How to derive relation for time derivative in a rotating reference frame

I am looking for an appropriate derivation of the $(\frac{d}{dt})_{\text{laboratory}} = (\frac{d}{dt})_{\text{rotating}} + \omega \times$ relationship that enables one to calculate all desired quantities in a rotating referece frame. Does anybody know a good way how to understand this transformation?

• Your equation looks incomplete. – udiboy1209 Aug 24 '13 at 16:00
• Could you be more specific? – Xin Wang Aug 24 '13 at 17:00
• $\frac d{dt}$ of what? and $\omega \times$ what? – udiboy1209 Aug 25 '13 at 15:37
• it is an operator relationship, so you would apply it to a vector in a particular reference system. – Xin Wang Aug 25 '13 at 16:57

The components of any vector function can be written any any desired basis. In particular, let \begin{align} \mathbf A_L(t) = (A^1_L(t) , A^2_L(t), A^3_L(t)) \end{align} denote the components of a vector function as written in an orthonormal basis fixed in the laboratory, and let \begin{align} \mathbf A_R(t) = (A^1_R(t), A^2_R(t), A^3_R(t)) \end{align} denote the components of the same vector as written in a rotating orthonormal basis. These components will be related by a time-dependent special orthogonal matrix (rotation); \begin{align} \mathbf A_L(t) = R(t)\mathbf A_R(t) \end{align} In particular, note that taking time derivatives on both sides gives \begin{align} \dot{\mathbf A}_L(t) &= R(t) \dot{\mathbf A}_R(t) + \dot R(t) \mathbf A_R(t) \\ &= R(t) \dot{\mathbf A}_R(t) + \dot R(t) R(t)^T\mathbf A_L(t) \end{align} Since $R(t)$ is an orthogonal matrix, we have \begin{align} R(t)R(t)^T = I \end{align} and taking derivatives of both sides, and using the fact that time derivatives and matrix transposes commute, we find that \begin{align} \dot R(t) R(t)^T = -(\dot R(t)R(t)^T)^T \end{align} in other words, $\dot R(t) R(t)^T$ is an antisymmetric matrix. We lose no generality by therefore writing \begin{align} \dot R(t) R(t)^T = \Omega(t) \end{align} where \begin{align} \Omega(t) = \begin{pmatrix} 0 & -\omega^3(t) & \omega^2(t) \\ \omega^3(t) & 0 & -\omega^1(t) \\ -\omega^2(t) & \omega^1(t) & 0 \\ \end{pmatrix} \end{align} for some vector of functions $\boldsymbol\omega=(\omega^1, \omega^2, \omega^3$). Therefore, we have \begin{align} \dot {\mathbf A}_L(t) = R(t)\dot{\mathbf A}_R(t) + \Omega(t)\mathbf A_L(t) \end{align} It is straightforward to explicitly show that multiplication by $\Omega(t)$ is equivalent to a cross product by $\boldsymbol\omega(t)$, so we can write \begin{align} \Omega(t) \mathbf A_L(t) = \boldsymbol\omega(t)\times\mathbf A_L(t) \end{align} and we are therefore let to the expression \begin{align} \boxed{\dot{\mathbf A}_L(t) = R(t)\dot{\mathbf A}_R(t) + \boldsymbol\omega(t)\times\mathbf A_L(t)} \end{align} If we make the identifications \begin{align} \dot{\mathbf A}_L(t) &=\left(\frac{d\mathbf A}{dt}\right)_\mathrm{laboratory} \\ R(t)\dot{\mathbf A}_R(t) &=\left(\frac{d\mathbf A}{dt}\right)_\mathrm{rotating} \end{align} then we see that this is equivalent to your formula. In my opinion, the physicist notation in the formula you wrote down is extremely confusing, and I prefer to use the more descriptive notation in the boxed expression above; I find that it leads to less errors and is more conceptually clear.

• Could You show the part where the multiplication with $\Omega(t)$ turns into a cross product with the angular velocity. – Alexander Cska Dec 22 '18 at 13:15
• @AlexanderCska That can be verified by explicitly computing the matrix product $\Omega \mathbf A$ and the cross product $\boldsymbol \omega \times \mathbf A$ and comparing the results. Have you tried that? – joshphysics Dec 22 '18 at 20:12
• i tried to understand the following approach. But failed to do so. damtp.cam.ac.uk/user/tong/dynamics/three.pdf – Alexander Cska Dec 23 '18 at 8:57

I'll consider two-dimensional reference frames, and i will use galilean transformation, adapted to your particular problem. Here is how i imagine this: Here, $\vec{r}$ is the position vector of a point, as seen from the laboratory frame; $\vec{r_0}$ is the position vector of the center of the other (moving) frame; and $\vec{r'}$ is the position of the point as seen from the primed frame. Note that $\vec{r'}$ will describe a circle, so we can write it as: $$\vec{r'}=r'\cos(\omega t) \vec{i}+r'\sin(\omega t) \vec{j}$$ (assuming that the particle starts at an angle $0$ when $t=0$).

Now, clearly, we have $\vec{r}=\vec{r_0}+\vec{r'}$. Differentiating this with respect to time, we get: $${\frac{d\vec{r}}{dt}}={\frac{d\vec{r_0}}{dt}}+{\frac{d\vec{r'}}{dt}}$$

which is the same as $$\vec{v}=\vec{v_0}+\vec{v'}$$ where $\vec{v}$ is the speed as seen from the laboratory frame, $\vec{v_0}$ is the speed of the moving frame, and $\vec{v'}$ is the speed in the moving frame.

But $\vec{v'}={\frac{d\vec{r'}}{dt}}=-\omega r'\sin(\omega t)\vec{i}+\omega r'\sin(\omega t)\vec{j}$, so that we finally get: $$\vec{v}=\vec{v_0}+\omega r' [-\sin(\omega t)\vec{i}+\cos(\omega t)\vec{j}]$$

To find the accelerations, you can differentiate again this relation.