# How to count degrees of freedom of metric in Newmann-Penrose formalism?

Usually a metric has 10 degrees of freedom. How to show the same in Newmann-Penrose formalism?

The Newman-Penrose formalism, ultimately is just fancy notation for a (null) tetrad formalism. The basic degree of freedom is a tetrad, a set of four 4-vectors. We thus start with 16 degrees of freedom. Given a tetrad $$(e^a)_\mu$$ (latin indices count the tetrad legs, greek indices are spacetime indices), one obtain the metric by

$$g_{\mu\nu} = (e^a)_\mu(e^b)_\nu \eta_{ab},$$

where $$\eta$$ is tetrad metric given by

$$\eta = \pm \begin{pmatrix}0 & 1 & 0 & 0\\ 1 & 0 & 0 & 0\\ 0 & 0 & 0 & -1\\ 0 & 0 & -1 & 0\\ \end{pmatrix}$$

with $$\pm$$ depending on your choice of signature.

Not all different tetrads lead to different metrics. More specifically any transformation on the tetrad legs that preserves $$\eta$$ will also presvere the spacetime metric. These transformations are called tetrad rotations, and form an SO(3,1) group. This group has 6 dimensions, bringing us back to the 10 degrees of freedom of the metric.

• @mment If you look at equation 4.3 of this link web.math.princeton.edu/~aretakis/columbiaGR.pdf, you will see a metric in double-null foliated spacetime. This metric seems to have only 6 degrees of freedom. If you can explain why is that so, it will be really helpful. Commented Nov 3, 2019 at 6:30
• @user44690 Not all different metric describe different spacetimes, due to the freedom to choose coordinates. Consequently, 4 of the 10 degrees of freedom of the metric are gauge degrees of freedom, and we are left with 6 real degrees of freedom. Commented Nov 3, 2019 at 10:53
• @mment So, what is the maximum possible extent to which we can set degrees of freedom of the metric to gauge degrees of freedom. Certainly we cannot set all degrees of freedom of the metric to gauge degrees of freedom ? Commented Nov 3, 2019 at 18:30