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I am trying to use the output of the sandwich product, P’=QVQ, Q being the conjugate of Q

Say I wanted to rotate a point at $(1,0,0)$ a full 180 degrees around the y axis, so that it would end up at $(1,0,0)$, from my understanding first I find the quaternion of rotation, which is cosine of the angle/2 for the real part, and each vector part is sine of the angle over 2

So $q_0=\cos(a/2)$

$q_1=x\sin(a/2)$

$q_2=y\sin(a/2)$

$q_3=z\sin(a/2)$

As cosine of 90=0, the real part is gone As this is a rotation around purely the y axis, y=1, x and z are = 0

Sine 90=1, 1•1=1, so my quaternions is just 1j, or j

My point as a quaternions should just be i

j times i is -k The conjugate of j is -j -k times -j= -(-i)=i But that’s just the same point we started with, so no rotation happened

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2 Answers 2

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Too long for a comment.

You use the right quaternion $\mathbf{j}$ except that you do not multiply quaternions correctly. They follow the rules $$ \mathbf{i\,j=k} $$ where a minus sign must be added when an uneven permutation is applied. In particular, $$ \mathbf{j\,i=-k}\,. $$ Using $\mathbf{j}^2=-1$ we also get $$ \quad\mathbf{k\,j=-i}\,. $$ Then, rotating $V=(1,0,0)$ around the $y$-axis by 180 degrees is: $$ V=\mathbf{i} $$ \begin{align} a&=\pi\,, &Q&=\mathbf{j}\,,&Q^\dagger&=-\mathbf{j}\,,&Q^\dagger VQ=-\mathbf{j}\,\mathbf{i}\,\mathbf{j}=\mathbf{k}\,\mathbf{j}=-\mathbf{i}=-V \end{align} as expected.

To address your comment: In this case it does not matter on which $Q$ we put the $\dagger\,:$ $$ QVQ^\dagger=\mathbf{j\,i\,(-j)}=-\mathbf{j\,i\,j}=-\mathbf{i}\,. $$

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  • $\begingroup$ Thank you! Used the formula you used Q̅PQ instead of QPQ̅ and it’s worked for other things I’ve used too. All the different things I’ve found online use the equation QPQ̅ an I’m not sure why. Thank you very much $\endgroup$
    – Lydia
    Nov 26, 2023 at 15:39
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The formula $V' = Q \cdot V \cdot \overline{Q}$ works. You just did a little mistake in the end. If done correctly it would be:

The rotation around the $y$-axis aka $\left( Q_{x},\, Q_{y},\, Q_{z} \right) = \left( 0,\, 1,\, 0 \right)$ by $\alpha = 180^{\circ}$ implies $$ \begin{align*} \vec{Q} ~\widehat{=} \begin{pmatrix} q_{0}\\ q_{1}\\ q_{2}\\ q_{3}\\ \end{pmatrix} = \begin{pmatrix} \cos\left( \frac{\alpha}{2} \right)\\ Q_{x} \cdot \sin\left( \frac{\alpha}{2} \right)\\ Q_{y} \cdot \sin\left( \frac{\alpha}{2} \right)\\ Q_{z} \cdot \sin\left( \frac{\alpha}{2} \right)\\ \end{pmatrix} = \begin{pmatrix} \cos\left( \frac{180^{\circ}}{2} \right)\\ 0 \cdot \sin\left( \frac{180^{\circ}}{2} \right)\\ 1 \cdot \sin\left( \frac{180^{\circ}}{2} \right)\\ 0 \cdot \sin\left( \frac{180^{\circ}}{2} \right)\\ \end{pmatrix} = \begin{pmatrix} 0\\ 0\\ 1\\ 0\\ \end{pmatrix} \end{align*} $$

aka $Q = 0 + 0 \cdot i + 1 \cdot j + 0 \cdot k = j \Longleftrightarrow \overline{Q} = -j$. The point we whant to rotate $V ~\widehat{=} \begin{pmatrix} 1\\ 0\\ 0\\ \end{pmatrix}$ wich implies $V = 1 \cdot i + 0 \cdot j + 0 \cdot k = i$.

This gives us: $$ \begin{align*} V' &= Q \cdot V \cdot \overline{Q}\\ V' &= j \cdot i \cdot \left( -j \right) = \left( j \cdot i \right) \cdot \left( -j \right)\\ V' &= -k \cdot \left( -j \right)\\ V' &= --\left( k \cdot j \right) = k \cdot j\\ V' &= -i\\ \end{align*} $$

aka $\vec{V'} ~\widehat{=} \begin{pmatrix} -1\\ 0\\ 0\\ \end{pmatrix}$ wich is $V$ when rotaded $180^{\circ}$ around the $y$-axis.

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