Derivation of rotation formula in a general coordinate system

[ShayanJ]

Homework Statement

[/B]
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

 

[Fred Wright]

You have:
r'_x = r_xcosφ + r_ysinφ
r'_y = -r_xsinφ + r_ycosφ
r'_z=r_z
n=k
rcosφ =( r_xi + r_yj + r_zk)cosφ
r'_xi =( r_xcosφ + r_ysinφ)i
r'_yj= (-r_xsinφ + r_ycosφ)j
r'_zk=r_z k
r' = ( r_xcosφ + r_ysinφ)i + (-r_xsinφ + r_ycosφ)j + r_zk
= rcosφ - r_ysinφi + r_xsinφj -r_zcosφk + r_zk
= rcosφ + (rxk)sinφ + (k*r)(1-cosφ)k

 

[ShayanJ]

Thanks Fred.
Can you generalize your proof for a general $\hat n$ ?

 

[samalkhaiat]

Homework Statement

[/B]
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).

 

[ShayanJ]

That was beautiful!
Thanks Sam!