This question comes from my own thinking, so I want to check with others to make sure I'm not led astray.
Let $R(t): [0, \infty)\rightarrow\textrm{SO}(3)$ be a differentiable path starting at $R(0) = I$. This represents some continuous rotation on $\mathbb{R}^{3}$. Suppose that at every $t\in [0, \infty)$ we are given the angular velocity vector $\vec{\omega}$ that describes the instantaneous rotation that is taking place via the "right-hand rule."
To elaborate what I mean by the "right-hand rule," I can give an example: if $\vec{\omega} = c\vec{e}_{z}$ for all $t$, then we have a rotation about the $z$-axis that is counterclockwise from the "top view looking down on the $xy$-plane" with angular frequency $c$.
If $\vec{\omega}(t):[0, \infty)\rightarrow\mathbb{R}^{3}$ is all we're given, how can I find (an expression for) $R(t)$ at any $t$?
My understanding is as follows. First, given that $\vec{\omega}$ is a pseudovector we convert
$$ \vec{\omega}(t) = \begin{pmatrix} \omega_{1}(t) \\ \omega_{2}(t) \\ \omega_{3}(t) \end{pmatrix}
\rightarrow
\widetilde{\omega}(t) = \begin{pmatrix} 0 & -\omega_{3}(t) & \omega_{2}(t) \\
\omega_{3}(t) & 0 & -\omega_{1}(t) \\
-\omega_{2}(t) & \omega_{1}(t) & 0 \end{pmatrix} $$
I take this as a lie algebra element, so the resulting integrated rotation matrix should be
$$ R(t) = e^{\textstyle \int_{0}^{t} \widetilde{\omega}(\tau)\, d\tau} \in\textrm{SO}(3). $$
Can anyone confirm or deny this formula is correct?
Best Answer
The idea does not work, since the elements of the Lie algebra at different times do not commute. the exponential sum is defined in powers of the integral, that are not time ordered.
One needs a time ordered exponential integral that satisfies $$e^{\int_0^{t+dt} \Omega(\tau)\, d\tau} = e^{\int_t^{t+dt} \Omega(\tau)\, d\tau }*e^{\int_0^{t} \Omega(\tau)\, d\tau } \approx (1 + d\tau \Omega(t) )\ * \ e^{\int_0^{t} \Omega(\tau)\, d\tau } $$ and a differential operation, that cuts off the leftmost factor of the last time. The usual way to define the time ordered exponential is, to use the classical product formula with
$$:e^{\int_0^t \Omega(\tau)\, d\tau} := \prod_{n=0}^{t/dt} (1+ dt\ \Omega(n\ dt))$$
Algebraically transparent is the normal order, that is to sort the time ordered product into a Lie-algebra basis ordered product by commuting the basis matrices in each monomial set of permutations according to their indices. Only by normal order identities or approximations can be demonstrated.