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In the book of Goldstein, Classical Mechanics, at the end of the page 171, it is stated that

A relation between the two differential changes in $G$ can be derived on the basis of physical arguments.We can write that the only difference between the two is the effect of rotation of the body axes:

$$d(G)_s = d(G)_b + d(G)_{rot},$$

but the does not give any argument; just states that it can be given. Moreover, in the next page, it argues that

Now consider a vector fixed in the rigid body.As the body moves there is of course no change in the components of this vector as seen by the body observer, i.e relative to body axes. The only contribution to $d(G)_s$ is then the effect of the rotation ofthe body. But since the vector is fixed in rhe body system, it rotates with it counterclockwise ,and the change in the vector. as observed in space is that given by $$dr' = d\Omega \times r,$$ hence $$d(G)_{rot} = d \Omega \times G$$

However, in general, $r'$, i.e $r+ dr$, is not fixed in the body frame, and it also moves, so this argument might fail because in general there will be contributions from the rotation of the point itself wrt to the body frame, so how such an argument can be valid ?

Question:

  1. What is the physical argument the existence of the relation $$d(G)_s = d(G)_b + d(G)_{rot} \quad ?$$
  2. How can the argument given for $d(G)_{rot} = d \Omega \times G$ be valid for any vector $G$ ?
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$G$ must be a vector:

$$d(\vec{G})_s=d(\vec{G})_b+d(\vec{G})_r$$

the components of the vector $\vec{G}$ can be given in space coordinate system

$$\vec{G}_s=R(\,\vec{\phi})\,\vec{G}_b$$ where $R$ is the transformation matrix between body coordinate system and space system and $\vec{\phi}$ is then vector containing the euler angles $(\alpha\,,\beta\,,\gamma)$

thus: $$\vec{\dot{G}}_s=R(\,\vec{\phi})\,\vec{\dot{G}}_b+\dot{R}(\,\vec{\phi})\,{{G}}_b$$

with $\dot{R}=R\,\tilde{\omega}$

$$\vec{\dot{G}}_s=R(\,\vec{\phi})\,\vec{\dot{G}}_b+R(\,\vec{\phi})\,\tilde{\omega}\,\vec{{G}}_b\tag 1$$

multiply equation (1) from the left with $R^T$

$$R^T\,\vec{\dot{G}}_s=\vec{\dot{G}}_b+\tilde{\omega}\,\vec{{G}}_b\tag 2$$

$$R^T\,\vec{d{G}}_s=\vec{d{G}}_b+\tilde{\omega}\,\vec{{G}}_b\,dt\tag 3$$

$$d(\vec{G})_s=d(\vec{G})_b+d(\vec{G})_{\text{rot}}$$

thus:

$$d(\vec{G})_s=R^T\,\vec{d{G}}_s$$

$$d(\vec{G})_{\text{rot}}=\tilde{\omega}\,\vec{{G}}_b\,dt=\left(A(\vec{\phi})\,d\vec{\phi}\right)\times \vec{G}_b$$

comment

$\tilde{\omega}\,\vec{G}=\vec{\omega}\,\times\,\vec{G}$ with $\tilde{\omega}$ a skew anti symmetric matrix

$\begin{bmatrix} 0 & -\omega_z & \omega_y \\ \omega_z & 0 & -\omega_x \\ -\omega_y & \omega_x & 0 \\ \end{bmatrix}$

From:

$\dot{R}=R(\vec{\phi})\,\tilde{\omega}\quad$ we get $\quad\vec{\omega}=A(\vec{\phi})\,\vec{\dot{\phi}}$

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  • $\begingroup$ How can you write $\dot R = R \tilde \omega ?$, and what is $\tilde \omega$ ? $\endgroup$ – onurcanbektas Apr 4 at 11:42
  • $\begingroup$ Note: I just didn't put vector sign to write more quickly; otherwise, even the cross product is not defined for scalars. $\endgroup$ – onurcanbektas Apr 4 at 11:44
  • $\begingroup$ @onurcanbektas see my comment $\endgroup$ – Eli Apr 4 at 12:12
  • $\begingroup$ Ok, but you don't provide any justification for why $\dot R (\phi)$ is $R(\phi) \tilde \omega$. Is it something that is generally known ? $\endgroup$ – onurcanbektas Apr 4 at 12:17
  • $\begingroup$ @onurcanbektas this is well known matrix equation $\endgroup$ – Eli Apr 4 at 12:21

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