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Consider a single particle moving around the origin in a circle. If the particle's coordinates are $(x,y)$ at some time, it will reach $(x,-y)$ after some time as they lie on the circle.

The position vector of the particle is rotating about the origin. Now rotation of a vector by an angle $Z$ clockwise is equivalent is to rotating the coordinate axes anticlockwise by $Z$.

The components in the new axes rotated anticlockwise are given by : $$ x' = x\cos(Z) + y\sin(Z)$$ $$ y' = -x\sin(Z) + y\cos(Z)$$

These are also the component transformations if I rotate the vector clockwise and keep my axes the same.

The question is : By rotation of the axes , we cannot transform the vector $(x,y)$ into $(x,-y)$ since there is no angle $Z$ that satisfies the equations for this pair ( x and y not 0 ). But on rotating the vector we get $(x,-y)$. This seems to be a paradox. What am I missing here?

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  • $\begingroup$ Whatever $+Z$ works with one case, $-Z$ should also work on the other. A rotation about an angle $+Z$ is equivalent to a coordinate transformation about an angle $-Z$. $\endgroup$ Nov 8, 2017 at 20:39
  • $\begingroup$ $Z=\pi$ doesn't return (0,-1) for input (0,1)? $\endgroup$
    – Kyle Kanos
    Nov 8, 2017 at 20:39
  • $\begingroup$ @KyleKanos it doesnt return (2,-3) for (2,3) $\endgroup$
    – cobra121
    Nov 8, 2017 at 20:41
  • $\begingroup$ Try $Z\sim225.25^\circ$ for that case. $\endgroup$
    – Kyle Kanos
    Nov 8, 2017 at 20:46
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    $\begingroup$ I don't really see how you guys are getting a passive rotation that maps $(x,y)\rightarrow (x,-y)$. This is a reflection of the $y$ axis about the $x$ axis and therefore is not a rotation. $\endgroup$
    – Diracology
    Nov 8, 2017 at 21:29

1 Answer 1

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Indeed you will not get the transformation $(x,y)\rightarrow(x,-y)$ by a rotation of the coordinate system (passive rotation) since this transformation actually corresponds to a reflection, namely $$S= \begin{pmatrix} 1&0\\0&-1 \end{pmatrix},\tag1$$ which does not belong to the group of rotations in the plane, $SO(2)$.

On the other hand, by an active rotation $R(Z)$, $$R(Z)= \begin{pmatrix} \cos Z&-\sin Z\\ \sin Z&\cos Z \end{pmatrix},\tag2 $$ of a definite angle $Z$ you definitely can obtain the transformation $(x,y)\rightarrow(x,-y)$ for some definite vector $(x,y)$. The paradox you mentioned could be restated as following: How can a reflection matrix (in the passive view) be identified with a rotation matrix (in the active view)? There is no paradox though, since (1) and (2) are not supposed to be identified as operators. For example, both matrices $$S= \begin{pmatrix} 1&0\\0&-1 \end{pmatrix}, \quad\mathrm{and}\quad R(3\pi/2)= \begin{pmatrix} 0&1\\ -1&0 \end{pmatrix} $$ take the vector $(1/\sqrt 2,1/\sqrt 2)$ to $(1/\sqrt 2,-1/\sqrt 2)$, but $S$ takes $(0,1)$ to $(0,-1)$ whereas $R$ maps $(0,1)$ to $(1,0)$. The explicit representation of the operators above already shows they are different.

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