# Derivation of rotation formula in a general coordinate system

Gold Member

## 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.

## 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

You have:
r'x = rxcosφ + rysinφ
r'y = -rxsinφ + rycosφ
r'z=rz
n=k
r
cosφ =( rxi + ryj + rzk)cosφ
r'xi =( rxcosφ + rysinφ)i
r'yj= (-rxsinφ + rycosφ)j
r'zk=rz k
r
' = ( rxcosφ + rysinφ)i + (-rxsinφ + rycosφ)j + rzk
= rcosφ - rysinφi + rxsinφj -rzcosφk + rzk
= rcosφ + (rxk)sinφ + (k*r)(1-cosφ)k

ShayanJ
Gold Member
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.

## 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).

BvU and ShayanJ
Gold Member
That was beautiful!
Thanks Sam!