I would like to know if the following proof of the constancy of the scalar product of a Killing vector with the geodesic tangent vector along the geodesic is correct. I already found a coordinate-dependent proof but my rationale seems to be rather different from the coordinate-dependent proof.

Be $K$ a Killing vector and $\gamma$ the geodesic and $\dot{\gamma}$ its tangent vector. $g(\cdot, \cdot)$ is the scalar product based on the metric $g$. $\nabla_V X$ is the covariant derivative.

$$d g(K,\dot{\gamma})(\dot{\gamma}) = \nabla_\dot{\gamma}g( K,\dot{\gamma}) = g( \nabla_\dot{\gamma}K, \dot{\gamma}) + g( K, \nabla_\dot{\gamma}\dot{\gamma}) = g( \nabla_\dot{\gamma}K, \dot{\gamma}) $$

where the second term vanishes because of $\nabla_\dot{\gamma}\dot{\gamma}=0$ along the geodesic. The first term also vanishes since I can conclude from the coordinate-independent form of the Killing vector equation:

$$g(\nabla_V K,W) + g(V,\nabla_W K)=0$$



So $$d g(K,\dot{\gamma})(\dot{\gamma})=0$$

which shows the constancy of the scalar product of a Killing vector with the geodesic tangent vector along the geodesic. Is this derivation correct?

  • $\begingroup$ it looks alright, which step are you feeling fishy? $\endgroup$ – chichi Aug 19 at 13:46
  • $\begingroup$ @chichi : the sources on differential geometry I dispose of only attest : $\nabla g(X,Y) = g(\nabla X, Y) + g(X, \nabla Y)$ instead of $\nabla_U g(X,Y) = g(\nabla_U X, Y) + g(X, \nabla_U Y)$ so I don't know if the latter is correct. $\endgroup$ – Frederic Thomas Aug 19 at 13:51
  • $\begingroup$ I think the latter equation you wrote is the definition of covariant derivative, no? $\endgroup$ – chichi Aug 19 at 14:03
  • $\begingroup$ Well, I would say, at least in my differential geometry book the definition is $\nabla_U (X\otimes Y) = \ldots $ (plus some addtional rules), I guess since $g$ already is an additional structure. $\endgroup$ – Frederic Thomas Aug 19 at 14:14

I'll give another proof, just for fun. We know that in a pseudo-Riemannian manifold $(M,g)$, geodesics are critical points of the energy functional $$E[\gamma] = \frac{1}{2}\int_I g_{\gamma(t)}(\dot{\gamma}(t),\dot{\gamma}(t))\,{\rm d}t$$Since $K$ is Killing, the flow of $K$ (consisting of isometries) leaves the Lagrangian we're integrating invariant. So Noether's Theorem says that the charge $$\mathscr{J}(x,v) = \mathbb{F}L(x,v)\left(\frac{{\rm d}}{{\rm d}s}\bigg|_{s=0} \Phi_{s,K}(x)\right) = \mathbb{F}L(x,v)K_x= \frac{\rm d}{{\rm d}t}\bigg|_{t=0} \frac{1}{2}g_x(v+tK_x,v+tK_x) = g_x(v,K_x) $$is constant along geodesics. But $\mathscr{J}(\gamma(t),\dot{\gamma}(t)) = g_{\gamma(t)}\big(\dot{\gamma}(t), K_{\gamma(t)}\big)$, and so we're done. Here $(x,v) \in TM$ (which is to say that $x \in M$ and $v \in T_xM$) and $\mathbb{F}L$ denotes the fiber derivative of $L$.

  • $\begingroup$ Thank you for the interest. Actually I don't know what the fiber derivative is. Furthermore, $L(x,v)$ seems to be a Lagrangian, is actually the energy functional $E[\gamma]$ meant by this? And finally, the expression $(\frac{d}{ds}|_{s=0} \Phi_{s,K}(x))$ is this an argument of $\mathbb{F}L$ or is it supposed to be multiplied to $\mathbb{F}L$ ? Is $\Phi_{s,K}$ the flow of $K$? I am not familiar with this notation. $\endgroup$ – Frederic Thomas Oct 3 at 11:05
  • $\begingroup$ 1) the fiber derivative of $L$ is $\mathbb{F}L(x,v)w = \frac{{\rm d}}{{\rm d}t}\bigg|_{t=0} L(x,v+tw)$; 2) yes; 3) argument, see 1); 4) it is the flow. $\endgroup$ – Ivo Terek Oct 3 at 13:22
  • $\begingroup$ I think, I found your approach as homework in a book. It's nice to see it her demonstrated. If your don't mind vote up my approach as it is apparently correct. I gonna vote up yours. $\endgroup$ – Frederic Thomas Oct 4 at 18:31
  • $\begingroup$ Yes, sure. It's probably the first proof I'd think of. $\endgroup$ – Ivo Terek Oct 4 at 20:46

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