復數
\[\begin{aligned} a &= x+yi\\ ii&=-1 \end{aligned} \]
四元數基礎
定義
\[\begin{aligned} ii&=jj=kk=-1\\ ij&=-ji=k\\ jk&=-kj=i\\ ki&=-ik=j\\ p&=p_w+p_xi+p_yj+p_zk\\ q&=q_w+q_xi+q_yj+q_zk\\ p\pm q:&=(p_w\pm q_w+(p_x\pm q_x)i+(p_y\pm q_y)j+(p_z\pm q_z)k\\ p*q:&=(p_w+p_xi+p_yj+p_zk)(q_w+q_xi+q_yj+q_zk)\\ &=p_wq_w - p_xq_x - p_yq_y - p_zq_z \\ &+(p_wq_x + q_wp_x + p_yq_z-p_zq_y)i\\ &+(p_wq_y + q_wp_y + p_zq_x-p_xq_z )j\\ &+(p_wq_z + q_wp_z + p_xq_y-p_yq_x )k\\ |p| :&= \sqrt{w^2+x^2+y^2+z^2} \end{aligned} \]
其中：
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\(s,w,x,y,z\in R\)
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單位四元數: \(|p| = \sqrt{w^2+x^2+y^2+z^2}=1\)
顯然，四元數集在上述加法與數乘的定義下滿足向量空間的公理，四元數集合是\(R^4\)向量空間。同時四元數也滿足乘法定義，所以四元數也是\(R\)代數。
標量、向量型記法
令\(xi+yj+zk=\vec{v}, q = w + \vec{v}\), 其中\(w,x,y,z\in R\)
\[\begin{aligned} p&=w_p+\vec{v_p}\\ q&=w_q+\vec{v_q}\\ p\pm q:&=w_p\pm w_q + (\vec{v_p}\pm \vec{v_q})\\ p* q:&=w_pw_q - \vec{v_p}\cdot \vec{v_q} + w_p\vec{v_q} + w_q\vec{v_p} + \vec{v_p}\times \vec{v_q}\\ \end{aligned} \]
繞任意軸的旋轉
以下的推導基於右手坐標系。
向量式
設任意軸單位向量為\(n(x,y,z),|n|=1,x,y,z\in R\)，逆時針旋轉為\(\theta\)，則向量\(p(p_x,p_y,p_z)\)旋轉后的向量\(p'\)為：
\[\begin{aligned} p_{\parallel}&=(n\cdot p)n\\ p_{\perp}&=p-p_{\parallel}\\ |p_{\perp}|&=|p|sin(\alpha)\\ |n\times p|=|n||p|sin(\alpha)&=|p|sin(\alpha)=|p_{\perp}|\\ \end{aligned} \]
\(\therefore p_{\perp},n\times p\)構成正交基，\(p'_⊥\)可由\(p_{\perp},n\times p\)線性表示:
\[\begin{aligned} p'&=p'_{\perp}+p_{\parallel}\\ &=p_{\perp}cos\theta+(v\times p)sin\theta+p_{\parallel}\\ &=(p-p_{\parallel})cos\theta+(v\times p)sin\theta+p_{\parallel}\\ &=pcos\theta+p_{\parallel}(1-cos\theta)+(v\times p)sin\theta\\ &=pcos\theta+(n\cdot p)n(1-cos\theta)+(v\times p)sin\theta \end{aligned} \]
矩陣式
由
\[v\times p = \begin{pmatrix}0 & -z & y\\ z & 0 &-x\\ -y & x & 0 \end{pmatrix}p \]
\[(n\cdot p)n= \begin{pmatrix}x^2 & xy & xz\\ xy & y^2 & yz\\ xz & yz & z^2 \end{pmatrix}p \]
代入向量式可得：
