# Minkowski spacetime in Newman Penrose formalism

I have a rather basic question for which (surprisingly!) I cannot find a short and clear answer anywhere:

I'm currently looking at the Newman Penrose (NP) formalism (I use primarily Chandrasekhar's "Mathematical Theory of Black Holes").

My question is: what exactly is Minkowski spacetime in NP formalism? That is, which are the vanishing/non-vanishing spin-coefficients?

In particular, is just vanishing of some of those coefficients sufficient to characterize the Minkowski spacetime (in a similar way that Goldberg Sachs theorem characterizes the algebraically special spacetimes).

My thoughts so far:

Clearly, we should have $\kappa=\sigma=\mu=\lambda=0$ since that gives us Type D by Goldberg Sachs theorem, but Minkowski is more than that. Say we also put $\epsilon=0$ for a suitable tetrad scaling. But what about the rest?

• Hint: spin coefficients are proportional to the derivatives of the tetrad vectors. So, which tetrad would you use in Minkowski space? – Igor Khavkine Aug 27 '15 at 7:26
• @IgorKhavkine the tetrad I could find in the literature is $\sqrt{2}l=\partial_{t}+\cos x_{3}\partial_{1}+\sin x_{3}\partial_{3}$, $\sqrt{2}n=\partial_{t}-\cos x_{3}\partial_{1}+\sin x_{3}\partial_{3}$, $\sqrt{2}m=-\sin x_{3}\partial_{1}+\cos x_{3}\partial 3+i\partial_{2}$. But I know neither how to obtain it nor how to proceed with it – GregVoit Aug 27 '15 at 7:35
• That's an odd choice... Why not just take all the tetrad vectors to be constant with respect to inertial coordinates? Then the answer is quite simple. As you can see, asking for the spin coefficients of a given spacetime is an ill-posed question. Those coefficients are only defined once a tetrad is chosen and that's a huge additional set of free parameters. – Igor Khavkine Aug 27 '15 at 8:14
• I see. So are you suggesting then the null tetrad $\sqrt{2}l=\partial_{1}+\partial_{4}$, $\sqrt{2}n=-\partial_{1}+\partial_{4}$, $\sqrt{2}m=\partial_{2}+i\partial_{3}$? Did I understand you correctly then, that the fact which spin coefficients vanish and which don't, in case of Minkowski depends on which null tetrad I pick, right? @IgorKhavkine – GregVoit Aug 27 '15 at 8:19
• I'm not sure I'd have a definite answer to the question, most likely because I don't actually use the tetrad formalism when I think about GR. A definite statement is that Minkowski spacetime is locally characterized precisely by the statement that the Riemann tensor is identically zero. Perhaps you can think of how to translate that statement to the tetrad language for your own purposes. – Igor Khavkine Aug 27 '15 at 18:56

The one thing which I think you are missing is that

Unlike the Riemann curvature, the spin coefficients are not tensors (they are not co- or contra-variant).

In particular, the spin coefficients heavily depend on the choice of the tetrad.

In Minkowski space you can construct "accelerated frames"; these (non-inertial) tetrads in which the spin coefficients do not identically vanish are somewhat important in terms of the interpretation of the equivalence principle. What you can observe, however, is that with regards to any tetrad, Minkowski space satisfies the condition that the curvature scalars $\Psi_0, \ldots, \Psi_4$ all vanish identically. This fact uniquely characterizes Minkowski space among Ricci-flat solutions, and if you also include conditions on the Ricci curvature scalars (that they all vanish), this uniquely (locally) characterizes Minkowski space.

In general, however, if you are willing to pick very wild tetrads, you can get very wild values of the spin coefficients.

You mentioned Golberg Sachs. Pay attention that it does not say "Algebraically special iff such and such coefficients vanish." It says "Algebraically special iff there exists a tetrad in which such and such coefficients vanish." One can certainly consider tetrads for Type D spacetimes which are not geodesic!

• you wrote that vanishing of all Weyl scalars uniquely characterizes Minkowski. Could you give a reference to where the proof of this fact is written? @Willie Wong – GregVoit Dec 3 '15 at 7:07
• @GregVoit: curvature scalars vanishing implies the Weyl tensor vanishes. That plus Ricci flat implies Riemann curvature vanishes. Riemann curvature vanish implies one can extend an orthonormal frame in $T_p M$ to a parallel orthonormal frame in $TM$ over an open neighborhood $U$ of $p$, and thereby explicitly construct the local isometry to Minkowski. The first two steps are simple algebra; the third step is classical (going back to approximately Riemann) and is frequently found in textbooks in Riemannian geometry. – Willie Wong Dec 3 '15 at 14:12
• For example, Choquet-Bruhat and DeWitt-Morette, Analysis, Manifolds, and Physics, volume 1, Section V.B.2. (I think the proof is written for Riemannian metrics, but the proof itself if you read it does not depend on the signature of the metric, only that it is non-degenerate.) – Willie Wong Dec 3 '15 at 14:23