# Why four velocity under covariant differential is considered to be zero?

In Einstein's general theory of relativity the elements of four velocity $U^{\mu} (\gamma c, \gamma v)$ under covariant differential is considered to be zero, why?

$$\mathcal{D} U^{\mu}=0$$

in other word:

Why the four velocity of a geodesic has zero directional covariant derivative?

wikipedia four acceleration

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Do you mean to ask why the four velocity of a geodesic has zero directional covariant derivative? –  joshphysics Apr 23 '13 at 16:53

$$\mathcal{D}_\nu U^\mu=\partial_\nu U^\mu+\Gamma^{\mu}_{\rho\nu}U^\rho=0$$
where I am using the Christoffel symbols to explicitly write the connection in the tangent bundle (notice this is now a $(1,1)$-tensor). This is what we mean by "directional derivative", since this is now saying that "the velocity does not change in the direction $\nu$."