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I understand that a Killing vector $K^{\mu}$ satisfies,

$$ K^{(\mu;\nu)} = 0 $$

I also know that along a geodesic, the quantity

$$ p_{\mu} K^{\mu}$$

is conserved, where $p_{\mu}$ is the photon 4 momentum, or more generally a tangent vector.

I want to understand why it should be that this quantity is conserved. i.e I want to show that,

$$ \frac{d}{d\lambda} (p_{\mu} K^{\mu}) = 0$$

but cannot see how to even start, given our knowledge of the property of the Killing vector.

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  • $\begingroup$ The covariant derivative of a scalar in a particular direction is just the ordinary derivative along the curve pointing in that direction. So: $$ \frac{\mathrm{d}}{\mathrm{d}\lambda} = p^\mu\nabla_\mu \,.$$ $\endgroup$ – gj255 Jun 16 '17 at 10:52
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Let's consider a Killing field $K_\mu$. Now the product of the Killing field and the tangent vector is $Q_K = K_\mu \dfrac{dx^\mu}{d\lambda}$.

Now, along a geodesic parametrized by an affine parameter $\lambda$,

$\dfrac{d}{d\lambda}Q_K = \dfrac{d}{d\lambda}\big(K_\mu \dfrac{dx^\mu}{d\lambda}\big)$

$=\dfrac{\partial K_\mu}{\partial x^\nu} \dfrac{dx^\nu}{d\lambda}\dfrac{dx^\mu}{d\lambda} + K_\mu \dfrac{d^2x^\mu}{d\lambda^2}$

$=K_{\alpha;\nu}+K_\alpha \Gamma^\alpha _{\mu\nu}\dfrac{dx^\nu}{d\lambda}\dfrac{dx^\mu}{d\lambda}+ K_\mu \dfrac{d^2x^\mu}{d\lambda^2} $

$=K_\mu \bigg(\dfrac{d^2x^\mu}{d\lambda^2}+\Gamma^\mu _{\alpha\nu}\dfrac{dx^\alpha}{d\lambda}\dfrac{dx^\nu}{d\lambda}\bigg) = 0$

So, the quantity $Q_K$ is conserved along the geodesic.

Edit The fact that a conserved quantity is apparent along the geodesic is beautifully related to Noether's theorems as illustrated by childofsaturn in this answer.

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  • $\begingroup$ The term $K_{\mu;\nu}=-K_{\nu;\mu}$, when summed with symmetrized $\frac{dx^\mu}{d\lambda}\frac{dx^\nu}{d\lambda}$, gives zero. $\endgroup$ – gamebm Feb 7 '19 at 7:34
  • $\begingroup$ Why is that the case? $\endgroup$ – fielder Dec 18 '19 at 11:53
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One can directly show that it's conserved along geodesics as Dvij has done, but it's also illuminating to note that this is just a special case of Noether's theorem. In particular whenever a metric has an isometry generated by a vector field $K^{a}$ the Lagrangian $L = \frac{1}{2}g_{ab}\dot{x}^a \dot{x}^b$ is invariant under the transformation $\delta x^{a} = \epsilon K^{a}$. Now recall that when a Lagrangian is invariant under some transformation $\delta x$, the corresponding Noether charge is $J = \delta x \frac{\partial L}{\partial \dot{x}}$. Here this just becomes $g_{ab}K^{a}\dot{x}^b.$

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