# Velocity of a particle at time t in a rotating frame

• I
Imagine two frames one inertial (x,y,z) and the other rotating (x',y',z'), their origins are always coincident. The rotating frame is rotating as seen from the inertial frame with a time-dependent angular velocity ##\boldsymbol{\Omega}(t)=(\Omega_x(t),\Omega_y(t),\Omega_z(t))##. In the rotating frame, a particle is moving in a circular fashion about the rotating frame's origin in the x-y plane at a radius ##r## in the form ##\boldsymbol{r}'(t)=r(\cos(\phi(t)),\sin(\phi(t)),0)##, where ##\phi(t)## is the angle between the position vector ##\boldsymbol{r}'(t)## in the rotating frame and the x' axis and ##r=|\boldsymbol{r}'|##.

I would like to know what the velocity of the particle is as seen from the inertial frame at time t, I have started this by considering the relation between velocities of particles in two frames, one inertial the other rotating

\frac{d\boldsymbol{r}}{dt}=\frac{d\boldsymbol{r}'}{dt}+\boldsymbol{\Omega}\times\boldsymbol{r}'

Where the term on the R.H.S is the velocity of the particle as seen from the inertial frame, and the first term on the L.H.S is the velocity of the particle as seen from the rotating frame, the second term being the rotation of the rotating frame as seen from the inertial frame.

I have some doubts about using this though, Eq.(1) seems to be an instantaneous velocity, it doesn't describe the evolution of the velocity as seen from the inertial frame for all times t. For example, if I give the position vector ##\boldsymbol{r}'## some time dependence ##\boldsymbol{r}'(t)## substitute into Eq.(1), what I get is the instantaneous velocity of the particle as if I started at the position ##\boldsymbol{r}'(t)##? And similarly, giving a time dependence to ##\boldsymbol{\Omega}##.

Or have I missed understood this equation?

vanhees71
Gold Member
I don't know, what bothers you. Your calculation is correct, and it describes the velocity in the inertial frame and relates it to the velocity in the rotating frame. What else are you looking for?

Dale
Mentor
2020 Award
it doesn't describe the evolution of the velocity as seen from the inertial frame
Why not? The evolution of the velocity in the inertial frame is: ##v(t)=\frac{dr(t)}{dt}##

I don't know, what bothers you. Your calculation is correct, and it describes the velocity in the inertial frame and relates it to the velocity in the rotating frame. What else are you looking for?

Specifically I am trying to look at the difference of the absolute velocities squared of two particles, one going clockwise (-) around a circuit in the rotating-frame and the other anticlockwise (+). When plugging in for ##\boldsymbol{r}_{\pm}'=r(\cos(\phi(t)),\pm\sin(\phi(t)),0)## into Eq.(1) I am not getting something I can make sense of when thinking about particular scenarios. Explicitly, I am calculating

\left|\dot{\boldsymbol{r}_\pm}\right|^2=\left|\dot{\boldsymbol{r}_\pm'}\right|^2+2\dot{\boldsymbol{r}_\pm'}\bullet\Omega\times\boldsymbol{r}_{\pm}'+\left|\boldsymbol{\Omega}\times\boldsymbol{r}_{\pm}'\right|^2

And looking at ##\left|\dot{\boldsymbol{r}}_{+}\right|^2-\left|\dot{\boldsymbol{r}}_{-}\right|^2##.

\left|\dot{\boldsymbol{r}}_{+}\right|^2-\left|\dot{\boldsymbol{r}}_{-}\right|^2=\Omega_x\Omega_y\sin(2\phi)-2\dot{\phi}\Omega_z

However, as FactChecker has mentioned, if ##\Omega## is time dependent then I need to consider the product rule in my thought scenario, which is, if I initially only have rotation about the x-axis which comes to a halt when a 90 deg rotation has occurred and at this point the two particles have also gone through a 90 deg rotation about the circuit, which puts them on the z-axis of the inertial frame one directly below the other, if at this point I have rotation about the z-axis I would expect the difference above to be 0, I do not see this scenario in the above equation.

So considering that was a scenario having time dependence in ##\Omega## I need to consider a product term on ##\boldsymbol{\Omega}\times\boldsymbol{r}##?

Last edited:
Would the correct equation to use for time-dependent ##\boldsymbol{\Omega}## be

\boldsymbol{a} = \boldsymbol{a}'+2\boldsymbol{\Omega}\times\boldsymbol{v}' + \boldsymbol{\Omega}\times(\boldsymbol{\Omega}\times\boldsymbol{r}') + \frac{d\boldsymbol{\Omega}}{dt}\times\boldsymbol{r}'

Where the ##\boldsymbol{a}##'s are now the acceleration in the respective frames. And then integrate with respect to time to solve for the ##\boldsymbol{v}##'s?

