# The time derivatives of vectors in rotating frames

The time rate of change of the vector $$\mathbf{G}$$ as seen by the two observers is then obtained by dividing the terms in Eq. (4.81) by the differential time element $$dt$$ under consideration; $$\bbox[yellow]{ \left(\frac{d\mathbf{G}}{dt}\right)_\text{space} =\left(\frac{d\textbf{G}}{dt}\right)_\text{body} +\boldsymbol{\omega}\times\mathbf{G}. } \tag{4.82}$$

The quote is from Goldstein's book. Here he is saying that the vector in LHS is observed in space frame and the $$\frac{d\textbf{G}}{dt}$$ in RHS is in body or rotaing frame. By using this he proved the expressions for Coriolis force and Euler's equations (rigid body dynamics). With his interpretation the $$\frac{d\textbf{L}}{dt}$$ in RHS should be zero(as angular momentum in the rotating frame is zero since everything ia at rest.). I read this answer and thought that Goldstein's interpretation is wrong and RHS time derivative is just expressed in rotating basis and is not the vector seen in rotating frame. During the proof of time derivatives formula itself we took that the vector is just expressed in rotating basis. But they generally say that Coriolis force is a force observed in rotating frame. Is that wrong? Or in this case somehow acceleration expressed in rotating basis coincides with that in rotating frame?

• You haven't defined the quantity dL/dt Nov 20, 2020 at 17:20
• @Buraian It's the angular momentum. Nov 20, 2020 at 17:20
• I think you should put some context behind it Nov 20, 2020 at 17:22
• @Buraian I added a wiki link to Euler's angles now. Should I add any other thing? Nov 20, 2020 at 17:24
• I had the same thoughts, the way I resolved this is by understanding that everything needs to be on a inertial frame. This way $\omega$ isn't zero on a body frame, but just rotated to align to the body coordinates. Nov 20, 2020 at 18:59

You have two coordinate system , the inertial system index I, and the body fixed system index B, thus the components of a vector in I-system are:

$$\left(\vec G\right)_I=R\,\left(\vec G\right)_B\tag 1$$ where $$R$$ is orthonormal transformation matrix between B-system and I-system

the derivative of eq. (1) is:

$$\frac{d}{dt}\left(\vec G\right)_I=R\,\frac {d}{dt}\left(\vec G\right)_B+ \dot R\,\left(\vec G\right)_B$$

with $$\dot R\,=R\, \left[ \begin {array}{ccc} 0&-\omega_{{z}}&\omega_{{y}} \\ \omega_{{z}}&0&-\omega_{{x}}\\ -\omega_{{y}}&\omega_{{x}}&0\end {array} \right] _B$$

you obtain:

$$\frac{d}{dt}\left(\vec G\right)_I=R\,\frac {d}{dt}\left(\vec G\right)_B+ R\,\left[\vec\omega_B\times \left(\vec G\right)_B\right]\tag 2$$

or :

$$R^T\,\frac{d}{dt}\left(\vec G\right)_I=\frac {d}{dt}\left(\vec G\right)_B+ \vec\omega_B\times \left(\vec G\right)_B\tag 3$$

you can compare now eq. (3) with Goldstein notation:

$$\left(\frac{d\vec G}{dt}\right)_{\text{space}}=\left(\frac{d\vec G}{dt}\right)_{\text{body}}+\vec{\omega}\,\times \vec{G}$$

Angular momentum is a vector, thus

$$\left(\vec L\right)_I=R\,\left(\vec L\right)_B$$ with $$\left(\vec L\right)_B=I_B\,\vec{\omega}_B$$

$$\vec{ \dot{ L}}_I=R\,I_B\,\vec{\dot\omega}_B+\dot{R}\,I_B\,\vec{\omega}_B=\left(\vec{\tau}\right)_I$$

$$R\,I_B\,\vec{\dot\omega}_B+R\left(\omega_B\,\times\,I_B\,\vec{\omega}_B\right)=\left(\vec{\tau}\right)_I$$

thus the Euler equation

$$I_B\,\vec{\dot\omega}_B+\omega_B\,\times\,I_B\,\vec{\omega}_B=R^T\,\left(\vec{\tau}\right)_I=\left(\vec{\tau}\right)_B$$

