Euler angle: space-fixed vs body-fixed axes

Question

I am sooo confused!! Between active and passive, intrinsic and extrinsic, vectors and basis ....

Stipulate that we stick to active rotations only. Then

Standard derivation of $R(\alpha, \beta,\gamma)=R_{z^{\prime\prime}}(\gamma)R_{y^\prime}(\beta)R_{z}(\alpha)$ uses intermediate frame $(x^\prime,y^\prime,z^\prime)$ in transformation from space-fixed axes $(x,y,z)$ to the body-fixed axes $(x^{\prime\prime},y^{\prime\prime},z^{\prime\prime})$ to derive $$ R(\alpha, \beta,\gamma) = \left(\begin{array}{ccc} ~~\cos{\gamma}&-\sin{\gamma} & 0 \\ \sin{\gamma}&\cos{\gamma}& 0 \\ 0 & 0& 1\end{array}\right) \left(\begin{array}{ccc} \cos{\beta} & 0 &\sin{\beta} \\ 0 &1& 0 \\ -\sin{\beta}& 0&~~\cos{\beta} \end{array}\right) \left(\begin{array}{ccc} ~~\cos{\alpha}&-\sin{\alpha} & 0 \\ \sin{\alpha}&\cos{\alpha}& 0 \\ 0 & 0& 1\end{array}\right) $$ But when rewriting in terms of spaced-fixed axes (Sakurai pg 172, e.g.), fairly straightforward arguments (mathematically, just similarity transformations), take us to $R(\alpha, \beta,\gamma)=R_z(\alpha)R_y(\beta)R_z(\gamma)$. But this does NOT multiply out as the same matrix -- despite the use of = everywhere! So I figured the former applies to the basis, the latter the vector components (since they transform inversely to one another). But the results are not transposes of one another. And even so, what of their purported equality?

As you can see, I'm really tied in knots!! Anyone have a sword?

Answer 1

Well, here's my epee:

The problem is that basis vectors transform oppositely to a vector's components. Sticking to an active approach — wherein there is only one basis $\{\hat{e}_i\}$, and starting with the well-worn $r^\prime_i=R_{ij}r_j$, $$ \vec{r}^\prime=R\vec{r} = R\left(\hat{e}_jr_j\right)= \hat{e}_i R_{ij} r_j ~, $$ we see that a column of basis vectors transforms as $R^T(\varphi)=R(-\varphi)$. (If you want, call this passive — but you stipulated active transformations only.)

Thus, when an active-convention text write $R(\alpha,\beta,\gamma)=R_3(\alpha)R_2(\beta)R_3(\gamma)$, since it's to be applied to the basis, they actually MEAN $R_i$'s of negative angles. Moreover, the canonical form of the individual rotations that you give is correct, because the elements of an operator are basis dependent, and each individual $R$ changes the basis. So the $R(\alpha,\beta,\gamma)$ above is expressed in a mixture of bases. That may be fine if you're only going to be acting on bases (passive), but an active approach demands expressing this operator in a single (and fixed) basis.

Now the relationship between $R(\alpha,\beta,\gamma)$ in the mixed basis and $R(\alpha,\beta,\gamma)$ in the space-fixed basis is straightforward — the very first equation, above, says they're related by transposition. So: \begin{eqnarray*} R(-\alpha,-\beta,-\gamma)^T &=& \left[R_3(-\alpha)R_2(-\beta)R_3(-\gamma)\right]^T \\ &=& R_3^T(-\gamma)R_2^T(-\beta)R_3^T(-\alpha) \\ &=& R_3(\gamma)R_2(\beta)R_3(\alpha)~. \end{eqnarray*} The operator is the same and has the same physical effect; the matrix elements differ because of the choice of bases.

Answer 2

What this refers to is the Rotation Reversal Theorem - rotating first about axis z with angle az , and then about the rotated y axis by angle ay , followed by rotation by the now twice rotated z axis by angle bz is the same as rotating first about the original z axis by bz, followed by rotation about the ORIGINAL y axis by ay and then finally about the ORIGINAL z axis by az. This remarkable theorem works for any number of rotations and for other axis sequences than the Euler angles. There is a reference on https://www.researchgate.net/profile/Edward_Barile called Rotation Dyads and Coordinate Transformations for Moving Radar Platforms. It also references some books as well Shuh, Jung Yang "Advanced Dynamics" and E Neal Moore "Theoretical Mechanics" The Rotation Sequence Theorem is treated in my reference on pg 14 but my notation is not to everyone's liking so the other references may be better