# Question about derivation of four-velocity vector

In order to describe a notion of rate of change of positon, in four-dimensional spacetime, we have to introduce the concept of four-velocity.

So, consider 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:

$$\frac{dx'^{\mu}}{dt'} = \Lambda^{\mu'}_{\nu}\frac{dx^{\nu}}{dt} \frac{dt}{dt'} = \frac{\Lambda^{\mu'}_{\nu}}{\Lambda^{0'}_{\nu}x^{\nu}}\frac{dx^{\nu}}{dt} \neq \Lambda^{\mu'}_{\nu}\frac{dx^{\nu}}{dt}\tag{2}$$

Well, I simply do not understand one elementary derivation:

--> It's not difficult to know the motivation for this definition:

$$v^{\mu} := \frac{dx^{\mu}}{dt} \equiv (c,\vec{v})$$

I mean, we need a four-vector and this is certainly a intuitive candidate, but then we realized that this object are not invariant under lorentz transformation, ok. But I'm struggling to derive the expression:

$$\frac{dx^{\mu}}{dt'} = \frac{dx^{\mu}}{dt} \frac{dt}{dt'} \tag{3}$$

Which is important to make the analysis of coordinate transformation as in $$(2)$$. I know that this is simply the chain rule structure, but I simply do not see how to derive it! The classical chain rule is then:

$$\frac{df[x(t),y(t),z(t)]}{dt} = \frac{\partial f }{\partial x }\frac{dx }{d t}+\frac{\partial f }{\partial y }\frac{d y}{d t}+\frac{\partial f }{\partial y }\frac{d y}{d t}$$

How can I derive $$(3)$$ from chain rule,explicitly?

You're misunderstanding the chain rule. For a function of a single argument, $$\frac{df}{dx} = \frac{df}{du} \frac{du}{dx}.$$ You only get multiple terms on the right-hand side if the function has multiple arguments. Now, the position $$x^\mu(t)$$ is simply a collection of four functions, each of one argument, $$t$$. So you just use the ordinary chain rule. For example, for the first function, $$\frac{dx^0}{dt} = \frac{dx^0}{dt'} \frac{dt'}{dt}.$$ The same holds for $$dx^1/dt$$, $$dx^2/dt$$, and $$dx^3/dt$$, so we conclude $$\frac{dx^\mu}{dt} = \frac{dx^\mu}{dt'} \frac{dt'}{dt}$$ for $$\mu \in \{0, 1, 2, 3\}$$.
• Yes, I agree. As I understand the whole thing is derived as the following (in 1+1 dimensions; \mu = {0,1}): $$x^{\mu} = x^{\mu}(t'(t)) = [x^{0}(t'(t)), x^{1}(t'(t))] =$$ Then: $$\frac{dx^{\mu}}{dt}\vec{e}_{\mu} = \frac{dx^{0}}{dt}\vec{e}_{0} + \frac{dx^{1}}{dt}\vec{e}_{1} \implies$$ $$\frac{dx^{\mu}}{dt}\vec{e}_{\mu} = \frac{dx^{0}}{dt'}\frac{dt'}{dt}\vec{e}_{0} + \frac{dx^{1}}{dt'}\frac{dt'}{dt}\vec{e}_{1} \implies$$ Dec 1, 2018 at 21:53
When there are multiple variables in play, such as here $$(t,x,y,z)$$ it can be easy to get muddled on when we need partial derivatives, and when it is legitimate to give a total derivative. The central concept you need here is the concept of the worldline. This traces one particular line through spacetime. When we write a total derivative such as $$dx^\mu /dt$$ we are implicitly referring to this worldline. It is a statement about the difference $$dx^\mu$$ between adjacent events on the worldline. $$dt$$ is the coordinate time difference between those two events. It is only because we have this line, i.e. a one-dimensional or single-parameter entity, that total derivatives can be used. Once you know you have a single-parameter function, you can use expressions such as $$\frac{df}{dr} = \frac{df}{ds} \frac{ds}{dr}$$ and thus, in your example, your eq (3).
$$dx^\mu$$ is a 4-vector (an infinitesimal one) but $$dt$$ is not a Lorentz invariant scalar. To get a 4-vector you need to divide $$dx^\mu$$ by a Lorentz-invariant scalar. In the case of 4-velocity you divide $$dx^\mu$$ by the Lorentz invariant proper $$d\tau$$ time between the two adjacent events. Thus you get $$\frac{dx^\mu}{d\tau}.$$