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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

1 Answer 1

Intuitively, geodesics are paths that objects under only the action of the gravitational field follow. The derivative of the four velocity being zero roughly means there is "no acceleration" in the covariant sense; ie no external forces. These objects are essentially just rolling around in spacetime, moving in the direction specified by the curvature.

There is a lot more to your question; for instance, you probably mean something more like the following local expression

$$\mathcal{D}_\nu U^\mu=\partial_\nu U^\mu+\Gamma^{\mu}_{\rho\nu}U^\rho=0$$

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 $\nu$."

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