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.

  • $\begingroup$ 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, 2017 at 10:52

2 Answers 2


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.

  • $\begingroup$ 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, 2019 at 7:34
  • $\begingroup$ Why is that the case? $\endgroup$ Dec 18, 2019 at 11:53

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 $g_{ab}K^{a}\dot{x}^b.$


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.