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I will begin by stating the question and then I will explain my doubt. The relation between time-derivatives of a vector $\vec{u}$ observed from fixed and rotating frames (with a common origin) is

$$ \left[\frac{d\vec{u}}{dt}\right]_f = \left[\frac{d\vec{u}}{dt}\right]_r + \vec{w} \times \vec{u}$$

Question: I don't understand, in the derivation of this equation, why the first term obtained in the RHS is actually $ \left[\frac{d\vec{u}}{dt}\right]_r $

That is the question, now I explain myself.

I have read a few articles where this equation is derived, Wikipedia for example. It starts by defining the unit vectors in the rotating frame $\hat{i} = (\textrm{cos}(w t), \textrm{sin}(w t), 0)$ and $\hat{j} = (-\textrm{sin}(w t), \textrm{cos}(w t), 0)$ where $w = ||\vec{w}||$ is the magnitude of the angular velocity of the rotating frame, assuming the rotation is performed around the $z$ axis. It is clear that this description of $\hat{i}$ and $\hat{j}$ is made from the fixed axis point of view (from the rotating axis, it would be $\left[\hat{i}\right]_r = (1,0,0)$ and $\left[\hat{j}\right]_r = (0,1,0)$, right?)

Then, the differentiation is made:

$$ \left[\frac{d\vec{u}}{dt}\right]_f = \frac{d (u_x \hat{i} + u_y \hat{j} + u_z \hat{k})}{dt} $$

In this last expression, it seems to me (and perhaps I am wrong here) that $u_x, u_y, u_z$ are the coordinates of the vector $\vec{u}$ as seen from the rotating frame. In other words: $\left[\vec{u}\right]_r = (u_x, u_y, u_z)$. On the other hand $\left[\vec{u}\right]_f = u_x \hat{i} + u_y \hat{j} + u_z \hat{k}$. Of course $\left[\vec{u}\right]_r \neq \left[\vec{u}\right]_f$ except at the times when both frames are aligned. I have a feeling that my confusion has something to do with what I've written in this paragraph.

I will omit the next steps in the calculations, the result is:

$$ \left[\frac{d\vec{u}}{dt}\right]_f = \left(\frac{du_x}{dt} \hat{i} + \frac{du_y}{dt} \hat{j} + \frac{du_z}{dt} \hat{k}\right) + \vec{w} \times \vec{u}$$

It is then stated that the term between parentheses on the RHS is $\left[\frac{d\vec{u}}{dt}\right]_r$. But this confuses me, I would have said that $\left[\frac{d\vec{u}}{dt}\right]_r = \left(\frac{du_x}{dt} , \frac{du_y}{dt}, \frac{du_z}{dt}\right)$ and again, by multiplying each component by its corresponding basis vector, I get the corresponding vector in the fixed frame, i.e. $ \left[\frac{d\vec{u}}{dt}\right]_f = \frac{du_x}{dt} \hat{i} + \frac{du_y}{dt} \hat{j} + \frac{du_z}{dt} \hat{k}$

I would really appreciate if someone could point out where is it that my confusion arises.

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    $\begingroup$ Related : Velocity in a turning reference frame. Especially @Emilio Pisanty's answer. $\endgroup$
    – Frobenius
    Commented Sep 18, 2020 at 21:50
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    $\begingroup$ @Frobenius thank you, your answer in that question is great. I like the fact that you emphasized for each vector, w.r.t. which reference frame AND in coordinates of which reference frame, I think this is what troubled me. I believe that I was getting confused by notation, now I believe that the subindices $r $ and $f $ specify w.r.t. which ref frame, while the coordinates used are always the fixed frame ones. Do you agree with this? $\endgroup$
    – Javi
    Commented Sep 18, 2020 at 22:21
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    $\begingroup$ Yes, precisely. In my answer equations (05),(14),(22) for $\mathbf{R},\mathbf{U},\mathbf{A}$ respectively are expressed by the fixed frame $Oxyz$ coordinates although they contain the vectors $\mathbf{r},\mathbf{u},\mathbf{a}$ - variables with respect to the rotating frame $O'x'y'z'$. I must admit that at the very begining I was very confused too. $\endgroup$
    – Frobenius
    Commented Sep 18, 2020 at 22:38

2 Answers 2

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Think of a vector $\vec{g}_r$ resting on a rotating frame, which without loss of generality has its axis of rotation out of the plane as seen below:

fig

The rotation angle is $\theta$ at any instant and the vector $\vec{g}_r$ is expressed in terms of the local coordinate vectors $\hat{i}_r$ and $\hat{j}_r$ as

$$ \vec{g}_r = x_r \hat{i}_r + y_r \hat{j}_r \tag{1}$$

This vector changes with time, and thus the rate of change is tracked by the rate of change of the components

$$ \dot{\vec{g}}_r = \dot{x}_r \hat{i}_r + \dot{y}_r \hat{j}_r \tag{2}$$

Now from a fixed frame of reference you have

$$ \hat{i}_f = \begin{bmatrix} \cos \theta & -\sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1 \end{bmatrix} \hat{i}_r $$

as well as

$$ \hat{j}_f = \begin{bmatrix} \cos \theta & -\sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1 \end{bmatrix} \hat{j}_r $$

