Gary
GaryVasco
Posts: 3,352 
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Post by Gary on Sept 6, 2016 4:29:59 GMT
Vasco,
Perhaps you can help with pp. 292-293, regarding the quaternion formula for the effect of $R^{\psi}_{\textbf{v}}$ on the position vector of a point in space. Before reading this, I would have guessed that one could write $\tilde{\mathbb{P}} = \mathbb{R}^{\psi}_{\textbf{v}} \mathbb{P}$. I would base that on the 2nd sentence in paragraph 3: “Suppose that $R^{\psi}_{\textbf{v}}$ rotates $\textbf{P}$ to $\tilde{\textbf{P}}$." I follow the sequence illustrated on the sphere in Figure [17], so I see the derivation of (29), but it’s not clear what it represents. My guess is that one would apply (29) to any point z or w on $\Sigma$, but that seems to conflict with the statement that we are seeking the effect on the position vector $\textbf{P} = X\textbf{i} + Y\textbf{j} + Z\textbf{k}$ of a point in space, because that position vector is then written as a pure quaternion and incorporated into the formula (rather than being used as an argument to the RHS of (29)).
I hope this is not total gibberish.
Gary
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Post by Admin on Sept 6, 2016 5:16:03 GMT
Gary
I will have a look at this today and get back to you.
13:20 GMT:
There is a difference between the transformation $\mathcal{R}$ and the quaternion $\mathbb{R}$. You can apply the transformation $\mathcal{R}$ to a point in space, but $\mathbb{R}\mathbb{P}$ is a multiplication of two quaternions - an entirely different thing. The way I understand things at the moment is that quaternions are not 'applied' to anything.
14:00 UK time
You may already know this but here goes anyway:
If you substitute for $\textbf{V}$ in (28) and then substitute the binary rotation matrices at the bottom of p. 291 for $\textbf{I},\textbf{J},\textbf{K}$ you obtain (21) on page 289. So the quaternion $\mathbb{R}$ in (28) contains all the information about the transformation $\mathcal{R}$.
Similarly, the quaternion $\mathbb{P}$ in (29) (if you do the same substitutions as above), produces a matrix from which it is possible to extract the values of $X,Y,Z$ and so find $\textbf{P}$. The quaternion $\widetilde{\mathbb{P}}$ is of the same form and so we can extract $\widetilde{\textbf{P}}$.
GMT 14:40 UK time
I would suggest at this point that you do exercise 6. I am sure that the thinking processes you have to go through to do this exercise will be a great help in understanding what (29) is all about. It's not the full story but a good part of it I think. I need to finish off exercise 6 myself.
GMT 19:50 UK time
I notice in part (vii) of exercise 6 that Needham refers to (29) as a transformation sending $\mathbb{P}$ to $\widetilde{\mathbb{P}}$. However this is in the sense that the two quaternions $\mathbb{P}$ and $\widetilde{\mathbb{P}}$ contain the information about the position vectors $\textbf{P}$ and $\widetilde{\textbf{P}}$.
I think that's probably all for today.
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Sept 7, 2016 0:42:57 GMT
Vasco,
Thank you. I will study all that and do ex. 6 and get back.
Gary
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Post by Admin on Sept 9, 2016 16:52:31 GMT
Vasco, Perhaps you can help with pp. 292-293, regarding the quaternion formula for the effect of $R^{\psi}_{\textbf{v}}$ on the position vector of a point in space. Before reading this, I would have guessed that one could write $\tilde{\mathbb{P}} = \mathbb{R}^{\psi}_{\textbf{v}} \mathbb{P}$. I would base that on the 2nd sentence in paragraph 3: “Suppose that $R^{\psi}_{\textbf{v}}$ rotates $\textbf{P}$ to $\tilde{\textbf{P}}$." I follow the sequence illustrated on the sphere in Figure [17], so I see the derivation of (29), but it’s not clear what it represents. My guess is that one would apply (29) to any point z or w on $\Sigma$, but that seems to conflict with the statement that we are seeking the effect on the position vector $\textbf{P} = X\textbf{i} + Y\textbf{j} + Z\textbf{k}$ of a point in space, because that position vector is then written as a pure quaternion and incorporated into the formula (rather than being used as an argument to the RHS of (29)). I hope this is not total gibberish. Gary Gary I have been thinking a bit more about your original post above. Did you notice the distinction between $R$ and $\mathcal{R}$ on the pages you refer to? In the sentence beginning "Suppose" that you quote above, the $R$ that rotates $\mathbb{P}$ does so via an induced Mobius transformation of $\mathbb{C}$. It seems to me that this is the reason that you cannot multiply $\mathbb{P}$ by $R$. $\mathbb{P}$ is 3 dimensional on the sphere. What do you think? Vasco |
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Sept 9, 2016 22:46:40 GMT
Vasco,
You could be right about the R's, but I'm not absolutely sure.
On page 292, $R^{\psi}_{\mathbf{v}}$ refers to "the effect of $R^{\psi}_{\mathbf{v}}$ on a point in space", so one would think that $R^{\psi}_\mathbf{v}$ was/is a rotation in space, but there could be a tacit implication that $R^{\psi}_\mathbf{v}$ is on the plane and induces a rotation in space. I think this $R$ must be the same as in (20), p. 288, where it indicates a $M\ddot{o}bius$ transformation, which puts it on the $\mathbb{C}$ plane. Notice that the "a" in (20) is on $\Sigma$, so it has the same algebraic structure as $\mathbf{v}$. So perhaps using the subscripts "a" and "$\mathbf{v}$" is just a way of suggesting that the function of the expression is to induce a rotation in $\Sigma$ by stereographic projection.
In Figure [17] we find $\mathcal{R}^{\psi}_{\hat{a}}$ referring to a rotation of the sphere about $\hat{a}$. So perhaps $R^{\psi}_{\mathbf{v}}$ and $\mathcal{R}^{\psi}_{\hat{a}}$ are different entities as you suggested. $R^{\psi}_\mathbf{v}$ is a $M\ddot{o}bius$ transformation and $\mathcal{R}^{\psi}_{\hat{a}}$ is a rotation of the sphere.
It would be nice to be sure about these things, but I don't know that it sheds light on my perplexity over (29), but then I haven't tried Ex. 6 (vi) and (vii) yet.
Gary
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