2
$\begingroup$

The Kinematic transport theorem is a very basic theorem relating time derivatives of vectors between a non rotating frame and another one that's rotating with respect to it with a uniform angular velocity.

I was trying to prove it for the special case of $3$ dimensions, and everything seems straightforward apart from the fact that I'm getting into a difficulty with obtaining the exact form of the final expression.

Following is my attempted proof:

Let's assume without loss of generality that we have a frame $\widetilde{O}$ rotating with angular velocity $\mathbf{\Omega} = (0,0,\omega)$ about the $z$ axis of another frame $O$. Let $\bf{f}$ be a vector seen by frame $O$ and $\widetilde{\mathbf{f}}$ the same vector as seen by an observer in the $\widetilde{O}$ frame, and further assume that this observer is located at position $\widetilde{\bf{r}}$ with respect to this rotating frame, at a distance $R$ away from the origin. Note that this distance $R$ is correct for both frames since their origins coincide by assumption. It is clear that this observer, as seen from the $O$ frame is given by the position vector $\bf{r}$ as follows:

\begin{equation} \textbf{r} = R(\cos{\omega t}, \sin{\omega t}, 0) \end{equation}

Hence the vector $\widetilde{\bf{f}}$ is given by:

\begin{equation} \widetilde{\textbf{f}} = \textbf{f} - \textbf{r} = (f_x - R\cos(\omega t), f_y - R\sin(\omega t), f_z) \end{equation}

Differentiating the above with respect to time we get:

\begin{align} \dot{\widetilde{\textbf{f}}} &= (\dot{f_x} + \omega R\sin(\omega t), \dot{f_y} - \omega R\cos(\omega t), \dot{f_z}) \\ & = \dot{\textbf{f}} + \omega R (\sin(\omega t), -\cos(\omega t), 0) \\ & = \dot{\textbf{f}} + \omega (r_y, -r_x, 0) \end{align}

Isolating $\dot{\mathbf{f}}$ we get:

\begin{equation} \dot{\textbf{f}} = \dot{\widetilde{\textbf{f}}} + \omega (-r_y, r_x, 0) \end{equation}

Now as you can see, what I'm getting is slightly different than what I'm supposed to get according to the theorem, the above in fact reads:

$$\left(\frac{d\mathbf{f}}{dt}\right)_O = \left( \frac{d\mathbf{f}}{dt}\right )_\widetilde{O} + \mathbf{\Omega} \times \mathbf{r} $$

Where I've replaced $\dot{\mathbf{f}}$ and $\dot{\widetilde{\mathbf{f}}}$ with the "abstract" notation that denotes differentiating with respect to each frame $O$ and $\widetilde{O}$, just to make it look more similar to how the theorem is usually stated.

However, what I'm supposed to get according to the theorem is in fact:

$$\left(\frac{d\mathbf{f}}{dt}\right)_O = \left( \frac{d\mathbf{f}}{dt}\right )_\widetilde{O} + \mathbf{\Omega} \times \mathbf{f} $$

Where is my error? I suspect that it may have to do with how I am interpreting the operation of differentiating the vector "in the rotating frame", for example I'm not totally sure it's correct to say that: $\dot{\widetilde{\textbf{f}}} = \left( \frac{d\mathbf{f}}{dt}\right )_\widetilde{O}$

Also it's very weird that the final expression I got depends on the position of the observer in the rotating frame, but I can't find what causes this error either.

$\endgroup$
1
  • 1
    $\begingroup$ Dear @amit, i think the theorem doesn't account for the translation of one frame with respect to another. The most often seen approach is to write down $\vec f = f'_{1} e'_{1}+f'_{2} e'_{2}+f'_{3} e'_{3}$ where $ e'_{i}$ are the coordinates of the rotating frame. And then we make use of the product rule of differentiation and differentiate both sides. I think this is the argument, that at a given moment, they observe the same vector (like, the same "physical vector", since laws of physics are not coordinate dependent), but due to the non inertial nature,the time derivatives are different $\endgroup$ Commented Apr 4, 2023 at 8:08

1 Answer 1

2
$\begingroup$

As mentioned in a comment, I've noticed that my method is inadequate once I saw that putting $R=0$ will lead to no rotation at all of the supposedly rotating frame. The proper way to describe vectors in different frames is, as @nickbros123 mentioned, via expressing the same vector in different coordinate bases. Once this is done correctly, it seems that the derivation is quite simple:

Let $\mathbf{b_1},\mathbf{b_2},\mathbf{b_3}$ be any orthonormal basis of the non rotating frame $O$ and $\widetilde{\mathbf{b_1}},\widetilde{\mathbf{b_2}},\widetilde{\mathbf{b_3}}$ those of the rotating frame $\widetilde{O}$, that's rotating with angular velocity $\mathbf{\Omega} = (0,0,\omega)$. So we have that:

\begin{align} \widetilde{\mathbf{b_1}} &= \cos{\omega t}\mathbf{b_1} + \sin{\omega t}\mathbf{b_2} \\ \widetilde{\mathbf{b_2}} &= -\sin{\omega t}\mathbf{b_1} + \cos{\omega t}\mathbf{b_2} \\ \widetilde{\mathbf{b_3}} &= \mathbf{b_3} \end{align}