FactChecker
Gold Member
The equation for acceleration does not make your original equation for velocity wrong. It seems like you are concerned that your original equation is wrong when the rotation rate is a function of time. But that does not make the instantaneous velocity equation wrong.

vanhees71
Yes, I am concerned that it is wrong when having a time-dependent rotation, this is because the result I get Eq.(3), doesn't seem to make sense when I consider particular examples of rotation with a time dependence. Admittedly this could just be a mistake in my workings or reasoning about the example I'm going through. I think I will go over my workings on this and try to make sense of it a bit more.

Dale
Mentor
2020 Award
Both your equation in post 1 and in post 5 are good. Of course, ##\Omega## should be continuous.

FactChecker
Gold Member
Of course, you must always be careful in this to distinguish between an acceleration due to a force and a derivative of velocity wrt time. They are not the same.

An extreme would be a pilot in an airplane traveling at 100 mph in straight and level, unaccelerated flight. There is no force-caused acceleration in any frame because there is no force. But if we looked at a pilot head-centered-head-fixed coordinate system, there would be a huge ##velocity_{head-centered-head-fixed}## time derivative any time he turned his head. That velocity could go from 100 MPH in the direction of his nose to 0 MPH in a second if he turns his head to the side.

Dale
Ok, so I've convinced myself that the result is fine. However, im now confused about coordinate transformations. If I write ##\dot{\boldsymbol{r}}=\boldsymbol{v}## as

\boldsymbol{v}=v_{x'}\boldsymbol{\hat{i}'}+v_{y'}\boldsymbol{\hat{j}'}+v_{z'}\boldsymbol{\hat{k}'}

and I want to look at the velocity component ##v_z\boldsymbol{\hat{k}}## in the inertial frame I have

v_z=v_{x'}\boldsymbol{\hat{i}'}\bullet\boldsymbol{\hat{k}}+v_{y'}\boldsymbol{\hat{j}'}\bullet\boldsymbol{\hat{k}}+v_{z'}\boldsymbol{\hat{k}'}\bullet\boldsymbol{\hat{k}}

I'm confused about how to evaluate the projection terms, for example, my first attempt was to just put ##\boldsymbol{\hat{i}'}## into Eq.(1) and go from there, but doing that gives me

\dot{\boldsymbol{\hat{i}'}}=\Omega_z\boldsymbol{\hat{j}'}-\Omega_y\boldsymbol{\hat{k}'}

which just seems circular?

vanhees71
Gold Member
The primed and unprimed basis vectors are related by (time dependent!) rotations. Let me switch notation and use the Einstein summation convention. The basis vectors fixed to the inertial frame are time-independent by definition, while the basis vectors in the rotating frame are timedependent, and the rotation matrix is given by
$$\vec{e}_j'(t) \cdot \vec{e}_k=D_{jk}(t).$$
Now let's calculate the time derivative of an arbitrary vector ##\vec{V}(t)##. We have
$$\vec{V}(t)=V_k(t) \vec{e}_k =V_j'(t) \vec{e}_j'(t) = V_j'(t) D_{jk}(t) \vec{e}_k.$$
Then you get
$$\dot{\vec{V}}(t) = \dot{V}_k(t) \vec{e}_k = [\dot{V}_j'(t) D_{jk}(t) + V_{j}'(t) \dot{D}_{jk}(t)] \vec{e}_k = \dot{V}_j'(t) \vec{e}_j'(t) + V_{j}'(t) \dot{D}_{jk} D_{lk} \vec{e}_l'(t).$$
Now
$$D_{jk}(t) D_{lk}(t)=\delta_{jl} \; \Rightarrow \; \omega_{jl}'=\dot{D}_{jk}(t) D_{kl}(t) =-D_{jk} \dot{D}_{lk}(t)=-\omega_{lj}'(t).$$
That means that ##\omega_{jl}'## is antisymmetric, and we can write
$$\omega_{jl}'(t)=\epsilon_{mjl} \Omega_m'(t).$$
$$\dot{\vec{V}}= [\dot{V}_l'(t) + \epsilon_{mjl} \Omega_m'(t) V_{j}'(t)] \vec{e}_l'.$$
Now using the matrix notation, with bold symbols standing for column ##\mathbb{R}^3## vectors, you can define the "covariant time derivative" in the rotating frame as
$$\mathrm{D}_t \boldsymbol{V}'=\dot{\boldsymbol{V}}' + \boldsymbol{\Omega}' \times \boldsymbol{V}'.$$
For the position vector you get
$$\boldsymbol{v}'=\mathrm{D}_t \boldsymbol{r}'=\dot{\boldsymbol{r}}'+\boldsymbol{\Omega}' \times \boldsymbol{r}'.$$
For the acceleration you finally find
$$\boldsymbol{a}'=\mathrm{D}_t \boldsymbol{v}' = \dot{\boldsymbol{v}}' + \boldsymbol{\Omega}' \times \boldsymbol{v}' = \ddot{\boldsymbol{r}}' + 2 \boldsymbol{\Omega}' \times \dot{\boldsymbol{r}}' + \boldsymbol{\Omega}' \times (\boldsymbol{\Omega}' \times \boldsymbol{r}') + \dot{\boldsymbol{\Omega}}' \times (\dot{\boldsymbol{r}}' +\boldsymbol{\Omega}' \times \boldsymbol{r}').$$

