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EDIT: I found an easier way to do (iii).

Question

Hi, I need help on part (i)[I know what the map is doing but can't describe it] and (iii). I found $ M_{ik}= (a\ \delta _{ik} + b\ \varepsilon_{ijk}\ n_j ) $ (summation Convention).

$M = \begin{pmatrix} a & -bn_{3} & bn_{2} \\bn_{3} & a & -bn_{1} \\-bn_{2} & bn_{1} & a \end{pmatrix} $

For part (iii) one way would be to invert the matrix M. I was wondering if there is some other quicker way, maybe using suffix notation.

Thanks

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2 Answers 2

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Part (iii) is easier once you've thought through part (i). Write $\bf x = \bf x_\parallel+\bf x_\perp$ with $\bf x_\parallel\parallel\bf n$ and $\bf x_\perp\perp\bf n$, then show that $\mathcal M$ factors into $\bf x_\parallel\to\bf x_\parallel'$ and $\bf x_\perp\to\bf x_\perp'$, and write down the separate transformation laws. In the transformation law for $\mathbf x_\perp$, you can take the cross product with $\mathbf n$ and then eliminate $\mathbf n\times\mathbf x_\perp$ from the two equations after using $\mathbf n\times(\mathbf n\times \mathbf x_\perp)=-\mathbf x_\perp$.

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I have done part (iii) without using part (i). Can you help me with part (i)? I still can't quite figure out the geometrical description. –  Ramanujan Apr 21 '12 at 16:19
    
@Ramanujan: $\mathbf n\times\mathbf x$ is $\mathbf x$ rotated by $\pi/2$ about $\mathbf n$. Thus, $\mathbf x_\parallel$ is merely scaled by a factor $a$, whereas $\mathbf x_\perp$ is scaled by a factor $a$ and also gets a component $b\mathbf n\times\mathbf x_\perp$ added to it, which is orthogonal to both $\mathbf n$ and $\mathbf x_\perp$. –  joriki Apr 21 '12 at 16:43

About your question (i) the trasformation $M$ is $\Lambda+R$,
where $\Lambda:x\to ax$ is the homothety of factor $a,$ and $R:x\to b\mathbf{n}\times x$ is the infintesimal generator of the counter-clockwise rotations around $b\mathbf{n}.$

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infinitesimal generator? –  Ramanujan Apr 21 '12 at 16:47
    
@Ramanujan: On infinitesimal generators, see e.g. math.stackexchange.com/a/116650. –  joriki Apr 21 '12 at 20:44

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