Confusion about the derivation of acceleration relative to rotating frames

[Leo Liu]

This derivation is found in Kleppner's mechanics book. It shows how to find the acceleration in rotating coordinates by differentiating $\vec v_{in}=\vec v_{rot}+\vec\Omega\times\vec r$; subscripts IN and ROT stand for inertial and rotation respectively.
My question is what the term $\left(\frac{d\vec v_{in}}{dt}\right)_{rot}$ means. Is it the same thing as $\frac{d\vec v_{rot}}{dt}$? Thank you.

Edit 1: It appears that someone has asked a similar question 12 years ago. So the question is now closed except for better explanations. Thanks!
https://www.physicsforums.com/threads/deriving-acceleration-in-rotating-reference-frames.286812/
Edit 2: Meh, it was of no help ¯\_(ツ)_/¯ .

 

[anuttarasammyak]

In Inertial frame it is interpreted as the quantity
$$(\frac{dv_{in}}{dt})_{in}-\Omega \times v_{in}$$
In rotationg frame it is interpreted as the quantity
$$(\frac{dv_{rot}}{dt})_{rot}+\Omega \times v_{rot}$$
With the second term in RHS, it is not genuine simple quantity neither in inertial frame or in rotating frame.

 

[wrobel]

Let $xyz$ be an inertial frame and let $\xi\eta\zeta$ stand for a rotating frame. Take any vector $\boldsymbol u(t)$ It follows that
$$\boldsymbol u(t)=u_x(t)\boldsymbol e_x+ u_y(t)\boldsymbol e_y+u_z(t)\boldsymbol e_z=U_\xi(t)\boldsymbol e_\xi+ U_\eta(t)\boldsymbol e_\eta+U_\zeta(t)\boldsymbol e_\zeta$$
Definition.
$$\frac{d}{dt}\Big|_{in}\boldsymbol u=\dot u_x(t)\boldsymbol e_x+ \dot u_y(t)\boldsymbol e_y+\dot u_z(t)\boldsymbol e_z$$
$$\frac{d}{dt}\Big|_{rot}\boldsymbol u=\dot U_\xi(t)\boldsymbol e_\xi+\dot U_\eta(t)\boldsymbol e_\eta+\dot U_\zeta(t)\boldsymbol e_\zeta$$
Theorem. Let $\boldsymbol\omega$ stand for the angular velocity of $\xi\eta\zeta$ relative xyz.
Then
$$\frac{d}{dt}\Big|_{in}\boldsymbol u=\frac{d}{dt}\Big|_{rot}\boldsymbol u+\boldsymbol\omega\times\boldsymbol u.$$

 

[Leo Liu]

Thanks for your help! Could I ask a few questions?

$$\boldsymbol u(t)=u_x(t)\boldsymbol e_x+ u_y(t)\boldsymbol e_y+u_z(t)\boldsymbol e_z=U_\xi(t)\boldsymbol e_\xi+ U_\eta(t)\boldsymbol e_\eta+U_\zeta(t)\boldsymbol e_\zeta$$

Does this equation represent the position vectors of a particle in two different frames and relate them with each other?
Is this equation the same as $\vec r=\vec r',\, d\vec r\neq d\vec r'$?

 

[etotheipi]

In that equation, $\mathbf{u}$ is any arbitrary vector, just written in terms of two sets of basis vectors which are rotating w.r.t. each other

 

[wrobel]

Let we have a frame Oxyz and a frame Aξηζ. The position vector of the point M relative the first frame is OM and for the second frame is AM. It is clear OM=OA+AM. Correspondingly,
ddt|OxyzOM=ddt|OxyzOA+ddt|OxyzAM.
You can also write
ddt|OxyzAM=ddt|AξηζAM+ω×AM
Here
ddt|AξηζAM is the velocity of M relative Aξηζ

 

[wrobel]

I will not retype it. This forum converts text spontaneously

 

[etotheipi]

It's a known bug. I fixed it up for you

Let we have a frame $Oxyz$ and a frame $A \xi \eta \zeta$ . The position vector of the point M relative the first frame is $\boldsymbol{OM}$ and for the second frame is $\boldsymbol{AM}$. It is clear $\boldsymbol{OM}=\boldsymbol{OA}+\boldsymbol{AM}$. Correspondingly,$$\frac{d}{dt} \Big|_{Oxyz} \boldsymbol{OM}= \frac{d}{dt}\Big|_{Oxyz} \boldsymbol{OA}+\frac{d}{dt}\Big|_{Oxyz} \boldsymbol{AM}$$You can also write$$ \frac{d}{dt}\Big|_{Oxyz} \boldsymbol{AM}=\frac{d}{dt}\Big|_{A \xi \eta \zeta} \boldsymbol{AM} + \boldsymbol{\omega} \times \boldsymbol{AM}$$
Here $\frac{d}{dt}\Big|_{A \xi \eta \zeta} \boldsymbol{AM}$ is the velocity of M relative $A \xi \eta \zeta$

