# I Relation between vectors in body coordinates and space coordinates

1. Apr 7, 2016

### Happiness

Why is $a_{ji}dG_j'=dG_i'$ ?
from the third last line below.

$G_i=a_{ji}G_j'$ because a vector labelled by the space axes is related to the same vector labelled by the body axes via a rotation transformation.

If $a_{ji}dG_j'=dG_i'$, then we are saying a vector $dG'$ labelled by the body axes is related to the same vector labelled similarly via a rotation transformation. This doesn't make sense.

Last edited: Apr 7, 2016
2. Apr 7, 2016

They are doing the case where instantaneously the $a_{ij}$ is the identity matrix, but it will still have a derivative. I think the equation should read, (in this case), $dG_i=dG_i '+ da_{ji} G_j '$. The author was trying to say that $dG_i$ is not equal to $dG_i '$ , but it appeared he might have written down something that isn't correct.

3. Apr 7, 2016

### Happiness

It doesn't seem like the equation is a mistake because he substituted it into (4.84) to get the equation you wrote and (4.85).

I think it's just because $a_{ji}=\delta_{ji}$.

Last edited: Apr 8, 2016
4. Apr 8, 2016

### Happiness

It seems like using $a_{ji}dG_j'=dG_i'$ I can prove that $dG_i=dG_i'$, a contradiction.

Let $G_{j1}'$ and $G_{j2}'$ be the vector $G$ at time $t=0$ and $t=dt$ respectively, labelled using the body axes.

Then, $dG_j'=G_{j2}'-G_{j1}'$.

$a_{ji}dG_j'=a_{ji}(G_{j2}'-G_{j1}')=G_{i2}-G_{i1}=dG_i$.

Thus, $dG_i'=dG_i$, a contradiction.

What's wrong?

EDIT: I found the mistake. $a_{ji}(G_{j2}'-G_{j1}')=G_{i2}-G_{i1}$ is wrong.

Last edited: Apr 8, 2016
5. Apr 8, 2016