Mechanics of Continua and Structures
We know from Eq. 6 in the section Computing the Acceleration Field that
\begin{equation} \boldsymbol{\Omega}_{0}= \boldsymbol{Q}^{\textsf{T}}\dot{\boldsymbol{Q}} \end{equation}
Rearranging gives
\begin{equation} \dot{\boldsymbol{Q}} = \boldsymbol{Q}\boldsymbol{\Omega}_{0} \end{equation}
Differentiating \(\dot{\boldsymbol{Q}}\),
\begin{equation} \ddot{\boldsymbol{Q}} = \dot{\boldsymbol{Q}}\boldsymbol{\Omega}_0 + \boldsymbol{Q}\dot{\boldsymbol{\Omega}}_0 \end{equation}
\begin{equation} \ddot{\boldsymbol{Q}} = \boldsymbol{Q}\boldsymbol{\Omega}_0^2 + \boldsymbol{Q}\dot{\boldsymbol{\Omega}}_0 =\boldsymbol{Q}\left(\boldsymbol{\Omega}_0^2 + \dot{\boldsymbol{\Omega}}_0 \right) \end{equation}
\begin{equation} \boldsymbol{Q}^{\textsf{T}}\ddot{\boldsymbol{Q}} = \boldsymbol{\Omega}_0^2 +\dot{\boldsymbol{\Omega}}_0 \end{equation}
equals
\begin{equation} \boldsymbol{P} = \boldsymbol{\Omega}_0^2 + \dot{\boldsymbol{\Omega}}_0 \end{equation}
We showed previously that \begin{equation} \textsf{sym}(\boldsymbol{P}) = \boldsymbol{\Omega}_0^2 \end{equation}
so substituting that into Eq.5, we get
\begin{equation} \boldsymbol{P}=\textsf{sym}(\boldsymbol{P}) +\dot{\boldsymbol{\Omega}}_0 \end{equation}
\begin{equation} \dot{\boldsymbol{\Omega}}_{0} = P-\textsf{sym}(\boldsymbol{P}) = \boldsymbol{P}- \left(\frac{\boldsymbol{P}+\boldsymbol{P}^{\textsf{T}}}{2}\right) = \left(\frac{\boldsymbol{P}-\boldsymbol{P}^{\textsf{T}}}{2}\right) = \textsf{skew}(\boldsymbol{P}) \end{equation}
where \(\textsf{skew}(\boldsymbol{P})\) is the skew-symmetric tensor of \(\boldsymbol{P}\).
Now that \begin{equation} \dot{\boldsymbol{\Omega}}_{0} = \textsf{skew}(\boldsymbol{P}) \end{equation},
integrating both sides will result in
\[\begin{align} \boldsymbol{\Omega}_{0}(t)=\int_{z=0}^{t}\textsf{skew}(\boldsymbol{P})(z)dz, \\ + \boldsymbol{\Omega}_{0}(0) \end{align}\]