At any instant the vector is

$$ \vec{g}_f = x_f \hat{i}_f + y_f \hat{j}_f \tag{3}$$

The time rate of the unit vectors come out to be

$$ \dot{\hat{i}}_f = \begin{bmatrix} 0 & -\dot{\theta} & 0 \\ \dot{\theta} & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} \cos \theta & -\sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1 \end{bmatrix} \hat{j}_r = \begin{bmatrix} 0 & -\dot{\theta} & 0 \\ \dot{\theta} & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \hat{j}_f $$

If you do the same for rotations about x and y and consider the arbitrary rotational motion $$\vec{\omega} = \pmatrix{\omega_x \\ \omega_y \\ \omega_z } $$

then you find that

$$ \dot{\hat{i}}_f = \begin{bmatrix} 0 & -\omega_z & \omega_y \\ \omega_z & 0 & -\omega_z \\ -\omega_y & \omega_x & 0 \end{bmatrix} \hat{i}_f = \vec{\omega} \times \hat{i}_f $$

and also

$$ \dot{\hat{j}}_f = \vec{\omega} \times \hat{j}_f $$

Finally, to find $\dot{\vec{g}}_f$ from (3) and the product rule $\tfrac{{\rm d}}{{\rm d}t} ( x_f \hat{i}_f ) = (\tfrac{{\rm d}}{{\rm d}t} x_f) \hat{i}_f + x_f ( \tfrac{{\rm d}}{{\rm d}t} \hat{i}_f )$

$$ \dot{\vec{g}}_f = \dot{x}_f \hat{i}_f + \dot{y}_f \hat{j}_f + \vec{\omega} \times ( {x}_f \hat{i}_f + {y}_f \hat{j}_f ) $$

or the more shorthand notation

$$ \tfrac{{\rm d}}{{\rm d}t} \vec{g}_f = \tfrac{\partial }{\partial t} \vec{g}_f + \vec{\omega} \times \vec{g}_f \tag{4}$$

where $\tfrac{\partial }{\partial t} \vec{g}_f$ stands for the vector of coefficient rates $\dot{x}_f \hat{i}_f + \dot{y}_f \hat{j}_f$ and $\vec{\omega} \times \vec{g}_f$ the effect of the rotation on the vector.

The first part accounts for the variability of the vector itself and the second part of the change of orientation due to the rotation.

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  • $\begingroup$ I really appreciate this answer. Nevertheless, I was not asking for a derivation of the formula, but for an explanation about why is the first term on the RHS the derivative w.r.t. the rotating frame. As I explain in my own answer, I was confusing the reference frame with the basis for selecting coordinates. However, there is one thing that I find confusing in your eq (4). Shouldn't the $\frac {\partial}{\partial t}$ vector be expressed w.r.t. to the rotating frame (subindex $r $) in the coefficient rates? $\endgroup$
    – Javi
    Commented Sep 22, 2020 at 2:55
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    $\begingroup$ @Javi - no, as you can see below (4) the first term corresponds to the changing vector if the frame was not rotating. Since $\vec{g}_f = \mathbf{R} \vec{g}_r$ then the partial with time is also $ \tfrac{\partial}{\partial t} \vec{g}_f = \mathbf{R} \tfrac{\partial}{\partial t} \vec{g}_r$ $\endgroup$ Commented Sep 22, 2020 at 12:11
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    $\begingroup$ in your post - The first RHS term is not in the rotating frame (it cannot be due to consistency), but it means the derivative as seen by an observer that is co-rotating. It is confusing notation, and this is the reason most people express this in terms of partial derivatives ($\partial t$) on the RHS and total derivatives (${\rm d}t$) on the LHS. $\endgroup$ Commented Sep 22, 2020 at 12:15
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I'll answer myself as this could be helpful for someone else, but the credit belongs to @Frobenius who has pointed me to his useful answer in the following link:

https://physics.stackexchange.com/a/252265/190100

It is now clear to me that it was notation that got me confused. All vectors here ($\vec {u}, \frac {d\vec {u}}{dt}$, unit rotating vectors) are expressed in coordinates of the fixed frame. The subindices $f $ and $r$ specify w.r.t. which frame a vector has been measured and have nothing to do with the coordinate system being used.

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    $\begingroup$ I must note, @Javi, that you can write down any equation by the rotating frame $O'x'y'z'$ coordinates too. But you must not express any equation mixing the two set of coordinates. $\endgroup$
    – Frobenius
    Commented Sep 19, 2020 at 8:32

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