弹性波动力学笔记(一) 旋转矩阵简介

Introduction of Rotation Transformation And Rotation Matrix

1.1 Summary of vector analysis

A vector is defined as a directed line segment, having both magnitude and direction. The magnitude, or length, of a vector a will be represented by $|\mathbf{a}|$. The sum and the difference of two vectors, and the multiplication of a vector by a scalar (real number) are defined using geometric rules. Given two vectors a and b, two products between them have been defined.

Scalar, or dot, product:

$${\bf{a}}{\bf{b}} = \left| \bf{a} \right|\left| \bf{b} \right|\cos \alpha \tag{1.1.1}$$

where $\alpha$ is the angle between the vectors.

Vector,or cross, product:

$${\bf{a}} \times {\bf{b}} = \left( {\left| {\bf{a}} \right|\left| {\bf{b}} \right|\sin \alpha } \right){\bf{n}} \tag{1.1.2}$$

where $\alpha$ is as before, and n is a unit vector (its length is equal to 1) perpendicular to both a and b such that the three vectors form a right-handed system.

An important property of the vector product, derived using geometric arguments, is the distributive law:

$$\left( {{\bf{a + b}}} \right){\bf{ \times c = a \times c + b \times c}} \tag{1.1.3}$$

By introducing a rectangular Cartesian coordinate system it is possible to write a vector in terms of three components. Let $\bf{e_{1}}=(1,0,0)^{T},\bf{e_{2}}=(0,1,0)^{T}$ ,and$\bf{e_{3}}=(0,0,1)^{T}$ be the three unit vectors along the $x_{1}$ ,$x_{2}$ and $x_{3}$ are axes of the coordinate. Then any vector $\bf{v}$ can be written as

$${\bf{v}} = {\left( {{v_1},{v_2},{v_3}} \right)^T} = {v_1}{{\bf{e}}_{\bf{1}}} + {v_2}{{\bf{e}}_{\bf{2}}} + {v_3}{{\bf{e}}_{\bf{3}}} \tag{1.1.4}$$

The components $v_{1},v_{2}$,and $v_{3}$ are the orthogonal projections of $\bf{v}$ in the directions of the three axes.

Before proceeding, a few words concerning the notation are necessary. A vector will be denoted by a bold-face letter, while its components will be denoted by the same letter in italics with subindex (literal or numerical). A bold-face letter with a subindex represents a vector, not a vector component. The three unit vectors defined above are examples of the latter. If we want to write the k-th component of the unit vector $\bf{e_{j}}$ we will write $(\bf{e_{j}})_{k}$ . For example, $(\bf{e_{2}})_{1}=0$ ,$(\bf{e_{2}})_{2}=0$,and $(\bf{e_{2}})_{3}=0$ . In addition , although vectors will usually be written in column form, when they are involved in matrix operations they should be considered as column vectors, as matrices of one column and three rows. For example, the matrix form of the scalar product $\bf{a}\cdot\bf{b}$is $\bf{a^T}\bf{b}$ , where T indicates transposition.

When the scalar product is applied to the unit vectors we find

$${{\bf{e}}_1} \cdot {{\bf{e}}_2} = {{\bf{e}}_1} \cdot {{\bf{e}}_3} = {{\bf{e}}_2} \cdot {{\bf{e}}_3} = 0 \tag{1.1.5}$$

$${{\bf{e}}_1} \cdot {{\bf{e}}_1} = {{\bf{e}}_2} \cdot {{\bf{e}}_2} = {{\bf{e}}_3} \cdot {{\bf{e}}_3} = 1 \tag{1.1.6}$$

Equations(5) and (6) can be summarized as follows:

$${{\bf{e}}_i} \cdot {{\bf{e}}_j} = {\delta _{ij}} = \left\{ \begin{array}{l} 1,i = j\\ 0,i \ne j \end{array} \right. \tag{1.1.7}$$

The symbol $\delta_{ij}$ is known as Kronecker delta, which is an example of a second-order tensor, and will play an important role in this book. As an example of equation (7) ,$\bf{e_{2}}\cdot\bf{e_{k}}=0$ unless $k=2$, in which case the scalar product is equal to 1.

Next we derive an alternative expression for a vector $\bf{v}$ .Using equation (4), the scalar product of $\bf{v}$ and $\bf{e_{i}}$ is

$${\bf{v}} \cdot {{\bf{e}}_i} = \left( {\sum\limits_{k = 1}^3 {{v_k}{{\bf{e}}_k}} } \right) \cdot {{\bf{e}}_i} = \sum\limits_{k = 1}^3 {{v_k}{{\bf{e}}_k}} \cdot {{\bf{e}}_i} = \sum\limits_{k = 1}^3 {{v_k}\left( {{{\bf{e}}_k} \cdot {{\bf{e}}_i}} \right)} = {v_i}\tag{1.1.8}$$

Note that when applying equation(4) the subindex in the summation must be different from $i$. To obtain equation(8) the following were used : the distributive law of the scalar product, the law of the product by a scalar and equation(7). Equation(8) shows that the $i-th$ component of $\bf{v}$ can be written as

$${v_i} = {\bf{v}} \cdot {{\bf{e}}_i} \tag{1.1.9}$$

When equation (9) is introduced in equation (4) we find

$${\bf{v}} = \sum\limits_{i = 1}^3 {\left( {{\bf{v}} \cdot {{\bf{e}}_i}} \right)} {{\bf{e}}_i} \tag{1.1.10}$$

This expression will be used in the discussion of dyadic.

In terms of its components the length of the vector is given by

$$\left| {\bf{v}} \right| = \sqrt {v_1^2 + v_2^2 + v_3^2} = {\left( {{\bf{v}} \cdot {\bf{v}}} \right)^{1/2}}\tag{1.1.11}$$

Using purely geometric arguments it is found that the scalar and vector products can be written in component form as follows :

$${\bf{u}} \cdot {\bf{v}} = {u_1}{v_1} + {u_2}{v_2} + {u_3}{v_3}\tag{1.1.12}$$

and

$${\bf{u}} \times {\bf{v}} = \left( {{u_2}{v_3} - {u_3}{v_2}} \right){{\bf{e}}_1} + \left( {{u_3}{v_1} - {u_1}{v_3}} \right){{\bf{e}}_2} + \left( {{u_1}{v_2} - {u_2}{v_1}} \right){{\bf{e}}_3}\tag{1.1.13}$$

The last expression is based on the use of equation(2) and equation(3).

