1
$\begingroup$

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:

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

that

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

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?

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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 $L$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.