Killing Symmetries and Geodesic Constants

Every Killing field produces a scalar that is constant along every affinely parametrized geodesic: contract the field with the geodesic tangent. The proof is one line once its two hypotheses are visible—Killing symmetry and affine geodesic motion—and it also shows exactly what fails for a nonaffine parameter or for a vector field that preserves only the conformal class.

Required background. Tensor Fields and Natural Operations supplies contractions and symmetrization; Affine Connections and Covariant Derivatives supplies the general covariant-derivative framework, while Parallel Transport Along Curves supplies differentiation along a curve.

Helpful background. Calculus and Linear Algebra for Relativity supplies the chain rule and bilinear algebra; Affine and Variational Geodesics develops affine parameters and geodesic equations.

The linear geodesic integral

Let $\gamma(\lambda)$ be an affinely parametrized geodesic with tangent $u^a$, so

$$u^b\nabla_bu^a=0.$$

Let $X^a$ be a Killing field, $\nabla_{(a}X_{b)}=0$. Define the scalar

$$J_X=X_au^a.$$

Differentiation along the curve gives

$$\begin{aligned} \frac{dJ_X}{d\lambda} &=u^b\nabla_b(X_au^a) \\ &=u^au^b\nabla_bX_a+X_a u^b\nabla_bu^a \\ &=u^au^b\nabla_{(b}X_{a)}=0. \end{aligned}$$

The antisymmetric part of $\nabla_bX_a$ disappears against the symmetric product $u^au^b$. The argument does not depend on whether the geodesic is timelike, null, or spacelike. Carroll gives the symmetry and geodesic calculation in Carroll 1997, §5, pp. 139–141, eqs. (5.40)–(5.46).

For a massive test curve one may use $p^a=mu^a$ and write $X_ap^a$. Calling this scalar an energy or angular momentum requires a specified symmetry and normalization; the geometric theorem itself supplies only a first integral.

The same result appears on the cotangent bundle. With geodesic Hamiltonian

$$H=\frac12 g^{ab}p_ap_b, \qquad J_X=X^ap_a,$$

the canonical bracket $\{F,G\}=\partial_{x^a}F\,\partial_{p_a}G-\partial_{p_a}F\,\partial_{x^a}G$ satisfies

$$\{J_X,H\}=-\frac12(\mathcal L_Xg^{ab})p_ap_b.$$

It vanishes when $X$ is Killing. This is an independent derivation: the tangent-bundle proof uses covariant differentiation along one curve, whereas the Hamiltonian proof states that $J_X$ is constant along the geodesic flow on phase space. The overall Poisson-sign convention can reverse both sides, but the zero condition is invariant.

Why affine parametrization matters

Suppose the same unparameterized geodesic is described by a parameter for which

$$u^b\nabla_bu^a=f(\lambda)u^a.$$

Then the previous calculation becomes

$$\frac{dJ_X}{d\lambda}=f(\lambda)J_X.$$

Thus $J_X$ is not generally constant, even though the curve and the Killing field have not changed. The rescaled combination

$$\exp\!\left(-\int^\lambda f(s)\,ds\right)J_X$$

is constant. Equivalently, choose an affine parameter. This check separates a genuine failure of symmetry from an artifact of parametrization.

Converse and its limits

There is a useful local converse. Let $X$ be smooth and assume $X_au^a$ is constant along every affinely parametrized geodesic through every point, for every initial tangent $u^a$. At the initial point,

$$u^au^b\nabla_{(a}X_{b)}=0$$

for every vector $u$. Polarization then implies $\nabla_{(a}X_{b)}=0$, so $X$ is Killing. The quantifier “every initial tangent” is essential. Conservation along a single geodesic, or even a restricted congruence, constrains only selected contractions of the symmetric derivative.

Restricting the test to null directions is also weaker. In Lorentzian dimension at least three, a symmetric bilinear form that vanishes on every null vector is proportional to the metric. The resulting condition is conformal Killing rather than necessarily Killing. This is why null geodesics alone detect the conformal structure.

Several generators give several linear integrals, but their algebra carries further information. With the canonical bracket declared above,

$$\{J_X,J_Y\} =\left(Y^b\partial_bX^a-X^b\partial_bY^a\right)p_a =-J_{[X,Y]}.$$

This minus sign is fixed by the choice $J_X=X^ap_a$; replacing every generator by $-J_X$ produces the corresponding homomorphism convention. Noncommuting first integrals can all be conserved without being in involution. Conservation and the mutually commuting integrals needed for Liouville integrability are separate questions.

In a chart adapted to a symmetry, the result has the familiar cyclic-coordinate form. If $X=\partial_q$ and the metric components are independent of $q$, then $J_X=g_{qa}u^a$ is the momentum conjugate to $q$. The invariant proof explains why this remains true after changing to a chart where the symmetry is no longer a coordinate vector. It also prevents a false converse: a metric coefficient that happens to be independent of one coordinate only on a submanifold does not establish a Killing field on an open region.

Explicit boost check

In two-dimensional Minkowski space, with $ds^2=-dt^2+dx^2$, the boost generator

$$K=x\partial_t+t\partial_x$$

is Killing. Along an affine geodesic $t=t_0+u^t\lambda$, $x=x_0+u^x\lambda$,

$$J_K=K_au^a=-x u^t+t u^x =-x_0u^t+t_0u^x.$$

The terms proportional to $\lambda$ cancel. This direct calculation checks both the metric sign and the abstract proof. It also makes clear that the constant depends on the chosen generator; rescaling $K$ rescales $J_K$.

For timelike motion, fixing $g(u,u)=-1$ removes affine rescalings but still permits an affine shift. For null motion there is no unit normalization, so $u\mapsto c\,u$ rescales $J_X$. Any comparison of numerical constants on distinct null geodesics must therefore specify how their affine tangents were normalized.

That normalization issue changes numerical labels, never the statement that the derivative vanishes along each chosen affine curve.

Exercises

Let $X=\partial_t$ in Minkowski space and reparametrize a straight geodesic by $\lambda=e^s$. Show directly that $X_au^a$ is not constant as a function of $s$, and identify the function $f(s)$ in the nonaffine geodesic equation.

Solution

Write the affine curve as $x^a(\lambda)=x_0^a+v^a\lambda$. In the $s$ parameter, $u^a=dx^a/ds=e^s v^a$, so $J_X=-e^s v^t$. Because $du^a/ds=u^a$ and the Cartesian connection vanishes, $\nabla_u u=u$, hence $f=1$. The predicted equation $dJ_X/ds=fJ_X$ is therefore satisfied. Multiplying by $e^{-s}$ recovers the affine constant $-v^t$.

References

• Sean M. Carroll, Lecture Notes on General Relativity, arXiv:gr-qc/9712019 (1997), arXiv.