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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.

Setup

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

Question

  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?
Sito
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1 Answers1

2

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.

Sam Gralla
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