Angular displacement cannot be adequately represented by a vector because rotations in 3 dimensions do not commute. However, one can easily define an angular velocity vector $\overrightarrow{\omega}$, and one could integrate the angular velocity vector with respect to time. What does $\int{\overrightarrow{\omega}dt}$ represent?
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Preliminaries and a few definitions
If we have a position vector that is going through a pure rotation relative to the origin, like in the case of any point on a rigid body in its body-fixed frame, we can express its value at time $t$, $\vec{r}(t)$, by a time dependent rotation matrix $\mathbf{R}(t)$ that acts on the initial position vector: \begin{equation*} \mathbf{R}(t)\vec{r}(0)=\vec{r}(t) \tag{1} \end{equation*} Differentiating with respect to time we find: $$ \dot{\mathbf{R}}(t)\vec{r}(0) = \dot{\vec{r}}(t)$$ Now $\mathbf{R}$ is an orthogonal matrix, so acting with the inverse/transpose on $\vec{r}(t)$ we can rewrite the left hand side as: $$ \dot{\mathbf{R}}(t)\mathbf{R}^{T}(t)\vec{r}(t) = \dot{\vec{r}}(t)$$This is a standard result that I include here mainly for completeness: the time derivative of a vector going through pure rotation is given by an application of $\dot{\mathbf{R}}\mathbf{R}^T$ to the vector at the same time $t$, where $\mathbf{R}$ is the rotation matrix as given in $(1)$.
Furthermore, we observe the following property of this matrix product, since $ \mathbf{RR}^T = \mathbf{I} $ we have: \begin{align*} \frac{\mathrm{d}}{\mathrm{d}t}(\mathbf{RR}^T) &= \dot{\mathbf{R}}\mathbf{R}^T + \mathbf{R}\dot{\mathbf{R}}^T \\ &= \dot{\mathbf{R}}\mathbf{R}^T + \left(\dot{\mathbf{R}}\mathbf{R}^T\right)^T \\ &= 0 \end{align*}
This tells us that the matrix $\dot{\mathbf{R}}\mathbf{R}^T$ is antisymmetric.
We give this antisymmetric matrix its traditional name then, define:
\begin{equation*} \mathbf{\Omega}(t) \equiv \dot{\mathbf{R}}(t)\mathbf{R}^T(t) \tag{2}\end{equation*}
Because of the antisymmetry, and the fact that we have a cross product in $\mathbb{R}^3$ we can write, via some unique (pseudo-)vector $\vec{\omega}$:
\begin{equation*} \mathbf{\Omega}\vec{r} = \vec{\mathbf{\omega}}\times\vec{r} \tag{3} \end{equation*}
So $\mathbf{\Omega}$ is the angular velocity matrix, which is completely equivalent to $\vec{\omega}$. This will be important in what follows.
Heuristic handwaving, and an example to feel better about it...
Staring at equations $(2)$ and $(3)$ a bit and thinking, how can they help us to integrate $\vec{\omega}$ (and indeed, what would this integration physically mean anyway), we may stumble upon the following, for the moment handwavy idea:
If we have a pair of real functions $f,g:\mathbb{R}\rightarrow\mathbb{R}$ that satisfy:
\begin{equation*}\dot{f}(t) = \frac{\dot{g}(t)}{g(t)} = \dot{g}(t)g^{-1}(t) \tag{4} \end{equation*}
We would have immediately said to ourselves, that equation is easy enough to integrate with respect to time as follows:
$$ f(t) = \int_0^t \dot{g}(t')g^{-1}(t') \mathrm{d}t' = \ln(g(t))-\ln(g(0)), $$
If we also assume $g(0)=1$ we can write:
$$ g(t) = e^{f(t)}. $$
Now, what I am going to write will probably be crystal clear to those of us who have mastered the Lie theory behind $\mathrm{SO(3)}$ (I definitely have not), but for the rest of us mortals, I want to make the heuristic suggestion (that turns out to be correct) that we can apply the same idea to equation $(2)$. Remembering that $\mathbf{R}^T=\mathbf{R}^{-1}$ because rotation matrices are orthogonal, we see that $(2)$ has a form that closely resembles $(4)$, apart from $(2)$ being a matrix equation and $(4)$ an equation over the reals.
So that the heuristic suggestion is to integrate $(2)$ analogously:
\begin{equation*}\mathbf{R}(t) = \mathcal{T}\left\{\large{ e^{\int_0^t \mathbf{\Omega}(t')\mathrm{d}t'}}\right\} \tag{5} \end{equation*}
(For now, ignore the added notation $\mathcal{T}$, I'll explain it in just a bit.)
In writing $(5)$ I have taken $\mathbf{R}(0)=\mathbf{I}$, which is generally a good convention, by which the rotation matrix induces no overall rotation at the initial time $t=0$.