• What about that angular momentum question I asked? Nov 22, 2020 at 16:51
• @KasiReddySreemanReddy i wrote you the derivation of the Euler equations
– Eli
Nov 22, 2020 at 17:21

Consider a any vector $$\mathbf{B}$$ that undergoes pure rotation around axis $$\mathbf{\hat{n}}$$ at rate $$\Omega$$.

$$\frac{d\mathbf{B}}{dt}=\mathbf{\Omega}\times\mathbf{B}$$

which can be proved with elementary math (see Section 9.5.1 in Introduction to Mechanics Klepner).

Consider any vector $$\mathbf{C}$$ that is changing at rate $$(d\mathbf{C}/{dt})_{in}$$ as observed in an inertial system. The problem is to find the time derivative $$(d\mathbf{C}/dt)_{rot}$$ as observed in a system rotating at rate $$\mathbf{\Omega}$$. Now what do we mean by this?

Vector is physically measurable, so its magnitude and direction remain the same regardless of what coordinate system we choose to assign its components. In calculating the time derivative of $$(d\mathbf{C}/dt)_{rot}$$ we must take into account that the base vectors in the rotating system all rotate with angular velocity $$\Omega$$. So There will be two-part One due to the rotating vector and the other due to the rotating basis. As $$\frac{d\hat{i'}}{dt}=\mathbf{\Omega}\times \hat{i}'$$ similarly for other two, so that $$\frac{d\mathbf{C}}{dt}=\left(\frac{dC'_x}{dt}+\cdots\right) +\left( C'_x\frac{d\hat{i}'}{dt}+\cdots\right)$$ The first one is $$(d\mathbf{C}/dt)_{rot}$$, the time derivative of $$\mathbf{C}$$ that would be measured by an observer in the rotating frame. You can work out further to show $$\left( \frac{d\mathbf{C}}{dt}\right)_{in}=\left( \frac{d\mathbf{C}}{dt}\right)_{rot}+\Omega\times\mathbf{C}$$ So that's the meaning what Goldstein says.

But they generally say that Coriolis force is a force observed in a rotating frame. Is that wrong?

Coriolis force is not some new force that only acts on the rotating frame. You can do the problem in an inertial frame and you will see the Coriolis term as $$\mathbf{\ddot{r}}=(\ddot{r}-r\dot{\theta}^2)\hat{r}+(r\ddot{\theta}+2\dot{r}\dot{\theta})\hat{\theta}$$

the second term of $$\hat{\theta}$$ is what Coriolis force is which come so naturally due to the fact we are using polar coordinate so it doesn't have that new name. But when you use a non-inertial frame pseudo force is a debt that you pay. You need to assure them to balance the fact that you are in a non-inertial frame.

Or in this case somehow acceleration expressed in rotating basis coincides with that in rotating frame?

Yeah! This is true because the observer is in a rotating frame like earth so its basis actually coincides with the angular velocity of the system (earth in this case). If you consider a two spherical system with the same center, both rotating with different angular velocity then the basis in one will not consider with the other one.

• "The first one is (dC /dt)rot, the time derivative of C that would be measured by an observer in the rotating frame. You can work out further to show"- I don't think that is true. If you observe in the rotating frame the angular momentum is always zero and so is its rate of change. So in the RHS (dC/dt)rot is not the vector observed in the rotating frame, it is measured in inertial frame but in a basis rotating with the rotating frame. Nov 20, 2020 at 18:22
• I'm not talking of angular momentum in some specific problem , I'm talking of any arbitrary vector that is rotating about some axis as Goldstein did. Nov 20, 2020 at 18:51
• I know you meant a general vector. But as an example I am considering the angular momentum. Nov 20, 2020 at 18:53
• So you are considering some vector that not rotating in inertial frame? Nov 20, 2020 at 18:58
• @ Kasi Reddy Sreeman Reddy is correct. Ang Mom in space frame L, is different vector from ang mom in body frame H which is zero. 4.82 applied to L evaluates L in body frame coordinates, it does not evaluate H. L and H are different vectors. Please see my answer. Aug 9, 2022 at 16:02

