Four-velocity

Consider Minkowski space with coordinates $(t, X^{1}, X^{2}, X^{3})$ . When we have a worldline with the particular form $γ (t) = (t, X^{1}, . . .)$ we can calculate the usual tangent vector $v_{γ} = (1, {\dot{X}}^{1}, . . .)^{T}$ , and this is the classical notion of velocity. We will call it the "coordinate velocity".
Now, we can consider another inertial coordinates $(s, Y^{1}, Y^{2}, Y^{3})$ , obtained from the previous coordinates by means of the Lorentz boost $Λ (v)$ , and the coordinate velocity of the corresponding worldline

\tilde{γ} = Λ (v) γ

is $v_{\tilde{γ}} = (1, {\dot{Y}}^{1}, . . .)^{T}$ . It turns out that

v_{\tilde{γ}} \neq Λ (v) v_{γ},

so the coordinate velocity, since it depends on the chosen coordinate, is not a well-defined magnitude. We can start with the coordinate velocity in $X$ and define it the other inertial coordinates as $Λ (v) v_{γ}$ , but the coordinate system $X$ should not be a distinguished one...

The solution is to consider, for every event $γ (t)$ in the worldline $γ$ , a set of inertial coordinates $(τ, Z^{1}, Z^{2}, Z^{3})$ obtained by applying to the coordinates $(t, X^{p})^{T}$ the element of the Poincare group obtained by composition of the translation to $γ (t)$ and the Lorentz boost in the direction given by the coordinate velocity $(1, {\dot{X}}^{1}, . . .)$ ("comoving frame" or "instantaneous rest frame"). Suppose, WLOG, that the movement is made in the $X^{1}$ direction only, so we have the Lorentz boost $Λ ({\dot{X}}^{1})$ . The (puntual) coordinate velocity in this "distinguished" frame is $(1, 0, 0, 0)^{T}$ , and we can rewrite this vector in any other inertial coordinates as $Λ (v) \cdot (1, 0, 0, 0)^{T}$ . In particular, in $(t, X^{1}, X^{2}, X^{3})$ is given by

Λ (- {\dot{X}}^{1}) (1, 0, 0, 0)^{T} = {(\frac{1}{\sqrt{1 - ({\dot{X}}^{1})^{2}}}, \frac{{\dot{X}}^{1}}{\sqrt{1 - ({\dot{X}}^{1})^{2}}}, 0, 0)}^{T}

It can be checked that in $(s, Y^{1}, Y^{2}, Y^{3})$ this vector is transformed as

{(\frac{1}{\sqrt{1 - ({\dot{Y}}^{1})^{2}}}, \frac{{\dot{Y}}^{1}}{\sqrt{1 - ({\dot{Y}}^{1})^{2}}}, 0, 0)}^{T} .

So we can use this expression to define a well-defined vector quantity, called the four-velocity.

Of course, this also works for more general Lorentz boosts, and it can be checked that the four velocity is

U = \frac{1}{\sqrt{1 - {\dot{X}}_{p} {\dot{X}}^{p}}} (1, {\dot{X}}^{1}, {\dot{X}}^{2}, {\dot{X}}^{3})^{T}

In practice, the four-velocity is usually defined as $U = \frac{d}{d τ} γ = \frac{d t}{d γ} v_{γ}$ .
This vector satisfies the normalization condition $η_{ν μ} U^{ν} U^{μ} = - 1$ .

With the usual units, we have $η_{ν μ} U^{ν} U^{μ} = - c$ .

Related: four-momentum.