# General Relativity - Four Velocity Derivative Question

I am trying to get my head around a small point used in a book I am reading about General Relativity. The book states that because $u_au^a = c^2$ it follows that $u_a \nabla_b u^a = 0$

The first part $u_au^a = c^2$ I'm fine with from the definition of the four velocity but I can't seem to see how to get to the final result. I have tried going through $\nabla_a u_a u^b = \nabla_a c^2 = 0$ but can't seem to get anywhere.

• $u_a u^b = c^2$ is wrong, it should be $u_a u^a = c^2$. Apr 25, 2015 at 21:39
• Have you tried brute-force? Often the worst method, but it can be successful. Apr 25, 2015 at 22:09
• @KyleKanos What do you mean? How can I brute-force my way to the given result? Apr 25, 2015 at 22:41
• Brute-force it by expanding the tensor induces (0-3) and playing around. Apr 25, 2015 at 23:21
• physics.stackexchange.com/questions/168492/…
– MBN
Apr 29, 2015 at 14:55

$\nabla_a(u^b u_b)=\nabla_a (c^2)=0=u_b\nabla_a(u^b )+u^b\nabla_a(u_b )$
$0 = u_b\nabla_a(g_{bc}u_c )+u^b\nabla_a(u_b ) = g_{bc}u_b\nabla_a(u_c )+u^b\nabla_a(u_b ) = u^c\nabla_a(u_c )+u^b\nabla_a(u_b)= 2u^b\nabla_a(u_b)$