This is very handwavy, because these are matrices after all, not our usual real valued functions. But I can at least convince you this works for the simple case where $\mathbf{\Omega}$ is constant (and consequently, $\vec{\omega}$ is constant as well).
We'll take the case of a uniform angular rotation around the $z$ axis, so that: $$ \vec{\omega} = \omega\hat{z} $$
$$ \mathbf{\Omega} = \begin{pmatrix} 0 & -\omega & 0 \\ \omega & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} $$ This is easy to integrate: $$ \int_0^t \mathbf{\Omega} \mathrm{d}t' = \mathbf{\Omega}t $$ So that we can recover $\mathbf{R}(t)$ now by applying the rule for matrix exponentiation: \begin{align*} \mathbf{R}(t) &= e^{\mathbf{\Omega}t} = \mathbf{I}+\mathbf{\Omega}t+\frac{1}{2!}\mathbf{\Omega}^2 t^2 + \frac{1}{3!}\mathbf{\Omega}^3t^3 + \dots \\ &= \mathbf{I}+ \sum_{n=1}^{\infty} \frac{1}{n!}(\mathbf{\Omega} t)^n \end{align*}
It's not too hard to convince yourself that for $n\geq1$:
\begin{equation*} (\mathbf{\Omega}t)^n= \begin{cases} \begin{pmatrix} (\omega t)^n & 0 & 0\\ 0 & (\omega t)^n & 0\\ 0 & 0 & 0 \end{pmatrix}(-1)^{n/2}, & \text{If } n \text{ even}\\ \begin{pmatrix} 0 & (\omega t)^n & 0 \\ -(\omega t)^n & 0 & 0\\ 0 & 0 & 0 \end{pmatrix}(-1)^{(n+1)/2},& \text{If } n \text{ odd} \end{cases} \end{equation*}
So that we have, putting $n=2k$ for even $n$ and $n=2k-1$ for odd $n$, where $k>0$:
$$ \left[\mathbf{R}(t)\right]_{11} = \left[ \mathbf{I}+ \sum_{n=1}^{\infty} \frac{1}{n!}(\mathbf{\Omega} t)^n \right]_{11} = 1+\sum_{k=1}^{\infty}\frac{(-1)^{k}}{2k!}(\omega t)^{2k} = \cos(\omega t) $$ $$ \left[\mathbf{R}(t)\right]_{21} = \left[ \mathbf{I}+ \sum_{n=1}^{\infty} \frac{1}{n!}(\mathbf{\Omega} t)^n \right]_{21} = \sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{(2k-1)!}(\omega t)^{2k-1} = \sin(\omega t) $$
From the above we also immediately observe that $\mathbf{R}_{11}=\mathbf{R}_{22}$, $\mathbf{R}_{21}=-\mathbf{R}_{12}$, $\mathbf{R}_{33}=1$ and all other entries of $\mathbf{R}(t)$ vanish. Therefore, this gives us then precisely, and unsurprisingly, the rotation matrix for rotating by an angle $\omega t$ in time $t$ around the $z$ axis:
$$ \mathbf{R}(t) = \begin{pmatrix} \cos(\omega t) & -\sin(\omega t) & 0 \\ \sin(\omega t) & \cos(\omega t) & 0 \\ 0 & 0 & 1 \end{pmatrix} = \mathbf{R}_z(\omega t) $$
Enter noncommutativity...
But here we come inevitably to the following question: what happens if the angular velocity matrix $\mathbf{\Omega}$ is not constant in time? Let's attempt to "digest" this idea in baby-steps. Suppose $\mathbf{\Omega}$ is given by:
\begin{equation*} \mathbf{\Omega}(t)= \begin{cases} \mathbf{\Omega}_1, & t\in[0,t_1)\\ \mathbf{\Omega}_2, & t\in[t_1,\infty) \end{cases} \end{equation*}
Where both $\mathbf{\Omega}_1$ and $\mathbf{\Omega}_2$ are constant matrices. Going back to $(4)$ then, we can "split" the integral into two as follows: \begin{align*} \int_0^{t} \mathbf{\Omega}(t') \mathrm{d}t' &= \int_0^{t_1} \mathbf{\Omega}_1\mathrm{d}t' + \int_{t_1}^t \mathbf{\Omega}_2\mathrm{d}t' \\&= \mathbf{\Omega}_1 t_1 + \mathbf{\Omega}_2 (t-t_1) \end{align*}
Now as you may have guessed, this is where noncommutativity must make its entrance. When we exponentiate the matrix sum of the above integral, we must ensure that the order of exponents is $\large{e^{\mathbf{\Omega}_2 (t-t_1)}e^{\mathbf{\Omega}_1 t_1}}$ (for cases where $t>t_1$), so that the rotation matrix resulting from the action of constant angular velocity $\mathbf{\Omega}_1$ is applied first, and likewise the one resulting from the action of $\mathbf{\Omega}_2$ is applied second. This also explains the notation $\mathcal{T}$ introduced in $(5)$, it means we're taking a time ordered exponential, so that the matrix that gets "built up" from equation $(5)$ can only be built up by a specific ordering of matrices, given by their temporal order.
