Wikipedia states that:
Any extrinsic rotation is equivalent to an intrinsic rotation by the same angles but with inverted order of elemental rotations, and vice-versa. For instance, the intrinsic rotations $x-y’-z''$ by angles $\alpha, \beta,\gamma$ are equivalent to the extrinsic rotations $z-y-x$ by angles $\gamma, \beta, \alpha$.
Is there a simple proof why this is so?
$\endgroup$ 32 Answers
$\begingroup$I googled on the Internet, but also found not much material on this topic. The most relevant references are from Wikipedia as @Apoo suggested, and a blog the proof is at bottom of the page. Though the arguments are quite complete, neither of them could convince me. In order to prove the statement, I have to first introduce changing a transformation between coordinate systems. There is a derivation from blender stackexchange, and I excerpt the equation as follows:$$ T_{\text {world}}=S_{\text {world}} \times T_{s} \times S_{\text {world}}^{-1}, \tag{1} $$where $T_{\text {world}}$ is the transformation matrix in world coordinate system,$S_{\text {world}}$ is the world matrix of the referenced object $s$, $T_{s}$ is the local transformation matrix based on $s $.
Assume that an intrinsic Euler angle can be represented as a product of three rotation matrices$$ R = Z^{\prime \prime}(\gamma) Y^{\prime}(\beta) X(\alpha). $$
The goal is to prove, there exists an extrinsic rotation sequence, s.t.
$$ Z^{\prime \prime}(\gamma) Y^{\prime}(\beta) X(\alpha) = X(\alpha)Y(\beta)Z(\gamma). $$
Consider $Y^\prime(\beta)$ as a rotation matrix around $y^\prime$ of angle $\beta$ relative to $x^\prime - y^\prime - z^\prime$ coordinate system, we can obtain the corresponding rotation matrix in $x-y-z$ coordinate system using conversion eq. $(1)$:$$ Y^\prime = X Y X^{-1}, $$it follows that$$ Y = X^{-1} Y^\prime X . \tag{2} $$Through a similar argument, we can also get the world rotation matrix of angle $\gamma$ around $z^{\prime \prime}$ in $x-y-z$ coordinate system:$$ Z^{\prime \prime} = (Y^\prime X) Z (Y^\prime X)^{-1}, $$and it follows that$$ Z = (Y^\prime X)^{-1} Z^{\prime \prime} (Y^\prime X). \tag{3} $$
Multiplying $X$ by eq. $(2)$ and eq. $(3)$, we obtain$$ \begin{align} X Y Z &= X X^{-1} Y^\prime X (Y^\prime X)^{-1} Z^{\prime \prime} (Y^\prime X) \\ &= Z^{\prime \prime} Y^{\prime} X , \end{align} $$which completes the proof.
$\endgroup$ 2 $\begingroup$The statement is valid not only for rotations. According to relativity,
there is no preferred frame of reference / coordinate system.
Therefore, in kinematics, all (times and) positions and motions are relative. (Dynamics - with forces - is another matter).
Consider a two-dimensional example, as has been found in a book about
Computer Graphics (to be precise: J.D. Foley, A. van Dam, Fundamentals
of Interactive Computer Graphics, 1982). There are two frames of reference,
one attached to the observer (world), one attached to an object (chair):
With a transformation of coordinates, the only thing that is important is therelative position of the object with respect ot the observer. This means that the end result of a coordinate transformation can be achieved in at least two ways. As is displayed in the example:
- Extrinsic.
Rotate $\,R\,$ the chair in the world coordinate system over an angle of $45^o$
and then Translate $\,T\,$ it over a distance $(4,10)$. Thus resulting in a transformation $\,TR$ . - Intrinsic. Translate $\,T^{-1}\,$ the observer in the chair coordinate system over a distance $(-4,-10)$ and then rotate $\,R^{-1}\,$ the world coordinate system over an angle $-45^o$. Thus resulting in a transformation $\,R^{-1}T^{-1}$ .
Hope you get the idea. Generalizing this to three dimensions is expected to be a matter of filling in the (somewhat more involved) technicalities.
Update. Hmm, "not quite a proof". Then perhaps this.
Let the coordinate system of the object be called $O$ and the world coordinate
system be called $W$. Both coordinate systems are coincident in the beginning.
The first step is to apply a transformation $R$ to $O$
(just as in the example, but in general now).
The second step is to make $W$
coincident again with $O$, which is done by applying the same transformation
$R$ to $W$ as has been done in the first step with $O$.
Then effectively nothing
has changed and we have the original configuration again: the product of step (1)
and step (2) is the identity.
It is thus obvious that
the first step could also have been accomplished by applying the inverse $R^{-1}$
transformation to $W$ instead of $O$.
Common
properties of inverse operations like $(AB)^{-1}= B^{-1}A^{-1}$ are assumed throughout to be well known. This completes the proof.
