Let there be an arbitrary vector $\mathbf{A}$ and let $S$ be an inertial set of axes $\hat{i},\hat{j},\hat{k}$ and $S'$ a rotating system with the same origin as $S$ and basis vectors $\hat{i'},\hat{j'},\hat{k'}$ . Now I've seen that the relation between the derivatives of $\mathbf{A}$ in the two frames is given by: $$\left(\frac{d\mathbf{A}}{dt}\right)_S=\left(\frac{d\mathbf{A}}{dt}\right)_{S'}+\boldsymbol{\omega}\times\mathbf{A}$$ What I don't understand is whether the $\mathbf{A}$ in the $\boldsymbol{\omega}\times\mathbf{A}$ term is in the coordinates of $S$ or in those of $S'$. Also, if the $\boldsymbol{\omega}\times\mathbf{A}$ is in the coordinates of $S'$ and so is the $\left(\frac{d\mathbf{A}}{dt}\right)_{S'}$ term, I get an equation of the form $\left(\frac{d\mathbf{A}}{dt}\right)_S=\dot{a_1}\hat{i}+\dot{a_2}\hat{j}+\dot{a_3}\hat{k}=a _1'\hat{i'}+a_2'\hat{j'}+a_3'\hat{k'}$ where $a_i'=\left(\dot{a_i} \right)_{S'}+(\boldsymbol{ \omega} \times \mathbf{A})_i$. My question is, how do I express $\left(\frac{d\mathbf{A}}{dt}\right)_S$ in the basis of $S$. If the rotation of $S'$ is about one of the principal axes then the transformation between the sets of axes is pretty straightforward and I understand it well enough. However, what about an arbitrary axis? If for example $\boldsymbol{\omega}$ creates an angle $\alpha$ with the $\hat{k'}$ axis? How can I find the rotation matrix for such a case?


Like anything else with vectors in mechanics, all quantities have to be expressed in the same coordinate system in order to be able to do standard vector calculus things like add or scale vectors.

So the answer is that $\boldsymbol{A}$ can be in either coordinate system, as long as $\boldsymbol{\omega}$ is also in the same orientation. The above equation is not a transformation of coordinates, but rather a statement on how to combine the local change of a vector $\left( \tfrac{\rm d}{{\rm d}t} A \right)_{S'}$ with the general motion of a body/frame $\boldsymbol{\omega}$.

What the notation $\left( \cdot \right)_S$ and $\left( \cdot \right)_{S'}$ means is not the resolution of vectors on different coordinate systems, but rather changes as seen by such frame.

A vector fixed on the body will seem unchanging by somebody riding on the body and hence $\left( \tfrac{\rm d}{{\rm d}t} \boldsymbol{A} \right)_{S'} = \boldsymbol{0}$, but the motion of the body will make the vector change as seen by someone on an inertial frame.

In the more general case, the expression below is a statement about kinematics, and not a transformation of coordinates.

$$ \underbrace{ \left( \tfrac{\rm d}{{\rm d}t} \boldsymbol{A} \right)}_{\text{as seen by S}} = \underbrace{ \left( \tfrac{\rm d}{{\rm d}t} \boldsymbol{A} \right)}_{\text{as seen by S'}} + \boldsymbol{\omega} \times \boldsymbol{A} \tag{1}$$

In the above scenario $\boldsymbol{A}$ is expressed in the same coordinate orientation as $\boldsymbol{\omega}$, and $S$ and $S'$ both have the same orientation at some instant, but one is rotating $S'$ and the other isn't.

If you want to do the above, but consider the changing rotation matrix $\mathbf{R}(t)$ between the two vectors expressed at different orientations, where ' indicates rotated frame, and no ' indicates word coordinate frame.

$$ \boldsymbol{A}(t) = \mathbf{R}(t)\,\boldsymbol{A}'(t) \tag{2}$$

We can omit the $(t)$ since everything can change with time. Then from geometry we know that the columns of the rotation matrix change with the following rule: $\tfrac{{\rm d}}{{\rm d}t}\mathbf{R} = \boldsymbol{\omega}\times\mathbf{R}$. So the differentiation of the above (using the derivative product rule) is

$$ \begin{aligned}\tfrac{{\rm d}}{{\rm d}t}\boldsymbol{A} & =\left(\tfrac{{\rm d}}{{\rm d}t}\mathbf{R}\right)\boldsymbol{A}'+\mathbf{R}\,\left(\tfrac{{\rm d}}{{\rm d}t}\boldsymbol{A}'\right)\\ & =\left(\boldsymbol{\omega}\times\mathbf{R}\right)\boldsymbol{A}'+\mathbf{R}\,\left(\tfrac{{\rm d}}{{\rm d}t}\boldsymbol{A}'\right)\\ & =\mathbf{R}\,\left(\tfrac{{\rm d}}{{\rm d}t}\boldsymbol{A}'\right)+\boldsymbol{\omega}\times\left(\mathbf{R}\boldsymbol{A}'\right)\\ & =\mathbf{R}\,\left(\tfrac{{\rm d}}{{\rm d}t}\boldsymbol{A}'\right)+\boldsymbol{\omega}\times\boldsymbol{A} \end{aligned} \tag{3}$$

which is the form of the rotating derivative you are inquiring about, that considers the rotation matrix $\mathbf{R}$ in addition to the motion $\boldsymbol{\omega}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.