To understand that a bit more physically, recall that a rotation with constant angular velocity always implies a rotation around a fixed axis, that can be also seen as a consequence of angular momentum conservation. When the angular velocity changes, that implies a net torque was applied to the body, and that (generally) changes the axis of rotation to a new one, which can also be seen as fixed for some (arbitrarily sized) interval of time. But now, we know that rotations around different axes do not commute, and that's why the only way that makes sense would be to apply them according to their temporal order.
Now, we can compute $\mathbf{R}_1(t)$ and $\mathbf{R}_2(t)$ separately, by applying the same method as before:
$$ \mathbf{R}_1(t) = e^{\mathbf{\Omega}_1 t} = \sum_{n=0}^{\infty} \frac{1}{n!}(\mathbf{\Omega}_1 t)^n $$ $$ \mathbf{R}_2(t-t_1) = e^{\mathbf{\Omega}_2 (t-t_1)} = \sum_{n=0}^{\infty} \frac{1}{n!}(\mathbf{\Omega}_2 (t-t_1))^n $$
and then recover $\mathbf{R}(t)$ as:
\begin{equation*} \mathbf{R}(t) = \begin{cases} \mathbf{R}_1(t), & t \in [0,t_1)\\ \mathbf{R}_2(t-t_1)\mathbf{R}_1(t_1), & t \in [t_1,\infty) \end{cases} \end{equation*}
It is easy to see how this approach can be generalized, and in fact, it turns out this is formalized as a mathematical operation with the fitting name of an Ordered exponential, which is indeed defined in the context of noncommutative algebras.
Summing up
Overall, I think that to the extent to which it is possible, the above answers the question. We can integrate the angular velocity $\vec{\omega}$ in a meaningful way, not by just applying a straightforward integral like your question suggests, but by treating it as the equivalent angular velocity matrix to which we apply a time ordered integral as given in $(5)$. Equation $(5)$ gives us a "recipe" by which we can compute the rotation matrix $\mathbf{R}(t)$, that relates the initial (at $t=0$) orientation of the vector/body to the one attained at some subsequent $t$ of our choice. Its right hand side is a time ordered integral/exponential, over the different values the angular velocity matrix $\mathbf{\Omega}(t)$ takes as it varies in time.
If $\mathbf{\Omega}(t)$ changes continuously, I believe solving this integral is only amenable to numerical techniques, over "small" enough time slices $\Delta t$ in which we can take $\mathbf{\Omega}$ to be nearly constant.
Additional resources and references
Amit
- 6,024
5
you obtain the angular velocity vector $~\vec \omega~$ from the rotation matrix $~\mathbf R~$ between body system and inertial system. The rotation matrix is depending on the rotation angles .
assuming you have 3 rotation angles thus, $~\mathbf R(\phi)=\mathbf R(~\phi_1~,\phi_2~,\phi_3~)$
from here the angular velocity
$$\vec \omega=\underbrace{\mathbf J_R(\mathbf \phi)}_{3\times 3}\,\mathbf{\dot{\phi}}\tag 1$$
with equation (1)
$$ \frac{d\mathbf\phi}{dt}=[\mathbf J_R(\mathbf \phi(t))]^{-1}\,\vec \omega(t) \tag 2$$
solving the differential equations (2) for a given $~\vec \omega (t)~$ you obtain $~\mathbf\phi(t)~$
for a small angles $~\phi \ll ~$ the rotation matrix is:
$$\mathbf R=\left[ \begin {array}{ccc} 1&-\phi_{{3}}&\phi_{{2}} \\ \phi_{{3}}&1&-\phi_{{1}}\\ - \phi_{{2}}&\phi_{{1}}&1\end {array} \right] \quad\Rightarrow~ \vec\omega= \left[ \begin {array}{ccc} 1&0&-\phi_{{2}}\\ -\phi_ {{3}}&1&0\\ 0&-\phi_{{1}}&1\end {array} \right]\, \begin{bmatrix} \dot{\phi}_1 \\ \dot{\phi}_2 \\ \dot{\phi}_3 \\ \end{bmatrix}\\ $$
