 Solution

Use an affine parameter $s$ on the geodesic, so its tangent vector $U$ obeys ${\nabla }_{U}U=0$. The product rule, metric compatibility, and the Conformal Killing equation giveHenceIt is conserved on every null geodesic with an affine parameter, but generally not on other geodesics. On a timelike geodesic parametrized by proper time, $g(U,U)=-1$, so the derivative is $-\lambda$. A Killing vector field, or a geodesic on which $\lambda$ vanishes, supplies the exceptional conservation. This is the conformal Killing geodesic identity.

For an arbitrary parameter with ${\nabla }_{U}U=fU$, the right-hand side acquires $f(X\cdot U)$. Thus even for a null geodesic, the simple conserved expression presupposes an affine parameter.