# Proving a relation with Four-velocity tensor [duplicate]

I'm trying to show that:

$U^a_{\space\space;b}U^bU_a = 0$ (Where U is four-velocity)

and I'm stuck on how to go about it. I tried expanding it out into the Christoffel symbols, but that didn't seem to get me anywhere. Would really appreciate advice on what I should do to proceed.

Thanks.

• I am not sure if this is a duplicate nor answered in the cited question. In this question the field is not necessarily geodesic. – MBN Mar 5 '15 at 17:06

In order to avoid Christoffel symbols, use the fact that your expression is tensorial to evaluate it in a locally flat inertial frame of reference, where the metric reduces to Minkowski's. Then your equation is simply the time derivative of $u_\mu u^\mu = c^2$, since $u^\mu {u^\nu}_{;\mu} = u^\mu {u^\nu}_{,\mu} = \dot u^\nu$ in this case.
If the four velocity is normalized $$U^aU_a=-1$$ then taking the covariant derivative will give you $$2U^bU^a_{;b}U_a=0$$