= Geodesic conserved quantity from a Killing vector
{title2=$C_K=g(K,U)$}
= Geodesic conserved quantities from Killing vectors
{synonym}
Each <Killing vector field> supplies a <conserved quantity> for freely moving <test particles>: it turns a geometric <symmetry> into a first integral of the <geodesic equation>. Let $\gamma(\lambda)$ be a <geodesic> with <affine parameter> $\lambda$, <tangent vector> $U^\mu=dx^\mu/d\lambda$, and $\nabla_UU=0$. Wherever a <Killing vector field> $K$ is defined along the <geodesic>,
$$
C_K=g(K,U)=K_\mu U^\mu,\qquad \frac{dC_K}{d\lambda}=0.
$$
The statement applies to <Riemannian manifolds> and to <timelike geodesics>, <null geodesics> and <spacelike geodesics> in a <Lorentzian manifold>. It requires an <affine parameter>; for a massive <test particle>, <proper time> is one choice. See https://davidtong.org/pdfs/teaching/general-relativity/gr.pdf#page=171[Tong, section 4.3.2, equation (4.32)].
For a worked example in the <Euclidean plane>, take $\gamma(t)=(1+t,2-t)$ with <affine parameter> $t$ and <velocity> $U=(1,-1)$. The <Euclidean Killing vector fields> for <translation> and <rotation> give
$$
C_{\mathrm{tr}}=v_x=1,\qquad
C_{\mathrm{rot}}=xv_y-yv_x=-(1+t)-(2-t)=-3.
$$
The rotational field along the <geodesic> is $K_{\mathrm{rot}}(\gamma(t))=(t-2,1+t)$: it changes from $(-2,1)$ at $t=0$ to $(-1,2)$ at $t=1$, yet its pairing with $U$ stays $-3$. With toy <mass> $m=1$ and $t$ interpreted as time, the first constant is the $x$ component of <linear momentum>, and the second is the signed $z$ component of <angular momentum> about the origin. Their numerical values use toy units. The <geodesic> is straight; the <rotation> flow orbits are <circles>.
\Image[/killing-geodesic-conserved-quantity.png]
{title=One straight Euclidean geodesic: the rotational Killing field varies while translation momentum and angular momentum stay constant for toy unit mass}
This <conservation law> follows from <metric compatibility> and the <Killing equation>:
$$
\frac{d}{d\lambda}(K_\mu U^\mu)
=U^\mu U^\nu\nabla_\mu K_\nu+K_\nu(\nabla_UU)^\nu
=U^\mu U^\nu\nabla_{(\mu}K_{\nu)}=0.
$$
The <geodesic equation> kills the acceleration term, and the symmetric product $U^\mu U^\nu$ selects the vanishing symmetric derivative. This <tensor contraction> is the whole proof outline; see https://arxiv.org/pdf/gr-qc/9712019#page=147[Carroll, equation (5.43)].
In a <stationary spacetime>, a chosen time-translation <Killing vector field> $K$ gives <Killing energy>. With <metric signature> $(-+++)$, $c=1$, and a massive <test particle>'s <four-momentum> $p^\mu=mU^\mu$, the physical normalization is
$$
E=-p_\mu K^\mu,\qquad L_z=p_\mu\Phi^\mu.
$$
Here $\Phi$ generates the axial <rotational symmetry> when one exists. In coordinates with $K=\partial_t$ and $\Phi=\partial_\phi$, these are $E=-p_t$ and $L_z=p_\phi$. Both are <conserved quantities> along the <geodesic>, even if the metric has a time-angle cross term. A future timelike $K$ makes $E$ positive for a future massive particle; the scale of $E$ depends on the normalization of $K$. Not every <spacetime> has a global timelike <Killing vector field>, and a stationary field can become spacelike in an <ergoregion of a stationary spacetime>. Conservation on the field's domain does not imply a universal global positive <energy>; see https://davidtong.org/pdfs/teaching/general-relativity/gr.pdf#page=170[Tong, sections 4.3.1-4.3.2 on the timelike-domain qualification and conserved charges].
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