# Transport Theorem in analytical dynamics: basis of the vectors

If we have two reference systems, $$N$$ and $$B$$, with common origins $$O_N=O_B$$ and $$B$$ being rotating around $$N$$ with angular velocity $$\omega_{B|N}$$, the time derivates of any vector $$\vec{u}$$ in both systems are related by the next theorem:

$$\frac{d \vec{u} }{dt}_N=\frac{d \vec{u} }{dt} _{B}+ \vec{\omega}_{B|N} \wedge \vec{u}$$

In wich reference frame are expressed the vectors $$\vec{\omega}$$ and $$\vec{u}$$ from the second term of the RHS: in the basis of $$N$$ or in the basis of $$B$$?

• surely your textbook will make this clear. – ZeroTheHero Apr 12 '18 at 15:39
• No, it doesn't, I ask here because of that – Quaerendo May 4 '18 at 7:25

The angular velocity $$\vec{\omega}$$ is that of the $$B$$ rotating about $$N$$. The vector $$\vec{u}$$ on the right-hand side is expressed in terms of the basis of $$B$$.
One way to convince yourself about it could be by considering $$\vec{u} = u_x\hat{i} + u_y\hat{j} + u_z\hat{k}$$ with $$u_x, u_y, u_z$$ as constants. Then, in the basis of $$B$$, the vector $$\vec{u}$$ remains a constant. However, it appears time-varying in the basis of $$N$$. For, $$\frac{d\tilde{u}_x}{dt} = \omega_y u_z - \omega_z u_y,$$ where I have chosen to write the components of $$\vec{u}$$ in the basis of $$N$$ with a tilde on top.
$$\frac{d {}^{B}\vec{u} }{dt}_N=\frac{d {}^{B}\vec{u} }{dt} _{B}+ {}^{B}\vec{\omega}_{B|N} \wedge {}^{B}\vec{u}$$
The angular velocity vector $$\vec{\omega}_{B|N}$$ is typically written in the B frame.
However, it is not necessary for the vector $$\vec{u}$$ to be written in the B coordinate frame, because $$\vec{u}$$ is simply one of the infinity of possible components of the unique vector $$\vec{u}$$. Rather, components can be written in any arbitrary coordinate frame.