Last edited:
weirdoguy
If I understand correctly what you have shown, and if I can write it down in a way I find easier to read, I have

\begin{pmatrix}
v_x \\ v_y \\ v_z
\end{pmatrix}
=
\begin{pmatrix}
v_x' \\ v_y' \\ v_z'
\end{pmatrix}
+
\begin{pmatrix}
0&-\Omega_z&\Omega_y\\ \Omega_z&0&-\Omega_x \\ -\Omega_y&\Omega_x&0
\end{pmatrix}
\begin{pmatrix}
r_x' \\ r_y' \\ r_z'
\end{pmatrix}

And so what I'm looking for is just

v_z=v_z'+r_y'\Omega_x-r_x'\Omega_y

vanhees71
Gold Member
No (8) is not correct. There's always a rotation matrix between the primed and unprimed components, and your angular-velocity tensor must get primed indices. As you can see from the 2nd formula in #11 for any vector ##\vec{V}## (symbols with arrows are the invariant vectors) you have (bold symbols are the components wrt. the inertial basis (unprimed) or the rotating basis (primed)). From the said formula you get
$$\boldsymbol{V}=\hat{D}^{\text{T}} \boldsymbol{V}' \; \Rightarrow \; \boldsymbol{V}'=\hat{D} \boldsymbol{V}.$$
Thus for the velocity components you get
$$\boldsymbol{v}=\hat{D}^{\text{T}} \boldsymbol{v}'=\hat{D}^{\text{T}} (\dot{\boldsymbol{r}}' + \boldsymbol{\Omega}' \times \boldsymbol{r}').$$

jamie.j1989 and FactChecker
I think this explains a lot of my confusion, so in my example, I know the rotation of the rotating frame about the inertial which I have labelled ##\boldsymbol{\Omega}## which I need to apply ##\hat{D}## to, to get ##\boldsymbol{\Omega}'=\hat{D}\boldsymbol{\Omega}##. So would this also be correct

\boldsymbol{v}=\hat{D}^T(\dot{\boldsymbol{r}}'+\hat{D}\boldsymbol{\Omega}\times\boldsymbol{r}')=\hat{D}^T\dot{\boldsymbol{r}}'+\boldsymbol{\Omega}\times\boldsymbol{r}'

And to find the components of ##\hat{D}## could I use the rotation matrix found here
https://en.wikipedia.org/wiki/Rotation_matrix

Where ##\hat{D}## is the ##R## under the 'Rotation matrix from axis angle section'?

FactChecker
Gold Member
Last edited:
I was under the impression that the matrix ##R## I suggested I should use was a quaternion? However, on closer inspection, it says for a constant rotation axis. Does this need a numerical approach? It doesn't seem obvious how to define ##\theta## (the amount rotated about the axis of rotation) in the rotation matrix if the axis of rotation is time-dependent.

vanhees71
Gold Member
Well, I'd not use quaternions. Nowadays physicists usually work with the much more flexible vectors and tensors. If the rotation is around a constant rotation axis, there's no need for Euler angles or even more complicated stuff.

Then just choose you Cartesian basis such that the ##x_3## axis is the axis of rotation, directed according to the right-hand rule. Then let ##\phi(t)## the time-dependent angle between ##\vec{e}_1## and ##\vec{e}_1'## (I use the notation of my previous postings, with the unprimed basis vectors the time-independent basis of the inertial frame and the primed basis vectors the time-dependent ones rotating wrt. the inertial frame). Then with a little drawing, you easily see that
$$\vec{e}_1' = \vec{e}_1 \cos \phi + \vec{e}_2 \sin \phi, \quad \vec{e}_2'=-\vec{e}_1 \sin \phi + \vec{e}_2 \cos \phi.$$
Using again $$R_{jk}=\vec{e}_j' \cdot \vec{e}_k$$, you get the matrix
$$\hat{R}=\begin{pmatrix} \cos \phi & \sin \phi & 0 \\ -\sin \phi & \cos \phi & 0 \\ 0 & 0 & 1 \end{pmatrix}$$
and the angular-rotation-velocity matrix becomes
$$\hat{\omega}'=\dot{\hat{D}} \hat{D}^{\text{T}}=\begin{pmatrix}0 & \dot{\phi} & 0 \\ -\dot{\phi} & 0 & 0 \\ 0 & 0 &0\end{pmatrix},$$
from which you get
$$\boldsymbol{\Omega}'=\begin{pmatrix} 0 \\ 0 \\ \dot{\phi} \end{pmatrix}.$$

Last edited: