Vector quantities and their transformation laws

Two positions and one displacement. On the page they are the same kind of object: an arrow, with a tip and a length. If the picture were the definition, they would be interchangeable. Watch what happens when the frame moves. They are not.

Slide the origin, then turn the axes. Notice the two arrows behave differently. The positions $r_1$ and $r_2$ are measured from the origin, so they jump when it moves. The free vector $A$ is not: it only turns with the frame. One measurement changed for two completely different reasons. Which of them is allowed in an equation?

Try the difference first. The arrow from $P_2$ to $P_1$ does not care where the origin is, so subtracting the positions cancels the translation: $x_i^{(1)} - x_i^{(2)} = \alpha_{ij}(x_j^{(1)'} - x_j^{(2)'})$. What is left transforms by the vector law $A_i = \alpha_{ij}A_j'$. This is why a displacement is a vector, and why velocity, a displacement over a time interval, is one too.

Now try the sum. The translation appears twice: $x_i^{(1)} + x_i^{(2)} = 2a_i + \alpha_{ij}(x_j^{(1)'} + x_j^{(2)'})$. That leftover $2a_i$ is the tell. Two observers with different origins get different answers, so the object describes nothing. Here is the key: the problem is not that the sum is hard to compute. It is that the sum is not defined.

So we can finally say what a vector is. Not an arrow: a rule. A quantity is a vector when its components change by $A_i = \alpha_{ij}A_j'$. A scalar transforms as $c = c'$. A position transforms as $x_i = a_i + \alpha_{ij}x_j'$. A tensor of second rank carries the rotation on both indices, $\Phi_{ij} = \alpha_{ik}\alpha_{jl}\Phi_{kl}'$. The rule is the definition, and the arrow is only a picture of the third case.

Here is a test you can run on anything. Pretend to change the frame and ask which law it follows. If it follows one, its value is a fact and physics may use it. If it follows none, it is an accident of your axes. The difference of two positions survives. Their sum does not. Next we ask how two vectors may be multiplied, and the same test will decide which product is admissible.