Question on Galilean transformation Let $a$ be a scalar, $D$ a rotation matrix and $b$ and $v$ are $1\times 3$-vectors.
We had the following Galiean transformation:
$(t, x(t)) \to (t + a, Dx + b + v\cdot t)$
But why is it not
$(t, x(t)) \to (t + a, Dx(t+a) + b + v\cdot (t+a))$
Or are both equivalent?
 A: Mathematically there is no difference, just call $\tilde{{\bf b}} = {\bf b} + a{\bf v}$, and you get the same form of the transformation
A: It seems you are confusing how the Galilean transformations act. There are three main subset of the Galilean transformations:
Translations: ${g_1(a,b);g_1(a,b)(t,x)=(t+a,x+b)}$,
Rotations: ${g_2(D);g_2(D)(t,x)=(t,Dx)}$,
Boosts: ${g_3(v);g_3(v)(t,x)=(t,x+vt)}$.
Note that we do not write $x$ as a function of time, it is a completely independent coordinate. 
You are doing transformations in the following way, which is not correct,
$$g_3(v)g_1(a,b)g_2(D)(t,x)=g_3(v)g_1(a,b)(t,Dx(t))=g_3(v)(t+a,Dx(t+a)+b)=(t+a,Dx(t+a)+b+vt),$$
$$g_1(a,b)g_3(v)g_2(D)(t,x)=g_1(a,b)g_3(v)(t,Dx(t))=g_3(v)(t,Dx(t)+vt)=(t+a,Dx(t+a)+v(t+a)+b).$$
The correct is
$$g_3(v)g_1(a,b)g_2(D)(t,x)=g_3(v)g_1(a,b)(t,Dx)=g_3(v)(t+a,Dx+b)=(t+a,Dx+b+vt),$$
$$g_1(a,b)g_3(v)g_2(D)(t,x)=g_1(a,b)g_3(v)(t,Dx)=g_3(v)(t,Dx+vt)=(t+a,Dx+vt+b).$$
Boosts and translations commute. The order only matters with respect to rotations.