 

[Ibix]

I will not retype it. This forum converts text spontaneously

It's a known bug. I think keeping the editor in plain text mode (click the cogwheel at the right hand end of the editor toolbar so all other icons are disabled) helps, since then the forum doesn't try to translate BBCODE/LaTeX in the editor.

 

[etotheipi]

(Double edit: I've mightily scuffed the chronological ordering of this thread, but I'm going to leave this reply up because I don't want to delete any of my sexy $\LaTeX$ )

Let's say we have the two bases, the body-fixed $\{ \mathbf{e}_a \}$ and the space-fixed $\{ \tilde{\mathbf{e}}_a \}$, as per Tong's notation. Any vector $\mathbf{u}$, you can express as$$\mathbf{u} = u^a \mathbf{e}_a = \tilde{u}^a \tilde{\mathbf{e}}_a$$We can take the derivative, treating the space-fixed $\tilde{\mathbf{e}}_a$ as constant, i.e.$$ \left( \frac{d\mathbf{u}}{dt} \right)_{\text{space}} = \frac{du^a}{dt} \mathbf{e}_a + u^a \frac{d\mathbf{e}_a}{dt} = \frac{d\tilde{u}^a}{dt} \tilde{\mathbf{e}}_a$$Tong used the example where the body is undergoing rotation about a fixed point which is the origin of both frames, and where the vector $\mathbf{u} := \mathbf{r}$ is the position vector, in which case the space-fixed coordinates are constant $\frac{d r^a}{dt} = 0$. That's all perfectly fine.

Kleppner is going one step further here, and thinking about taking the time derivative, treating the body-fixed $\mathbf{e}_a$ as constant, i.e.$$ \left( \frac{d\mathbf{u}}{dt} \right)_{\text{body}} = \frac{du^a}{dt} \mathbf{e}_a = \frac{d\tilde{u}^a}{dt} \tilde{\mathbf{e}}_a + \tilde{u}^a \frac{d\tilde{\mathbf{e}}_a}{dt}$$That's all the bracket notation means, it's telling you which basis vectors you're treating as constant. We can combine these two equations,$$\left( \frac{d\mathbf{u}}{dt} \right)_{\text{space}} = \left( \frac{d\mathbf{u}}{dt} \right)_{\text{body}} + u^a \frac{d\mathbf{e}_a}{dt}$$but notice that since we can express$$\frac{d\mathbf{e}_a}{dt}= \boldsymbol{\omega} \times \mathbf{e}_a \implies u^a \frac{d\mathbf{e}_a}{dt} = \boldsymbol{\omega} \times u^a \mathbf{e}_a = \boldsymbol{\omega} \times \mathbf{u}$$we have$$\left( \frac{d\mathbf{u}}{dt} \right)_{\text{space}} = \left( \frac{d\mathbf{u}}{dt} \right)_{\text{body}} + \boldsymbol{\omega} \times \mathbf{u}$$

 

[Leo Liu]

