij}\hfill \\ & =& {a}_{i}{b}_{i}\hfill \end{array}$

The scalar product of a vector with itself is the square of the magnitude of the vector.

$\begin{array}{ccc}\mathbf{a}•\mathbf{a}& =& |\mathbf{a}||\mathbf{a}|\mathrm{cos}0\hfill \\ & =& {|\mathbf{a}|}^{2}\hfill \\ \mathbf{a}•\mathbf{a}& =& {a}_{i}{a}_{i}\hfill \\ & =& {|\mathbf{a}|}^{2}\hfill \end{array}$

The most common application of the scalar product is the projection or component of a vector in the direction of another vector. For example, suppose n is a unit vector (e.g., the normal to a surface) the component of a in the direction of n is as follows.

$\mathbf{a}•\mathbf{n}=|\mathbf{a}|\mathrm{cos}\theta$

Directional cosines for coordinate transformation

The properties of the directional cosines for the rotation of the Cartesian coordinate reference frame can now be easily illustrated. Suppose the unit vectors in the original system is ${\mathbf{e}}_{\left(i\right)}$ and in the rotated system is ${\stackrel{_}{\mathbf{e}}}_{\left(j\right)}$ . The components of the unit vector, ${\stackrel{_}{\mathbf{e}}}_{\left(j\right)}$ , in the original reference frame is ${l}_{ij}$ . This can be expressed as the scalar product.

$\begin{array}{l}{\overline{e}}_{\left(j\right)}={l}_{1j}{e}_{\left(1\right)}+{l}_{2j}{e}_{\left(2\right)}+{l}_{3j}{e}_{\left(3\right)},\text{}j=1,2,3\\ {e}_{\left(i\right)}•{\overline{e}}_{\left(j\right)}={l}_{ij},\text{}i,j=1,2,3\end{array}$

Since ${\stackrel{_}{\mathbf{e}}}_{\left(j\right)}$ is a unit vector, it has a magnitude of unity.

${\overline{e}}_{\left(j\right)}•{\overline{e}}_{\left(j\right)}=1={l}_{i\left(j\right)}{l}_{i\left(j\right)}={l}_{1\left(j\right)}{l}_{1\left(j\right)}+{l}_{2\left(j\right)}{l}_{2\left(j\right)}+{l}_{3\left(j\right)}{l}_{3\left(j\right)},\text{}j=1,2,3$

Also, the axis of a Cartesian system are orthorgonal.

$\begin{array}{l}{\mathbf{e}}_{\left(i\right)}•{\mathbf{e}}_{\left(j\right)}=\left\{\begin{array}{c}0,\text{}\text{if}i\ne j\\ 1,\text{}\text{if}i=j\end{array}\\ \text{thus}\\ {\mathbf{e}}_{\left(i\right)}•{\mathbf{e}}_{\left(j\right)}={d}_{ij}\end{array}$
$\begin{array}{ccc}{\overline{e}}_{\left(i\right)}•{\overline{e}}_{\left(j\right)}& =& {l}_{ki}{l}_{kj}={l}_{1i}{l}_{1j}+{l}_{2i}{l}_{2j}+{l}_{3i}{l}_{3j},i,j=1,2,3\hfill \\ & =& {\delta }_{ij}\hfill \end{array}$

Vector product

The vector product (or cross product ) of two vectors, a and b , denoted as a × b , is a vector that is perpendicular to the plane of a and b such that a , b , and a × b form a right-handed system. (i.e., a , b , and a × b have the orientation of the thumb, first finger, and third finger of the right hand.) It has the following magnitude where θ is the angle between a and b .

$|\mathbf{a}×\mathbf{b}|=|\mathbf{a}|\text{}|\mathbf{b}|\text{}\mathrm{sin}\theta$

The magnitude of the vector product is equal to the area of a parallelogram two of whose sides are the vectors a and b .

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