- #1
Gabriel Maia
- 72
- 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.
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.