Now the statement that for any vector $\mathbf{f}$ it must hold that $\left(\mathbf{f}\right)_O = \left(\mathbf{f}\right)_{\widetilde{O}}$ is simply that the vector is the same regardless of the basis it is expressed by, that is, if it has components $(f_x,f_y,f_z)$ with respect to $O$ and components $(\tilde{f_x},\tilde{f_y},\tilde{f_z})$ with respect to $\widetilde{O}$ then it must hold that:

$$ f_x\mathbf{b_1}+f_y\mathbf{b_2}+f_z\mathbf{b_3} = \tilde{f_x}\widetilde{\mathbf{b_1}}+\tilde{f_y}\widetilde{\mathbf{b_2}}+\tilde{f_z}\widetilde{\mathbf{b_3}} $$

Differentiating the left hand side simply yields $\left(\frac{d\mathbf{f}}{dt}\right)_O = (\dot{f_x},\dot{f_y},\dot{f_z})$ and there are no additional terms in that basis simply because by assumption $\dot{\mathbf{b_i}} = 0$ for $i=1,2,3$. Differentiating also the right hand side, we get:

\begin{align} \left(\frac{d\mathbf{f}}{dt}\right)_O &= \dot{\tilde{f_x}}\widetilde{\mathbf{b_1}}+\dot{\tilde{f_y}}\widetilde{\mathbf{b_2}}+\dot{\tilde{f_z}}\widetilde{\mathbf{b_3}}+\tilde{f_x}\dot{\widetilde{\mathbf{b_1}}}+\tilde{f_y}\dot{\widetilde{\mathbf{b_2}}}+\tilde{f_z}\dot{\widetilde{\mathbf{b_3}}} \\ &= \left(\frac{d\mathbf{f}}{dt}\right)_\widetilde{O} + \tilde{f_x}\dot{\widetilde{\mathbf{b_1}}},+\tilde{f_y}\dot{\widetilde{\mathbf{b_2}}}+\tilde{f_z}\dot{\widetilde{\mathbf{b_3}}} \end{align}

Now since we note that:

$$\dot{\widetilde{\mathbf{b_1}}} = -\omega\sin{\omega t}\mathbf{b_1}+\omega\cos{\omega t}\mathbf{b_2} = \omega\widetilde{\mathbf{b_2}} \\ \dot{\widetilde{\mathbf{b_2}}} = -\omega\cos{\omega t}\mathbf{b_1}-\omega\sin{\omega t}\mathbf{b_2} = -\omega\widetilde{\mathbf{b_1}} \\ \dot{\widetilde{\mathbf{b_3}}} = 0$$

We obtain:

\begin{align} \left(\frac{d\mathbf{f}}{dt}\right)_O &= \left(\frac{d\mathbf{f}}{dt}\right)_\widetilde{O} + \tilde{f_x}\dot{\widetilde{\mathbf{b_1}}},+\tilde{f_y}\dot{\widetilde{\mathbf{b_2}}}+\tilde{f_z}\dot{\widetilde{\mathbf{b_3}}} \\ &= \left(\frac{d\mathbf{f}}{dt}\right)_\widetilde{O} + \omega\tilde{f_x}\widetilde{\mathbf{b_2}} -\omega\tilde{f_y}\widetilde{\mathbf{b_1}} \\ &= \left(\frac{d\mathbf{f}}{dt}\right)_\widetilde{O} + \left(\mathbf{\Omega} \times \mathbf{f}\right)_{\widetilde{O}} \end{align}

Which appears to be the correct result.


Additional observations and useful resources

  1. Despite the fact that the above argument assume a very special looking form for $\mathbf{\Omega}$ it can in fact be easily generalized, by considering that the orthonormal basis which is our starting point $\mathbf{b_1},\mathbf{b_2},\mathbf{b_3}$ can be related to any other fixed orthonormal basis by a fixed rotation. With respect to this other basis then, $\mathbf{\Omega}$ will no longer have this special form, and yet such a fixed rotation clearly leaves the rotated basis vectors as time independent, and hence does not affect the rest of the derivation.

  2. Since writing this, I have found two very good posts that are very much related and worth reading. The first one is a derivation of the centrifugal and Coriolis force terms that appear in a rotating reference frame, which basically derives the same theorem (without naming it, which is why it took me a long time to find!) in a slightly different way. The second one is a truly beautiful mathematical treatment that much more generally derives the existence of all the known fictitious forces that arise in a non inertial frame of reference, the kinematic transport theorem can also be derived by applying the same techniques used there.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.