# Derivative of angular momentum in a rotating frame of reference

I keep seeing that when taking the time derivative of the angular momentum in a rotating reference frame, we get: $$\frac{d\vec{L}}{dt} = \vec{\tau} + \vec{\omega} \times \vec{L}$$ meaning the torque as the rotating frame sees it, plus another term.

Is it derived somewhere in the internet? I couldn't find it. I like to understand what I am doing, but in this instance I have no idea why the derivative gives this extra term.

• What do you mean? the units are $kg*\frac{m^2}{sec^2}$ in mks – Darkenin Mar 12 '20 at 11:07

## 3 Answers

Consider total derivative of $$\vec{A}$$ in an inertial frame, $$\frac{d}{dt}\vec{A}=\hat{i}\frac{d}{dt}A_x+\hat{j}\frac{d}{dt}A_y+\hat{k}\frac{d}{dt}A_z+A_x\frac{d\hat{i}}{dt}+A_y\frac{d\hat{j}}{dt}+A_z\frac{d\hat{k}}{dt}.$$ In an inertial system $$\frac{d\hat{i}}{dt}=\frac{d\hat{j}}{dt}=\frac{d\hat{k}}{dt}=0$$. Thus we have, $$\frac{d\vec{A}}{dt}_{inertial}=\hat{i}\frac{d}{dt}A_x+\hat{j}\frac{d}{dt}A_y+\hat{k}\frac{d}{dt}A_z.$$ Now consider in a rotating frame one has $$\vec{A}=A_x^\prime\hat{i}^\prime+A_y^\prime\hat{j}^\prime+A_z^\prime\hat{k}^\prime$$. Taking the total derivative, $$\frac{d}{dt}\vec{A}=\hat{i}^\prime\frac{d}{dt}A_x^\prime+\hat{j}^\prime\frac{d}{dt}A_y^\prime+\hat{k}^\prime\frac{d}{dt}A_z^\prime+A_x^\prime\frac{d\hat{i^\prime}}{dt}+A_y^\prime\frac{d\hat{j^\prime}}{dt}+A_z^\prime\frac{d\hat{k}^\prime}{dt}.$$ Note that in a rotating frame the quantities like $$\frac{d\hat{i}^\prime}{dt}$$ do not vanish. Therefore the total time derivative in a rotating frame is,

$$\frac{d\vec{A}}{dt}=\frac{d\vec{A}}{dt}_{rotating}+A_x^\prime\frac{d\hat{i^\prime}}{dt}+A_y^\prime\frac{d\hat{j^\prime}}{dt}+A_z^\prime\frac{d\hat{k}^\prime}{dt}$$ $$\frac{d\vec{A}}{dt}_{rotating}=\hat{i}^\prime\frac{d}{dt}A_x^\prime+\hat{j}^\prime\frac{d}{dt}A_y^\prime+\hat{k}^\prime\frac{d}{dt}A_z^\prime$$ is the apparent time derivative of the vector in a rotating frame and the combination $$A_x^\prime\frac{d\hat{i^\prime}}{dt}+A_y^\prime\frac{d\hat{j^\prime}}{dt}+A_z^\prime\frac{d\hat{k}^\prime}{dt}$$ captures the effects of rotation. Now, we know the relation between the linear velocity and angular velocity, $$\vec{v}=\frac{d\vec{r}}{dt}=\omega\times\vec{r}$$. Choosing $$\vec{r}$$ to be equal to $$\hat{i}^\prime$$, $$\hat{j}^\prime$$, and $$\hat{k}^\prime$$ respectively we have, \begin{align}\frac{d\hat{i}^\prime}{dt}&=\omega\times\hat{i}^\prime\\\frac{d\hat{j}^\prime}{dt}&=\omega\times\hat{j}^\prime\\\frac{d\hat{k}^\prime}{dt}&=\omega\times\hat{k}^\prime\end{align} Using these we get in a rotating frame, $$\frac{d\vec{A}}{dt}=\frac{d\vec{A}}{dt}_{rotating}+\omega\times\vec{A}$$ Choosing $$\vec{A}=\vec{L}$$ we get, $$\frac{d\vec{L}}{dt}=\frac{d\vec{L}}{dt}_{rotating}+\omega\times\vec{L}=\vec{\tau}+\omega\times\vec{L}.$$

• Great answer, I'd also like to briefly note the parallels to the material derivative, $D/Dt = (\partial_t + v \cdot \nabla)$, and this "rotating" derivative: $D/Dt = (\partial_t + \omega \times)$. Thinking about this further may help solidify this result in your brain, as it did for me back when (also note the linking of inertial/rotating frames to lagrangian/eulerian frames). – Novice C Mar 12 '20 at 14:39
• That's an interesting thought @NoviceC. However, I think the quantities $\omega \times (\cdot)$ and $v \cdot \nabla (\cdot)$ are conceptually more distinct than similar, with the only simlarity in form arising from the application of the chain rule. On the other hand, I there is a similarity in the product rule of complex numbers and the cross product of vectors. I think this is the key idea underlying the Hamiltonian product rule, but I have not mastered that topic yet. Let me know what you think. – kbakshi314 Mar 12 '20 at 17:06
• I do not mean to imply any mathematical similarity between the cross product and dot product of the gradient. I am merely trying to impart physical intuition by expanding upon what the chain rule gives you. From the fluid side, consider what context the material derivative can be used in. Euler's equation written with the material derivative looks like Newton's Second Law: $Dp/Dt =$ forces. (1/2) – Novice C Mar 21 '20 at 19:46
• We also know that in the rotating problem $\omega \times$ is what gives rise to the non-inertial terms, and likewise $v \cdot \nabla$ is what deviates Euler's equation from Newton's Second law for fluids. These terms are "forces" that arise from infinitesimal changes in the "frame," either by rotating the frame or advecting it through a fluid with gradients. (2/2) – Novice C Mar 21 '20 at 19:47

I prefer to use this notation.

The components of a arbitrary vector $$\vec{x'}$$ in rotating system are transformed to inertial system by this equation:

$$\vec{x}=R\,\vec{x'}\tag 1$$

where R is the transformation matrix between the rotating system and inertial system

The time derivative of equation (1) is:

$$\vec{\dot{x}}=R\,\vec{\dot{x}'}+\dot{R}\,\vec{x'}\tag 2$$

with

$$\dot{R}=R\,\tilde{\omega}\quad$$ and $$\tilde{\omega}=\left[ \begin {array}{ccc} 0&-\omega_{{z}}&\omega_{{y}} \\ \omega_{{z}}&0&-\omega_{{x}}\\ -\omega_{{y}}&\omega_{{x}}&0\end {array} \right]$$ thus: $$\vec{\dot{x}}=R\,\vec{\dot{x}'}+R\,\tilde{\omega}\,\vec{x'}\tag 3$$

multiply equation (3) from the left with $$R^T$$

$$R^T\,\vec{\dot{x}}=\vec{\dot{x}'}+\vec{\omega}\times \vec{x'}$$

thus

$$\boxed{\left(\vec{\dot{x}}\right)_R=\left(\vec{\dot{x}'}\right)_R+ \left(\vec{\omega}\times \vec{x'}\right)_R}$$

where index R means the components are given in the rotating system.

I like to understand what I am doing, but in this instance I have no idea why the derivative gives this extra term.

The extra term is pretty clear to understand on physical grounds as follows:

Consider a gyroscope in the absence of any external torque. The gyroscope maintains a fixed angular momentum with respect to the axes of an inertial frame.

The axes of the rotating frame rotate with respect to the inertial frame, so the gyroscope must rotate with respect to the rotating frame. This torque-free rotation is the additional term.