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Derivation of rotation formula in a general coordinate system

  1. May 22, 2016 #1

    ShayanJ

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    Gold Member

    1. The problem statement, all variables and given/known data

    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.

    2. Relevant equations


    3. 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
     
  2. jcsd
  3. May 24, 2016 #2
    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
     
  4. May 24, 2016 #3

    ShayanJ

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    Gold Member

    Thanks Fred.
    Can you generalize your proof for a general ##\hat n## ?
     
  5. May 26, 2016 #4

    samalkhaiat

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    First note that [itex]R(\hat{n} , \phi)[/itex] leaves the unit vector [itex]\hat{n}[/itex] invariant, i.e., [tex]R_{ij}n_{j} = n_{i} . \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)[/tex]
    Also note the following trivial identities
    [tex]\delta_{ij}n_{j} = n_{i} , \ \ \ n_{i} n_{j} n_{j} = n_{i} (\hat{n} \cdot \hat{n}) = n_{i} ,[/tex] and [tex]\epsilon_{ijk}n_{j}n_{k} = 0 .[/tex] So, covariance requires [itex]R_{ij}[/itex] to be of the form
    [tex]R_{ij} = a(\phi) \ \delta_{ij} + b(\phi) \ n_{i}n_{j} + c(\phi) \ \epsilon_{ijk}n_{k} , \ \ \ \ (2)[/tex] where [itex]a,b,c[/itex] are functions of the only available scalar [itex]\phi[/itex], the rotation angle. Now, if you substitute (2) in (1) and use the above identities, you find [tex]a + b = 1.[/tex]
    Next, consider the special case where [tex]\hat{n} = \hat{e}_{3} = (0,0,1) .[/tex]
    So, [tex]R_{11}(e_{3},\phi) = a(\phi) = \cos \phi ,[/tex] and
    [tex]R_{12}(e_{3},\phi) = c(\phi) \epsilon_{123}n_{3} = c(\phi) = - \sin \phi .[/tex]
    Substituting these in the general form (2), we obtain
    [tex]R_{ij} = \delta_{ij} \ \cos \phi + n_{i}n_{j} \left(1 - \cos \phi \right) - \epsilon_{ijk} n_{k} \sin \phi .[/tex] Now, contracting this with [itex]x_{j}[/itex] gives you
    [tex]\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 .[/tex]
    You can also do it geometrically by decomposing the vector [itex]\vec{x}[/itex] into the sum of two vectors: one parallel to [itex]\hat{n}[/itex] and the other perpendicular to [itex]\hat{n}[/itex]
    [tex]\vec{x} = \left( \hat{n} \cdot \vec{x}\right) \hat{n} + \vec{x}_{\perp} . \ \ \ (3)[/tex] So, rotation by an angle [itex]\phi[/itex] will leaves the parallel vector invariant and sends [itex]\vec{x}_{\perp}[/itex] into
    [tex]R \vec{x}_{\perp} = \vec{x}_{\perp} \cos \phi + \left( \hat{n} \times \vec{x}_{\perp} \right) \sin \phi .[/tex]
    Since [tex]\hat{n} \times \vec{x} = \hat{n} \times \vec{x}_{\perp},[/tex] we find
    [tex]\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 .[/tex] Now, the final result follow from substituting (3).
     
  6. May 26, 2016 #5

    ShayanJ

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    Gold Member

    That was beautiful!
    Thanks Sam!
     
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