Vectors, and vector operations such as the scalar and vector products, among others, are defined independently of any coordinate system. Vector relations derived without recourse to vector components will be valid when written in component form regardless of the coordinate system used. Of course, the same vector may (and generally will) have different components in different coordinate systems, but they will represent the same geometric entity. This is true for Cartesian and more general coordinate systems, such as spherical and cylindrical ones, but in the following we will consider the former only.

Now suppose that we want to define new vector entities based on operations on the components of other vectors. In view of the comments in 1.1 it is reasonable to expect that not every arbitrary definition will represent a vector, an entity intrinsically independent of the coordinate system used to represent the space. To see this consider the following example, which for simplicity refers to vectors in two-dimensional (2-D) space.

1.2 Rotation of Cartesian coordinates. Definition of a vector

Let $Ox_{1},Ox_{2},Ox_{3}$ represent a Cartesian coordinate system and $Ox_1^{'}，Ox_2^{'}，Ox_3^{'}$ another system obtained from the previous one by a rotation about their common origin $O$. Let $\bf{e_{1}},\bf{e_{2}},\bf{e_{3}}$ and $\bf{e_1^{'}},\bf{e_2^{'}},\bf{e_3^{'}}$ be the unit vectors along the three axes in the original and rotated systems. Finally, Let $a_{ij}$ denote the cosine of the angle between $Ox_i^{'}$ and $Ox_j$ . The $a_{ij}$ are known as direction cosines, and are related to $\bf{e_i^{'}}$ and $\bf{e_j}$ by

$$\bf{e_i^{'}}\cdot \bf{e_j}=a_{ij} \tag{1.2.1}$$

Given an arbitrary vector $\bf{v}$ with components $v_{1},v_{2},v_{3}$ in the original system, we are interested in finding the new components $v_1^{'},v_2^{'},v_3^{'}$ in the rotated system. To find the relation between the sets of components we will consider first the relation between the corresponding unit vectors. Using (1.2.1) equation $\bf{e_i{'}}$ can be written as

$$\bf{e_i^{'}}=a_{i1}\bf{e_1}+a_{i2}\bf{e_2}+a_{i3}\bf{e_3}= \sum\limits_{j = 1}^3 {{a_{ij}}{{\bf{e}}_{\bf{j}}}} \tag{1.2.2}$$

Furthermore, in original and rotated systems $\bf{v}$ can be written as

$$\bf{v}=\sum\limits_{j = 1}^3v_j\bf{e_j} \tag{1.2.3}$$

and

$$\bf{v}=\sum\limits_{j = 1}^3v_i^{'}\bf{e_i^{'}} \tag{1.2.4}$$

Now introduce (1.2.2) in (1.2.4)

$$\mathbf{v}=\sum_{i=1}^3 v_i^{\prime} \sum_{i=1}^3 a_{i j} \mathbf{e}_{\mathbf{j}} \equiv \sum_{i=1}^3\left(\sum_{i=1}^3 a_{i j} v_i^{\prime}\right) \mathbf{e_j} \tag{1.2.5}$$

Since(1.2.3) and (1.2.5) represent the same vector, and the three unit vector $\mathbf{e_1},\mathbf{e_2}$ and $\bf{e_{3}}$ are independent of each other, we conclude that

$$v_j = \sum\limits_{i = 1}^3 a_{ij}v_j^{'} \tag{1.2.6}$$

If we write the $\bf{e_j}$ is in terms of the $\bf{e_i^{'}}$ and replace them in equation (1.2.6) we find that

$$v_i^{'}=\sum\limits_{j= 1}^3a_{ij}v_{j} \tag{1.2.7}$$

Note that equation(1.2.6) the sum is over the first subindex of $a_{ij}$ , while in (1.2.7) equation the sum is over the second subindex of $a_{ij}$ . This distinction is critical and must be respected.

Now we are ready to introduce the following definition of a vector:

Three scalars are the components of a vector if under a rotation of coordinate they transform according to equation(1.2.7).

What this definition means is that if we want to define a vector by some set of rules, we have to verify that the vector components satisfy the transformation equations.

Before proceeding we will introduce a summation convention (due to Einstein) that will simplify the mathematical manipulations significantly. The convention applies to monomial expressions (such as a single term in an equation) and consists of dropping the sum symbol and summing over repeated index. This convention requires that the same index should appear no more than twice in the same term.

Repeated indexes are known as dummy indices , while those that are not repeated. Using this convention, while those that are not repeated are called free indices. Using this convention, we will write, for example,

$$\mathbf{v}=\sum\limits_{j= 1}^3v_j\mathbf{e_{j}}=v_j\mathbf{e_j} \tag{1.2.8}$$

$$v_j=\sum\limits_{j= 1}^3a_{ij}v_i^{'}=a_{ij} \tag{1.2.9}$$

$$v_{i}^{'}=\sum\limits_{j= 1}^3a_{ij}v_j \tag{1.2.10}$$

It is important to have a clear idea of the difference between free and dummy indices. A particular dummy index can be changed at will as long as it is replaced (in its two occurrences) by some other index not equal to any other existing indices in the same term. Free indices, on the other hand, are fixed and cannot be changed inside a single term. However, a free index can be replaced by another as long as the change is effected in all the terms in an equation, and the new index is different from all the other indices in the equation. In equation(1.2.9) $i$ is a dummy index and $j$ is a free index, while in equation (1.2.10) their role is reversed. The examples below show legal and illegal index manipulations.

The following relationship, derived from equation(1.2.9) , are true

$$v_j=a_{ij}v_i^{'}=a_{kj}v_k^{'}=a_{lj}v_l^{'} \tag{1.2.11}$$

because the repeated index $i$ was replaced by a different repeated index(equal to $k$ or $l$). However, it would not be correct to replace $i$ by $j$ because $j$ is already present in the equation. If $i$ were replaced by $j$ we would have

$$v_j=a_{jj}v_j^{'} \tag{1.2.12}$$

which would not be correct because the index $j$ has been appeared more than twice in the right-hand term, which is not allowed. Neither would it be correct to write

$$v_j=a_{ik}v_i^{'} \tag{1.2.13}$$

because the free index $j$ has been changed to $k$ only in the right-hand term. On the other hand, (1.3.9) can be written as

$$v_k=a_{ik}v_{i}^{'} \tag{1.2.14}$$

because the free index $j$ has been replaced by $k$ on both sides of the equation.

