I know that this question has been asked and answerd already (see for example here and here) and although the second answer comes pretty close to what my problem is (even touching upon my question one, it just brushes it of with a "by definition"), I still don't see where my reasoning exactly fails.


Let $\gamma : I \to M$, for $I\subset\mathbb{R}$, be a smooth curve, then $\dot\gamma(\lambda)$ is an element of the tangent space at $\gamma(\lambda)$, which we will denote by $T_{\gamma(\lambda)}M$. We can now choose a local chart $x$ on some open subset $U$ of $M$, such that we can express $\dot\gamma(\lambda)$ as $\dot\gamma(\lambda)\equiv \dot x^\mu(\lambda)\partial_\mu$.

If $\nabla$ is the Riemannian connection on $M$, then we have defined a geodesic to be a curve in $M$ that satisfies $\nabla_{\dot\gamma}\dot\gamma=0$.

Let $K$ be a Killing vector field. Then apprently the following is true: $$\frac{d}{d\lambda}(K_\mu \dot x^\mu)=0.$$

If tried to compute this straightforwardly: $$\begin{align*}\frac{d}{d\lambda}(K_\mu\dot x^\mu)&\equiv\nabla_{\dot\gamma}(K_\mu \dot x^\mu) = \dot x^\nu \nabla_{\partial_\nu}(K_\mu\dot x^\mu) = \dot x^\nu (K_\mu \nabla_{\partial_\nu}\dot x^\mu +K_{\mu,\nu}\dot x^\mu)\\ &= (\nabla_{\partial_\nu}\dot x^\mu)\dot x^\nu K_\mu + \frac{1}{2}(K_{\mu,\nu}+K_{\nu,\mu})\dot x^\mu\dot x^\nu\\ &\overset{(*)}{=} (\nabla_{\partial_\nu}\dot x^\mu)\dot x^\nu K_\mu + \frac{1}{2}(\underbrace{K_{\mu;\nu}+K_{\nu;\mu}}_{=0})\dot x^\mu\dot x^\nu + \Gamma^\lambda_{\mu\nu}K_\lambda\dot x^\mu\dot x^\nu\\ &= (\nabla_{\partial_\nu}\dot x^\mu)\dot x^\nu K_\mu+ \Gamma^\lambda_{\mu\nu}K_\lambda\dot x^\mu\dot x^\nu.\end{align*}$$


  1. In the expression $(\nabla_{\partial_\nu}\dot x^\mu)\dot x^\nu K_\mu$ the only part that can be zero for every $\lambda$ is $\nabla_{\partial_\nu}\dot x^\mu$. I don't see why. This has persumably something to do with the fact that we are talking about geodesics here, but I don't see how the condtion $\nabla_{\dot\gamma}\dot\gamma=0$ leads to this.
  2. In the step $(*)$ I switched the partial derivatives to covariant derivatives $,\to\, ;$ to make use of the Killing equation. This created the term proportional to $\Gamma^\lambda_{\mu\nu}K_\lambda$. Can somebody explain why this term is supposed to be zero?
  • $\begingroup$ Since $\frac{d}{d\lambda}=\dot x^{\mu}\partial_{\mu}$ is the tangent vector, then $\dot x^{\mu}\partial_{\mu}(K_{\nu}\dot x^{\nu})= \dot x^{\mu}(\partial_{\mu}K_{\nu})\dot x_{\nu}+\dot x^{\mu}K_{\mu}\partial_{\mu}\dot x^{\nu})$ where the last term on the right $K_{\mu}\partial_{\mu}\dot x^{\nu}=0$ since the affine curve is along $\dot x^{\mu}$. Using $\dot x^{\mu}\partial_{\mu}(K_{\nu}\dot x^{\nu})= \dot x^{\mu}(\partial_{\mu}K_{\nu})\dot x_{\nu}$ you should be able to use the Killing condition and play with the indices to show the result. It may not be trivial. $\endgroup$ Dec 23, 2019 at 1:05
  • $\begingroup$ And. incidently, since $\langle K,\dot\gamma\rangle=(K_\nu \dot x^\nu)$ is a scalar, then $\Gamma^\lambda_{\mu\nu}=0$. $\endgroup$ Dec 23, 2019 at 6:07
  • $\begingroup$ @CinaedSimson Could you maybe explain in bit more detail why exactly the $\Gamma^\mu_{\lambda\nu}=0$? $\endgroup$
    – Sito
    Dec 23, 2019 at 14:26
  • $\begingroup$ There are no basis vectors $(K_\nu \dot x^\nu)$ is a contraction - it's a scalar function. Hence, the covariant derivative reduces to a partial derivative. Since the Christoffel symbols are defined as $\nabla_{\partial_{i}}(\partial_{j})=\Gamma_{ij}^{k}\partial_{k}$, this implies $\Gamma^{k}_{ij}=0.$ $\endgroup$ Dec 23, 2019 at 19:23

1 Answer 1


Your Christoffel term is off by a factor of 2 and missing some x-dots. If you fix it then the last line will be the covariant derivative on the worldline:

$$ \dot{x}^\nu \nabla_\nu \dot{x}^\mu = \dot{x}^\nu \partial_\nu \dot{x}^\mu + \Gamma^\mu_{\alpha \beta} \dot{x}^\alpha \dot{x}^\beta $$.

There's no reason to ever use coordinate derivatives though:

$$ \frac{d}{d\lambda} (K_\mu \dot{x}^\mu) = \dot{x}^\nu \nabla_\nu (K_\mu \dot{x}^\mu) = (\dot{x}^\nu \nabla_\nu K_\mu) \dot{x}^\mu + K_\mu(\dot{x}^\nu \nabla_\nu \dot{x}^\mu) = 0 $$

The first term vanishes by Killing's equation and the second by the affinely parameterized geodesic equation.

  • $\begingroup$ Thank you for pointing out the missing $x$ (I think I corrected it properly now). Could you maybe elaborate on "the last line will be the covariant derivative on the worldline"? I'm not really sure what you exactly mean here or why the result then would be zero. Also thanks for pointing out the alternative way to calculate the result. Even though I don't really need to use coordinate derivatives, I'd still like to know how it could be done using them. $\endgroup$
    – Sito
    Dec 23, 2019 at 14:34
  • 1
    $\begingroup$ You still have an extra factor of 2 in front of the Gamma terms -- did you forget about the 1/2 from symmetrizing? I'll edit the post to show what I mean by covariant derivatives on the worldline $\endgroup$
    – Sam Gralla
    Dec 23, 2019 at 15:18

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.