and with equation (2)
$$ \frac{d\mathbf\phi}{dt}=[~\mathbf I_3~]\,\vec \omega(t)\quad, \phi(t)=\int \vec\omega\, dt $$
conclusion from the integration of the angular you obtain the angular displacement only if angular displacement is small a
The angular velocity equation
$$ \left[ \begin {array}{ccc} 0&-\omega_{{3}}&\omega_{{2}} \\ \omega_{{3}}&0&-\omega_{{1}}\\ -\omega_{{2}}&\omega_{{1}}&0\end {array} \right] = \mathbf R^T\,\mathbf{\dot{R}}$$
example
$$\mathbf R= \left[ \begin {array}{ccc} \cos \left( \phi_{{3}} \right) &-\sin \left( \phi_{{3}} \right) &0\\ \sin \left( \phi_{{3 }} \right) &\cos \left( \phi_{{3}} \right) &0\\ 0&0& 1\end {array} \right] \quad,\mathbf{\dot{R}}=\dot\phi_3\,\left[ \begin {array}{ccc} -\sin \left( \phi_{{3}} \right) &-\cos \left( \phi_{{3}} \right) &0\\ \cos \left( \phi_{{3 }} \right) &-\sin \left( \phi_{{3}} \right) &0\\ 0&0 &0\end {array} \right] $$
$$\left[ \begin {array}{ccc} 0&-\omega_{{3}}&\omega_{{2}} \\ \omega_{{3}}&0&-\omega_{{1}}\\ -\omega_{{2}}&\omega_{{1}}&0\end {array} \right] = \mathbf R^T\,\mathbf{\dot{R}}=\dot\phi_3\left[ \begin {array}{ccc} 0&- \left( \cos \left( \phi_{{3}} \right) \right) ^{2}- \left( \sin \left( \phi_{{3}} \right) \right) ^{2}&0 \\\left( \sin \left( \phi_{{3}} \right) \right) ^ {2}+ \left( \cos \left( \phi_{{3}} \right) \right) ^{2}&0&0 \\ 0&0&0\end {array} \right] \\ \vec\omega=\begin{bmatrix} 0\\ 0 \\ \dot{\phi}_3 \\ \end{bmatrix} $$ in this case the contribution matrix $~\mathbf J_R=I_3 ~$
Take this rotation matrix
$$\mathbf R=\mathbf R_z(\phi_3)\,\mathbf R_y(\phi_2)\,\mathbf R_z(\phi_1)$$
you obtain
\begin{align*} &\vec\omega= \left[ \begin {array}{c} \sin \left( \phi_{{1}} \right) \dot\phi _{{2}}- \sin \left( \phi_{{2}} \right) \cos \left( \phi_{{1}} \right) \dot\phi _{ {3}}\\ \cos \left( \phi_{{1}} \right) \dot\phi _{{2}}+ \sin \left( \phi_{{2}} \right) \sin \left( \phi_{{1}} \right) \dot\phi _{ {3}}\\ \dot\phi _{{1}}+\cos \left( \phi_{{2}} \right) \dot\phi _{{3}}\end {array} \right] = \underbrace{\left[ \begin {array}{ccc} 0&\sin \left( \phi_{{1}} \right) &-\sin \left( \phi_{{2}} \right) \cos \left( \phi_{{1}} \right) \\ 0&\cos \left( \phi_{{1}} \right) &\sin \left( \phi_{{2}} \right) \sin \left( \phi_{{1}} \right) \\ 1&0&\cos \left( \phi_{{2}} \right) \end {array} \right]}_{\mathbf J_R(\phi)=\mathbf J_R(\phi_1,\phi_2,\phi_3)} \, \underbrace{\left[ \begin {array}{c} \dot\phi _{{1}}\\ \dot\phi _{{2} }\\ \dot\phi _{{3}}\end {array} \right]}_{\dot\phi} \end{align*}
from here \begin{align*} \begin{bmatrix} \dot{\phi}_1 \\ \dot{\phi}_2 \\ \dot{\phi}_3 \\ \end{bmatrix}= \underbrace{\left[ \begin {array}{ccc} {\frac {\cos \left( \phi_{{2}} \right) \cos \left( \phi_{{1}} \right) }{\sin \left( \phi_{{2}} \right) }}&-{ \frac {\cos \left( \phi_{{2}} \right) \sin \left( \phi_{{1}} \right) } {\sin \left( \phi_{{2}} \right) }}&1\\ \sin \left( \phi_{{1}} \right) &\cos \left( \phi_{{1}} \right) &0 \\ -{\frac {\cos \left( \phi_{{1}} \right) }{\sin \left( \phi_{{2}} \right) }}&{\frac {\sin \left( \phi_{{1}} \right) } {\sin \left( \phi_{{2}} \right) }}&0\end {array} \right]}_{\mathbf{[J_R]}^{-1}} \,\begin{bmatrix} \omega_1 \\ \omega_2 \\ \omega_3 \\ \end{bmatrix} \end{align*}
Eli
- 13,829