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I was going over my notes on classical mechanics and just started to review rotation matrices which is the first topic the book starts with. On page 3

enter image description here

The rotation matrix associated with 1.2a and 1.2b is

\begin{pmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \\ \end{pmatrix}

but when I try to derive the matrix by following the unit vectors $\hat i$ and $\hat j$

I get

\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \\ \end{pmatrix}

The one that the book derives would be clockwise rotation, and the one I got would be for counter-clockwise rotation correct?

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  • $\begingroup$ There are two points of view : The point $\: \mathrm P\:$ is rotated clockwise and the coordinate system remains fixed or the coordinate system is rotated anticlockwise and the point $\: \mathrm P\:$ remains fixed. $\endgroup$
    – Frobenius
    Commented Apr 21, 2018 at 6:36
  • $\begingroup$ Yea I was just realizing that. I think what the book does is rotate the coordinate system counter-clockwise but then it measures lengths in the new coordinate system. Meaning that they effectively changed basis while the matrix I was computing was in the old basis. Is it correct to think of it in this way? $\endgroup$
    – Elvis
    Commented Apr 21, 2018 at 6:39
  • $\begingroup$ Yes, precisely. $\endgroup$
    – Frobenius
    Commented Apr 21, 2018 at 6:41
  • $\begingroup$ In two dimensions you could solve your problems with complex numbers. Suppose you want to rotate the point $\:\mathrm P : z=x+iy\:$ by an angle $\:\theta\:$ to a new position $\: \mathrm P' : z'=x'+iy'\:$. Don't bother yourself if clock- or anticlock- wise, the sign of $\:\theta\:$ cares for this. \begin{align} z'=e^{i\theta}z\: \Longrightarrow \:x'+iy' & =\left(\cos\theta+i\sin\theta\right)\left(x+iy\right) \tag{01}\\ &=\left(x\cos\theta-y\sin\theta \right)+i\left(x\sin\theta+y\cos\theta \right) \nonumber \end{align} $\endgroup$
    – Frobenius
    Commented Apr 21, 2018 at 7:15
  • $\begingroup$ ...so \begin{equation} \begin{bmatrix} x'\\ y' \end{bmatrix} = \begin{bmatrix} \cos\theta & -\sin\theta\\ \sin\theta & \hphantom{-}\cos\theta \end{bmatrix} \begin{bmatrix} x\\ y \end{bmatrix} \tag{02} \end{equation} $\endgroup$
    – Frobenius
    Commented Apr 21, 2018 at 7:16

1 Answer 1

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You can construct the rotation matrix by finding the direction cosines defined as $\lambda_{i,j}=\cos(x'_{i},x_{j})$ so:

$\lambda_{11}=\cos(x'_{1},x_{1})=\cos\theta$

$\lambda_{12}=\cos(x'_{1},x_{2})=\cos(\frac{\pi}{2}-\theta)=\sin\theta$

$\lambda_{21}=\cos(x'_{2},x_{1})=\cos(\frac{\pi}{2}+\theta)=-\sin\theta$

$\lambda_{22}=\cos(x'_{2},x_{2})=\cos\theta$

The book is right.

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    $\begingroup$ Sorry but your post doesn't explain anything. My concern is that the book presents a rotation matrix that seems to represent clockwise rotation while what they illustrate is counter-clockwise rotation; is this incorrect? $\endgroup$
    – Elvis
    Commented Apr 21, 2018 at 4:57
  • $\begingroup$ Oh, the problem the book represents a rotation from $(x'_{1},x'_{2})$ system to $(x_{1},x_{2})$ system. $\endgroup$ Commented Apr 21, 2018 at 5:02
  • $\begingroup$ Note that your rotation matrix, which is the inverse of book's matrix, represents the rotation from $(x_{1},x_{2})$ to $(x'_{1},x'_{2})$, maybe that is your confusion. $\endgroup$ Commented Apr 21, 2018 at 5:11
  • $\begingroup$ Correct, I'm going from $(x_{1}, x_{2})$ to $(x'_{1}, x'_{2})$ so shouldn't that transformation be $\begin{pmatrix} x'_{1} \\ x'_{2} \\ \end{pmatrix} = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \\ \end{pmatrix} \begin{pmatrix} x_{1} \\ x_{2} \\ \end{pmatrix} $? $\endgroup$
    – Elvis
    Commented Apr 21, 2018 at 5:44
  • $\begingroup$ To do what the book is doing wouldn't it be $\begin{pmatrix} x_{1} \\ x_{2} \\ \end{pmatrix} = \begin{pmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \\ \end{pmatrix} \begin{pmatrix} x'_{1} \\ x'_{2} \\ \end{pmatrix} $? $\endgroup$
    – Elvis
    Commented Apr 21, 2018 at 5:50

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