- #1

Kashmir

- 468

- 74

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)"

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)"

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