As equations (1.2.9) and (1.3.10) are of fundamental importance, it is necessary to pay attention to the fact that in the former the sum is over the first index of $a_{ij}$ while in the latter the sum is over the second index of $a_{ij}$. Also note that (1.2.10) can be written as product of a matrix and a vector:

$$\mathbf{v}^{\prime}=\left(\begin{array}{l} v_1^{\prime} \\ v_2^{\prime} \\ v_3^{\prime} \end{array}\right)=\left(\begin{array}{lll} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array}\right)\left(\begin{array}{l} v_1 \\ v_2 \\ v_3 \end{array}\right) \equiv \mathbf{A} \mathbf{v} \tag{1.2.15}$$

Where $\mathbf{A}$ is the matrix with elements $a_{ij}$ . It is clear that (1.2.9) can be written as

$$\mathbf{v}=\mathbf{A^T}\mathbf{v^{'}} \tag{1.2.16}$$

where the superscript $\mathbf{T}$ indicates transposition.

Now we will derive an important property of $\mathbf{A}$ . By introducing (1.2.10) in (1.2.9) we obtain

$$v_{j}=a_{ij}a_{ik}v_{k} \tag{1.2.17}$$

Note that it was necessary to change the dummy index in equation（1.2.10）to satisfy the summation convention. Equation(1.2.17) implies that any of the three components of $\mathbf{v}$ is a combination of all three components. However, this cannot be generally true because $\mathbf{v}$ is an arbitrary vector.

Therefore, the right-hand side of equation (1.2.17) must be equal to $v_{j}$ , which in turn implies that the product $a_{ij}a_{ik}$ must be equal to unity when $j=k$, and equal to zero when $j\neq i$ . This happens to be the definition of the Kronecker delta $\delta_{jk}$ introduced in equation (1.2.7), so that

$$a_{ij}a_{ik}=\delta_{jk} \tag{1.2.18}$$

If equation (1.2.9) is introduced in equation (1.2.10) we obtain

$$a_{ij}a_{kj}=\delta_{ik} \tag{1.2.19}$$

Setting $i=k$ in equation (1.2.19) and writing in full gives

$$1=a_{i1}^2+a_{i2}^2+a_{i3}^2=|\mathbf{e_{i}^{'}}|^2; (i=1,2,3) \tag{1.2.20}$$

where the equality on the right-hand side follows from(1.2.2).

When$i\neq j$ ,equation(1.2.19) transform into

$$0=a_{i1}a_{k1}+a_{i2}a_{k2}+a_{i3}a_{k3}=\mathbf{e_{i}^{'}}\mathbf{e_{k}^{'}} \tag{1.2.21}$$

where the equality on the right-hand side also follows from (1.2.2). Therefore, (1.2.19) summarizes the fact that the $\mathbf{e_{j}^{'}}$ are unit vectors orthogonal to each other, while (1.2.18) does the same thing for the $\mathbf{e_i}$s. Any set of vectors having these properties is known as an orthonormal set.

In matrix form, equation (1.2.18) and (1.2.19) can be written as

$$\mathbf{A^T}\mathbf{A}=\mathbf{A}\mathbf{A^T}=\mathbf{I} \tag{1.2.22}$$

Where $\mathbf{I}$ is the identity matrix.

Equation(1.2.22) can be written in the following useful way:

$$\mathbf{A^T}=\mathbf{A^{-1}} \tag{1.2.23}$$

$$(\mathbf{A^T})^{-1}=\mathbf{A} \tag{1.2.24}$$

where the superscript-1 indicates matrix inversion. From (1.2.23) we also find

$$|\mathbf{A}\mathbf{A^T}|=|\mathbf{A}||\mathbf{A^T}|=|\mathbf{I}|=1,\tag{1.2.25}$$

where vertical bars indicate the determinant of a matrix.

Linear transformations with a matrix such that its determinant squared is equal to 1 are known as orthogonal transformations. When $|\mathbf{A}|=1$ , the transformation of one coordinate axis in a coordinate plane. An example of reflection is the transformation that leave the $x_{1}$ and $x_{2}$ axes unchanged and replaces the $x_{3}$ axis by $-x_{3}$ . Reflections change the orientation of the space: if the original system is right-handed, then the new system is left-handed , and vice versa.

​ figure 1. Rotation of Axes. Primed and unprimed quantities refer to the original and rotated coordinate systems, respectively. Both

1.3 Rotation Matrices in two, three and n-dimensions space

A matrix is a representation of a linear transformation described in 1.2 chapter, which can be viewed as a machine that consumes a vector and spits out another vector. A rotation is a transformation with the property that the vector consumed by the machine and the vector spit out by the machine have the same length. That is, physically rotating a vector by an angle $\theta$ leaves the length of the vector unchanged. As a matrix equation, if $\mathbf{A}$ is a rotation and $\mathbf{v}$ is a vector ,then

$$\mathbf{v^{'}}=\mathbf{A}\mathbf{v}, where ||\mathbf{v}||_{L2}=||\mathbf{v^{'}}||_{L2} \tag{1.3.1}$$

A. Property of Two Dimension Rotation Matrices

In the case of $n=2$ . it is easy to work out the form of the most general real orthogonal matrix. set $A$ is 2D rotation matrix

$$\begin{array}{l} A = \left[ {\begin{array}{*{20}{c}} a&b\\ c&d \end{array}} \right]\\ \end{array} \tag{1.3.2}$$

According to property of rotation matrix $\mathbf{AA^T}=I$ , we can obtain

$$\left( {\begin{array}{*{20}{c}} a&b\\ c&d \end{array}} \right)\left( {\begin{array}{*{20}{c}} a&c\\ b&d \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 1&0\\ 0&1 \end{array}} \right) \tag{1.3.3}$$

