# Rotation systems. Problem interpreting an equation

In this equation: $$\mathbf a_i\overset{\rm def}{=}\left(\frac{d^2\mathbf r}{dt^2}\right)_i=\left(\frac{d\mathbf v}{dt}\right)_i=\left[\left(\frac{d}{dt}\right)_r+\boldsymbol\Omega\times\right]\left[\left(\frac{d\mathbf r}{dt}\right)_r+\boldsymbol\Omega\times\mathbf r\right]$$ (from Wikipedia), why is $$\left(\frac{d}{dt}\right)_r \boldsymbol{\Omega} \times \mathbf{r}=\frac{d\boldsymbol{\Omega}}{dt}\times\bf{r}+\bf{\Omega}\times \bf{V_r}$$

In particular, I have qualms with the term $$\frac{d\bf{\Omega}}{dt}\times\bf{r}$$

Why are we deriving the angular velocity? Why is it a derivative not of the rotational type (Namely $(\frac{d}{dt})_r$ )

Other sources do not point out that term I have problems with. In any case, I want to know how you evaluate that derivative.

• Deleted my previous answer as I misunderstood the question but I can direct you to where the answer is. Get the V.I. Arnold's book (mathematical methods of classical mechanics) if you don't have it already, and take a look at the theorem on page 130. There's no mathemagic in this book, he proves every step and everything is well defined. Commented Nov 26, 2014 at 7:15
• I'm not understanding their notation :(
– DLV
Commented Nov 26, 2014 at 17:07
• I am trying to understand your problem. The calculi in Wikipedia seem to me very simple, so, where do you have difficulty? You ask: "Why are we deriving the angular velocity?" Because in general it is not constant. Do I misunderstand you problem? Next you ask: " Why is it a derivative not of the rotational time (Namely (d/dt)_r )?" What you mean by "rotational time" ? Commented Nov 26, 2014 at 19:39
• He wants to know why $(\frac{d\bf{\Omega}}{dt})_r =(\frac{d\bf{\Omega}}{dt})$ Commented Nov 26, 2014 at 23:41
• David: I looked at the proof in Wikipedia, from the beginning. What I saw is that the angles θ are defined in the Inertial frame. Then, Ω too is defined in the inertial frame. I see also that the origins of the two frames are taken the same. By the way, for a frame rotating with angular velocity Ω, the angular velocity with respect to this frame should be zero. But, it's better that you follow the proof, step by step, to be sure in which frame is defined each quantity. Commented Nov 27, 2014 at 11:39

I see now your problem and I believe that I can help.

Let's begin from the velocity formula

$$v_i = v_r + Ω \times r .\tag{1}$$

Let's take the derivative of $v_i$ IN THE INERTIAL frame,

$$a_i = \left(\frac{dv_r}{dt}\right)_i + \left(\frac{dΩ}{dt}\right)_i \times r + Ω \times v_i .$$

Here we use as much as we can our formula $(dF/dt)_i = (dF/dt)_r + Ω \times F$, and also substitute $v_i$ in the last term, by its formula (1). So,

$$a_i = \left(\frac{dv_r}{dt}\right)_r + Ω \times v_r + \left(\frac{dΩ}{dt}\right)_i \times r + Ω \times v_r + Ω \times Ω \times r .$$

Gathering identical terms,

$$a_i = a_r + 2Ω \times v_r + \left(\frac{dΩ}{dt}\right)_i \times r + Ω \times Ω \times r .$$

This is what they got in Wikipedia.

• Hmm. Vector $r$ in your second equation is measured in the non-inertial frame, which makes it hard to take a derivative in the inertial frame. Is this a problem, or is it not? Thanks.
– DLV
Commented Nov 27, 2014 at 22:07
• Would you first make clear where disappeared your question? You did not mark your it as answered, but I don't see it in the list of unanswered questions. Anyway, it is a useful question, it may serve other people too. Commented Nov 28, 2014 at 9:42
• Now, to your comment/doubt: thank you for calling my attention and for following attentively my proof. I will post an additional answer. For clarity let me denote by capital letters quantities measured in the inertial frame, including derivatives (i.e. D/Dt) and by small letters quantities measured in the rotating frame. It will spare me subscripts Commented Nov 28, 2014 at 9:56
• The way you do a subscript with Latex is exactly what you said. F_b surrounded in dollar signs is $F_b$. Commented Nov 28, 2014 at 14:10
• To Lionel: I know that I should use LaTex. But I don't have patience for that, please believe me. People tell me that it's simple, but I am under big pressure of work. I will look at how to use LaTeX when I have a bit more time. Commented Nov 28, 2014 at 20:27

Other sources do not point out that term I have problems with.

Other sources explicitly assume a constant angular velocity and thus ignore that component. The wikipedia article you cited is correct.

In any case, I want to know how you evaluate that derivative.

