Let $u^m\rightarrow u^m+\frac{d}{d\tau}k^m$.

How does $S=m\int d\tau\sqrt{-g_{mn}u^mu^n}$ change to first order under this transformation if $k$ is a Killing vector?

My attempt:
Using $\frac{d}{d\tau}k^m=(\nabla_lk^m)u^l$, I thought \begin{align} S'&=m\int d\tau\sqrt{-g_{mn}(u^m+\frac{d}{d\tau}k^m)(u^n+\frac{d}{d\tau}k^n)}\\ &=m\int d\tau\sqrt{-g_{mn}(u^mu^n+(\nabla_lk^m)u^lu^n+(\nabla_lk^n)u^lu^m)} \end{align} Then I think I can use $\nabla_{(l}k_{m)}=0$ to show that $S'=S$. But how? What is the next step?


HINT:(I hope)

So what I think you do is the following: begin by multiplying the metric tensor in to the final two terms to get the following expression (obviously this is under a square root and there is another term. However, I've omitted for clarity)

$$ \left( \nabla_l g_{mn}k^m \right)u^lu^n+\left(\nabla_l g_{mn}k^n \right)u^lu^m.$$

Next, I would contract the metric tensor with the Killing vector to get

$$ \left( \nabla_l k_n \right)u^lu^n+\left(\nabla_l k_m \right)u^lu^m.$$

Finally, changing the $l$ index in both expressions we can then use Killing's equation

$$ \nabla_\mu k_\nu + \nabla_\nu k_\mu \equiv 0,$$

and from there you should be able to show $S=S'$.

  • $\begingroup$ What do you mean by "similar operation on the $l$ index"? $\endgroup$ – Thomas Penney Oct 23 '17 at 11:21
  • $\begingroup$ So what I mean is change the $l$ index as its repeated. I've slightly updated the Phrasing in the answer. $\endgroup$ – Rumplestillskin Oct 23 '17 at 11:32

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