式展開: 四元数において，可逆元とその逆元を左右から作用させる

四元数において，可逆元とその逆元を左右から作用させる

$q p q^{- 1}$

\begin{array}{rcl} q p q^{- 1} & = & q p \frac{\overset{―}{q}}{| q |^{2}} = \frac{1}{| q |^{2}} q p \overset{―}{q} \\ = & \frac{1}{w^{2} + | V |^{2}} (w, V) (w_{p}, V_{p}) (w, - V) \dots \overset{―}{q} = w - x i - y j - z k = (w, - V), (実 部, i j k 部), V = x i + y j + z k \\ = & \frac{1}{w^{2} + | V |^{2}} (w w_{p} - V \cdot V_{p}, w V_{p} + w_{p} V + V \times V_{p}) (w, - V) \dots q_{1} q_{2} = (w_{1}, V_{1}) (w_{2}, V_{2}) = (w_{1} w_{2} - V_{1} \cdot V_{2}, w_{1} V_{2} + w_{2} V_{1} + V_{1} \times V_{2}) \\ = & \frac{1}{w^{2} + | V |^{2}} ((w w_{p} - V \cdot V_{p}) w - (w V_{p} + w_{p} V + V \times V_{p}) \cdot (- V), (w w_{p} - V \cdot V_{p}) (- V) + w (w V_{p} + w_{p} V + V \times V_{p}) + (w V_{p} + w_{p} V + V \times V_{p}) \times (- V)) \dots q_{1} q_{2} = (w_{1}, V_{1}) (w_{2}, V_{2}) = (w_{1} w_{2} - V_{1} \cdot V_{2}, w_{1} V_{2} + w_{2} V_{1} + V_{1} \times V_{2}) \\ = & \frac{1}{w^{2} + | V |^{2}} (w^{2} w_{p} - w V \cdot V_{p} + w V_{p} \cdot V + w_{p} V \cdot V + (V \times V_{p}) \cdot V), - w w_{p} V + (V \cdot V_{p}) V + w^{2} V_{p} + w w_{p} V + w V \times V_{p} + V \times (w V_{p} + w_{p} V + V \times V_{p})) \\ \dots A \times B = - B \times A \\ = & \frac{1}{w^{2} + | V |^{2}} (w^{2} w_{p} + w_{p} | V |^{2} + (V \times V) \cdot V_{p}, + (V \cdot V_{p}) V + w^{2} V_{p} + w V \times V_{p} + w V \times V_{p} + w_{p} V \times V + V \times (V \times V_{p})) \\ \dots (A \times B) \cdot C = (B \times C) \cdot A = (C \times A) \cdot B, A \times A = 0, A \cdot A = | A |^{2} \\ = & \frac{1}{w^{2} + | V |^{2}} (w_{p} (w^{2} + | V |^{2}), (V \cdot V_{p}) V + w^{2} V_{p} + 2 w V \times V_{p} + (V \cdot V_{p}) V - (V \cdot V) V_{p}) \\ \dots A \times (B \times C) = (A \cdot C) B - (A \cdot B) C \\ = & \frac{1}{w^{2} + | V |^{2}} (w_{p} (w^{2} + | V |^{2}), 2 (V \cdot V_{p}) V + w^{2} V_{p} + 2 w V \times V_{p} - | V |^{2} V_{p}) \\ \dots A \cdot A = | A |^{2} \\ = & \frac{1}{w^{2} + | V |^{2}} (w_{p} (w^{2} + | V |^{2}), 2 (V \cdot V_{p}) V + (w^{2} - | V |^{2}) V_{p} + 2 w V \times V_{p}) \end{array}

特殊なp, qを考える

ここで，

q = (\cos (\frac{θ}{2}), \sin (\frac{θ}{2}) V) (た だ し | V | = 1), p = (0, V_{p})

とする．

\begin{array}{rcl} q p q^{- 1} & = & \frac{1}{\cos^{2} (\frac{θ}{2}) + {| \sin (\frac{θ}{2}) V |}^{2}} (0 (\cos^{2} (\frac{θ}{2}) + {| \sin (\frac{θ}{2}) V |}^{2}), 2 (\sin (\frac{θ}{2}) V \cdot V_{p}) \sin (\frac{θ}{2}) V + (\cos^{2} (\frac{θ}{2}) - {| \sin (\frac{θ}{2}) V |}^{2}) V_{p} + 2 \cos (\frac{θ}{2}) \sin (\frac{θ}{2}) V \times V_{p}) \\ = & \frac{1}{\cos^{2} (\frac{θ}{2}) + \sin (\frac{θ}{2}) {| V |}^{2}} (0, 2 \sin^{2} (\frac{θ}{2}) (V \cdot V_{p}) V + (\cos^{2} (\frac{θ}{2}) - \sin^{2} (\frac{θ}{2}) {| V |}^{2}) V_{p} + 2 \cos (\frac{θ}{2}) \sin (\frac{θ}{2}) V \times V_{p}) \\ = & \frac{1}{\cos^{2} (\frac{θ}{2}) + \sin (\frac{θ}{2}) \cdot 1} (0, 2 \sin^{2} (\frac{θ}{2}) (V \cdot V_{p}) V + (\cos^{2} (\frac{θ}{2}) - \sin^{2} (\frac{θ}{2}) \cdot 1) V_{p} + 2 \cos (\frac{θ}{2}) \sin (\frac{θ}{2}) V \times V_{p}) \\ = & \frac{1}{1} (0, 2 \sin^{2} (\frac{θ}{2}) (V \cdot V_{p}) V + (\cos^{2} (\frac{θ}{2}) - \sin^{2} (\frac{θ}{2})) V_{p} + 2 \cos (\frac{θ}{2}) \sin (\frac{θ}{2}) V \times V_{p}) \\ = & (0, (1 - \cos (θ)) (V \cdot V_{p}) V + \cos (θ) V_{p} + \sin (θ) V \times V_{p}) \\ \dots 2 \sin^{2} (\frac{θ}{2}) = \sin^{2} (\frac{θ}{2}) + \sin^{2} (\frac{θ}{2}) = (1 - \cos^{2} (\frac{θ}{2})) + \sin^{2} (\frac{θ}{2}) = 1 - (\cos^{2} (\frac{θ}{2}) - \sin^{2} (\frac{θ}{2})) = 1 - \cos (θ) \\ \dots \cos^{2} (\frac{θ}{2}) - \sin^{2} (\frac{θ}{2}) = \cos (\frac{θ}{2} + \frac{θ}{2}) = \cos (θ) \\ \dots 2 \cos (\frac{θ}{2}) \sin (\frac{θ}{2}) = \sin (\frac{θ}{2} + \frac{θ}{2}) = \sin (θ) \\ = & (0, (V \cdot V_{p}) V - \cos (θ) (V \cdot V_{p}) V + \cos (θ) V_{p} + \sin (θ) V \times V_{p}) \\ = & (0, (V \cdot V_{p}) V + \cos (θ) (V_{p} - (V \cdot V_{p}) V) + \sin (θ) V \times V_{p}) \end{array}

位置ベクトル

V_{p}

を任意の回転軸

V

θ

だけ回転させる計算に対応する．

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