Given any vector quantity $\mathbf q$ that is the same (other than component representation) in the inertial and rotating frame, the time derivative of that vector from the perspective of an inertial versus rotating observer is $$\left(\frac{d\mathbf q}{dt}\right)_I = \left(\frac{d\mathbf q}{dt}\right)_R + \boldsymbol\Omega \times \mathbf q$$ In dynamics, this is sometimes called the transport theorem (but there are a number of other things called the transport theorem).

Applying the transport theorem to the angular velocity vector yields $$\left(\frac{d\boldsymbol\Omega}{dt}\right)_I = \left(\frac{d \boldsymbol\Omega}{dt}\right)_R + \boldsymbol\Omega \times \boldsymbol\Omega = \left(\frac{d \boldsymbol\Omega}{dt}\right)_R$$ In other words, angular acceleration is fundamentally the same vector in the inertial and rotating frame.

Applying the transport theorem instead to angular momentum yields $$\left(\frac{d\mathbf L}{dt}\right)_I = \left(\frac{d\mathbf L}{dt}\right)_R + \boldsymbol\Omega \times \mathbf L$$ The rotational analog of Newton's second law provides an alternative representation of the left-hand side of the above: $\frac {d\mathbf L}{dt} = \boldsymbol \tau_{\text{ext}}$ where the derivative is calculated from the perspective of an inertial frame and $\boldsymbol \tau_{\text{ext}}$ is the external torque on the system. If the system is a rigid body, the angular momentum is given $\mathbf L = \mathrm I \boldsymbol\Omega$ where $\mathrm I$ is the object's inertia tensor. Since the inertia tensor of a rigid body is constant in a frame rotating with the body, the time derivative of the angular momentum vector from the perspective of an observer rotating with the object simplifies to $\left(\frac{d\mathbf L}{dt}\right)_R = \mathrm I\left(\frac{d \boldsymbol\Omega}{dt}\right)_R$. Putting all of the above together yields $$\boldsymbol\tau_{\text{ext}} = \mathbf I \frac{d \boldsymbol\Omega}{dt} + \mathbf \Omega \times (\mathrm I \, \boldsymbol\Omega)$$ or $$\frac {d \boldsymbol\Omega}{dt} = {\mathbf I}^{-1} \left( \boldsymbol\tau_{\text{ext}} - \boldsymbol \Omega \times (\mathrm I \, \boldsymbol \Omega) \right)$$

This yields a way to calculate $\frac {d\boldsymbol \Omega}{dt}$ at any point in time for a rigid body. Whether this is integrable via elementary methods is a different question. In most cases, it isn't. It's rather challenging to find a non-trivial rotational system that has an analytic solution. One typically has to revert to numerical methods to determine the rotational behavior of an object.

If you just follow your nose, then... $$\left(\frac{d}{dt}\right)_{rotating} {\bf{\Omega}} =\frac{d\bf{\Omega}}{dt}+\bf{\Omega}\times {\bf{\Omega}}$$ Do you know what the second term is equal to? Hopefully this clears up the problem you have.

• Lionel, what is simple for you may be complicated for other people. I am not in the position to give advice, but just as a non-formal comment, I know that a detailed explanation, step by step, is welcome. Commented Nov 28, 2014 at 20:30
• Thanks. I'll try to keep that in mind, but at this level of physics, people need to be given opportunity to work things out for themselves (with hints, of course). Commented Nov 28, 2014 at 20:54
• Thank you for considering my comment. Let me though add something: books, articles, class lectures, are full of mathematics, a rain of formulas. The physical/phenomenological meaning of the issues, of the formulas, is left in shadow. This is a chronic disease. This is why people are eager for very simple explanations which stress the intuitive part. Commented Nov 28, 2014 at 22:14

$$\mathbf V = \mathbf v + \boldsymbol \Omega \times \mathbf r$$

The derivative of $\mathbf V$ in the inertial frame is indeed,

$$\mathbf A = \frac{\mathrm D \mathbf v}{\mathrm Dt} + \frac{\mathrm D \boldsymbol \Omega}{\mathrm D t} \times \mathbf r + \boldsymbol \Omega \times \frac{\mathrm D \mathbf r}{\mathrm Dt}.$$

You are right, both $\mathbf v$ and $\mathbf r$ are described according to the rotating axes. Though, for the observer in the inertial frame they rotate together with those axes. So, for doing the derivative according to the inertial frame, you derivate again according to the rotating axes and then derivate the movement of the rotating axes. Let's write that:

\begin{align} \mathbf A &= \frac{\mathrm d \mathbf v}{\mathrm dt} + \boldsymbol \Omega \times \mathbf v + \frac{\mathrm D \boldsymbol \Omega}{\mathrm Dt} \times \mathbf r + \boldsymbol \Omega \times (\mathbf v + \boldsymbol \Omega \times \mathbf r) \\ & = \mathbf a + 2\boldsymbol \Omega \times \mathbf v + \frac{\mathrm D \boldsymbol \Omega}{\mathrm Dt} \times \mathbf r + \boldsymbol \Omega \times \boldsymbol \Omega \times \mathbf r . \end{align}