\[\begin{aligned} p'&=[cos\theta I_3+sin\theta\begin{pmatrix}0 & -z & y\\ z & 0 &-x\\ -y & x & 0 \end{pmatrix}+(1-cos\theta)\begin{pmatrix} x^2 & xy & xz\\ xy & y^2 & yz\\ xz & yz & z^2 \end{pmatrix}]p \\ &=\begin{pmatrix} cos\theta+x^2(1-cos\theta) & -zsin\theta+xy(1-cos\theta) & ysin\theta+xz(1-cos\theta)\\ zsin\theta+xy(1-cos\theta) & cos\theta+y^2(1-cos\theta) & -xsin\theta+yz(1-cos\theta)\\ -ysin\theta+xz(1-cos\theta) & xsin\theta+yz(1-cos\theta) & cos\theta+z^2(1-cos\theta) \end{pmatrix}p \end{aligned} \]
從而得到繞\(\vec{n}\)軸旋轉\(\theta\)的旋轉矩陣
\[R_n(\theta)=\begin{pmatrix} cos\theta+x^2(1-cos\theta) & -zsin\theta+xy(1-cos\theta) & ysin\theta+xz(1-cos\theta)\\ zsin\theta+xy(1-cos\theta) & cos\theta+y^2(1-cos\theta) & -xsin\theta+yz(1-cos\theta)\\ -ysin\theta+xz(1-cos\theta) & xsin\theta+yz(1-cos\theta) & cos\theta+z^2(1-cos\theta) \end{pmatrix} \]
令\(c= cos\theta, s= sin\theta, n(n_x,n_y,n_z)\)，代入上式得：
\[R_n(\theta)=\begin{pmatrix} c+n_x^2(1-c) & -n_zs+n_xn_y(1-c) & n_ys+n_xn_z(1-c)\\ n_zs+n_xn_y(1-c) & c+n_y^2(1-c) & -n_xs+n_yn_z(1-c)\\ -n_ys+n_xn_z(1-c) & n_xs+n_yn_z(1-c) & c+n_z^2(1-c) \end{pmatrix} \]
四元數與旋轉
設\(p=s+\vec{v}\)為單位四元數，則\(q^{-1}=s-\vec{v}\)，令向量\(p(p_x,p_y,p_z)\)旋轉后的向量\(p'\)為：
\[\begin{aligned} p'&=q\vec{p}q^{-1}\\ &= (s+\vec{v}) \vec{p}(s-\vec{v})\\ &= (s\vec{p}-\vec{v}\cdot \vec{p}+\vec{v}\times \vec{p})(s-\vec{v})\\ &= s^2\vec{p}-\underline{s\vec{v}\cdot \vec{p}}+s\vec{v}\times \vec{p}+\underline{s\vec{p}\cdot \vec{v}}-+\underbrace{s\vec{p}\times \vec{v}}_{-s\vec{v}\times \vec{p}}+(\vec{v}\cdot \vec{p})\vec{v}+\underbrace{(\vec{v}\times \vec{p})\cdot\vec{v}}_0-\vec{v}\times \vec{p}\times\vec{v}\\ &=s^2\vec{p}+2s(\vec{v}\times \vec{p})+(\vec{v}\cdot \vec{p})\vec{v}-\underbrace{\vec{v}\times \vec{p}\times\vec{v}}_{\vec{v}^2\vec{p}-(\vec{v}\cdot\vec{p})\vec{v}}\\ &=(s^2-\vec{v}^2)\vec{p}+2s(\vec{v}\times \vec{p})+2(\vec{v}\cdot \vec{p})\vec{v} \end{aligned} \]
用單位向量表示\(v\)，令\(\vec{v}=t\vec{n}\)，代入上式得
\[p'=(s^2-t^2)\vec{p}+2st(\vec{n}\times \vec{p})+2t^2(\vec{n}\cdot \vec{p})\vec{n} \]
四元數表達式與向量式之間的關系
聯立向量式\(p'=pcos\theta+(n\cdot p)n(1-cos\theta)+(v\times p)sin\theta\)和四元數表達式可得：
\[\begin{cases} s^2-t^2=cos\theta \\ 2st=sin\theta\\ 2t^2 = 1-cos\theta \end{cases} \]
解得：\(t=sin(\theta/2), s=cos(\theta/2)\)，從而得到由任意軸單位向量為\(n(x,y,z),|n|=1,x,y,z\in R\)，逆時針旋轉為\(\theta\)，的單位四元數：
\[p=(s,t\vec{n})=(cos(\theta/2),(n_xsin\theta, n_ysin\theta, n_zsin\theta )) \]
