Changing the order of two rotations around perpendicular axes

Question

From here I take that the rotation around one vector $u$ can be decomposed into two rotations around $v$ and $w$, i.e. $$ R^u_\alpha = R^v_\beta R^w_\gamma $$ but then I can obviously also decompose the rotation as $$ R^u_\alpha = R^w_{\gamma'} R^v_{\beta'} $$ where, in general, $\beta \not = \beta'$ and $\gamma \not = \gamma'$.

Thus $$ R^w_{\gamma'} R^v_{\beta'} = R^v_\beta R^w_\gamma $$ Now apply this rotation to a vector which happens to be $w$, then \begin{align} R^w_{\gamma'} R^v_{\beta'}w & = R^v_\beta R^w_\gamma w \\ &=R^v_\beta w \end{align} since rotating $w$ around $w$ isn't doing anything.

However, this result seems very counter intuitive (if not wrong) to me. It seems like the "single rotation" can not express all the rotations that the "double rotation" can.

Where is the error in my argument or am I fooled by my intuition?

Answer 1

You are bothered by $$R^{w}_{\gamma'}R^{v}_{\beta'}w=R^{v}_{\beta}w$$ because it looks like there are two degrees of freedom on the left, and only one on the right. But $\gamma'$ and $\beta'$ came together as a package. You don't have some freedom to choose $\gamma'$ and $\beta'$ separately. With $w$ fixed, a choice of $v$ induces a value for $\beta$, and also a choice of $(\gamma',\beta')$ as a package deal.

So I think it is your intuition misleading you with that objection. There isn't anything wrong with the conclusion.

An observation about comments elsewhere here: people are confusing the equality of two rotations $$R^w_{\gamma'} R^v_{\beta'} =R^v_{\beta}$$ with the equality of two vectors $$R^w_{\gamma'} R^v_{\beta'}w =R^v_{\beta}w$$ Just because $R^w_{\gamma'} R^v_{\beta'}w =R^v_{\beta}w$ that does not mean the rotations $R^w_{\gamma'} R^v_{\beta'}$ and $R^v_{\beta}$ are equal. People are making that conclusion and that is leading them to think the premise must be wrong. But they shouldn't be inferring that the rotations are equal.

Answer 2

The following counterexample is possible.

Consider the standard coordinate plane $xOy$ in 3-dimensional rectangle coordinates. As the first rotation, consider the rotation about the abscissa axis 90 degrees clockwise, the second as a 180-degree rotation around the y-axis (equivalent to symmetric reflection with respect to the $O$ point). The starting point is placed inside the first quadrant.

Easy to see that the original sequence of rotations raises a point above the plane $ xoy $, and the inverse sequence - omits.

Answer 3

Consider the problem of constructing the one radian rotation about the $z$-axis from a combination of rotations about the $x$-axis and $y$-axis respectively. By noticing that the rotation we are trying to construct fixes the $z$-axis we know that no non-trivial rotation about the $x$-axis axis which fixes the $z$-axis and likewise with the $y$-axis. This means we need two rotations, one about each axis, that are both non-trivial and fix the $z$-axis under composition.

Now if we follow the point $(0,0,1)$ on the unit sphere under some non-trivial rotation about the $x$-axis it will lie on some other point on the unit circle in the $y$-$z$ plane, call it $P$. Now if we look at the rotations about the $y$-axis now we can see that $P$ traces out a circle parallel to the $x$-$z$ plane which will be trivial if and only if $P$ is on the $y$-axis and will be in the $x$-$z$ plane through the origin if and only if we rotate by exactly a half or trivially However we now see that since $(0,0,1)$ is in the $x$-$y$ plane at the origin that no rotation about the $y$-axis can take $P$ back to $(0,0,1)$ unless the rotation about the $x$-axis is trivial or exactly a half turn and so the only rotation about the $y$-axis that can fix $(0,0,1)$ are the trivial rotations and the half rotations by $x$ and $y$. In either case we can't produce the rotation of one radian about the $z$-axis so we must need an additional rotation.

Note however if you allow me to do an $x$-axis rotation, then a $y$-axis rotation then undo the original $x$-axis rotation I can make an arbitrary rotation about the $z$-axis by rotating all the work to an orthogonal plane then rotating back afterwards.

Answer 4

[Disclaimer: I deleted my previous answer because Steven Stadnicki's and WRSomsky's comments made me realize that it had some wrong statements, I hope this one is more rigorous]

TL;DR: I think you're being fooled by your intuition. I agree with the core point of alex.jordan's answer, I just went on rambling about the details of rotation decomposition.