Thus, we can obtain three constraint equations, as expected from

$$\left( {\begin{array}{*{20}{c}} {{a^2} + {b^2}}&{ac + bd}\\ {ac + bd}&{{c^2} + {d^2}} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 1&0\\ 0&1 \end{array}} \right) \tag{1.3.4}$$

Thus, we can obtain three constraint equations:

$$a^{2}+b^{2}=1 \tag{1.3.5} \\ c^{2}+d^{2}=1 \\ ac+bd=0$$

We now consider separately the cases of proper and improper rotations. In the case of proper rotations, we add a fourth equation (1.3.5)

$$det\mathbf{A}=ad-bc=1 \tag{1.3.6}$$

Suppose that $b\neq0$ and $c\neq0$ . From equation(1.3.5) ,it follows that $d=-ac/b$. Substituting this into equation (1.3.6) yields,

$$a^{2}+b^{2}=-b/c \tag{1.3.7}$$

Combining equation. and(), we conclude that $b=-c$. Plugging this result back into yields $a=d$ . With these results, is automatically satisfied in light of . Hence，the most general $2\times2$ real orthogonal matrix with determinant equal to 1 is given by,

$$\mathbf{A} = \left( {\begin{array}{*{20}{c}} a&b\\ { - b}&a \end{array}} \right),{\rm{ }}with{\rm{ }}{a^2} + {b^2} = 1{\rm{ and }}\left| a \right| \le 1,{\rm{ }}\left| b \right| \le 1\tag{1.3.8}$$

Without loss of generality, we can set $a=cos(\theta)$ and $b=\sqrt {1-{a^2}}= \pm \sin \theta$ ,since $cos(\theta)$ and $sin(\theta)$ satisfy same properties as $a$ and $b$, namely, $sin^{2}\theta+cos^{2}\theta=1$ and $|sin(\theta)|\leq1$ ,$|cos(\theta)|\leq1$ for $0\leq\theta\leq2\pi$ . Indeed, the rotation matrix,

$$\mathbf{A} = \left( {\begin{array}{*{20}{c}} {\cos \theta }&{ - \sin \theta }\\ {\sin \theta }&{\cos \theta } \end{array}} \right) \tag{1.3.9}$$

represent a proper counterclockwise rotation by an angle $\theta$ in the plane, as discussed .

In the case of improper rotations， we start with $\mathbf{\bar A}=[a,b;c,d]$ and impose $\mathbf{\bar A}\mathbf{\bar A^T}=\mathbf{I}$ , which again we can obtain follow equations:

$$det\mathbf{\bar A}=ad-bc=-1 \tag{1.3.10}$$

Suppose that $b\neq0$ and $c\neq0$ . From equation (1.3.5) , it follows that $d=-ac/b$. Substituting this into equation(1.3.5) :

$$a^2+b^2=b/c \tag{1.3.11}$$

we can conclude that $b=c$. Plugging this result into equation (1.3.5) yields $a=-d$ .With these results, eq(1.3.11) is automatically satisfied in light of .Hence, the most general $2\times2$ real orthogonal matrix with determinant equal to -1 is given by,

$$\mathbf{\bar A} = \left( {\begin{array}{*{20}{c}} a&b\\ b&{ - a} \end{array}} \right) {\rm{with}} {a^2} + {b^2} = 1{\rm{ and }}\left| a \right| \le 1,\left| b \right| \le 1 \tag{1.3.12}$$

we can again set $a=cos\theta$ and $b=\pm\sqrt{1-a^2}=\pm sin\theta$. Thus, an example of an improper rotation matrix is,

$$\mathbf{\bar A}= \left( {\begin{array}{*{20}{c}} {\cos \theta }&{ \sin \theta }\\ {\sin \theta }&{-\cos \theta } \end{array}} \right) where (0\leq\theta\lt2\pi)\tag{1.3.13}$$

This matrix satisfies the property, $\bar R^2=I$ . Note that we can express $\mathbf{A}$ given equation (1.3.9) as the product of a proper rotation and a reflection,

B. Property of Three Dimension Rotation Matrices

It becomes quickly evident that the methods used in the previous section become much less practical for n=3. In this section I shall present the explicit form of the $3\times 3$ proper rotation matrix along with its most important properties. A derivation of the rotation matrix is given in an Appendix to these notes.

As previous noted, the most general three-dimensional rotation, which we henceforth by $\mathbf{A}(\widehat{\mathbf{n}},\theta)$, can be specified by an-axis of rotation pointing in the direction of the unit vector $\widehat{\mathbf{n}}$ , and a rotation angle $\theta$ . Conventionally, a positive rotation angle $\theta$ corresponds to a counterclockwise rotation. The direction of the axis is determined by the right hand rule. Namely, curl the fingers of your right hand around the axis of rotation, where your fingers point in the $\theta$ direction. Then, your thumb points perpendicular to the plane of rotation in the direction of $\widehat{\mathbf{n}}$ . In general, rotation matrices do not commute under multiplication. However, if both rotations are taken with respect to the same fixed axis, then

$$\mathbf{A(\widehat{n},\theta_{1})}\mathbf{A(\widehat{n},\theta_{2})}=\mathbf{A(\widehat{n},\theta_{1}+\theta_{2})}\tag{1.3.14}$$

Simple geometric considerations will convince you that the following relations are satisfied:

$$\mathbf{A}(\mathbf{\widehat{n}},\theta+2k\pi)=\mathbf{A}(\mathbf{\widehat{n}},\theta) (k=\pm1,\pm2,...)\tag{1.3.15}$$

$$[\mathbf{A}(\mathbf{\widehat{n}},\theta)]^{-1}=\mathbf{A}(-\mathbf{\widehat{n}},\theta)=\mathbf{A}(\mathbf{\widehat{n}},-\theta) \tag{1.3.16}$$

Combining these two results, it follows that

$$\mathbf{A}(\mathbf{\widehat{n}},2\pi-\theta)=\mathbf{A}(-\mathbf{\widehat{n}},\theta) \tag{1.3.17}$$

which implies that any three-dimensional rotation can be described by a counterclockwise rotation by an angle $\theta$ about an arbitrary axis $\widehat{\mathbf{n}}$ , where $0\leq\theta\leq \pi$. However, if we substitute $\theta=\pi$ , we conclude that

$$\mathbf{A}(\widehat{\mathbf{n}},\pi)=\mathbf{A}(-\widehat{\mathbf{n}},\pi) \tag{1.3.18}$$

which means that for the special case of $\theta=\pi$ , $A(\widehat{n},\pi)$ and $A(-\widehat{n},\pi)$ represent the same rotation. In particular, note that

$$[\mathbf{A}(\widehat{n},\pi)]^2=\mathbf{I} \tag{1.3.19}$$

Indeed for any choice of $\mathbf{\widehat{n}}$ , the $A(\mathbf{\widehat{n}},\pi)$ are the only non-trivial rotation matrices whose square is equal to the identity matrix. Finally, if $\theta=0$ then $R(\widehat{n},0)=I$ is the identity matrix (sometimes called the trivial rotation), independently of the direction of $\widehat{n}$.

