How to prove that $\Delta r^2 =\Delta x^2+\Delta y^2 +\Delta z^2$ is invariant under Galilean transformation Consider events A and B with coordinates $(t_A, x_A, y_A,z_A)$ and $(t_B, x_B, y_B,z_B)$ respectively. 
I am trying to prove that the quantity
$$\Delta r^2 =\Delta x^2+\Delta y^2 +\Delta z^2$$
where $\Delta x = x_A-x_B$ etc.
is invariant under Galilean transformation.
The Galilean transformation I used was $$t'=t,$$ $$x' =x-vt,$$ $$y'=y,$$ $$z'=z,$$
where $v$ is the velocity of the  frame $S'$ moving away from frame $S$.
To prove that $\Delta x^2 = (x_A-x_B)^2 =(x'_A-x'_B)^2=\Delta x'^2$, I started with $(x'_A-x'_B)^2$ and applied the Galilean transformation:
$$(x'_A-x'_B)^2=[(x_A-vt_A)-(x_B-vt_B)]^2$$
$$=[(x_A-x_B) +v(t_B-t_A)]^2$$
$$=(x_A-x_B)^2 +2(x_A-x_B)v(t_B-t_A) +v^2(t_B-t_A)^2$$
So it seems that for $\Delta x'^2=(x'_A-x'_B)^2 = (x_A-x_B)^2 =\Delta x^2$ to be true, the expression $$2(x_A-x_B)v(t_B-t_A) +v^2(t_B-t_A)^2$$
has be zero. 
How can this expression be shown to be  zero? If it is expanded, there will be  terms like $v^2t_B^2$ that cannot be cancelled. 
 A: Length only makes sense if you are comparing measurements of the ends of the object ($x_A$ and $x_B$) at the same instant $t_A = t_B$. 
From which your spurious terms cancel.
A: It's best to work in differentials :
$$
\begin{align}
\left(\frac {\partial}{\partial t} r^\prime \right)^2 - \left(\frac {\partial}{\partial t} r \right)^2 &= \left(\frac {\partial}{\partial t} \left(r-vt\right) \right)^2 - \left(\frac {\partial}{\partial t} r \right)^2
\\&=\left(\frac {\partial}{\partial t} r -v\frac {\partial t}{\partial t}  \right)^2 - \left(\frac {\partial}{\partial t} r \right)^2
\\&= \left( u-v \right)^2 - u^2
\\&=\left(u^2+v^2-2\,u\,v\right)-u^2
\\&=v^2-2\,u\,v
\end{align}
$$
A: 
In general, $\Delta r^2$ is not invariant under Galilean transformation. This is a direct consequence of the velocity addition theorem.

This is basic classical kinetics. From your transformation, one can quickly write $y_A=y_B$, $z_A=z_B$, $y’_A=y’_B$, $z’_A=z’_B$,  therefore $\Delta y= \Delta z= \Delta y’= \Delta z’=0$,
This means that $\Delta r^2$ only depends on $\Delta x^2$ in both systems. Now, we just need to show that $\Delta x$ ≠ $\Delta x’$.
Since $t=t’$ derivation on either one is the same. Now, derivating the transformation for $x’$ with respect to time, one gets the so-called velocity addition theorem,
$$v’=\frac{dx’}{dt’}=\frac{dx’}{dt}=\frac{dx}{dt}-v\frac{dt}{dt}= u-v \tag{1}$$
Where $v’$ is the speed of the measuring phenomenon in relation to $S’$, $u$ is the speed of the measuring phenomenon in relation to $S$, and $v$ is as you defined the speed of the  $S’$ frame in relation to $S$. From this we get
$$\Delta x=u \Delta t \tag{2}$$
$$\Delta x’=(u-v) \Delta t \tag{3}$$
$$\Delta r^2-\Delta r’^2=v(2u-v) \Delta t^2 \tag{4}$$
Equation (4) shows that only in three situations $\Delta r^2$ is invariant under Galilean transform.

*

*$v=0$ which is obvious as both systems are the same;

*In case of an instantaneous measurement ($\Delta t=0$), which is the time equivalent to 1;

*The measuring object is moving with half the boost speed.

For a simple example, lets consider that an observer in $S’$ takes a lamp bulb and departs from $S$ with speed $v$ along the $S$’s $x$ axis. With no surprise after the experiment, $S’$ will measure that in between time $t’_A =t_A$ and $t’_B =t_B$ a $\Delta x’=0$, since the bulb was with him and going nowhere. $S$ in the other hand, will measure that in between time $t_A$ and $t_B$ the bulb was displaced by the amount $\Delta x=v \Delta t= v \Delta t’≠\Delta x’$.
Another way of showing that $\Delta r^2$ is not invariant in general, is using the theory of relativity: It is well known (arguably the most fundamental consequence of the theory of relativity), that the quantity
$$\Delta s^2=-c\Delta t^2+\Delta x^2+\Delta y^2+\Delta z^2 \tag{5}$$
Is invariant, i.e., $\Delta s^2=\Delta s’^2$.   This could be consider a generalization of what you are trying to prove. However, this invariance only occurs because of the four-dimensional term of $t$. You see, as the observers move, time dilation and space contraction will perfectly balance to the point of $\Delta s^2$ being an invariant. To show that $\Delta r^2$ cannot be invariant in general using relativity, one can check that, in order for $\Delta s^2$ to collapse into $\Delta r^2$, $\Delta t$ and $\Delta t’$ should both be $0$, which is a simultaneous measurement. This only happens if both observers are in the same point at the same time, meaning, they are the same.