Update: I finally understand the logic behind this derivation after reading Tong's notes. I coundln't grasp the idea because the notation $(d/dt)_{in/rot}$ was novel and confusing to me. Tong dispeled my confusing by sticking with the old-school notation. This is what I leant:
We first describe the position of a particle in two frames.
Inertial:
$$\vec r=r_1\hat e_1+r_2\hat e_2+r_3\hat e_3$$
Rotating:
$$\vec r=r_1'\hat e_1'+r_2'\hat e_2'+r_3'\hat e_3'$$
To find the velocity, we have to differentiate them. The first one is quite staightforward, while we need to apply chain rule to the second one because the base vectors of a rotating frame could be moving.
Therefore,
$$\begin{aligned}
\dot{\vec r}=\underbrace{\sum_{i=1}^3r_i\hat e_i}_\text{v in stationary frame}&=\underbrace{(\dot r_1',\dot r_2',\dot r_3') \langle \hat e_1',\hat e_2',\hat e_3'\rangle+(r_1',r_2',r_3') \langle \dot{\hat e_1'},\dot{\hat e_2'},\dot{\hat e_3'}\rangle}_\text{v in rotating frame}\\
&=\sum_{i=1}^3 \dot r_i'\hat e_i'+r_i\cdot\underbrace{\langle \omega_1,\omega_2,\omega_3\rangle\times\langle \hat e_1',\hat e_2',\hat e_3'\rangle}_{\because\,\vec v=\vec\omega\times\vec r}
\end{aligned}$$
This is the true velocity in two frames.
We have to differentiate it again to find out the acceleration.
$$\ddot{\vec r}=\frac{d\sum_{i=1}^3 \dot r_i'\hat e_i'+\omega_i\times r_i\hat e_i'}{dt}$$
$$
\begin{aligned}
a_{in}=&\sum_{i=1}^3 \ddot{r_i'}\hat e_i'+\underbrace{\dot r_i'\dot{\hat e_i'}}_{=\vec\omega_i\times\dot r_i'\dot{\hat e_i'}}+\underbrace{\underbrace{\dot{\vec\omega}\times r_i'\hat e_i'}_{=0\text{ if frame rotates at constant speed}}+\vec\omega\times r_i'\dot{\vec e_i'}+\vec\omega\times\dot r_i'\hat e_i'}_\text{3 terms chain rule}\\
=&\vec a_{rot}+2\vec\omega\times\vec v_{rot}+\vec\omega\times(\vec\omega\times\vec r)
\end{aligned}$$
This is the equation that relates acc in rotating frame and acc in inertial frame; the condition is that the frame rotates at a constant rate.
I hope this post can help someone who may have the same question in the future.

 

[Leo Liu]

(Edit: this is a reply to a post that was deleted, but hopefully it will still be useful )

@Leo Liu it's worth noting that what Tong is doing is slightly different to what Kleppner is doing. Let's say we have the two bases, the body-fixed $\{ \mathbf{e}_a \}$ and the space-fixed $\{ \tilde{\mathbf{e}}_a \}$, as per Tong's notation. Any vector $\mathbf{u}$, you can express as$$\mathbf{u} = u^a \mathbf{e}_a = \tilde{u}^a \tilde{\mathbf{e}}_a$$We can take the derivative, treating the space-fixed $\tilde{\mathbf{e}}_a$ as constant, i.e.$$ \left( \frac{d\mathbf{u}}{dt} \right)_{\text{space}} = \frac{du^a}{dt} \mathbf{e}_a + u^a \frac{d\mathbf{e}_a}{dt} = \frac{d\tilde{u}^a}{dt} \tilde{\mathbf{e}}_a$$Tong used the example where the body is undergoing rotation about a fixed point which is the origin of both frames, and where the vector $\mathbf{u} := \mathbf{r}$ is the position vector, in which case the space-fixed coordinates are constant $\frac{d r^a}{dt} = 0$. That's all perfectly fine.

Meh, how could you type latex equations that fast ?

 

[vanhees71]

The confusion is just to make it more complicated than it is. You have an inertial reference frame with a (by definition) time-independent cartesian basis $\vec{e}_j$ and a cartesian basis fixed in the rotating frame, which is time dependen, $\vec{e}_k'(t)$. Of course there's a time-dependent rotation matrix $D_{jk}(t)$ connecting the two bases (here and in the following Einstein summation convention applies):
$$\vec{e}_k'(t)=\vec{e}_j D_{jk}(t).$$
Now we have
$$\dot{\vec{e}}_k'=\vec{e}_j \dot{D}_{jk} = \vec{e}_l' D_{jl} \dot{D}_{jk}.$$
Further
$$D_{jl} D_{jk}=\delta_{jk},$$
because $(D_{jk}) \in \mathrm{SO}(3)$. and thus
$$\Omega_{lk}'=D_{jl} \dot{D}_{jk}=-\Omega_{kl}'=\epsilon_{klm} \omega_m'=\epsilon_{lmk} \omega_m'.$$
So for the derivative of a vector you get
$$\dot{\vec{V}}=\mathrm{d}_t (V_j' \vec{e}_j')=\dot{V}_j' \vec{e}_j' + V_j' \dot{\vec{e}}_j' = \dot{V}_j' \vec{e}_j' + V_j' D_{kj} \dot{D}_{kl} \vec{e}_l'=(\dot{V}_l' + \epsilon_{lmj} \omega_m' V_j')\vec{e}_l'.$$
Now you define a "covariant time derivative" for vector components wrt. the rotating basis by
$$\dot{\vec{V}}=\vec{e}_l' \mathrm{D}_t V_l',$$
leading to
$$\mathrm{D}_t V_l' = \dot{V}_l' + (\vec{\omega}' \times \vec{V}')_l.$$