How would I show the following equation for the acceleration vector

$$ a^\mu = u^\nu \nabla_\nu u^\mu = g^{\mu\nu}\nabla_\nu \ln V $$

for an observer instantaneously 'at rest,' where $u^\mu = dx^\mu/d\tau$ and $V^2 = -\xi_\mu \xi^\mu$ ($\xi^\mu$ is a time-like killing vector field). I guess we're assuming that we have a stationary spacetime metric.

I'm looking at Sean Carroll's Spacetime and Geometry, page 274 equation 6.15. They leave out the derivation for this. So $V$ is the redshift factor.

  • $\begingroup$ $\nabla^a\ln V$ looks like acceleration with velocity $\xi^a$, but $u^a$ and $\xi^a$ are different vectors. Can you give us some context? $\endgroup$
    – auxsvr
    Apr 25, 2014 at 12:57
  • 2
    $\begingroup$ Are you trying to find the surface gravity of a static black hole? $\endgroup$
    – auxsvr
    Apr 25, 2014 at 13:52

1 Answer 1


Never mind, I got it. For the instruction of others, here it is:

Using four velocity $u^\mu u_\mu=-1$ and then $∇_\nu(u^\mu u_\mu)=0$ $u^\mu∇_\nu u_\mu + u_\mu∇_\nu u^\mu = 0 ⇒ 2u_\mu∇_\nu u^\mu = 0 ⇒ u_\mu∇_\nu u^\mu = 0.$

Using this, the Killing equation, $∇_\mu K_\nu + ∇_\nu K_\mu = 0$ and $K_\alpha = V(x)u_\alpha$, we have $(∇_\mu V)u_\nu + V(∇_\mu u_\nu) + (∇_\nu V)u_\mu + V(∇_\nu u_\mu) = 0.$ $(∇_\mu V)u^\nu u_\nu + V u^\nu(∇_\mu u_\nu) + (∇_\nu V)u^\nu u_\mu + V u^\nu(∇_\nu u_\mu) = 0.$ $−∇_\nu V + u^\nu u_\mu(∇_\nu V) + V u^\nu(∇_\nu u_\mu) = 0.$

$−u^\mu ∇_\nu V + u^\nu u^\mu u_\mu(∇_\nu V) + V u^\nu u^\mu(∇_\nu u_\mu) = 0$

$−u^\mu∇_\mu V − u^\nu ∇_\nu V = 0$

$u^\mu ∇_\mu V = 0$

So $−∇_\mu V + Vu ^\nu (∇_\nu u_\mu) = 0$ and $a_\mu = (∇_\mu V)/V = ∇_\mu lnV.$

Thanks, paper by Ying-Ming Huang.

  • $\begingroup$ Where does that killing field come from...how do we know the timelike killing field has the same form as the four velocity? $\endgroup$ Nov 22, 2015 at 22:12
  • $\begingroup$ How did you get the last line? i.e. $-\nabla_\mu V + V u^\nu(\nabla_\nu u_\mu)=0$ $\endgroup$
    – abhijit975
    Oct 29, 2020 at 6:01

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