We now present an explicit form for $A(\mathbf{\widehat{n}},\theta)$. Since $A(\mathbf{\widehat{n}},\theta)$ describes a rotation by an angle $\theta$ about an axis $\widehat{\mathbf{n}}$ , the formula for $A(\mathbf{\widehat{n}},\theta)$ will depend on the angle $\theta$ and on the coordinates of $\mathbf{\widehat{n}}=(n_{1},n_{2},n_{3})$ with respect to a fixed Cartesian coordinate system. Note that since $\widehat{\mathbf{n}}$ is a unit vector, it follows that,

$$n_{1}^2+n_{2}^2+n_{3}^2=1 \tag{1.3.20}$$

We can also express $\widehat{\mathbf{n}}$ in terms of its polar and azimuthal angles ($\theta_{n}$ and $\phi_{n}$, respectively) with respect to a fixed $z-axis$. In particular,

$$n_{1}=\sin\theta_{n}cos\phi_{n}, \space n_{2}=\sin\theta_{n}sin\phi_{n}, \space n_{3} =\cos\theta_{n} \tag{1.3.21}$$

One can check that equation (1.3.20) is indeed satisfied. Thus, $\widehat{\mathbf{n}}$ depends on two independent parameters, $\theta_{n}$ and $\phi_{n}$ , which together with the rotation angle $\theta$ constitute the three independent parameters that describe a three dimensional rotation.

The explicit formula for the real orthogonal $3\times3$ matrix $\mathbf{A}(\widehat{n},\theta)$ with determinant equal to 1 is given by,

$$\mathbf{A}(\hat{\boldsymbol{n}}, \theta)=\left(\begin{array}{ccc} \cos \theta+n_1^2(1-\cos \theta) & n_1 n_2(1-\cos \theta)-n_3 \sin \theta & n_1 n_3(1-\cos \theta)+n_2 \sin \theta \\ n_1 n_2(1-\cos \theta)+n_3 \sin \theta & \cos \theta+n_2^2(1-\cos \theta) & n_2 n_3(1-\cos \theta)-n_1 \sin \theta \\ n_1 n_3(1-\cos \theta)-n_2 \sin \theta & n_2 n_3(1-\cos \theta)+n_1 \sin \theta & \cos \theta+n_3^2(1-\cos \theta) \end{array}\right) \tag{1.3.22}$$

One can easily check that equations (1.3.15) and(1.3.16), the rotations $A(\widehat{n},\pi)$ represent the same rotation, and are writen as tensor form

$$\mathbf{A}(\hat{\boldsymbol{n}}, \pi)=\left(\begin{array}{ccc} 2n_1^2-1 & 2n_1 n_2 & 2n_1 n_3 \\2n_1 n_2 & 2n_2^2-1 & 2n_2 n_3 \\ 2n_1 n_3 & 2n_2 n_3 & 2n_3^2-1 \end{array}\right)=2n_in_j-\delta_{ij}\tag{1.3.23}$$

where the Kronecker $\delta$ symbol was introduced in equation . Finally,as expected, $\mathbf{A}(\mathbf{\widehat{n}},0)=\delta_{ij}$ , independently of the direction of $\widehat{\mathbf{n}}$ . I leave it as an exercise to the student to verify explicitly that $\mathbf{A}=\mathbf{A}(\widehat{\mathbf{n}},\theta)$ given in equation (1.3.22) satisfies the conditions $AA^{T}=\mathbf{I}$ and $\det{\mathbf{A}}=1$ .

Although equation (1.3.22) looks complicated, one can present an elegant expression for the matrix elements of $\mathbf{A}(\mathbf{\widehat{n}},\theta)$, denoted below by $R_{ij}$. Indeed, it is not difficult to check that the following expression for $R_{ij}$ is equivalent to the matrix elements of $\mathbf{A}(\widehat{\mathbf{n}},\theta)$ exhibited in equation (1.3.22)

$$A_{ij}(\widehat{\mathbf{n}},\theta)=\delta_{ij}\cos\theta+(1-\cos\theta)n_{i}n_{j}-\epsilon_{ijk}n_{k}\sin\theta \tag{1.3.24}$$

where $\epsilon_{ijk}$ is the Levi-Civita epsilon symbol, which is defined as follows,

$$\epsilon_{i j k}=\left\{\begin{aligned} +1, & \text { if }\{i, j, k\} \text { is an even permutation of }\{1,2,3\}, \\ -1, & \text { if }\{i, j, k\} \text { is an odd permutation of }\{1,2,3\}, \\ 0, & \text { if not all the integers } 1,2,3 \text { are distinct. } \end{aligned}\right. \tag{1.3.25}$$

Note that $\epsilon_{ijk}$ is the $n=3$ version of the Levi-Civita symbol introduced in the class handout entitled Determinant and the Adjugate.

It is instructive to check special cases of equation(1.3.22). For example, suppose we choose $\widehat{\mathbf{n}}=\mathbf{k}$ corresponding to a rotation axis that points along the positive $z-direction$. In this case, $n_1=n_2=0$ and $n_3=1$ , and equation(1.3.22) becomes

$$\mathbf{A}\left( {\mathbf{k},\theta } \right) = \left( {\begin{array}{*{20}{c}} {\cos \theta }&{ - \sin \theta }&0\\ {\sin \theta }&{\cos \theta }&0\\ 0&0&1 \end{array}} \right) \tag{1.3.26}$$

Of course, equation (1.3.26) is the expected result given the form of the two-dimension rotation matrix given in equation (1.3.9).