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First of all, you cannot decompose a given $R_\alpha^\mathbf{u}$ into two rotations around arbitrary axes $\mathbf{v}$ and $\mathbf{w}$, as other users have noticed, because usually you don't have enough parameters (2 rotations) to cover all degrees of freedom (3 - dimension of the rotation group $\small SO(3)$).

A decomposition like the one in your question could still happen, but it would be a special case of the general one I'm gonna deal with.

Rotation decomposition

Given a fixed 3D coordinate system $\mathbf{xyz}$ and a rotation $R_\theta^\mathbf{u}$, you can decompose it into three extrinsic rotations (around the axes $\mathbf{xyz}$). For example, using Tait-Bryan angles (but there are other choices) you get $$R_\theta^\mathbf{u} \,p=R_\alpha^\mathbf{x}R_\beta^\mathbf{y}R_\gamma^\mathbf{z} \,p \qquad\forall p\in\mathbb{R}^3$$

Now, if you want, you can set $R_\phi^\mathbf{v}=R_\alpha^\mathbf{x}R_\beta^\mathbf{y}$ and $\mathbf{w}=\mathbf{z}$, and get back to the problem you proposed $$R_\theta^\mathbf{u}=R_\phi^\mathbf{v}R_\gamma^\mathbf{w}$$ but, please, notice that you cannot choose axis $\mathbf{v}$ a priori. This means that not every couple $(\mathbf{v},\mathbf{w})$ is suitable for decomposing $R_\theta^\mathbf{u}$. Moreover, even if $(\mathbf{v},\mathbf{w})$ is a valid choice for your decomposition, it doesn't mean that $(\mathbf{w},\mathbf{v})$ will be!

Anyway, let's go back where we left.

You can rearrange the order of the axes (any permutation is valid) and obtain another decomposition, let's say $$R_\theta^\mathbf{u} \,p=R_{\gamma'}^\mathbf{z}R_{\beta'}^\mathbf{y}R_{\alpha'}^\mathbf{x} \,p \qquad\forall p\in\mathbb{R}^3$$ where, of course, new angles $(\alpha',\beta',\gamma')$ usually differ from previous ones $(\alpha,\beta,\gamma)$.

Every rotation $R_\theta^\mathbf{u}$ has a set of fixed points, which is the axis of rotation $\mathbf{u}$. So, you're right when you say $$R_\theta^\mathbf{u} \,p = p \qquad \forall p\in\mathbf{u}$$

and, if you decompose $R_\theta^\mathbf{u}$, the same holds true for $R_\gamma^\mathbf{z}$ hence $$R_\theta^\mathbf{u} \,p=R_\alpha^\mathbf{x}R_\beta^\mathbf{y}R_\gamma^\mathbf{z} \,p=R_\alpha^\mathbf{x}R_\beta^\mathbf{y} \,p \qquad\forall p\in\mathbf{z}$$

We seem to have reached a situation similar to the one you proposed, indeed $$R_{\gamma'}^\mathbf{z}R_{\beta'}^\mathbf{y}R_{\alpha'}^\mathbf{x} \,p = R_\alpha^\mathbf{x}R_\beta^\mathbf{y} \,p \qquad\forall p\in\mathbf{z}$$

Wrong or counter-intuitive?

Notice that, while the choice of the coordinate system $\mathbf{xyz}$ is arbitrary (Euler's rotation theorem states that any two Cartesian coordinate systems with common origin are equivalent up to a rotation), the angles of rotation $\alpha$, $\beta$ and $\gamma$ are not (the same applies to $\alpha'$, $\beta'$ and $\gamma'$). So for every rotation $R_\alpha^\mathbf{u}$ of your choice, you get three precise rotations $R_\alpha^\mathbf{x}$, $R_\beta^\mathbf{y}$ and $R_\gamma^\mathbf{z}$.

The converse is not true: not every triple of rotations $(R_\alpha^\mathbf{x},R_\beta^\mathbf{y},R_\gamma^\mathbf{z})$ will give you $R_\alpha^\mathbf{u}$.

And this is the point where I think your intuition fails you: what you call "double rotation" (in this case "triple rotation") is in fact just a single rotation and not three distinct (arbitrary) rotations (i.e. you don't have three degrees of freedom).

A single (or "double") rotation cannot express all the rotations, but when you write $$R_{\gamma'}^\mathbf{z}R_{\beta'}^\mathbf{y}R_{\alpha'}^\mathbf{x} \,p = R_\alpha^\mathbf{x}R_\beta^\mathbf{y} \,p \qquad\forall p\in\mathbf{z}$$ keep in mind that none of the angles are arbitrary (again, no degrees of freedom): you have just one rotation on each side (remember that Euler's theorem tells you that a composition of two or more rotations is also a rotation).