Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Passive and Active Transformations

  1. Jun 12, 2015 #1
    Alright. I was looking into a 2D rotation matrix and there are two equations: one is through the transformation of the component of p (always with respect to x,y), x,y into x',y' and the other is through the transformation of the unit vectors i,j into i',j'.

    In a sense 1 is passive the other is an active transformation. I do understand the difference between the two, but my question is that why do we get the different matrices if when we derive them, we do so with a clockwise orientation?

    Hope its clear, If not--I'd be happy to clarify.
     
  2. jcsd
  3. Jun 12, 2015 #2
    Can you clarify. I guess I understand what you mean, but just to be sure.
     
  4. Jun 12, 2015 #3
    LOL Okay. I got the answer. Welp it seems like they way you derive the matrix you have to say consistent, otherwise you fall in a a trap.
    https://en.wikipedia.org/wiki/Active_and_passive_transformation#/media/File:PassiveActive.JPG
    Staring at this image for a bit and making sense of this reasoning: http://math.stackexchange.com/quest...erent-representations-of-2d-rotation-matrices

    helped. You need to use a point relative to the rotated basis once you rotate the basis. You cannot use a rotated basis on a point relative to a the old basis, but with rotational matrices you only an inverted answer and nothing any more alarming.
     
  5. Jun 12, 2015 #4

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    See this thread for some general comments about "active" vs. "passive" rotations.

    The convention that I'm familiar with is to rotate vectors counterclockwise. A rotation is linear, so if we know how the basis vectors are rotated, we will know how all vectors are rotated. If you rotate (1,0) by an angle ##\theta## counterclockwise, the result is ##(\cos\theta,\sin\theta)##. The same rotation applied to (0,1) yields ##(-\sin\theta,\cos\theta)##. It follows immediately from this that the matrix of the rotation operator with respect to the standard ordered basis is
    $$\begin{pmatrix}\cos\theta & -\sin\theta\\ \sin\theta & \cos\theta\end{pmatrix}.$$ The rotated basis vectors end up as the columns of the matrix. If you don't understand this, I recommend the FAQ post that I linked to in the post I linked to above.
     
    Last edited: Jun 12, 2015
  6. Jun 12, 2015 #5
    Hehe alright, thanks. Looking back at it, I was being pretty dense.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Passive and Active Transformations
  1. Linear transformations (Replies: 6)

  2. Linear transformation (Replies: 5)

  3. Linear Transformations (Replies: 5)

Loading...