As Goldstein points out, only after the differentiation has been carried out in relationship (4.82) of the question, can the components be taken along another set of coordinate axes. That is $${d\vec G \over dt})_{space}$$ must be evaluated using space coordinates before it can be expressed in body coordinates, and $${d\vec G \over dt})_{body}$$ must be evaluated using body coordinates before it can be expressed in space coordinates.

Consider two coordinate systems $$S$$ and $$B$$, assumed to have a common origin. $$S$$ is a fixed, inertial system and $$B$$ is a non-inertial system rotating at angular velocity $$\vec \omega$$ about the common origin with respect to $$S$$. See Figure 1. (For a rigid body, $$S$$ denotes the space frame and $$B$$ the body frame.)

We wish to evaluate any vector $$\vec G$$ from the perspective of both coordinate systems. As developed in Goldstein, we can consider the rotation of the $$B$$ frame as a passive rotation of the coordinate system, and $$\vec G$$ is the same vector in both coordinate systems. [Ref 1] Alternatively, we can consider an active rotation that rotates the vector. Equivalent results are obtained using either the passive or the active rotation approach, as subsequently discussed. Let the vector $$(e_{1S}, e_{2S}, e_{3S})$$ represent the axes unit vectors in the $$S$$ system and the vector $$(e_{1B}, e_{2B}, e_{3B})$$ the axes unit vectors in the $$B$$ system. We can describe $$\vec G$$ in terms of the $$S$$ and $$B$$ frame coordinates as ;$$(1) \vec G = \sum_{i=1}^{3} G_{iS} \hat e{i_S}$$ $$(2) \vec G = \sum_{i=1}^{3} G_{iB} \hat e{i_B}$$ $$G_{iS}$$ and $$G_{iB}$$ are the magnitudes of the components of $$\vec G$$ along the $$i^{th}$$ axis in the $$S$$ and $$B$$ systems, respectively. The time derivatives of (1) and (2) with respect to the inertial frame are equal so: $$(3)\sum_{i=1}^{3} \dot G_{iS} \hat e{i_S} = \sum_{i=1}^{3} \dot G_{iB} \hat e{i_B} + \sum_{i=1}^{3} G_{iB} \dot {\hat e{i_B}}$$ [Ref 2] Note that in $$B$$ coordinates the time derivative has to account for the change in direction of the $$(e_{1B}, e_{2B}, e_{3B})$$ unit vectors in time. For consistency with the Goldstein nomenclature, Let $${d\vec G \over dt})_S$$ be defined as $$\sum_{i=1}^{3} \dot G_{iS} \hat e{i_S}$$, and let $${d\vec G \over dt})_B$$ be defined as $$\sum_{i=1}^{3} \dot G_{iB} \hat e{i_B}$$. [Ref 1] It can be shown that $$\sum_{i=1}^{3} G_{iB} \dot {\hat e{i_B}} = \vec \omega(t) \times \vec G$$. [Ref 2] Therefore, we can express (3) as $$(4) {d\vec G \over dt})_S = {d\vec G \over dt})_B + \vec \omega(t) \times \vec G$$ (4) is the relationship used in Goldstein and is the relationship addressed in your question. [Ref. 1] Since (4) is equivalent (3), we interpret $${d\vec G \over dt})_S$$ to be in $$S$$ coordinates , and $${d\vec G \over dt})_B$$ and $$= \vec \omega(t) \times \vec G$$ to both be in $$B$$ coordinates. However, as Goldstein points out, the vectors in (4) can be resolved along either the $$S$$ or $$B$$ axes, only after the differentiation has been carried out. [ Ref 1] This is addressed in subsequent examples.