Likewise, one can choose either $\mathbf{\widehat{n}}=\mathbf{i}$ or $\mathbf{\widehat{n}}=\mathbf{j}$ corresponding to rotation axes that point along the positive $x-direction$ (i.e. , $n_2=n_3=1 and n_1=1$) or along the positive y-direction (i.e., $n_1=n_3=0$ and $n_2=1$), respectively. In these cases, equation (1.3.22) yields,

$$\mathbf{A}\left( {\mathbf{i},\theta } \right) = \left( {\begin{array}{*{20}{c}} {1}&{0}&0\\ {0}&{\cos \theta }&{-\sin\theta}\\ 0&{\sin\theta}&{\cos\theta} \end{array}} \right) \tag{1.3.27}$$

$$\mathbf{A}\left( {\mathbf{j},\theta } \right) = \left( {\begin{array}{*{20}{c}} {\cos\theta}&{0}&{\sin\theta}\\ {0}&{1}&{0}\\ {-\sin\theta}&{0}&{\cos\theta} \end{array}} \right) \tag{1.3.28}$$

To learn more about the properties of a general three-dimensional rotation, consider the matrix representation $\mathbf{A}(\widehat{\mathbf{n}},\theta)$, with respect to the standard basis $B_s=(\mathbf{i},\mathbf{j},\mathbf{k})$ . We can define a new coordinate system in which the unit vector $\widehat{\mathbf{n}}$ points in the direction of the new $z-axis$; the corresponding new basis will be denoted by $B_s^{'}$ . The matrix representation of the rotation with respect to $B^{'}$ is then given by equation(1.3.26);

Using the formalism developed in the class handout, Vector coordinates, matrix elements，changes of basis, and matrix diagonalization, there exists a real $3\times 3$ special orthogonal matrix $P$ such that

$$\mathbf{A}(\mathbf{\widehat{n},\theta})=\mathbf{P}\mathbf{A}(\mathbf{k},\theta)\mathbf{P^{-1}} \tag{1.3.29}$$

where $\mathbf{A}(\mathbf{k},\theta)$ is given by equation (1.3.26). In Appendix A, we will determine an explicit form of the matrix $P$ . However, the mere existence of the matrix $P$ in equation (1.3.29) is sufficient to provide a simple algorithm for determining the rotation axis $\widehat{\mathbf{n}}$ (up to an overall sign) and the rotation angle $\theta$ that characterize a general three-dimensional rotation matrix.

C. Determining the rotation axis and the rotation angle

Given a general three-dimensional rotation matrix, $A(\widehat{\mathbf{n}},\theta)$, we can determine the angle of rotation $\theta$ and the axis of rotation $\widehat{\mathbf{n}}$ . Using equation (1.3.22), the trace of $\mathbf{A}(\widehat{\mathbf{n}},\theta)$ is given by:

$$Tr\mathbf{A}(\widehat{\mathbf{n}},\theta)=1+2\cos\theta \tag{1.3.30}$$

which coincides with our previous result obtain in equation (1.3.30). Thus, yields,

$$\cos\theta=\frac{1}{2}(Tr A-1)\\ \sin\theta=(1-\cos^2\theta)=\frac{1}{2}\sqrt{(3-Tr A)(1+Tr{A})} \tag{1.3.31}$$

where $0\leq\sin\theta\leq 1$ is a consequence of the range of the rotation angle, $0\leq\theta\leq\pi$.

To determine $\widehat{\mathbf{n}}$, we multiply equation (1.3.24) by $\epsilon_{ijk}$ and sum over $i$ and $j$. Noting that

$$\epsilon_{ijm}\delta_{ij}=\epsilon_{ijm}n_in_j=0\\ \epsilon_{ijk}\epsilon_{ijm}=2\delta_{km}\tag{1.3.32}$$

it follows that

$$2n_{m}\sin\theta=-A_{ij}\epsilon_{ijm} \tag{1.3.33}$$

If $A$ is a symmetric matrix (i.e. $R_{ij}=R_{ji}$), then $R_{ij}\epsilon_{ijm}=0$ automatically since $\epsilon_{ijk}$ is antisymmetric under the interchange of the indices $i$ and $j$. In this case $\sin\theta=0$, and we must seek other means to determine $\widehat{\mathbf{n}}$ . If $\sin\theta\neq0$, then one can divide both sides of equation (1.3.33) by $\sin\theta$. Using equation (1.3.33), we obtain:

$$n_m=-\frac{A_{ij}\epsilon_{ijm}}{2\sin\theta}=\frac{-A_{ij}\epsilon_{ijm}}{\sqrt{(3-Tr A)(1+Tr A)}}\tag{1.3.34}$$

More explicitly,

$$\widehat{\mathbf{n}}=\frac{1}{\sqrt{(3-Tr A)(1+Tr A)}}(A_{32}-A_{23},A_{13}-A_{31},A_{21}-A_{12})\\ Tr A\neq-1,3\tag{1.3.35}$$

If we multiply equation (1.3.33) by $n_{m}$ and sum over $m$, then

$$\sin\theta=-\frac{1}{2}\epsilon_{ijm}A_{ij}n_m \tag{1.3.36}$$

after using $n_mn_m=1$. This provides an additional check on the determination of the rotation angle.

