Sign up ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

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?

share|cite|improve this question
The three matrices you give are under space-fixed axes. – Siyuan Ren Aug 10 '12 at 5:22
But those are the three matrices everyone uses to construct $R(\alpha,\beta,\gamma)$. The first rotation has space- and body-fixed axes coincident. The presumption is that the 2nd and 3rd rotations (R to L, of course), are rotating around temporarily-fixed body axes; if so, the form given is correct. – gilonik Aug 10 '12 at 15:06

2 Answers 2

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.

share|cite|improve this answer
Actually, it's even easier than that. Forget all about active/passive; just pick a convention and stick with it. The secret is $$ \vec{r}^\prime=R\vec{r} = R\left(\hat{e}_jr_j\right)= \hat{e}_i R_{ij} r_j ~. $$ So components transform by (the usual) left multiplication, the basis by right multiplication. When texts construct the Euler rotations $R(\alpha,\beta,\gamma)=R_3(\alpha)R_2(\beta)R_3(\gamma)$ they're implicitly right-multiplying. And the order is correct. Interpreted as a left multiplication, the matrix acts on the column of components. The rest follows. – GeekyCool Aug 11 '12 at 2:01

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 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

share|cite|improve this answer
Dear 1946dodge: For your information, Physics.SE has a policy that it is OK to cite oneself, but it should be stated clearly and explicitly in the answer itself, not in attached links. – Qmechanic Dec 3 '14 at 0:38

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.