In variational mechanics, an action integral $\mathcal{S}$ is constructed from: $ \begin{aligned} \mathcal{S} = \int _{{t_{0}}} ^{t_{f}} \mathcal{L} \, dt \end{aligned} \tag{1} $ where $\mathcal{L}$ is the Lagrangian. For a constant mass system, we can write: $ \mathcal{L} = \frac{1}{2} \hat{\boldsymbol \omega} \cdot J \cdot \hat{\boldsymbol \omega} \tag{2} $ As in Manchester and Peck's paper, $\hat{\boldsymbol \omega}$ is indicates the formation of a quaternion with zero scalar part from the vector ${\boldsymbol \omega}$: $ \hat{\boldsymbol \omega} = \begin{bmatrix} 0 \\ {\boldsymbol \omega} \end{bmatrix} $ $^{\epsilon} \hat{\boldsymbol \omega}$ is the variation of the angular velocity and is given by: $ ^{\epsilon} \hat{\boldsymbol \omega} = 2 {^\epsilon q^{\dagger}} {^\epsilon {\dot{q}}} \approx \hat{\boldsymbol \omega} + \epsilon (\hat{\boldsymbol \omega}\hat{\boldsymbol \eta} - \hat{\boldsymbol \eta} \hat{\boldsymbol \omega} + 2 \dot{\hat{\boldsymbol \eta}}) \tag{3} $ ```ad-important Please see Equation ($16$) of Manchester's paper, which I believe has a typo; Equation ($2$) above is from your handwritten result. ``` Now, if Equation ($3$) is substituted into Equation ($2$), we should be getting what I think is called the varied Lagrangian $^\epsilon \mathcal{L}$. Further, if you substitute the varied Lagrangian into Equation ($1$) and then compute the variation of the action, you get Equation ($4$) below (**Please correct this phrasing as I imagine this is wrongly or poorly done on my part.**) $ \delta{S} = \frac{\partial}{\partial \epsilon}\Bigr|_{\epsilon=0} \int _{{t_{0}}} ^{t_{f}} \frac{1}{2} {^{\epsilon} \hat{\boldsymbol \omega}} \cdot J \cdot {^{\epsilon} \hat{\boldsymbol \omega}} \, dt \tag{4} $ Manchester and Peck shows that this eventually becomes: $ \delta{S} = \int _{{t_{0}}} ^{t_{f}} \hat{\boldsymbol \omega} \cdot J \cdot (2\hat{\boldsymbol \omega} \hat{\boldsymbol \eta} + 2 \dot{\hat{\boldsymbol \eta}}) \, dt \tag{5} $ #### Angadh's work so far The integrand of Equation (4) is: $ ^\epsilon \mathcal{L} = \frac{1}{2} {^{\epsilon} \hat{\boldsymbol \omega}} \cdot J \cdot {^{\epsilon} \hat{\boldsymbol \omega}} \tag{6} $ Using Equation $(3)$ in ($6$) is: $ ^\epsilon \mathcal{L} = \frac{1}{2} \bigg(\hat{\boldsymbol \omega} + \epsilon (\hat{\boldsymbol \omega}\hat{\boldsymbol \eta} - \hat{\boldsymbol \eta} \hat{\boldsymbol \omega} + 2 \dot{\hat{\boldsymbol \eta}})\bigg) \cdot J \cdot \bigg(\hat{\boldsymbol \omega} + \epsilon (\hat{\boldsymbol \omega}\hat{\boldsymbol \eta} - \hat{\boldsymbol \eta} \hat{\boldsymbol \omega} + 2 \dot{\hat{\boldsymbol \eta}})\bigg) \tag{7} $ And we can expand this as: $ \begin{align} ^\epsilon \mathcal{L} = \frac{1}{2} \hat{\boldsymbol \omega} &\cdot J \cdot \hat{\boldsymbol \omega}\\ +\frac{1}{2} &\hat{\boldsymbol \omega} \cdot J \cdot \bigg(\epsilon (\hat{\boldsymbol \omega}\hat{\boldsymbol \eta} - \hat{\boldsymbol \eta} \hat{\boldsymbol \omega} + 2 \dot{\hat{\boldsymbol \eta}})\bigg)\\ + \frac{1}{2} &\bigg(\epsilon (\hat{\boldsymbol \omega}\hat{\boldsymbol \eta} - \hat{\boldsymbol \eta} \hat{\boldsymbol \omega} + 2 \dot{\hat{\boldsymbol \eta}})\bigg) \cdot J \cdot \hat{\boldsymbol \omega} \end{align} \tag{8} $ So.... how does Equation ($8$) eventually become Equation ($5$) above...? ```ad-todo title: Steven's Mission If-Possible I was unable to show myself how Equation ($4$) becomes Equation $(5)$. Can you derive this below, please? ```