Change of a vector in a rotating coordinate system

Question

Goldstein 3 ed, pg 171, under" rate of change of a vector " :

The author derives the relationship between the change of a vector in a stationary and rotating coordinate system.

In the process he uses this assumption :

It is no loss of generality to take the space and body axes as instantaneously coincident at the time $t$.

And after more steps we get that at $t=t +dt$:

$(d \mathbf{G})_{\text {space }}=(d \mathbf{G})_{\text {body }}+d \Omega \times \mathbf{G}--------- (4-119)$

Hence

$\left(\frac{d \mathbf{G}}{d t}\right)_{\text {space }}=\left(\frac{d \mathbf{G}}{d t}\right)_{\text {body }}+\omega \times \mathbf{G} ------ -- (4-120)$

The above equation should only work in a coordinate system that was aligned with the body axis $dt$ time earlier, however I think this equation is used without that restriction.

Why is that so?

Here is the proof that the author uses if you might need it :

"A more formal derivation of the basic Eq. $(4-120)$ can be given in terms of the orthogonal matrix of transformation between the space and body coordinates. The component of $\mathbf{G}$ along the $i$ th space axis is related to the components along the body axes: $$ G_{i}=\tilde{a}_{i j} G_{j}^{\prime}=a_{j i} G_{j}^{\prime} $$ As the body moves in time the components $G_{j}^{\prime}$ will change as will also the elements $a_{i j}$ of the transformation matrix. Hence the change in $G_{i}$ in a differential time element $d t$ is $$ d G_{i}=a_{j i} d G_{j}^{\prime}+d a_{j i} G_{j}^{\prime} $$ It is no loss of generality to take the space and body axes as instantaneously coincident at the time $t$. Components in the two systems will then be the same instantaneously, but differentials will not be the same, since the two systems are moving relative to each other. Thus $G_{j}^{\prime}=G_{j}$ but $a_{j i} d G_{j}^{\prime}=d G_{i}^{\prime}$, the prime emphasizing the differential is measured in the body axis system. The change in the matrix $\mathbf{A}$ in the time $d t$ is thus a change from the unit matrix and therefore corresponds to the matrix $\boldsymbol{\epsilon}$ of the infinitesimal rotation. Hence $$ d a_{j i}=(\overline{\boldsymbol{\epsilon}})_{i j}=-\mathbf{\epsilon}_{i j} $$ using the antisymmetry property of $\epsilon$. In terms of the permutation symbol $\epsilon_{i j k}$ the elements of $\epsilon$ are such that (cf. Eq. 4-105) $$ -\epsilon_{i j}=-\epsilon_{i j k} d \Omega_{k}=\epsilon_{i k j} d \Omega_{k} $$ Equation (4-122) can now be written $$ d G_{i}=d G_{i}^{\prime}+\epsilon_{i k j} d \Omega_{k} G_{j} $$ The last term on the right will be recognized as the expression for the $i$ th component of a cross product, so that the final expression for the relation between differentials in the two systems is $$ d G_{i}=d G_{i}^{\prime}+(d \Omega \times G)_{i} $$ which is the same as the $i$ th component of Eq. (4-119)"

Answer 1

Perhaps this can help you ?

The initial components of the vector $~G~$ to the center of mass are

$$\mathbf G_I=[_B^I\,R]\,\mathbf G_B\tag 1$$

where the index "I" states for initial coordinate system "B" for body system and

$~\mathbf R~$ is the transformation matrix from "B" to "I"

from here the derivative

$$\frac{d}{dt}\left(\mathbf G_I\right)=[_B^I\,R]\,\mathbf{\dot{G}}_B+[_B^I\,\dot R]\,\mathbf G_B \tag 2$$

with

$$[_B^I\,\dot R]= [_B^I\,R]\,\tilde\omega_B\quad , \tilde\omega= \left[ \begin {array}{ccc} 0&-\omega_{{z}}&\omega_{{y}} \\ \omega_{{z}}&0&-\omega_{{x}}\\ -\omega_{{y}}&\omega_{{x}}&0\end {array} \right] $$

hence equation (2)

$$\frac{d}{dt}\left(\mathbf G_I\right)=[_B^I\,R]\,\mathbf{\dot{G}}_B+[_B^I\, R]\left(\omega_B\times \mathbf G_B\right) \tag 3$$

multiply from the left with $~[_I^B\, R]~$

$$\underbrace{[_I^B\, R]\,\frac{d}{dt}\left(\mathbf G_I\right)}_{\left(\frac{d}{dt}\,G_I\right)_B }=\mathbf{\dot{G}}_B+\left(\omega_B\times \mathbf G_B\right) \tag 4$$

instead of "B" components you can use the "I" components

hence

$$\left(\frac{d}{dt}\,G_I\right)=\frac{d}{d\tau}\,G+\omega\times \mathbf G $$

Goldstein notation

$$\left(\frac{d \mathbf{G}}{d t}\right)_{\text {space }}=\left(\frac{d \mathbf{G}}{d t}\right)_{\text {body }}+\omega \times \mathbf{G} $$

where space and body are the derivatives not the components

Answer 2

“It is no loss of generality to take the space and body axes as instantaneously coincident at the time ”

In other words, the equation still holds when this isn’t true. We are only computing a rate of change. Imagine a frame in which they don’t coincide. Define an “offset vector” so that $$G_{space} = G_{body} + A$$ At time $t.$ Then when calculating the difference at time $t+dt$, the vector $A$ is a constant so is irrelevant.