Consider,again (Question about derivation of four-velocity vector) the following:
For a massive particle with position $x^{\mu}(t) = (x^{0},x^{1},x^{2},x^{3}) \equiv (x^{0},\vec{x})$ we define the coodinate velocity as:$$ v^{\mu} := \frac{dx^{\mu}}{dt} \equiv (c,\vec{v})\tag{1}$$ Where the spatial components of $(1)$ coincide with classical velocity vector and t is the coordinate time.
But, $(1)$ is not a vector object indeed, because the components didn't transforms as vectors under a lorentz transformation:
$$\frac{dx'^{\mu}}{dt'} = \Lambda^{\mu'}_{\nu}\frac{dx^{\nu}}{dt} \frac{dt}{dt'} = \frac{\Lambda^{\mu'}_{\nu}}{\Lambda^{0'}_{\nu}x^{\nu}}\frac{dx^{\nu}}{dt} \neq \Lambda^{\mu'}_{\nu}\frac{dx^{\nu}}{dt}\tag{2}$$
So, I'm still in doubt about $(2)$. We know that $(1,0)-$tensors in Minkowski space transforms like:
$$ A'^{\mu} = \Lambda^{\mu'}_{\nu} A^{\nu}$$
where $\Lambda^{\mu'}_{\nu}$ is the boost matrix:
$$\Lambda^{\mu'}_{\nu} = \begin{bmatrix} \gamma & -v\gamma & 0 & 0 \\ -v\gamma & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} $$
So, firstly, Einstein have derived the kinematics of Relativity in terms of 3D Euclidean geometry and with the postulates of special relativity, showing the very fundamental concepts of physics near to speed of light (e.g. time dilation). Then, Minkowski introduced the spacetime and all the whole of a invariant interval, i.e. the metric formalism of special relativity. Furthermore, with Minkowski's formalism a new type of object was needed to ensamble the kinematical quantities into spacetime formalism. One of these quantities was the 4-velocity vector, which required a concept of the 4-vector. But even if the one defines a true 4-vector, this vector must transform properly under a lorentz transformation; that's the case of four-velocity.
So, consider ,then, an attempt to 4-four velocity, defined as:
$$v^{\mu} := (\frac{dx^{0}}{dt},\frac{dx^{1}}{dt},\frac{dx^{2}}{dt},\frac{dx^{3}}{dt}) \tag{3}$$
Where $t$, is the coodinate time in $S$ reference frame [with $x^{\mu} = (ct,x(t),y(t),z(t))$ coordintes]. Then, if we want a true Minkowski we must to check the tensor character:
$$\frac{dx'^{\mu}}{dt'} = \Lambda^{\mu'}_{\nu}\frac{dx^{\nu}}{dt} \tag{4}$$
As you might expect this isn't right (and the true behaviour is by introducing the proper time). And that is the point, I do not understand this, what is $t'$? Propertime or just coodinate time in $S'$?Well, actually I'm lost in what really is happening in $(2)$, why we use the chain rule, what is the dependence of the times [ I mean if is $t(t')$ or $t'(t)$]and so on.
