Change of a vector in a rotating coordinate system

[Kashmir]

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* :

"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)"

 

[vanhees71]

$$\newcommand{\uvec}[1]{\underline{#1}}$$
At this point it is important to distinguish between vectors $\vec{G}$, which are independent of any choice of a basis and components of the vector with respect to bases.

In Newtonian physics everything is expressed with respect to inertial reference frames. So we define an arbitrary inertial reference frame with a Cartesian right-handed basis $\vec{e}_j$. These are fixed once and for all, i.e., time-independent.

Now we want to formulate the physical laws in terms of observables of an observer at rest in a reference frame that is accelerated wrt. the inertial frame. Particularly it can rotate. Such an observer will use another Cartesian right-handed basis $\vec{e}_k'$. As all vectors we can write
$$\vec{e}_k'(t)=D_{jk}(t) \vec{e}_j.$$
The $\vec{e}_k'$ are thus time-dependent. Since by choice both the inertial and the rotating basis vectors are right-handed orthonormal vectors, the matrix $\hat{D}=(D_{jk})$ must be a rotation matrix (or SO(3) matrix), i.e., it fulfills
$$\hat{D} \hat{D}^T=\hat{D}^T \hat{D}=\hat{1}, \quad \mathrm{det} \hat{D}=+1.$$
For the arbitrary vector $\vec{G}$ you have
$$\vec{G}=G_j \vec{e}_j = G_k' \vec{e}_k'.$$
Here and in the following the Einstein summation convention applies, i.e., over any index which appears twice in an equation you have to sum from 1 to 3.

Now we want to express the time derivative of this invariant vector in terms of the components $G_k'$ wrt. the rotating basis. Using the product rule we get
$$\vec{G}=\dot{G}_k' \vec{e}_k' + G_k' \dot{\vec{e}}_k'. \qquad (1)$$
Next we need to express the time derivatives of the rotating basis vectors in term of these basis vectors. For that we use the rotation matrix introduced above
$$\vec{e}_l'=D_{jl} \vec{e}_j \; \Rightarrow \; \dot{\vec{e}}_l' = \dot{D}_{jl} \vec{e}_j,$$
where we have used that $\vec{e}_j$ is time-independent. Now we need the inverse transformation. Since $\hat{D}^{-1}=\hat{D}^{T}$ we get
$$\vec{e}_j=D_{jk} \vec{e}_k'$$
and thus
$$\dot{\vec{e}}_l'=\dot{D}_{jl} D_{jk} \vec{e}_k'.$$
Now
$$\dot{D}_{jl} D_{jk} =(\hat{D}^T \dot{D})_{kl}=\Omega_{kl}.$$
From $\hat{D}^T \hat{D}=\hat{1}$ we get by deriving wrt. to time,
$$\dot{\hat{D}}^T \hat{D}+\hat{D}^T \dot{\hat{D}}=0 \; \Rightarrow ; \hat{\Omega} = -\hat{\Omega}^T,$$
i.e., $\hat{\Omega}$ is an antisymmetric matrix. We thus can write
$$\Omega_{kl} = \epsilon_{kml} \omega_m'.$$
Plugging all this in (1) finally gives
$$\dot{\vec{G}}=\dot{G}_l' \vec{e}_l' + G_l' \epsilon_{kml} \omega_m' \vec{e}_k'.$$
So the components of $\dot{\vec{G}}$ wrt. the rotating basis is given by the "covariant time derivative"
$$\mathrm{D}_t G_l'=\dot{G}_l' - \epsilon_{kml} \omega_m' G_l'$$
or for the column vector $\uvec{G}'=(G_1',G_2',G_3')^T$
$$\mathrm{D}_t \uvec{G} = \dot{\uvec{G}} + \uvec{\omega}' \times \uvec{G}'.$$