# Commutator of intrinsic derivatives in NP formalism when timelike and null congruence are both given

Suppose we have a congruence of affinely parametrized null geodesics (light rays), with tangent vector $\ell^a$, and a congruence of timelike curves (observers), with tangent vector $u^a$, such that the observers measure a circular frequency of $\omega$. This, of course, means that $u^a\ell_a = \omega$, and we can construct a null vector dual to $\ell^a$ by letting $n^a = \omega^{-1}u^a - \frac{1}{2}\omega^{-2}\ell^a$. Letting $r$ be the null parameter and $t$ be the timelike parameter, we thus have \begin{align} D \equiv \ell^a\partial_a &= \partial_r, \\ \Delta \equiv n^a\partial_a &= \omega^{-1}\partial_t - \frac{1}{2}\omega^{-2}\partial_r, \end{align} where $D$ and $\Delta$ are the intrinsic derivatives of the NP formalism. Thus, when acting on a scalar field we find that $$[\Delta,D] = \frac{D\omega}{\omega}\Delta - \frac{D\omega}{2\omega^3}D,$$ where $[\Delta,D]$ is the commutator. However, it is common knowledge that in terms of the spin coefficients $$[\Delta,D] = (\gamma + \bar{\gamma})D + (\epsilon + \bar{\epsilon})\Delta - (\bar{\tau} + \pi)\delta - (\tau + \bar{\pi})\bar{\delta},$$ when acting on a scalar field. The crux lies in that $(\epsilon + \bar{\epsilon}) = 0$ whenever the null geodesics are affinely parametrized, seemingly implying that $D\omega = 0$, in effect claiming that $\omega$ is constant along the lightrays, regardless of observer velocity and curvature.

Obviously, I am making some fundamental error above, but where is it?

## 1 Answer

As usual when it comes to apparent contradictions like this, the error lies in an assumption. Specifically, when I say that $r$ is the null parameter of a congruence of null curves $c_1$ and thus $D = \partial_r$, what I am actually doing is taking a chart such that $c_1^{-1}$ is one of the coordinate maps. So far so good (as long as there are no self-intersections), but when we wish to simultaneously do this with two different congruences of curves, $c_1$ and $c_2$, we run into a problem. Because this requires that we have $(c_1^{-1} \circ c_2)(t) = \text{constant}$, and conversely $(c_2^{-1} \circ c_1)(r) = \text{constant}$, which obviously requires us to fix the parametrization.

If we wish to retain the affine and proper time parametrizations, or indeed either of them, we can use either $r$ or $t$ as a coordinate, but not both.