# Time derivative of a "general" vector $\vec A$ in an accelerating frame: what about e.g. velocity $\vec v$?

According to Morin "Classical Mechanics" (Section 10.1, page 459), the derivative of a general vector $$\vec A$$ in an accelerating frame may be given as $$\frac{d\vec A}{dt}=\frac{\delta \vec A}{\delta t}+\vec \omega\times \vec A,$$ where $$\frac{\delta \vec A}{\delta t}$$ denotes the rate of change of $$\vec A$$ measured in the rotating frame. I understand the derivation, but I am somewhat confused about the "general vector" part. The derivation assumes you have position vector, or that's how I interpreted it, because then $$\vec r=\vec R+\vec r'$$, where $$\vec r$$ is the position in the inertial frame, $$\vec R$$ is the position of the origin of the accelerating frame in the inertial frame and $$\vec r'$$ is the position in the accelerating frame. This is all fine and good, but surely if I take any other vector, say the velocity $$\vec v$$, this rule no longer applies? So my question: what is different about this "general vector" $$\vec A$$, or what am I not quite grasping?

• It’s not clear to me what your question is. If you’re saying that perhaps Morin phrased the argument only for the position vector but not other types of vectors, then my response is simply that you have to repeat the similar steps and see that the argument goes through in general. This formula is completely general, and holds for any vector-valued map. See here for the general formulation. Jan 1 at 13:41
• If one has a velocity vector ${\vec v}^\prime$ in a non-inertial frame which has an origin with velocity, $\vec V$ with respect to some inertial frame; then in the inertial frame the velocity is given by:$\vec v=\vec V+\vec v^{\prime}$. Such reasoning follows from the ordinary notion of addition of vectors. Jan 1 at 15:36
• Although my above statement about vectors is true, it doesn't really help when one wants to determine an explicit form for the time derivative of a vector in a non-inertial frame. Jan 1 at 15:56

I struggled with this exact question before and let's be clear with something from the start. $${\bf A}$$ can be any vector and this formula (10.5) works.
The second group arises because the coordinate axes are moving. In what manner are they moving? We have already extracted the motion of the origin of the accelerating system by introducing the vector $${\bf R}$$, so the only thing left is a rotation about some axis ω through this origin (see Theorem 9.1).
The $${\bf\hat{x}}$$, $${\bf\hat{y}}, {\bf\hat{z}}$$ and $${\bf R}$$ are defined in (10.3) in such a way that$${\bf\hat{x}}$$, $${\bf\hat{y}}$$ and $${\bf\hat{z}}$$ they will only have rotational motion (no linear acceleration) and so we apply Thm 9.2 to $$\frac{d{\bf\hat{x}}}{dt}$$ etc... This is what the expression $${\bf r}_I$$ = $${\bf R}$$ + $${\bf r}$$ is used for in the derivation. So let me be clear about that, the expression $${\bf r}_I$$ = $${\bf R}$$ + $${\bf r}$$ is used in the general derivation for the vector $${\bf A}$$.
For the general vector $${\bf A}$$, he doesn't need to write down a similar relationship because of how he is using $${\bf\hat{x}}$$, $${\bf\hat{y}}, {\bf\hat{z}}$$ which were indeed defined in a (useful) way that required position vector $${\bf R}$$.
To answer your question concretely, he could've done this with some other vector (e.g. $${\bf V}= \frac{d{\bf R}}{dt}$$, $${\bf v}_I$$ = $${\bf V}$$ + $${\bf v}$$)but what's important is that he wrote down $${\bf\hat{x}}$$, $${\bf\hat{y}}$$ and $${\bf\hat{z}}$$ in a way that they don't have linear acceleration.