Alternatively, we can define a matrix $S$ whose matrix elements are given by:

$$S_{jk}=A_{jk}+A_{kj}+(1-Tr A)\delta_{jk}=2(1-cos\theta)n_jn_k=(3-Tr R)n_jn_k \tag{1.3.37}$$

after using equation (1.3.24) for $A_{jk}$. Hence，

$$n_jn_k=\frac{S_{jk}}{3-Tr A}，Tr A\neq3 \tag{1.3.38}$$

To determine $\widehat{\mathbf{n}}$ up to an overall sign, we simply set $j=k$ (no sum) in equation (1.3.38), which fixes the value of

$$n_j^2=\frac{S_{jj}}{3-Tr A},Tr{A}\neq 3 \tag{1.3.39}$$

If $\sin \theta\neq0$, the overall sign of $\widehat{\mathbf{n}}$ is fixed by equation (1.3.34) by $n_m$ . Note that equation (1.3.37) implies that $Tr S=3-Tr A$. Summing over $j$ in equation (1.3.39) then yields

$$\widehat{\mathbf{n}}\cdot\widehat{\mathbf{n}}=\frac{TrS}{3-Tr R}=1 \tag{1.3.40}$$

as required for a unit vector. As noted above, if $A$ is a symmetric matrix (i.e. $A_{ij}=A_{ji}$), then $\sin\theta=0$ and $\widehat{\mathbf{n}}$ cannot be determined from equation (1.3.34) . In this case, equation (1.3.30) determines whether $\cos\theta=+1$ or $cos=-1$. For $cos\theta=-1$ , equation (1.3.39) yields $n_jn_k=\frac{1}{4}S_{ik}$ , which determines $\widehat{\mathbf{n}}$ up to an overall sign. Equivalently, one can use equation (1.3.22) to derive

$$\widehat{\mathbf{n}}=\left(\epsilon_1\sqrt{\frac{1}{2}(1+A_{11})} , \epsilon_2\sqrt{\frac{1}{2}(1+A_{22})},\epsilon_3\sqrt{\frac{1}{2}(1+A_{33})} \right), if \space Tr A=-1 \tag{1.3.41}$$

where the individual signs $\epsilon_i=\pm1$ are determined up to an overall sign via

$$\epsilon_i\epsilon_j=\frac{A_{ij}}{\sqrt{(1+A_{ii})(1+A_{jj})}}, for \space fixed \space i\neq j,A_{ii}\neq-1,A_{jj}\neq-1 \tag{1.3.42}$$

where the individual signs of $\widehat{\mathbf{n}}$ is immaterial. Finally, in the case of $\cos \theta=1$(which corresponds to $Tr A=3$), $A(\widehat{\mathbf{n}},0)=\mathbf{I}$ is the $3\times 3$ identity matrix, which is independent of the direction of $\widehat{\mathbf{n}}$ .To summarize, equation(1.3.31), (1.3.3.35), (1.3.38) provide a simple algorithm for determining the unit vector $\widehat{\mathbf{n}}$ and the rotation angle $\theta$ for any rotation matrix $A(\widehat{\mathbf{n}},\theta)\neq \mathbf{I}$.

D. Euler angle representation of $\mathbf{R}(\mathbf{\widehat{n}},\theta)$

An arbitrary rotation matrix can be written as:

$$\mathbf{A}(\mathbf{\widehat{n}},\theta)=\mathbf{A}(\widehat{\mathbf{z}},\alpha)\mathbf{A}(\widehat{\mathbf{y}},\beta)\mathbf{A}(\widehat{\mathbf{z}},\gamma) \tag{1.3.30}$$

where $\alpha$, $\beta$ and $\gamma$ are called the Euler angles. The ranges of the Euler angles are: $0\leq \alpha,\gamma<2\pi$ and $0\leq \beta\leq \pi$. We shall prove these statements “by construction.” That is, we shall explicitly derive the relations between the Euler angles and the angles $\theta,\theta_{n}$ and $\phi_{n}$ that characterize the rotation $\mathbf{R}(\widehat{\mathbf{n}},\theta)$, where $\theta_{n}$ and $\phi_{n}$ are the polar and azimuthal angle that define the axis of rotation $\widehat{\mathbf{n}}$ ,

$$\widehat{\mathbf{n}}=(\sin\theta_{n}\cos\phi_{n},\sin\theta_{n}\sin\phi_{n},\cos\theta_{n}) \tag{1.3.31}$$

Multiplying out the three matrices on the right-hand side of equation (1.3.30), we obtain the Euler angle parameterization of the three-dimensional rotation matrix.

$${\bf{R}}\left( {{\bf{\hat n}},\theta } \right) = \left( {\begin{array}{*{20}{c}} {\cos \alpha \cos \beta \cos \gamma - \sin \alpha \sin \gamma }&{ - \cos \alpha \cos \beta \sin \gamma - \sin \alpha \cos \gamma }&{\cos \alpha \sin \beta }\\ {\cos \alpha \cos \beta \cos \gamma + \cos \alpha \sin \gamma }&{ - \sin \alpha \cos \beta \sin \gamma + \cos \alpha \cos \gamma }&{\sin \alpha \sin \beta }\\ { - \sin \beta \cos \gamma }&{\sin \beta \sin \gamma }&{\cos \beta } \end{array}} \right) \tag{1.3.32}$$

One can now make use of the results of section 4 to obtain $\theta$ and $\mathbf{\widehat{n}}$ in terms of the Euler angles $\alpha,\beta$ and $\gamma$ .

Appendix A. An explicit formula for the Rotation Matrix

Suppose we wish to determine the explicit form of the rotation matrix $\mathbf{A}(\widehat{\mathbf{n}},\theta)$. Here is one possible strategy. The matrix $\mathbf{A}(\widehat{\mathbf{n}},\theta)$ is specified with respect to the standard $B_{s}=(\mathbf{i},\mathbf{j},\mathbf{k})$ . Given that the explicit form for $\mathbf{A}(\widehat{\mathbf{n}},\mathbf{k})$ is known according to equation (1.3.29) suggests that we should transform to a new orthonormal basis, $B_{s}^{'}=(\mathbf{i^{'}},\mathbf{j^{'}},\mathbf{k^{'}})$ , in which new positive $z-axis$ points the direction of $\mathbf{\widehat{n}}$ . That is,

$$\mathbf{k^{'}}=\mathbf{\widehat{n}}\equiv(n_{1},n_{2},n_{3}) \tag{A-1}$$

$$n_1^{2}+n_2^{2}+n_3^{2}=1 \tag{A-2}$$

The new positive y-axis can be chosen to lie along

$$\mathbf{j}^{'}= \left( {\frac{{ - {n_2}}}{{\sqrt {n_1^2 + n_2^2} }},\frac{{{n_1}}}{{\sqrt {n_1^2 + n_2^2} }},0} \right) \tag{A-3}$$

since by construction, $\mathbf{j}^{'}$ is a unit vector orthogonal to $\mathbf{k}^{'}$ . We complete the new right-handed coordinate system by choosing:

$$\boldsymbol{i}^{\prime}=\boldsymbol{j}^{\prime} \times \boldsymbol{k}^{\prime}=\left|\begin{array}{ccc} \boldsymbol{i} & \boldsymbol{j} & \boldsymbol{k} \\ \frac{-n_2}{\sqrt{n_1^2+n_2^2}} & \frac{n_1}{\sqrt{n_1^2+n_2^2}} & 0 \\ n_1 & n_2 & n_3 \end{array}\right|=\left(\frac{n_3 n_1}{\sqrt{n_1^2+n_2^2}}, \frac{n_3 n_2}{\sqrt{n_1^2+n_2^2}},-\sqrt{n_1^2+n_2^2}\right) \tag{A-4}$$