Comparing (4) to (3), we can provide the following physical interpretations of each term in (4). The term on the left-hand side of (4) $${d\vec G \over dt})_S$$ is the time derivative of $$\vec G$$ as seen by an observer fixed in the inertial $$S$$ frame. The first term on the right-hand side of (4), $${d\vec G \over dt})_B$$, can be considered as the time derivative of $$\vec G$$ as seen by an observer rotating along with (fixed in) the $$B$$ system; or this term can be considered as the time derivative of $$\vec G$$ if $$B$$ is not rotating. The second term on the right-hand side of (4), $$\vec \omega(t) \times \vec G$$, accounts for the change in the $$B$$ system unit vector directions with time; that is, this second term is due to the rotation of the axes in the $$B$$ system.

An important point is that the same vector $$\vec G$$ is used on both sides of (4). For application to the rotational motion of a rigid body about its center of mass, let S be the space frame and B be the body frame. The body rotates in the space frame but is fixed in the body frame. Let $$\vec L$$ be the angular momentum in the space frame. Using (5), $${d \vec {L} \over dt})_S = {d \vec {L} \over dt})_B + \vec \omega \times \vec L$$. The first term on the right-hand side of this relationship is the time rate of change of the space frame angular momentum $$\vec L$$ expressed using body frame coordinates. Let $$\vec H$$ denote the angular momentum in the body frame; $$\vec H$$ is zero. $$\vec L$$ is NOT the angular momentum in the body frame; $$\vec L$$ is a different vector than $$\vec H$$. $$\vec L$$ is the same vector in both frames, and $$\vec H$$ is the same vector in both frames, but $$\vec L\ne \vec H$$. An earlier answer by @Eli provides the relationships for $$\vec L$$.

We can express the transformation of $$\vec G$$ from the $$S$$ system to the $$B$$ system using a rotation matrix $$\bf A$$. Let $$\vec G_S$$ represent $$\vec G$$ as expressed in (1) and let $$\vec G_B$$ represent $$\vec G$$ as expressed in (2). Specifically, $$\vec G_B = {\bf A} \vec G_S$$. $$\bf A$$ represents a passive rotation of the coordinate system as used in Ref 1. (Alternatively, we use an active rotation and define the rotation matrix $$\bf R$$ as transforming $$B$$ to $$S$$ components: $$\vec G_S = {\bf R} \vec G_B$$. $$\bf R$$ and $$\bf A$$ are related as follows: $${\bf R} = \bf A^T$$ where $$\bf A^T$$ is the transpose of $$\bf A$$.)

To express all three vectors in (4) in $$S$$ coordinates using the passive rotation matrix $${\bf A}$$ we have
$$(5){d\vec G \over dt})_S = {\bf A^T} {d\vec G \over dt})_B + {\bf A^T} (\vec \omega(t) \times \vec G)$$ To express all three vectors in (4) in $$B$$ coordinates using the passive rotation matrix $${\bf A}$$ we have
$$(6){\bf A}{d\vec G \over dt})_S = {d\vec G \over dt})_B + \vec \omega(t) \times \vec G$$ To express all three vectors in (4) in $$S$$ coordinates using the active rotation matrix $${\bf R}$$ we have
$$(7){d\vec G \over dt})_S = {\bf R} {d\vec G \over dt})_B + {\bf R} (\vec \omega(t) \times \vec G)$$ To express all three vectors in (4) in $$B$$ coordinates using the active rotation matrix $${\bf R}$$ we have
$$(8){\bf R^T}{d\vec G \over dt})_S = {d\vec G \over dt})_B + \vec \omega(t) \times \vec G$$ Results (7) and (8) agree with the results provided in an earlier answer by @Eli as relationships (2) and (3) in that answer.

