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In the book I am studying the standard matrices for counterclockwise rotation around the $x$-axis through an angle $\alpha$ is:

$R_x = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos(\alpha) & -\sin(\alpha)\\ 0 & \sin(\alpha) & cos(\alpha) \end{bmatrix}$

I hope I can explain my question without using very much graphics:
For example, as an explanation for why $R_x$ looks like it does, we consider
$R_x$v.
The first column in $R_x$ makes sure $v_1$ stays the same throught the matrix transformation (rotation), this I understand. However, as for the explanation of the other two columns, the book says that one can view it as the 2-D case, where the unit circle is used to derive the coordnates for the standard matrix.
But this I can only imagine will work if you know how the coordinate system looks like, because the directions of the $y$- and $z$-axes can be interchanged. If we interchange the directions of the $y$- and $z$-axes, the points that the rotation travels between will still be the same, but the position of these points will be different relative to the $y$- and $z$-axes, if we interchange the directions of the $y$- and $z$-axes.

Therefore, I imagine that the rotation matrix around the $x$-axis could look in two different ways. One way would be as $R_x$ above, but also like:

$R^{´}_x = \begin{bmatrix} 1 & 0 & 0 \\ 0 & -\sin(\alpha) & \cos(\alpha)\\ 0 & cos(\alpha) & \sin(\alpha) \end{bmatrix}$

And I feel like this makes sense since the resulting matrix will still be the same. But I also feel like then the standard matrix is not very "standard", so am I doing something wrong here?

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  • $\begingroup$ One trick is the definition of counterclockwise. If you look from one end at a spinning cylinder, it might be spinning clockwise, look at it from the other end, and it is spinning counterclockwise. So, from whose vantage is the spin "counterclockwise?" In the case of spinning around the $x$-axis, I assume it is the vantage point of the point $(1,0,0)$. $\endgroup$ – Thomas Andrews Oct 7 '15 at 13:02
  • $\begingroup$ So basically, "clockwise" doesn't have a definition without the coordinate system. $\endgroup$ – Thomas Andrews Oct 7 '15 at 13:04
  • $\begingroup$ @ThomasAndrews Yes you are right, the rotation is viewed as if you where sitting on the x-axis with the vector pointing towards you. $\endgroup$ – hampadampadoo Oct 7 '15 at 13:05
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Consider that when $\alpha =0$, you should get the identity matrix. That should at least convince you that your matrix is wrong, if not why it is wrong.

Swapping the order of $y$ and $z$ does get you a slightly different matrix:

$$R_x'(\alpha) = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos(\alpha) & \sin(\alpha)\\ 0 &-\sin(\alpha) & cos(\alpha) \end{bmatrix}=R_x(-\alpha)$$

That is, if we swap $y$ and $z$, what was counterclockwise is now clockwise rotation by $\alpha$, and visa versa.

The above definition essentially defines clockwise as well as rotation. As you say, it is impossible to define clockwise around the $x$-axis unless you know whether the rotation of $\pi/4$ clockwise sends $(0,1,0)$ to $(0,0,1)$ or visa versa. So the above definition is giving you that. Alternatively, it is assuming a picture of the axes that makes it clear.

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  • $\begingroup$ I see now that my matrix is not the right one, I was thinking in terms of multiplication of the coordinates rather than addition for some strange reason. What I do not understand though, is how the definition of "counterclockwise" can change if we swap the $y$- and $z$-axes, because I feel like the definition should depend solely on our vantage point, and the vantage point is the same regardless of the directions of the $y$- and $z$-axes ? $\endgroup$ – hampadampadoo Oct 7 '15 at 13:16
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    $\begingroup$ Just consider a rotation in the plene, if we switch $x$ and $y$. A clockwise rotation in one frame is counterclockwise in another. It is as if we are looking at the plane "from the other side." $\endgroup$ – Thomas Andrews Oct 7 '15 at 13:20
  • $\begingroup$ Oh I think I get it now, the definition of counterclockwise is dependent on the basis, If we do not define it geometrically as "in that direction". Thank you so much for your help and your patience! $\endgroup$ – hampadampadoo Oct 7 '15 at 13:21
  • $\begingroup$ It's a fairly standard confusion, and one books don't often do a good job of clarifying. When the author is saying "counter-clockwise," he probably is talking in reference to a particular picture of the $x,y,z$-axes. $\endgroup$ – Thomas Andrews Oct 7 '15 at 13:25
  • $\begingroup$ @hampadampadoo: If you swap the $y-$ and $z-$axes you are essentially replacing the old set of coordinate axes by its mirror image. Take a look at the second hand of an analog clock in a mirror - you'll see that it appears to be moving anticlockwise. $\endgroup$ – PM 2Ring Oct 7 '15 at 13:26

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