Following the class handout entitled, Vector coordinates, matrix elements, changes of basis, and matrix elements, changes of basis, and matrix diagonalization, we determine the matrix $\mathbf{P}$ whose matrix elements are defined by

$$\mathbf{b}_j^{'}=\sum\limits_{i = 1}^n \mathbf{P}_{ij}\mathbf{\widehat{e}}_i \tag{A-5}$$

where the $\mathbf{\widehat{e}}_i=(\mathbf{i},\mathbf{j},\mathbf{k})$ are the basis vectors of $B_s$ and the $\mathbf{b}_j^{'}$ are the basis vectors of $B_{s}^{'}$ .The columns of $P$ are the coefficients of expansion of the new basis vectors in terms of the old basis vectors. Thus,

$$P=\left(\begin{array}{ccc} \frac{n_3 n_1}{\sqrt{n_1^2+n_2^2}} & \frac{-n_2}{\sqrt{n_1^2+n_2^2}} & n_1 \\ \frac{n_3 n_2}{\sqrt{n_1^2+n_2^2}} & \frac{n_1}{\sqrt{n_1^2+n_2^2}} & n_2 \\ -\sqrt{n_1^2+n_2^2} & 0 & n_3 \end{array}\right) \tag{A-6}$$

The inverse $P^{-1}$ is easily computed since the columns of $P$ are orthonormal, which implies that $P$ is an orthogonal matrix, i.e. $P^{-1}=P^{T}$ .

According to equation (16), of the class handout, Vector coordinates, matrix elements, changes of basis and matrix diagonalization,

$$[\mathbf{A}]_{B^{'}}=\mathbf{P}^{-1}[\mathbf{A}]_{B_{s}}\mathbf{P} \tag{A-7}$$

where $[A]_{B^{'}}$ is the matrix $A$ with respect to the standard basis, and $[A]_{B^{'}}$ is the matrix $A$ with respect to the new basis (in which $\widehat{\mathbf{n}}$ points along the new positive $z-axis$). In particular,

$$[A]_{B}=\mathbf{A}(\widehat{\mathbf{n}},\theta) \tag{A-8}$$

$$[\mathbf{R}]_{B^{'}}=\mathbf{R}(\mathbf{k},\theta)=\left( {\begin{array}{*{20}{c}} {\cos \theta }&{ - \sin \theta }&0\\ {\sin \theta }&{\cos \theta }&0\\ 0&0&1 \end{array}} \right) \tag{A-9}$$

Hence, equation (A-6) yields

$$\mathbf{R}(\widehat{\mathbf{n}})=\mathbf{P}\mathbf{A}(\mathbf{k},\theta)\mathbf{P}^{-1} \tag{A-10}$$

where $P$ is given by equation (A-7) and $P^{-1}=P^{T}$ .

For ease of notation, we define

$$N_{12}=\sqrt{n_1^2+n_2^2} \tag{A-11}$$

Note that $N_{12}^2+n_{3}^2=1$, since $\widehat{\mathbf{n}}$ is a unit vector. Writing out the matrices :

$$\begin{aligned} R(\hat{\boldsymbol{n}}, \theta)= & \left(\begin{array}{ccc} n_3 n_1 / N_{12} & -n_2 / N_{12} & n_1 \\ n_3 n_2 / N_{12} & n_1 / N_{12} & n_2 \\ -N_{12} & 0 & n_3 \end{array}\right)\left(\begin{array}{ccc} \cos \theta & -\sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1 \end{array}\right)\left(\begin{array}{ccc} n_3 n_1 / N_{12} & n_3 n_2 / N_{12} & -N_{12} \\ -n_2 / N_{12} & n_1 / N_{12} & 0 \\ n_1 & n_2 & n_3 \end{array}\right) \\ & =\left(\begin{array}{ccc} n_3 n_1 / N_{12} & -n_2 / N_{12} & n_1 \\ n_3 n_2 / N_{12} & n_1 / N_{12} & n_2 \\ -N_{12} & 0 & n_3 \end{array}\right)\left(\begin{array}{ccc} \frac{n_3 n_1 \cos \theta+n_2 \sin \theta}{N_{12}} & \frac{n_3 n_2 \cos \theta-n_1 \sin \theta}{N_{12}} & -N_{12} \cos \theta \\ \frac{n_3 n_1 \sin \theta-n_2 \cos \theta}{N_{12}} & \frac{n_3 n_2 \sin \theta+n_1 \cos \theta}{N_{12}} & -N_{12} \sin \theta \\ n_1 & n_2 & n_3 \end{array}\right) . \end{aligned} \tag{A-12}$$

Using $N_{12}^2=n_{1}^2+n_{2}^2$ and $n_{3}^2=1-N_{12}^2$, the final matrix multiplication then yields,

$$\mathbf{A}(\hat{\boldsymbol{n}}, \theta)=\left(\begin{array}{ccc}\cos \theta+n_1^2(1-\cos \theta) & n_1 n_2(1-\cos \theta)-n_3 \sin \theta & n_1 n_3(1-\cos \theta)+n_2 \sin \theta \\n_1 n_2(1-\cos \theta)+n_3 \sin \theta & \cos \theta+n_2^2(1-\cos \theta) & n_2 n_3(1-\cos \theta)-n_1 \sin \theta \\n_1 n_3(1-\cos \theta)-n_2 \sin \theta & n_2 n_3(1-\cos \theta)+n_1 \sin \theta & \cos \theta+n_3^2(1-\cos \theta)\end{array}\right) \tag{A-13}$$

which coincides with the result previously exhibited in the class handout entitled Rotation matrix.