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.

  • 3
    $\begingroup$ $u_a u^b = c^2$ is wrong, it should be $u_a u^a = c^2$. $\endgroup$
    – ACuriousMind
    Apr 25, 2015 at 21:39
  • $\begingroup$ Have you tried brute-force? Often the worst method, but it can be successful. $\endgroup$
    – Kyle Kanos
    Apr 25, 2015 at 22:09
  • $\begingroup$ @KyleKanos What do you mean? How can I brute-force my way to the given result? $\endgroup$
    – msd27
    Apr 25, 2015 at 22:41
  • $\begingroup$ Brute-force it by expanding the tensor induces (0-3) and playing around. $\endgroup$
    – Kyle Kanos
    Apr 25, 2015 at 23:21
  • $\begingroup$ physics.stackexchange.com/questions/168492/… $\endgroup$
    – MBN
    Apr 29, 2015 at 14:55

1 Answer 1


I was just making a slight mistake:

$ \nabla_a(u^b u_b)=\nabla_a (c^2)=0=u_b\nabla_a(u^b )+u^b\nabla_a(u_b )$

Using the raising and lowering properties of the metric:

$ 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) $

From which the result follows.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.