或
\[\begin{cases} s=cos(\theta/2)\\ v_x=n_xsin(\theta/2)\\ v_y=n_ysin(\theta/2)\\ v_z=n_zsin(\theta/2) \end{cases} \]
四元數表達式與矩陣式之間的關系
利用旋轉四元數中的分量\(w^2+v_x^2+v_y^2+v_z^2=1\)對旋轉矩陣進行化簡：
\[\begin{aligned} cos(\theta/2)&=w\\ n_xsin(\theta/2)&=v_x=x\\ n_ysin(\theta/2)&=v_y=y\\ n_zsin(\theta/2)&=v_z=z\\ c=cos\theta&=2cos^2(\theta/2)-1=2w^2-1\\ 1-cos\theta&=2sin^2(\theta/2)\\ n_x^2(1-cos\theta)&=2n_x^2sin^2(\theta/2)=2v_x^2\\ n_y^2(1-cos\theta)&=2n_y^2\sim =2v_y^2\\ n_z^2(1-cos\theta)&=2n_z^2\sim =2v_z^2\\ sin\theta&=2sin(\theta/2)cos(\theta/2)=2sin(\theta/2)w\\ n_xsin(\theta)&=n_x2sin(\theta/2)w=2wv_x\\ n_ysin(\theta)&=n_y\sim=2wv_y\\ n_zsin(\theta)&=n_z\sim=2wv_z\\ n_xn_y(1-cos(\theta))&=2n_xn_ysin^2(\theta/2)=2v_xv_y\\ n_xn_z(1-cos(\theta))&=2n_xn_z\sim=2v_xv_z\\ n_yn_z(1-cos(\theta))&=2n_yn_z\sim=2v_yv_z\\ \end{aligned} \]
\[\begin{aligned} R_n(\theta)&=\begin{pmatrix} c+n_x^2(1-c) & -n_zs+n_xn_y(1-c) & n_ys+n_xn_z(1-c)\\ n_zs+n_xn_y(1-c) & c+n_y^2(1-c) & -n_xs+n_yn_z(1-c)\\ -n_ys+n_xn_z(1-c) & n_xs+n_yn_z(1-c) & c+n_z^2(1-c) \end{pmatrix}\\ &=\begin{pmatrix} 2w^2+2x^2-1 & 2xy-2wz & 2xz+2wy\\ 2xy+2wz & 2w^2+2y^2-1 & 2yz-2wx\\ 2xz-2wy & 2yz+2wx & 2w^2+2z^2-1 \end{pmatrix}\\ &=\begin{pmatrix} 1-2y^2-2z^2 & 2xy-2wz & 2xz+2wy\\ 2xy+2wz & 1-2x^2-2z^2 & 2yz-2wx\\ 2xz-2wy & 2yz+2wx & 1-2x^2-2y^2 \end{pmatrix} \end{aligned} \]
分開助記：
\[\begin{aligned} R_n(\theta)&=[cos\theta I_3+sin\theta\begin{pmatrix}0 & -n_z & n_y\\ n_z & 0 &-n_x\\ -n_y & n_x & 0 \end{pmatrix}+(1-cos\theta)\begin{pmatrix} n_x^2 & n_xn_y & n_xn_z\\ n_xn_y & n_y^2 & n_yn_z\\ n_xn_z & n_yn_z & n_z^2 \end{pmatrix}]\\ &=(2w^2-1)\begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{pmatrix}+2w\begin{pmatrix} 0 & -z & y\\ z & 0 &-x\\ -y & x & 0 \end{pmatrix}+\begin{pmatrix} 0 & 2xy & 2xz\\ 2xy & 0 & 2yz\\ 2xz & 2yz & 0 \end{pmatrix}\\ &=\begin{pmatrix} 2w^2-1 & 0 & 0\\ 0 & 2w^2-1 & 0\\ 0 & 0 & 2w^2-1 \end{pmatrix}+\begin{pmatrix} 0 & -2wz & 2wy\\ 2wz & 0 & -2wx\\ -2wy & 2wx & 0 \end{pmatrix}+\begin{pmatrix} 0 & 2xy & 2xz\\ 2xy & 0 & 2yz\\ 2xz & 2yz & 0 \end{pmatrix} \end{aligned} \]
通過四元數，旋轉矩陣用了竟然是（對角陣+反稱矩陣+對稱矩陣），神奇也優美！