Examples

For the following two examples, consider rotation in a plane about the $$\hat e_{3S} = \hat e_{3B}$$ axis. $$\vec \omega(t)$$ is $$\left\{0,0,\theta '(t)\right\}$$ where $$\theta^{'} = \omega = {d \theta \over dt}$$. For this situation, the passive rotation matrix $$\bf A$$ is $$\left( \begin{array}{ccc} \cos (\theta (t)) & \sin (\theta (t)) & 0 \\ -\sin (\theta (t)) & \cos (\theta (t)) & 0 \\ 0 & 0 & 1 \\ \end{array} \right)$$ The following results were calculated using Mathematica.

Example 1

Let $$\vec G$$ be constant in both time and magnitude in the S frame. $$\vec G_S = (c1, c2, c3)$$ where $$c1, c2,$$ and $$c3$$ are constant in time.

Using either (5) or (7), the results in space coordinates are as follows:

$$({d \vec G \over dt})_S$$ expressed in $$S$$ coordinates is: $$\{0,0,0\}$$.

$$({d \vec G \over dt})_B$$ expressed in $$S$$ coordinates is: $$\left\{\text{c2} \theta '(t),-\text{c1} \theta '(t),0\right\}$$.

$$\vec \omega \times G$$ expressed in $$S$$ coordinates is: $$\left\{-\text{c2} \theta '(t),\text{c1} \theta '(t),0\right\}$$.

Using either (6) or (8), the results in body coordinates are as follows:

$$({d \vec G \over dt})_S$$ expressed in $$B$$ coordinates is: $${0,0,0}$$.

$$({d \vec G \over dt})_B$$ expressed in $$B$$ coordinates is: $$\left\{\theta '(t) (\text{c2} \cos (\theta (t))-\text{c1} \sin (\theta (t))),\theta '(t) (-(\text{c1} \cos (\theta (t))+\text{c2} \sin (\theta (t)))),0\right\}$$.

$$\vec \omega \times G$$ expressed in $$B$$ coordinates is: $$\left\{\theta '(t) (\text{c1} \sin (\theta (t))-\text{c2} \cos (\theta (t))),\theta '(t) (\text{c1} \cos (\theta (t))+\text{c2} \sin (\theta (t))),0\right\}$$.

Example 2

Let $$\vec G$$ be constant in direction but have magnitude increasing linearly with time in the S frame. $$\vec G_S = {c t, c t, 0}$$ where c is a constant.

Using either (5) or (7), the results in space coordinates are as follows:

$$({d \vec G \over dt})_S$$ expressed in $$S$$ coordinates is: $$\{c,c,0\}$$.

$$({d \vec G \over dt})_B$$ expressed in $$S$$ coordinates is: $$\left\{c t \theta '(t)+c,c-c t \theta '(t),0\right\}$$.

$$\vec \omega \times G$$ expressed in $$S$$ coordinates is: $$\left\{-c t \theta '(t),c t \theta '(t),0\right\}$$.

Using either (6) or (8), the results in body coordinates are as follows:

$$({d \vec G \over dt})_S$$ expressed in $$B$$ coordinates is $$\{c \sin (\theta (t))+c \cos (\theta (t)),c \cos (\theta (t))-c \sin (\theta (t)),0\}$$

$$({d \vec G \over dt})_B$$ expressed in $$B$$ coordinates is: $$\left\{c \left(t \theta '(t) (\cos (\theta (t))-\sin (\theta (t)))+\sin (\theta (t))+\cos (\theta (t))\right),-c \left(t \theta '(t) (\sin (\theta (t))+\cos (\theta (t)))+\sin (\theta (t))-\cos (\theta (t))\right),0\right\}$$.

$$\vec \omega \times G$$ expressed in $$B$$ coordinates is: $$\left\{c t \theta '(t) (\sin (\theta (t))-\cos (\theta (t))),c t \theta '(t) (\sin (\theta (t))+\cos (\theta (t))),0\right\}$$.

References

[Ref 1] Goldstein, Classical Mechanics

[Ref 2] Symon, Mechanics