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enter image description here

The image 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?

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  • $\begingroup$ You haven't defined the quantity dL/dt $\endgroup$ – Buraian Nov 20 at 17:20
  • $\begingroup$ @Buraian It's the angular momentum. $\endgroup$ – Kasi Reddy Sreeman Reddy Nov 20 at 17:20
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    $\begingroup$ I think you should put some context behind it $\endgroup$ – Buraian Nov 20 at 17:22
  • $\begingroup$ @Buraian I added a wiki link to Euler's angles now. Should I add any other thing? $\endgroup$ – Kasi Reddy Sreeman Reddy Nov 20 at 17:24
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    $\begingroup$ 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. $\endgroup$ – JAlex Nov 20 at 18:59
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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.

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  • $\begingroup$ "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. $\endgroup$ – Kasi Reddy Sreeman Reddy Nov 20 at 18:22
  • $\begingroup$ 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. $\endgroup$ – Young Kindaichi Nov 20 at 18:51
  • $\begingroup$ I know you meant a general vector. But as an example I am considering the angular momentum. $\endgroup$ – Kasi Reddy Sreeman Reddy Nov 20 at 18:53
  • $\begingroup$ So you are considering some vector that not rotating in inertial frame? $\endgroup$ – Young Kindaichi Nov 20 at 18:58
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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

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

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  • $\begingroup$ What about that angular momentum question I asked? $\endgroup$ – Kasi Reddy Sreeman Reddy Nov 22 at 16:51
  • $\begingroup$ @KasiReddySreemanReddy i wrote you the derivation of the Euler equations $\endgroup$ – Eli Nov 22 at 17:21

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