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Minkowski spacetime in Newman Penrose formalism

+ 3 like - 0 dislike
102 views

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?

This post imported from StackExchange MathOverflow at 2015-08-30 08:06 (UTC), posted by SE-user GregVoit
asked Aug 27, 2015 in Theoretical Physics by GregVoit (115 points) [ no revision ]
retagged Aug 30, 2015
Hint: spin coefficients are proportional to the derivatives of the tetrad vectors. So, which tetrad would you use in Minkowski space?

This post imported from StackExchange MathOverflow at 2015-08-30 08:06 (UTC), posted by SE-user Igor Khavkine
@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

This post imported from StackExchange MathOverflow at 2015-08-30 08:06 (UTC), posted by SE-user GregVoit
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.

This post imported from StackExchange MathOverflow at 2015-08-30 08:06 (UTC), posted by SE-user Igor Khavkine
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

This post imported from StackExchange MathOverflow at 2015-08-30 08:06 (UTC), posted by SE-user GregVoit
That is, we cannot make a statement "spacetime is Minkowski iff such and such spin coeffs vanish"?

This post imported from StackExchange MathOverflow at 2015-08-30 08:06 (UTC), posted by SE-user GregVoit
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.

This post imported from StackExchange MathOverflow at 2015-08-30 08:06 (UTC), posted by SE-user Igor Khavkine

1 Answer

+ 3 like - 0 dislike

Well, it certainly is surprising that the answer does not seem to be anywhere out there. Let's see

  1. If any of the Weyl scalars is nonzero (in $\Lambda$-vacuum), then from the NP field equations we see that the spin coefficients cannot be zero.
  2. Hence, the only space-times which could have vanishing spin coefficients are Type O (all Weyl scalars vanish): AdS, Minkowski and dS.
  3. Now it would seem that any light-like congruence will have non-zero expansion (contraction) in dS (AdS) because the Raychaudhuri equation and the like involves projections of the full curvature tensor (not only the vanished Weyl tensor) which is non-zero for $\Lambda \neq 0$.

This could be also seen from the not-decomposed geodesic deviation equation of the null congruence; the only space-time where curvature vanishes globally is Minkowski so there will always be some geodesic deviation in $\Lambda \neq 0$. The optical scalars offer a complete characterization of the geodesic deviation, hence they cannot be globally zero in space-times with non-zero curvature.

Come to think of it, the global Riemann flatness of Minkowski+good definition of optical scalars makes this easier than expected: Minkowski is simply the unique space-time where all the optical scalars of a null geodesic congruence globally vanish, and the NP spin coefficients can be made zero by setting the tetrad thus.

answered Aug 30, 2015 by Void (1,505 points) [ no revision ]

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