Trying to work through a textbook derivation of the geodesic deviation equation, I've calculated this identity:$$u_{;\beta}^{\alpha}u_{\alpha}=u_{\alpha;\beta}u^{\alpha}.$$

If this is true, I'm making progress. If it isn't, it's back to the drawing board. Does anyone know if it is correct? I've tried writing it out as$$\left(\frac{\partial u^{\alpha}}{\partial x^{\beta}}+u{}^{\gamma}\Gamma_{\gamma\beta}^{\alpha}\right)u_{\alpha}=\left(\frac{\partial u_{\alpha}}{\partial x^{\beta}}-u_{\gamma}\Gamma_{\alpha\beta}^{\gamma}\right)u^{\alpha},$$ but I'm none the wiser.

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    $\begingroup$ It is true assuming that the connection is the Levi-Civita one, as it is a ``metric'' connection... $\endgroup$ Nov 17, 2014 at 19:31
  • $\begingroup$ For TeX purposes, you can use \, to create a space: $u^\alpha_{\,;\beta}$ creates $u^\alpha_{\,;\beta}$. $\endgroup$
    – Kyle Kanos
    Nov 17, 2014 at 19:38
  • $\begingroup$ @ValterMoretti Any hints as to how I can show it's true? $\endgroup$
    – Peter4075
    Nov 17, 2014 at 19:48
  • $\begingroup$ To get the correct spacing regardless of the number of up indices, $u^\alpha{}_{;\beta}: $u^\alpha{}_{;\beta}$. $\endgroup$ Nov 17, 2014 at 19:49
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    $\begingroup$ For a metric connection $g_{ab}\:_{;c} = 0$ $\endgroup$ Nov 17, 2014 at 19:49

1 Answer 1


The identity in question is given as,

$$(\nabla_\beta u^\alpha) u_\alpha=(\nabla_\beta u_\alpha)u^\alpha$$

Expanding the left-hand side, we find,

$$(\nabla_{\beta}u^\alpha)u_\alpha = (\nabla_\beta g^{\alpha \delta}u_\delta)u_\alpha = (u_\delta \nabla_\beta g^{\alpha\delta} + g^{\alpha\delta}\nabla_\beta u_\delta)u_\alpha$$

The Levi-Civita connection is precisely the connection which is compatible with the metric structure; as such parallel transporting two tangent vectors preserves the inner product, so we have,

$$\nabla_\beta g^{\alpha\delta} = 0$$

Hence, for the left-hand side, we are left with a term $u^\alpha \nabla_\beta u_\alpha$. Just a general tip regarding these computations: try to avoid expanding the covariant derivative explicitly in terms of the Christoffel symbols as you have done - there are usually better paths to take, just as one is usually advised not to fully expand commutators most of the time.


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