Special relativity was not born as a 4-covariant theory. Instead, Einstein derivated the kinematical quantities without spacetime,therefore without 4-vectors.

Consider then the following:

First Einstein derivated the kinematics of Relativity in terms of 3D Euclidean geometry and with the postulates of special relativity, showing the very fundamental concepts of physics in near to speed of light (e.g. time dilation). Then, Minkowski introduced the spacetime and invariant interval;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 velocity vector which required a concept of the 4-velocity vector. But in the transition to spacetime formalism, the velocity vector do not transforms (with a lorentz transformation) as a true vector . In order to construct the concept of velocity in spacetime we must consider proper time.

Well, this paragraph is my conclusion about the topic, but this is a intuitive conclusion. The formal explanation is then that 3-velocity is not a vector object, because the components didn't transforms as vectors under a lorentz transformation:

$$\frac{dx'^{\mu}}{dt'} = \Lambda^{\mu'}_{\nu}\frac{dx^{\nu}}{dt} \frac{dt}{dt'} = \frac{\Lambda^{\mu'}_{\nu}}{\Lambda^{0'}_{\nu}x^{\nu}}\frac{dx^{\nu}}{dt} \neq \Lambda^{\mu'}_{\nu}\frac{dx^{\nu}}{dt}\tag{1}$$

Even so, I still do not understand the following:

1) We can write the 4-velocity as, $$u(\tau) = (\frac{dx^{0}}{d\tau},\frac{dx^{1}}{d\tau},\frac{dx^{2}}{d\tau},\frac{dx^{3}}{d\tau})$$

but with relation,

$$d\tau = \frac{1}{\gamma}dt$$

we can write the 4-velocity in terms of 3-velocity $$u(t) = \gamma(c,\vec{v}(t))$$

I mean, if we already have a invariant object,the four velocity, why the one wants to consider 4-velocity in terms of 3-velocity?

  • 1
    $\begingroup$ I imagine it's largely related to people being taught 3-vectors before 4-vectors, so writing the 4-vector in terms of the 3-vector is likely just due to familiarity. $\endgroup$
    – Kyle Kanos
    Jan 7, 2019 at 18:11

2 Answers 2


I'm not sure I understand your question. But it seems like your saying why is it useful to define the 4-velocity in terms of the 3 velocity $\vec{v}$? My response is quite short, but I think it is sufficient to address your concern.


Because it is the entire goal! That is, the whole goal of the Minkowski formalism was precisely to generalize the notion of of 3-position, velocity etc into Lorentz covariant quantitates.

How else would one construct a Minkowski generalization of an objects velocity $\vec{v}$. If we want it to be anything meaningful, then it actually ought to incorporate our "old notion" of velocity with this new one.

Given that an object is moving at a velocity $\vec{v}$ I can't think of any other way to construct a 4$^{th}$ component with the same dimensions that transforms covariantly other than

$$ V^\mu \equiv \frac{dx^\mu}{dx^\tau} $$


I mean, if we already have a invariant object,the four velocity, why the one wants to consider 4-velocity in terms of 3-velocity?

Elaborating on @Kyle 's comment...

While geometric quantities and invariant calculations are often neat, precise, and concise (demoting the need for components and for transformation formulae),
one often expresses geometrical quantities in terms of space and time components because these are related to measurements done with an observer's apparatuses... and these are better tied to our physical interpretation. In some sense, the 3-quantities are our crutches for the more geometrical 4-quantities.

The 3-velocity represents the relative-velocity between observers.

In a similar way, one expresses the electromagnetic field tensor in terms of electric and magnetic fields, as measured by an observer.

and so on.


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