Vectors in inertial and non-inertial frames

Question

Trying to understand how vectors change in inertial and non-inertial frames

Am I right in saying vectors are defined by their invariance under coordinate transformations? My main question is are vectors invariant when shifting between inertial and non-inertial frames of reference.

My textbook says: $$\vec{A} = A_{i}\vec{e}_{i}=A_j\vec{e}_{j}$$

where the i basis vectors are an inertial frame, K, and the j vectors are non-inertial, K', therefore time dependent inside the K frame.

Taking the rate of change: $$\vec{\dot{A}|_{k}} = \vec{\dot{A}|_{k'}} + \omega\times\vec{A} $$

The first term on LHS corresponds to rate of change of A vector in K', but for the second term, which frame is A being measured in. From the first equation, it seems like it can be either since the vector is the same, but is a vector in an inertial frame the same as one in a non-inertial frame?

My guess would be it can't be due to the existence of inertial forces.

If so, can you explain how the first equation makes sense?

Thanks, I don't have much experience with linear algebra.

Answer 1

This is a subtlety in the definitions. Let the inertial and rotating frames be $K$ and $K'$ respectively. Consider any vector $\mathbf{A}$ other than the angular velocity. The vector $\mathbf{A}$ itself is the same vector throughout and can be decomposed with respect to either the inertial or rotating basis, yielding different components. However, when we speak of the rate of change of $\mathbf{A}$ in the inertial and rotating frames, denoted by $\dot{\mathbf{A}}|_K$ and $\dot{\mathbf{A}}|_{K'}$ respectively, we are referring to two genuinely different vectors. They are the observed rates of change in the two frames which differ precisely because one frame is rotating and the other isn't. They are two different vectors each admitting different decompositions with respect to the inertial and rotating bases. The equation $\dot{\mathbf{A}}|_K = \dot{\mathbf{A}}|_{K'} + \boldsymbol{\omega}\times\mathbf{A}$ is a vector equation which means that it holds in any basis.

To make it more precise, let the inertial and rotating bases be $(\hat{\mathbf{i}}, \hat{\mathbf{j}}, \hat{\mathbf{k}})$ and $(\hat{\mathbf{i}}', \hat{\mathbf{j}}', \hat{\mathbf{k}}')$ respectively, located at the common fixed origin of both frames. We have $$\mathbf{A} = A_1\hat{\mathbf{i}} + A_2\hat{\mathbf{j}} + A_3 \hat{\mathbf{k}} \\ = {A_1}'\hat{\mathbf{i}}' + {A_2}'\hat{\mathbf{j}}' + {A_3}' \hat{\mathbf{k}}'.$$ Taking the time derivative in the inertial frame, we have $$\dot{\mathbf{A}}|_K = \frac{\mathrm{d}\mathbf{A}}{\mathrm{d}t} = \dot{A_1}\hat{\mathbf{i}} + \dot{A_2}\hat{\mathbf{j}} + \dot{A_3} \hat{\mathbf{k}} \\ = \dot{{A_1}'}\hat{\mathbf{i}}' + \dot{{A_2}'}\hat{\mathbf{j}}' + \dot{{A_3}'} \hat{\mathbf{k}}' + {A_1}' \frac{\mathrm{d}\hat{\mathbf{i}}'}{\mathrm{d}t} + {A_2}' \frac{\mathrm{d}\hat{\mathbf{j}}'}{\mathrm{d}t} + {A_3}' \frac{\mathrm{d}\hat{\mathbf{k}}'}{\mathrm{d}t}$$ However, taking the time derivative in the rotating frame, $$\dot{\mathbf{A}}|_{K'} = \dot{{A_1}'}\hat{\mathbf{i}}' + \dot{{A_2}'}\hat{\mathbf{j}}' + \dot{{A_3}'} \hat{\mathbf{k}}'$$ which is only the first half of the last line above because $(\hat{\mathbf{i}}', \hat{\mathbf{j}}', \hat{\mathbf{k}}')$ are constant in the rotating frame. The latter half is what we attribute to $\boldsymbol{\omega} \times \mathbf{A}$.

Answer 2

$A$ is usually measured in the inertial frame however, notice that it can be measured in either of them since $A=e_jA^j=e'_iA'^i$. However, you cannot do the same with the velocities since in the inertial frame $\dot{r}=e_ir^i$ while in the rotating frame $\dot{r'}=\dot{e}'_ir'^i+e'_i\dot{r}'^i$.