Question about velocities in different reference frames

Question

Suppose $\hat{x^{'}}, \hat{y^{'}}, \hat{z^{'}} $ are the unit vectors of an inertial frame and $\hat{x}, \hat{y}, \hat{z} $ are the unit vectors of a frame which maybe accelerating, rotating, whatever. Consider a particle which is moving. Let $\mathbf{R}$ be the position of the origin of the the 2nd frame wrt to the 1st, and the particles position wrt both the frames is $\mathbf{r^{'}}$ and $\mathbf{r}$ respectively. Then $\mathbf{r^{'}} = \mathbf{R} + \mathbf{r}. $ I read in Morin that the velocity wrt the second frame is $\frac{dr_x}{dt}\hat{x} + \frac{dr_y}{dt}\hat{y} + \frac{dr_z}{dt}\hat{z} .$ I understand why that would be, since $r_x, r_y, r_z$ are the coordinates wrt the second frame and those are the ones measured. But I don't understand why $\frac{d\mathbf{r}}{dt}$ isn't the velocity wrt the second frame. And note that this and the earlier expression are different, if $\frac{d\mathbf{r}}{dt}$ is expanded using chain rule, then the earlier expression is one part of it. I am confused by this and any answer would be much appreciated. Chapter 10 of morin, accelerating frames of reference, section 10.1.

Answer 1

Short answer. The velocity and the acceleration w.r.t. a reference frames are defined here to be the product of the time derivatives of the components and the vectors of the basis. There is no derivative of the vectors of the basis, since each observer is not measuring any change of the vectors of the basis he's moving with: if the observer doesn't look to other reference frames in relative motion, he just can't know anything about the (relative) motion of the vectors of the basis he's moving with.

Some details. So, for simplicity using two Cartesian reference frames $\{O, \mathbf{\hat{x}}, \mathbf{\hat{y}}, \mathbf{\hat{z}}\}$ and $\{O', \mathbf{\hat{x}}', \mathbf{\hat{y}}', \mathbf{\hat{z}}'\}$,

• the position of a point w.r.t. the origins of these reference frame are defined as $\mathbf{r} := x_i \mathbf{\hat{x}}_i$, and $\mathbf{r}' := x'_i \mathbf{\hat{x}}'_i$
• the velocity $\mathbf{v} = v_i \mathbf{\hat{x}}_i = \dot{x_i} \mathbf{\hat{x}}_i$, and $\mathbf{v}' = v'_i \mathbf{\hat{x}}'_i = \dot{x_i}' \mathbf{\hat{x}}'_i$; sometimes the time derivative acting only on the components is indicated with a letter or a symbol identifying the reference frame considered as fixed (so that no time derivative of the vectors of the basis): we could then write \begin{equation} \begin{aligned} \mathbf{v} & = \dfrac{d \mathbf{r}}{dt} = \dfrac{d}{dt}(x_i \mathbf{\hat{x}}_i) = \dot{x}_i \mathbf{\hat{x}}_i \\ \mathbf{v}'& = \dfrac{'d \mathbf{r}'}{dt} = \dfrac{'d}{dt}(x'_i \mathbf{\hat{x}}'_i) = \dot{x}'_i \mathbf{\hat{x}}'_i \ , \end{aligned} \end{equation} where the prime before the time derivative means ``with constant vectors of the basis $\{O', \mathbf{\hat{x}}', \mathbf{\hat{y}}', \mathbf{\hat{z}}'\}$''
• the acceleration $\mathbf{a} = a_i \mathbf{x}_i = \dot{v}_i \mathbf{\hat{x}}_i = \ddot{x_i} \mathbf{\hat{x}}_i$, and $\mathbf{a}' = a'_i \mathbf{x}'_i = \dot{v}'_i \mathbf{\hat{x}}'_i = \ddot{x_i}' \mathbf{\hat{x}}'_i$, or \begin{equation} \mathbf{a} = \dfrac{d^2 \mathbf{r}}{dt^2} \qquad \text{and} \qquad \mathbf{a}' = \dfrac{'d^2 \mathbf{r}'}{dt^2} \ , \end{equation} Relative position. Using the rule for the sum of vectors, we can simply write for the position of a point $P$ \begin{equation} \mathbf{r}_P = \mathbf{r}_{O'} + \mathbf{r'}_{P} \ , \end{equation} being the $\mathbf{r}_P$, $\mathbf{r}_{O'}$ the position of the points $P$ and $O'$ (origin of the second reference frame) as seen by the first reference frame, and $\mathbf{r}'$ the position of $P$ as seen by the second reference frame. We can write this using coordinates and vectors of the bases as \begin{equation} x_{P,i} \mathbf{\hat{x}}_i = x_{P,i} \mathbf{\hat{x}}_i \ , \qquad x_{O',i} \mathbf{\hat{x}}_i = x_{O',i} \mathbf{\hat{x}}_i \ , \qquad x'_{P,i} \mathbf{\hat{x}}'_i = x'_{P,i} \mathbf{\hat{x}}'_i \end{equation}

Velocity. Taking the time derivative $\frac{d}{dt}$, thus considering the vectors of the first basis as constant, of $\mathbf{r}_P = \mathbf{r}_{O'} + \mathbf{r'}_{P}$ we get \begin{equation} \underbrace{\dfrac{d}{dt}{x_{P,i}} \mathbf{\hat{x}}_i}_{\mathbf{v}_P} = \underbrace{\dfrac{d}{dt}{x_{O',i}} \mathbf{\hat{x}}_i}_{\mathbf{v}_{O'}} + \underbrace{\dfrac{d}{dt}{x'_{P,i}} \mathbf{\hat{x}}'_i}_{\mathbf{v}'_P} + \underbrace{{x'_{P,i}} \dfrac{d}{dt} \mathbf{\hat{x}}'_i}_{\mathbf{\Omega}' \times \mathbf{r}'} \ , \end{equation} having used the formula for the time derivative of a unit vector $\frac{d}{dt} \mathbf{\hat{x}}'_i = \mathbf{\Omega}' \times \mathbf{\hat{x}}'_i$, being $\mathbf{\Omega}'$ the angular velocity of the vector $\mathbf{\hat{x}}'$ with the un-primed reference frame.

Acceleration.

Answer 2

There are two things which are changing.
First the position vector as measured in the two frames and secondly the coordinate system in the moving frame relative to the coordinate system in the stationary frame.

Thus, Morin writes a general vector in the moving frame as $\vec A = A_{\rm x} \hat x+ A_{\rm y} \hat y+ A_{\rm z} \hat z$ and then differentiates the expression with respect to time which gives two terms.
The first is $ \frac {dA_{\rm x}}{dt} \hat x+ \frac {dA_{\rm y}}{dt} \hat y+ \frac {dA_{\rm z}}{dt} \hat z$ which is the rate of change of $\vec A$ as measured in the moving frame, ie using the moving frame coordinates.
However to a stationary observer the moving frame is moving! and so are the unit vectors $\hat x,\,\hat y,\,\hat z$ in the stationary frame.
Morin's second term, $A_{\rm x}\frac {d\hat x}{dt}+A_{\rm y}\frac {d\hat y}{dt}+A_{\rm z}\frac {d\hat z}{dt}$ accounts for this.