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Active and Passive rotations

  1. Mar 11, 2014 #1
    Hi This is the problem... I'm reading a paper where the author says

    Transformation between the laboratory and spherical frames can be represented by the product of two rotations through the angles θ and [itex]\varphi[/itex], Û[itex]_{S}[/itex]=[itex]\hat{R}[/itex][itex]_{y}[/itex](θ)[itex]\hat{R}[/itex][itex]_{z}[/itex]([itex]\varphi[/itex]):

    |E>[itex]_{S}[/itex]=Û[itex]_{S}[/itex]|E>[itex]_{L}[/itex].

    The matrix Û[itex]_{S}[/itex] is


    Û[itex]_{S}[/itex] = [itex] \left[\begin{matrix} \cosθ\cos\varphi & \cosθ\sin\varphi & -\sinθ \\ -\sin\varphi & \cos\varphi & 0 \\ \sinθ\cos\varphi & \sinθ\sin\varphi & \cosθ \end{matrix}\right][/itex]

    Now... I was trying to understand this transformation from cartesian laboratory coordinates to spherical ones and I've obtained something similar but not the same. What I did was a concatenation of two rotations. First I considered a point p with projection q on the xy-plane. Then I rotated the frame about the z axis by [itex]\varphi[/itex] (counterclock-wise) obtaining a second frame. This second frame has its x-axis aligned to the projection q. Finally I rotated the second frame about the y-axis by θ, aligning the z-axis of the third frame with the direction of p. Let's the original frame be called A, the second B and the third C. The matrix I used to go from A to B is

    R[itex]^{B}_{A}[/itex]=[itex] \left[\begin{matrix} \cos\varphi & -\sin\varphi & 0 \\ \sin\varphi & \cos\varphi & 0 \\ 0 & 0 & 1\end{matrix}\right][/itex]

    And to go from B to C

    R[itex]^{C}_{B}[/itex]=[itex] \left[\begin{matrix} \cos\theta & 0 & \sin\theta \\ 0 & 1 & 0 \\-\sin\theta & 0 & \cos\theta \end{matrix}\right][/itex]

    So, the matrix which takes us from A to C is

    R[itex]^{C}_{A}[/itex] = R[itex]^{B}_{A}[/itex]*R[itex]^{C}_{B}[/itex] = R[itex]^{C}_{B}[/itex]=[itex] \left[\begin{matrix} \cos\theta\cos\varphi & -\sin\varphi & \sin\theta\cos\varphi \\ cos\theta\sin\varphi & \cos\varphi & \sin\theta\sin\varphi \\-\sin\theta & 0 & \cos\theta \end{matrix}\right][/itex]

    This matrix is the transpose of Û[itex]_{S}[/itex] so my doubt is what is going on? I expected the to be the same. I know that active and passive transformations have matrices that are the transpose of one another but I'm failing to see which case is which here.


    Thank you.
     
  2. jcsd
  3. Mar 11, 2014 #2

    A.T.

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    Gold Member

    Two things to check here:
    - Which convention is used: R * v or vT * R ?
    - Do the matrices represent the rotation of the local system relative to lab system or the coordiante transformation from the lab system to the local system ?
     
  4. Mar 12, 2014 #3

    D H

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    You've fallen prey to the "active transformation versus passive transformation" trap. I personally am not thrilled with those two terms. "Active transformation" versus "passive transformation": what does that mean? I use the terms "rotation" and "transformation" instead. Admittedly this doesn't helps that much. With my nomenclature, there's a "rotation versus transformation" trap into which the uninitiated can be lured. "Rotation" versus "transformation": what does that mean?

    So, whether it's active transformation vs passive transformation or rotation vs transformation, what exactly do those terms mean?

    Active transformation, which I call rotation, answers the question "Suppose I have some vector quantity ##\vec q## and I physically rotate it about some axis by some angle. Where does that vector point now?" Note that only one coordinate system is involved in this question. Passive transformation, which I just call transformation, answers the question "Suppose I have two coordinate systems U and V with a common origin. What is the relationship between the representation of some vector quantity ##\vec q## in frame U versus it's representation in frame V?"

    These two questions are closely related. With the first question, you can "actively transform" the x, y, and z axes and create a rotated reference frame. Now you can answer the second question using this new frame. With the second question, you can ask about the relation between the representations of x, y, and z axes in the two frames. This is equivalent to an "active transformation" or "rotation".

    The relationship between the two concepts is that the same matrix represents the "active transformation" that rotates frame U to form frame V and the "passive transformation" that transforms a vector expressed in frame V to frame U are the same matrix. Another way to look at it: The matrix that represents the "passive transformation" from frame U to frame V is the transpose of the matrix that represents the "active transformation" that rotates frame U to form frame V.
     
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