Derivation of body force in $SO(3)$

T = \frac{\partial L}{\partial R}

R = (e_{1} e_{2} e_{3})

Let $T_{i} = \frac{\partial L}{\partial e _{i}}$ , then

T = (T_{1} T_{2} T_{3})^{T}

T R = T_{1} \cdot e_{1} T_{2} \cdot e_{1} T_{3} \cdot e_{1} T_{1} \cdot e_{2} T_{2} \cdot e_{2} T_{3} \cdot e_{2} T_{1} \cdot e_{3} T_{2} \cdot e_{3} T_{3} \cdot e_{3}

\hat{b}_{1} \hat{b}_{2} \hat{b}_{3} = 000001 0 - 1 0 = 00 - 1 000100 = 010 - 1 00 000

Computing $T_{i} = tr (T R \hat{b}_{i})$ , you get

T_{1} T_{2} T_{3} = T_{2} \cdot e_{3} - T_{3} \cdot e_{2} = - T_{1} \cdot e_{3} + T_{3} \cdot e_{1} = T_{1} \cdot e_{2} - T_{2} \cdot e_{1}

Now note that using $e_{3} = e_{1} \times e_{2}$ we have that

T_{2} \cdot e_{3} = T_{2} \cdot (e_{1} \times e_{2}) = - e_{1} \cdot (T_{2} \times e_{2})

where we used the fact that the triple product is unchanged under a circular shift.

So we get

T_{1} T_{2} T_{3} = - e_{1} (T_{2} \times e_{2} + T_{3} \times e_{3}) = - e_{1} \cdot i \sum T_{i} \times e_{i} = - e_{2} (T_{1} \times e_{1} + T_{3} \times e_{3}) = - e_{2} \cdot i \sum T_{i} \times e_{i} = - e_{3} (T_{2} \times e_{2} + T_{1} \times e_{1}) = - e_{3} \cdot i \sum T_{i} \times e_{i}

Finally

T = - R^{T} i \sum T_{i} \times e_{i}

Lukas Kikuchi