Shyan said:
Homework Statement
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In a set of axes where the z axis is the axis of rotation of a finite rotation, the rotation matrix is given by
## \left[ \begin{array}{lcr} &\cos\phi \ \ \ &\sin\phi \ \ \ &0 \\ & -\sin\phi \ \ \ &\cos\phi \ \ \ &0 \\ &0 \ \ \ &0 \ \ \ &1 \end{array} \right]##.
Derive the rotation formula
##\vec{r'}=\vec r \cos\phi+\hat n(\hat n \cdot \vec r)(1-\cos\phi)+(\vec r \times \hat n) \sin\phi ##
by transforming to an arbitrary coordinate system, expressing the orthogonal matrix of transformation in terms of the direction cosines of the axis of the finite rotation.
Homework Equations
The Attempt at a Solution
As far as I know, the transformation is ## S'=A^{-1} S A ## where A is a general rotation matrix and its such a big matrix that even writing it down is a tedious thing to do, let alone multiplying it several times with another matrix! Is there any other way to do this?
Thanks
First note that R(\hat{n} , \phi) leaves the unit vector \hat{n} invariant, i.e., R_{ij}n_{j} = n_{i} . \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)
Also note the following trivial identities
\delta_{ij}n_{j} = n_{i} , \ \ \ n_{i} n_{j} n_{j} = n_{i} (\hat{n} \cdot \hat{n}) = n_{i} , and \epsilon_{ijk}n_{j}n_{k} = 0 . So, covariance requires R_{ij} to be of the form
R_{ij} = a(\phi) \ \delta_{ij} + b(\phi) \ n_{i}n_{j} + c(\phi) \ \epsilon_{ijk}n_{k} , \ \ \ \ (2) where a,b,c are functions of the only available scalar \phi, the rotation angle. Now, if you substitute (2) in (1) and use the above identities, you find a + b = 1.
Next, consider the special case where \hat{n} = \hat{e}_{3} = (0,0,1) .
So, R_{11}(e_{3},\phi) = a(\phi) = \cos \phi , and
R_{12}(e_{3},\phi) = c(\phi) \epsilon_{123}n_{3} = c(\phi) = - \sin \phi .
Substituting these in the general form (2), we obtain
R_{ij} = \delta_{ij} \ \cos \phi + n_{i}n_{j} \left(1 - \cos \phi \right) - \epsilon_{ijk} n_{k} \sin \phi . Now, contracting this with x_{j} gives you
\bar{x}_{i} = x_{i} \cos \phi + n_{i} \left( \hat{n} \cdot \vec{x}\right) \left( 1 - \cos \phi \right) + \left( \hat{n} \times \vec{x} \right)_{i} \sin \phi .
You can also do it geometrically by decomposing the vector \vec{x} into the sum of two vectors: one parallel to \hat{n} and the other perpendicular to \hat{n}
\vec{x} = \left( \hat{n} \cdot \vec{x}\right) \hat{n} + \vec{x}_{\perp} . \ \ \ (3) So, rotation by an angle \phi will leaves the parallel vector invariant and sends \vec{x}_{\perp} into
R \vec{x}_{\perp} = \vec{x}_{\perp} \cos \phi + \left( \hat{n} \times \vec{x}_{\perp} \right) \sin \phi .
Since \hat{n} \times \vec{x} = \hat{n} \times \vec{x}_{\perp}, we find
\vec{x} \to R \vec{x} = \left( \hat{n} \cdot \vec{x}\right) \hat{n} + \vec{x}_{\perp} \cos \phi + \left( \hat{n} \times \vec{x} \right) \sin \phi . Now, the final result follow from substituting (3).
